Se v e n th E d itio n
ENGINEERING
ECONOMY
Se v e n th E d itio n
ENGINEERING
ECONOMY
Leland Blank, P. E.
Texas A & M University
American University of Sharjah, United Arab Emirates
Anthony Tarquin, P. E.
University of Texas at El Paso
TM
TM
ENGINEERING ECONOMY: SEVENTH EDITION
Published by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New
York, NY 10020. Copyright © 2012 by The McGraw-Hill Companies, Inc. All rights reserved. Previous editions
© 2005, 2002, and 1998. No part of this publication may be reproduced or distributed in any form or by any means, or
stored in a database or retrieval system, without the prior written consent of The McGraw-Hill Companies, Inc.,
including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance
learning.
Some ancillaries, including electronic and print components, may not be available to customers outside the United
States.
This book is printed on recycled, acid-free paper containing 10% postconsumer waste.
1 2 3 4 5 6 7 8 9 0 QDB/QDB 1 0 9 8 7 6 5 4 3 2 1
ISBN 978-0-07-337630-1
MHID 0-07-337630-2
Vice President & Editor-in-Chief: Marty Lange
Vice President EDP/Central Publishing Services: Kimberly Meriwether David
Global Publisher: Raghothaman Srinivasan
Sponsoring Editor: Peter E. Massar
Senior Marketing Manager: Curt Reynolds
Development Editor: Lorraine K. Buczek
Senior Project Manager: Jane Mohr
Design Coordinator: Brenda A. Rolwes
Cover Designer: Studio Montage, St. Louis, Missouri
Cover Image: © Brand X Pictures/PunchStock RF
Buyer: Kara Kudronowicz
Media Project Manager: Balaji Sundararaman
Compositor: MPS Limited, a Macmillan Company
Typeface: 10/12 Times
Printer: Quad/Graphics-Dubuque
All credits appearing on page or at the end of the book are considered to be an extension of the copyright page.
Library of Congress Cataloging-in-Publication Data
Blank, Leland T.
Engineering economy / Leland Blank, Anthony Tarquin. — 7th ed.
p. cm.
Includes bibliographical references and index.
ISBN-13: 978-0-07-337630-1 (alk. paper)
ISBN-10: 0-07-337630-2
1. Engineering economy. I. Tarquin, Anthony J. II. Title.
TA177.4.B58 2012
658.15—dc22
2010052297
www.mhhe.com
This book is dedicated to Dr. Frank W. Sheppard, Jr. His lifelong
commitment to education, fair financial practices, international
outreach, and family values has been an inspiration to many—one
person at a time.
M C GRAW-HILL DIGITAL OFFERINGS
McGraw-Hill Create™
Craft your teaching resources to match the way you teach! With McGraw-Hill Create™, www
.mcgrawhillcreate.com, you can easily rearrange chapters, combine material from other content
sources, and quickly upload content you have written like your course syllabus or teaching notes.
Find the content you need in Create by searching through thousands of leading McGraw-Hill
textbooks. Arrange your book to fit your teaching style. Create even allows you to personalize
your book’s appearance by selecting the cover and adding your name, school, and course information. Order a Create book and you’ll receive a complimentary print review copy in 3–5 business days or a complimentary electronic review copy (eComp) via email in minutes. Go to www.
mcgrawhillcreate.com today and register to experience how McGraw-Hill Create™ empowers
you to teach your students your way.
McGraw-Hill Higher Education and Blackboard Have
Teamed Up
Blackboard, the Web-based course-management system, has partnered with McGraw-Hill to better allow students and faculty to use online materials and activities to complement face-to-face
teaching. Blackboard features exciting social learning and teaching tools that foster more logical,
visually impactful and active learning opportunities for students. You’ll transform your closeddoor classrooms into communities where students remain connected to their educational experience 24 hours a day.
This partnership allows you and your students access to McGraw-Hill’s Create™ right from
within your Blackboard course – all with one single sign-on. McGraw-Hill and Blackboard can
now offer you easy access to industry leading technology and content, whether your campus
hosts it, or we do. Be sure to ask your local McGraw-Hill representative for details.
Electronic Textbook Options
This text is offered through CourseSmart for both instructors and students. CourseSmart is an
online resource where students can purchase the complete text online at almost half the cost of a
traditional text. Purchasing the eTextbook allows students to take advantage of CourseSmart’s
web tools for learning, which include full text search, notes and highlighting, and email tools for
sharing notes between classmates. To learn more about CourseSmart options, contact your sales
representative or visit www.CourseSmart.com.
CONTENTS
Preface to Seventh Edition
LEARNING
STAGE 1
Chapter 1
xiii
THE FUNDAMENTALS
Foundations of Engineering Economy
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
Chapter 2
Factors: How Time and Interest Affect Money
PE
2.1
2.2
2.3
2.4
2.5
2.6
2.7
Chapter 3
Progressive Example—The Cement Factory Case
Single-Amount Factors (F兾P and P兾F )
Uniform Series Present Worth Factor and Capital Recovery Factor (P兾A and A兾P)
Sinking Fund Factor and Uniform Series Compound Amount Factor (A兾F and F兾A)
Factor Values for Untabulated i or n Values
Arithmetic Gradient Factors (P兾G and A兾G)
Geometric Gradient Series Factors
Determining i or n for Known Cash Flow Values
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Time Marches On; So Does the Interest Rate
Combining Factors and Spreadsheet Functions
3.1
3.2
3.3
Chapter 4
Engineering Economics: Description and
Role in Decision Making
Performing an Engineering Economy Study
Professional Ethics and Economic Decisions
Interest Rate and Rate of Return
Terminology and Symbols
Cash Flows: Estimation and Diagramming
Economic Equivalence
Simple and Compound Interest
Minimum Attractive Rate of Return
Introduction to Spreadsheet Use
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Renewable Energy Sources for Electricity Generation
Case Study—Refrigerator Shells
Calculations for Uniform Series That Are Shifted
Calculations Involving Uniform Series and Randomly Placed Single Amounts
Calculations for Shifted Gradients
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Preserving Land for Public Use
Nominal and Effective Interest Rates
PE
4.1
4.2
4.3
4.4
4.5
Progressive Example—The Credit Card Offer Case
Nominal and Effective Interest Rate Statements
Effective Annual Interest Rates
Effective Interest Rates for Any Time Period
Equivalence Relations: Payment Period and Compounding Period
Equivalence Relations: Single Amounts with PP ⱖ CP
2
3
4
7
10
13
15
19
21
25
27
31
31
35
36
37
38
39
39
43
46
48
50
58
61
64
64
69
70
72
73
76
80
86
86
92
93
94
95
96
99
105
106
107
viii
Contents
4.6
4.7
4.8
4.9
LEARNING
STAGE 2
Chapter 5
Present Worth Analysis
5.1
5.2
5.3
5.4
5.5
Advantages and Uses of Annual Worth Analysis
Calculation of Capital Recovery and AW Values
Evaluating Alternatives by Annual Worth Analysis
AW of a Permanent Investment
Life-Cycle Cost Analysis
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—The Changing Scene of an Annual Worth Analysis
Rate of Return Analysis: One Project
7.1
7.2
7.3
7.4
7.5
7.6
Chapter 8
Progressive Example—Water for Semiconductor Manufacturing Case
Formulating Alternatives
Present Worth Analysis of Equal-Life Alternatives
Present Worth Analysis of Different-Life Alternatives
Future Worth Analysis
Capitalized Cost Analysis
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Comparing Social Security Benefits
Annual Worth Analysis
6.1
6.2
6.3
6.4
6.5
Chapter 7
109
112
114
116
117
118
122
124
BASIC ANALYSIS TOOLS
PE
Chapter 6
Equivalence Relations: Series with PP ⱖ CP
Equivalence Relations: Single Amounts and Series with PP ⬍ CP
Effective Interest Rate for Continuous Compounding
Interest Rates That Vary over Time
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Is Owning a Home a Net Gain or Net Loss over Time?
128
129
129
131
133
137
138
142
142
147
149
150
151
153
155
157
160
164
164
169
171
172
Interpretation of a Rate of Return Value
Rate of Return Calculation Using a PW or AW Relation
Special Considerations When Using the ROR Method
Multiple Rate of Return Values
Techniques to Remove Multiple Rates of Return
Rate of Return of a Bond Investment
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Developing and Selling an Innovative Idea
173
175
179
180
184
190
193
193
198
200
Rate of Return Analysis: Multiple Alternatives
202
8.1
8.2
8.3
8.4
8.5
8.6
Why Incremental Analysis Is Necessary
Calculation of Incremental Cash Flows for ROR Analysis
Interpretation of Rate of Return on the Extra Investment
Rate of Return Evaluation Using PW: Incremental and Breakeven
Rate of Return Evaluation Using AW
Incremental ROR Analysis of Multiple Alternatives
203
203
206
207
213
214
Contents
8.7
Chapter 9
All-in-One Spreadsheet Analysis (Optional)
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—ROR Analysis with Estimated Lives That Vary
Case Study—How a New Engineering Graduate Can Help His Father
Benefit/Cost Analysis and Public Sector Economics
PE
9.1
9.2
9.3
9.4
9.5
9.6
Progressive Example—Water Treatment Facility #3 Case
Public Sector Projects
Benefit/Cost Analysis of a Single Project
Alternative Selection Using Incremental B/C Analysis
Incremental B/C Analysis of Multiple, Mutually Exclusive Alternatives
Service Sector Projects and Cost-Effectiveness Analysis
Ethical Considerations in the Public Sector
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Comparing B/C Analysis and CEA of Traffic Accident Reduction
LEARNING
STAGE 2
EPILOGUE: SELECTING THE BASIC ANALYSIS TOOL
LEARNING
STAGE 3
MAKING BETTER DECISIONS
Chapter 10
Project Financing and Noneconomic Attributes
10.1
10.2
10.3
10.4
10.5
10.6
10.7
Chapter 11
MARR Relative to the Cost of Capital
Debt-Equity Mix and Weighted Average Cost of Capital
Determination of the Cost of Debt Capital
Determination of the Cost of Equity Capital and the MARR
Effect of Debt-Equity Mix on Investment Risk
Multiple Attribute Analysis: Identification and Importance of Each Attribute
Evaluation Measure for Multiple Attributes
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Which Is Better—Debt or Equity Financing?
Replacement and Retention Decisions
PE
11.1
11.2
11.3
11.4
11.5
11.6
Progressive Example—Keep or Replace the Kiln Case
Basics of a Replacement Study
Economic Service Life
Performing a Replacement Study
Additional Considerations in a Replacement Study
Replacement Study over a Specified Study Period
Replacement Value
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Will the Correct ESL Please Stand?
ix
218
219
220
225
226
227
228
229
230
235
238
242
246
250
251
252
258
259
266
267
269
271
273
275
278
282
283
284
289
290
292
293
294
296
302
306
307
312
312
313
319
321
x
Contents
Chapter 12
Independent Projects with Budget Limitation
12.1
12.2
12.3
12.4
12.5
Chapter 13
Breakeven and Payback Analysis
13.1
13.2
13.3
13.4
LEARNING
STAGE 4
Chapter 14
Effects of Inflation
Understanding the Impact of Inflation
Present Worth Calculations Adjusted for Inflation
Future Worth Calculations Adjusted for Inflation
Capital Recovery Calculations Adjusted for Inflation
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Inflation versus Stock and Bond Investments
Cost Estimation and Indirect Cost Allocation
15.1
15.2
15.3
15.4
15.5
15.6
15.7
15.8
Chapter 16
Breakeven Analysis for a Single Project
Breakeven Analysis Between Two Alternatives
Payback Analysis
More Breakeven and Payback Analysis on Spreadsheets
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Water Treatment Plant Process Costs
322
323
325
327
329
332
334
334
338
340
341
345
348
352
355
355
361
363
ROUNDING OUT THE STUDY
14.1
14.2
14.3
14.4
Chapter 15
An Overview of Capital Rationing among Projects
Capital Rationing Using PW Analysis of Equal-Life Projects
Capital Rationing Using PW Analysis of Unequal-Life Projects
Capital Budgeting Problem Formulation Using Linear Programming
Additional Project Ranking Measures
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Understanding How Cost Estimation Is Accomplished
Unit Method
Cost Indexes
Cost-Estimating Relationships: Cost-Capacity Equations
Cost-Estimating Relationships: Factor Method
Traditional Indirect Cost Rates and Allocation
Activity-Based Costing (ABC) for Indirect Costs
Making Estimates and Maintaining Ethical Practices
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Indirect Cost Analysis of Medical Equipment Manufacturing Costs
Case Study—Deceptive Acts Can Get You in Trouble
Depreciation Methods
16.1
16.2
16.3
16.4
16.5
Depreciation Terminology
Straight Line (SL) Depreciation
Declining Balance (DB) and Double Declining Balance (DDB) Depreciation
Modified Accelerated Cost Recovery System (MACRS)
Determining the MACRS Recovery Period
366
367
369
374
377
378
379
384
385
386
387
390
391
394
395
397
401
403
404
404
410
411
412
414
415
418
419
422
426
Contents
16.6
Depletion Methods
Chapter Summary
Appendix
16A.1 Sum-of-Years-Digits (SYD) and Unit-of-Production (UOP) Depreciation
16A.2 Switching between Depreciation Methods
16A.3 Determination of MACRS Rates
Problems
Additional Problems and FE Exam Review Questions
Appendix Problems
Chapter 17
After-Tax Economic Analysis
17.1
17.2
17.3
17.4
17.5
17.6
17.7
17.8
17.9
Chapter 18
Sensitivity Analysis and Staged Decisions
18.1
18.2
18.3
18.4
18.5
18.6
Chapter 19
Determining Sensitivity to Parameter Variation
Sensitivity Analysis Using Three Estimates
Estimate Variability and the Expected Value
Expected Value Computations for Alternatives
Staged Evaluation of Alternatives Using a Decision Tree
Real Options in Engineering Economics
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Sensitivity to the Economic Environment
Case Study—Sensitivity Analysis of Public Sector Projects—Water Supply Plans
More on Variation and Decision Making under Risk
19.1
19.2
19.3
19.4
19.5
Appendix A
Income Tax Terminology and Basic Relations
Calculation of Cash Flow after Taxes
Effect on Taxes of Different Depreciation Methods and Recovery Periods
Depreciation Recapture and Capital Gains (Losses)
After-Tax Evaluation
After-Tax Replacement Study
After-Tax Value-Added Analysis
After-Tax Analysis for International Projects
Value-Added Tax
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—After-Tax Analysis for Business Expansion
Interpretation of Certainty, Risk, and Uncertainty
Elements Important to Decision Making under Risk
Random Samples
Expected Value and Standard Deviation
Monte Carlo Sampling and Simulation Analysis
Chapter Summary
Problems
Additional Problems and FE Exam Review Questions
Case Study—Using Simulation and Three-Estimate Sensitivity Analysis
Using Spreadsheets and Microsoft Excel©
A.1
A.2
A.3
A.4
A.5
A.6
Introduction to Using Excel
Organization (Layout) of the Spreadsheet
Excel Functions Important to Engineering Economy
Goal Seek—A Tool for Breakeven and Sensitivity Analysis
Solver—An Optimizing Tool for Capital Budgeting, Breakeven, and Sensitivity Analysis
Error Messages
xi
427
429
430
430
432
435
438
442
443
444
445
448
450
453
456
462
465
468
470
472
473
481
482
484
485
490
491
492
494
498
503
503
509
510
511
514
515
518
523
526
533
540
540
543
544
547
547
549
550
558
559
560
xii
Contents
Appendix B
Basics of Accounting Reports and Business Ratios
B.1
B.2
B.3
The Balance Sheet
Income Statement and Cost of Goods Sold Statement
Business Ratios
561
561
562
563
Appendix C
Code of Ethics for Engineers
566
Appendix D
Alternate Methods for Equivalence Calculations
569
D.1
D.2
Appendix E
Glossary of Concepts and Terms
E.1
E.2
Reference Materials
Factor Tables
Photo Credits
Index
Using Programmable Calculators
Using the Summation of a Geometric Series
579
581
610
611
Important Concepts and Guidelines
Symbols and Terms
569
570
573
573
576
PREFACE TO SEVENTH EDITION
This edition includes the time-tested approach and topics of previous editions and introduces significantly new print and electronic features useful to learning about and successfully applying the exciting field of engineering economics. Money makes a huge difference in the life of a corporation, an
individual, and a government. Learning to understand, analyze, and manage the money side of any
project is vital to its success. To be professionally successful, every engineer must be able to deal with
the time value of money, economic facts, inflation, cost estimation, tax considerations, as well as
spreadsheet and calculator use. This book is a great help to the learner and the instructor in accomplishing these goals by using easy-to-understand language, simple graphics, and online features.
What's New and What's Best
This seventh edition is full of new information and features. Plus the supporting online materials
are new and updated to enhance the teaching and learning experience.
New topics:
• Ethics and the economics of engineering
• Service sector projects and their evaluation
• Real options development and analysis
• Value-added taxes and how they work
• Multiple rates of return and ways to eliminate them using spreadsheets
• No tabulated factors needed for equivalence computations (Appendix D)
New features in print and online:
• Totally new design to highlight important terms, concepts, and decision guidelines
• Progressive examples that continue throughout a chapter
• Downloadable online presentations featuring voice-over slides and animation
• Vital concepts and guidelines identified in margins; brief descriptions available (Appendix E)
• Fresh spreadsheet displays with on-image comments and function details
• Case studies (21 of them) ranging in topics from ethics to energy to simulation
Retained features:
• Many end-of-chapter problems (over 90% are new or revised)
• Easy-to-read language to enhance understanding in a variety of course environments
• Fundamentals of Engineering (FE) Exam review questions that double as additional or
review problems for quizzes and tests
• Hand and spreadsheet solutions presented for many examples
• Flexible chapter ordering after fundamental topics are understood
• Complete solutions manual available online (with access approval for instructors)
How to Use This Text
This textbook is best suited for a one-semester or one-quarter undergraduate course. Students
should be at the sophomore level or above with a basic understanding of engineering concepts
and terminology. A course in calculus is not necessary; however, knowledge of the concepts in
advanced mathematics and elementary probability will make the topics more meaningful.
Practitioners and professional engineers who need a refresher in economic analysis and cost
estimation will find this book very useful as a reference document as well as a learning medium.
Chapter Organization and Coverage Options
The textbook contains 19 chapters arranged into four learning stages, as indicated in the flowchart
on the next page, and five appendices. Each chapter starts with a statement of purpose and a specific learning outcome (ABET style) for each section. Chapters include a summary, numerous
xiv
Preface to Seventh Edition
CHAPTERS IN EACH LEARNING STAGE
Composition by level
Chapter 1
Foundations of
Engineering Economy
Chapter 2
Factors: How Time and
Interest Affect Money
Learning
Stage 1:
The
Fundamentals
Chapter 3
Combining Factors and
Spreadsheet Functions
Chapter 4
Nominal and Effective
Interest Rates
Learning
Stage 2:
Basic
Analysis
Tools
Chapter 5
Present Worth
Analysis
Chapter 6
Annual Worth
Analysis
Chapter 7
Rate of Return
Analysis:
One Project
Chapter 8
Rate of Return
Analysis: Multiple
Alternatives
Chapter 9
Benefit/Cost Analysis
and Public Sector
Economics
Learning Stage 2 Epilogue
Selecting the Basic
Analysis Tool
Learning
Stage 3:
Making
Better
Decisions
Learning
Stage 4:
Rounding
Out the
Study
Chapter 10
Project Financing and
Noneconomic Attributes
Chapter 14
Effects of
Inflation
Chapter 11
Replacement and
Retention Decisions
Chapter 15
Cost Estimation and
Indirect Cost Allocation
Chapter 12
Independent Projects
with Budget Limitation
Chapter 13
Breakeven and
Payback Analysis
Chapter 16
Depreciation
Methods
Chapter 18
Sensitivity Analysis
and Staged Decisions
Chapter 17
After-Tax Economic
Analysis
Chapter 19
More on Variation
and Decision Making
under Risk
end-of-chapter problems (essay and numerical), multiple-choice problems useful for course review and FE Exam preparation, and a case study.
The appendices are important elements of learning for this text:
Appendix A
Appendix B
Appendix C
Appendix D
Appendix E
Spreadsheet layout and functions (Excel is featured)
Accounting reports and business ratios
Code of Ethics for Engineers (from NSPE)
Equivalence computations using calculators and geometric series; no tables
Concepts, guidelines, terms, and symbols for engineering economics
There is considerable flexibility in the sequencing of topics and chapters once the first six
chapters are covered, as shown in the progression graphic on the next page. If the course is designed to emphasize sensitivity and risk analysis, Chapters 18 and 19 can be covered immediately
Chapter Organization and Coverage Options
CHAPTER AND TOPIC PROGRESSION OPTIONS
Topics may be introduced at the point indicated or any point thereafter
(Alternative entry points are indicated by
)
Numerical progression
through chapters
Inflation
Cost
Estimation
Taxes and
Depreciation
Sensitivity, Staged
Decisions, and Risk
1. Foundations
2. Factors
3. More Factors
4. Effective i
5. Present Worth
6. Annual Worth
7. Rate of Return
8. More ROR
9. Benefit/Cost
10. Financing and
Noneconomic Attributes
11. Replacement
12. Capital Budgeting
13. Breakeven and
Payback
14. Inflation
15. Estimation
16. Depreciation
17. After-Tax
18. Sensitivity, Decision
Trees, and Real Options
19. Risk and Simulation
after Learning Stage 2 (Chapter 9) is completed. If depreciation and tax emphasis are vitally
important to the goals of the course, Chapters 16 and 17 can be covered once Chapter 6 (annual
worth) is completed. The progression graphic can help in the design of the course content and
topic ordering.
xv
SAMPLE OF RESOURCES FOR
LEARNING OUTCOMES
L E A R N I N G
Each chapter begins with a purpose, list
of topics, and learning outcomes
(ABET style) for each corresponding
section. This behavioral-based
approach sensitizes the reader to what
is ahead, leading to improved
understanding and learning.
O U T C O M E S
Purpose: Use multiple factors and spreadsheet functions to find equivalent amounts for cash flows that have nonstandard placement.
SECTION
TOPIC
LEARNING OUTCOME
3.1
Shifted series
• Determine the P, F or A values of a series
starting at a time other than period 1.
3.2
Shifted series and single cash
flows
• Determine the P, F, or A values of a shifted series
and randomly placed single cash flows.
3.3
Shifted gradients
• Make equivalence calculations for shifted
arithmetic or geometric gradient series that
increase or decrease in size of cash flows.
CONCEPTS AND GUIDELINES
Time value of money
It is a well-known fact that money makes money. The time value of money explains the change
in the amount of money over time for funds that are owned (invested) or owed (borrowed).
This is the most important concept in engineering economy.
IN-CHAPTER EXAMPLES
To highlight the fundamental building
blocks of the course, a checkmark and title
in the margin call attention to particularly
important concepts and decision-making
guidelines. Appendix E includes a brief
description of each fundamental concept.
EXAMPLE 4.6
Numerous in-chapter examples
throughout the book reinforce the
basic concepts and make
understanding easier. Where
appropriate, the example is solved
using separately marked hand and
spreadsheet solutions.
A dot-com company plans to place money in a new venture capital fund that currently returns
18% per year, compounded daily. W hat effective rate is this ( a ) yearly and ( b ) semiannually?
Solution
( a ) Use Equation [4.7], with
r
⫽ 0.18 and m ⫽ 365.
(
( b ) Here
r
0.18
—
Effective i % per year ⫽ 1 ⫹ —
365
⫽ 0.09 per 6 months and m ⫽ 182 days.
(
)
365
0.09
—
Effective i % per 6 months ⫽ 1 ⫹ —
182
⫺ 1 ⫽ 19.716%
)
182
⫺ 1 ⫽ 9.415%
PROGRESSIVE EXAMPLES
PE
Water for Semiconductor Manufacturing Case: The worldwide contribution of
semiconductor sales is about $250 billion
per year, or about 10% of the world’s
GDP (gross domestic product). This industry produces the microchips used in many
of the communication, entertainment,
transportation, and computing devices
we use every day. Depending upon the
type and size of fabrication plant (fab),
the need for ultrapure water (UPW) to
manufacture these tiny integrated circuits
is high, ranging from 500 to 2000 gpm
(gallons per minute). Ultrapure water is
obtained by special processes that commonly include reverse osmosis兾deionizing
resin bed technologies. Potable water
obtained from purifying seawater or
brackish groundwater may cost from
$2 to $3 per 1000 gallons, but to obtain
UPW on-site for semiconductor manufacturing may cost an additional $1 to $3 per
1000 gallons.
A fab costs upward of $2.5 billion to
construct, with approximately 1% of this
total, or $25 million, required to provide
the ultrapure water needed, including
the necessary wastewater and recycling
equipment.
A newcomer to the industry, Angular
Enterprises, has estimated the cost profiles for two options to supply its anticipated fab with water. It is fortunate to
have the option of desalinated seawater
or purified groundwater sources in the
location chosen for its new fab. The initial cost estimates for the UPW system are
given below.
Source
Equipment first
cost, $M
AOC, $M per year
Seawater
(S)
⫺20
Groundwater
(G)
⫺22
⫺0.5
⫺0.3
Salvage value, % of
first cost
5
10
Cost of UPW, $ per
1000 gallons
4
5
Angular has made some initial estimates
for the UPW system.
Life of UPW equipment 10 years
UPW needs
1500 gpm
Operating time
16 hours per
day for 250 days
per year
This case is used in the following topics
(Sections) and problems of this chapter:
PW analysis of equal-life alternatives
(Section 5.2)
PW analysis of different-life alternatives (Section 5.3)
Capitalized cost analysis (Section 5.5)
Problems 5.20 and 5.34
Several chapters include a progressive
example—a more detailed problem statement
introduced at the beginning of the chapter and
expanded upon throughout the chapter in
specially marked examples. This approach
illustrates different techniques and some
increasingly complex aspects of a real-world
problem.
INSTRUCTORS AND STUDENTS
xvii
Contents
ONLINE PRESENTATIONS
An icon in the margin indicates the
availability of an animated voice-over slide
presentation that summarizes the material in
the section and provides a brief example for
learners who need a review or prefer videobased materials. Presentations are keyed to
the sections of the text.
3.1 Calculations for Uniform Series That Are Shifted
W hen a un i form ser i es beg i ns at a t i m e other than at the end of per i od 1, i t i s called a shifted
series. In th i s case several m ethods can be used to fi nd the equ i valent present w orth P . For
exam ple, P of the un i form ser i es show n i n F i gure 3–1 could be determ i ned by any of the
follow i ng m ethods :
• Use the P 兾F factor to fi nd the present worth of each d i sbursement at year 0 and add them .
• Use the F 兾P factor to fi nd the future worth of each d i sbursement i n year 13, add them, and
then fi nd the present worth of the total, us i ng P ⫽ F ( P 兾F ,i ,13) .
• Use the F 兾A factor to fi nd the future amount F ⫽ A ( F 兾A ,i ,10), and then compute the present
worth, us i ng P ⫽ F ( P 兾F ,i ,13) .
• Use the P 兾A factor to compute the “present worth” P 3 ⫽ A ( P 兾A ,i ,10) (wh i ch w i ll be located
i n year 3, not year 0), and then fi nd the present worth i n year 0 by us i ng the ( P 兾F , i ,3) factor.
MARR
Filter 1 ROR ⬇ 25%
Filter 2 ROR ⬇ 23%
MARR
Breakeven ROR ⬇ 17%
EXAMPLE 13.8
Breakeven
C h ris a n d h e r fa th e r ju st pIncremental
u rc h a se d a ROR
sm a ll⬇o17%
ffi c e b u ild in g fo r $ 1 6 0 ,0 0 0 th a t is in n e e d o f a lo t
o f re p a irs, b u t is lo c a te d in a p rim e c o m m e rc ia l a re a o f th e c ity. T h e e stim a te d c o sts e a c h y e a r
fo r re p a irs, in su ra n c e , e tc . a re $ 1 8 ,0 0 0 th e fi rst y e a r, in c re a sin g b y $ 1 0 0 0 p e r y e a r th e re a fte r.
A t a n 8–6
e x p e c te d 8 % p e r y e a r re tu rn , u se sp re a d sh e e t a n a ly sis to d e te rm in e th e p a y b a c k p e rio d
Figure
if versus
th e b uiild
in g and
is ( a
) k eversus
p t fo rincremental
2 y e a rs a nidgraph,
so ld fo
r $ 2 9 08.4
,0 0. 0 so m e tim e b e y o n d y e a r 2 o r ( b) k e p t
PW
graph
PW
Example
fo r 3 y e a rs a n d so ld fo r $ 3 7 0 ,0 0 0 so m e tim e b e y o n d 3 y e a rs.
Solution
F ig u re 1 3 – 11 sh o w s th e a n n u a l c o sts (c o lu m n B ) a n d th e sa le s p ric e s if th e b u ild in g is k e p t 2
o r 3 y e a rs (c o lu m n s C a n d E , re sp e c tiv e ly ). T h e N P V fu n c tio n is a p p lie d (c o lu m n s D a n d F ) to
d e te rm in e w h e n th e P W c h a n g e s sig n fro m p lu s to m in u s. T h e se re su lts b ra c k e t th e p a y b a c k
p e rio d fo r e a c h re te n tioFigure
n p e rio d 7–12
a n d sa le s p ric e . W h e n P W ⬎ 0 , th e 8 % re tu rn is e x c e e d e d .
Goal
Seek,
7.6.
p a y Spreadsheet
b a c k p e rio d is application
b e tw e e n 3 a nof
d 4ROIC
y e a rs method
(c o lu m n using
D ). If th
e b u ild
in g isExample
so ld
( a) T h e 8 % re tu rn
a fte r e x a c tly 3 y e a rs fo r $ 2 9 0 ,0 0 0 , th e p a y b a c k p e rio d w a s n o t e x c e e d e d ; b u t a fte r 4 y e a rs
it is e x c e e d e d .
( b) A t a sa le s p ric e o f $ 3 7 0 ,0 0 0 , th e 8 % re tu rn p a y b a c k p e rio d is b e tw e e n 5 a n d 6 y e a rs (c o lu m n F ). If th e b u ild in g is so ld a fte r 4 o r 5 y e a rs, th e p a y b a c k is n o t e x c e e d e d ; h o w e v e r, a
sa le a fte r 6 y e a rs is b e y o n d th e 8 % -re tu rn p a y b a c k p e rio d .
If k e p t 2 y e a rs a n d
so ld , p a y b a c k is
b e tw e e n 3 a n d 4
If k e p t 3 y e a rs a n d
so ld , p a y b a c k is
b e tw e e n 5 a n d 6
SPREADSHEETS
The text integrates spreadsheets to show
how easy they are to use in solving virtually
any type of engineering economic analysis
problem. Cell tags or full cells detail
built-in functions and relations developed
to solve a specific problem.
⫽ N P V (8 % ,$ B $ 4 :B 7 )+ $ B $ 3 ⫺ P V (8 % ,A 7 ,,2 9 0 0 0 0 )
Figure 13–11
P a y b a c k p e rio d a n a ly sis, E x a m p le 1 3 .8
FE EXAM AND COURSE
REVIEWS
Each chapter concludes with several
multiple-choice, FE Exam–style
problems that provide a simplified
review of chapter material. Additionally,
these problems cover topics for test
reviews and homework assignments.
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
8.38 When conducting a rate of return (ROR) analysis
involving multiple mutually exclusive alternatives, the first step is to:
(a) Rank the alternatives according to decreasing initial investment cost
(b) Rank the alternatives according to increasing
initial investment cost
(c) Calculate the present worth of each alternative using the MARR
(d) Find the LCM between all of the alternatives
8.39 In comparing mutually exclusive alternatives by
the ROR method, you should:
(a) Find the ROR of each alternative and pick
the one with the highest ROR
(b) S l
h
l
i
h
i
l
8.43 For these alternatives, the sum of the incremental
cash flows is:
Year
A
B
0
1
2
3
4
5
⫺10,000
⫹2,500
⫹2,500
⫹2,500
⫹2,500
⫹2,500
⫺14,000
⫹4,000
⫹4,000
⫹4,000
⫹4,000
⫹4,000
(a)
(b)
(c)
(d)
$2500
$3500
$6000
$8000
CASE STUDIES
CASE STUDY
RENEWABLE ENERGY SOURCES FOR ELECTRICITY GENERATION
PEC average cost to residential customers: 10.27 ¢/kWh
(from primary sources) and 10.92 ¢/kWh (renewable sources)
B ack grou n d
Pedernales Electric Cooperative (PEC) is the largest
member-owned electric co-op in the United States with over
232,000 meters in 12 Central Texas counties. PEC has a capacity of approximately 1300 MW (megawatts) of power, of
which 277 MW, or about 21%, is from renewable sources.
The latest addition is 60 MW of power from a wind farm in
south Texas close to the city of Corpus Christi. A constant
question is how much of PEC’ s generation capacity sh ould be
from renewable sources, especially given the environmental
issues with coal-generated electricity and the rising costs of
hydrocarbon fuels.
Wind and nuclear sources are the current consideration for
the PEC leadership as Texas is increasing its generation by
nuclear power and the state is the national leader in wind
farm–produced electricity.
Consider yourself a member of the board of directors of
PEC. You are an engineer who has been newly elected by the
PEC membership to serve a 3-year term as a director-at-large.
As such, you do not represent a specifi c district within the
entire service area; all other directors do represent a specifi c
district. You have many questions about the operations of
PEC, plus you are interested in the economic and societal
benefi ts of pursuing more renewable source generation
capacity.
Expected life of a generation facility: 20 to 40 years (it is
likely closer to 20 than 40)
Time to construct a facility: 2 to 5 years
Capital cost to build a generation facility: $900 to $1500
per kW
You have also learned that the PEC staff uses the wellrecognized l e v e l i z e d e n e r g y c o s t (LEC) method to determine
the price of electricity that must be charged to customers to
break even. The formula takes into account the capital cost of
the generation facilities, the cost of capital of borrowed
money, annual maintenance and operation (M&O) costs, and
the expected life of the facility. The LEC formula, expressed
in dollars per kW h for ( t ⫽ 1, 2, . . . , n ), is
t ⫽n
⌺ — —⫹(1—⫹—⫹—) —
P
t ⫽1
where
P
t
A
t
t
E
t
4 to 9
4 to 10.5
4.8 to 9.1
4.5 to 15.5
Reasonable Average
7.4
8.6
8.2
8.8
National average cost of electricity to residential customers: 11¢/kW h
t
i
t
t
⫽ fuel costs for year t
⫽ amount of electricity generated in year t
⫽ expected life of facility
⫽ discount rate (cost of capital)
C a s e S tu d y E x e r c is e s
1.
Generation Cost, ¢/kWh
Coal
Natural gas
Wind
Solar
t
t
⫽ capital investments made in year t
⫽ annual maintenance and operating (M&O) costs
for year
C
i
Likely Range
C
t
——
⌺—
(1 ⫹ )
E
n
Fuel Source
A
i
I n fo r m a tio n
Here are some data that you have obtained. The information
is sketchy, as this point, and the numbers are very approximate. Electricity generation cost estimates are national
in scope, not PEC-specifi c, and are provided in cents per
kilowatt-hour (¢/kW h).
t
t ⫽1
—————
LEC ⫽ — —
t ⫽n
2.
If you wanted to know more about the new arrangement with the wind farm in south Texas for the additional 60 M W per year, what types of questions would
you ask of a staff member in your fi rst meeting with
him or her?
Much of the current generation capacity of PEC facilities
utilizes coal and natural gas as the primary fuel source.
What about the ethical aspects of the government’ sallowance for these plants to continue polluting the atmosphere
with the emissions that may cause health problems for
citizens and further the effects of global warming? What
types of regulations, if any, should be developed for PEC
(and other generators) to follow in the future?
New and updated case studies at the
end of most chapters present realworld, in-depth treatments and
exercises in the engineering
profession. Each case includes a
background, relevant information,
and an exercise section.
ACKNOWLEDGMENT OF CONTRIBUTORS
It takes the input and efforts of many individuals to make significant improvements in a textbook.
We wish to give special thanks to the following persons for their contributions to this edition.
Paul Askenasy, Texas Commission on Environmental Quality
Jack Beltran, Bristol-Myers Squibb
Robert Lundquist, Ohio State University
William Peet, Infrastructure Coordination, Government of Niue
Sallie Sheppard, Texas A&M University
We thank the following individuals for their comments, feedback, and review of material to assist
in making this edition a real success.
Ahmed Alim, University of Houston
Alan Atalah, Bowling Green State University
Fola Michael Ayokanmbi, Alabama A&M University
William Brown, West Virginia University at Parkersburg
Hector Carrasco, Colorado State University–Pueblo
Robert Chiang, California State University, Pomona
Ronald Cutwright, Florida State University
John F. Dacquisto, Gonzaga University
Houshang Darabi, University of Illinois at Chicago
Freddie Davis, West Texas A&M University
Edward Lester Dollar, Southern Polytechnic State University
Ted Eschenbach, University of Alaska
Clara Fang, University of Hartford
Abel Fernandez, University of the Pacific
Daniel A. Franchi, California Polytechnic State University, San Luis Obispo
Mark Frascatore, Clarkson University
Benjamin M. Fries, University of Central Florida
Nathan Gartner, University of Massachusetts–Lowell
Johnny R. Graham, University of North Carolina–Charlotte
Liling Huang, Northern Virginia Community College
David Jacobs, University of Hartford
Adam Jannik, Northwestern State University
Peter E. Johnson, Valparaiso University
Justin W. Kile, University of Wisconsin–Platteville
John Kushner, Lawrence Technological University
Clifford D. Madrid, New Mexico State University
Saeed Manafzadeh, University of Illinois at Chicago
Quamrul Mazumder, University of Michigan–Flint
Deb McAvoy, Ohio University
Gene McGinnis, Christian Brothers University
Bruce V. Mutter, Bluefield State College
Hong Sioe Oey, University of Texas at El Paso
Richard Palmer, University of Massachusetts
Michael J. Rider, Ohio Northern University
John Ristroph, University of Louisiana at Lafayette
Saeid L. Sadri, Georgia Institute of Technology
Scott Schultz, Mercer University
Kyo D. Song, Norfolk State University
James Stevens, University of Colorado at Colorado Springs
John A. Stratton, Rochester Institute of Technology
Mathias J. Sutton, Purdue University
Pete Weiss, Valparaiso University
xx
Acknowledgment of Contributors
Greg Wiles, Southern Polytechnic State University
Richard Youchak, University of Pittsburgh at Johnstown
William A. Young, II, Ohio University
If you discover errors that require correction in the next printing of the textbook or in updates of
the online resources, please contact us. We hope you find the contents of this edition helpful in
your academic and professional activities.
Leland Blank
Anthony Tarquin
lelandblank@yahoo.com
atarquin@utep.edu
L E A R N I N G S TA G E 1
The Fundamentals
LEARNING STAGE 1
The Fundamentals
CHAPTER
1
Foundations of
Engineering Economy
CHAPTER
2
Factors: How Time
and Interest Affect
Money
CHAPTER
3
Combining Factors
and Spreadsheet
Functions
CHAPTER
4
Nominal and Effective
Interest Rates
T
he fundamentals of engineering economy are introduced in
these chapters. When you have completed stage 1, you will be
able to understand and work problems that account for the
time value of money, cash flows occurring at different times with
different amounts, and equivalence at different interest rates. The
techniques you master here form the basis of how an engineer in
any discipline can take economic value into account in virtually any
project environment.
The factors commonly used in all engineering economy computations are introduced and applied here. Combinations of these factors assist in moving monetary values forward and backward through
time and at different interest rates. Also, after these chapters, you
should be comfortable using many of the spreadsheet functions.
Many of the terms common to economic decision making are
introduced in learning stage 1 and used in later chapters. A checkmark icon in the margin indicates that a new concept or guideline
is introduced at this point.
CHAPTER 1
Foundations
of Engineering
Economy
L E A R N I N G
O U T C O M E S
Purpose: Understand and apply fundamental concepts and use the terminology of engineering economics.
SECTION
TOPIC
LEARNING OUTCOME
1.1
Description and role
• Define engineering economics and describe its
role in decision making.
1.2
Engineering economy study
approach
• Understand and identify the steps in an
engineering economy study.
1.3
Ethics and economics
• Identify areas in which economic decisions can
present questionable ethics.
1.4
Interest rate
• Perform calculations for interest rates and rates
of return.
1.5
Terms and symbols
• Identify and use engineering economic
terminology and symbols.
1.6
Cash flows
• Understand cash flows and how to graphically
represent them.
1.7
Economic equivalence
• Describe and calculate economic equivalence.
1.8
Simple and compound interest
• Calculate simple and compound interest
amounts for one or more time periods.
1.9
MARR and opportunity cost
• State the meaning and role of Minimum
Attractive Rate of Return (MARR) and
opportunity costs.
1.10
Spreadsheet functions
• Identify and use some Excel functions
commonly applied in engineering economics.
T
he need for engineering economy is primarily motivated by the work that engineers
do in performing analyses, synthesizing, and coming to a conclusion as they work on
projects of all sizes. In other words, engineering economy is at the heart of making
decisions. These decisions involve the fundamental elements of cash flows of money, time,
and interest rates. This chapter introduces the basic concepts and terminology necessary for
an engineer to combine these three essential elements in organized, mathematically correct
ways to solve problems that will lead to better decisions.
1.1 Engineering Economics: Description and
Role in Decision Making
Decisions are made routinely to choose one alternative over another by individuals in everyday
life; by engineers on the job; by managers who supervise the activities of others; by corporate
presidents who operate a business; and by government officials who work for the public good.
Most decisions involve money, called capital or capital funds, which is usually limited in
amount. The decision of where and how to invest this limited capital is motivated by a primary
goal of adding value as future, anticipated results of the selected alternative are realized.
Engineers play a vital role in capital investment decisions based upon their ability and experience
to design, analyze, and synthesize. The factors upon which a decision is based are commonly a
combination of economic and noneconomic elements. Engineering economy deals with the
economic factors. By definition,
Engineering economy involves formulating, estimating, and evaluating the expected economic
outcomes of alternatives designed to accomplish a defined purpose. Mathematical techniques
simplify the economic evaluation of alternatives.
Because the formulas and techniques used in engineering economics are applicable to all
types of money matters, they are equally useful in business and government, as well as for
individuals. Therefore, besides applications to projects in your future jobs, what you learn
from this book and in this course may well offer you an economic analysis tool for making
personal decisions such as car purchases, house purchases, major purchases on credit, e.g.,
furniture, appliances, and electronics.
Other terms that mean the same as engineering economy are engineering economic analysis,
capital allocation study, economic analysis, and similar descriptors.
People make decisions; computers, mathematics, concepts, and guidelines assist people in
their decision-making process. Since most decisions affect what will be done, the time frame of
engineering economy is primarily the future. Therefore, the numbers used in engineering economy are best estimates of what is expected to occur. The estimates and the decision usually
involve four essential elements:
Cash flows
Times of occurrence of cash flows
Interest rates for time value of money
Measure of economic worth for selecting an alternative
Since the estimates of cash flow amounts and timing are about the future, they will be somewhat different than what is actually observed, due to changing circumstances and unplanned
events. In short, the variation between an amount or time estimated now and that observed
in the future is caused by the stochastic (random) nature of all economic events. Sensitivity
analysis is utilized to determine how a decision might change according to varying estimates, especially those expected to vary widely. Example 1.1 illustrates the fundamental
nature of variation in estimates and how this variation may be included in the analysis at a
very basic level.
EXAMPLE 1.1
An engineer is performing an analysis of warranty costs for drive train repairs within the first
year of ownership of luxury cars purchased in the United States. He found the average cost (to
the nearest dollar) to be $570 per repair from data taken over a 5-year period.
4
Foundations of Engineering Economy
Chapter 1
Year
Average Cost, $/repair
2006
525
2007
430
2008
619
2009
650
2010
625
What range of repair costs should the engineer use to ensure that the analysis is sensitive to
changing warranty costs?
Solution
At first glance the range should be approximately –25% to ⫹15% of the $570 average cost to
include the low of $430 and high of $650. However, the last 3 years of costs are higher and
more consistent with an average of $631. The observed values are approximately ⫾3% of this
more recent average.
If the analysis is to use the most recent data and trends, a range of, say, ⫾5% of $630 is recommended. If, however, the analysis is to be more inclusive of historical data and trends, a range
of, say, ⫾20% or ⫾25% of $570 is recommended.
The criterion used to select an alternative in engineering economy for a specific set of estimates
is called a measure of worth. The measures developed and used in this text are
Present worth (PW)
Future worth (FW)
Annual worth (AW)
Rate of return (ROR)
Benefit/cost (B/C)
Capitalized cost (CC)
Payback period
Economic value added (EVA)
Cost Effectiveness
All these measures of worth account for the fact that money makes money over time. This is the
concept of the time value of money.
Time value of money
It is a well-known fact that money makes money. The time value of money explains the change
in the amount of money over time for funds that are owned (invested) or owed (borrowed).
This is the most important concept in engineering economy.
The time value of money is very obvious in the world of economics. If we decide to invest
capital (money) in a project today, we inherently expect to have more money in the future than
we invested. If we borrow money today, in one form or another, we expect to return the original
amount plus some additional amount of money.
Engineering economics is equally well suited for the future and for the analysis of past cash
flows in order to determine if a specific criterion (measure of worth) was attained. For example,
assume you invested $4975 exactly 3 years ago in 53 shares of IBM stock as traded on the New
York Stock Exchange (NYSE) at $93.86 per share. You expect to make 8% per year appreciation,
not considering any dividends that IBM may declare. A quick check of the share value shows it
is currently worth $127.25 per share for a total of $6744.25. This increase in value represents a
rate of return of 10.67% per year. (These type of calculations are explained later.) This past
investment has well exceeded the 8% per year criterion over the last 3 years.
1.2 Performing an Engineering Economy Study
An engineering economy study involves many elements: problem identification, definition of the
objective, cash flow estimation, financial analysis, and decision making. Implementing a structured procedure is the best approach to select the best solution to the problem.
The steps in an engineering economy study are as follows:
1.
2.
3.
4.
Identify and understand the problem; identify the objective of the project.
Collect relevant, available data and define viable solution alternatives.
Make realistic cash flow estimates.
Identify an economic measure of worth criterion for decision making.
Performing an Engineering Economy Study
1.2
5. Evaluate each alternative; consider noneconomic factors; use sensitivity analysis as needed.
6. Select the best alternative.
7. Implement the solution and monitor the results.
Technically, the last step is not part of the economy study, but it is, of course, a step needed to
meet the project objective. There may be occasions when the best economic alternative
requires more capital funds than are available, or significant noneconomic factors preclude the
most economic alternative from being chosen. Accordingly, steps 5 and 6 may result in selection
of an alternative different from the economically best one. Also, sometimes more than one project may be selected and implemented. This occurs when projects are independent of one another.
In this case, steps 5 through 7 vary from those above. Figure 1–1 illustrates the steps above for
one alternative. Descriptions of several of the elements in the steps are important to understand.
Problem Description and Objective Statement A succinct statement of the problem and
primary objective(s) is very important to the formation of an alternative solution. As an illustration, assume the problem is that a coal-fueled power plant must be shut down by 2015 due to the
production of excessive sulfur dioxide. The objectives may be to generate the forecasted electricity
Step in
study
1
Problem
description
Objective
statement
Available data
2
Alternatives for
solution
3
Cash flows and
other estimates
4
Measure of worth
criterion
5
Engineering
economic analysis
6
Best alternative
selection
7
Implementation
and monitoring
One or more approaches
to meet objective
Expected life
Revenues
Costs
Taxes
Project financing
PW, ROR, B/C, etc.
Consider:
• Noneconomic factors
• Sensitivity analysis
• Risk analysis
Time
passes
1
New problem
description
Figure 1–1
Steps in an engineering economy study.
New engineering
economy study
begins
5
6
Chapter 1
Foundations of Engineering Economy
needed for 2015 and beyond, plus to not exceed all the projected emission allowances in these
future years.
Alternatives These are stand-alone descriptions of viable solutions to problems that can meet
the objectives. Words, pictures, graphs, equipment and service descriptions, simulations, etc.
define each alternative. The best estimates for parameters are also part of the alternative. Some
parameters include equipment first cost, expected life, salvage value (estimated trade-in, resale,
or market value), and annual operating cost (AOC), which can also be termed maintenance and
operating (M&O) cost, and subcontract cost for specific services. If changes in income (revenue)
may occur, this parameter must be estimated.
Detailing all viable alternatives at this stage is crucial. For example, if two alternatives are
described and analyzed, one will likely be selected and implementation initiated. If a third, more
attractive method that was available is later recognized, a wrong decision was made.
Cash Flows All cash flows are estimated for each alternative. Since these are future expenditures and revenues, the results of step 3 usually prove to be inaccurate when an alternative is
actually in place and operating. When cash flow estimates for specific parameters are expected to
vary significantly from a point estimate made now, risk and sensitivity analyses (step 5) are
needed to improve the chances of selecting the best alternative. Sizable variation is usually expected in estimates of revenues, AOC, salvage values, and subcontractor costs. Estimation of
costs is discussed in Chapter 15, and the elements of variation (risk) and sensitivity analysis are
included throughout the text.
Engineering Economy Analysis The techniques and computations that you will learn and
use throughout this text utilize the cash flow estimates, time value of money, and a selected
measure of worth. The result of the analysis will be one or more numerical values; this can be
in one of several terms, such as money, an interest rate, number of years, or a probability. In
the end, a selected measure of worth mentioned in the previous section will be used to select
the best alternative.
Before an economic analysis technique is applied to the cash flows, some decisions about
what to include in the analysis must be made. Two important possibilities are taxes and
inflation. Federal, state or provincial, county, and city taxes will impact the costs of every
alternative. An after-tax analysis includes some additional estimates and methods compared to
a before-tax analysis. If taxes and inflation are expected to impact all alternatives equally, they
may be disregarded in the analysis. However, if the size of these projected costs is important,
taxes and inflation should be considered. Also, if the impact of inflation over time is important
to the decision, an additional set of computations must be added to the analysis; Chapter 14
covers the details.
Selection of the Best Alternative The measure of worth is a primary basis for selecting
the best economic alternative. For example, if alternative A has a rate of return (ROR) of
15.2% per year and alternative B will result in an ROR of 16.9% per year, B is better economically. However, there can always be noneconomic or intangible factors that must be
considered and that may alter the decision. There are many possible noneconomic factors;
some typical ones are
•
•
•
•
•
Market pressures, such as need for an increased international presence
Availability of certain resources, e.g., skilled labor force, water, power, tax incentives
Government laws that dictate safety, environmental, legal, or other aspects
Corporate management’s or the board of director’s interest in a particular alternative
Goodwill offered by an alternative toward a group: employees, union, county, etc.
As indicated in Figure 1–1, once all the economic, noneconomic, and risk factors have been
evaluated, a final decision of the “best” alternative is made.
At times, only one viable alternative is identified. In this case, the do-nothing (DN) alternative may be chosen provided the measure of worth and other factors result in the alternative being
a poor choice. The do-nothing alternative maintains the status quo.
1.3
Professional Ethics and Economic Decisions
Whether we are aware of it or not, we use criteria every day to choose between alternatives.
For example, when you drive to campus, you decide to take the “best” route. But how did you
define best? Was the best route the safest, shortest, fastest, cheapest, most scenic, or what? Obviously, depending upon which criterion or combination of criteria is used to identify the best, a
different route might be selected each time. In economic analysis, financial units (dollars or
other currency) are generally used as the tangible basis for evaluation. Thus, when there are
several ways of accomplishing a stated objective, the alternative with the lowest overall cost or
highest overall net income is selected.
1.3 Professional Ethics and Economic Decisions
Many of the fundamentals of engineering ethics are intertwined with the roles of money and
economics-based decisions in the making of professionally ethical judgments. Some of these
integral connections are discussed here, plus sections in later chapters discuss additional aspects
of ethics and economics. For example, Chapter 9, Benefit/Cost Analysis and Public Sector Economics, includes material on the ethics of public project contracts and public policy. Although it
is very limited in scope and space, it is anticipated that this coverage of the important role of
economics in engineering ethics will prompt further interest on the part of students and instructors of engineering economy.
The terms morals and ethics are commonly used interchangeably, yet they have slightly
different interpretations. Morals usually relate to the underlying tenets that form the character
and conduct of a person in judging right and wrong. Ethical practices can be evaluated by
using a code of morals or code of ethics that forms the standards to guide decisions and
actions of individuals and organizations in a profession, for example, electrical, chemical,
mechanical, industrial, or civil engineering. There are several different levels and types of
morals and ethics.
Universal or common morals These are fundamental moral beliefs held by virtually all people. Most people agree that to steal, murder, lie, or physically harm someone is wrong.
It is possible for actions and intentions to come into conflict concerning a common moral.
Consider the World Trade Center buildings in New York City. After their collapse on September 11,
2001, it was apparent that the design was not sufficient to withstand the heat generated by the
firestorm caused by the impact of an aircraft. The structural engineers who worked on the design
surely did not have the intent to harm or kill occupants in the buildings. However, their design
actions did not foresee this outcome as a measurable possibility. Did they violate the common
moral belief of not doing harm to others or murdering?
Individual or personal morals These are the moral beliefs that a person has and maintains
over time. These usually parallel the common morals in that stealing, lying, murdering, etc. are
immoral acts.
It is quite possible that an individual strongly supports the common morals and has excellent
personal morals, but these may conflict from time to time when decisions must be made. Consider the engineering student who genuinely believes that cheating is wrong. If he or she does not
know how to work some test problems, but must make a certain minimum grade on the final
exam to graduate, the decision to cheat or not on the final exam is an exercise in following or
violating a personal moral.
Professional or engineering ethics Professionals in a specific discipline are guided in their
decision making and performance of work activities by a formal standard or code. The code
states the commonly accepted standards of honesty and integrity that each individual is expected
to demonstrate in her or his practice. There are codes of ethics for medical doctors, attorneys,
and, of course, engineers.
Although each engineering profession has its own code of ethics, the Code of Ethics for
Engineers published by the National Society of Professional Engineers (NSPE) is very commonly used and quoted. This code, reprinted in its entirety in Appendix C, includes numerous
sections that have direct or indirect economic and financial impact upon the designs, actions,
7
8
Chapter 1
Foundations of Engineering Economy
and decisions that engineers make in their professional dealings. Here are three examples from
the Code:
“Engineers, in the fulfillment of their duties, shall hold paramount the safety, health, and welfare of the public.” (section I.1)
“Engineers shall not accept financial or other considerations, including free engineering designs, from material or equipment suppliers for specifying their product.” (section III.5.a)
“Engineers using designs supplied by a client recognize that the designs remain the property
of the client and may not be duplicated by the engineer for others without express permission.”
(section III.9.b)
As with common and personal morals, conflicts can easily rise in the mind of an engineer
between his or her own ethics and that of the employing corporation. Consider a manufacturing
engineer who has recently come to firmly disagree morally with war and its negative effects on
human beings. Suppose the engineer has worked for years in a military defense contractor’s
facility and does the detailed cost estimations and economic evaluations of producing fighter
jets for the Air Force. The Code of Ethics for Engineers is silent on the ethics of producing and
using war materiel. Although the employer and the engineer are not violating any ethics code,
the engineer, as an individual, is stressed in this position. Like many people during a declining
national economy, retention of this job is of paramount importance to the family and the engineer. Conflicts such as this can place individuals in real dilemmas with no or mostly unsatisfactory
alternatives.
At first thought, it may not be apparent how activities related to engineering economics may
present an ethical challenge to an individual, a company, or a public servant in government service. Many money-related situations, such as those that follow, can have ethical dimensions.
In the design stage:
• Safety factors are compromised to ensure that a price bid comes in as low as possible.
• Family or personal connections with individuals in a company offer unfair or insider information that allows costs to be cut in strategic areas of a project.
• A potential vendor offers specifications for company-specific equipment, and the design engineer does not have sufficient time to determine if this equipment will meet the needs of the
project being designed and costed.
While the system is operating:
• Delayed or below-standard maintenance can be performed to save money when cost overruns
exist in other segments of a project.
• Opportunities to purchase cheaper repair parts can save money for a subcontractor working on
a fixed-price contract.
• Safety margins are compromised because of cost, personal inconvenience to workers, tight
time schedules, etc.
A good example of the last item—safety is compromised while operating the system—is the
situation that arose in 1984 in Bhopal, India (Martin and Schinzinger 2005, pp. 245–8). A Union
Carbide plant manufacturing the highly toxic pesticide chemical methyl isocyanate (MIC) experienced a large gas leak from high-pressure tanks. Some 500,000 persons were exposed to inhalation of this deadly gas that burns moist parts of the body. There were 2500 to 3000 deaths within
days, and over the following 10-year period, some 12,000 death claims and 870,000 personal
injury claims were recorded. Although Union Carbide owned the facility, the Indian government
had only Indian workers in the plant. Safety practices clearly eroded due to cost-cutting measures, insufficient repair parts, and reduction in personnel to save salary money. However, one of
the surprising practices that caused unnecessary harm to workers was the fact that masks, gloves,
and other protective gear were not worn by workers in close proximity to the tanks containing
MIC. Why? Unlike in plants in the United States and other countries, there was no air conditioning in the Indian plant, resulting in high ambient temperatures in the facility.
Many ethical questions arise when corporations operate in international settings where the
corporate rules, worker incentives, cultural practices, and costs in the home country differ from
those in the host country. Often these ethical dilemmas are fundamentally based in the economics
that provide cheaper labor, reduced raw material costs, less government oversight, and a host of
1.3
Professional Ethics and Economic Decisions
other cost-reducing factors. When an engineering economy study is performed, it is important for
the engineer performing the study to consider all ethically related matters to ensure that the cost
and revenue estimates reflect what is likely to happen once the project or system is operating.
It is important to understand that the translation from universal morals to personal morals and
professional ethics does vary from one culture and country to another. As an example, consider the
common belief (universal moral) that the awarding of contracts and financial arrangements for services to be performed (for government or business) should be accomplished in a fair and transparent
fashion. In some societies and cultures, corruption in the process of contract making is common and
often “overlooked” by the local authorities, who may also be involved in the affairs. Are these immoral or unethical practices? Most would say, “Yes, this should not be allowed. Find and punish the
individuals involved.” Yet, such practices do continue, thus indicating the differences in interpretation of common morals as they are translated into the ethics of individuals and professionals.
EXAMPLE 1.2
Jamie is an engineer employed by Burris, a United States–based company that develops subway and surface transportation systems for medium-sized municipalities in the United States
and Canada. He has been a registered professional engineer (PE) for the last 15 years. Last
year, Carol, an engineer friend from university days who works as an individual consultant,
asked Jamie to help her with some cost estimates on a metro train job. Carol offered to pay for
his time and talent, but Jamie saw no reason to take money for helping with data commonly
used by him in performing his job at Burris. The estimates took one weekend to complete, and
once Jamie delivered them to Carol, he did not hear from her again; nor did he learn the identity of the company for which Carol was preparing the estimates.
Yesterday, Jamie was called into his supervisor’s office and told that Burris had not received
the contract award in Sharpstown, where a metro system is to be installed. The project estimates were prepared by Jamie and others at Burris over the past several months. This job was
greatly needed by Burris, as the country and most municipalities were in a real economic
slump, so much so that Burris was considering furloughing several engineers if the Sharpstown
bid was not accepted. Jamie was told he was to be laid off immediately, not because the bid was
rejected, but because he had been secretly working without management approval for a prime
consultant of Burris’ main competitor. Jamie was astounded and angry. He knew he had done
nothing to warrant firing, but the evidence was clearly there. The numbers used by the competitor to win the Sharpstown award were the same numbers that Jamie had prepared for Burris
on this bid, and they closely matched the values that he gave Carol when he helped her.
Jamie was told he was fortunate, because Burris’ president had decided to not legally charge
Jamie with unethical behavior and to not request that his PE license be rescinded. As a result,
Jamie was escorted out of his office and the building within one hour and told to not ask anyone
at Burris for a reference letter if he attempted to get another engineering job.
Discuss the ethical dimensions of this situation for Jamie, Carol, and Burris’ management.
Refer to the NSPE Code of Ethics for Engineers (Appendix C) for specific points of concern.
Solution
There are several obvious errors and omissions present in the actions of Jamie, Carol, and
Burris’ management in this situation. Some of these mistakes, oversights, and possible code
violations are summarized here.
Jamie
• Did not learn identity of company Carol was working for and whether the company was to
be a bidder on the Sharpstown project
• Helped a friend with confidential data, probably innocently, without the knowledge or approval of his employer
• Assisted a competitor, probably unknowingly, without the knowledge or approval of his
employer
• Likely violated, at least, Code of Ethics for Engineers section II.1.c, which reads, “Engineers shall not reveal facts, data, or information without the prior consent of the client or
employer except as authorized or required by law or this Code.”
9
10
Foundations of Engineering Economy
Chapter 1
Carol
• Did not share the intended use of Jamie’s work
• Did not seek information from Jamie concerning his employer’s intention to bid on the
same project as her client
• Misled Jamie in that she did not seek approval from Jamie to use and quote his information
and assistance
• Did not inform her client that portions of her work originated from a source employed by a
possible bid competitor
• Likely violated, at least, Code of Ethics for Engineers section III.9.a, which reads, “Engineers shall, whenever possible, name the person or persons who may be individually responsible for designs, inventions, writings, or other accomplishments.”
Burris’ management
• Acted too fast in dismissing Jamie; they should have listened to Jamie and conducted an
investigation
• Did not put him on administrative leave during a review
• Possibly did not take Jamie’s previous good work record into account
These are not all ethical considerations; some are just plain good business practices for Jamie,
Carol, and Burris.
1.4 Interest Rate and Rate of Return
Interest is the manifestation of the time value of money. Computationally, interest is the difference
between an ending amount of money and the beginning amount. If the difference is zero or negative, there is no interest. There are always two perspectives to an amount of interest—interest paid
and interest earned. These are illustrated in Figure 1–2. Interest is paid when a person or organization borrowed money (obtained a loan) and repays a larger amount over time. Interest is earned
when a person or organization saved, invested, or lent money and obtains a return of a larger
amount over time. The numerical values and formulas used are the same for both perspectives, but
the interpretations are different.
Interest paid on borrowed funds (a loan) is determined using the original amount, also called
the principal,
Interest ⫽ amount owed now ⫺ principal
[1.1]
When interest paid over a specific time unit is expressed as a percentage of the principal, the result is called the interest rate.
interest accrued per time unit
Interest rate (%) ⴝ ————————————— ⴛ 100%
principal
[1.2]
The time unit of the rate is called the interest period. By far the most common interest period
used to state an interest rate is 1 year. Shorter time periods can be used, such as 1% per month.
Thus, the interest period of the interest rate should always be included. If only the rate is stated,
for example, 8.5%, a 1-year interest period is assumed.
Loan
Bank
Repayment
⫹ interest
Loan
Borrower
Investor
Repayment
⫹ interest
(a)
Figure 1–2
(a) Interest paid over time to lender. (b) Interest earned over time by investor.
(b)
Corporation
Interest Rate and Rate of Return
1.4
EXAMPLE 1.3
An employee at LaserKinetics.com borrows $10,000 on May 1 and must repay a total of
$10,700 exactly 1 year later. Determine the interest amount and the interest rate paid.
Solution
The perspective here is that of the borrower since $10,700 repays a loan. Apply Equation [1.1]
to determine the interest paid.
Interest paid ⫽ $10,700 ⫺ 10,000 ⫽ $700
Equation [1.2] determines the interest rate paid for 1 year.
$700
Percent interest rate ⫽ ———— ⫻ 100% ⫽ 7% per year
$10,000
EXAMPLE 1.4
Stereophonics, Inc., plans to borrow $20,000 from a bank for 1 year at 9% interest for new
recording equipment. (a) Compute the interest and the total amount due after 1 year. (b) Construct a column graph that shows the original loan amount and total amount due after 1 year
used to compute the loan interest rate of 9% per year.
Solution
(a) Compute the total interest accrued by solving Equation [1.2] for interest accrued.
Interest ⫽ $20,000(0.09) ⫽ $1800
The total amount due is the sum of principal and interest.
Total due ⫽ $20,000 ⫹ 1800 ⫽ $21,800
(b) Figure 1–3 shows the values used in Equation [1.2]: $1800 interest, $20,000 original loan
principal, 1-year interest period.
$
$21,800
Interest = $1800
$20,000
Original
loan
amount
Interest rate
$1800
⫻ 100%
$20,000
= 9% per year
Now
1 year
later
Interest
period is
1 year
Figure 1–3
Values used to compute an interest rate of 9% per year. Example 1.4.
Comment
Note that in part (a), the total amount due may also be computed as
Total due ⫽ principal(1 ⫹ interest rate) ⫽ $20,000(1.09) ⫽ $21,800
Later we will use this method to determine future amounts for times longer than one interest
period.
11
12
Foundations of Engineering Economy
Chapter 1
From the perspective of a saver, a lender, or an investor, interest earned (Figure 1–2b) is the
final amount minus the initial amount, or principal.
Interest earned ⫽ total amount now ⫺ principal
[1.3]
Interest earned over a specific period of time is expressed as a percentage of the original amount
and is called rate of return (ROR).
interest accrued per time unit
Rate of return (%) ⴝ ————————————— ⴛ 100%
principal
[1.4]
The time unit for rate of return is called the interest period, just as for the borrower’s perspective. Again, the most common period is 1 year.
The term return on investment (ROI) is used equivalently with ROR in different industries and
settings, especially where large capital funds are committed to engineering-oriented programs.
The numerical values in Equations [1.2] and [1.4] are the same, but the term interest rate paid
is more appropriate for the borrower’s perspective, while the rate of return earned is better for
the investor’s perspective.
EXAMPLE 1.5
(a) Calculate the amount deposited 1 year ago to have $1000 now at an interest rate of 5%
per year.
(b) Calculate the amount of interest earned during this time period.
Solution
(a) The total amount accrued ($1000) is the sum of the original deposit and the earned interest.
If X is the original deposit,
Total accrued ⫽ deposit ⫹ deposit(interest rate)
$1000 ⫽ X ⫹ X(0.05) ⫽ X(1 ⫹ 0.05) ⫽ 1.05X
The original deposit is
1000 ⫽ $952.38
X ⫽ ———
1.05
(b) Apply Equation [1.3] to determine the interest earned.
Interest ⫽ $1000 ⫺ 952.38 ⫽ $47.62
In Examples 1.3 to 1.5 the interest period was 1 year, and the interest amount was calculated
at the end of one period. When more than one interest period is involved, e.g., the amount of interest after 3 years, it is necessary to state whether the interest is accrued on a simple or compound
basis from one period to the next. This topic is covered later in this chapter.
Since inflation can significantly increase an interest rate, some comments about the fundamentals of inflation are warranted at this early stage. By definition, inflation represents a decrease
in the value of a given currency. That is, $10 now will not purchase the same amount of gasoline
for your car (or most other things) as $10 did 10 years ago. The changing value of the currency
affects market interest rates.
Inflation
In simple terms, interest rates reflect two things: a so-called real rate of return plus the expected
inflation rate. The real rate of return allows the investor to purchase more than he or she could
have purchased before the investment, while inflation raises the real rate to the market rate that
we use on a daily basis.
The safest investments (such as government bonds) typically have a 3% to 4% real rate of
return built into their overall interest rates. Thus, a market interest rate of, say, 8% per year on a
bond means that investors expect the inflation rate to be in the range of 4% to 5% per year.
Clearly, inflation causes interest rates to rise.
From the borrower’s perspective, the rate of inflation is another interest rate tacked on to the
real interest rate. And from the vantage point of the saver or investor in a fixed-interest account,
Terminology and Symbols
1.5
inflation reduces the real rate of return on the investment. Inflation means that cost and revenue
cash flow estimates increase over time. This increase is due to the changing value of money that
is forced upon a country’s currency by inflation, thus making a unit of currency (such as the dollar) worth less relative to its value at a previous time. We see the effect of inflation in that money
purchases less now than it did at a previous time. Inflation contributes to
•
•
•
•
•
A reduction in purchasing power of the currency
An increase in the CPI (consumer price index)
An increase in the cost of equipment and its maintenance
An increase in the cost of salaried professionals and hourly employees
A reduction in the real rate of return on personal savings and certain corporate investments
In other words, inflation can materially contribute to changes in corporate and personal economic
analysis.
Commonly, engineering economy studies assume that inflation affects all estimated values
equally. Accordingly, an interest rate or rate of return, such as 8% per year, is applied throughout
the analysis without accounting for an additional inflation rate. However, if inflation were explicitly taken into account, and it was reducing the value of money at, say, an average of 4% per year,
then it would be necessary to perform the economic analysis using an inflated interest rate. (The
rate is 12.32% per year using the relations derived in Chapter 14.)
1.5 Terminology and Symbols
The equations and procedures of engineering economy utilize the following terms and symbols.
Sample units are indicated.
P ⫽ value or amount of money at a time designated as the present or time 0. Also P is
referred to as present worth (PW), present value (PV), net present value (NPV), discounted cash flow (DCF), and capitalized cost (CC); monetary units, such as dollars
F ⫽ value or amount of money at some future time. Also F is called future worth (FW)
and future value (FV); dollars
A ⫽ series of consecutive, equal, end-of-period amounts of money. Also A is called the
annual worth (AW) and equivalent uniform annual worth (EUAW); dollars per
year, euros per month
n ⫽ number of interest periods; years, months, days
⫽ interest rate per time period; percent per year, percent per month
t ⫽ time, stated in periods; years, months, days
i
The symbols P and F represent one-time occurrences: A occurs with the same value in each interest period for a specified number of periods. It should be clear that a present value P represents a
single sum of money at some time prior to a future value F or prior to the first occurrence of an
equivalent series amount A.
It is important to note that the symbol A always represents a uniform amount (i.e., the same
amount each period) that extends through consecutive interest periods. Both conditions must
exist before the series can be represented by A.
The interest rate i is expressed in percent per interest period, for example, 12% per year. Unless stated otherwise, assume that the rate applies throughout the entire n years or interest periods. The decimal equivalent for i is always used in formulas and equations in engineering economy computations.
All engineering economy problems involve the element of time expressed as n and interest
rate i. In general, every problem will involve at least four of the symbols P, F, A, n, and i, with at
least three of them estimated or known.
Additional symbols used in engineering economy are defined in Appendix E.
EXAMPLE 1.6
Today, Julie borrowed $5000 to purchase furniture for her new house. She can repay the loan
in either of the two ways described below. Determine the engineering economy symbols and
their value for each option.
13
14
Foundations of Engineering Economy
Chapter 1
(a) Five equal annual installments with interest based on 5% per year.
(b) One payment 3 years from now with interest based on 7% per year.
Solution
(a) The repayment schedule requires an equivalent annual amount A, which is unknown.
P ⫽ $5000
i ⫽ 5% per year
n ⫽ 5 years
A⫽?
(b) Repayment requires a single future amount F, which is unknown.
P ⫽ $5000
i ⫽ 7% per year
n ⫽ 3 years
F⫽?
EXAMPLE 1.7
You plan to make a lump-sum deposit of $5000 now into an investment account that pays 6%
per year, and you plan to withdraw an equal end-of-year amount of $1000 for 5 years, starting
next year. At the end of the sixth year, you plan to close your account by withdrawing the remaining money. Define the engineering economy symbols involved.
Solution
All five symbols are present, but the future value in year 6 is the unknown.
P ⫽ $5000
A ⫽ $1000 per year for 5 years
F ⫽ ? at end of year 6
i ⫽ 6% per year
n ⫽ 5 years for the A series and 6 for the F value
EXAMPLE 1.8
Last year Jane’s grandmother offered to put enough money into a savings account to generate
$5000 in interest this year to help pay Jane’s expenses at college. (a) Identify the symbols, and
(b) calculate the amount that had to be deposited exactly 1 year ago to earn $5000 in interest
now, if the rate of return is 6% per year.
Solution
(a) Symbols P (last year is ⫺1) and F (this year) are needed.
P⫽?
i ⫽ 6% per year
n ⫽ 1 year
F ⫽ P ⫹ interest ⫽ ? ⫹ $5000
(b) Let F ⫽ total amount now and P ⫽ original amount. We know that F – P ⫽ $5000 is
accrued interest. Now we can determine P. Refer to Equations [1.1] through [1.4].
F ⫽ P ⫹ Pi
The $5000 interest can be expressed as
Interest ⫽ F – P ⫽ (P ⫹ Pi) – P
⫽ Pi
$5000 ⫽ P(0.06)
$5000
P ⫽ ——— ⫽ $83,333.33
0.06
15
Cash Flows: Estimation and Diagramming
1.6
1.6 Cash Flows: Estimation and Diagramming
As mentioned in earlier sections, cash flows are the amounts of money estimated for future projects
or observed for project events that have taken place. All cash flows occur during specific time periods, such as 1 month, every 6 months, or 1 year. Annual is the most common time period. For
example, a payment of $10,000 once every year in December for 5 years is a series of 5 outgoing
cash flows. And an estimated receipt of $500 every month for 2 years is a series of 24 incoming cash
flows. Engineering economy bases its computations on the timing, size, and direction of cash flows.
Cash inflows are the receipts, revenues, incomes, and savings generated by project and business
activity. A plus sign indicates a cash inflow.
Cash outflows are costs, disbursements, expenses, and taxes caused by projects and business
activity. A negative or minus sign indicates a cash outflow. When a project involves only costs,
the minus sign may be omitted for some techniques, such as benefit/cost analysis.
Cash flow
Of all the steps in Figure 1–1 that outline the engineering economy study, estimating cash flows
(step 3) is the most difficult, primarily because it is an attempt to predict the future. Some examples of cash flow estimates are shown here. As you scan these, consider how the cash inflow
or outflow may be estimated most accurately.
Cash Inflow Estimates
Income: ⫹$150,000 per year from sales of solar-powered watches
Savings: ⫹$24,500 tax savings from capital loss on equipment salvage
Receipt: ⫹$750,000 received on large business loan plus accrued interest
Savings: ⫹$150,000 per year saved by installing more efficient air conditioning
Revenue: ⫹$50,000 to ⫹$75,000 per month in sales for extended battery life iPhones
Cash Outflow Estimates
Operating costs: ⫺$230,000 per year annual operating costs for software services
First cost: ⫺$800,000 next year to purchase replacement earthmoving equipment
Expense: ⫺$20,000 per year for loan interest payment to bank
Initial cost: ⫺$1 to ⫺$1.2 million in capital expenditures for a water recycling unit
All of these are point estimates, that is, single-value estimates for cash flow elements of an
alternative, except for the last revenue and cost estimates listed above. They provide a range estimate,
because the persons estimating the revenue and cost do not have enough knowledge or experience
with the systems to be more accurate. For the initial chapters, we will utilize point estimates. The use
of risk and sensitivity analysis for range estimates is covered in the later chapters of this book.
Once all cash inflows and outflows are estimated (or determined for a completed project), the
net cash flow for each time period is calculated.
Net cash flow ⴝ cash inflows ⴚ cash outflows
NCF ⴝ R ⴚ D
[1.5]
[1.6]
where NCF is net cash flow, R is receipts, and D is disbursements.
At the beginning of this section, the timing, size, and direction of cash flows were mentioned
as important. Because cash flows may take place at any time during an interest period, as a matter
of convention, all cash flows are assumed to occur at the end of an interest period.
The end-of-period convention means that all cash inflows and all cash outflows are assumed to
take place at the end of the interest period in which they actually occur. When several inflows
and outflows occur within the same period, the net cash flow is assumed to occur at the end of
the period.
End-of-period convention
16
Foundations of Engineering Economy
Chapter 1
Figure 1–4
Year 1
Year 5
A typical cash flow time
scale for 5 years.
0
1
2
3
4
5
Time scale
Figure 1–5
F=?
+
Example of positive and
negative cash flows.
Cash flow
i = 4% per year
1
2
3
4
5
Year
–
In assuming end-of-period cash flows, it is important to understand that future (F) and uniform
annual (A) amounts are located at the end of the interest period, which is not necessarily
December 31. If in Example 1.7 the lump-sum deposit took place on July 1, 2011, the withdrawals will take place on July 1 of each succeeding year for 6 years. Remember, end of the period
means end of interest period, not end of calendar year.
The cash flow diagram is a very important tool in an economic analysis, especially when the
cash flow series is complex. It is a graphical representation of cash flows drawn on the y axis with
a time scale on the x axis. The diagram includes what is known, what is estimated, and what is
needed. That is, once the cash flow diagram is complete, another person should be able to work
the problem by looking at the diagram.
Cash flow diagram time t ⫽ 0 is the present, and t ⫽ 1 is the end of time period 1. We assume
that the periods are in years for now. The time scale of Figure 1–4 is set up for 5 years. Since the
end-of-year convention places cash flows at the ends of years, the “1” marks the end of year 1.
While it is not necessary to use an exact scale on the cash flow diagram, you will probably
avoid errors if you make a neat diagram to approximate scale for both time and relative cash flow
magnitudes.
The direction of the arrows on the diagram is important to differentiate income from outgo. A
vertical arrow pointing up indicates a positive cash flow. Conversely, a down-pointing arrow indicates a negative cash flow. We will use a bold, colored arrow to indicate what is unknown
and to be determined. For example, if a future value F is to be determined in year 5, a wide,
colored arrow with F ⫽ ? is shown in year 5. The interest rate is also indicated on the diagram.
Figure 1–5 illustrates a cash inflow at the end of year 1, equal cash outflows at the end of years 2
and 3, an interest rate of 4% per year, and the unknown future value F after 5 years. The arrow
for the unknown value is generally drawn in the opposite direction from the other cash flows;
however, the engineering economy computations will determine the actual sign on the F value.
Before the diagramming of cash flows, a perspective or vantage point must be determined so
that ⫹ or – signs can be assigned and the economic analysis performed correctly. Assume you
borrow $8500 from a bank today to purchase an $8000 used car for cash next week, and you plan
to spend the remaining $500 on a new paint job for the car two weeks from now. There are several perspectives possible when developing the cash flow diagram—those of the borrower (that’s
you), the banker, the car dealer, or the paint shop owner. The cash flow signs and amounts for
these perspectives are as follows.
Perspective
Activity
Cash flow with Sign, $
You
Borrow
Buy car
Paint job
Lender
Car sale
Paint job
⫹8500
−8000
−500
−8500
⫹8000
⫹500
Banker
Car dealer
Painter
Time, week
0
1
2
0
1
2
Cash Flows: Estimation and Diagramming
1.6
$8500
1
2
Week
0
$500
$8000
Figure 1–6
Cash flows from perspective of borrower for loan and purchases.
One, and only one, of the perspectives is selected to develop the diagram. For your perspective,
all three cash flows are involved and the diagram appears as shown in Figure 1–6 with a time scale
of weeks. Applying the end-of-period convention, you have a receipt of ⫹$8500 now (time 0) and
cash outflows of ⫺$8000 at the end of week 1, followed by ⫺$500 at the end of week 2.
EXAMPLE 1.9
Each year Exxon-Mobil expends large amounts of funds for mechanical safety features
throughout its worldwide operations. Carla Ramos, a lead engineer for Mexico and Central
American operations, plans expenditures of $1 million now and each of the next 4 years just
for the improvement of field-based pressure-release valves. Construct the cash flow diagram to
find the equivalent value of these expenditures at the end of year 4, using a cost of capital estimate for safety-related funds of 12% per year.
Solution
Figure 1–7 indicates the uniform and negative cash flow series (expenditures) for five periods,
and the unknown F value (positive cash flow equivalent) at exactly the same time as the fifth
expenditure. Since the expenditures start immediately, the first $1 million is shown at time 0,
not time 1. Therefore, the last negative cash flow occurs at the end of the fourth year, when F
also occurs. To make this diagram have a full 5 years on the time scale, the addition of the
year ⫺1 completes the diagram. This addition demonstrates that year 0 is the end-of-period
point for the year ⫺1.
F=?
i = 12%
⫺1
0
1
2
3
4
Year
A = $1,000,000
Figure 1–7
Cash flow diagram, Example 1.9.
EXAMPLE 1.10
An electrical engineer wants to deposit an amount P now such that she can withdraw an equal
annual amount of A1 ⫽ $2000 per year for the first 5 years, starting 1 year after the deposit, and
a different annual withdrawal of A2 ⫽ $3000 per year for the following 3 years. How would the
cash flow diagram appear if i ⫽ 8.5% per year?
17
18
Foundations of Engineering Economy
Chapter 1
Solution
The cash flows are shown in Figure 1–8. The negative cash outflow P occurs now. The withdrawals (positive cash inflow) for the A1 series occur at the end of years 1 through 5, and A2
occurs in years 6 through 8.
A2 = $3000
A1 = $2000
0
2
1
3
4
6
5
7
8 Year
i = 8.5%
P=?
Figure 1–8
Cash flow diagram with two different A series, Example 1.10.
EXAMPLE 1.11
A rental company spent $2500 on a new air compressor 7 years ago. The annual rental income
from the compressor has been $750. The $100 spent on maintenance the first year has increased each year by $25. The company plans to sell the compressor at the end of next year for
$150. Construct the cash flow diagram from the company’s perspective and indicate where the
present worth now is located.
Solution
Let now be time t ⫽ 0. The incomes and costs for years ⫺7 through 1 (next year) are tabulated
below with net cash flow computed using Equation [1.5]. The net cash flows (one negative,
eight positive) are diagrammed in Figure 1–9. Present worth P is located at year 0.
End of Year
−7
−6
−5
−4
−3
−2
−1
0
1
Income
Cost
Net Cash Flow
$
$2500
100
125
150
175
200
225
250
275
$−2500
650
625
600
575
550
525
500
625
0
750
750
750
750
750
750
750
750 ⫹ 150
P=?
$650
–7 –6
$625
–5
$600
–4
$625
$575
–3
$2500
Figure 1–9
Cash flow diagram, Example 1.11.
$550
–2
$525
–1
$500
0
1 Year
Economic Equivalence
1.7
19
1.7 Economic Equivalence
Economic equivalence is a fundamental concept upon which engineering economy computations
are based. Before we delve into the economic aspects, think of the many types of equivalency we
may utilize daily by transferring from one scale to another. Some example transfers between
scales are as follows:
Length:
12 inches ⫽ 1 foot
3 feet ⫽ 1 yard
39.370 inches ⫽ 1 meter
100 centimeters ⫽ 1 meter
1000 meters ⫽ 1 kilometer
1 kilometer ⫽ 0.621 mile
Pressure:
1 atmosphere ⫽ 1 newton/meter2 ⫽ 103 pascal ⫽ 1 kilopascal
Often equivalency involves two or more scales. Consider the equivalency of a speed of 110 kilometers per hour (kph) into miles per minute using conversions between distance and time scales
with three-decimal accuracy.
Speed:
1 mile ⫽ 1.609 kilometers
1 hour ⫽ 60 minutes
110 kph ⫽ 68.365 miles per hour (mph)
68.365 mph ⫽ 1.139 miles per minute
Four scales—time in minutes, time in hours, length in miles, and length in kilometers—are
combined to develop these equivalent statements on speed. Note that throughout these statements, the fundamental relations of 1 mile ⫽ 1.609 kilometers and 1 hour ⫽ 60 minutes are
applied. If a fundamental relation changes, the entire equivalency is in error.
Now we consider economic equivalency.
Economic equivalence is a combination of interest rate and time value of money to determine the different amounts of money at different points in time that are equal in economic
value.
As an illustration, if the interest rate is 6% per year, $100 today (present time) is equivalent to
$106 one year from today.
Amount accrued ⫽ 100 ⫹ 100(0.06) ⫽ 100(1 ⫹ 0.06) ⫽ $106
If someone offered you a gift of $100 today or $106 one year from today, it would make no difference which offer you accepted from an economic perspective. In either case you have $106
one year from today. However, the two sums of money are equivalent to each other only when the
interest rate is 6% per year. At a higher or lower interest rate, $100 today is not equivalent to $106
one year from today.
In addition to future equivalence, we can apply the same logic to determine equivalence for
previous years. A total of $100 now is equivalent to $100兾1.06 ⫽ $94.34 one year ago at an
interest rate of 6% per year. From these illustrations, we can state the following: $94.34 last
year, $100 now, and $106 one year from now are equivalent at an interest rate of 6% per year.
The fact that these sums are equivalent can be verified by computing the two interest rates for
1-year interest periods.
$6
$100
——— ⫻ 100% ⫽ 6% per year
and
$5.66
$94.34
——— ⫻ 100% ⫽ 6% per year
The cash flow diagram in Figure 1–10 indicates the amount of interest needed each year to make
these three different amounts equivalent at 6% per year.
Economic equivalence
20
Foundations of Engineering Economy
Chapter 1
i = 6% per year
Amount, $
100
94.34
$6.00 interest
$5.66 interest
50
0
⫺1
Last
year
0
Now
⫹1
Next
year
Time
Figure 1–10
Equivalence of money at 6% per year interest.
EXAMPLE 1.12
Manufacturers make backup batteries for computer systems available to Batteries⫹ dealers
through privately owned distributorships. In general, batteries are stored throughout the year,
and a 5% cost increase is added each year to cover the inventory carrying charge for the distributorship owner. Assume you own the City Center Batteries⫹ outlet. Make the calculations
necessary to show which of the following statements are true and which are false about battery
costs.
(a) The amount of $98 now is equivalent to a cost of $105.60 one year from now.
(b) A truck battery cost of $200 one year ago is equivalent to $205 now.
(c) A $38 cost now is equivalent to $39.90 one year from now.
(d) A $3000 cost now is equivalent to $2887.14 one year earlier.
(e) The carrying charge accumulated in 1 year on an investment of $20,000 worth of
batteries is $1000.
Solution
(a) Total amount accrued ⫽ 98(1.05) ⫽ $102.90 ⫽ $105.60; therefore, it is false. Another
way to solve this is as follows: Required original cost is 105.60兾1.05 ⫽ $100.57 ⫽ $98.
(b) Equivalent cost 1 year ago is 205.00兾1.05 ⫽ $195.24 ⫽ $200; therefore, it is false.
(c) The cost 1 year from now is $38(1.05) ⫽ $39.90; true.
(d) Cost now is 2887.14(1.05) ⫽ $3031.50 ⫽ $3000; false.
(e) The charge is 5% per year interest, or $20,000(0.05) ⫽ $1000; true.
Comparison of alternative cash flow series requires the use of equivalence to determine when
the series are economically equal or if one is economically preferable to another. The keys to the
analysis are the interest rate and the timing of the cash flows. Example 1.13 demonstrates how
easy it is to be misled by the size and timing of cash flows.
EXAMPLE 1.13
Howard owns a small electronics repair shop. He wants to borrow $10,000 now and repay it
over the next 1 or 2 years. He believes that new diagnostic test equipment will allow him to
work on a wider variety of electronic items and increase his annual revenue. Howard received
2-year repayment options from banks A and B.
Simple and Compound Interest
1.8
Amount to pay, $ per year
Year
Bank A
Bank B
1
−5,378.05
−5,000.00
2
−5,378.05
−5,775.00
Total paid
−10,756.10
−10,775.00
After reviewing these plans, Howard decided that he wants to repay the $10,000 after only
1 year based on the expected increased revenue. During a family conversation, Howard’s
brother-in-law offered to lend him the $10,000 now and take $10,600 after exactly 1 year.
Now Howard has three options and wonders which one to take. Which one is economically
the best?
Solution
The repayment plans for both banks are economically equivalent at the interest rate of 5% per
year. (This is determined by using computations that you will learn in Chapter 2.) Therefore,
Howard can choose either plan even though the bank B plan requires a slightly larger sum of
money over the 2 years.
The brother-in-law repayment plan requires a total of $600 in interest 1 year later plus the
principal of $10,000, which makes the interest rate 6% per year. Given the two 5% per year
options from the banks, this 6% plan should not be chosen as it is not economically better than
the other two. Even though the sum of money repaid is smaller, the timing of the cash flows
and the interest rate make it less desirable. The point here is that cash flows themselves, or
their sums, cannot be relied upon as the primary basis for an economic decision. The interest
rate, timing, and economic equivalence must be considered.
1.8 Simple and Compound Interest
The terms interest, interest period, and interest rate (introduced in Section 1.4) are useful in calculating equivalent sums of money for one interest period in the past and one period in the future.
However, for more than one interest period, the terms simple interest and compound interest become important.
Simple interest is calculated using the principal only, ignoring any interest accrued in preceding interest periods. The total simple interest over several periods is computed as
Simple interest ⫽ (principal)(number of periods)(interest rate)
I ⫽ Pni
[1.7]
where I is the amount of interest earned or paid and the interest rate i is expressed in decimal form.
EXAMPLE 1.14
GreenTree Financing lent an engineering company $100,000 to retrofit an environmentally
unfriendly building. The loan is for 3 years at 10% per year simple interest. How much money
will the firm repay at the end of 3 years?
Solution
The interest for each of the 3 years is
Interest per year ⫽ $100,000(0.10) ⫽ $10,000
Total interest for 3 years from Equation [1.7] is
Total interest ⫽ $100,000(3)(0.10) ⫽ $30,000
21
22
Foundations of Engineering Economy
Chapter 1
The amount due after 3 years is
Total due ⫽ $100,000 ⫹ 30,000 ⫽ $130,000
The interest accrued in the first year and in the second year does not earn interest. The interest due each year is $10,000 calculated only on the $100,000 loan principal.
In most financial and economic analyses, we use compound interest calculations.
For compound interest, the interest accrued for each interest period is calculated on the
principal plus the total amount of interest accumulated in all previous periods. Thus,
compound interest means interest on top of interest.
Compound interest reflects the effect of the time value of money on the interest also. Now the
interest for one period is calculated as
Compound interest ⫽ (principal ⫹ all accrued interest)(interest rate)
[1.8]
In mathematical terms, the interest It for time period t may be calculated using the relation.
(
j⫽t⫺1
It ⫽ P ⫹
⌺
j⫽1
)
IJ (i)
[1.9]
EXAMPLE 1.15
Assume an engineering company borrows $100,000 at 10% per year compound interest and
will pay the principal and all the interest after 3 years. Compute the annual interest and total
amount due after 3 years. Graph the interest and total owed for each year, and compare with
the previous example that involved simple interest.
Solution
To include compounding of interest, the annual interest and total owed each year are calculated
by Equation [1.8].
Interest, year 1:
Total due, year 1:
Interest, year 2:
Total due, year 2:
Interest, year 3:
Total due, year 3:
100,000(0.10) ⫽ $10,000
100,000 ⫹ 10,000 ⫽ $110,000
110,000(0.10) ⫽ $11,000
110,000 ⫹ 11,000 ⫽ $121,000
121,000(0.10) ⫽ $12,100
121,000 ⫹ 12,100 ⫽ $133,100
The repayment plan requires no payment until year 3 when all interest and the principal, a total
of $133,100, are due. Figure 1–11 uses a cash flow diagram format to compare end-of-year
(a) simple and (b) compound interest and total amounts owed. The differences due to compounding are clear. An extra $133,100 – 130,000 ⫽ $3100 in interest is due for the compounded
interest loan.
Note that while simple interest due each year is constant, the compounded interest due
grows geometrically. Due to this geometric growth of compound interest, the difference between simple and compound interest accumulation increases rapidly as the time frame increases. For example, if the loan is for 10 years, not 3, the extra paid for compounding interest
may be calculated to be $59,374.
23
Simple and Compound Interest
1.8
0
1
2
3
⫺10
I
I
I
0
Year
⫺10
I is constant
1
2
3
I increases
geometrically
I
⫺11
I
⫺12
I
⫺100
Amount owed (⫻ $1000)
Amount owed (⫻ $1000)
⫺100
⫺110
⫺120
⫺110
⫺120
Geometric
increase
Arithmetic
increase
⫺130
⫺130
⫺140
⫺140
(a)
(b)
Figure 1–11
Interest I owed and total amount owed for (a) simple interest (Example 1.14) and (b) compound interest
(Example 1.15).
A more efficient way to calculate the total amount due after a number of years in Example 1.15
is to utilize the fact that compound interest increases geometrically. This allows us to skip the
year-by-year computation of interest. In this case, the total amount due at the end of each
year is
Year 1: $100,000(1.10)1 ⫽ $110,000
Year 2: $100,000(1.10)2 ⫽ $121,000
Year 3: $100,000(1.10)3 ⫽ $133,100
This allows future totals owed to be calculated directly without intermediate steps. The general
form of the equation is
Total due after n years ⴝ principal(1 ⴙ interest rate)n years
ⴝ P(1 ⴙ i)n
Year
[1.10]
where i is expressed in decimal form. Equation [1.10] was applied above to obtain the
$133,100 due after 3 years. This fundamental relation will be used many times in the upcoming chapters.
We can combine the concepts of interest rate, compound interest, and equivalence to demonstrate that different loan repayment plans may be equivalent, but differ substantially in amounts
paid from one year to another and in the total repayment amount. This also shows that there are
many ways to take into account the time value of money.
24
Foundations of Engineering Economy
Chapter 1
EXAMPLE 1.16
Table 1–1 details four different loan repayment plans described below. Each plan repays a
$5000 loan in 5 years at 8% per year compound interest.
• Plan 1: Pay all at end. No interest or principal is paid until the end of year 5. Interest accumulates each year on the total of principal and all accrued interest.
• Plan 2: Pay interest annually, principal repaid at end. The accrued interest is paid each
year, and the entire principal is repaid at the end of year 5.
• Plan 3: Pay interest and portion of principal annually. The accrued interest and one-fifth
of the principal (or $1000) are repaid each year. The outstanding loan balance decreases
each year, so the interest (column 2) for each year decreases.
• Plan 4: Pay equal amount of interest and principal. Equal payments are made each year
with a portion going toward principal repayment and the remainder covering the accrued
interest. Since the loan balance decreases at a rate slower than that in plan 3 due to the equal
end-of-year payments, the interest decreases, but at a slower rate.
TABLE 1–1
(1)
End of
Year
Different Repayment Schedules Over 5 Years for $5000 at 8% Per Year
Compound Interest
(2)
Interest Owed
for Year
(3)
Total Owed at
End of Year
(4)
End-of-Year
Payment
(5)
Total Owed
After Payment
Plan 1: Pay All at End
0
1
2
3
4
5
$400.00
432.00
466.56
503.88
544.20
$5400.00
5832.00
6298.56
6802.44
7346.64
Total
—
—
—
—
$ – 7346.64
$5000.00
5400.00
5832.00
6298.56
6802.44
$ – 7346.64
Plan 2: Pay Interest Annually; Principal Repaid at End
0
1
2
3
4
5
$400.00
400.00
400.00
400.00
400.00
$5400.00
5400.00
5400.00
5400.00
5400.00
Total
$⫺400.00
⫺400.00
⫺400.00
⫺400.00
– 5400.00
$5000.00
5000.00
5000.00
5000.00
5000.00
$⫺7000.00
Plan 3: Pay Interest and Portion of Principal Annually
0
1
2
3
4
5
$400.00
320.00
240.00
160.00
80.00
$5400.00
4320.00
3240.00
2160.00
1080.00
Total
$⫺1400.00
⫺1320.00
⫺1240.00
⫺1160.00
– 1080.00
$5000.00
4000.00
3000.00
2000.00
1000.00
$⫺6200.00
Plan 4: Pay Equal Annual Amount of Interest and Principal
0
1
2
3
4
5
Total
$400.00
331.82
258.18
178.65
92.76
$5400.00
4479.54
3485.43
2411.80
1252.28
$−1252.28
−1252.28
−1252.28
−1252.28
– 1252.28
$−6261.40
$5000.00
4147.72
3227.25
2233.15
1159.52
Minimum Attractive Rate of Return
1.9
(a) Make a statement about the equivalence of each plan at 8% compound interest.
(b) Develop an 8% per year simple interest repayment plan for this loan using the same
approach as plan 2. Comment on the total amounts repaid for the two plans.
Solution
(a) The amounts of the annual payments are different for each repayment schedule, and
the total amounts repaid for most plans are different, even though each repayment
plan requires exactly 5 years. The difference in the total amounts repaid can be explained by the time value of money and by the partial repayment of principal prior to
year 5.
A loan of $5000 at time 0 made at 8% per year compound interest is equivalent to each
of the following:
Plan 1 $7346.64 at the end of year 5
Plan 2 $400 per year for 4 years and $5400 at the end of year 5
Plan 3 Decreasing payments of interest and partial principal in years 1 ($1400)
through 5 ($1080)
Plan 4 $1252.28 per year for 5 years
An engineering economy study typically uses plan 4; interest is compounded, and a constant amount is paid each period. This amount covers accrued interest and a partial
amount of principal repayment.
(b) The repayment schedule for 8% per year simple interest is detailed in Table 1–2. Since
the annual accrued interest of $400 is paid each year and the principal of $5000 is repaid
in year 5, the schedule is exactly the same as that for 8% per year compound interest, and
the total amount repaid is the same at $7000. In this unusual case, simple and compound
interest result in the same total repayment amount. Any deviation from this schedule will
cause the two plans and amounts to differ.
TABLE 1–2
A 5-Year Repayment Schedule of $5000 at 8% per Year Simple Interest
End of
Year
Interest Owed
for Year
Total Owed at
End of Year
End-of-Year
Payment
Total Owed
After Payment
0
1
2
3
4
5
$400
400
400
400
400
$5400
5400
5400
5400
5400
$⫺400
⫺400
⫺400
⫺400
– 5400
$5000
5000
5000
5000
5000
0
Total
$⫺7000
1.9 Minimum Attractive Rate of Return
For any investment to be profitable, the investor (corporate or individual) expects to receive more
money than the amount of capital invested. In other words, a fair rate of return, or return on investment, must be realizable. The definition of ROR in Equation [1.4] is used in this discussion,
that is, amount earned divided by the principal.
Engineering alternatives are evaluated upon the prognosis that a reasonable ROR can be
expected. Therefore, some reasonable rate must be established for the selection criteria
(step 4) of the engineering economy study (Figure 1–1).
25
26
Foundations of Engineering Economy
Chapter 1
Minimum Attractive Rate
of Return (MARR)
Cost of capital
The Minimum Attractive Rate of Return (MARR) is a reasonable rate of return established
for the evaluation and selection of alternatives. A project is not economically viable unless
it is expected to return at least the MARR. MARR is also referred to as the hurdle rate,
cutoff rate, benchmark rate, and minimum acceptable rate of return.
Figure 1–12 indicates the relations between different rate of return values. In the United
States, the current U.S. Treasury Bill return is sometimes used as the benchmark safe rate. The
MARR will always be higher than this, or a similar, safe rate. The MARR is not a rate that is
calculated as a ROR. The MARR is established by (financial) managers and is used as a criterion against which an alternative’s ROR is measured, when making the accept/reject investment decision.
To develop a foundation-level understanding of how a MARR value is established and used
to make investment decisions, we return to the term capital introduced in Section 1.1. Although
the MARR is used as a criterion to decide on investing in a project, the size of MARR is fundamentally connected to how much it costs to obtain the needed capital funds. It always costs
money in the form of interest to raise capital. The interest, expressed as a percentage rate per
year, is called the cost of capital. As an example on a personal level, if you want to purchase a
new widescreen HDTV, but do not have sufficient money (capital), you could obtain a bank loan
for, say, a cost of capital of 9% per year and pay for the TV in cash now. Alternatively, you might
choose to use your credit card and pay off the balance on a monthly basis. This approach will
probably cost you at least 15% per year. Or, you could use funds from your savings account that
earns 5% per year and pay cash. This approach means that you also forgo future returns
from these funds. The 9%, 15%, and 5% rates are your cost of capital estimates to raise the
capital for the system by different methods of capital financing. In analogous ways, corporations
estimate the cost of capital from different sources to raise funds for engineering projects and
other types of projects.
Rate of return,
percent
Expected rate of return on
a new proposal
Range for the rate of return on
accepted proposals, if other
proposals were rejected
for some reason
All proposals must offer
at least MARR to
be considered
MARR
Rate of return on
“safe investment”
Figure 1–12
Size of MAAR relative to other rate of return values.
27
Introduction to Spreadsheet Use
1.10
In general, capital is developed in two ways—equity financing and debt financing. A combination of these two is very common for most projects. Chapter 10 covers these in greater detail, but
a snapshot description follows.
Equity financing The corporation uses its own funds from cash on hand, stock sales, or retained
earnings. Individuals can use their own cash, savings, or investments. In the example above, using
money from the 5% savings account is equity financing.
Debt financing The corporation borrows from outside sources and repays the principal and interest according to some schedule, much like the plans in Table 1–1. Sources of debt capital may be
bonds, loans, mortgages, venture capital pools, and many others. Individuals, too, can utilize debt
sources, such as the credit card (15% rate) and bank options (9% rate) described above.
Combinations of debt-equity financing mean that a weighted average cost of capital (WACC)
results. If the HDTV is purchased with 40% credit card money at 15% per year and 60% savings
account funds earning 5% per year, the weighted average cost of capital is 0.4(15) ⫹ 0.6(5) ⫽
9% per year.
For a corporation, the established MARR used as a criterion to accept or reject an investment
alternative will usually be equal to or higher than the WACC that the corporation must bear to
obtain the necessary capital funds. So the inequality
ROR ⱖ MARR ⬎ WACC
[1.11]
must be correct for an accepted project. Exceptions may be government-regulated requirements
(safety, security, environmental, legal, etc.), economically lucrative ventures expected to lead to
other opportunities, etc.
Often there are many alternatives that are expected to yield a ROR that exceeds the MARR as
indicated in Figure 1–12, but there may not be sufficient capital available for all, or the project’s
risk may be estimated as too high to take the investment chance. Therefore, new projects that are
undertaken usually have an expected return at least as great as the return on another alternative
that is not funded. The expected rate of return on the unfunded project is called the opportunity
cost.
The opportunity cost is the rate of return of a forgone opportunity caused by the inability to
pursue a project. Numerically, it is the largest rate of return of all the projects not accepted
(forgone) due to the lack of capital funds or other resources. When no specific MARR is
established, the de facto MARR is the opportunity cost, i.e., the ROR of the first project not
undertaken due to unavailability of capital funds.
As an illustration of opportunity cost, refer to Figure 1–12 and assume a MARR of 12% per
year. Further, assume that a proposal, call it A, with an expected ROR ⫽ 13% is not funded due
to a lack of capital. Meanwhile, proposal B has a ROR ⫽ 14.5% and is funded from available
capital. Since proposal A is not undertaken due to the lack of capital, its estimated ROR of 13%
is the opportunity cost; that is, the opportunity to make an additional 13% return is forgone.
1.10 Introduction to Spreadsheet Use
The functions on a computer spreadsheet can greatly reduce the amount of hand work for equivalency computations involving compound interest and the terms P, F, A, i, and n. The use of a
calculator to solve most simple problems is preferred by many students and professors as described in Appendix D. However, as cash flow series become more complex, the spreadsheet
offers a good alternative. Microsoft Excel is used throughout this book because it is readily
available and easy to use. Appendix A is a primer on using spreadsheets and Excel. The functions used in engineering economy are described there in detail, with explanations of all the
Opportunity cost
28
Foundations of Engineering Economy
Chapter 1
parameters. Appendix A also includes a section on spreadsheet layout that is useful when the
economic analysis is presented to someone else—a coworker, a boss, or a professor.
A total of seven Excel functions can perform most of the fundamental engineering economy
calculations. The functions are great supplemental tools, but they do not replace the understanding of engineering economy relations, assumptions, and techniques. Using the symbols P, F, A,
i, and n defined in the previous section, the functions most used in engineering economic analysis
are formulated as follows.
To find the present value P: ⴝ PV(i%, n, A, F)
To find the future value F: ⴝ FV(i%, n, A, P)
To find the equal, periodic value A: ⴝ PMT(i%, n, P, F)
To find the number of periods n: ⴝ NPER(i%, A, P, F)
To find the compound interest rate i: ⴝ RATE( n, A, P, F)
To find the compound interest rate i: ⴝ IRR(first_cell:last_cell)
To find the present value P of any series: ⴝ NPV(i%, second_cell:last_cell) ⴙ first_cell
If some of the parameters don’t apply to a particular problem, they can be omitted and zero is
assumed. For readability, spaces can be inserted between parameters within parentheses. If the
parameter omitted is an interior one, the comma must be entered. The last two functions require
that a series of numbers be entered into contiguous spreadsheet cells, but the first five can be used
with no supporting data. In all cases, the function must be preceded by an equals sign (⫽) in the
cell where the answer is to be displayed.
To understand how the spreadsheet functions work, look back at Example 1.6a, where the
equivalent annual amount A is unknown, as indicated by A ⫽ ?. (In Chapter 2, we learn how
engineering economy factors calculate A, given P, i, and n.) To find A using a spreadsheet
function, simply enter the PMT function ⫽ PMT(5%,5,5000). Figure 1–13 is a screen image
of a spreadsheet with the PMT function entered into cell B4. The answer ($1154.87) is displayed. The answer may appear in red and in parentheses, or with a minus sign on your screen
to indicate a negative amount from the perspective of a reduction in the account balance. The
right side of Figure 1–13 presents the solution to Example 1.6b. The future value F is determined by using the FV function. The FV function appears in the formula bar; and many examples throughout this text will include cell tags, as shown here, to indicate the format of
important entries.
The following example demonstrates the use of a spreadsheet to develop relations (not
built-in functions) to calculate interest and cash flows. Once set up, the spreadsheet can be
used to perform sensitivity analysis for estimates that are subject to change. We will illustrate the use of spreadsheets throughout the chapters. (Note: The spreadsheet examples may
be omitted, if spreadsheets are not used in the course. A solution by hand is included in virtually all examples.)
⫽ PMT(5%,5,5000)
Figure 1–13
Use of spreadsheet functions PMT and FV, Example 1.6.
⫽ FV(7%,3,,5000)
Introduction to Spreadsheet Use
1.10
EXAMPLE 1.17
A Japan-based architectural firm has asked a United States–based software engineering group
to infuse GPS sensing capability via satellite into monitoring software for high-rise structures
in order to detect greater than expected horizontal movements. This software could be very
beneficial as an advance warning of serious tremors in earthquake-prone areas in Japan and the
United States. The inclusion of accurate GPS data is estimated to increase annual revenue over
that for the current software system by $200,000 for each of the next 2 years, and by $300,000
for each of years 3 and 4. The planning horizon is only 4 years due to the rapid advances made
internationally in building-monitoring software. Develop spreadsheets to answer the questions
below.
(a) Determine the total interest and total revenue after 4 years, using a compound rate of
return of 8% per year.
(b) Repeat part (a) if estimated revenue increases from $300,000 to $600,000 in years 3 and 4.
(c) Repeat part (a) if inflation is estimated to be 4% per year. This will decrease the real rate
of return from 8% to 3.85% per year (Chapter 14 shows why).
Solution by Spreadsheet
Refer to Figure 1–14a to d for the solutions. All the spreadsheets contain the same information,
but some cell values are altered as required by the question. (Actually, all the questions can be
answered on one spreadsheet by changing the numbers. Separate spreadsheets are shown here
for explanation purposes only.)
The Excel functions are constructed with reference to the cells, not the values themselves, so that sensitivity analysis can be performed without function changes. This approach treats the value in a cell as a global variable for the spreadsheet. For example, the
8% rate in cell B2 will be referenced in all functions as B2, not 8%. Thus, a change in the
rate requires only one alteration in the cell B2 entry, not in every relation where 8% is used.
See Appendix A for additional information about using cell referencing and building
spreadsheet relations.
(a) Figure 1–14a shows the results, and Figure 1–14b presents all spreadsheet relations for
estimated interest and revenue (yearly in columns C and E, cumulative in columns D
and F). As an illustration, for year 3 the interest I3 and revenue plus interest R3 are
I3 ⫽ (cumulative revenue through year 2)(rate of return)
⫽ $416,000(0.08)
⫽ $33,280
R3 ⫽ revenue in year 3 ⫹ I3
⫽ $300,000 ⫹ 33,280
⫽ $333,280
The detailed relations shown in Figure 1–14b calculate these values in cells C8 and E8.
Cell C8 relation for I3: ⫽ F7*B2
Cell E8 relation for CF3: ⫽ B8 ⫹ C8
The equivalent amount after 4 years is $1,109,022, which is comprised of $1,000,000 in
total revenue and $109,022 in interest compounded at 8% per year. The shaded cells in
Figure 1–14a and b indicate that the sum of the annual values and the last entry in the cumulative columns must be equal.
(b) To determine the effect of increasing estimated revenue for years 3 and 4 to $600,000,
use the same spreadsheet and change the entries in cells B8 and B9 as shown in
Figure 1–14c. Total interest increases 22%, or $24,000, from $109,222 to $133,222.
(c) Figure 1–14d shows the effect of changing the original i value from 8% to an inflationadjusted rate of 3.85% in cell B2 on the first spreadsheet. [Remember to return to the
$300,000 revenue estimates for years 3 and 4 after working part (b).] Inflation has now
reduced total interest by 53% from $109,222 to $51,247, as shown in cell C10.
29
30
Foundations of Engineering Economy
Chapter 1
(a) Total interest and revenue for base case, year 4
(b) Spreadsheet relations for base case
Revenue changed
(c) Totals with increased revenue in years 3 and 4
Rate of
return
changed
(d) Totals with inflation of 4% per year considered
Figure 1–14
Spreadsheet solutions with sensitivity analysis, Example 1.17a to c.
Comment
Later we will learn how to utilize the NPV and FV Excel financial functions to obtain the same
answers determined in Figure 1–14, where we developed each basic relation.
When you are working with an Excel spreadsheet, it is possible to display all of the entries
and functions on the screen as shown in Figure 1–14b by simultaneously touching the <Ctrl>
and < `> keys, which may be in the upper left of the keyboard on the key with <~>.
Problems
31
CHAPTER SUMMARY
Engineering economy is the application of economic factors and criteria to evaluate alternatives,
considering the time value of money. The engineering economy study involves computing a
specific economic measure of worth for estimated cash flows over a specific period of time.
The concept of equivalence helps in understanding how different sums of money at different
times are equal in economic terms. The differences between simple interest (based on principal
only) and compound interest (based on principal and interest upon interest) have been described
in formulas, tables, and graphs. This power of compounding is very noticeable, especially over
extended periods of time, and for larger sums of money.
The MARR is a reasonable rate of return established as a hurdle rate to determine if an alternative is economically viable. The MARR is always higher than the return from a safe investment
and the cost to acquire needed capital.
Also, we learned a lot about cash flows:
End-of-year convention for cash flow location
Net cash flow computation
Different perspectives in determining the cash flow sign
Construction of a cash flow diagram
Difficulties with estimating future cash flows accurately
PROBLEMS
Basic Concepts
1.1 List the four essential elements involved in decision making in engineering economic analysis.
1.2 What is meant by (a) limited capital funds and
(b) sensitivity analysis?
1.3 List three measures of worth that are used in engineering economic analysis.
1.4 Identify the following factors as either economic
(tangible) or noneconomic (intangible): first cost,
leadership, taxes, salvage value, morale, dependability, inflation, profit, acceptance, ethics, interest rate.
Ethics
1.5 Stefanie is a design engineer with an international
railroad locomotive manufacturing company in
Illinois. Management wants to return some of the
engineering design work to the United States
rather than export all of it to India, where the primary design work has been accomplished for the
last decade. This transfer will employ more people
locally and could improve the economic conditions for families in and around Illinois.
Stefanie and her design team were selected as a
test case to determine the quality and speed of the
design work they could demonstrate on a more
fuel-efficient diesel locomotive. Neither she nor
any of her team members have done such a significant design job, because their jobs had previously
entailed only the interface with the subcontracted
engineers in India. One of her team members had a
great design idea on a key element that will improve fuel efficiency by approximately 15%. She
told Stefanie it came from one of the Indiangenerated documents, but that it would probably be
okay for the team to use it and remain silent as to its
origin, since it was quite clear the U.S. management was about to cancel the foreign contract.
Although reluctant at first, Stefanie did go forward
with a design that included the efficiency improvement, and no mention of the origin of the idea was
made at the time of the oral presentation or documentation delivery. As a result, the Indian contract
was canceled and full design responsibility was
transferred to Stefanie’s group.
Consult the NSPE Code of Ethics for Engineers
(Appendix C) and identify sections that are points
of concern about Stefanie’s decisions and actions.
1.6 Consider the common moral precept that stealing
is wrong. Hector is with a group of friends in a
local supermarket. One of Hector’s buddies takes a
high-energy drink from a six-pack on the shelf,
opens it, drinks it, and returns the empty can to the
package, with no intention of paying for it. He then
invites the others to do the same, saying, “It’s only
32
Chapter 1
one drink. Others do it all the time.” All the others,
except Hector, have now consumed a drink of their
choice. Personally, Hector believes this is a form
of stealing. State three actions that Hector can
take, and evaluate them from the personal moral
perspective.
1.7 While going to work this morning off site from his
office, an engineer accidently ran a stop sign and
was in a car accident that resulted in the death of a
5-year-old child. He has a strong belief in the universal moral that it is wrong to do serious harm to
another person. Explain the conflict that can arise
for him between the universal moral and his personal moral about doing serious harm, given the
accident was deemed his fault.
1.8 Claude is a fourth-year engineering university student who has just been informed by his instructor
that he made a very low grade on his Spanish language final test for the year. Although he had a
passing score prior to the final, his final grade was
so low that he has now flunked the entire year and
will likely have to extend his graduation another
semester or two.
Throughout the year, Claude, who hated the
course and his instructor, has copied homework,
cheated on tests, and never seriously studied for
anything in the course. He did realize during the
semester that he was doing something that even
he considered wrong morally and ethically. He
knew he had done badly on the final. The classroom was reconfigured for the final exam in a way
that he could not get any answers from classmates,
and cell phones were collected prior to the exam,
thus removing texting possibilities to friends outside the classroom who might help him on the
final exam. Claude is now face to face with the
instructor in her office. The question to Claude is,
“What have you been doing throughout this year
to make passing scores repeatedly, but demonstrate such a poor command of Spanish on the
final exam?”
From an ethical viewpoint, what options does
Claude have in his answer to this question? Also,
discuss some of the possible effects that this experience may have upon Claude’s future actions and
moral dilemmas.
Interest Rate and Rate of Return
1.9 RKI Instruments borrowed $3,500,000 from a private equity firm for expansion of its manufacturing
facility for making carbon monoxide monitors/
controllers. The company repaid the loan after
1 year with a single payment of $3,885,000. What
was the interest rate on the loan?
Foundations of Engineering Economy
1.10 Emerson Processing borrowed $900,000 for installing energy-efficient lighting and safety equipment in its La Grange manufacturing facility. The
terms of the loan were such that the company could
pay interest only at the end of each year for up to 5
years, after which the company would have to pay
the entire amount due. If the interest rate on the
loan was 12% per year and the company paid only
the interest for 4 years, determine the following:
(a) The amount of each of the four interest
payments
(b) The amount of the final payment at the end of
year 5
1.11 Which of the following 1-year investments has the
highest rate of return?
(a) $12,500 that yields $1125 in interest,
(b) $56,000 that yields $6160 in interest, or
(c) $95,000 that yields $7600 in interest.
1.12 A new engineering graduate who started a consulting business borrowed money for 1 year to furnish
the office. The amount of the loan was $23,800,
and it had an interest rate of 10% per year. However, because the new graduate had not built up a
credit history, the bank made him buy loan-default
insurance that cost 5% of the loan amount. In addition, the bank charged a loan setup fee of $300.
What was the effective interest rate the engineer
paid for the loan?
1.13 When the inflation rate is expected to be 8% per
year, what is the market interest rate likely to be?
Terms and Symbols
1.14 The symbol P represents an amount of money at a
time designated as present. The following symbols
also represent a present amount of money and require similar calculations. Explain what each symbol stands for: PW, PV, NPV, DCF, and CC.
1.15 Identify the four engineering economy symbols
and their values from the following problem statement. Use a question mark with the symbol whose
value is to be determined.
Thompson Mechanical Products is planning to
set aside $150,000 now for possibly replacing its
large synchronous refiner motors whenever it becomes necessary. If the replacement is not needed
for 7 years, how much will the company have in its
investment set-aside account, provided it achieves
a rate of return of 11% per year?
1.16 Identify the four engineering economy symbols and
their values from the following problem statement.
33
Problems
Use a question mark with the symbol whose value is
to be determined.
Atlas Long-Haul Transportation is considering installing Valutemp temperature loggers in all
of its refrigerated trucks for monitoring temperatures during transit. If the systems will reduce
insurance claims by $100,000 two years from
now, how much should the company be willing to
spend now, if it uses an interest rate of 12% per
year?
1.17 Identify the four engineering economy symbols
and their values from the following problem statement. Use a question mark with the symbol whose
value is to be determined.
A green algae, Chlamydomonas reinhardtii,
can produce hydrogen when temporarily deprived
of sulfur for up to 2 days at a time. A small
company needs to purchase equipment costing
$3.4 million to commercialize the process. If the
company wants to earn a rate of return of 10% per
year and recover its investment in 8 years, what
must be the net value of the hydrogen produced
each year?
1.18 Identify the four engineering economy symbols
and their values from the following problem statement. Use a question mark with the symbol whose
value is to be determined.
Vision Technologies, Inc., is a small company
that uses ultra-wideband technology to develop
devices that can detect objects (including people)
inside of buildings, behind walls, or below ground.
The company expects to spend $100,000 per year
for labor and $125,000 per year for supplies before
a product can be marketed. At an interest rate of
15% per year, what is the total equivalent future
amount of the company’s expenses at the end of
3 years?
Cash Flows
1.21 Many credit unions use semiannual interest periods
to pay interest on customer savings accounts. For a
credit union that uses June 30 and December 31 as
its semiannual interest periods, determine the endof-period amounts that will be recorded for the deposits shown in the table.
Deposit, $
Jan
Feb
Mar
Apr
May
June
July
Aug
Sept
Oct
Nov
Dec
50
70
—
120
20
—
150
90
—
—
40
110
1.22 For a company that uses a year as its interest period, determine the net cash flow that will be recorded at the end of the year from the cash flows
shown.
Month
Receipts,
$1000
Disbursements,
$1000
Jan
Feb
Mar
Apr
May
June
July
Aug
Sept
Oct
Nov
Dec
500
800
200
120
600
900
800
700
900
500
400
1800
300
500
400
400
500
600
300
300
500
400
400
700
1.23 Construct a cash flow diagram for the following
cash flows: $25,000 outflow at time 0, $9000 per
year inflow in years 1 through 5 at an interest rate
of 10% per year, and an unknown future amount in
year 5.
1.24 Construct a cash flow diagram to find the present
worth in year 0 at an interest rate of 15% per year
for the following situation.
1.19 What is meant by end-of-period convention?
1.20 Identify the following as cash inflows or outflows
to commercial air carriers: fuel cost, pension plan
contributions, fares, maintenance, freight revenue, cargo revenue, extra-bag charges, water and
sodas, advertising, landing fees, seat preference
fees.
Month
Year
Cash Flow, $
0
1–4
⫺19,000
⫹8,100
1.25 Construct a cash flow diagram that represents the
amount of money that will be accumulated in
15 years from an investment of $40,000 now at an
interest rate of 8% per year.
Equivalence
1.26 At an interest rate of 15% per year, an investment
of $100,000 one year ago is equivalent to how
much now?
34
Chapter 1
1.27 During a recession, the price of goods and services
goes down because of low demand. A company
that makes Ethernet adapters is planning to expand
its production facility at a cost of $1,000,000 one
year from now. However, a contractor who needs
work has offered to do the job for $790,000 if the
company will do the expansion now instead of
1 year from now. If the interest rate is 15% per year,
how much of a discount is the company getting?
1.28 As a principal in the consulting firm where you
have worked for 20 years, you have accumulated
5000 shares of company stock. One year ago, each
share of stock was worth $40. The company has
offered to buy back your shares for $225,000. At
what interest rate would the firm’s offer be equivalent to the worth of the stock last year?
1.29 A design/build engineering company that usually
gives year-end bonuses in the amount of $8000 to
each of its engineers is having cash flow problems. The company said that although it could not
give bonuses this year, it would give each engineer two bonuses next year, the regular one of
$8000 plus an amount equivalent to the $8000
that each engineer should have gotten this year.
If the interest rate is 8% per year, what will be the
total amount of bonus money the engineers should
get next year?
1.30 University tuition and fees can be paid by using
one of two plans.
Early-bird: Pay total amount due 1 year in
advance and get a 10% discount.
On-time: Pay total amount due when classes start.
The cost of tuition and fees is $10,000 per year.
(a) How much is paid in the early-bird plan?
(b) What is the equivalent amount of the savings
compared to the on-time payment at the time
that the on-time payment is made?
Simple and Compound Interest
1.31 If a company sets aside $1,000,000 now into a
contingency fund, how much will the company
have in 2 years, if it does not use any of the money
and the account grows at a rate of 10% per year?
1.32 Iselt Welding has extra funds to invest for future
capital expansion. If the selected investment pays
simple interest, what interest rate would be
required for the amount to grow from $60,000 to
$90,000 in 5 years?
1.33 To finance a new product line, a company that
makes high-temperature ball bearings borrowed
$1.8 million at 10% per year interest. If the com-
Foundations of Engineering Economy
pany repaid the loan in a lump sum amount after
2 years, what was (a) the amount of the payment
and (b) the amount of interest?
1.34 Because market interest rates were near all-time
lows at 4% per year, a hand tool company decided
to call (i.e., pay off ) the high-interest bonds that it
issued 3 years ago. If the interest rate on the bonds
was 9% per year, how much does the company
have to pay the bond holders? The face value
(principal) of the bonds is $6,000,000.
1.35 A solid waste disposal company borrowed money
at 10% per year interest to purchase new haulers
and other equipment needed at the companyowned landfill site. If the company got the loan
2 years ago and paid it off with a single payment of
$4,600,000, what was the principal amount P of
the loan?
1.36 If interest is compounded at 20% per year, how
long will it take for $50,000 to accumulate to
$86,400?
1.37 To make CDs look more attractive than they really
are, some banks advertise that their rates are higher
than their competitors’ rates; however, the fine
print says that the rate is a simple interest rate. If a
person deposits $10,000 at 10% per year simple
interest, what compound interest rate would yield
the same amount of money in 3 years?
MARR and Opportunity Cost
1.38 Give three other names for minimum attractive
rate of return.
1.39 Identify the following as either equity or debt financing: bonds, stock sales, retained earnings,
venture capital, short-term loan, capital advance
from friend, cash on hand, credit card, home equity loan.
1.40 What is the weighted average cost of capital for a
corporation that finances an expansion project
using 30% retained earnings and 70% venture capital? Assume the interest rates are 8% for the equity financing and 13% for the debt financing.
1.41 Managers from different departments in Zenith
Trading, a large multinational corporation, have offered six projects for consideration by the corporate
office. A staff member for the chief financial officer
used key words to identify the projects and then
listed them in order of projected rate of return as
shown below. If the company wants to grow rapidly through high leverage and uses only 10%
equity financing that has a cost of equity capital of
9% and 90% debt financing with a cost of debt
Additional Problems and FE Exam Review Questions
capital of 16%, which projects should the company
undertake?
Project ID
Inventory
Technology
Warehouse
Products
Energy
Shipping
Projected ROR,
% per year
30
28.4
19
13.1
9.6
8.2
Spreadsheet Functions
1.42 State the purpose for each of the following built-in
spreadsheet functions.
(a) PV(i%,n,A,F)
(b) FV(i%,n,A,P)
(c) RATE(n,A,P,F)
(d) IRR(first_cell:last_cell)
(e) PMT(i%,n,P,F)
(f ) NPER(i%,A,P,F)
1.43 What are the values of the engineering economy
symbols P, F, A, i, and n in the following functions?
Use a question mark for the symbol that is to be
35
determined.
(a) NPER(8%,⫺1500,8000,2000)
(b) FV(7%,102000,⫺9000)
(c) RATE(10,1000,⫺12000,2000)
(d ) PMT(11%,20,,14000)
(e) PV(8%,15,⫺1000,800)
1.44 Write the engineering economy symbol that corresponds to each of the following spreadsheet
functions.
(a) PMT (b) FV (c) NPER (d) PV (e) IRR
1.45 In a built-in spreadsheet function, if a certain parameter is not present, (a) under what circumstances can it be left blank and (b) when must a
comma be entered in its place?
1.46 Sheryl and Marcelly both invest $1000 at 10% per
year for 4 years. Sheryl receives simple interest and
Marcelly gets compound interest. Use a spreadsheet
and cell reference formats to develop relations that
show a total of $64 more interest for Marcelly at the
end of the 4 years. Assume no withdrawals or further deposits are made during the 4 years.
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
1.47 The concept that different sums of money at different points in time can be said to be equal to each
other is known as:
(a) Evaluation criterion
(b) Equivalence
(c) Cash flow
(d ) Intangible factors
1.48 The evaluation criterion that is usually used in an
economic analysis is:
(a) Time to completion
(b) Technical feasibility
(c) Sustainability
(d ) Financial units (dollars or other currency)
1.49 All of the following are examples of cash outflows,
except:
(a) Asset salvage value
(b) Income taxes
(c) Operating cost of asset
(d ) First cost of asset
1.50 In most engineering economy studies, the best alternative is the one that:
(a) Will last the longest time
(b) Is most politically correct
(c) Is easiest to implement
(d ) Has the lowest cost
1.51 The following annual maintenance and operation
(M&O) costs for a piece of equipment were collected over a 5-year period: $12,300, $8900,
$9200, $11,000, and $12,100. The average is
$10,700. In conducting a sensitivity analysis, the
most reasonable range of costs to use (i.e., percent
from the average) is:
(a) ⫾5% (b) ⫾11% (c) ⫾17% (d) ⫾25%
1.52 At an interest rate of 10% per year, the equivalent
amount of $10,000 one year ago is closest to:
(a) $8264 (b) $9091 (c) $11,000 (d) $12,000
1.53 Assume that you and your best friend each have
$1000 to invest. You invest your money in a fund
that pays 10% per year compound interest. Your
friend invests her money at a bank that pays 10%
per year simple interest. At the end of 1 year, the
difference in the total amount for each of you is:
(a) You have $10 more than she does
(b) You have $100 more than she does
(c) You both have the same amount of money
(d ) She has $10 more than you do
1.54 The time it would take for a given sum of money to
double at 4% per year simple interest is closest to:
(a) 30 years (b) 25 years
(c) 20 years (d) 10 years
36
Foundations of Engineering Economy
Chapter 1
1.55 All of the following are examples of equity financing, except:
(a) Mortgage
(b) Money from savings
(c) Cash on hand
(d ) Retained earnings
1.56 To finance a new project costing $30 million, a
company borrowed $21 million at 16% per year
interest and used retained earnings valued at 12%
per year for the remainder of the investment. The
company’s weighted average cost of capital for the
project was closest to:
(a) 12.5% (b) 13.6% (c) 14.8% (d) 15.6%
CASE STUDY
RENEWABLE ENERGY SOURCES FOR ELECTRICITY GENERATION
Background
Pedernales Electric Cooperative (PEC) is the largest
member-owned electric co-op in the United States with over
232,000 meters in 12 Central Texas counties. PEC has a capacity of approximately 1300 MW (megawatts) of power, of
which 277 MW, or about 21%, is from renewable sources.
The latest addition is 60 MW of power from a wind farm in
south Texas close to the city of Corpus Christi. A constant
question is how much of PEC’s generation capacity should be
from renewable sources, especially given the environmental
issues with coal-generated electricity and the rising costs of
hydrocarbon fuels.
Wind and nuclear sources are the current consideration for
the PEC leadership as Texas is increasing its generation by
nuclear power and the state is the national leader in wind
farm–produced electricity.
Consider yourself a member of the board of directors of
PEC. You are an engineer who has been newly elected by the
PEC membership to serve a 3-year term as a director-at-large.
As such, you do not represent a specific district within the
entire service area; all other directors do represent a specific
district. You have many questions about the operations of
PEC, plus you are interested in the economic and societal
benefits of pursuing more renewable source generation
capacity.
Information
Here are some data that you have obtained. The information
is sketchy, as this point, and the numbers are very approximate. Electricity generation cost estimates are national
in scope, not PEC-specific, and are provided in cents per
kilowatt-hour (¢/kWh).
Generation Cost, ¢/kWh
Fuel Source
Coal
Natural gas
Wind
Solar
Likely Range
4 to 9
4 to 10.5
4.8 to 9.1
4.5 to 15.5
Reasonable Average
7.4
8.6
8.2
8.8
National average cost of electricity to residential customers: 11¢/kWh
PEC average cost to residential customers: 10.27 ¢/kWh
(from primary sources) and 10.92 ¢/kWh (renewable sources)
Expected life of a generation facility: 20 to 40 years (it is
likely closer to 20 than 40)
Time to construct a facility: 2 to 5 years
Capital cost to build a generation facility: $900 to $1500
per kW
You have also learned that the PEC staff uses the wellrecognized levelized energy cost (LEC) method to determine
the price of electricity that must be charged to customers to
break even. The formula takes into account the capital cost of
the generation facilities, the cost of capital of borrowed
money, annual maintenance and operation (M&O) costs, and
the expected life of the facility. The LEC formula, expressed
in dollars per kWh for (t ⫽ 1, 2, . . . , n), is
t⫽n
LEC ⫽
P ⫹A ⫹C
⌺ ——————
(1 ⫹ i)
t
t
t
t
t⫽1
———————
t⫽n
E
⌺ ———
(1 ⫹ i)
t
t⫽1
t
where Pt ⫽ capital investments made in year t
At ⫽ annual maintenance and operating (M&O) costs
for year t
Ct ⫽ fuel costs for year t
Et ⫽ amount of electricity generated in year t
n ⫽ expected life of facility
i ⫽ discount rate (cost of capital)
Case Study Exercises
1. If you wanted to know more about the new arrangement with the wind farm in south Texas for the additional 60 MW per year, what types of questions would
you ask of a staff member in your first meeting with
him or her?
2. Much of the current generation capacity of PEC facilities
utilizes coal and natural gas as the primary fuel source.
What about the ethical aspects of the government’s allowance for these plants to continue polluting the atmosphere
with the emissions that may cause health problems for
citizens and further the effects of global warming? What
types of regulations, if any, should be developed for PEC
(and other generators) to follow in the future?
Case Study
3. You developed an interest in the LEC relation and
the publicized cost of electricity of 10.27¢/kWh for
this year. You wonder if the addition of 60 MW of
wind-sourced electricity will make any difference in
the LEC value for this next year. You did learn the
following:
This is year t ⫽ 11 for LEC computation purposes
n ⫽ 25 years
i ⫽ 5% per year
E11 ⫽ 5.052 billion kWh
LEC last year was 10.22 ¢/kWh (last year’s breakeven
cost to customers)
From these sketchy data, can you determine the value of unknowns in the LEC relation for this year? Is it possible to
determine if the wind farm addition of 60 MW makes any
difference in the electricity rate charged to customers? If not,
what additional information is necessary to determine the
LEC with the wind source included?
CASE STUDY
REFRIGERATOR SHELLS
Background
Large refrigerator manufacturers such as Whirlpool, General
Electric, Frigidaire, and others may subcontract the molding of
their plastic liners and door panels. One prime national subcontractor is Innovations Plastics. Because of improvements in mechanical properties, the molded plastic can sustain increased vertical and horizontal loading, thus significantly reducing the need
for attached metal anchors for some shelving. However, improved molding equipment is needed to enter this market now.
The company president wants a recommendation on whether
Innovations should offer the new technology to the major manufacturers and an estimate of the necessary capital investment
to enter this market.
You work as an engineer for Innovations. At this stage,
you are not expected to perform a complete engineering economic analysis, for not enough information is available. You
are asked to formulate reasonable alternatives, determine
what data and estimates are needed for each one, and ascertain what criteria (economic and noneconomic) should be
utilized to make the final decision.
Information
Some information useful at this time is as follows:
• The technology and equipment are expected to last about
10 years before new methods are developed.
• Inflation and income taxes will not be considered in the
analysis.
• The expected returns on capital investment used for the
last three new technology projects were compound rates of
15%, 5%, and 18%. The 5% rate was the criterion for
enhancing an employee-safety system on an existing
chemical-mixing process.
• Equity capital financing beyond $5 million is not possible.
The amount of debt financing and its cost are unknown.
• Annual operating costs have been averaging 8% of first
cost for major equipment.
• Increased annual training costs and salary requirements
for handling the new plastics and operating new equipment can range from $800,000 to $1.2 million.
There are two manufacturers working on the new-generation
equipment. You label these options as alternatives A and B.
Case Study Exercises
1. Use the first four steps of the decision-making process
to generally describe the alternatives and identify what
economic-related estimates you will need to complete
an engineering economy analysis for the president.
2. Identify any noneconomic factors and criteria to be considered in making the alternative selection.
3. During your inquiries about alternative B from its manufacturer, you learn that this company has already produced
a prototype molding machine and has sold it to a company
in Germany for $3 million (U.S. dollars). Upon inquiry,
you further discover that the German company already
has unused capacity on the equipment for manufacturing
plastic shells. The company is willing to sell time on the
equipment to Innovations immediately to produce its own
shells for U.S. delivery. This could allow immediate market entry into the United States. Consider this as alternative C, and develop the estimates necessary to evaluate C
at the same time as alternatives A and B.
37
CHAPTER 2
Factors: How
Time and
Interest Affect
Money
L E A R N I N G
O U T C O M E S
Purpose: Derive and use the engineering economy factors to account for the time value of money.
SECTION
TOPIC
LEARNING OUTCOME
2.1
F兾P and P兾F factors
• Derive and use factors for single amounts—
compound amount (F兾P) and present worth (P兾F)
factor.
2.2
P兾A and A兾P factors
• Derive and use factors for uniform series—present
worth (P兾A) and capital recovery (A兾P) factors.
2.3
F兾A and A兾F factors
• Derive and use factors for uniform series—
compound amount (F兾A) and sinking fund (A兾F)
factors.
2.4
Factor values
• Use linear interpolation in factor tables or
spreadsheet functions to determine factor values.
2.5
Arithmetic gradient
• Use the present worth (P兾G) and uniform annual
series (A兾G) factors for arithmetic gradients.
2.6
Geometric gradient
• Use the geometric gradient series factor (P兾A,g)
to find present worth.
2.7
Find i or n
• Use equivalence relations to determine i (interest
rate or rate of return) or n for a cash flow series.
T
he cash flow is fundamental to every economic study. Cash flows occur in many
configurations and amounts—isolated single values, series that are uniform, and
series that increase or decrease by constant amounts or constant percentages.
This chapter develops derivations for all the commonly used engineering economy factors
that take the time value of money into account.
The application of factors is illustrated using their mathematical forms and a standard notation format. Spreadsheet functions are used in order to rapidly work with cash flow series
and to perform sensitivity analysis.
If the derivation and use of factors are not covered in the course, alternate ways to perform time value of money calculations are summarized in Appendix D.
PE
The Cement Factory Case: Votorantim
Cimentos North America, Inc., is a subsidiary of a Brazil-based company that
recently announced plans to develop a
new cement factory in Houston County
in the state of Georgia. The plant will
be called Houston American Cement, or
HAC. The location is ideal for cement
making because of the large deposit of
limestone in the area.
The plant investment, expected
to amount to $200 million, has been
planned for 2012; however, it is currently
delayed due to the economic downturn
in construction. When the plant is completed and operating at full capacity,
based upon the projected needs and
cost per metric ton, it is possible that
the plant could generate as much as
$50,000,000 annually in revenue. All
analysis will use a planning horizon of
5 years commencing when the plant
begins operation.
This case is used in the following
topics (and sections) of this chapter:
Single-amount factors (2.1)
Uniform series factors (2.2 and 2.3)
Arithmetic gradient factors (2.5)
Geometric gradient factors (2.6)
Determining unknown n values (2.7)
2.1 Single-Amount Factors (F兾P and P兾F )
The most fundamental factor in engineering economy is the one that determines the amount
of money F accumulated after n years (or periods) from a single present worth P, with interest
compounded one time per year (or period). Recall that compound interest refers to interest paid
on top of interest. Therefore, if an amount P is invested at time t ⫽ 0, the amount F1 accumulated
1 year hence at an interest rate of i percent per year will be
F1 ⫽ P ⫹ Pi
⫽ P(1 ⫹ i)
where the interest rate is expressed in decimal form. At the end of the second year, the amount
accumulated F2 is the amount after year 1 plus the interest from the end of year 1 to the end of
year 2 on the entire F1.
F 2 ⫽ F 1 ⫹ F 1i
⫽ P(1 ⫹ i) ⫹ P(1 ⫹ i)i
[2.1]
The amount F2 can be expressed as
F2 ⫽ P(1 ⫹ i ⫹ i ⫹ i2 )
⫽ P(1 ⫹ 2i ⫹ i2 )
⫽ P(1 ⫹ i)2
Similarly, the amount of money accumulated at the end of year 3, using Equation [2.1], will be
F 3 ⫽ F 2 ⫹ F 2i
40
Factors: How Time and Interest Affect Money
Chapter 2
F=?
F = given
i = given
0
1
2
n–2
i = given
n–1
n
0
P = given
1
2
n–2
n–1
n
P=?
(a)
(b)
Figure 2–1
Cash flow diagrams for single-payment factors: (a) find F, given P, and (b) find P, given F.
Substituting P(1 ⫹ i)2 for F2 and simplifying, we get
F3 ⫽ P(1 ⫹ i)3
From the preceding values, it is evident by mathematical induction that the formula can be generalized for n years. To find F, given P,
F ⴝ P(1 ⴙ i)n
[2.2]
The factor (1 ⴙ i)n is called the single-payment compound amount factor (SPCAF), but it is usually referred to as the F兾P factor. This is the conversion factor that, when multiplied by P, yields
the future amount F of an initial amount P after n years at interest rate i. The cash flow diagram
is seen in Figure 2–1a.
Reverse the situation to determine the P value for a stated amount F that occurs n periods
in the future. Simply solve Equation [2.2] for P.
[
]
1
P ⴝ F ————
ⴝF(1 ⴙ i)ⴚn
(1 ⴙ i)n
[2.3]
The expression (1 ⴙ i)⫺n is known as the single-payment present worth factor (SPPWF), or the
P兾F factor. This expression determines the present worth P of a given future amount F after n
years at interest rate i. The cash flow diagram is shown in Figure 2–1b.
Note that the two factors derived here are for single payments; that is, they are used to find the
present or future amount when only one payment or receipt is involved.
A standard notation has been adopted for all factors. The notation includes two cash flow symbols, the interest rate, and the number of periods. It is always in the general form (X兾Y,i,n). The
letter X represents what is sought, while the letter Y represents what is given. For example, F兾P
means find F when given P. The i is the interest rate in percent, and n represents the number of
periods involved.
Using this notation, (F兾P,6%,20) represents the factor that is used to calculate the future
amount F accumulated in 20 periods if the interest rate is 6% per period. The P is given. The
standard notation, simpler to use than formulas and factor names, will be used hereafter.
Table 2–1 summarizes the standard notation and equations for the F兾P and P兾F factors. This
information is also included inside the front cover.
TABLE 2–1
F兾P and P兾F Factors: Notation and Equations
Factor
Notation
Name
Find/Given
Standard Notation
Equation
Equation
with Factor Formula
Excel
Function
(F兾P,i,n)
Single-payment
compound amount
Single-payment
present worth
F兾P
F ⫽ P(F兾P,i,n)
F ⫽ P(1 ⫹ i)n
⫽ FV(i%,n,,P)
P兾F
P ⫽ F(P兾F,i,n)
P ⫽ F (1 ⫹ i)⫺n
⫽ PV(i%,n,,F)
(P兾F,i,n)
Single-Amount Factors (F兾P and P兾F )
2.1
To simplify routine engineering economy calculations, tables of factor values have been prepared for interest rates from 0.25% to 50% and time periods from 1 to large n values, depending
on the i value. These tables, found at the rear of the book, have a colored edge for easy identification. They are arranged with factors across the top and the number of periods n down the left side.
The word discrete in the title of each table emphasizes that these tables utilize the end-of-period
convention and that interest is compounded once each interest period. For a given factor, interest
rate, and time, the correct factor value is found at the intersection of the factor name and n. For
example, the value of the factor (P兾F,5%,10) is found in the P兾F column of Table 10 at period 10
as 0.6139. This value is determined by using Equation [2.3].
1
(P兾F,5%,10) ⫽ ————
(1 ⫹ i)n
1
⫽ ————
(1.05)10
1 ⫽ 0.6139
⫽ ———
1.6289
For spreadsheets, a future value F is calculated by the FV function using the format
ⴝ FV(i%,n,,P)
[2.4]
A present amount P is determined using the PV function with the format
ⴝ PV(i%,n,,F)
[2.5]
These functions are included in Table 2–1. Refer to Appendix A or Excel online help for more
information on the use of FV and PV functions.
EXAMPLE 2.1
Sandy, a manufacturing engineer, just received a year-end bonus of $10,000 that will be invested
immediately. With the expectation of earning at the rate of 8% per year, Sandy hopes to take the
entire amount out in exactly 20 years to pay for a family vacation when the oldest daughter is due
to graduate from college. Find the amount of funds that will be available in 20 years by using
(a) hand solution by applying the factor formula and tabulated value and (b) a spreadsheet function.
Solution
The cash flow diagram is the same as Figure 2–1a. The symbols and values are
P ⫽ $10,000
F⫽?
i ⫽ 8% per year
n ⫽ 20 years
(a) Factor formula: Apply Equation [2.2] to find the future value F. Rounding to four decimals, we have
F ⫽ P(1 ⫹ i)n ⫽ 10,000(1.08)20 ⫽ 10,000(4.6610)
⫽ $46,610
Standard notation and tabulated value: Notation for the F兾P factor is (F兾P,i%,n).
F ⫽ P(F兾P,8%,20) ⫽ 10,000(4.6610)
⫽ $46,610
Table 13 provides the tabulated value. Round-off errors can cause a slight difference in
the final answer between these two methods.
(b) Spreadsheet: Use the FV function to find the amount 20 years in the future. The format is
that shown in Equation [2.4]; the numerical entry is ⫽ FV(8%,20,,10000). The spreadsheet will appear similar to that in the right side of Figure 1–13, with the answer
($46,609.57) displayed. (You should try it on your own computer now.) The FV function
has performed the computation in part (a) and displayed the result.
The equivalency statement is: If Sandy invests $10,000 now and earns 8% per year every year
for 20 years, $46,610 will be available for the family vacation.
41
42
Factors: How Time and Interest Affect Money
Chapter 2
PE
EXAMPLE 2.2 The Cement Factory Case
As discussed in the introduction to this chapter, the Houston American Cement factory will
require an investment of $200 million to construct. Delays beyond the anticipated implementation year of 2012 will require additional money to construct the factory. Assuming that the cost
of money is 10% per year, compound interest, use both tabulated factor values and spreadsheet functions to determine the following for the board of directors of the Brazilian company
that plans to develop the plant.
(a) The equivalent investment needed if the plant is built in 2015.
(b) The equivalent investment needed had the plant been constructed in the year 2008.
Solution
Figure 2–2 is a cash flow diagram showing the expected investment of $200 million ($200 M)
in 2012, which we will identify as time t ⫽ 0. The required investments 3 years in the future
and 4 years in the past are indicated by F3 ⫽ ? and P⫺4 ⫽ ?, respectively.
Figure 2–2
F3 = ?
Cash flow diagram for
Example 2.2a and b.
P−4 = ?
−4 −3
2008
2009
−2
−1
0
1
2
2010
2011
2012
2013
2014
3
2015
t
Year
$200 M
(a) To find the equivalent investment required in 3 years, apply the F兾P factor. Use $1 million units and the tabulated value for 10% interest (Table 15).
F3 ⫽ P(F兾P,i,n) ⫽ 200(F兾P,10%,3) ⫽ 200(1.3310)
⫽ $266.2 ($266,200,000)
Now, use the FV function on a spreadsheet to find the same answer, F3 ⫽ $266.20 million.
(Refer to Figure 2–3, left side.)
⫽ FV(10%,3,,200)
⫽ PV(10%,4,,200)
Figure 2–3
Spreadsheet functions for Example 2.2.
(b) The year 2008 is 4 years prior to the planned construction date of 2012. To determine the
equivalent cost 4 years earlier, consider the $200 M in 2012 (t ⫽ 0) as the future value F
and apply the P兾F factor for n ⫽ 4 to find P⫺4. (Refer to Figure 2–2.) Table 15 supplies
the tabulated value.
P⫺4 ⫽ F(P兾F,i,n) ⫽ 200(P兾F,10%,4) ⫽ 200(0.6830)
⫽ $136.6 ($136,600,000)
The PV function ⫽ PV(10%,4,,200) will display the same amount as shown in Figure 2–3, right side.
This equivalence analysis indicates that at $136.6 M in 2008, the plant would have cost about
68% as much as in 2012, and that waiting until 2015 will cause the price tag to increase about
33% to $266 M.
43
Uniform Series Present Worth Factor and Capital Recovery Factor (P兾A and A兾P)
2.2
2.2 Uniform Series Present Worth Factor and
Capital Recovery Factor (P兾A and A兾P)
The equivalent present worth P of a uniform series A of end-of-period cash flows (investments)
is shown in Figure 2–4a. An expression for the present worth can be determined by considering
each A value as a future worth F, calculating its present worth with the P兾F factor, Equation [2.3],
and summing the results.
[
] [ ] [
] [ ]
]
1
1
1
P ⫽ A ————
⫹ A ————
⫹ A ————
⫹...
(1 ⫹ i)1
(1 ⫹ i)2
(1 ⫹ i)3
1
1
⫹ A ————
⫹ A ————
(1 ⫹ i)n
(1 ⫹ i)n⫺1
[
The terms in brackets are the P兾F factors for years 1 through n, respectively. Factor out A.
[
1
1
1
1
1
⫹ ————
⫹ ————
⫹ . . . ⫹ ————
⫹ ————
P ⫽ A ————
n
(1 ⫹ i)1 (1 ⫹ i)2 (1 ⫹ i)3
(1 ⫹ i)n⫺1 (1 ⫹ i)
]
[2.6]
To simplify Equation [2.6] and obtain the P兾A factor, multiply the n-term geometric progression
in brackets by the (P兾F,i%,1) factor, which is 1兾(1 ⫹ i). This results in Equation [2.7]. Now
subtract the two equations, [2.6] from [2.7], and simplify to obtain the expression for P when
i ⫽ 0 (Equation [2.8]).
[ (1 ⫹ i)
1
1
1
1
1
P ⫽ A ————
———
⫹ ————
⫹ ————
⫹ . . . ⫹ ————
n ⫹ ————
2
3
4
n⫹1
1⫹i
(1 ⫹ i)
(1 ⫹ i)
(1 ⫹ i)
[ (1 ⫹ i) (1 ⫹ i)
(1 ⫹ i)
⫺
P⫽A
[ (1 ⫹1 i) ⫹ (1 ⫹1 i) ⫹ . . . ⫹ (1 ⫹1i)
⫺i P ⫽ A
[ (1 ⫹1i) ⫺ (1 ⫹1 i) ]
1⫹i
(1 ⫹ i)
1
1
1
1
1 P ⫽ A ————
———
⫹ ————
⫹ . . . ⫹ ————
n ⫹ ————
2
3
n⫹1
1⫹i
————1
———
(1 ⫹ i)
[2.7]
]
1
————
⫹ ————
n
n⫺1
————2
————
n⫹1
(1 ⫹ i)
]
]
————1
[
A ————
1
⫺1
P ⫽ ——
⫺i (1 ⫹ i)n
]
[
(1 ⴙ i)n ⴚ 1
P ⴝ A ——————
i(1 ⴙ i)n
]
i⫽0
[2.8]
The term in brackets in Equation [2.8] is the conversion factor referred to as the uniform series
present worth factor (USPWF). It is the P兾A factor used to calculate the equivalent P value in
year 0 for a uniform end-of-period series of A values beginning at the end of period 1 and extending for n periods. The cash flow diagram is Figure 2–4a.
P = given
P=?
i = given
0
1
2
n–2
i = given
n–1
n
0
1
A = given
(a)
Figure 2–4
Cash flow diagrams used to determine (a) P, given a uniform series A, and (b) A, given a present worth P.
2
n–2
A=?
(b)
n–1
n
44
Factors: How Time and Interest Affect Money
Chapter 2
TABLE 2–2
Notation
(P兾A,i,n)
(A兾P,i,n)
P兾A and A兾P Factors: Notation and Equations
Factor
Name
Find/Given
Uniform series
present worth
Capital recovery
P兾A
Factor
Formula
Standard
Notation Equation
Excel
Function
(1 ⫹ i)n ⫺ 1
i(1 ⫹ i)
i(1 ⫹ i)n
—————
(1 ⫹ i)n − 1
P ⫽ A(P兾A,i,n)
⫽ PV(i%,n,A)
A ⫽ P(A兾P,i,n)
⫽ PMT(i%,n,P)
—————
n
A兾P
To reverse the situation, the present worth P is known and the equivalent uniform series
amount A is sought (Figure 2–4b). The first A value occurs at the end of period 1, that is, one
period after P occurs. Solve Equation [2.8] for A to obtain
[
i(1 ⴙ i)n
A ⴝ P ——————
(1 ⴙ i)n ⴚ 1
]
[2.9]
The term in brackets is called the capital recovery factor (CRF), or A兾P factor. It calculates the
equivalent uniform annual worth A over n years for a given P in year 0, when the interest
rate is i.
Placement of P
The P兾A and A兾P factors are derived with the present worth P and the first uniform annual
amount A one year (period) apart. That is, the present worth P must always be located one
period prior to the first A.
The factors and their use to find P and A are summarized in Table 2–2 and inside the front cover.
The standard notations for these two factors are (P兾A,i%,n) and (A兾P,i%,n). Tables at the end of
the text include the factor values. As an example, if i ⫽ 15% and n ⫽ 25 years, the P兾A factor
value from Table 19 is (P兾A,15%,25) ⫽ 6.4641. This will find the equivalent present worth at
15% per year for any amount A that occurs uniformly from years 1 through 25.
Spreadsheet functions can determine both P and A values in lieu of applying the P兾A and A兾P
factors. The PV function calculates the P value for a given A over n years and a separate F value
in year n, if it is given. The format, is
ⴝ PV(i%,n,A,F)
[2.10]
Similarly, the A value is determined by using the PMT function for a given P value in year 0 and
a separate F, if given. The format is
ⴝ PMT(i%, n,P,F)
[2.11]
Table 2–2 includes the PV and PMT functions.
EXAMPLE 2.3
How much money should you be willing to pay now for a guaranteed $600 per year for 9 years
starting next year, at a rate of return of 16% per year?
Solution
The cash flows follow the pattern of Figure 2–4a, with A ⫽ $600, i ⫽ 16%, and n ⫽ 9. The
present worth is
P ⫽ 600(P兾A,16%,9) ⫽ 600(4.6065) ⫽ $2763.90
The PV function ⫽ PV(16%,9,600) entered into a single spreadsheet cell will display the
answer P ⫽ ($2763.93).
2.2
Uniform Series Present Worth Factor and Capital Recovery Factor (P兾A and A兾P)
PE
EXAMPLE 2.4 The Cement Factory Case
As mentioned in the chapter introduction of this case, the Houston American Cement plant
may generate a revenue base of $50 million per year. The president of the Brazilian parent
company Votorantim Cimentos may have reason to be quite pleased with this projection for
the simple reason that over the 5-year planning horizon, the expected revenue would total
$250 million, which is $50 million more than the initial investment. With money worth
10% per year, address the following question from the president: Will the initial investment
be recovered over the 5-year horizon with the time value of money considered? If so, by how
much extra in present worth funds? If not, what is the equivalent annual revenue base required
for the recovery plus the 10% return on money? Use both tabulated factor values and spreadsheet functions.
Solution
Tabulated value: Use the P兾A factor to determine whether A ⫽ $50 million per year for
n ⫽ 5 years starting 1 year after the plant’s completion (t ⫽ 0) at i ⫽ 10% per year is equivalently less or greater than $200 M. The cash flow diagram is similar to Figure 2–4a, where the
first A value occurs 1 year after P. Using $1 million units and Table 15 values,
P ⫽ 50(P兾A,10%,5) ⫽ 50(3.7908)
⫽ $189.54
($189,540,000)
The present worth value is less than the investment plus a 10% per year return, so the president
should not be satisfied with the projected annual revenue.
To determine the minimum required to realize a 10% per year return, use the A兾P factor.
The cash flow diagram is the same as Figure 2–4b, where A starts 1 year after P at t ⫽ 0 and
n ⫽ 5.
A ⫽ 200(A兾P,10%,5) ⫽ 200(0.26380)
⫽ $52.76 per year
The plant needs to generate $52,760,000 per year to realize a 10% per year return over
5 years.
Spreadsheet: Apply the PV and PMT functions to answer the question. Figure 2–5 shows
the use of ⫽ PV(i%,n,A,F) on the left side to find the present worth and the use of
⫽ PMT(i%,n,P,F) on the right side to determine the minimum A of $52,760,000 per year.
Because there are no F values, it is omitted from the functions. The minus sign placed before
each function name forces the answer to be positive, since these two functions always display
the answer with the opposite sign entered on the estimated cash flows.
⫽ ⫺PV(10%,5,50)
Figure 2–5
Spreadsheet functions to find P and A for the cement factory case, Example 2.4.
45
⫽ ⫺PMT(10%,5,200)
46
Factors: How Time and Interest Affect Money
Chapter 2
2.3 Sinking Fund Factor and Uniform Series Compound
Amount Factor (A兾F and F兾A)
The simplest way to derive the A兾F factor is to substitute into factors already developed. If P
from Equation [2.3] is substituted into Equation [2.9], the following formula results.
][
[
i(1 ⫹ i)n
1
A ⫽ F ————
n —————
(1 ⫹ i) (1 ⫹ i)n − 1
[
i
A ⴝ F —————
(1 ⴙ i)n ⴚ 1
]
]
[2.12]
The expression in brackets in Equation [2.12] is the A兾F or sinking fund factor. It determines
the uniform annual series A that is equivalent to a given future amount F. This is shown graphically in Figure 2–6a, where A is a uniform annual investment.
The uniform series A begins at the end of year (period) 1 and continues through the year of
the given F. The last A value and F occur at the same time.
Placement of F
Equation [2.12] can be rearranged to find F for a stated A series in periods 1 through n (Figure 2–6b).
[
(1 ⴙ i)n ⴚ 1
F ⴝ A ——————
i
]
[2.13]
The term in brackets is called the uniform series compound amount factor (USCAF), or F兾A factor.
When multiplied by the given uniform annual amount A, it yields the future worth of the uniform
series. It is important to remember that the future amount F occurs in the same period as the last A.
Standard notation follows the same form as that of other factors. They are (F兾A,i,n) and
(A兾F,i,n). Table 2–3 summarizes the notations and equations, as does the inside front cover.
As a matter of interest, the uniform series factors can be symbolically determined by using an
abbreviated factor form. For example, F兾A ⫽ (F兾P)(P兾A), where cancellation of the P is correct.
Using the factor formulas, we have
(1 ⫹ i)n ⫺ 1
(1 ⫹ i)n ⫺ 1
(F兾A,i,n) ⫽ [(1 ⫹ i)n] ——————
⫽ ——————
n
i
i(1 ⫹ i)
[
]
For solution by spreadsheet, the FV function calculates F for a stated A series over n years.
The format is
ⴝ FV(i%,n,A,P)
[2.14]
The P may be omitted when no separate present worth value is given. The PMT function determines the A value for n years, given F in year n and possibly a separate P value in year 0. The
format is
ⴝ PMT(i%,n,P,F)
[2.15]
If P is omitted, the comma must be entered so the function knows the last entry is an F value.
F = given
F=?
i = given
i = given
0
1
2
n–2
n–1
n
A=?
(a)
Figure 2–6
Cash flow diagrams to (a) find A, given F, and (b) find F, given A.
0
1
2
n–2
A = given
(b)
n–1
n
Sinking Fund Factor and Uniform Series Compound Amount Factor (A兾F and F兾A)
2.3
TABLE 2–3
Notation
(F兾A,i,n)
(A兾F,i,n)
47
F兾A and A兾F Factors: Notation and Equations
Factor
Name
Find/Given
Uniform series
compound amount
Sinking fund
Factor
Formula
Standard Notation
Equation
(1 ⫹ i)n ⫺ 1 F ⫽ A(F兾A,i,n)
i
A ⫽ F(A兾F,i,n)
i
—————
(1 ⫹ i)n − 1
F兾A
—————
A兾F
Excel
Functions
⫽ FV(i%,n,A)
⫽ PMT(i%,n,F)
EXAMPLE 2.5
The president of Ford Motor Company wants to know the equivalent future worth of a $1 million capital investment each year for 8 years, starting 1 year from now. Ford capital earns at a
rate of 14% per year.
Solution
The cash flow diagram (Figure 2–7) shows the annual investments starting at the end of year 1
and ending in the year the future worth is desired. In $1000 units, the F value in year 8 is found
by using the F兾A factor.
F ⫽ 1000(F兾A,14%,8) ⫽ 1000(13.2328) ⫽ $13,232.80
F=?
i = 14%
0
1
2
3
4
5
6
7
8
A = $1000
Figure 2–7
Diagram to find F for a uniform series, Example 2.5.
EXAMPLE 2.6 The Cement Factory Case
Once again, consider the HAC case presented at the outset of this chapter, in which a projected
$200 million investment can generate $50 million per year in revenue for 5 years starting
1 year after start-up. A 10% per year time value of money has been used previously to determine
P, F, and A values. Now the president would like the answers to a couple of new questions
about the estimated annual revenues. Use tabulated values, factor formulas, or spreadsheet
functions to provide the answers.
(a) What is the equivalent future worth of the estimated revenues after 5 years at 10% per year?
(b) Assume that, due to the economic downturn, the president predicts that the corporation
will earn only 4.5% per year on its money, not the previously anticipated 10% per year.
What is the required amount of the annual revenue series over the 5-year period to be economically equivalent to the amount calculated in (a)?
Solution
(a) Figure 2–6b is the cash flow diagram with A ⫽ $50 million. Note that the last A value and
F ⫽ ? both occur at the end of year n ⫽ 5. We use tabulated values and the spreadsheet
function to find F in year 5.
Tabulated value: Use the F兾A factor and 10% interest factor table. In $1 million units, the
future worth of the revenue series is
F ⫽ 50(F兾A,10%,5) ⫽ 50(6.1051)
⫽ $305.255 ($305,255,000)
PE
48
Factors: How Time and Interest Affect Money
Chapter 2
⫽ ⫺PMT(4.5%,5,,B5)
⫽ ⫺FV(10%,5,50)
Figure 2–8
Spreadsheet functions to find F and A at i ⫽ 4.5% for the cement factory case,
Example 2.6.
If the rate of return on the annual revenues were 0%, the total amount after 5 years would
be $250,000,000. The 10% per year return is projected to grow this value by 22%.
Spreadsheet: Apply the FV factor in the format ⫽ ⫺FV(10%,5,50) to determine F ⫽
$305.255 million. Because there is no present amount in this computation, P is omitted
from the factor. See Figure 2–8, left side. (As before, the minus sign forces the FV function
to result in a positive value.)
(b) The president of the Brazilian company planning to develop the cement plant in Georgia
is getting worried about the international economy. He wants the revenue stream to generate the equivalent that it would at a 10% per year return, that is, $305.255 million, but
thinks that only a 4.5% per year return is achievable.
Factor formula: The A兾F factor will determine the required A for 5 years. Since the factor
tables do not include 4.5%, use the formula to answer the question. In $1 million units,
0.045
⫽ 305.255(0.18279)
A ⫽ 305.255(A兾F,4.5%,5) ⫽ 305.255 ——————
(1.045)5 ⫺ 1
⫽ $55.798
[
]
The annual revenue requirement grows from $50 million to nearly $55,800,000. This is a
significant increase of 11.6% each year.
Spreadsheet: It is easy to answer this question by using the ⫽ PMT(i%,n,,F) function with
i ⫽ 4.5% and F ⫽ $305.255 found in part (a). We can use the cell reference method
(described in Appendix A) for the future amount F. Figure 2–8, right side, displays the required A of $55.798 per year (in $1 million units).
2.4 Factor Values for Untabulated i or n Values
Often it is necessary to know the correct numerical value of a factor with an i or n value that is
not listed in the compound interest tables in the rear of the book. Given specific values of i and
n, there are several ways to obtain any factor value.
• Use the formula listed in this chapter or the front cover of the book,
• Use an Excel function with the corresponding P, F, or A value set to 1.
• Use linear interpolation in the interest tables.
When the formula is applied, the factor value is accurate since the specific i and n values are
input. However, it is possible to make mistakes since the formulas are similar to each other, especially when uniform series are involved. Additionally, the formulas become more complex
when gradients are introduced, as you will see in the following sections.
A spreadsheet function determines the factor value if the corresponding P, A, or F argument in the function is set to 1 and the other parameters are omitted or set to zero. For example, the P兾F factor is determined using the PV function with A omitted (or set to 0) and
F ⫽ 1, that is, PV(i%,n,,1) or PV(i%,n,0,1). A minus sign preceding the function identifier
causes the factor to have a positive value. Functions to determine the six common factors are
as follows.
Factor Values for Untabulated i or n Values
2.4
Factor
P兾F
F兾P
P兾A
A兾P
F兾A
A兾F
To Do This
Excel Function
Find P, given F.
Find F, given P.
Find P, given A.
Find A, given P.
Find F, given A.
Find A, given F.
⫽ ⫺PV(i%,n,,1)
⫽ ⫺FV(i%,n,,1)
⫽ ⫺PV(i%,n,1)
⫽ ⫺PMT(i%,n,1)
⫽ ⫺FV(i%,n,1)
⫽ ⫺PMT(i%,n,,1)
Figure 2–9 shows a spreadsheet developed explicitly to determine these factor values. When it is
made live in Excel, entering any combination of i and n displays the exact value for all six factors. The
values for i ⫽ 3.25% and n ⫽ 25 years are shown here. As we already know, these same functions will
determine a final P, A, or F value when actual or estimated cash flow amounts are entered.
Linear interpolation for an untabulated interest rate i or number of years n takes more time
to complete than using the formula or spreadsheet function. Also interpolation introduces some
level of inaccuracy, depending upon the distance between the two boundary values selected for
i or n, as the formulas themselves are nonlinear functions. Interpolation is included here for individuals who wish to utilize it in solving problems. Refer to Figure 2–10 for a graphical description of the following explanation. First, select two tabulated values (x1 and x2) of the parameter
for which the factor is requested, that is, i or n, ensuring that the two values surround and are not
too distant from the required value x. Second, find the corresponding tabulated factor values
(f1 and f2). Third, solve for the unknown, linearly interpolated value f using the formulas below,
where the differences in parentheses are indicated in Figure 2–10 as a through c.
Figure 2–9
Enter requested i and n
Figure 2–10
Factor value
axis
f2
c
f
Linear interpolation in factor
value tables.
Table
Linear
assumption
Unknown
d
f1
Use of Excel functions to display
factor values for any i and n
values.
Table
a
Known
x1
Required
x
b
Known i or n
x2
axis
49
50
Factors: How Time and Interest Affect Money
Chapter 2
(x – x1)
(f – f )
f ⫽ f1 ⫹ ————
(x2 – x1) 2 1
ac ⫽ f ⫹ d
f ⫽ f1 ⫹ —
1
b
[2.16]
[2.17]
The value of d will be positive or negative if the factor is increasing or decreasing, respectively,
in value between x1 and x2 .
EXAMPLE 2.7
Determine the P兾A factor value for i ⫽ 7.75% and n ⫽ 10 years, using the three methods described previously.
Solution
Factor formula: Apply the formula from inside the front cover of the book for the P兾A factor.
Showing 5-decimal accuracy,
(1 ⫹ i)n ⫺ 1
(1.0775)10 ⫺ 1
1.10947
(P兾A,7.75%,10) ⫽ —————
⫽ ————
⫽ ———————
n
i(1 ⫹ i)
0.0775(1.0775)10 0.16348
⫽ 6.78641
Spreadsheet: Utilize the spreadsheet function in Figure 2–9, that is, ⫽ ⫺PV(7.75%,10,1), to
display 6.78641.
Linear interpolation: Use Figure 2–10 as a reference for this solution. Apply the Equation [2.16] and [2.17] sequence, where x is the interest rate i, the bounding interest rates are
i1 ⫽ 7% and i2 ⫽ 8%, and the corresponding P兾A factor values are f1 ⫽ (P兾A,7%,10) ⫽ 7.0236
and f2 ⫽ (P兾A,8%,10) ⫽ 6.7101. With 4-place accuracy,
(i ⫺ i1)
(7.75 ⫺ 7)
(f – f ) ⫽ 7.0236 ⫹ ————— (6.7101 ⫺ 7.0236)
f ⫽ f1 ⫹ ———
( i 2 – i 1) 2 1
(8 ⫺ 7)
⫽ 7.0236 ⫹ (0.75)(−0.3135) ⫽ 7.0236 − 0.2351
⫽ 6.7885
Comment
Note that since the P兾A factor value decreases as i increases, the linear adjustment is negative
at ⫺0.2351. As is apparent, linear interpolation provides an approximation to the correct factor
value for 7.75% and 10 years, plus it takes more calculations than using the formula or spreadsheet function. It is possible to perform two-way linear interpolation for untabulated i and n
values; however, the use of a spreadsheet or factor formula is recommended.
2.5 Arithmetic Gradient Factors (P兾G and A兾G)
Assume a manufacturing engineer predicts that the cost of maintaining a robot will increase by
$5000 per year until the machine is retired. The cash flow series of maintenance costs involves a
constant gradient, which is $5000 per year.
An arithmetic gradient series is a cash flow series that either increases or decreases by a constant amount each period. The amount of change is called the gradient.
Formulas previously developed for an A series have year-end amounts of equal value. In the
case of a gradient, each year-end cash flow is different, so new formulas must be derived. First,
assume that the cash flow at the end of year 1 is the base amount of the cash flow series and,
therefore, not part of the gradient series. This is convenient because in actual applications, the
base amount is usually significantly different in size compared to the gradient. For example, if
you purchase a used car with a 1-year warranty, you might expect to pay the gasoline and insurance costs during the first year of operation. Assume these cost $2500; that is, $2500 is the base
amount. After the first year, you absorb the cost of repairs, which can be expected to increase
51
Arithmetic Gradient Factors (P兾G and A兾G)
2.5
0
1
2
3
4
n
n–1
Time
Figure 2–11
Cash flow diagram of an
arithmetic gradient series.
$2500
$2700
$2900
$3100
$2500
+ (n – 2)200
0
1
2
3
4
5
$2500
+ (n – 1)200
n–1
n
Time
G
2G
3G
4G
(n – 2)G
(n – 1)G
each year. If you estimate that total costs will increase by $200 each year, the amount the second
year is $2700, the third $2900, and so on to year n, when the total cost is 2500 ⫹ (n ⫺ 1)200. The
cash flow diagram is shown in Figure 2–11. Note that the gradient ($200) is first observed between year 1 and year 2, and the base amount ($2500 in year 1) is not equal to the gradient.
Define the symbols G for gradient and CFn for cash flow in year n as follows.
G ⫽ constant arithmetic change in cash flows from one time period to the next; G may be positive
or negative.
CFn ⴝ base amount ⴙ (n ⴚ 1)G
[2.18]
It is important to realize that the base amount defines a uniform cash flow series of the size A that
occurs eash time period. We will use this fact when calculating equivalent amounts that involve
arithmetic gradients. If the base amount is ignored, a generalized arithmetic (increasing) gradient cash flow diagram is as shown in Figure 2–12. Note that the gradient begins between years
1 and 2. This is called a conventional gradient.
EXAMPLE 2.8
A local university has initiated a logo-licensing program with the clothier Holister, Inc. Estimated fees (revenues) are $80,000 for the first year with uniform increases to a total of $200,000
by the end of year 9. Determine the gradient and construct a cash flow diagram that identifies
the base amount and the gradient series.
Solution
The year 1 base amount is CF1 ⫽ $80,000, and the total increase over 9 years is
CF9 ⫺ CF1 ⫽ 200,000 – 80,000 ⫽ $120,000
Equation [2.18], solved for G, determines the arithmetic gradient.
(CF9 ⫺ CF1) 120,000
⫽ ————
G ⫽ ——————
n⫺1
9⫺1
⫽ $15,000 per year
Figure 2–12
Conventional arithmetic
gradient series without
the base amount.
52
Factors: How Time and Interest Affect Money
Chapter 2
CF9 =
$200,000
$185,000
$170,000
$155,000
$140,000
$125,000
$110,000
G = $15,000
CF1 = $95,000
$80,000
0
1
2
3
4
5
6
7
8
9
Year
Figure 2–13
Diagram for gradient series, Example 2.8.
The cash flow diagram (Figure 2–13) shows the base amount of $80,000 in years 1 through 9
and the $15,000 gradient starting in year 2 and continuing through year 9.
The total present worth PT for a series that includes a base amount A and conventional arithmetic gradient must consider the present worth of both the uniform series defined by A and the
arithmetic gradient series. The addition of the two results in PT.
PT ⴝ PA ⴞ PG
[2.19]
where PA is the present worth of the uniform series only, PG is the present worth of the gradient
series only, and the ⫹ or ⫺ sign is used for an increasing (⫹G) or decreasing (⫺G) gradient,
respectively.
The corresponding equivalent annual worth AT is the sum of the base amount series annual
worth AA and gradient series annual worth AG, that is,
AT ⴝ AA ⴞ AG
[2.20]
Three factors are derived for arithmetic gradients: the P兾G factor for present worth, the A兾G
factor for annual series, and the F兾G factor for future worth. There are several ways to derive
them. We use the single-payment present worth factor (P兾F,i,n), but the same result can be obtained by using the F兾P, F兾A, or P兾A factor.
In Figure 2–12, the present worth at year 0 of only the gradient is equal to the sum of the present worths of the individual cash flows, where each value is considered a future amount.
P ⫽ G(P兾F,i,2) ⫹ 2G(P兾F,i,3) ⫹ 3G(P兾F,i,4) ⫹ . . .
⫹ [(n ⫺ 2)G](P兾F,i,n ⫺ 1) ⫹ [(n ⫺ 1)G](P兾F,i,n)
Factor out G and use the P兾F formula.
[
3
n ⫺ 2 ⫹ ————
n⫺1
1
2
⫹ ————
⫹ ————
⫹ . . . ⫹ ————
P ⫽ G ————
n
(1 ⫹ i)2
(1 ⫹ i)3 (1 ⫹ i)4
(1 ⫹ i)n⫺1 (1 ⫹ i)
]
[2.21]
Multiplying both sides of Equation [2.21] by (1 ⫹ i)1 yields
[
]
3
n ⫺ 2 ⫹ ————
n⫺1
1
2
P (1 ⫹ i)1 ⫽ G ————
[2.22]
⫹ ————
⫹ ————
⫹ . . . ⫹ ————
(1 ⫹ i)1 (1 ⫹ i)2 (1 ⫹ i)3
(1 ⫹ i)n⫺2 (1 ⫹ i)n⫺1
Subtract Equation [2.21] from Equation [2.22] and simplify.
[
] [
n
1
1
1
1
iP ⫽ G ————
⫺ G ————
⫹ ————
⫹ . . . ⫹ ————
⫹ ————
(1 ⫹ i)n
(1 ⫹ i)n
(1 ⫹ i)1 (1 ⫹ i)2
(1 ⫹ i)n⫺1
]
[2.23]
The left bracketed expression is the same as that contained in Equation [2.6], where the
P兾A factor was derived. Substitute the closed-end form of the P兾A factor from Equation [2.8]
53
Arithmetic Gradient Factors (P兾G and A兾G)
2.5
PG = ?
i = given
0
1
2
3
4
n
n–1
0
1
2
3
4
n–1
n
G
2G
3G
(n – 2)G
(n – 1)G
(b)
(a)
Figure 2–14
Conversion diagram from an arithmetic gradient to a present worth.
into Equation [2.23] and simplify to solve for P G , the present worth of the gradient series
only.
[
(1 ⴙ i)n ⴚ 1
G ——————
n
ⴚ ————
PG ⴝ —
i
i(1 ⴙ i)n
(1 ⴙ i)n
]
[2.24]
Equation [2.24] is the general relation to convert an arithmetic gradient G (not including the
base amount) for n years into a present worth at year 0. Figure 2–14a is converted into the
equivalent cash flow in Figure 2–14b. The arithmetic gradient present worth factor, or P兾G
factor, may be expressed in two forms:
[
(1 + i)n ⫺ 1
n
1 —————
⫺ ————
(P兾G,i,n) ⫽ —
i i(1 ⫹ i)n
(1 ⫹ i)n
]
(1 ⫹ i)n ⫺ in ⫺ 1
(P兾G,i,n) ⫽ ————————
i2(1 ⫹ i)n
or
[2.25]
Remember: The conventional arithmetic gradient starts in year 2, and P is located in year 0.
Placement of
gradient PG
Equation [2.24] expressed as an engineering economy relation is
PG ⫽ G(P兾G,i,n)
[2.26]
which is the rightmost term in Equation [2.19] to calculate total present worth. The G carries a
minus sign for decreasing gradients.
The equivalent uniform annual series AG for an arithmetic gradient G is found by multiplying
the present worth in Equation [2.26] by the (A兾P,i,n) formula. In standard notation form, the
equivalent of algebraic cancellation of P can be used.
AG ⫽ G(P兾G,i,n)(A兾P,i,n)
⫽ G(A兾G,i,n)
In equation form,
[
][
(1 ⫹ i)n ⫺ 1
i(1 ⫹ i)n
G —————
n
⫺ ————
AG ⫽ —
n
n ——————
i
i(1 ⫹ i)
(1 ⫹ i)
(1 ⫹ i)n ⫺ 1
[
n
1 ⴚ ——————
AG ⴝ G —
i (1 ⴙ i)n ⴚ 1
]
]
[2.27]
which is the rightmost term in Equation [2.20]. The expression in brackets in Equation [2.27] is
called the arithmetic gradient uniform series factor and is identified by (A兾G,i,n). This factor
converts Figure 2–15a into Figure 2–15b.
The P兾G and A兾G factors and relations are summarized inside the front cover. Factor values
are tabulated in the two rightmost columns of factor values at the rear of this text.
54
Factors: How Time and Interest Affect Money
Chapter 2
AG = ?
i = given
0
1
2
3
4
n–1
n
0
1
2
3
4
n–1
n
G
2G
3G
(n – 2)G
(n – 1)G
(a)
(b)
Figure 2–15
Conversion diagram of an arithmetic gradient series to an equivalent uniform annual series.
There is no direct, single-cell spreadsheet function to calculate PG or AG for an arithmetic
gradient. Use the NPV function to display PG and the PMT function to display AG after entering
all cash flows (base and gradient amounts) into contiguous cells. General formats for these functions are
ⴝ NPV(i%, second_cell:last_cell) ⴙ first_cell
ⴝ PMT(i%, n, cell_with_PG)
[2.28]
[2.29]
The word entries in italic are cell references, not the actual numerical values. (See Appendix A,
Section A.2, for a description of cell reference formatting.) These functions are demonstrated in
Example 2.10.
An F兾G factor (arithmetic gradient future worth factor) to calculate the future worth FG of a
gradient series can be derived by multiplying the P兾G and F兾P factors. The resulting factor,
(F兾G,i,n), in brackets, and engineering economy relation is
[( ) (
) ]
(1 ⫹ i)n – 1
1 —————
FG ⫽ G —
⫺n
i
i
EXAMPLE 2.9
Neighboring parishes in Louisiana have agreed to pool road tax resources already designated for bridge refurbishment. At a recent meeting, the engineers estimated that a total of
$500,000 will be deposited at the end of next year into an account for the repair of old and
safety-questionable bridges throughout the area. Further, they estimate that the deposits
will increase by $100,000 per year for only 9 years thereafter, then cease. Determine the
equivalent (a) present worth and (b) annual series amounts, if public funds earn at a rate
of 5% per year.
Solution
(a) The cash flow diagram of this conventional arithmetic gradient series from the perspective of the parishes is shown in Figure 2–16. According to Equation [2.19], two computations must be made and added: the first for the present worth of the base amount PA
and the second for the present worth of the gradient PG. The total present worth PT
occurs in year 0. This is illustrated by the partitioned cash flow diagram in Figure 2–17.
In $1000 units, the total present worth is
PT ⫽ 500(P兾A,5%,10) ⫹ 100(P兾G,5%,10)
⫽ 500(7.7217) ⫹ 100(31.6520)
⫽ $7026.05
($7,026,050)
Arithmetic Gradient Factors (P兾G and A兾G)
2.5
0
1
2
$500
3
$600
4
$700
5
$800
6
$900
7
$1000
8
$1100
9
$1200
10
$1300
$1400
Figure 2–16
Cash flow series with a conventional arithmetic gradient (in $1000 units),
Example 2.9.
PG = ?
PA = ?
A = $500
1 2
9 10
G = $100
1 2
9 10
+
$100
Base
Gradient $900
PT = ?
PT = PA + PG
1
2
$500
$600
3
$700
4
5
$800
$900
6
$1000
7
$1100
8
$1200
9
$1300
10
$1400
Figure 2–17
Partitioned cash flow diagram (in $1000 units), Example 2.9.
(b) Here, too, it is necessary to consider the gradient and the base amount separately. The
total annual series AT is found by Equation [2.20] and occurs in years 1 through 10.
AT ⫽ 500 ⫹ 100(A兾G,5%,10) ⫽ 500 ⫹ 100(4.0991)
⫽ $909.91 per year
($909,910)
Comment
Remember: The P兾G and A兾G factors determine the present worth and annual series of the
gradient only. Any other cash flows must be considered separately.
If the present worth is already calculated [as in part (a)], PT can be multiplied by an A兾P
factor to get AT. In this case, considering round-off error,
AT ⫽ PT (A兾P,5%,10) ⫽ 7026.05(0.12950)
⫽ $909.873
($909,873)
55
56
Factors: How Time and Interest Affect Money
Chapter 2
PE
EXAMPLE 2.10 The Cement Factory Case
The announcement of the HAC cement factory states that the $200 million (M) investment is
planned for 2012. Most large investment commitments are actually spread out over several
years as the plant is constructed and production is initiated. Further investigation may determine, for example, that the $200 M is a present worth in the year 2012 of anticipated investments during the next 4 years (2013 through 2016). Assume the amount planned for 2013 is
$100 M with constant decreases of $25 M each year thereafter. As before, assume the time
value of money for investment capital is 10% per year to answer the following questions using
tabulated factors and spreadsheet functions, as requested below.
(a) In equivalent present worth values, does the planned decreasing investment series equal
the announced $200 M in 2012? Use both tabulated factors and spreadsheet functions.
(b) Given the planned investment series, what is the equivalent annual amount that will be
invested from 2013 to 2016? Use both tabulated factors and spreadsheet functions.
(c) (This optional question introduces Excel’s Goal Seek tool.) What must be the amount of
yearly constant decrease through 2016 to have a present worth of exactly $200 M in
2012, provided $100 M is expended in 2013? Use a spreadsheet.
Solution
(a) The investment series is a decreasing arithmetic gradient with a base amount of $100 M
in year 1 (2013) and G ⫽ $⫺25 M through year 4 (2016). Figure 2–18 diagrams the cash
flows with the shaded area showing the constantly declining investment each year. The PT
value at time 0 at 10% per year is determined by using tables and a spreadsheet.
Tabulated factors: Equation [2.19] with the minus sign for negative gradients determines
the total present worth PT. Money is expressed in $1 million units.
PT ⫽ PA ⫺ PG ⫽ 100(P兾A,10%,4) ⫺ 25(P兾G,10%,4)
[2.30]
⫽ 100(3.1699) – 25(4.3781)
⫽ $207.537
($207,537,000)
In present worth terms, the planned series will exceed the equivalent of $200 M in 2012
by approximately $7.5 M.
Spreadsheet: Since there is no spreadsheet function to directly display present worth for a
gradient series, enter the cash flows in a sequence of cells (rows or columns) and use the
NPV function to find present worth. Figure 2–19 shows the entries and function
NPV(i%,second_cell:last_cell). There is no first_cell entry here, because there is no
investment per se in year 0. The result displayed in cell C9, $207.534, is the total PT for
the planned series. (Note that the NPV function does not consider two separate series of
cash flows as is necessary when using tabulated factors.)
The interpretation is the same as in part (a); the planned investment series exceeds the
$200 M in present worth terms by approximately $7.5 M.
(b) Tabulated factors: There are two equally correct ways to find AT. First, apply Equation [2.20] that utilizes the A兾G factor, and second, use the PT value obtained above and
the A兾P factor. Both relations are illustrated here, in $1 million units,
Figure 2–18
PT ⫽ ?
Cash flow diagram for decreasing gradient in $1 million units, Example 2.10.
i = 10% per year
0
2013
2014
2015
2016
Year
1
2
3
4
Time
Base
A ⫽ $100
$50
$25
$100
Gradient
G ⫽ $⫺25
57
Arithmetic Gradient Factors (P兾G and A兾G)
2.5
Figure 2–19
Spreadsheet solution for
Example 2.10a and b.
Present worth of investments
⫽ NPV(10%,C5:C8)
Annual worth of investments
⫽ ⫺PMT(10%,4,C9)
Use Equation [2.20]:
AT ⫽ 100 – 25(A兾G,10%,4) ⫽ 100 ⫺ 25(1.3812)
⫽ $65.471
($65,471,000 per year)
Use PT :
AT ⫽ 207.537(A兾P,10%,4) ⫽ 207.537(0.31547)
⫽ $65.471 per year
Spreadsheet: Apply the PMT function in Equation [2.29] to obtain the same AT ⫽
$65.471 per year (Figure 2–19).
(c) (Optional) The Goal Seek tool is described in Appendix A. It is an excellent tool to apply
when one cell entry must equal a specific value and only one other cell can change. This
is the case here; the NPV function (cell C9 in Figure 2–19) must equal $200, and the gradient G (cell C1) is unknown. This is the same as stating PT ⫽ 200 in Equation [2.30] and
solving for G. All other parameters retain their current value.
Figure 2–20 (top) pictures the same spreadsheet used previously with the Goal Seek template added and loaded. When OK is clicked, the solution is displayed; G ⫽ $⫺26.721.
Refer to Figure 2–20 again. This means that if the investment is decreased by a constant
annual amount of $26.721 M, the equivalent total present worth invested over the 4 years
will be exactly $200 M.
Figure 2–20
Solution for arithmetic
gradient using Goal Seek,
Example 2.10c.
Present worth of investments:
⫽ NPV(10%,C5:C8)
Set up Goal Seek template
Solution for G = $⫺26.721 to make
present worth exactly $200
58
Factors: How Time and Interest Affect Money
Chapter 2
2.6 Geometric Gradient Series Factors
It is common for annual revenues and annual costs such as maintenance, operations, and labor to
go up or down by a constant percentage, for example, ⫹5% or ⫺3% per year. This change occurs
every year on top of a starting amount in the first year of the project. A definition and description
of new terms follow.
A geometric gradient series is a cash flow series that either increases or decreases by a constant
percentage each period. The uniform change is called the rate of change.
g ⫽ constant rate of change, in decimal form, by which cash flow values increase or decrease
from one period to the next. The gradient g can be ⫹ or ⫺.
A1 ⫽ initial cash flow in year 1 of the geometric series
P g ⫽ present worth of the entire geometric gradient series, including the initial amount
A1
Note that the initial cash flow A1 is not considered separately when working with geometric
gradients.
Figure 2–21 shows increasing and decreasing geometric gradients starting at an amount A1 in
time period 1 with present worth Pg located at time 0. The relation to determine the total present
worth Pg for the entire cash flow series may be derived by multiplying each cash flow in Figure 2–21a by the P兾F factor 1兾(1 ⫹ i)n.
A1
A1(1 ⫹ g) A1(1 ⫹ g)2 . . . A1(1 ⫹ g)n⫺1
————
Pg ⫽ ————
⫹
⫹ —————
⫹
⫹ ——————
(1 ⫹ i)n
(1 ⫹ i)1
(1 ⫹ i)2
(1 ⫹ i)3
[
(1 ⫹ g)2 . . . (1 ⫹ g)n⫺1
1⫹ g
1 ⫹ ————
————
⫽ A1 ———
⫹
⫹
⫹ —————
(1 ⫹ i)n
1 ⫹ i (1 ⫹ i)2 (1 ⫹ i)3
]
[2.31]
Multiply both sides by (1 ⫹ g)兾(1 ⫹ i), subtract Equation [2.31] from the result, factor out Pg,
and obtain
(
[
)
1⫹g
(1 ⫹ g)n
1
Pg ——— ⫺ 1 ⫽ A1 —————
⫺ ———
1⫹i
(1 ⫹ i)n⫹1 1 ⫹ i
]
Solve for Pg and simplify.
[
(
)
1⫹g n
1 ⫺ ———
1⫹i
Pg ⫽ A1 ———————
i⫺g
]
g⫽i
[2.32]
The term in brackets in Equation [2.32] is the (P兾A,g,i,n) or geometric gradient series present
worth factor for values of g not equal to the interest rate i. When g ⫽ i, substitute i for g in Equation [2.31] and observe that the term 1/(1 + i) appears n times.
Pg = ?
Pg = ?
i = given
g = given
0
1
A1
2
3
i = given
g = given
4
n
0
1
3
4
A1(1 – g)3
A1(1 – g)2
A1(1 – g)
A1(1 + g)
A1(1 + g)2
A1(1 + g)3
A1(1 + g)n – 1
(a)
2
A1
(b)
Figure 2–21
Cash flow diagram of (a) increasing and (b) decreasing geometric gradient series and present worth Pg.
n
A1(1 – g)n – 1
(
1 ⫹ ———
1 ⫹ . . . ⫹ ———
1
1 ⫹ ———
Pg ⫽ A1 ———
(1 ⫹ i) (1 ⫹ i) (1 ⫹ i)
(1 ⫹ i)
nA1
Pg ⫽ ———
(1 ⫹ i)
)
[2.33]
The (P兾A,g,i,n) factor calculates Pg in period t ⫽ 0 for a geometric gradient series starting in
period 1 in the amount A1 and increasing by a constant rate of g each period.
The equation for Pg and the (P兾A,g,i,n) factor formula are
Pg ⫽ A1(P兾A,g,i,n)
(
)
[2.34]
n
1ⴙg
1 ⴚ ———
1ⴙi
(P兾A,g,i,n) ⴝ ——————
i−g
n
———
1ⴙi
gⴝi
[2.35]
gⴝi
It is possible to derive factors for the equivalent A and F values; however, it is easier to determine
the Pg amount and then multiply by the A兾P or F兾P factor.
As with the arithmetic gradient series, there are no direct spreadsheet functions for geometric
gradient series. Once the cash flows are entered, P and A are determined using the NPV and PMT
functions, respectively.
EXAMPLE 2.11
A coal-fired power plant has upgraded an emission control valve. The modification costs only
$8000 and is expected to last 6 years with a $200 salvage value. The maintenance cost is expected to be high at $1700 the first year, increasing by 11% per year thereafter. Determine the
equivalent present worth of the modification and maintenance cost by hand and by spreadsheet
at 8% per year.
Solution by Hand
The cash flow diagram (Figure 2–22) shows the salvage value as a positive cash flow and all
costs as negative. Use Equation [2.35] for g ⫽ i to calculate Pg. Total PT is the sum of three
present worth components.
PT ⫽ ⫺8000 ⫺ Pg ⫹ 200(P兾F,8%,6)
1 ⫺ (1.11兾1.08)6
⫽ ⫺8000 ⫺ 1700 ——————— ⫹ 200(P兾F,8%,6)
0.08 ⫺ 0.11
⫽ ⫺8000 ⫺ 1700(5.9559) ⫹ 126 ⫽ $⫺17,999
[
]
Figure 2–22
PT = ?
Cash flow diagram of a geometric
gradient, Example 2.11.
Pg = ?
i = 8%
g = 11%
1
$1700
2
3
$200
5
4
6
$1700(1.11)
$1700(1.11)2
$8000
59
Geometric Gradient Series Factors
2.6
$1700(1.11)3
$1700(1.11)4
$1700(1.11)5
Placement of
Gradient Pg
60
Factors: How Time and Interest Affect Money
Chapter 2
Solution by Spreadsheet
Figure 2–23 details the spreadsheet operations to find the geometric gradient present worth Pg
and total present worth PT. To obtain PT ⫽ $⫺17,999, three components are summed—first
cost, present worth of estimated salvage in year 6, and Pg. Cell tags detail the relations for the
second and third components; the first cost occurs at time 0.
Comment
The relation that calculates the (P兾A,g,i%,n) factor is rather complex, as shown in the cell tag
and formula bar for C9. If this factor is used repeatedly, it is worthwhile using cell reference
formatting so that A1, i, g, and n values can be changed and the correct value is always obtained. Try to write the relation for cell C9 in this format.
Present worth of salvage
⫽ ⫺PV(8%,6,,200)
Present worth of maintenance costs, Eq. [2.35]
⫽ ⫺1700* ((1-((1.11)/(1.08))^6)/(0.08-0.11))
Figure 2–23
Geometric gradient and total present worth calculated via spreadsheet, Example 2.11.
PE
EXAMPLE 2.12 The Cement Factory Case
Now let’s go back to the proposed Houston American Cement plant in Georgia. The revenue
series estimate of $50 million annually is quite optimistic, especially since there are many
other cement product plants operating in Florida and Georgia on the same limestone deposit.
(The website for the HAC plant shows where they are located currently; it is clear that keen
competition will be present.) Therefore, it is important to be sensitive in our analysis to possibly declining and increasing revenue series, depending upon the longer-term success of the
plant’s marketing, quality, and reputation. Assume that revenue may start at $50 million by the
end of the first year, but then decreases geometrically by 12% per year through year 5. Determine the present worth and future worth equivalents of all revenues during this 5-year time
frame at the same rate used previously, that is, 10% per year.
Solution
The cash flow diagram appears much like Figure 2–21b, except that the arrows go up for revenues. In year 1, A1 ⫽ $50 M and revenues decrease in year 5 to
A1(1 ⫺ g)n⫺1 ⫽ 50 M(1 ⫺ 0.12)5–1 ⫽ 50 M(0.88)4 ⫽ $29.98 M
First, we determine Pg in year 0 using Eq. [2.35] with i ⫽ 0.10 and g ⫽ ⫺0.12, then we calculate
F in year 5. In $1 million units,
[
(
)
]
0.88 5
1 ⫺ ——
1.10
Pg ⫽ 50 ———————
⫽ 50[3.0560]
0.10 ⫺ (⫺0.12)
⫽ $152.80
F ⫽ 152.80(F兾P,10%,5) ⫽ 152.80(1.6105)
⫽ $246.08
This means that the decreasing revenue stream has a 5-year future equivalent worth of
$246.080 M. If you look back to Example 2.6, we determined that the F in year 5 for the
Determining i or n for Known Cash Flow Values
2.7
uniform revenue series of $50 M annually is $305.255 M. In conclusion, the 12% declining
geometric gradient has lowered the future worth of revenue by $59.175 M, which is a sizable
amount from the perspective of the owners of Votorantim Cimentos North America, Inc.
2.7 Determining i or n for Known Cash Flow Values
When all the cash flow values are known or have been estimated, the i value (interest rate or rate
of return) or n value (number of years) is often the unknown. An example for which i is sought
may be stated as follows: A company invested money to develop a new product. After the net
annual income series is known following several years on the market, determine the rate of return
i on the investment. There are several ways to find an unknown i or n value, depending upon the
nature of the cash flow series and the method chosen to find the unknown. The simplest case involves only single amounts (P and F) and solution utilizing a spreadsheet function. The most
difficult and complex involves finding i or n for irregular cash flows mixed with uniform and
gradient series utilizing solution by hand and calculator. The solution approaches are summarized below, followed by examples.
Single Amounts—P and F Only
Hand or Calculator Solution Set up the equivalence relation and (1) solve for the variable
using the factor formula, or (2) find the factor value and interpolate in the tables.
Spreadsheet Solution Use the IRR or RATE function to find i or the NPER function to find
n. (See below and Appendix A for details.)
Uniform Series—A Series
Hand or Calculator Solution Set up the equivalence relation using the appropriate factor
(P/A, A/P, F/A, or A/F), and use the second method mentioned above.
Spreadsheet Solution Use the IRR or RATE function to find i or the NPER function to find n.
Mixed A Series, Gradients, and/or Isolated Values
Hand or Calculator Solution Set up the equivalence relation and use (1) trial and error or
(2) the calculator functions.
Spreadsheet Solution Use the IRR or RATE function to find i or the NPER function to find
n. (This is the recommended approach.)
Besides the PV, FV, and NPV functions, other spreadsheet functions useful in determining i
are IRR (internal rate of return) and RATE, and NPER (number of periods) to find n. The formats
are shown here and the inside front cover with a detailed explanation in Appendix A. In all three
of these functions, at least one cash flow entry must have a sign opposite that of others in order
to find a solution.
ⴝ IRR(first_cell:last_cell)
[2.36]
To use IRR to find i, enter all cash flows into contiguous cells, including zero values.
ⴝ RATE(n,A,P,F)
[2.37]
The single-cell RATE function finds i when an A series and single P and/or F values are involved.
ⴝ NPER(i%,A,P,F)
NPER is a single-cell function to find n for single P and F values, or with an A series.
[2.38]
61
62
Factors: How Time and Interest Affect Money
Chapter 2
EXAMPLE 2.13
If Laurel made a $30,000 investment in a friend’s business and received $50,000 5 years later,
determine the rate of return.
Solution
Since only single amounts are involved, i can be determined directly from the P/F factor.
1
P ⫽ F(P兾F,i,n) ⫽ F ————
(1 ⫹ i)n
1
30,000 ⫽ 50,000 ————
(1 ⫹ i)5
1
0.600 ⫽ ————
(1 ⫹ i)5
1 0.2 ⫺ 1 ⫽ 0.1076 (10.76%)
i ⫽ ——
0.6
Alternatively, the interest rate can be found by setting up the standard P兾F relation, solving for
the factor value, and interpolating in the tables.
( )
P ⫽ F(P兾F,i,n)
30,000 ⫽ 50,000(P兾F,i,5)
(P兾F,i,5) ⫽ 0.60
From the interest tables, a P/F factor of 0.6000 for n ⫽ 5 lies between 10% and 11%. Interpolate between these two values to obtain i ⫽ 10.76%.
EXAMPLE 2.14
Pyramid Energy requires that for each of its offshore wind power generators $5000 per year
be placed into a capital reserve fund to cover unexpected major rework on field equipment.
In one case, $5000 was deposited for 15 years and covered a rework costing $100,000 in
year 15. What rate of return did this practice provide to the company? Solve by hand and
spreadsheet.
Solution by Hand
The cash flow diagram is shown in Figure 2–24. Either the A兾F or F兾A factor can be used.
Using A兾F,
A ⫽ F(A兾F,i,n)
5000 ⫽ 100,000(A兾F,i,15)
(A兾F,i,15) ⫽ 0.0500
From the A兾F interest tables for 15 years, the value 0.0500 lies between 3% and 4%. By interpolation, i ⫽ 3.98%.
F = $100,000 Figure 2–24
Diagram to determine the rate
of return, Example 2.14.
i= ?
0
1
2
3
4
5
6
7
8
A = $5000
9
10
11
12
13
14 15
Determining i or n for Known Cash Flow Values
2.7
63
Solution by Spreadsheet
Refer to the cash flow diagram (Figure 2–24) while completing the spreadsheet (Figure 2–25).
A single-cell solution using the RATE function can be applied since A ⫽ $⫺5000 occurs
each year and F ⫽ $100,000 takes place in the last year of the series. The function
⫽ RATE(15,⫺5000,,100000) displays the value i ⫽ 3.98%. This function is fast, but it allows
only limited sensitivity analysis because all the A values have to change by the same amount.
The IRR function is much better for answering “what if ” questions.
To apply the IRR function, enter the value 0 in a cell (for year 0), followed by –5000 for
14 years and in year 15 enter ⫹95,000 (Figure 2–25). In any cell enter the IRR function. The
answer i ⫽ 3.98% is displayed. It is advisable to enter the year numbers 0 through n (15 in
this example) in the column immediately to the left of the cash flow entries. The IRR function does not need these numbers, but it makes the cash flow entry activity easier and more
accurate. Now any cash flow can be changed, and a new rate will be displayed immediately
via IRR.
Figure 2–25
Use of RATE and IRR
functions to determine
i value for a uniform
series, Example 2.14.
i using RATE function
⫽ RATE(15,-5000,,100000)
i using IRR function
⫽ IRR(E2:E17))
EXAMPLE 2.15 The Cement Factory Case
From the introductory comments about the HAC plant, the annual revenue is planned to be
$50 million. All analysis thus far has taken place at 10% per year; however, the parent company has made it clear that its other international plants are able to show a 20% per year return
on the initial investment. Determine the number of years required to generate 10%, 15%, and
20% per year returns on the $200 million investment at the Georgia site.
Solution
If hand solution is utilized, the present worth relation can be established and the n values
interpolated in the tables for each of the three rate of return values. In $1 million units, the
relation is
P ⫽ ⫺200 ⫹ 50(P兾A,i%,n)
(i ⫽ 10%, 15%, 20%)
(P兾A,i%,n) ⫽ 4.00
This is a good opportunity to utilize a spreadsheet and repeated NPER functions from
Equation [2.38], since several i values are involved. Figure 2–26 shows the single-cell
⫽ NPER(i%,50,⫺200) function for each rate of return. The number of years (rounded up) to
produce at least the required returns are
Return, i%
Years
10
15
20
6
7
9
PE
64
Chapter 2
Factors: How Time and Interest Affect Money
Figure 2–26
Use of NPER function to find n values for various rate of return requirements,
Example 2.15.
CHAPTER SUMMARY
Formulas and factors derived and applied in this chapter perform equivalence calculations for
present, future, annual, and gradient cash flows. Capability in using these formulas and their
standard notation manually and with spreadsheets is critical to complete an engineering economy
study. Using these formulas and spreadsheet functions, you can convert single cash flows into
uniform cash flows, gradients into present worths, and much more. Additionally, you can solve
for rate of return i or time n.
PROBLEMS
Use of Interest Tables
2.1 Look up the numerical value for the following factors from the interest tables.
1. (P兾F,6%,8)
2. (A兾P,10%,10)
3. (A兾G,15%,20)
4. (A兾F,2%,30)
5. (P兾G,35%,15)
Determination of F, P, and A
2.2 How much can Haydon Rheosystems, Inc., afford
to spend now on an energy management system if
the software will save the company $21,300 per
year for the next 5 years? Use an interest rate of
10% per year.
to be FAA-certified by December 31, 2011. The
cost is $985,000, and a $100,000 deposit will hold
one of the first 100 “cars.” Assume a buyer pays
the $885,000 balance 3 years after making the
$100,000 deposit. At an interest rate of 10% per
year, what is the effective total cost of the PAV in
year 3?
2.5 A family that won a $100,000 prize on America’s
Funniest Home Videos decided to put one-half of
the money in a college fund for their child who
was responsible for the prize. If the fund earned
interest at 6% per year, how much was in the account 14 years after it was started?
2.3 A manufacturer of off-road vehicles is considering
the purchase of dual-axis inclinometers for installation in a new line of tractors. The distributor of the
inclinometers is temporarily overstocked and is offering them at a 40% discount from the regular cost
of $142. If the purchaser gets them now instead of
2 years from now, which is when they will be
needed, what is the present worth of the savings per
unit? The company would pay the regular price, if
purchased in 2 years. Assume the interest rate is
10% per year.
2.6 One of the biggest vulnerabilities in a control system is network devices, such as Ethernet-based
network switches that are located in unsecured
locations and accessible to everyone. DeltaX
switches, manufactured by Dahne Security, allow
the user to automatically lock and unlock the port
access to all switches in the network. The company is considering expanding its manufacturing
lines now or doing it in 3 years. If the cost now
would be $1.9 million, what equivalent amount
could the company afford to spend in 3 years? The
interest rate is 15% per year.
2.4 The Moller Skycar M400 is a flying car known
as a personal air vehicle (PAV) that is expected
2.7 A company that sells high-purity laboratory chemicals is considering investing in new equipment
Problems
that will reduce cardboard costs by better matching the size of the products to be shipped to the
size of the shipping container. If the new equipment will cost $220,000 to purchase and install,
how much must the company save each year for
3 years in order to justify the investment, if the
interest rate is 10% per year?
2.8 Red Valve Co. of Carnegie, Pennsylvania, makes a
control pinch valve that provides accurate, repeatable control of abrasive and corrosive slurries, outlasting gate, plug, ball, and even satellite coated
valves. How much can the company afford to
spend now on new equipment in lieu of spending
$75,000 four years from now? The company’s rate
of return is 12% per year.
2.9 If GHD Plastics purchases a new building now for
$1.3 million for its corporate headquarters, what
must the building be worth in 10 years? The company expects all expenditures to earn a rate of return of at least 18% per year.
2.10 CGK Rheosystems makes high-performance rotational viscometers capable of steady shear and
yield stress testing in a rugged, compact footprint.
How much could the company afford to spend now
on new equipment in lieu of spending $200,000
one year from now and $300,000 three years from
now, if the company uses an interest rate of 15%
per year?
2.11 Five years ago a consulting engineer purchased a
building for company offices constructed of bricks
that were not properly fired. As a result, some of
the bricks were deteriorated from their exposure to
rain and snow. Because of the problem with the
bricks, the selling price of the building was 25%
below the price of comparable, structurally sound
buildings. The engineer repaired the damaged
bricks and arrested further deterioration by applying an extra-strength solvent-based RTV elastomeric sealant. This resulted in restoring the building to its fair market value. If the depressed
purchase price of the building was $600,000 and
the cost of getting it repaired was $25,000, what is
the equivalent value of the “forced appreciation”
today, if the interest rate is 8% per year?
2.12 Metso Automation, which manufactures addressable quarter-turn electric actuators, is planning to
set aside $100,000 now and $150,000 one year
from now for possible replacement of the heating
and cooling systems in three of its larger manufacturing plants. If the replacement won’t be needed
for 4 years, how much will the company have in
the account, if it earns interest at a rate of 8% per
year?
65
2.13 Syringe pumps often fail because reagents adhere
to the ceramic piston and deteriorate the seal. Trident Chemical developed an integrated polymer
dynamic seal that provides a higher sealing force
on the sealing lip, resulting in extended seal life.
One of Trident’s customers expects to reduce
downtime by 30% as a result of the new seal design. If lost production would have cost the company $110,000 per year for the next 4 years, how
much could the company afford to spend now on
the new seals, if it uses an interest rate of 12% per
year?
2.14 China spends an estimated $100,000 per year on
cloud seeding efforts, which includes using antiaircraft guns and rocket launchers to fill the sky
with silver iodide. In the United States, utilities
that run hydroelectric dams are among the most
active cloud seeders, because they believe it is a
cost-effective way to increase limited water supplies by 10% or more. If the yields of cash crops
will increase by 4% each year for the next 3 years
because of extra irrigation water captured behind
dams during cloud seeding, what is the maximum
amount the farmers should spend now on the cloud
seeding activity? The value of the cash crops without the extra irrigation water would be $600,000
per year. Use an interest rate of 10% per year.
2.15 The Public Service Board (PSB) awarded two contracts worth a combined $1.07 million to improve
(i.e., deepen) a retention basin and reconstruct the
spillway that was severely damaged in a flood
2 years ago. The PSB said that, because of the weak
economy, the bids came in $950,000 lower than engineers expected. If the projects are assumed to
have a 20-year life, what is the annual worth of the
savings at an interest rate of 6% per year?
2.16 The National Highway Traffic Safety Administration raised the average fuel efficiency standard to
35.5 miles per gallon for cars and light trucks by
the year 2016. The rules will cost consumers an
average of $434 extra per vehicle in the 2012
model year. If a person purchases a new car in
2012 and keeps it for 5 years, how much must be
saved in fuel costs each year to justify the extra
cost? Use an interest rate of 8% per year.
2.17 In an effort to reduce childhood obesity by reducing the consumption of sugared beverages,
some states have imposed taxes on soda and
other soft drinks. A survey by Roland Sturm of
7300 fifth-graders revealed that if taxes averaged
4 cents on each dollar’s worth of soda, no real
difference in overall consumption was noticed.
However, if taxes were increased to 18 cents on
the dollar, Sturm calculated they would make a
66
Factors: How Time and Interest Affect Money
Chapter 2
significant difference. For a student who consumes 100 sodas per year, what is the future worth
of the extra cost from 4 cents to 18 cents per
soda? Assume the student consumes sodas from
grade 5 through graduation in grade 12. Use an
interest rate of 6% per year.
2.18 The Texas Tomorrow Fund (TTF) is a program
started in 1996 in Texas wherein parents could prepay their child's college tuition when the child was
young. Actuaries set the price based on costs and
investment earnings at that time. Later, the Texas
legislature allowed universities to set their own tuition rates; tuition costs jumped dramatically. The
cost for entering a newborn in 1996 was $10,500.
If the TTF fund grew at a rate of 4% per year,
while tuition costs increased at 7% per year, determine the state’s shortfall when a newborn enters
college 18 years later.
2.19 Henry Mueller Supply Co. sells tamperproof,
normally open thermostats (i.e., thermostat closes
as temperature rises). Annual cash flows are
shown in the table below. Determine the future
worth of the net cash flows at an interest rate of
10% per year.
Year
Income, $1000
Cost, $1000
1
2
3
4
5
6
7
8
200 200 200 200 200 200 200 200
90 90 90 90 90 90 90 90
2.20 A company that makes self-clinching fasteners expects to purchase new production-line equipment
in 3 years. If the new units will cost $350,000, how
much should the company set aside each year, if
the account earns 10% per year?
calculated values, assuming the formula-calculated
value is the correct one.
Arithmetic Gradient
2.25 Profits from recycling paper, cardboard, aluminum, and glass at a liberal arts college have increased at a constant rate of $1100 in each of the
last 3 years. If this year’s profit (end of year 1) is
expected to be $6000 and the profit trend continues through year 5, (a) what will the profit be at
the end of year 5 and (b) what is the present
worth of the profit at an interest rate of 8% per
year?
2.26 A report by the Government Accountability Office (GAO) shows that the GAO expects the
U.S. Postal Service to lose a record $7 billion at
the end of this year, and if the business model is
not changed, the losses will total $241 billion by
the end of year 10. If the losses increase uniformly over the 10-year period, determine the
following:
(a) The expected increase in losses each year
(b) The loss 5 years from now
(c) The equivalent uniform worth of the losses at
an interest rate of 8% per year
2.27 Rolled ball screws are suitable for high-precision
applications such as water jet cutting. Their total
manufacturing cost is expected to decrease because of increased productivity, as shown in the
table. Determine the equivalent annual cost at an
interest rate of 8% per year.
Year
Cost, $1000
1
2
3
4
5
6
7
8
200 195 190 185 180 175 170 165
Factor Values
2.21 Find the numerical value of the following factors
using (a) interpolation and (b) the formula.
1. (A兾P,13%,15)
2. (P兾G,27%,10)
2.22 Find the numerical value of the following factors
using (a) interpolation, (b) the formula, and (c) a
spreadsheet function.
1. (F兾P,14%,62)
2. (A兾F,1%,45)
2.23 For the factor (F兾P,10%,43), find the percent difference between the interpolated and formulacalculated values, assuming the formula-calculated
value is the correct one.
2.24 For the factor (F兾A,15%,52), find the percent difference between the interpolated and formula-
2.28 Western Hydra Systems makes a panel milling machine with a 2.7-m-diameter milling head that
emits low vibration and processes stress-relieved
aluminum panels measuring up to 6000 mm long.
The company wants to borrow money for a new
production/warehouse facility. If the company offers to repay the loan with $60,000 in year 1 and
amounts increasing by $10,000 each year through
year 5, how much can the company borrow at an
interest rate of 10% per year?
2.29 GKX Industries expects sales of its hydraulic seals
(in inch and metric sizes) to increase according to
the cash flow sequence $70 ⫹ 4k, where k is in
years and cash flow is in $1000.
(a) What is the amount of the cash flow in year 3?
(b) What is the future worth of the entire cash
flow series in year 10? Let i ⫽10% per year.
67
Problems
2.30 For the cash flows below, determine the amount in year 1, if the annual worth in years 1 through 9 is $601.17 and
the interest rate is 10% per year.
Year
1
2
3
4
5
6
7
8
9
Cost, $1000
A
A ⫹ 30
A ⫹ 60
A ⫹ 90
A ⫹ 120
A ⫹ 150
A ⫹ 180
A ⫹ 210
A ⫹ 240
2.31 Apple Computer wants to have $2.1 billion
available 5 years from now to finance production of a handheld “electronic brain” that, based
on your behavior, will learn how to control
nearly all the electronic devices in your home,
such as the thermostat, coffee pot, TV, and
sprinkler system. The company expects to set
aside uniformly increasing amounts of money
each year to meet its goal. If the amount set
aside at the end of year 1 is $50 million, how
much will the constant increase G have to be
each year? Assume the investment account
grows at a rate of 18% per year.
2.32 Tacozza Electric, which manufactures brush dc
servomotors, budgeted $75,000 per year to pay for
certain components over the next 5 years. If the
company expects to spend $15,000 in year 1, how
much of a uniform (arithmetic) increase each year
is the company expecting in the cost of this part?
Assume the company uses an interest rate of 10%
per year.
Geometric Gradient
2.33 There are no tables in the back of your book for the
geometric gradient series factors. Calculate the first
two annual worth factor values, that is, A values for
n ⫽ 1 and 2, that would be in a 10% interest table
for a growth rate of 4% per year.
2.34 Determine the present worth of a geometric gradient series with a cash flow of $50,000 in year 1 and
increases of 6% each year through year 8. The interest rate is 10% per year.
2.35 Determine the difference in the present worth values of the following two commodity contracts at
an interest rate of 8% per year.
Contract 1 has a cost of $10,000 in year 1; costs
will escalate at a rate of 4% per year for 10 years.
Contract 2 has the same cost in year 1, but costs
will escalate at 6% per year for 11 years.
2.36 El Paso Water Utilities (EPWU) purchases surface
water for treatment and distribution to EPWU customers from El Paso County Water Improvement
District during the irrigation season. A new contract
between the two entities resulted in a reduction in
future price increases in the cost of the water from
8% per year to 4% per year for the next 20 years. If
the cost of water next year (which is year 1 of the
new contract) will be $260 per acre-foot, what is
the present worth of the savings (in $/acre-ft) to the
utility between the old and the new contracts? Let
the interest rate equal 6% per year.
2.37 Determine the present worth of a maintenance
contract that has a cost of $30,000 in year 1 and
annual increases of 6% per year for 10 years. Use
an interest rate of 6% per year.
Interest Rate and Rate of Return
2.38 Gesky Industrial Products manufactures brushless
blowers for boilers, food service equipment, kilns,
and fuel cells. The company borrowed $18,000,000
for a plant expansion and repaid the loan in seven
annual payments of $3,576,420, with the first payment made 1 year after the company received the
money. What was the interest rate on the loan? Use
hand and spreadsheet solutions.
2.39 If the value of Jane’s retirement portfolio increased from $170,000 to $813,000 over a 15-year
period, with no deposits made to the account over
that period, what annual rate of return did she
make?
2.40 A person’s credit score is important in determining
the interest rate on a home mortgage. According to
Consumer Credit Counseling Service, a homeowner with a $100,000 mortgage and a 520 credit
score will pay $110,325 more in interest charges
over the life of a 30-year loan than a homeowner
with the same mortgage and a credit score of 720.
How much higher would the interest rate per year
have to be in order to account for this much difference in interest charges, if the $100,000 loan is
repaid in a single lump-sum payment at the end of
30 years?
2.41 During a period when the real estate market in
Phoenix, Arizona, was undergoing a significant
downturn, CSM Consulting Engineers made an
agreement with a distressed seller to purchase an
office building under the following terms: total
price of $1.2 million with a down payment of
$200,000 now and no payments for 4 years, after
which the remaining balance of $1 million would
68
Chapter 2
be paid. CSM was able to make this deal because
of poor market conditions at the time of purchase,
and, at the same time, planning to sell the building
in 4 years (when market conditions would probably be better) and move to a larger office building
in Scottsdale, Arizona. If CSM was able to sell the
building in exactly 4 years for $1.9 million, what
rate of return per year did the company make on
the investment?
2.42 A start-up company that makes hydraulic seals
borrowed $800,000 to expand its packaging and
shipping facility. The contract required the company to repay the investors through an innovative mechanism called faux dividends, a series of
uniform annual payments over a fixed period of
time. If the company paid $250,000 per year for
5 years, what was the interest rate on the loan?
2.43 Bessimer Electronics manufactures addressable
actuators in one of its Maquiladora plants in
Mexico. The company believes that by investing
$24,000 each year in years 1, 2, and 3, it will avoid
spending $87,360 in year 3. If the company does
make the annual investments, what rate of return
will it realize?
2.44 UV curable epoxy resins are used in sealing, in gap
filling, and as a clear coating. Your boss saw a report submitted by the chief financial officer (CFO)
that said the equivalent annual worth of maintaining the equipment used in producing the resins was
$48,436 over the last 5 years. The report showed
that the cost in year 1 was $42,000, and it increased
arithmetically by $4000 each year. Your boss
thought $48,436 was too high, so she asked you to
determine what interest rate the CFO used in making the calculations. What was the interest rate?
Number of Years
2.45 Acme Bricks, a masonry products company, wants
to have $600,000 on hand before it invests in new
conveyors, trucks, and other equipment. If the
company sets aside $80,000 per year in an account
that increases in value at a rate of 15% per year,
how many years will it be before Acme can purchase the equipment?
2.46 An engineer who was contemplating retirement
had $1.6 million in his investment portfolio. However, a severe recession caused his portfolio to decrease to only 55% of the original amount, so he
kept working. If he was able to invest his money at
a rate of return of 9% per year after the recession
ended, how many years did it take for his account
to get back to the $1.6 million value?
Factors: How Time and Interest Affect Money
2.47 You own a small engineering consulting company.
If you invest $200,000 of the company’s money in
a natural gas well that is expected to provide income of $29,000 per year, how long must the well
produce at that rate in order to get the money back
plus a rate of return of 10% per year?
2.48 A perceptive engineer started saving for her retirement 15 years ago by diligently saving $18,000
each year through the present time. She invested in
a stock fund that averaged a 12% rate of return over
that period. If she makes the same annual investment and gets the same rate of return in the future,
how long will it be from now (time zero) before she
has $1,500,000 in her retirement fund?
2.49 A mechanical engineering graduate who wanted
to have his own business borrowed $350,000
from his father as start-up money. Because he
was family, his father charged interest at only 4%
per year. If the engineer was able to pay his father
$15,000 in year 1, $36,700 in year 2, and amounts
increasing by $21,700 each year, how many
years did it take for the engineer to repay the
loan?
2.50 The energy costs of a company involved in powder
coating of outdoor furniture are expected to increase at a rate of $400 per year. The cost at the
end of the next year (year 1) is expected to be
$13,000. How many years will it be from now before the equivalent annual cost is $16,000 per year,
if interest is 8% per year?
2.51 In cleaning out some files that were left behind by
the engineer who preceded you in your current job,
you found an old report that had a calculation for
the present worth of certain maintenance costs for
state highways. The report contained the following
equation (with cost in $1 million):
12{1 ⫺ [(1 ⫹ 0.03)兾(1 ⫹ 0.06)]x}兾(0.06 ⫺ 0.03)
⫽ 140
The value of x that was used in the calculation was
illegible. What is its value?
2.52 The equivalent annual worth of an increasing
arithmetic gradient is $135,300. If the cash flow in
year 1 is $35,000 and the gradient amount is
$19,000, what is the value of n at an interest rate of
10% per year?
2.53 You are told that the present worth of an increasing
geometric gradient is $88,146. If the cash flow in
year 1 is $25,000 and the gradient increase is
18% per year, what is the value of n? The interest
rate is 10% per year.
Additional Problems and FE Exam Review Questions
69
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
2.54 The amount of money that Diamond Systems can
spend now for improving productivity in lieu of
spending $30,000 three years from now at an interest rate of 12% per year is closest to:
(a) $15,700
(b) $17,800
(c) $19,300
(d ) $21,350
2.55 A manufacturing company spent $30,000 on a new
conveyor belt. If the conveyor belt resulted in cost
savings of $4200 per year, the length of time it
would take for the company to recover its investment at 8% per year is closest to:
(a) Less than 9 years
(b) 9 to 10 years
(c) 11 to 12 years
(d ) Over 12 years
2.56 Levi Strauss has some of its jeans stone-washed
under a contract with independent U.S. Garment
Corp. If U.S. Garment’s operating cost per machine is $22,000 for year 1 and increases by a
constant $1000 per year through year 5, what is
the equivalent uniform annual cost per machine
for the 5 years at an interest rate of 8% per year?
(a) $23,850
(b) $24,650
(c) $25,930
(d ) Over $26,000
2.57 The F兾G factor values can be derived by multiplying:
(a) (P兾F) and (A兾G) factor values
(b) (F兾P) and (A兾G) factor values
(c) (P兾F) and (P兾G) factor values
(d ) (F兾P) and (P兾G) factor values
2.58 At i ⫽ 4% per year, A for years 1 through 6 of the
cash flows shown below is closest to:
(a) $300
(b) $560
(c) $800
(d ) $1040
0
1
2
3
4
5
6
$300
$400
Years
2.59 The value of the factor (P兾F,i,10) can be found by
getting the factor values for (P兾F,i,4) and (P兾F,i,6)
and:
(a) Adding the values for (P兾F,i,4) and (P兾F,i,6)
(b) Multiplying the values for (P兾F,i,4) and
(P兾F,i,6)
(c) Dividing the value for (P兾F,i,6) by the value
for (P兾F,i,4)
(d ) None of the above
2.60 A small construction company is considering the
purchase of a used bulldozer for $61,000. If the
company purchases the dozer now, the equivalent future amount in year 4 that the company is
paying for the dozer at 4% per year interest is
closest to:
(a) $52,143
(b) $65,461
(c) $71,365
(d ) Over $72,000
2.61 The cost of lighting and maintaining the tallest
smokestack in the United States (at a shuttered
ASARCO refinery) is $90,000 per year. At an
interest rate of 10% per year, the present worth
of maintaining the smokestack for 10 years is
closest to:
(a) $1,015,000
(b) $894,000
(c) $712,000
(d ) $553,000
2.62 An enthusiastic new engineering graduate plans to
start a consulting firm by borrowing $100,000 at
10% per year interest. The loan payment each year
to pay off the loan in 7 years is closest to:
(a) $18,745
(b) $20,540
(c) $22,960
(d ) $23,450
2.63 An engineer who believed in “save now and play
later” wanted to retire in 20 years with $1.5 million.
At 10% per year interest, to reach the $1.5 million
goal, starting 1 year from now, the engineer must
annually invest:
(a) $26,190
(b) $28,190
(c) $49,350
(d ) $89,680
$500
$600
$700
$800
2.64 The cost of a border fence is $3 million per
mile. If the life of such a fence is assumed to
be 10 years, the equivalent annual cost of a
70
Chapter 2
10-mile-long fence at an interest rate of 10%
per year is closest to:
(a) $3.6 million
(b) $4.2 million
(c) $4.9 million
(d ) Over $5.0 million
2.65 An investment of $75,000 in equipment that
will reduce the time for machining self-locking
fasteners will save $20,000 per year. At an interest rate of 10% per year, the number of years
required to recover the initial investment is
closest to:
(a) 6 years
(b) 5 years
(c) 4 years
(d) 3 years
2.66 The number of years required for an account to accumulate $650,000 if Ralph deposits $50,000 each
year and the account earns interest at a rate of 6%
per year is closest to:
(a) 13 years
(b) 12 years
(c) 11 years
(d) 10 years
2.67 Aero Serve, Inc., manufactures cleaning nozzles
for reverse-pulse jet dust collectors. The company
spent $40,000 on a production control system that
will increase profits by $13,400 per year for
5 years. The rate of return per year on the investment is closest to:
(a) 20%
(b) 18%
(c) 16%
(d) Less than 15%
Factors: How Time and Interest Affect Money
2.68 Energy costs for a green chemical treatment have
been increasing uniformly for 5 years. If the cost in
year 1 was $26,000 and it increased by $2000 per
year through year 5, the present worth of the costs
at an interest rate of 10% per year is closest to:
(a) $102,900
(b) $112,300
(c) $122,100
(d) $195,800
2.69 In planning for your retirement, you expect to
save $5000 in year 1, $6000 in year 2, and amounts
increasing by $1000 each year through year 20. If
your investments earn 10% per year, the amount
you will have at the end of year 20 is closest to:
(a) $242,568
(b) $355,407
(c) $597,975
(d) $659,125
2.70 Income from a precious metals mining operation
has been decreasing uniformly for 5 years. If income in year 1 was $300,000 and it decreased by
$30,000 per year through year 4, the annual worth
of the income at 10% per year is closest to:
(a) $310,500
(b) $258,600
(c) $203,900
(d) $164,800
2.71 If you are able to save $5000 in year 1, $5150 in
year 2, and amounts increasing by 3% each year
through year 20, the amount you will have at the
end of year 20 at 10% per year interest is closest to:
(a) $60,810
(b) $102,250
(c) $351,500
(d) Over $410,000
CASE STUDY
TIME MARCHES ON; SO DOES THE INTEREST RATE
Background
Information
During the last week, Sundara has read about different situations that involve money, interest rate, and different amounts
of time. She has gotten interested in the major effects that
time and interest rates have on the amount of money necessary to do things and the significant growth in the amount of
money when a large number of years are considered. In all
cases, the interest focuses on the amount of money at the end
of the time period.
The four situations are described here.
A. Manhattan Island was purchased in 1626 for $24. After
385 years in 2011, at 6% per year compounded interest,
the current value must be very large.
B. At the age of 22, if she saved only $2000 per year for
the next 10 years (starting next year) and made a return
of 6% per year, by today’s standards, she would have
accumulated a nice sum at the age of 70.
Case Study
C. A corporation invested $2 million in developing and
marketing a new product in 1945 (just after World
War II, this was a lot of money) and has made a steady
net cash flow of $300,000 per year for some 65 years.
Sundara estimated the annual rate of return must be
quite good, especially given that she is lucky to earn 4%
per year on her own investments these days.
D. A friend who is not good with money, went to a pawn
shop and borrowed $200 for one week and paid $30 in
interest. Sundara thought this might be a pretty good
deal, in case she ever ran low on cash. However, she did
not know whether the interest was simple or compounded monthly, and how much may be owed were
this loan not paid off for 1 year.
Case Study Exercises
1. What is the annual interest rate for each situation? Include both the annual simple and the compound rates
for situation D.
2. Calculate and observe the total amount of money involved in each situation at the end of the time periods
compared to the starting amount. Is the ending amount
larger or smaller than you would expect it to be prior to
making any computations?
3. Think of a situation for yourself that may be similar to
any of those above. Determine the interest rate, the
time period, and the starting and ending amounts of
money.
71
CHAPTER 3
Combining
Factors and
Spreadsheet
Functions
L E A R N I N G
O U T C O M E S
Purpose: Use multiple factors and spreadsheet functions to find equivalent amounts for cash flows that have nonstandard placement.
SECTION
TOPIC
LEARNING OUTCOME
3.1
Shifted series
• Determine the P, F or A values of a series
starting at a time other than period 1.
3.2
Shifted series and single cash
flows
• Determine the P, F, or A values of a shifted series
and randomly placed single cash flows.
3.3
Shifted gradients
• Make equivalence calculations for shifted
arithmetic or geometric gradient series that
increase or decrease in size of cash flows.
M
ost estimated cash flow series do not fit exactly the series for which the factors, equations, and spreadsheet functions in Chapter 2 were developed. For
a given sequence of cash flows, there are usually several correct ways to determine the equivalent present worth P, future worth F, or annual worth A. This chapter
explains how to combine engineering economy factors and spreadsheet functions to
address more complex situations involving shifted uniform series, gradient series, and
single cash flows.
3.1 Calculations for Uniform Series That Are Shifted
When a uniform series begins at a time other than at the end of period 1, it is called a shifted
series. In this case several methods can be used to find the equivalent present worth P. For
example, P of the uniform series shown in Figure 3–1 could be determined by any of the
following methods:
• Use the P兾F factor to find the present worth of each disbursement at year 0 and add them.
• Use the F兾P factor to find the future worth of each disbursement in year 13, add them, and
then find the present worth of the total, using P ⫽ F(P兾F,i,13).
• Use the F兾A factor to find the future amount F ⫽ A(F兾A,i,10), and then compute the present
worth, using P ⫽ F(P兾F,i,13).
• Use the P兾A factor to compute the “present worth” P3 ⫽ A(P兾A,i,10) (which will be located
in year 3, not year 0), and then find the present worth in year 0 by using the (P兾F,i,3) factor.
Typically the last method is used for calculating the present worth of a uniform series that does
not begin at the end of period 1. For Figure 3–1, the “present worth” obtained using the P兾A factor is located in year 3. This is shown as P3 in Figure 3–2. Note that a P value is always located
1 year or period prior to the beginning of the first series amount. Why? Because the P兾A factor
was derived with P in time period 0 and A beginning at the end of period 1. The most common
mistake made in working problems of this type is improper placement of P. Therefore, it is extremely important to remember:
The present worth is always located one period prior to the first uniform series amount when
using the P兾A factor.
Placement of P
To determine a future worth or F value, recall that the F兾A factor derived in Section 2.3 had
the F located in the same period as the last uniform series amount. Figure 3–3 shows the location
of the future worth when F兾A is used for Figure 3–1 cash flows.
The future worth is always located in the same period as the last uniform series amount when
using the F兾A factor.
Placement of F
It is also important to remember that the number of periods n in the P兾A or F兾A factor is equal
to the number of uniform series values. It may be helpful to renumber the cash flow diagram to
avoid errors in counting. Figures 3–2 and 3–3 show Figure 3–1 renumbered to determine n ⫽ 10.
0
1
2
3
4
5
6
7
8
9
10
11
12
13
Year
Figure 3–1
A uniform series that is
shifted.
A = $50
Figure 3–2
P3 = ?
0
1
2
3
4
1
5
2
6
3
7
4
8
5
A = $50
9
6
10
11
12
7
8
9
13
10
Year
n
Location of present worth
and renumbering for n for
the shifted uniform series
in Figure 3–1.
74
Combining Factors and Spreadsheet Functions
Chapter 3
Figure 3–3
Placement of F and
renumbering for n for the
shifted uniform series of
Figure 3–1.
F=?
0
1
2
3
4
1
5
2
6
7
3
4
8
5
9
6
10
11
12 13
Year
7
8
9
n
10
A = $50
As stated above, several methods can be used to solve problems containing a uniform series
that is shifted. However, it is generally more convenient to use the uniform series factors than the
single-amount factors. Specific steps should be followed to avoid errors:
1.
2.
3.
4.
5.
Draw a diagram of the positive and negative cash flows.
Locate the present worth or future worth of each series on the cash flow diagram.
Determine n for each series by renumbering the cash flow diagram.
Draw another cash flow diagram representing the desired equivalent cash flow.
Set up and solve the equations.
These steps are illustrated below.
EXAMPLE 3.1
The offshore design group at Bechtel just purchased upgraded CAD software for $5000 now
and annual payments of $500 per year for 6 years starting 3 years from now for annual upgrades. What is the present worth in year 0 of the payments if the interest rate is 8% per year?
Solution
The cash flow diagram is shown in Figure 3–4. The symbol PA is used throughout this chapter
to represent the present worth of a uniform annual series A, and PA' represents the present worth
at a time other than period 0. Similarly, PT represents the total present worth at time 0. The
correct placement of PA' and the diagram renumbering to obtain n are also indicated. Note that
PA' is located in actual year 2, not year 3. Also, n ⫽ 6, not 8, for the P兾A factor. First find the
value of P'A of the shifted series.
PA' ⫽ $500(P兾A,8%,6)
Since PA' is located in year 2, now find PA in year 0.
PA ⫽ PA' (P兾F,8%,2)
PT = ?
i = 8% per year
PA⬘ = ?
PA = ?
1
2
3
0
1
4
2
5
3
6
4
7
5
8 Year
6
n
A = $500
P0 = $5000
Figure 3–4
Cash flow diagram with placement of P values, Example 3.1.
The total present worth is determined by adding PA and the initial payment P0 in year 0.
PT ⫽ P0 ⫹ PA
⫽ 5000 ⫹ 500(P兾A,8%,6)(P兾F,8%,2)
⫽ 5000 ⫹ 500(4.6229)(0.8573)
⫽ $6981.60
Calculations for Uniform Series That Are Shifted
3.1
The more complex that cash flow series become, the more useful are the spreadsheet functions. When the uniform series A is shifted, the NPV function is used to determine P, and the
PMT function finds the equivalent A value. The NPV function, like the PV function, determines
the P values, but NPV can handle any combination of cash flows directly from the cells. As we
learned in Chapter 2, enter the net cash flows in contiguous cells (column or row), making sure
to enter “0” for all zero cash flows. Use the format
NPV(i%, second_cell:last_cell) ⴙ first_cell
First_cell contains the cash flow for year 0 and must be listed separately for NPV to correctly
account for the time value of money. The cash flow in year 0 may be 0.
The easiest way to find an equivalent A over n years for a shifted series is with the PMT function, where the P value is from the NPV function above. The format is the same as we learned
earlier; the entry for P is a cell reference, not a number.
PMT(i%, n, cell_with_P,F)
Alternatively, the same technique can be used when an F value was obtained using the FV function. Now the last entry in PMT is “cell_with_F.”
It is very fortunate that any parameter in a spreadsheet function can itself be a function. Thus,
it is possible to write the PMT function in a single cell by embedding the NPV function (and FV
function, if needed). The format is
PMT(i%, n, NPV(i%,second_cell:last_cell) ⴙ first_cell,F)
[3.1]
Of course, the answer for A is the same for the two-cell operation or a single-cell, embedded
function. All three of these functions are illustrated in Example 3.2.
EXAMPLE 3.2
Recalibration of sensitive measuring devices costs $8000 per year. If the machine will be recalibrated for each of 6 years starting 3 years after purchase, calculate the 8-year equivalent
uniform series at 16% per year. Show hand and spreadsheet solutions.
Solution by Hand
Figure 3–5a and b shows the original cash flows and the desired equivalent diagram. To convert the $8000 shifted series to an equivalent uniform series over all periods, first convert the
uniform series into a present worth or future worth amount. Then either the A兾P factor or the
A兾F factor can be used. Both methods are illustrated here.
Present worth method. (Refer to Figure 3–5a.) Calculate PA' for the shifted series in year 2,
followed by PT in year 0. There are 6 years in the A series.
P'A ⫽ 8000(P兾A,16%,6)
PT ⫽ PA' (P兾F,16%,2) ⫽ 8000(P兾A,16%,6)(P兾F,16%,2)
⫽ 8000(3.6847)(0.7432) ⫽ $21,907.75
The equivalent series A' for 8 years can now be determined via the A兾P factor.
A' ⫽ PT (A兾P,16%,8) ⫽ $5043.60
Future worth method. (Refer to Figure 3–5a.) First calculate the future worth F in year 8.
F ⫽ 8000(F兾A,16%,6) ⫽ $71,820
The A兾F factor is now used to obtain A' over all 8 years.
A' ⫽ F(A兾F,16%,8) ⫽ $5043.20
75
76
Combining Factors and Spreadsheet Functions
Chapter 3
PT = ?
0
F=?
PA⬘ = ?
1 2
3
4
5
6
7 8
i = 16% per year
0
1
2
3
4
A = $8000
A⬘ = ?
(a)
(b)
5
6
7
8
⫽ ⫺PMT(16%,8,B12)
⫽ ⫺PMT(16%,8,NPV(16%,B4:B11) ⫹ B3
⫽ NPV(16%,B4:B11) ⫹ B3
(c)
Figure 3–5
(a) Original and (b) equivalent cash flow diagrams; and (c) spreadsheet functions to determine
P and A, Example 3.2.
Solution by Spreadsheet
(Refer to Figure 3–5c.) Enter the cash flows in B3 through B11 with entries of “0” in the first
three cells. Use the NPV function to display P ⫽ $21,906.87.
There are two ways to obtain the equivalent A over 8 years. Of course, only one of these
PMT functions needs to be entered. (1) Enter the PMT function making direct reference to
the P value (see cell tag for D兾E5), or (2) use Equation [3.1] to embed the NPV function into
the PMT function (see cell tag for D兾E8).
3.2 Calculations Involving Uniform Series and Randomly
Placed Single Amounts
When a cash flow includes both a uniform series and randomly placed single amounts, the procedures of Section 3.1 are applied to the uniform series and the single-amount formulas are applied to the one-time cash flows. This approach, illustrated in Examples 3.3 and 3.4, is merely a
combination of previous ones. For spreadsheet solutions, it is necessary to enter the net cash
flows before using the NPV and other functions.
EXAMPLE 3.3
An engineering company in Wyoming that owns 50 hectares of valuable land has decided to
lease the mineral rights to a mining company. The primary objective is to obtain long-term
income to finance ongoing projects 6 and 16 years from the present time. The engineering
company makes a proposal to the mining company that it pay $20,000 per year for 20 years
beginning 1 year from now, plus $10,000 six years from now and $15,000 sixteen years from
now. If the mining company wants to pay off its lease immediately, how much should it pay
now if the investment is to make 16% per year?
Calculations Involving Uniform Series and Randomly Placed Single Amounts
3.2
Solution
The cash flow diagram is shown in Figure 3–6 from the owner’s perspective. Find the present
worth of the 20-year uniform series and add it to the present worth of the two one-time amounts
to determine PT'.
PT ⫽ 20,000(P兾A,16%,20) ⫹ 10,000(P兾F,16%,6) ⫹ 15,000(P兾F,16%,16)
⫽ $124,075
Note that the $20,000 uniform series starts at the end of year 1, so the P兾A factor determines the
present worth at year 0.
$15,000
$10,000
A = $20,000
0
1
2
3
PT = ?
4
5
6
7
15
16
17
18
19
20
Year
i = 16% per year
Figure 3–6
Diagram including a uniform series and single amounts, Example 3.3.
When you calculate the A value for a cash flow series that includes randomly placed single
amounts and uniform series, first convert everything to a present worth or a future worth.
Then you obtain the A value by multiplying P or F by the appropriate A兾P or A兾F factor.
Example 3.4 illustrates this procedure.
EXAMPLE 3.4
A design-build-operate engineering company in Texas that owns a sizable amount of land
plans to lease the drilling rights (oil and gas only) to a mining and exploration company. The
contract calls for the mining company to pay $20,000 per year for 20 years beginning 3 years
from now (i.e., beginning at the end of year 3 and continuing through year 22) plus $10,000 six
years from now and $15,000 sixteen years from now. Utilize engineering economy relations by
hand and by spreadsheet to determine the five equivalent values listed below at 16% per year.
1.
2.
3.
4.
5.
Total present worth PT in year 0
Future worth F in year 22
Annual series over all 22 years
Annual series over the first 10 years
Annual series over the last 12 years
Solution by Hand
Figure 3–7 presents the cash flows with equivalent P and F values indicated in the correct years
for the P兾A, P兾F, and F兾A factors.
1. PT in year 0: First determine P'A of the series in year 2. Then PT is the sum of three
P values: the series present worth value moved back to t ⫽ 0 with the P兾F factor, and the
two P values at t ⫽ 0 for the two single amounts in years 6 and 16.
77
78
Combining Factors and Spreadsheet Functions
Chapter 3
P'A ⫽ 20,000(P兾A,16%,20)
PT ⫽ P'A(P兾F,16%,2) ⫹ 10,000(P兾F,16%,6) ⫹ 15,000(P兾F,16%,16)
⫽ 20,000(P兾A,16%,20)(P兾F,16%,2) ⫹ 10,000(P兾F,16%,6)
⫹ 15,000(P兾F,16%,16)
⫽ $93,625
[3.2]
2. F in year 22: To determine F in year 22 from the original cash flows (Figure 3–7), find
F for the 20-year series and add the two F values for the two single amounts. Be sure to
carefully determine the n values for the single amounts: n ⫽ 22 ⫺ 6 ⫽ 16 for the $10,000
amount and n ⫽ 22 ⫺ 16 ⫽ 6 for the $15,000 amount.
F ⫽ 20,000(F兾A,16%,20) ⫹ 10,000(F兾P,16%,16) ⫹ 15,000(F兾P,16%,6)
[3.3]
⫽ $2,451,626
3. A over 22 years: Multiply PT ⫽ $93,625 from (1) above by the A兾P factor to determine
an equivalent 22-year A series, referred to as A1–22 here.
A1–22 ⫽ PT (A兾P,16%,22) ⫽ 93,625(0.16635) ⫽ $15,575
[3.4]
An alternate way to determine the 22-year series uses the F value from (2) above. In this
case, the computation is A1–22 ⫽ F(A兾F,16%,22) ⫽ $15,575.
$15,000
$10,000
$20,000
0
0
PT = ?
1
1 2
3
2
3
4
4
5
6
PA⬘ = ?
5
13
7
15
20
16
i = 16% per year
17
18
19
20
21 22
n
Year
F= ?
Figure 3–7
Diagram for Example 3.4.
4. A over years 1 to 10: This and (5), which follows, are special cases that often occur in
engineering economy studies. The equivalent A series is calculated for a number of years
different from that covered by the original cash flows. This occurs when a defined study
period or planning horizon is preset for the analysis. (More is mentioned about study periods later.) To determine the equivalent A series for years 1 through 10 only (call it A1–10),
the PT value must be used with the A兾P factor for n ⫽ 10. This computation transforms
Figure 3–7 into the equivalent series A1–10 in Figure 3–8a.
A1–10 ⫽ PT (A兾P,16%,10) ⫽ 93,625(0.20690) ⫽ $19,371
[3.5]
5. A over years 11 to 22: For the equivalent 12-year series for years 11 through 22 (call it
A11–22), the F value must be used with the A兾F factor for 12 years. This transforms Figure 3–7 into the 12-year series A11–22 in Figure 3–8b.
A11–22 ⫽ F(A兾F,16%,12) ⫽ 2,451,626(0.03241) ⫽ $79,457
[3.6]
Notice the huge difference of more than $60,000 in equivalent annual amounts that occurs
when the present worth of $93,625 is allowed to compound at 16% per year for the first
10 years. This is another demonstration of the time value of money.
Solution by Spreadsheet
Figure 3–9 is a spreadsheet image with answers for all five questions. The $20,000 series and
the two single amounts have been entered into separate columns, B and C. The zero cash flow
3.2
79
Calculations Involving Uniform Series and Randomly Placed Single Amounts
Figure 3–8
A1–10 = ?
1
2
3
4
5
6
7
8
9
10
11
21
22 Year
20
21 22
Cash flows of Figure 3–7
converted to equivalent
uniform series for
(a) years 1 to 10 and
(b) years 11 to 22.
i = 16%
PT = $93,625
(a)
A11–22 = ?
0
1
9
10
11
12
13
14
15
16
17
18
19
Year
i = 16%
F = $2,451,626
(b)
Figure 3–9
Spreadsheet using cell
reference format.
Example 3.4.
values are all entered so that the functions will work correctly. This is an excellent example
demonstrating the versatility of the NPV, FV, and PMT functions. To prepare for sensitivity
analysis, the functions are developed using cell reference format or global variables, as indicated in the column E function. This means that virtually any number—the interest rate, any
cash flow estimate in the series or the single amounts, and the timing within the 22-year time
frame—can be changed and the new answers will be immediately displayed.
1. Present worth values for the series and single amounts are determined in cells F6 and F10,
respectively, using the NPV function. The sum of these in F14 is PT ⫽ $93,622, which
corresponds to the value in Equation [3.2]. Alternatively, PT can be determined directly
via the sum of two NPV functions, shown in row 15.
2. The FV function in row 18 uses the P value in F14 (preceded by a minus sign) to determine F twenty-two years later. This is significantly easier than Equation [3.3].
3. To find the 22-year A series starting in year 1, the PMT function in row 21 references
the P value in cell F14. This is effectively the same procedure used in Equation [3.4] to
obtain A1–22.
80
Chapter 3
Combining Factors and Spreadsheet Functions
For the spreadsheet enthusiast, it is possible to find the 22-year A series value directly
by using the PMT function with embedded NPV functions. The cell reference format is
⫽ PMT(D1,22,−(NPV(D1,B6:B27)⫹B5 ⫹ NPV(D1,C6:C27)⫹C5)).
4. and 5. It is quite simple to determine an equivalent uniform series over any number of
periods using a spreadsheet, provided the series starts one period after the P value is located or ends in the same period that the F value is located. These are both true for the
series requested here. The results in F24 and F27 are the same as A1–10 and A11–22 in Equations [3.5] and [3.6], respectively.
Comment
Remember that round-off error will always be present when comparing hand and spreadsheet
results. The spreadsheet functions carry more decimal places than the tables during calculations. Also, be very careful when constructing spreadsheet functions. It is easy to miss a value,
such as the P or F in PMT and FV functions, or a minus sign between entries. Always check
your function entries carefully before touching <Enter>.
3.3 Calculations for Shifted Gradients
In Section 2.5, we derived the relation P ⫽ G(P兾G,i,n) to determine the present worth of the
arithmetic gradient series. The P兾G factor, Equation [2.25], was derived for a present worth in
year 0 with the gradient first appearing in year 2.
The present worth of an arithmetic gradient will always be located two periods before the
gradient starts.
Placement of gradient P
Refer to Figure 2–14 as a refresher for the cash flow diagrams.
The relation A ⫽ G(A兾G,i,n) was also derived in Section 2.5. The A兾G factor in Equation [2.27] performs the equivalence transformation of a gradient only into an A series from
years 1 through n (Figure 2–15). Recall that the base amount must be treated separately. Then the
equivalent P or A values can be summed to obtain the equivalent total present worth PT and total
annual series A T .
A conventional gradient series starts between periods 1 and 2 of the cash flow sequence. A
gradient starting at any other time is called a shifted gradient. The n value in the P兾G and A兾G
factors for a shifted gradient is determined by renumbering the time scale. The period in which
the gradient first appears is labeled period 2. The n value for the gradient factor is determined by
the renumbered period where the last gradient increase occurs.
Partitioning a cash flow series into the arithmetic gradient series and the remainder of the cash
flows can make very clear what the gradient n value should be. Example 3.5 illustrates this partitioning.
EXAMPLE 3.5
Fujitsu, Inc. has tracked the average inspection cost on a robotics manufacturing line for
8 years. Cost averages were steady at $100 per completed unit for the first 4 years, but have
increased consistently by $50 per unit for each of the last 4 years. Analyze the gradient increase, using the P兾G factor. Where is the present worth located for the gradient? What is the
general relation used to calculate total present worth in year 0?
Solution
The cash flow diagram in Figure 3–10a shows the base amount A ⫽ $100 and the arithmetic
gradient G ⫽ $50 starting between periods 4 and 5. Figure 3–10b and c partitions these two
series. Gradient year 2 is placed in actual year 5 of the entire sequence in Figure 3–10c. It is
clear that n ⫽ 5 for the P兾G factor. The PG ⫽ ? arrow is correctly placed in gradient year 0,
which is year 3 in the cash flow series.
Calculations for Shifted Gradients
3.3
0
1
2
3
4
5
A = $100
6
$150
7
$200
8
$250
G = $50
Year
$300
(a)
0
1
2
3
4
5
6
7
8
Year
5
6
7
8
Year
A = $100
(b)
PG = ?
0
1
2
3
4
0
1
2
3
$50
4
$100
5
$150
Gradient
n
$200
G = $50
(c)
Figure 3–10
Partitioned cash flow, (a) ⫽ (b) ⫹ (c), Example 3.5.
The general relation for PT is taken from Equation [2.19]. The uniform series A ⫽ $100
occurs for all 8 years, and the G ⫽ $50 gradient present worth PG appears in year 3.
PT ⫽ PA ⫹ PG ⫽ 100(P兾A,i,8) ⫹ 50(P兾G,i,5)(P兾F,i,3)
It is important to note that the A兾G factor cannot be used to find an equivalent A value in periods 1 through n for cash flows involving a shifted gradient. Consider the cash flow diagram of
Figure 3–11. To find the equivalent annual series in years 1 through 10 for the gradient series
only, first find the present worth PG of the gradient in actual year 5, take this present worth back
to year 0, and annualize the present worth for 10 years with the A兾P factor. If you apply the annual series gradient factor (A兾G,i,5) directly, the gradient is converted into an equivalent annual
series over years 6 through 10 only, not years 1 through 10, as requested.
To find the equivalent A series of a shifted gradient through all the n periods, first find the
present worth of the gradient at actual time 0, then apply the (A兾P, i, n) factor.
PG = ?
0
1
2
3
4
$10
$10
$10
$10
5
6
1
0
$10
7
2
8
3
9
4
Year
10
5
Gradient n
G = $15
$50
$65
$80
$95
$110
Figure 3–11
Determination of G and n values used in factors for a shifted gradient.
81
82
Combining Factors and Spreadsheet Functions
Chapter 3
EXAMPLE 3.6
Set up the engineering economy relations to compute the equivalent annual series in years 1
through 7 for the cash flow estimates in Figure 3–12.
0
1
2
3
$50
$50
$50
4
5
6
7
$70
$90
$110
$130
Figure 3–12
Diagram of a shifted gradient, Example 3.6.
Solution
The base amount annual series is AB ⫽ $50 for all 7 years (Figure 3–13). Find the present worth
PG in actual year 2 of the $20 gradient that starts in actual year 4. The gradient n is 5.
PG ⫽ 20(P兾G,i,5)
Bring the gradient present worth back to actual year 0.
P0 ⫽ PG(P兾F,i,2) ⫽ 20(P兾G,i,5)(P兾F,i,2)
Annualize the gradient present worth from year 1 through year 7 to obtain AG.
AG ⫽ P0(A兾P,i,7)
Finally, add the base amount to the gradient annual series.
AT ⫽ 20(P兾G,i,5)(P兾F,i,2)(A兾P,i,7) ⫹ 50
For a spreadsheet solution, enter the original cash flows into adjacent cells, say, B3 through
B10, and use an embedded NPV function in the PMT function. The single-cell function is
⫽ PMT(i%,7,−NPV(i%, B3:B10)).
P0 = ?
0
PG = ?
1
$50
2
AT = ?
3
0
1
$50
$50
AB = $50
4
5
2
3
6
7
4
Year
5
Gradient n
$70
$90
G = $20
$110
$130
Figure 3–13
Diagram used to determine A for a shifted gradient, Example 3.6.
In Section 2.6, we derived the relation Pg ⫽ A1(P兾A,g,i,n) to determine the present worth of a
geometric gradient series, including the initial amount A1. The factor was derived to find the
present worth in year 0, with A1 in year 1 and the first gradient appearing in year 2.
Placement of
gradient P
The present worth of a geometric gradient series will always be located two periods before
the gradient stars, and the initial amount is included in the resulting present worth. Refer
to Figure 2–21 as a refresher for the cash flows.
Equation [2.35] is the formula used for the factor. It is not tabulated.
Calculations for Shifted Gradients
3.3
EXAMPLE 3.7
Chemical engineers at a Coleman Industries plant in the Midwest have determined that a
small amount of a newly available chemical additive will increase the water repellency of
Coleman’s tent fabric by 20%. The plant superintendent has arranged to purchase the additive through a 5-year contract at $7000 per year, starting 1 year from now. He expects
the annual price to increase by 12% per year thereafter for the next 8 years. Additionally,
an initial investment of $35,000 was made now to prepare a site suitable for the contractor
to deliver the additive. Use i ⫽ 15% per year to determine the equivalent total present
worth for all these cash flows.
Solution
Figure 3–14 presents the cash flows. The total present worth PT is found using g ⫽ 0.12 and
i ⫽ 0.15. Equations [2.34] and [2.35] are used to determine the present worth Pg for the entire
geometric series at actual year 4, which is moved to year 0 using (P兾F,15%,4).
PT ⫽ 35,000 ⫹ A(P兾A,15%,4) ⫹ A1(P兾A,12%,15%,9)(P兾F,15%,4)
1 ⫺ (1.12兾1.15)9
⫽ 35,000 ⫹ 7000(2.8550) ⫹ 7000 ——————— (0.5718)
0.15 ⫺ 0.12
⫽ 35,000 ⫹ 19,985 ⫹ 28,247
[
]
⫽ $83,232
Note that n ⫽ 4 in the (P兾A,15%,4) factor because the $7000 in year 5 is the cash flow of the
initial amount A1.
For solution by spreadsheet, enter the cash flows of Figure 3–14. If cells B1 through B14
are used, the function to find P ⫽ $83,230 is
NPV(15%,B2:B14)⫹B1
The fastest way to enter the geometric series is to enter $7000 for year 5 (into cell B6) and set
up each succeeding cell multiplied by 1.12 for the 12% increase.
PT = ?
0
Pg = ?
1
2
3 4
0
$7000
i = 15% per year
5
6
7
8
9
10 11 12 13
1
2
3
4
5
6
7
8
9
Year
Geometric
gradient n
$7840
$35,000
$17,331
12% increase
per year
Figure 3–14
Cash flow diagram including a geometric gradient with g ⫽ 12%, Example 3.7.
Decreasing arithmetic and geometric gradients are common, and they are often shifted gradient series. That is, the constant gradient is –G or the percentage change is –g from one period to
the next, and the first appearance of the gradient is at some time period (year) other than year 2
of the series. Equivalence computations for present worth P and annual worth A are basically the
same as discussed in Chapter 2, except for the following.
83
84
Combining Factors and Spreadsheet Functions
Chapter 3
$800
$800
$700
$600
$500
$400
0
0
1
1
2
2
3
3
4
4
5
5
Gradient n
6
Year
PT = ?
(a)
AB = $800
$800
G = $100
$100
$200
$300
$400
–
0
1
2
3
4
5
6 Year
0
1
2
3
4
5
6 Year
PG = ?
PA = ?
(b)
(c)
Figure 3–15
Partitioned cash flow of a shifted arithmetic gradient, (a) ⫽ (b) ⫺ (c).
For shifted, decreasing gradients:
• The base amount A (arithmetic) or initial amount A1 (geometric) is the largest amount in the
first year of the series.
• The gradient amount is subtracted from the previous year’s amount, not added to it.
• The amount used in the factors is –G for arithmetic and –g for geometric gradient series.
• The present worth PG or Pg is located 2 years prior to the appearance of the first gradient;
however, a P兾F factor is necessary to find the present worth in year 0.
Figure 3–15 partitions a decreasing arithmetic gradient series with G ⫽ $−100 that is shifted
1 year forward in time. PG occurs in actual year 1, and PT is the sum of three components.
PT ⫽ $800(P兾F,i,1) ⫹ 800(P兾A,i,5)(P兾F,i,1) ⫺ 100(P兾G,i,5)(P兾F,i,1)
EXAMPLE 3.8
Morris Glass Company has decided to invest funds for the next 5 years so that development of
“smart” glass is well funded in the future. This type of new-technology glass uses electrochrome coating to allow rapid adjustment to sun and dark in building glass, as well as assisting
with internal heating and cooling cost reduction. The financial plan is to invest first, allow appreciation to occur, and then use the available funds in the future. All cash flow estimates are
in $1000 units, and the interest rate expectation is 8% per year.
Years 1 through 5: Invest $7000 in year 1, decreasing by $1000 per year through year 5.
Years 6 through 10: No new investment and no withdrawals.
Years 11 through 15: Withdraw $20,000 in year 11, decreasing 20% per year through year 15.
Calculations for Shifted Gradients
3.3
Determine if the anticipated withdrawals will be covered by the investment and appreciation
plans. If the withdrawal series is over- or underfunded, what is the exact amount available in
year 11, provided all other estimates remain the same?
Solution by Hand
Figure 3–16 presents the cash flow diagram and the placement of the equivalent P values used
in the solution. Calculate the present worth of both series in actual year 0 and add them to determine if the investment series is adequate to fund the anticipated withdrawals.
Pg,10 = ?
Pg,0 = ?
$20,000
A1 = $20,000
g = ⫺0.20
i = 8% per year
0 1
1
2
2
3
3
4
4
$8192
5
5
0
6
7
8
9
10
1
11
2
12
3
13
4
14
5
15
Gradient n
Year
A = $⫺7000
G = $⫺1000
PG = ?
Figure 3–16
Investment and withdrawal series, Example 3.8.
Investment series: Decreasing, conventional arithmetic series starting in year 2 with A ⫽
$⫺7000, G ⫽ $⫺1000, and gradient n ⫽ 5 years. The present worth in year 0 is
PG ⫽ ⫺[7000(P兾A,8%,5) ⫺ 1000(P兾G,8%,5)]
⫽ $⫺20,577
Withdrawal series: Decreasing, shifted geometric series starting in year 12 with A1 ⫽ $20,000,
g ⫽ ⫺0.20, and gradient n ⫽ 5 years. If the present worth in year 10 is identified as Pg,10, the
present worth in year 0 is Pg,0. Use Equation [2.35] for the (P兾A,⫺20%,8%,5) factor.
Pg,0 ⫽ Pg,10 (P兾F,i,n) ⫽ A1(P兾A,g,i,n)(P兾F,i,n)
{
[
]
[3.7]
}
1 ⫹ (⫺0.20) 5
1 ⫺ ——————
1 ⫹ 0.08
⫽ 20,000 ————————
(0.4632)
0.08 ⫺ (⫺0.20)
⫽ 20,000(2.7750)(0.4632)
⫽ $25,707
The net total present worth is
PT ⫽ ⫺20,577 ⫹ 25,707 ⫽ $⫹5130
This means that more is withdrawn than the investment series earns. Either additional funds
must be invested or less must be withdrawn to make the series equivalent at 8% per year.
To find the exact amount of the initial withdrawal series to result in PT ⫽ 0, let A1 be an
unknown in Equation [3.7] and set Pg,0 ⫽ ⫺PG ⫽ 20,577.
20,577 ⫽ A1(2.7750)(0.4632)
A1 ⫽ $16,009 in year 11
The geometric series withdrawal would be 20% less each year.
85
86
Combining Factors and Spreadsheet Functions
Chapter 3
⫽ NPV(8%,B4:B18) + B3
(a)
(b)
Figure 3–17
Spreadsheet solution, Example 3.8. (a) Cash flows and NPV function and (b) Goal Seek to determine initial withdrawal amount in year 11.
Solution by Spreadsheet
See Figure 3–17a. To determine if the investment series will cover the withdrawal series, enter
the cash flows (in column B using the functions shown in column C) and apply the NPV function shown in the cell tag to display PT ⫽ $⫹5130 directly. As above, the ⫹ sign indicates that,
from a time value of money perspective, there is more withdrawn than invested and earned.
The Goal Seek tool is very handy in determining the initial withdrawal amount that results
in PT ⫽ 0 (cell B19). Figure 3–17b shows the template and result A1 ⫽ $16,009 in year 11.
Each succeeding withdrawal is 80% of the previous one.
Comment
If the withdrawal series is fixed as estimated initially and the investment series base amount A
can be increased, the Goal Seek tool can be used to, again, set PT ⫽ 0 (cell B19). However, now
establish the entry in B4 as the changing cell. The response is A ⫽ $⫺8285, and, as before,
succeeding investments are $1000 less.
CHAPTER SUMMARY
In Chapter 2, we derived the equations to calculate the present, future, or annual worth of specific
cash flow series. In this chapter, we learned how to use these equations on cash flow series that
are shifted in time from those for which the basic relations are derived. For example, when a
uniform series does not begin in period 1, we still use the P兾A factor to find the “present worth”
of the series, except the P value is located one period ahead of the first A value, not at time 0. For
arithmetic and geometric gradients, the P value is two periods ahead of where the gradient starts.
With this information, it is possible to solve for P, A, or F for any conceivable cash flow series.
We have used the power of spreadsheet functions in determining P, F, and A values by singlecell entries and using cash flow estimates entered into a series of spreadsheet cells. Though
spreadsheet solutions are fast, they do remove some of the understanding of how the time value
of money and the factors change the equivalent value of money.
PROBLEMS
Present Worth Calculations
3.1 Industrial Electric Services has a contract with the
U.S. Embassy in Mexico to provide maintenance
for scanners and other devices in the building. If
the first payment of $12,000 is received now,
what is the present worth of the contract, provided
the company will receive a total of 10 payments
(i.e., years 0 through 9) and the interest rate is
10% per year?
87
Problems
3.2 Civil engineering consulting firms that provide services to outlying communities are vulnerable to a
number of factors that affect the financial condition
of the communities, such as bond issues and real
estate developments. A small consulting firm entered into a fixed-price contract with a large developer, resulting in a stable income of $260,000 per
year in years 1 through 3. At the end of that time, a
mild recession slowed the development, so the parties signed another contract for $190,000 per year
for 2 more years. Determine the present worth of the
two contracts at an interest rate of 10% per year.
3.3 The net cash flow associated with development
and sale of a new product is shown. Determine the
present worth at an interest rate of 12% per year.
The cash flow is in $1000 units. Show (a) hand and
(b) spreadsheet solutions.
Year
1
2
3
4
5
6
7
8
9
Cash
Flow, $ −120 −100 −40 ⫹50 ⫹50 ⫹80 ⫹80 ⫹80 ⫹80
3.4 Standby power for water utility pumps and other
electrical devices is provided by diesel-powered
generators. As an alternative, the utility can use
natural gas to power the generators, but it will be a
few years before the gas is available at remote
sites. The utility estimates that by switching to gas,
it will save $22,000 per year, starting 3 years from
now. At an interest rate of 8% per year, determine
the present worth in year 0 of the projected savings
that will occur in years 3 through 10.
3.5 Find the present worth at i ⫽ 10% per year for the
cash flow series shown below.
i = 10% per year
0
1
2
$200
3
$200
4
5
6
7
8
$90
$90
$90
Year
$200
3.6 The subsidized cost of producing water at the El
Paso Water Utilities (EPWU) Kay Bailey Hutchison
(KBH) desalting plant is $1.56 per 1000 gallons.
Under a contract that EPWU has with Fort Bliss,
the utility sells water to the Army post at a discounted price of $1.28 per 1000 gallons. (The
KBH plant was built on Army property.) If Fort
Bliss uses 2 billion gallons of water each year,
what is the present worth of the discount for a 20year period at an interest rate of 6% per year?
3.7 The rising cost of athletic programs at major universities has induced college presidents and athletic
directors to devise innovative ways to finance cashstrapped sports programs. One of the latest schemes
for big-time athletics is the “sports mortgage.” At
the University of Kansas, Jayhawk fans can sign up
to pay $105,000 now, or over a 10-year period, for
the right to buy top seats for football games during
the next 30 years. In return, the seats themselves
will stay locked in at current-year prices. Season
tickets in tier 1 are currently selling for $350 each.
A fan plans to purchase the sports mortgage along
with a current-season ticket and pay for both now,
then buy a ticket each year for the next 30 years.
What is the total present worth of the pricing plan at
an interest rate of 10% per year?
3.8 The cash flow associated with making self-locking
fasteners is shown below. Determine the net present worth (year 0) at an interest rate of 10% per
year.
Year
0
1
2
3
4
5
6
7
8
9
Income, $1000 20 20 20 20 30 30 30 30 30
Cost, $1000
8 8 8 8 12 12 12 12 12
30
25
3.9 Beckman Technologies, a relatively small manufacturer of precision laboratory equipment, borrowed $2 million to renovate one of its testing
labs. In an effort to pay off the loan quickly, the
company made four payments in years 1 through
4, with each payment being twice as large as the
preceding one. At an interest rate of 10% per year,
what was the size of the first payment?
Annual Worth Calculations
3.10 Revenue from the sale of ergonomic hand tools
was $300,000 in years 1 through 4 and $465,000 in
years 5 through 9. Determine the equivalent annual revenue in years 1 through 9 at an interest rate
of 10% per year.
3.11 Two engineering graduates who recently got married are planning for their early retirement 20 years
from now. They believe that they will need
$2,000,000 in year 20. Their plan is to live on one
of their salaries and invest the other. They already
have $25,000 in their investment account. (a) How
much will they have to invest each year if the account grows at a rate of 10% per year? (b) If the
maximum they have available to invest each year
is $40,000, will they reach their goal of $2 million
by year 20?
3.12 Costs associated with the manufacture of miniature high-sensitivity piezoresistive pressure transducers are $73,000 per year. A clever industrial
engineer found that by spending $16,000 now to
reconfigure the production line and reprogram
88
Combining Factors and Spreadsheet Functions
Chapter 3
two of the robotic arms, the cost will go down to
$58,000 next year and $52,000 in years 2 through 5.
Using an interest rate of 10% per year, determine
(a) the equivalent annual cost of the manufacturing
operations and (b) the equivalent annual savings in
years 1 through 5.
3.13 Calculate the equivalent annual cost in years 1
through 9 of the following series of disbursements.
Use an interest rate of 10% per year. Show (a)
hand and (b) spreadsheet solutions.
Year
Disbursement, $
0
1
2
3
4
8000
4000
4000
4000
4000
Year
Disbursement, $
5
6
7
8
9
4000
5000
5000
5000
5000
3.14 For the cash flows below, find the value of x that
makes the equivalent annual worth in years 1
through 7 equal to $300 per year. Use an interest
rate of 10% per year. Show solutions (a) by hand
and (b) using the Goal Seek tool.
Year
Cash Flow, $
Year
Cash Flow, $
0
1
2
3
200
200
200
200
4
5
6
7
x
200
200
200
3.15 Precision Instruments, Inc. manufactures highsensitivity mini accelerometers designed for
modal analysis testing. The company borrowed
$10,000,000 with the understanding that it would
make a $2,000,000 payment at the end of year 1
and then make equal annual payments in years 2
through 5 to pay off the loan. If the interest rate on
the loan was 9% per year, how much was each
payment in years 2 through 5?
3.16 A construction management company is examining
its cash flow requirements for the next 7 years. The
company expects to replace office machines and
computer equipment at various times over the
7-year planning period. Specifically, the company
expects to spend $6000 one year from now, $9000
three years from now, and $10,000 six years from
now. What is the annual worth (in years 1 through 7)
of the planned expenditures, at an interest rate of
10% per year?
3.17 Find the equivalent annual worth for the cash
flows shown, using an interest rate of 12% per
year. Monetary units are $1000.
Year
Cash Flow, $
1
2
3
4
5
6
7
8
9
20
20
20
20
60
60
60
60
60
3.18 Calculate the net annual worth in years 1 through
10 of the following series of incomes and expenses, if the interest rate is 10% per year.
Year
Income, $兾Year
Expense, $兾Year
0
1–4
5–10
0
700
2000
⫺2500
⫺200
⫺300
3.19 The city of El Paso gave the El Paso Tennis and
Swim Club a 20-year lease to continue using a
10-acre arroyo park for its facilities. The club
will pay $1000 per year plus make $350,000 in
improvements at the park. In addition, the club
will open its tennis courts to the public from 1
to 5 P.M. Monday through Thursday. If the club
makes $100,000 worth of improvements now
and then $50,000 worth each year for the next
5 years, what is the equivalent annual cost of
the lease to the club at an interest rate of 10%
per year?
3.20 Stadium Capital Financing Group is a Chicago
company that conceived the so-called sports
mortgage wherein fans agree to pay a relatively
large amount of money over a 10- to 30-year
period for the right to buy top seats for football
games for up to the next 50 years. In return,
season ticket prices stay locked in at currentyear prices, and the package can be sold in the
secondary market, while taking a tax write-off
for donating to a school. Assume a fan buys a
sports mortgage at West Virginia University for
$150,000 that is to be paid over a 10-year period with the right to buy two season tickets for
$300 each for the next 30 years. The first
payment is made now (i.e., beginning-of-year
payment), and an additional nine payments are
to be made at the end of each year for the next
9 years. Assume the fan also purchases the two
season tickets (also beginning-of-year payments). What is the total amount of the payment
each year in years 0 through 9? Use an interest
rate of 10% per year.
Future Worth Calculations
3.21 The expansion plans of Acme Granite, Stone &
Brick call for the company to add capacity for a
new product in 5 years. The company wants to
have $360,000 available before it announces the
product. If the company sets aside $55,000 now
and $90,000 in year 2, what uniform annual
amount will it have to put in an account in years 3
through 5 to have the $360,000? Assume the account earns interest at 8% per year.
89
Problems
3.22 For the cash flows shown, calculate the future
worth in year 8 using i ⫽ 10% per year.
Year
Cash Flow, $
0
1
2
3
4
5
6
100
100
100
100
100
300
300
3.23 Assume you plan to start an annuity plan by making your first deposit now. If you make annual
deposits of a uniform amount A into the account
that earns interest at a rate of 7% per year, how
many years from now will it be until the value of
the account is equal to 10 times the value of a
single deposit?
3.24 New actuator element technology enables engineers to simulate complex computer-controlled
movements in any direction. If the technology results in cost savings in the design of amusement
park rides, what is the future worth in year 5 of savings of $70,000 now and $20,000 per year in years
1 through 3 at an interest rate of 10% per year?
3.25 Austin Utilities is planning to install solar panels
to provide some of the electricity for its groundwater desalting plant. The project would be done in
two phases. The first phase will cost $4 million in
year 1 and $5 million in year 2. This investment
will result in energy savings (phase 2) of $540,000
in year 3, $546,000 in year 4, and amounts increasing by $6000 each year through year 10. Let i ⫽
10% per year.
(a) What is the future worth of the savings?
(b) Is the cost of the solar project justified by the
savings? (Hint: Calculate the difference between savings and cost).
3.26 For the cash flow diagram shown, determine the
value of W that will render the equivalent future
worth in year 8 equal to $⫺500 at an interest rate of
10% per year.
i = 10% per year
0
1
2
3
4
5
6
7
8
i = 10% per year
0
1
2
3
4
5
x
x
x
x
x
6
7
8
2x
2x
2x
Year
Random Single Amounts and Uniform Series
3.28 A small oil company is planning to replace its
Coriolis flowmeters with Emerson Hastelloy flowmeters. The replacement process will cost the
company $50,000 three years from now. How
much money must the company set aside each year
beginning now (year 0) in order to have the total
amount available immediately after making the
last deposit at the end of year 3? Assume the company can invest its funds at 15% per year.
3.29 A company that manufactures air-operated drain
valve assemblies budgeted $74,000 per year to pay
for plastic components over a 5-year period. If the
company spent only $42,000 in year 1, what uniform annual amount should the company expect to
spend in each of the next 4 years to expend the
entire budget? Assume the company uses an interest rate of 10% per year.
3.30 A recently hired chief executive officer wants to
reduce future production costs to improve the company’s earnings, thereby increasing the value of the
company’s stock. The plan is to invest $40,000
now and $40,000 in each of the next 2 years to improve productivity. By how much must annual
costs decrease in years 3 through 7 to recover the
investment plus a return of 12% per year?
3.31 Use the cash flow diagram to determine the single
amount of money Q4 in year 4 that is equivalent to
all of the cash flows shown. Use i ⫽ 10% per year.
Year
Q4 = ?
i = 10% per year
$40
$40
$40
$40
$40
$40
$40
⫺1
0
1
2
3
$25
$25
$25
$25
4
5
6
7
$50
$50
8 Year
W
3.27 For the cash flows shown in the diagram, determine the value of x that will make the future worth
in year 8 equal to $⫺70,000.
$25
$25
$25
$50
90
Combining Factors and Spreadsheet Functions
Chapter 3
3.32 For the following series of income and expenses,
find the equivalent value in year 9 at an interest rate
of 12% per year. Show (a) hand and (b) spreadsheet solutions.
3.36 For the cash flows shown, determine the present
worth in year 0, if the interest rate is 12% per year.
Year
1
9
10
Years
Income, $
Expense, $
Cash Flow, $
13 13 13 13 16 19 22 25 28
31
0
1–6
7–9
10–16
0
9,000
28,000
38,000
⫺70,000
⫺13,000
⫺14,000
⫺19,000
3.33 An investor just purchased property under a unique
financing agreement with the seller. The contract
price is $1.6 million. The payment plan is Z dollars
now, 2Z dollars in year 2, and 3Z dollars in years 3
through 5. If the interest rate on the transaction is
10% per year, how much is the payment in year 2?
3.34 A foursome of entrepreneurial electrical engineering graduates has a plan to start a new solar power
equipment company based on STE (solar thermal
electric) technology. They have recently approached
a group of investors with their idea and asked for a
loan of $5 million. Within the agreement, the loan is
to be repaid by allocating 80% of the company’s
profits each year for the first 4 years to the investors.
In the fifth year, the company will pay the balance
remaining on the loan in cash. The company’s business plan indicates that they expect to make no
profit for the first year, but in years 2 through 4, they
anticipate profits to be $1.5 million per year. If the
investors accept the deal at an interest rate of 15%
per year, and the business plan works to perfection,
what is the expected amount of the last loan payment (in year 5)?
Shifted Gradients
3.35 Find the present worth in year 0 for the cash flows
shown. Let i ⫽ 10% per year.
i = 10% per year
0
1
2
$50
$50
3
4
5
6
7
8
$70
Year
3.37
2
3
4
5
6
7
8
A low-cost noncontact temperature measuring tool
may be able to identify railroad car wheels that are in
need of repair long before a costly structural failure
occurs. If BNF Railroad saves $100,000 in years 1
through 5, $110,000 in year 6, and constant amounts
increasing by $10,000 each year through year 20,
what is the equivalent annual worth over the 20 years
of the savings? The interest rate is 10% per year.
3.38 Determine the present worth in year 0 of the cash
flows shown at an interest rate of 15% per year.
Year
1
Cash Flow, $
90 90 90 85 80 75 70 65 60 55
2
3
4
5
6
7
8
9
10
3.39 The Pedernales Electric Cooperative estimates that
the present worth now of income from an investment in renewable energy sources is $12,475,000.
There will be no income in years 1 and 2, but in
year 3 income will be $250,000, and thereafter it
will increase according to an arithmetic gradient
through year 15. What is the required gradient, if
the interest rate is 15% per year? Solve (a) by hand
and (b) using a spreadsheet.
3.40 Calculate the annual cost in years 1 through 9 of
the following series of disbursements. Use an interest rate of 10% per year.
Year
Disbursement, $
Year
Disbursement, $
0
1
2
3
4
5000
5500
6000
6500
7000
5
6
7
8
9
7500
8000
8500
9000
9500
3.41 The cost associated with manufacturing highperformance lubricants closely follows the cost of
crude oil. For the last 10 years, a small, independent refiner had a cost of $3.4 million in
years 1 through 3, after which the cost increased by
3% per year through this year. Determine the current equivalent worth (i.e., now) of the manufacturing cost, using an interest rate of 10% per year.
Show both (a) hand and (b) spreadsheet solutions.
$90
3.42 Find the future worth in year 10 of $50,000 in
year 0 and amounts increasing by 15% per year
through year 10 at an interest rate of 10% per year.
$130
$150
$170
$170
3.43 The cost of tuition at public universities has been
steadily increasing for many years. One Midwestern
university pledged to keep the tuition constant for
91
Problems
rolled steel was $1.8 million for the first 3 years.
An effective energy conservation program resulted
in a reduced cost to $1.77 million in year 4,
$1.74 million in year 5, and amounts decreasing by
$30,000 each year through year 9. What is the
equivalent annual worth of the pumping costs over
the 9 years at an interest rate of 12% per year?
4 years for all students who finished in the top 3%
of their class. One such student who liked research
planned to enroll at the university and continue
there until earning a PhD degree (a total time of
9 years). If the tuition for the first 4 years will be
$7200 per year and it increases by 5% per year for
the next 5 years, what is the present worth of the
tuition cost at an interest rate of 8% per year?
3.44 A private equity firm purchased a cable company
and assumed the company’s debt as part of the
transaction. The deal was structured such that the
private equity firm received $3 million immediately after the deal was closed (in year 0) through
the sale of some assets. This year (year 1) income
was $3.36 million, and it is projected to increase
by 12% each year through expansion of the customer base. What was the present worth in the year
of purchase of the income stream over a 10-year
period? The firm’s expected rate of return for any
purchase is 15% per year.
3.45 Calculate the present worth in year 0 of a series of
cash flows that starts at $150,000 in year 0 and increases by 10% per year through year 5. Assume
i ⫽ 10% per year.
3.49 Income from the mining of mineral deposits usually decreases as the resource becomes more difficult to extract. Determine the future worth in
year 10 of a mineral lease that yielded income of
$14,000 in years 1 through 4 and then amounts
that decreased by 5% per year through year 10.
Use an interest rate of 18% per year. Show (a) hand
and (b) spreadsheet solutions.
3.50 For the cash flows shown in the diagram, determine the future worth in year 8 at an interest rate of
10% per year.
i = 10% per year
0
1
2
3
4
5
6
7
8
Year
$220
Shifted Decreasing Gradients
$270
$320
3.46 For the cash flows shown, determine the value of
G such that the present worth in year 0 equals
$16,000 at an interest rate of 10% per year.
Year
0
1
2
3
Cash
Flow, $
0 8000 8000 8000⫺G
4
5
8000⫺2G
8000⫺3G
3.47 For these cash flows, find the equivalent annual
worth in years 1 through 7 at an interest rate of
10% per year.
Year
Cash Flow, $
Year
Cash Flow, $
0
1
2
3
850
800
750
700
4
5
6
7
650
600
550
500
3.48 The pumping cost for delivering water from the
Ohio River to Wheeling Steel for cooling hot-
$370
$420
$470
$470
3.51 Income from the sale of application software (apps)
is usually constant for several years and then decreases quite rapidly as the market gets close to
saturation. Income from one smart phone app was
$38,000 in years 1 through 3 and then decreased
geometrically by 15% per year through year 7. Determine the equivalent annual income in years 1
through 7, using an interest rate of 10% per year.
3.52 Determine the future worth in year 10 of a cash
flow series that starts in year 0 at $100,000 and
decreases by 12% per year. Use an interest rate of
12% per year.
92
Combining Factors and Spreadsheet Functions
Chapter 3
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
3.53 A manufacturer of toilet flush valves wants to have
$1,900,000 available 3 years from now so that a
new product line can be initiated. If the company
plans to deposit money each year, starting now, the
equation that represents the deposit each year at
8% per year interest is:
(a) 1,900,000(A兾F,8%,3)
(b) 1,900,000(A兾F,8%,4)
(c) 1,900,000 ⫹ 1,900,000(A兾F,8%,3)
(d) 1,900,000 ⫹ 1,900,000(A兾F,8%,2)
3.54 The present worth in year 0 of a lease that requires
a payment of $9000 now and amounts increasing
by 5% per year through year 10 at 8% per year
interest is closest to:
(a) $73,652
(b) $79,939
(c) $86,335
(d) Over $87,000
3.55 For the diagram shown, the respective values of n
for the following equation are:
P0 ⫽ 100(P兾A,10%,n)(P兾F,10%,n)
(a) 6 and 1
(b) 6 and 2
(c) 7 and 1
(d) 7 and 2
3.58 Cindy wants to deposit money for 4 consecutive
years starting 3 years from now so she can withdraw
$50,000 twelve years from now. Assume the interest
rate is 8% per year. The annual deposit is closest to:
(a) $6990
(b) $7670
(c) $8530
(d) $10,490
3.59 The net present worth in year 0 of the following
series of incomes and expenses at 8% per year is
closest to:
Years
(a)
(c)
2
3
4
5
6
7
3.56
$100
$100
$100
$100
Summit Metals is planning to expand its Wichita,
Kansas, manufacturing operation 5 years from now
at a cost of $10,000,000. If the company plans to deposit money into an account each year for 4 years
beginning 2 years from now (first deposit is in year 2)
to pay for the expansion, the equation that represents
the amount of the deposit at 9% per year interest is:
(a) A ⫽ 10,000,000(A兾F,9%,5)
(b) A ⫽ 10,000,000(A兾F,9%,4)
(c) A ⫽ 10,000,000(A兾P,9%,4)
(d) A ⫽ 10,000,000(A兾F,9%,4)(P兾F,9%,1)
3.57 The amount of money a person must deposit
3 years from now in order to withdraw $10,000 per
year for 10 years beginning 15 years from now at
an interest rate of 10% per year is closest to:
(a) $15,500
(b) $17,200
(c) $19,300
(d) $21,500
(b)
(d)
$15,500
$16,500
Annual Period
Amount, $
0
1
2
3
4
5
1000
1000
1000
1000
1000
1500
Year
$100
⫺1000
⫺100
⫺200
3.60 For the cash flows shown, the equivalent annual
worth in periods 1 through 5 at an interest rate of
10% per year is closest to:
(a)
(c)
$100
12,000
700
900
$14,300
$16,100
i = 10% per year
1
Expenses, $
0
1–6
7–11
P0 = ?
0
Income, $
$1120
$1350
(b)
(d)
$1240
$1490
3.61 For the cash flows shown, the value of X that will
make the present worth in year 0 equal to $5000 at
an interest rate of 10% per year is closest to:
Year
0
Cash
Flow, $
1
2
3
4
5
6
7
8
9
200 300 400 500 600 700 800 900 1000 X
(a)
(c)
$2895
$3305
(b)
(d)
$3125
$3765
3.62 In order to have cash available for unforeseen
emergencies, Baring Systems, a military contractor, wants to have $2,000,000 in a contingency
fund 4 years from now. The amount the company
must deposit each year in years 0 through 4 at an
interest rate of 10% per year is closest to:
(a) $420,100
(b) $327,600
(c) $284,600
(d) $206,900
Case Study
CASE STUDY
PRESERVING LAND FOR PUBLIC USE
Background and Information
The Trust for Public Land (TPL) is a national organization
that purchases and oversees the improvement of large land
sites for government agencies at all levels. Its mission is to
ensure the preservation of the natural resources, while providing necessary, but minimal, development for recreational
use by the public. All TPL projects are evaluated at 7% per
year, and TPL reserve funds earn 7% per year.
A southern U.S. state, which has long-term groundwater
problems, has asked the TPL to manage the purchase of
10,000 acres of aquifer recharge land and the development of
three parks of different use types on the land. The 10,000
acres will be acquired in increments over the next 5 years
with $4 million expended immediately on purchases. Total
annual purchase amounts are expected to decrease 25% each
year through year 5 and then cease for this particular project.
A city with 1.5 million citizens immediately to the southeast of this acreage relies heavily on the aquifer’s water. Its
citizens passed a bond issue last year, and the city government now has available $3 million for the purchase of land.
The bond interest rate is an effective 7% per year.
The engineers working on the park plan intend to complete all the development over a 3-year period starting in year
4, when the amount budgeted is $550,000. Increases in construction costs are expected to be $100,000 each year through
year 6.
At a recent meeting, the following agreements were made:
• Purchase the initial land increment now. Use the bond
issue funds to assist with this purchase. Take the remaining amount from TPL reserves.
• Raise the remaining project funds over the next 2 years in
equal annual amounts.
• Evaluate a financing alternative (suggested informally by
one individual at the meeting) in which the TPL provides
all funds, except the $3 million available now, until the
parks development is initiated in year 4.
Case Study Exercises
1. For each of the 2 years, what is the equivalent annual
amount necessary to supply the remaining project
funds?
2. If the TPL did agree to fund all costs except the $3 million bond proceeds now available, determine the equivalent annual amount that must be raised in years 4
through 6 to supply all remaining project funds. Assume
the TPL will not charge any extra interest over the 7% to
the state or city on the borrowed funds.
3. Review the TPL website (www.tpl.org). Identify some
economic and noneconomic factors that you believe
must be considered when the TPL is deciding to purchase land to protect it from real estate development.
93
CHAPTER 4
Nominal and
Effective
Interest Rates
L E A R N I N G
O U T C O M E S
Purpose: Make computations for interest rates and cash flows that are on a time basis other than a year.
SECTION
TOPIC
LEARNING OUTCOME
4.1
Statements
• Understand interest rate statements that include
nominal and effective rates.
4.2
Effective annual rate
• Derive and use the formula for an effective
annual interest rate.
4.3
Effective rate
• Determine the effective interest rate for any
stated time period.
4.4
Payment period and
compounding period
• Determine the payment period (PP) and
compounding period (CP) for equivalence
computations.
4.5
Single cash flows with PP ⱖ CP
• Perform equivalence calculations for singleamount cash flows and PP ⱖ CP.
4.6
Series cash flows with PP ⱖ CP
• Perform equivalence calculations for series and
gradient cash flows and PP ⱖ CP.
4.7
Single amounts and series with
PP ⬍ CP
• Perform equivalence calculations for cash flows
with PP ⬍ CP.
4.8
Continuous compounding
• Derive and use the effective interest rate formula
for interest rates that are compounded
continuously.
4.9
Varying rates
• Perform equivalency calculations for interest rates
that vary from one time period to another.
I
n all engineering economy relations developed thus far, the interest rate has been
a constant, annual value. For a substantial percentage of the projects evaluated by
professional engineers in practice, the interest rate is compounded more frequently
than once a year; frequencies such as semiannually, quarterly, and monthly are common. In fact,
weekly, daily, and even continuous compounding may be experienced in some project evaluations. Also, in our own personal lives, many of our financial considerations—loans of all types
(home mortgages, credit cards, automobiles, boats), checking and savings accounts, investments, stock option plans, etc.—have interest rates compounded for a time period shorter than
1 year. This requires the introduction of two new terms—nominal and effective interest rates.
This chapter explains how to understand and use nominal and effective interest rates
in engineering practice and in daily life situations. Equivalence calculations for any compounding frequency in combination with any cash flow frequency are presented.
PE
The Credit Card Offer Case: Today, Dave
received a special offer of a new credit
card from Chase Bank linked with the
major airline that he flies frequently.
It offers a generous bonus package
for signing up by a specific date about
60 days from now. The bonus package
includes extra airline points (once the
first purchase is made), priority airport
check-in services (for 1 year), several
free checked-bag allowances (for up
to 10 check-ins), extra frequent-flyer
points on the airline, access to airline
lounges (provided he uses the card
on a set time basis), plus several other
rewards (rental car discounts, cruise trip
amenities, and floral order discounts).
APR (annual percentage rate) for purchases and balance transfers*
APR for cash and overdraft advances*
Penalty APR for late minimum payment, exceeding credit limit, and
returned unpaid payments*†
Fees are listed as follows:
Annual membership
Balance transfers
Cash advances
Late payment
Over the credit limit
Returned check or payment
The annual fee of $85 for membership
does not start until the second year,
and balance transfers from other credit
cards have a low transfer fee, provided
they are made at the time of initial
membership.
Dave has a credit card currently with
a bank that he is planning to leave due
to its poor customer service and high
monthly fees. If he enrolls, he will transfer the $1000 balance on the current
card to the new Chase Bank card.
In the page that accompanies the
offer letter, “pricing information” is
included. This includes interest rates,
interest charges, and fees. A summary of
several of these rates and fees follows.
14.24% per year (sum of the current U.S.
Government prime rate of 3.25% and
10.99%, which is the APR added to determine
the balance transfer APR for Chase Bank)
19.24% per year
29.99% per year (maximum penalty APR)
$85; free the first year
$5 or 3% of each transfer, whichever is greater
$10 or 3% of each advance, whichever is greater
$39 each occurrence, if balance exceeds $250
$39 each occurrence
$39 each occurrence
*All APR rates are variable, based on a current 3.25% prime rate with 10.99% added to
determine purchase兾balance transfer APR; with 15.99% added to determine cash兾overdraft
APR; and with 26.99% added to determine penalty APR.
†The penalty APR applies indefinitely to future transactions. If no minimum payment is received
within 60 days, the penalty APR applies to all outstanding balances and all future transactions
on the account.
(Continued)
96
Nominal and Effective Interest Rates
Chapter 4
This case is used in the following topics (and sections) of this chapter:
Nominal and effective interest rate
statements (4.1)
Effective annual interest rates (4.2)
Equivalence relations: Series with
PP ⱖ CP (4.6)
4.1 Nominal and Effective Interest Rate Statements
In Chapter 1, we learned that the primary difference between simple interest and compound
interest is that compound interest includes interest on the interest earned in the previous period,
while simple interest does not. Here we discuss nominal and effective interest rates, which
have the same basic relationship. The difference here is that the concepts of nominal and effective must be used when interest is compounded more than once each year. For example, if an
interest rate is expressed as 1% per month, the terms nominal and effective interest rates must
be considered.
To understand and correctly handle effective interest rates is important in engineering practice
as well as for individual finances. The interest amounts for loans, mortgages, bonds, and stocks
are commonly based upon interest rates compounded more frequently than annually. The engineering economy study must account for these effects. In our own personal finances, we manage
most cash disbursements and receipts on a nonannual time basis. Again, the effect of compounding more frequently than once per year is present. First, consider a nominal interest rate.
Nominal interest rate r
A nominal interest rate r is an interest rate that does not account for compounding. By definition,
r ⴝ interest rate per time period ⴛ number of periods
[4.1]
A nominal rate may be calculated for any time period longer than the time period stated by using
Equation [4.1]. For example, the interest rate of 1.5% per month is the same as each of the following nominal rates.
Time Period
Nominal Rate by
Equation [4.1]
What This Is
24 months
12 months
6 months
3 months
1.5 ⫻ 24 ⫽ 36%
1.5 ⫻ 12 ⫽ 18%
1.5 ⫻ 6 ⫽ 9%
1.5 ⫻ 3 ⫽ 4.5%
Nominal rate per 2 years
Nominal rate per 1 year
Nominal rate per 6 months
Nominal rate per 3 months
Note that none of these rates mention anything about compounding of interest; they are all of the
form “r% per time period.” These nominal rates are calculated in the same way that simple rates
are calculated using Equation [1.7], that is, interest rate times number of periods.
After the nominal rate has been calculated, the compounding period (CP) must be included in the interest rate statement. As an illustration, again consider the nominal rate of 1.5%
per month. If we define the CP as 1 month, the nominal rate statement is 18% per year, compounded monthly, or 4.5% per quarter, compounded monthly. Now we can consider an effective
interest rate.
Effective interest rate i
An effective interest rate i is a rate wherein the compounding of interest is taken into
account. Effective rates are commonly expressed on an annual basis as an effective annual
rate; however, any time basis may be used.
The most common form of interest rate statement when compounding occurs over time periods
shorter than 1 year is “% per time period, compounded CP-ly,” for example, 10% per year, compounded monthly, or 12% per year, compounded weekly. An effective rate may not always
include the compounding period in the statement. If the CP is not mentioned, it is understood to
Nominal and Effective Interest Rate Statements
4.1
be the same as the time period mentioned with the interest rate. For example, an interest rate of
“1.5% per month” means that interest is compounded each month; that is, CP is 1 month. An
equivalent effective rate statement, therefore, is 1.5% per month, compounded monthly.
All of the following are effective interest rate statements because either they state they are
effective or the compounding period is not mentioned. In the latter case, the CP is the same as
the time period of the interest rate.
Statement
CP
What This Is
i ⫽ 10% per year
i ⫽ effective 10% per year,
compounded monthly
i ⫽ 1_12 % per month
CP not stated; CP ⫽ year
CP stated; CP ⫽ month
Effective rate per year
Effective rate per year
i ⫽ effective 1_12% per month,
compounded monthly
i ⫽ effective 3% per quarter,
compounded daily
CP not stated; CP ⫽ month
Effective rate per month
CP stated; CP ⫽ month
Effective rate per month; terms effective
and compounded monthly are redundant
Effective rate per quarter
CP stated; CP ⫽ day
All nominal interest rates can be converted to effective rates. The formula to do this is discussed
in the next section.
All interest formulas, factors, tabulated values, and spreadsheet functions must use an effective
interest rate to properly account for the time value of money.
The term APR (Annual Percentage Rate) is often stated as the annual interest rate for credit
cards, loans, and house mortgages. This is the same as the nominal rate. An APR of 15% is the
same as a nominal 15% per year or a nominal 1.25% on a monthly basis. Also the term APY
(Annual Percentage Yield) is a commonly stated annual rate of return for investments, certificates of deposit, and saving accounts. This is the same as an effective rate. The names are different, but the interpretations are identical. As we will learn in the following sections, the effective
rate is always greater than or equal to the nominal rate, and similarly APY ⱖ APR.
Based on these descriptions, there are always three time-based units associated with an interest rate statement.
Interest period (t)—The period of time over which the interest is expressed. This is the t in
the statement of r% per time period t, for example, 1% per month. The time unit of 1 year is
by far the most common. It is assumed when not stated otherwise.
Compounding period (CP)—The shortest time unit over which interest is charged or earned.
This is defined by the compounding term in the interest rate statement, for example, 8% per
year, compounded monthly. If CP is not stated, it is assumed to be the same as the interest
period.
Compounding frequency (m)—The number of times that compounding occurs within the
interest period t. If the compounding period CP and the time period t are the same, the compounding frequency is 1, for example, 1% per month, compounded monthly.
Consider the (nominal) rate of 8% per year, compounded monthly. It has an interest period t of
1 year, a compounding period CP of 1 month, and a compounding frequency m of 12 times per
year. A rate of 6% per year, compounded weekly, has t ⫽ 1 year, CP ⫽ 1 week, and m ⫽ 52, based
on the standard of 52 weeks per year.
In previous chapters, all interest rates had t and CP values of 1 year, so the compounding frequency was always m ⫽ 1. This made them all effective rates, because the interest period and
compounding period were the same. Now, it will be necessary to express a nominal rate as an
effective rate on the same time base as the compounding period.
An effective rate can be determined from a nominal rate by using the relation
r% per time period t
r
Effective rate per CP ⴝ ————————————— ⴝ —
m compounding periods per t m
[4.2]
As an illustration, assume r ⫽ 9% per year, compounded monthly; then m ⫽ 12. Equation [4.2] is used to obtain the effective rate of 9%兾12 ⫽ 0.75% per month, compounded monthly.
97
98
Nominal and Effective Interest Rates
Chapter 4
Note that changing the interest period t does not alter the compounding period, which is 1 month
in this illustration. Therefore, r ⫽ 9% per year, compounded monthly, and r ⫽ 4.5% per 6 months,
compounded monthly, are two expression of the same interest rate.
EXAMPLE 4.1
Three different bank loan rates for electric generation equipment are listed below. Determine
the effective rate on the basis of the compounding period for each rate.
(a) 9% per year, compounded quarterly.
(b) 9% per year, compounded monthly.
(c) 4.5% per 6 months, compounded weekly.
Solution
Apply Equation [4.2] to determine the effective rate per CP for different compounding periods.
The graphic in Figure 4–1 indicates the effective rate per CP and how the interest rate is distributed over time.
3
4
Quarter
.75%
2
.75%
1
.75%
2.25%
.75%
2.25%
.75%
0.75%
2.25%
.75%
12
2.25%
.75%
Month
2.25%
Distribution over Time Period t
.75%
(b) 9% per
year
4
( )
.75%
Quarter
Effective
Rate per
r
CP —
m
.75%
(a) 9% per
year
Compounding
Frequency (m)
.75%
Compounding
Period (CP)
.75%
Nominal
r % per t
1
2
3
4
5
6
7
8
9
10 11 12
Month
0.173%
(c) 4.5% per
6 months
Week
26
0.173%
1
12 14 16
26 Week
Figure 4–1
Relations between interest period t, compounding period CP, and effective interest rate per CP.
Sometimes it is not obvious whether a stated rate is a nominal or an effective rate. Basically
there are three ways to express interest rates, as detailed in Table 4–1. The right column includes
a statement about the effective rate. For the first format, a nominal interest rate is given and the
compounding period is stated. The effective rate must be calculated (discussed in the next sections). In the second format, the stated rate is identified as effective (or APY could also be used),
so the rate is used directly in computations.
In the third format, no compounding period is identified, for example, 8% per year. This rate
is effective over a compounding period equal to the stated interest period of 1 year in this case.
The effective rate for any other time period must be calculated.
TABLE 4–1
Various Ways to Express Nominal and Effective Interest Rates
Format of Rate Statement
Example of Statement
What about the Effective Rate?
Nominal rate stated,
compounding period stated
8% per year, compounded
quarterly
Find effective rate for any time
period (next two sections)
Effective rate stated
Effective 8.243% per year,
compounded quarterly
Use effective rate of 8.243% per year
directly for annual cash flows
Interest rate stated, no
compounding period stated
8% per year
Rate is effective for CP equal to stated
interest period of 1 year; find effective rate for all other time periods
Effective Annual Interest Rates
4.2
EXAMPLE 4.2 The Credit Card Offer Case
As described in the introduction to this case, Dave has been offered what is described as a credit
card deal that should not be refused—at least that is what the Chase Bank offer letter implies. The
balance transfer APR interest rate of 14.24% is an annual rate, with no compounding period mentioned. Therefore, it follows the format of the third entry in Table 4–1, that is, interest rate stated,
no CP stated. Therefore, we should conclude that the CP is 1 year, the same as the annual interest
period of the APR. However, as Dave and we all know, credit card payments are required monthly.
(a) First, determine the effective interest rates for compounding periods of 1 year and
1 month so Dave knows some effective rates he might be paying when he transfers the
$1000 balance from his current card.
(b) Second, assume that immediately after he accepts the card and completes the $1000 transfer,
Dave gets a bill that is due 1 month later. What is the amount of the total balance he owes?
Now, Dave looks a little closer at the fine print of the “pricing information” sheet and discovers
a small-print statement that Chase Bank uses the daily balance method (including new transactions) to determine the balance used to calculate the interest due at payment time.
(c) We will reserve the implication of this new finding until later, but for now help Dave by
determining the effective daily interest rate that may be used to calculate interest due at
the end of 1 month, provided the CP is 1 day.
Solution
(a) The interest period is 1 year. Apply Equation [4.2] for both CP values of 1 year (m ⫽ 1
compounding period per year) and 1 month (m ⫽ 12 compounding periods per year).
CP of year:
Effective rate per year ⫽ 14.24兾1 ⫽ 14.24%
CP of month:
Effective rate per month ⫽ 14.24兾12 ⫽ 1.187%
(b) The interest will be at the monthly effective rate, plus the balance transfer fee of 3%.
Amount owed after 1 month ⫽ 1000 ⫹ 1000(0.01187) ⫹ 0.03(1000)
⫽ 1000 ⫹ 11.87 ⫹ 30
⫽ $1041.87
Including the $30 fee, this represents an interest rate of (41.87兾1000)(100%) ⫽ 4.187%
for only the 1-month period.
(c) Again apply Equation [4.2], now with m ⫽ 365 compounding periods per year.
CP of day:
Effective rate per day ⫽ 14.24兾365 ⫽ 0.039%
4.2 Effective Annual Interest Rates
In this section, effective annual interest rates are calculated. Therefore, the year is used as the interest period t, and the compounding period CP can be any time unit less than 1 year. For example,
we will learn that a nominal 18% per year, compounded quarterly is the same as an effective rate of
19.252% per year.
The symbols used for nominal and effective interest rates are
r ⫽ nominal interest rate per year
CP ⫽ time period for each compounding
m ⫽ number of compounding periods per year
i ⫽ effective interest rate per compounding period ⫽ r兾m
ia ⫽ effective interest rate per year
The relation i ⫽ r兾m is exactly the same as Equation [4.2].
99
PE
100
Nominal and Effective Interest Rates
Chapter 4
Figure 4–2
P(1 + i)m = P(1 + ia)
Future worth calculation
at a rate i, compounded m
times in a year.
P(1 + i)m – 1
m–2
P(1 + i)
P(1 + i)3
P(1 +
Future worth amounts
i)2
P(1 + i)
1
2
m–2
3
m–1
m
Number of CPs
P
i
i
i
1
2
3
i
i
i
m–2
m–1
m
Effective i per
compounding period
Compounding period
As mentioned earlier, treatment for nominal and effective interest rates parallels that of
simple and compound interest. Like compound interest, an effective interest rate at any point
during the year includes (compounds) the interest rate for all previous compounding periods
during the year. Therefore, the derivation of an effective interest rate formula directly parallels
the logic used to develop the future worth relation F ⫽ P(1 ⫹ i)n. We set P ⫽ $1 for
simplification.
The future worth F at the end of 1 year is the principal P plus the interest P(i) through the
year. Since interest may be compounded several times during the year, use the effective annual
rate symbol ia to write the relation for F with P ⫽ $1.
F ⫽ P ⫹ Pia ⫽ 1(1 ⫹ ia)
Now consider Figure 4–2. The effective rate i per CP must be compounded through all m periods
to obtain the total effect of compounding by the end of the year. This means that F can also be
written as
F ⫽ 1(1 ⫹ i)m
Equate the two expressions for F and solve for ia. The effective annual interest rate formula
for ia is
ia ⴝ (1 ⴙ i)m ⴚ 1
[4.3]
Equation [4.3] calculates the effective annual interest rate ia for any number of compounding
periods per year when i is the rate for one compounding period.
If the effective annual rate ia and compounding frequency m are known, Equation [4.3] can be
solved for i to determine the effective interest rate per compounding period.
1
i ⫽ (1 ⫹ ia) 兾m ⫺ 1
[4.4]
As an illustration, Table 4–2 utilizes the nominal rate of 18% per year for different compounding periods (year to week) to determine the effective annual interest rate. In each case, the effective rate i per CP is applied m times during the year. Table 4–3 summarizes the effective annual
rate for frequently quoted nominal rates using Equation [4.3]. A standard of 52 weeks and
365 days per year is used throughout. The values in the continuous-compounding column are
discussed in Section 4.8.
TABLE 4–2
Effective Annual Interest Rates Using Equation [4.3]
r ⫽ 18% per year, compounded CP-ly
Compounding
Period, CP
Year
Times
Compounded
per Year, m
1
Rate per
Compound
Period, i%
Distribution of i over the Year of
Compounding Periods
18
Effective Annual
Rate, ia ⴝ (1 ⴙ i)m ⴚ 1
(1.18)1 ⫺ 1 ⫽ 18%
18%
1
6 months
Quarter
2
4
9
4.5
9%
9%
1
2
(1.09)2 ⫺ 1 ⫽ 18.81%
4.5%
4.5%
4.5%
4.5%
1
2
3
4
(1.045)4 ⫺ 1 ⫽ 19.252%
1.5% in each
Month
12
(1.015)12 ⫺ 1 ⫽ 19.562%
1.5
1
2
3
4
5
6
7
8
9
10 11 12
0.34615% in each
Week
52
(1.0034615)52 ⫺ 1 ⫽ 19.684%
0.34615
1 2 3
24
26
28
50
52
101
102
Nominal and Effective Interest Rates
Chapter 4
TABLE 4–3 Effective Annual Interest Rates for Selected Nominal Rates
Nominal
Rate r %
Semiannually
(m ⴝ 2)
Quarterly
(m ⴝ 4)
Monthly
(m ⴝ 12)
Weekly
(m ⴝ 52)
Daily
(m ⴝ 365)
0.25
0.50
1.00
1.50
2
3
4
5
6
7
8
9
10
12
15
18
20
25
30
40
50
0.250
0.501
1.003
1.506
2.010
3.023
4.040
5.063
6.090
7.123
8.160
9.203
10.250
12.360
15.563
18.810
21.000
26.563
32.250
44.000
56.250
0.250
0.501
1.004
1.508
2.015
3.034
4.060
5.095
6.136
7.186
8.243
9.308
10.381
12.551
15.865
19.252
21.551
27.443
33.547
46.410
60.181
0.250
0.501
1.005
1.510
2.018
3.042
4.074
5.116
6.168
7.229
8.300
9.381
10.471
12.683
16.076
19.562
21.939
28.073
34.489
48.213
63.209
0.250
0.501
1.005
1.511
2.020
3.044
4.079
5.124
6.180
7.246
8.322
9.409
10.506
12.734
16.158
19.684
22.093
28.325
34.869
48.954
64.479
0.250
0.501
1.005
1.511
2.020
3.045
4.081
5.126
6.180
7.247
8.328
9.417
10.516
12.745
16.177
19.714
22.132
28.390
34.968
49.150
64.816
Continuously
(m ⴝ ⴥ; er ⴚ 1)
0.250
0.501
1.005
1.511
2.020
3.046
4.081
5.127
6.184
7.251
8.329
9.417
10.517
12.750
16.183
19.722
22.140
28.403
34.986
49.182
64.872
EXAMPLE 4.3
Janice is an engineer with Southwest Airlines. She purchased Southwest stock for $6.90 per share
and sold it exactly 1 year later for $13.14 per share. She was very pleased with her investment
earnings. Help Janice understand exactly what she earned in terms of (a) effective annual rate and
(b) effective rate for quarterly compounding, and for monthly compounding. Neglect any commission fees for purchase and selling of stock and any quarterly dividends paid to stockholders.
Solution
(a) The effective annual rate of return ia has a compounding period of 1 year, since the
stock purchase and sales dates are exactly 1 year apart. Based on the purchase price of
$6.90 per share and using the definition of interest rate in Equation [1.2],
amount of increase per 1 year
6.24 ⫻ 100% ⫽ 90.43% per year
ia ⫽ ———————————— ⫻ 100% ⫽ ——
original price
6.90
(b) For the effective annual rates of 90.43% per year, compounded quarterly, and 90.43%,
compounded monthly, apply Equation [4.4] to find corresponding effective rates on the
basis of each compounding period.
Quarter:
Month:
m ⫽ 4 times per year
1兾4
i ⫽ (1.9043) ⫺ 1 ⫽ 1.17472 ⫺ 1 ⫽ 0.17472
This is 17.472% per quarter, compounded quarterly.
1兾12
m ⫽ 12 times per year
i ⫽ (1.9043) ⫺ 1 ⫽ 1.05514 ⫺ 1 ⫽ 0.05514
This is 5.514% per month, compounded monthly.
Comment
Note that these quarterly and monthly rates are less than the effective annual rate divided by
the number of quarters or months per year. In the case of months, this would be 90.43%兾12 ⫽
7.54% per month. This computation is incorrect because it neglects the fact that compounding
takes place 12 times during the year to result in the effective annual rate of 90.43%.
The spreadsheet function that displays the result of Equation [4.3], that is, the effective annual rate ia, is the EFFECT function. The format is
Effective Annual Interest Rates
4.2
103
ⴝ EFFECT(nominal_rate_per_year, compounding_frequency)
ⴝ EFFECT(r%, m)
[4.5]
Note that the rate entered in the EFFECT function is the nominal annual rate r% per year, not
the effective rate i% per compounding period. The function automatically finds i for use in Equation [4.3]. As an example, assume the nominal annual rate is r ⫽ 5.25% per year, compounded
quarterly, and you want to find the effective annual rate ia. The correct input is ⫽ EFFECT(5.25%,4)
to display ia ⫽ 5.354% per year. This is the spreadsheet equivalent of Equation [4.3] with
i ⫽ 5.25兾4 ⫽ 1.3125% per quarter with m ⫽ 4.
ia ⫽ (1 ⫹ 0.013125)4 ⫺1 ⫽ 0.05354
(5.354%)
The thing to remember about using the EFFECT function is that the nominal rate r entered
must be expressed over the same period of time as that of the effective rate required, which is
1 year here.
The NOMINAL spreadsheet function finds the nominal annual rate r. The format is
ⴝ NOMINAL(effective_rate, compounding_frequency_per_year)
ⴝ NOMINAL(ia%,m)
[4.6]
This function is designed to display only nominal annual rates. Accordingly, the m entered must
be the number of times interest is compounded per year. For example, if the effective annual rate
is 10.381% per year, compounded quarterly, and the nominal annual rate is requested, the function is ⫽ NOMINAL(10.381%,4) to display r ⫽ 10% per year, compounded quarterly. The
nominal rates for shorter time periods than 1 year are determined by using Equation [4.1]. For
example, the quarterly rate is 10%兾4 ⫽ 2.5%.
The things to remember when using the NOMINAL function are that the answer is always a
nominal annual rate, the rate entered must be an effective annual rate, and the m must equal the
number of times interest is compounded annually.
EXAMPLE 4.4 The Credit Card Offer Case
In our Progressive Example, Dave is planning to accept the offer for a Chase Bank credit card
that carries an APR (nominal rate) of 14.24% per year, or 1.187% per month. He will transfer
a balance of $1000 and plans to pay it and the transfer fee of $30, due at the end of the first
month. Let’s assume that Dave makes the transfer, and only days later his employer has a
1-year assignment for him in the country of the Cameroon in northwestern Africa. Dave accepts the employment offer, and in his hurried, excited departure, he forgets to send the credit
card service company a change of address. Since he is now out of mail touch, he does not pay
his monthly balance due, which we calculated in Example 4.2 to be $1041.87.
(a) If this situation continues for a total of 12 months, determine the total due after 12 months
and the effective annual rate of interest Dave has accumulated. Remember, the fine print on
the card’s interest and fee information states a penalty APR of 29.99% per year after one late
payment of the minimum payment amount, plus a late payment fee of $39 per occurrence.
(b) If there were no penalty APR and no late-payment fee, what effective annual interest rate
would be charged for this year? Compare this rate with the answer in part (a).
Solution
(a) Because Dave did not pay the first month’s amount, the new balance of $1041.87 and all
future monthly balances will accumulate interest at the higher monthly rate of
29.99%兾12 ⫽ 2.499% per month
Additionally, the $39 late-payment fee will be added each month, starting with the second
month, and interest will be charged on these fees also each month thereafter. The first
3 months and last 2 months are detailed below. Figure 4–3 details the interest and fees for
all 12 months.
PE
Nominal and Effective Interest Rates
Chapter 4
⫽ SUM(B6:D6)
⫽ J4/12
⫽ K4/12
Month 1: transfer fee
Months 2-12: late-payment fee
Interest rate per month
Month 1: 1.187%
Months 2-12: penalty 2.499%
Figure 4–3
Monthly amounts due for a credit card, Example 4.4.
Month 1:
Month 2:
Month 3:
1000 ⫹ 1000(0.01187) ⫹ 30 ⫽ $1041.87
1041.87 ⫹ 1041.87(0.02499) ⫹ 39 ⫽ $1106.91
1106.91 ⫹ 1106.91(0.02499) ⫹ 39 ⫽ $1173.57
...
104
Month 11:
Month 12:
1689.25 ⫹ 1689.25(0.02499) ⫹ 39 ⫽ $1770.46
1770.46 ⫹ 1770.46(0.02499) ⫹ 39 ⫽ $1853.71
The effective monthly rate is determined by using the F兾P factor to find the i value at
which $1000 now is equivalent to $1853.71 after 12 periods.
12
1853.71 ⫽ 1000(F兾P, i,12) ⫽ 1000(1 ⫹ i)
1兾12
1 ⫹ i ⫽ (1.85371)
⫽ 1.05278
i ⫽ 5.278% per month
Since the compounding period is 1 month, use Equation [4.3] to determine the effective
annual rate of 85.375% per year, compounded monthly.
12
ia ⫽ (1 ⫹ i )m ⫺ 1 ⫽ (1.05278) ⫺ 1
⫽ 0.85375
(85.375%)
(b) If there were no penalty fees for late payments and the nominal annual rate of 14.24% (or
1.187% per month) were applied throughout the 12 months, the effective annual rate
would be 15.207% per year, compounded monthly. By Equation [4.3], with a small
rounding error included,
ia ⫽ (1 ⫹ i)m ⫺ 1⫽ (1.01187)12 ⫺1 ⫽ 0.15207
First, Dave will not pay at the stated rate of 14.24%, because this is the APR (nominal
rate), not the APY (effective rate) of 15.207%. Second, and much more important, is the
huge difference made by (1) the increase in rate to an APR of 29.99% and (2) the monthly
fees of $39 for not making a payment. These large fees become part of the credit balance
and accumulate interest at the penalty rate of 29.99% per year. The result is an effective
annual rate jump from 15.207% to 85.375% per year, compounded monthly.
Comment
This is but one illustration of why the best advice to an individual or company in debt is to
spend down the debt. The quoted APR by credit card, loan, and mortgage institutions can be
quite deceiving; plus, the addition of penalty fees increases the effective rate very rapidly.
Effective Interest Rates for Any Time Period
4.3
When Equation [4.3] is applied to find ia the result is usually not an integer. Therefore, the
engineering economy factor cannot be obtained directly from the interest factor tables. There are
alternative ways to find the factor value.
• Use the factor formula with the ia rate substituted for i.
• Use the spreadsheet function with ia (as discussed in Section 2.4).
• Linearly interpolate between two tabulated rates (as discussed in Section 2.4).
4.3 Effective Interest Rates for Any Time Period
Equation [4.3] in Section 4.2 calculates an effective interest rate per year from any effective rate
over a shorter time period. We can generalize this equation to determine the effective interest
rate for any time period (shorter or longer than 1 year).
r m
Effective i per time period ⴝ ( 1 ⴙ —
m) ⴚ 1
[4.7]
where i ⴝ effective rate for specified time period (say, semiannual)
r ⴝ nominal interest rate for same time period (semiannual)
m ⴝ number of times interest is compounded per stated time period (times per
6 months)
The term r兾m is always the effective interest rate over a compounding period CP, and m is
always the number of times that interest is compounded per the time period on the left of the
equals sign in Equation [4.7]. Instead of ia, this general expression uses i as the symbol for the
effective interest rate, which conforms to the use of i throughout the remainder of this text.
Examples 4.5 and 4.6 illustrate the use of this equation.
EXAMPLE 4.5
Tesla Motors manufactures high-performance battery electric vehicles. An engineer is on a
Tesla committee to evaluate bids for new-generation coordinate-measuring machinery to be
directly linked to the automated manufacturing of high-precision vehicle components. Three
bids include the interest rates that vendors will charge on unpaid balances. To get a clear understanding of finance costs, Tesla management asked the engineer to determine the effective
semiannual and annual interest rates for each bid. The bids are as follows:
Bid 1:
Bid 2:
Bid 3:
9% per year, compounded quarterly
3% per quarter, compounded quarterly
8.8% per year, compounded monthly
(a) Determine the effective rate for each bid on the basis of semiannual periods.
(b) What are the effective annual rates? These are to be a part of the final bid selection.
(c) Which bid has the lowest effective annual rate?
Solution
(a) Convert the nominal rates to a semiannual basis, determine m, then use Equation [4.7] to
calculate the effective semiannual interest rate i. For bid 1,
r ⫽ 9% per year ⫽ 4.5% per 6 months
m ⫽ 2 quarters per 6 months
(
)
0.045 2 ⫺ 1 ⫽ 1.0455 ⫺ 1 ⫽ 4.55%
Effective i% per 6 months ⫽ 1 ⫹ ———
2
Table 4–4 (left section) summarizes the effective semiannual rates for all three bids.
105
106
Nominal and Effective Interest Rates
Chapter 4
TABLE 4–4 Effective Semiannual and Annual Interest Rates for Three Bid Rates, Example 4.5
Semiannual Rates
Annual Rates
Bid
Nominal r per
6 Months, %
CP per
6 Months, m
Equation [4.7],
Effective i, %
Nominal r
per Year, %
CP per
Year, m
Equation [4.7],
Effective i, %
1
2
3
4.5
6.0
4.4
2
2
6
4.55
6.09
4.48
9
12
8.8
4
4
12
9.31
12.55
9.16
(b) For the effective annual rate, the time basis in Equation [4.7] is 1 year. For bid 1,
r ⫽ 9% per year
m ⫽ 4 quarters per year
(
)
0.09 4 − 1 ⫽ 1.0931 ⫺ 1 ⫽ 9.31%
Effective i% per year ⫽ 1 ⫹ ——
4
The right section of Table 4–4 includes a summary of the effective annual rates.
(c) Bid 3 includes the lowest effective annual rate of 9.16%, which is equivalent to an effective semiannual rate of 4.48% when interest is compounded monthly.
EXAMPLE 4.6
A dot-com company plans to place money in a new venture capital fund that currently returns
18% per year, compounded daily. What effective rate is this (a) yearly and (b) semiannually?
Solution
(a) Use Equation [4.7], with r ⫽ 0.18 and m ⫽ 365.
0.18 365 ⫺ 1 ⫽ 19.716%
Effective i% per year ⫽ 1 ⫹ ——
365
(b) Here r ⫽ 0.09 per 6 months and m ⫽ 182 days.
0.09 182 ⫺ 1 ⫽ 9.415%
Effective i% per 6 months ⫽ 1 ⫹ ——
182
(
)
(
)
4.4 Equivalence Relations: Payment Period
and Compounding Period
Now that the procedures and formulas for determining effective interest rates with consideration
of the compounding period are developed, it is necessary to consider the payment period.
The payment period (PP) is the length of time between cash flows (inflows or outflows). It is
common that the lengths of the payment period and the compounding period (CP) do not coincide. It is important to determine if PP ⫽ CP, PP ⬎ CP, or PP ⬍ CP.
If a company deposits money each month into an account that earns at the nominal rate of 8%
per year, compounded semiannually, the cash flow deposits define a payment period of 1 month
and the nominal interest rate defines a compounding period of 6 months. These time periods are
shown in Figure 4–4. Similarly, if a person deposits a bonus check once a year into an account
that compounds interest quarterly, PP ⫽ 1 year and CP ⫽ 3 months.
r = nominal 8% per year, compounded semiannually
CP
6 months
Figure 4–4
One-year cash flow diagram for a monthly payment period (PP) and
semiannual compounding
period (CP).
0
1
PP
1 month
2
3
4
CP
6 months
5
6
7
8
9
10
11
12
Months
Equivalence Relations: Single Amounts with PP ⱖ CP
4.5
TABLE 4–5 Section References for Equivalence Calculations Based on
Payment Period and Compounding Period Comparison
Length
of Time
Involves
Single Amounts
(P and F Only)
Involves Uniform Series
or Gradient Series
(A, G, or g)
PP ⫽ CP
PP > CP
PP < CP
Section 4.5
Section 4.5
Section 4.7
Section 4.6
Section 4.6
Section 4.7
As we learned earlier, to correctly perform equivalence calculations, an effective interest rate
is needed in the factors and spreadsheet functions. Therefore, it is essential that the time periods
of the interest rate and the payment period be on the same time basis. The next three sections (4.5
to 4.7) describe procedures to determine correct i and n values for engineering economy factors
and spreadsheet functions. First, compare the length of PP and CP, then identify the cash flows as
only single amounts (P and F) or as a series (A, G, or g). Table 4–5 provides the section reference.
The section references are the same when PP ⫽ CP and PP > CP, because the procedures to
determine i and n are the same, as discussed in Sections 4.5 and 4.6.
A general principle to remember throughout these equivalence computations is that when
cash actually flows, it is necessary to account for the time value of money. For example, assume that cash flows occur every 6 months and that interest is compounded quarterly. After 3
months there is no cash flow and no need to determine the effect of quarterly compounding.
However, at the 6-month time point, it is necessary to consider the interest accrued during the
previous two quarters.
4.5 Equivalence Relations: Single Amounts with PP ⱖ CP
With only P and F estimates defined, the payment period is not specifically identified. In virtually
all situations, PP will be equal to or greater than CP. The length of the PP is defined by the interest
period in the stated interest rate. If the rate is 8% per year, for example, PP ⫽ CP ⫽ 1 year. However, if the rate is 10% per year, compounded quarterly, then PP is 1 year, CP is 1 quarter or
3 months, and PP > CP. The procedures to perform equivalence computations are the same for
both situations, as explained here.
When only single-amount cash flows are involved, there are two equally correct ways to determine i and n for P兾F and F兾P factors. Method 1 is easier to apply, because the interest tables
in the back of the text can usually provide the factor value. Method 2 likely requires a factor
formula calculation, because the resulting effective interest rate is not an integer. For spreadsheets, either method is acceptable; however, method 1 is usually easier.
Method 1: Determine the effective interest rate over the compounding period CP, and set n
equal to the number of compounding periods between P and F. The relations to calculate P and
F are
P ⫽ F(P兾F, effective i% per CP, total number of periods n)
[4.8]
F⫽ P(F兾P, effective i% per CP, total number of periods n)
[4.9]
For example, assume that the stated credit card rate is nominal 15% per year, compounded
monthly. Here CP is 1 month. To find P or F over a 2-year span, calculate the effective monthly
rate of 15%兾12 ⫽ 1.25% and the total months of 2(12) ⫽ 24. Then 1.25% and 24 are used in the
P兾F and F兾P factors.
Any time period can be used to determine the effective interest rate; however, the interest rate
that is associated with the CP is typically the best because it is usually a whole number. Therefore, the factor tables in the back of the text can be used.
Method 2: Determine the effective interest rate for the time period t of the nominal rate, and
set n equal to the total number of periods, using this same time period.
107
108
Nominal and Effective Interest Rates
Chapter 4
The P and F relations are the same as in Equations [4.8] and [4.9] with the term effective i% per t
substituted for the interest rate. For a credit card rate of 15% per year, compounded monthly, the
time period t is 1 year. The effective rate over 1 year and the n values are
(
0.15
Effective i% per year ⫽ 1 ⫹ ——
12
)
12
⫺ 1 ⫽ 16.076%
n ⫽ 2 years
The P兾F factor is the same by both methods: (P兾F,1.25%,24) ⫽ 0.7422 using Table 5 in the rear
of the text; and (P兾F,16.076%,2) ⫽ 0.7422 using the P兾F factor formula.
EXAMPLE 4.7
Over the past 10 years, Gentrack has placed varying sums of money into a special capital
accumulation fund. The company sells compost produced by garbage-to-compost plants in
the United States and Vietnam. Figure 4–5 is the cash flow diagram in $1000 units. Find the
amount in the account now (after 10 years) at an interest rate of 12% per year, compounded
semiannually.
Solution
Only P and F values are involved. Both methods are illustrated to find F in year 10.
Method 1: Use the semiannual CP to express the effective semiannual rate of 6% per 6-month
period. There are n ⫽ (2)(number of years) semiannual periods for each cash flow. Using tabulated factor values, the future worth by Equation [4.9] is
F ⫽ 1000(F兾P,6%,20) ⫹ 3000(F兾P,6%,12) ⫹ 1500(F兾P,6%,8)
⫽ 1000(3.2071) ⫹ 3000(2.0122) ⫹ 1500(1.5938)
⫽ $11,634
Method 2:
($11.634 million)
Express the effective annual rate, based on semiannual compounding.
(
)
2
0.12 ⫺ 1 ⫽ 12.36%
Effective i% per year ⫽ 1 ⫹ ——
2
The n value is the actual number of years. Use the factor formula (F兾P,i,n) ⫽ (1.1236)n and
Equation [4.9] to obtain the same answer as above.
F ⫽ 1000(F兾P,12.36%,10) ⫹ 3000(F兾P,12.36%,6) ⫹ 1500(F兾P,12.36%,4)
⫽ 1000(3.2071) ⫹ 3000(2.0122) ⫹ 1500(1.5938)
⫽ $11,634
($11.634 million)
F=?
0
1
2
3
4
5
6
$1000
$1500
$3000
Figure 4–5
Cash flow diagram, Example 4.7.
7
8
9
10
Year
Equivalence Relations: Series with PP ⱖ CP
4.6
TABLE 4–6 Examples of n and i Values Where PP ⫽ CP or PP > CP
What to Find;
What Is Given
Standard Notation
Find P; given A
P ⫽ 500(P兾A,8%,10)
Find F; given A
F ⫽ 75(F兾A,2%,36)
Find F; given A
F ⫽ 180(F兾A,5%,60)
1% per month
Find P; given G
P ⫽ 25(P兾G,1%,48)
1% per month
Find A; given P
A ⫽ 5000(A兾P,3.03%,24)
Cash Flow Series
Interest Rate
$500 semiannually
for 5 years
16% per year,
compounded
semiannually
24% per year,
compounded
monthly
5% per quarter
$75 monthly for
3 years
$180 quarterly for
15 years
$25 per month
increase for
4 years
$5000 per quarter
for 6 years
4.6 Equivalence Relations: Series with PP ⱖ CP
When uniform or gradient series are included in the cash flow sequence, the procedure is basically the same as method 2 above, except that PP is now defined by the length of time between
cash flows. This also establishes the time unit of the effective interest rate. For example, if cash
flows occur on a quarterly basis, PP is 1 quarter and the effective quarterly rate is necessary. The
n value is the total number of quarters. If PP is a quarter, 5 years translates to an n value of
20 quarters. This is a direct application of the following general guideline:
When cash flows involve a series (i.e., A, G, g) and the payment period equals or exceeds the
compounding period in length:
• Find the effective i per payment period.
• Determine n as the total number of payment periods.
In performing equivalence computations for series, only these values of i and n can be used in
interest tables, factor formulas, and spreadsheet functions. In other words, there are no other
combinations that give the correct answers, as there are for single-amount cash flows.
Table 4–6 shows the correct formulation for several cash flow series and interest rates. Note
that n is always equal to the total number of payment periods and i is an effective rate expressed
over the same time period as n.
EXAMPLE 4.8
For the past 7 years, Excelon Energy has paid $500 every 6 months for a software maintenance
contract. What is the equivalent total amount after the last payment, if these funds are taken
from a pool that has been returning 8% per year, compounded quarterly?
Solution
The cash flow diagram is shown in Figure 4–6. The payment period (6 months) is longer than
the compounding period (quarter); that is, PP > CP. Applying the guideline, we need to determine an effective semiannual interest rate. Use Equation [4.7] with r ⫽ 4% per 6-month
period and m ⫽ 2 quarters per semiannual period.
(
)
0.04 2 ⫺ 1 ⫽ 4.04%
Effective i% per 6 months ⫽ 1 ⫹ ——
2
The effective semiannual interest rate can also be obtained from Table 4–3 by using the r value
of 4% and m ⫽ 2 to get i ⫽ 4.04%.
109
110
Nominal and Effective Interest Rates
Chapter 4
The value i ⫽ 4.04% seems reasonable, since we expect the effective rate to be slightly
higher than the nominal rate of 4% per 6-month period. The total number of semiannual payment periods is n ⫽ 2(7) ⫽ 14. The relation for F is
F ⫽ A(F兾A,4.04%,14)
⫽ 500(18.3422)
⫽ $9171.09
To determine the F兾A factor value 18.3422 using a spreadsheet, enter the FV function from
Figure 2–9, that is, ⫽ ⫺FV(4.04%,14,1). Alternatively, the final answer of $9171.09 can be
displayed directly using the function ⫽ ⫺FV(4.04%,14,500).
F=?
i = 8% per year, compounded quarterly
0
1
2
3
4
5
6
7
Years
A = $500
Figure 4–6
Diagram of semiannual deposits used to determine F, Example 4.8.
PE
EXAMPLE 4.9 Credit Card Offer Case
In our continuing credit card saga of Dave and his job transfer to Africa, let’s assume he did
remember that the total balance is $1030, including the $30 balance transfer fee, and he wants
to set up a monthly automatic checking account transfer to pay off the entire amount in 2 years.
Once he learned that the minimum payment is $25 per month, Dave decided to be sure the
monthly transfer exceeds this amount to avoid any further penalty fees and the penalty APR of
29.99% per year. What amount should he ask to be transferred by the due date each month?
What is the APY he will pay, if this plan is followed exactly and Chase Bank does not change
the APR during the 2-year period? Also, assume he left the credit card at home and will charge
no more to it.
Solution
The monthly A series is needed for a total of n ⫽ 2(12) ⫽ 24 payments. In this case,
PP ⫽ CP ⫽ 1 month, and the effective monthly rate is i ⫽ 14.24%兾12 ⫽ 1.187% per month.
Solution by hand:
Use a calculator or hand computation to determine the A兾P factor value.
A ⫽ P(A兾P,i,n) ⫽ 1030(A兾P,1.187%,24) ⫽ 1030(0.04813)
⫽ $49.57 per month for 24 months
Solution by Spreadsheet: Use the function ⫽ ⫺PMT(1.187%,24,1) to determine the factor value 0.04813 to determine A for n ⫽ 24 payments. Alternatively use the function
⫽ PMT (1.187%,24,1030) to directly display the required monthly payment of A ⫽ $⫺49.57.
The effective annual interest rate or APY is computed using Equation [4.7] with r ⫽ 14.24%
per year, compounded monthly, and m ⫽ 12 times per year.
(
0.1424
Effective i per year ⫽ 1 ⫹ ———
12
)
12
– 1 ⫽ 1.15207 ⫺1
⫽ 15.207% per year
This is the same effective annual rate ia determined in Example 4.4b.
Equivalence Relations: Series with PP ⱖ CP
4.6
EXAMPLE 4.10
The Scott and White Health Plan (SWHP) has purchased a robotized prescription fulfillment
system for faster and more accurate delivery to patients with stable, pill-form medication for
chronic health problems, such as diabetes, thyroid, and high blood pressure. Assume this highvolume system costs $3 million to install and an estimated $200,000 per year for all materials,
operating, personnel, and maintenance costs. The expected life is 10 years. An SWHP biomedical engineer wants to estimate the total revenue requirement for each 6-month period that
is necessary to recover the investment, interest, and annual costs. Find this semiannual A value
both by hand and by spreadsheet, if capital funds are evaluated at 8% per year, using two different compounding periods:
Rate 1. 8% per year, compounded semiannually.
Rate 2. 8% per year, compounded monthly.
Solution
Figure 4–7 shows the cash flow diagram. Throughout the 20 semiannual periods, the annual
cost occurs every other period, and the capital recovery series is sought for every 6-month
period. This pattern makes the solution by hand quite involved if the P兾F factor, not the P兾A
factor, is used to find P for the 10 annual $200,000 costs. The spreadsheet solution is recommended in cases such as this.
Solution by hand—rate 1: Steps to find the semiannual A value are summarized below:
PP ⫽ CP at 6 months; find the effective rate per semiannual period.
Effective semiannual i ⫽ 8%兾2 ⫽ 4% per 6 months, compounded semiannually.
Number of semiannual periods n ⫽ 2(10) ⫽ 20.
Calculate P, using the P兾F factor for n ⫽ 2, 4, . . . , 20 periods because the costs are annual,
not semiannual. Then use the A兾P factor over 20 periods to find the semiannual A.
P ⫽ 3,000,000 ⫹ 200,000
[
20
⌺ (P兾F,4%,k)
k⫽2,4
]
⫽ 3,000,000 ⫹ 200,000(6.6620) ⫽ $4,332,400
A ⫽ $4,332,400(A兾P,4%,20) ⫽ $318,778
Conclusion: Revenue of $318,778 is necessary every 6 months to cover all costs and interest
at 8% per year, compounded semiannually.
2
4
6
8
1
2
3
4
A per 6 months = ?
10
12
14
16
18
20
8
9
10
0
6 months
5
6
7
$200,000 per year
i1 = 8%, compounded semiannually
i2 = 8%, compounded monthly
P = $3 million
Figure 4–7
Cash flow diagram with two different compounding periods, Example 4.10.
Years
111
112
Nominal and Effective Interest Rates
Chapter 4
Solution by hand—rate 2: The PP is 6 months, but the CP is now monthly; therefore, PP ⬎ CP.
To find the effective semiannual rate, the effective interest rate Equation [4.7] is applied with
r ⫽ 4% and m ⫽ 6 months per semiannual period.
(
0.04
Effective semiannual i ⫽ 1 ⫹ ———
6
P ⫽ 3,000,000 ⫹ 200,000
[
)
6
⫺ 1 ⫽ 4.067%
20
⌺ (P兾F,4.067%,k)
k⫽2,4
]
⫽ 3,000,000 ⫹ 200,000(6.6204) ⫽ $4,324,080
A ⫽ $4,324,080(A兾P,4.067%,20) ⫽ $320,064
Now, $320,064, or $1286 more semiannually, is required to cover the more frequent compounding of the 8% per year interest. Note that all P兾F and A兾P factors must be calculated
with factor formulas at 4.067%. This method is usually more calculation-intensive and errorprone than the spreadsheet solution.
Solution by spreadsheet—rates 1 and 2: Figure 4–8 presents a general solution for the problem
at both rates. (Several rows at the bottom of the spreadsheet are not printed. They continue the
cash flow pattern of $200,000 every other 6 months through cell B32.) The functions in C8 and
E8 are general expressions for the effective rate per PP, expressed in months. This allows some
sensitivity analysis to be performed for different PP and CP values. Note the functions in C7
and E7 to determine m for the effective rate relations. This technique works well for spreadsheets once PP and CP are entered in the time unit of the CP.
Each 6-month period is included in the cash flows, including the $0 entries, so the NPV and
PMT functions work correctly. The final A values in D14 ($318,784) and F14 ($320,069) are
the same (except for rounding) as those above.
⫽ C5/C6
⫽ E5/E6
⫽ +((1+((E2/(12/E5))/E7))∧E7)⫺1
⫽ PMT(E8,E4,⫺F9
⫽ NPV(E8,B13:B32)+B12
Figure 4–8
Spreadsheet solution for semiannual A series for different compounding periods, Example 4.10.
4.7 Equivalence Relations: Single Amounts
and Series with PP ⬍ CP
If a person deposits money each month into a savings account where interest is compounded
quarterly, do all the monthly deposits earn interest before the next quarterly compounding time?
If a person's credit card payment is due with interest on the 15th of the month, and if the full payment is made on the 1st, does the financial institution reduce the interest owed, based on early
payment? The usual answers are no. However, if a monthly payment on a $10 million, quarterly
Equivalence Relations: Single Amounts and Series with PP ⬍ CP
4.7
compounded, bank loan were made early by a large corporation, the corporate financial officer
would likely insist that the bank reduce the amount of interest due, based on early payment.
These are examples of PP < CP. The timing of cash flow transactions between compounding
points introduces the question of how interperiod compounding is handled. Fundamentally,
there are two policies: interperiod cash flows earn no interest, or they earn compound interest.
For a no-interperiod-interest policy, negative cash flows (deposits or payments, depending on the
perspective used for cash flows) are all regarded as made at the end of the compounding period,
and positive cash flows (receipts or withdrawals) are all regarded as made at the beginning.
As an illustration, when interest is compounded quarterly, all monthly deposits are moved to the end
of the quarter (no interperiod interest is earned), and all withdrawals are moved to the beginning (no
interest is paid for the entire quarter). This procedure can significantly alter the distribution of cash
flows before the effective quarterly rate is applied to find P, F, or A. This effectively forces the cash
flows into a PP ⫽ CP situation, as discussed in Sections 4.5 and 4.6. Example 4.11 illustrates this
procedure and the economic fact that, within a one-compounding-period time frame, there is no
interest advantage to making payments early. Of course, noneconomic factors may be present.
EXAMPLE 4.11
Last year AllStar Venture Capital agreed to invest funds in Clean Air Now (CAN), a start-up
company in Las Vegas that is an outgrowth of research conducted in mechanical engineering at
the University of Nevada–Las Vegas. The product is a new filtration system used in the process
of carbon capture and sequestration (CCS) for coal-fired power plants. The venture fund manager
generated the cash flow diagram in Figure 4–9a in $1000 units from AllStar’s perspective. Included are payments (outflows) to CAN made over the first year and receipts (inflows) from CAN
to AllStar. The receipts were unexpected this first year; however, the product has great promise,
and advance orders have come from eastern U.S. plants anxious to become zero-emission
coal-fueled plants. The interest rate is 12% per year, compounded quarterly, and AllStar uses the
no-interperiod-interest policy. How much is AllStar in the “red” at the end of the year?
Receipts from CAN
$90
$90
$120
$45
1
0
0
1
2
3
4
5
6
$75
$100
$150
7
8
9
10
$50
11
12
Year
Month
Payments to CAN
$200
(a)
$180
0
1
0
1
2
3
$165
2
4
5
6
3
7
8
9
4
10
11
12
$50
Quarter
Month
$150
$175
$200
F=?
(b)
Figure 4–9
(a) Actual and (b) moved cash flows (in $1000) for quarterly compounding periods using no interperiod
interest, Example 4.11.
113
114
Nominal and Effective Interest Rates
Chapter 4
Solution
With no interperiod interest considered, Figure 4–9b reflects the moved cash flows. All negative cash flows (payments to CAN) are moved to the end of the respective quarter, and all
positive cash flows (receipts) are moved to the beginning of the respective quarter. Calculate
the F value at 12%兾4 ⫽ 3% per quarter.
F ⫽ 1000[⫺150(F兾P,3%,4) ⫺ 200(F兾P,3%,3) ⫹ (⫺175 ⫹ 180)(F兾P,3%,2)
⫹ 165(F兾P,3%,1) ⫺50]
⫽ $⫺262,111
AllStar has a net investment of $262,111 in CAN at the end of the year.
If PP ⬍ CP and interperiod compounding is earned, then the cash flows are not moved, and the
equivalent P, F, or A values are determined using the effective interest rate per payment period.
The engineering economy relations are determined in the same way as in the previous two sections
for PP ⱖ CP. The effective interest rate formula will have an m value less than 1, because there is
only a fractional part of the CP within one PP. For example, weekly cash flows and quarterly compounding require that m ⫽ 1兾13 of a quarter. When the nominal rate is 12% per year, compounded
quarterly (the same as 3% per quarter, compounded quarterly), the effective rate per PP is
Effective weekly i% ⫽ (1.03)1兾13 ⫺ 1 ⫽ 0.228% per week
4.8 Effective Interest Rate for Continuous
Compounding
If we allow compounding to occur more and more frequently, the compounding period becomes
shorter and shorter and m, the number of compounding periods per payment period, increases.
Continuous compounding is present when the duration of CP, the compounding period, becomes
infinitely small and m, the number of times interest is compounded per period, becomes infinite.
Businesses with large numbers of cash flows each day consider the interest to be continuously
compounded for all transactions.
As m approaches infinity, the effective interest rate Equation [4.7] must be written in a new
form. First, recall the definition of the natural logarithm base.
(
)
1 h ⫽ e ⫽ 2.71828⫹
lim 1 ⫹ —
h
⬁
h
[4.10]
The limit of Equation [4.7] as m approaches infinity is found by using r兾m ⫽ 1兾h, which makes
m ⫽ hr.
r m
lim i ⫽ lim ( 1 ⫹ —
m) ⫺ 1
m
⬁
m
⬁
(
1
⫽ lim 1 ⫹ —
h
h ⬁
)
hr
[(
1
⫺ 1 ⫽ lim 1 ⫹ —
h ⬁
h
i ⴝ er ⴚ 1
) ] ⫺1
h r
[4.11]
Equation [4.11] is used to compute the effective continuous interest rate, when the time periods
on i and r are the same. As an illustration, if the nominal annual r ⫽ 15% per year, the effective
continuous rate per year is
i% ⫽ e0.15 ⫺ 1 ⫽ 16.183%
For convenience, Table 4–3 includes effective continuous rates for the nominal rates listed.
To find an effective or nominal interest rate for continuous compounding using the spreadsheet functions EFFECT or NOMINAL, enter a very large value for the compounding frequency m in Equation [4.5] or [4.6], respectively. A value of 10,000 or higher provides sufficient
accuracy. Both functions are illustrated in Example 4.12.
Effective Interest Rate for Continuous Compounding
4.8
EXAMPLE 4.12
(a) For an interest rate of 18% per year, compounded continuously, calculate the effective
monthly and annual interest rates.
(b) An investor requires an effective return of at least 15%. What is the minimum annual
nominal rate that is acceptable for continuous compounding?
Solution by Hand
(a) The nominal monthly rate is r ⫽ 18%兾12 ⫽ 1.5%, or 0.015 per month. By Equation [4.11], the effective monthly rate is
i% per month ⫽ er − 1 ⫽ e0.015 − 1 ⫽ 1.511%
Similarly, the effective annual rate using r ⫽ 0.18 per year is
i% per year ⫽ er − 1 ⫽ e0.18 − 1 ⫽ 19.722%
(b) Solve Equation [4.11] for r by taking the natural logarithm.
er − 1 ⫽ 0.15
er ⫽ 1.15
ln er ⫽ ln 1.15
r ⫽ 0.13976
Therefore, a rate of 13.976% per year, compounded continuously, will generate an effective 15% per year return. The general formula to find the nominal rate, given the effective
continuous rate i, is r ⫽ ln(1 ⫹ i).
Solution by Spreadsheet
(a) Use the EFFECT function with the nominal monthly rate r ⫽ 1.5% and annual rate
r ⫽ 18% with a large m to display effective i values. The functions to enter on a spreadsheet and the responses are as follows:
Monthly:
Annual:
⫽ EFFECT(1.5%,10000)
⫽ EFFECT(18%,10000)
effective i ⫽ 1.511% per month
effective i ⫽ 19.722% per year
(b) Use the function in Equation [4.6] in the format ⫽ NOMINAL(15%,10000) to display
the nominal rate of 13.976% per year, compounded continuously.
EXAMPLE 4.13
Engineers Marci and Suzanne both invest $5000 for 10 years at 10% per year. Compute the
future worth for both individuals if Marci receives annual compounding and Suzanne receives
continuous compounding.
Solution
Marci: For annual compounding the future worth is
F ⫽ P(F兾P,10%,10) ⫽ 5000(2.5937) ⫽ $12,969
Suzanne: Using Equation [4.11], first find the effective i per year for use in the F兾P factor.
Effective i% ⫽ e0.10 ⫺ 1 ⫽ 10.517%
F ⫽ P(F兾P,10.517%,10) ⫽ 5000(2.7183) ⫽ $13,591
Continuous compounding causes a $622 increase in earnings. For comparison, daily compounding yields an effective rate of 10.516% (F ⫽ $13,590), only slightly less than the
10.517% for continuous compounding.
115
116
Nominal and Effective Interest Rates
Chapter 4
For some business activities, cash flows occur throughout the day. Examples of costs are energy and water costs, inventory costs, and labor costs. A realistic model for these activities is to
increase the frequency of the cash flows to become continuous. In these cases, the economic
analysis can be performed for continuous cash flow (also called continuous funds flow) and the
continuous compounding of interest as discussed above. Different expressions must be derived
for the factors for these cases. In fact, the monetary differences for continuous cash flows relative
to the discrete cash flow and discrete compounding assumptions are usually not large. Accordingly, most engineering economy studies do not require the analyst to utilize these mathematical
forms to make a sound economic decision.
4.9 Interest Rates That Vary over Time
Real-world interest rates for a corporation vary from year to year, depending upon the financial
health of the corporation, its market sector, the national and international economies, forces of
inflation, and many other elements. Loan rates may increase from one year to another. Home
mortgages financed using ARM (adjustable-rate mortgage) interest is a good example. The mortgage rate is slightly adjusted annually to reflect the age of the loan, the current cost of mortgage
money, etc.
When P, F, and A values are calculated using a constant or average interest rate over the life
of a project, rises and falls in i are neglected. If the variation in i is large, the equivalent values
will vary considerably from those calculated using the constant rate. Although an engineering
economy study can accommodate varying i values mathematically, it is more involved computationally to do so.
To determine the P value for future cash flow values (Ft) at different i values (it) for each year t,
we will assume annual compounding. Define
it ⫽ effective annual interest rate for year t
(t ⫽ years 1 to n)
To determine the present worth, calculate the P of each Ft value, using the applicable it, and sum
the results. Using standard notation and the P兾F factor,
P ⴝ F1(P兾F,i1,1) ⴙ F2(P兾F,i1,1)(P兾F,i2,1) ⴙ . . .
ⴙ Fn(P兾F,i1,1)(P兾F,i2,1) . . . (P兾F,in,1)
[4.12]
When only single amounts are involved, that is, one P and one F in the final year n, the last term
in Equation [4.12] is the expression for the present worth of the future cash flow.
P ⴝ Fn(P兾F,i1,1)(P兾F,i2,1) . . . (P兾F,in,1)
[4.13]
If the equivalent uniform series A over all n years is needed, first find P, using either of the last
two equations; then substitute the symbol A for each Ft symbol. Since the equivalent P has been
determined numerically using the varying rates, this new equation will have only one unknown,
namely, A. Example 4.14 illustrates this procedure.
EXAMPLE 4.14
CE, Inc., leases large earth tunneling equipment. The net profit from the equipment for each of
the last 4 years has been decreasing, as shown below. Also shown are the annual rates of return
on invested capital. The return has been increasing. Determine the present worth P and equivalent uniform series A of the net profit series. Take the annual variation of rates of return into
account.
Year
Net Profit
Annual Rate
1
2
3
4
$70,000
7%
$70,000
7%
$35,000
9%
$25,000
10%
117
Chapter Summary
Solution
Figure 4–10 shows the cash flows, rates for each year, and the equivalent P and A. Equation [4.12] is used to calculate P. Since for both years 1 and 2 the net profit is $70,000 and the
annual rate is 7%, the P兾A factor can be used for these 2 years only.
P ⫽ [70(P兾A,7%,2) ⫹ 35(P兾F,7%,2)(P兾F,9%,1)
⫹ 25(P兾F,7%,2)(P兾F,9%,1)(P兾F,10%,1)](1000)
⫽ [70(1.8080) ⫹ 35(0.8013) ⫹ 25(0.7284)](1000)
⫽ $172,816
[4.14]
$70,000
A=?
$35,000
$25,000
0
0
1
2
3
4
1
i = 7%
7%
i = 7%
2
7%
3
9%
4
10%
i = 9%
i = 10%
P=?
$172,816
Figure 4–10
Equivalent P and A values for varying interest rates, Example 4.14.
To determine an equivalent annual series, substitute the symbol A for all net profit values
on the right side of Equation [4.14], set it equal to P ⫽ $172,816, and solve for A. This
equation accounts for the varying i values each year. See Figure 4–10 for the cash flow diagram
transformation.
$172,816 ⫽ A[(1.8080) ⫹ (0.8013) ⫹ (0.7284)] ⫽ A[3.3377]
A ⫽ $51,777 per year
Comment
If the average of the four annual rates, that is, 8.25% is used, the result is A ⫽ $52,467. This is
a $690 per year overestimate of the equivalent annual net profit.
When there is a cash flow in year 0 and interest rates vary annually, this cash flow must be
included to determine P. In the computation for the equivalent uniform series A over all years,
including year 0, it is important to include this initial cash flow at t ⫽ 0. This is accomplished by
inserting the factor value for (P兾F,i0,0) into the relation for A. This factor value is always 1.00. It
is equally correct to find the A value using a future worth relation for F in year n. In this case, the
A value is determined using the F兾P factor, and the cash flow in year n is accounted for by including the factor (F兾P,in,0) ⫽ 1.00.
CHAPTER SUMMARY
Since many real-world situations involve cash flow frequencies and compounding periods other
than 1 year, it is necessary to use nominal and effective interest rates. When a nominal rate r is
stated, the effective interest rate i per payment period is determined by using the effective interest
rate equation.
r m
Effective i ⴝ ( 1 ⴙ —
m) ⴚ 1
Year
118
Chapter 4
Nominal and Effective Interest Rates
The m is the number of compounding periods (CP) per interest period. If interest compounding
becomes more and more frequent, then the length of a CP approaches zero, continuous compounding results, and the effective i is er ⴚ 1.
All engineering economy factors require the use of an effective interest rate. The i and n values placed in a factor depend upon the type of cash flow series. If only single amounts (P and F)
are present, there are several ways to perform equivalence calculations using the factors. However, when series cash flows (A, G, and g) are present, only one combination of the effective rate
i and number of periods n is correct for the factors. This requires that the relative lengths of PP
and CP be considered as i and n are determined. The interest rate and payment periods must have
the same time unit for the factors to correctly account for the time value of money.
From one year (or interest period) to the next, interest rates will vary. To accurately perform
equivalence calculations for P and A when rates vary significantly, the applicable interest rate
should be used, not an average or constant rate.
PROBLEMS
Nominal and Effective Rates
4.1 From the interest statement 18% per year, compounded monthly, determine the values for interest
period, compounding period, and compounding
frequency.
4.2 From the interest statement 1% per month, determine the values for interest period, compounding
period, and compounding frequency.
4.3 Determine the number of times interest would be
compounded in 6 months from the interest statements (a) 18% per year, compounded monthly,
(b) 1% per month, and (c) 2% per quarter.
4.4 For an interest rate of 1% per 2 months, determine
the number of times interest would be compounded
in (a) 2 months, (b) two semiannual periods, and
(c) 3 years.
4.5 Identify the compounding period for the following
interest statements: (a) 3% per quarter; (b) 10%
per year, compounded semiannually; (c) nominal
7.2% per year, compounded monthly; (d) effective
3.4% per quarter, compounded weekly; and (e) 2%
per month, compounded continuously.
Given Interest Rate
Desired Interest Rate
1% per month
3% per quarter
2% per quarter
0.28% per week
6.1% per 6 months
Nominal rate per year
Nominal rate per 6 months
Nominal rate per year
Nominal rate per quarter
Nominal rate per 2 years
4.8 What nominal interest rate per year is equivalent to
11.5% per year, compounded monthly?
4.9 For a Federal Credit Union that offers an interest
rate of 8% per year, compounded quarterly, determine the nominal rate per 6 months.
4.10 The Second National Bank of Fullertum advertises
an APR of 14% compounded monthly for student
loans. Determine the APY. Show hand and spreadsheet solutions.
4.11 For an effective annual rate ia of 15.87% compounded quarterly, determine (a) the effective
quarterly rate and (b) the nominal annual rate.
(c) What is the spreadsheet function to find the
nominal annual rate above?
4.6 Identify the following interest rate statements as
either nominal or effective: (a) 1.5% per month,
compounded daily; (b) 17% per year, compounded
quarterly; (c) effective 15% per year, compounded
monthly; (d) nominal 0.6% per month, compounded weekly; (e) 0.3% per week, compounded
weekly; and (f) 8% per year.
4.12 High-tech companies such as IBM, AMD, and
Intel have been using nanotechnology for several
years to make microchips with faster speeds while
using less power. A less well-known company in
the chip business has been growing fast enough
that the company uses a minimum attractive rate of
return of 60% per year. If this MARR is an effective annual rate compounded monthly, determine
the effective monthly rate.
4.7 Convert the given interest rates in the left-hand
column into the nominal rates listed in the righthand column. (Assume 4 weeks兾month.)
4.13 An interest rate of 21% per year, compounded
every 4 months, is equivalent to what effective rate
per year? Show hand and spreadsheet solutions.
Problems
4.14 An interest rate of 8% per 6 months, compounded
monthly, is equivalent to what effective rate per
quarter?
4.15 A small company that makes modular bevel gear
drives with a tight swing ratio for optimizing pallet
truck design was told that the interest rate on a
mortgage loan would be an effective 4% per quarter, compounded monthly. The owner was confused by the terminology and asked you to help.
What are (a) the APR and (b) the APY?
4.16 In ‘N Out Payday Loans advertises that for a fee of
only $10, you can immediately borrow up to $200
for one month. If a person accepts the offer, what
are (a) the nominal interest rate per year and
(b) the effective rate per year?
4.17 A government-required truth-in-lending document
showed that the APR was 21% and the APY was
22.71%. Determine the compounding frequency at
which the two rates are equivalent.
4.18 Julie has a low credit rating, plus she was furloughed from her job 2 months ago. She has a new
job starting next week and expects a salary to start
again in a couple of weeks. Since she is a little
short on money to pay her rent, she decided to borrow $100 from a loan company, which will charge
her only $10 interest if the $110 is paid no more
than 1 week after the loan is made. What are the
(a) nominal annual and (b) effective annual interest rates that she will pay on this loan?
Equivalence When PP ⱖ CP
4.19 Assume you deposit 25% of your monthly check
of $5500 into a savings account at a credit union
that compounds interest semiannually. (a) What
are the payment and compounding periods? (b) Is
the payment period greater than or less than the
compounding period?
4.20 Interest is compounded quarterly, and singlepayment cash flows (that is, F and P) are separated
by 5 years. What must the compounding period be
on the interest rate, if the value of n in the P兾F or
F兾P equation is (a) n ⫽ 5, (b) n ⫽ 20, or (c) n ⫽ 60?
4.21 When interest is compounded quarterly and a
uniform series cash flow of $4000 occurs every
6 months, what time periods on i and n must be
used?
4.22 Pinpoint Laser Systems is planning to set aside
$260,000 now for possibly replacing its 4-pole,
38-MW synchronous motors when it becomes
necessary. If the replacement is expected to take
119
place in 3 years, how much will the company have
in its investment set-aside account? Assume the
company can achieve a rate of return of 12% per
year, compounded quarterly.
4.23 Wheeling-Pittsburgh Steel is investigating whether
it should replace some of its basic oxygen furnace
equipment now or wait to do it later. The cost
later (i.e., 3 years from now) is estimated to be
$1,700,000. How much can the company afford to
spend now, if its minimum attractive rate of return
is 1.5% per month?
4.24 How much can Wells Fargo lend to a developer
who will repay the loan by selling 6 view lots at
$190,000 each 2 years from now? Assume the
bank will lend at a nominal 14% per year, compounded semiannually.
4.25 How much will be in a high-yield account at the
National Bank of Arizona 12 years from now if
you deposit $5000 now and $7000 five years from
now? The account earns interest at a rate of 8% per
year, compounded quarterly.
4.26 Loadstar Sensors is a company that makes load兾
force sensors based on capacitive sensing technology. The company wants to have $28 million for a
plant expansion 4 years from now. If the company
has already set aside $12 million in an investment
account for the expansion, how much more must
the company add to the account next year (i.e.,
1 year from now) so that it will have the $28 million 4 years from now? The account earns interest
at 12% per year, compounded quarterly.
4.27 A structural engineering consulting company is
examining its cash flow requirements for the next
6 years. The company expects to replace office
machines and computer equipment at various
times over the 6-year planning period. Specifically, the company expects to spend $21,000 two
years from now, $24,000 three years from now,
and $10,000 five years from now. What is the present worth of the planned expenditures at an interest
rate of 10% per year, compounded semiannually?
4.28 Irvin Aerospace of Santa Ana, California, was
awarded a 5-year contract to develop an advanced
space capsule airbag landing attenuation system
for NASA’s Langley Research Center. The company’s computer system uses fluid structure interaction modeling to test and analyze each airbag
design concept. What is the present worth of the
contract at 16% per year, compounded quarterly, if
the quarterly cost in years 1 through 5 is $2 million
per quarter?
120
Chapter 4
4.29 Heyden Motion Solutions ordered $7 million
worth of seamless tubes for its drill collars from
the Timken Company of Canton, Ohio. (A drill
collar is the heavy tubular connection between a
drill pipe and a drill bit.) At 12% per year, compounded semiannually, what is the equivalent uniform cost per semiannual period over a 5-year
amortization period?
4.30 In 2010, the National Highway Traffic Safety Administration raised the average fuel efficiency standard to 35.5 miles per gallon (mpg) for cars and
light trucks by the year 2016. The rules will cost
consumers an average of $926 extra per vehicle in
the 2016 model year. Assume Yolanda will purchase a new car in 2016 and plans to keep it for
5 years. How much will the monthly savings in the
cost of gasoline have to be to recover Yolanda’s
extra cost? Use an interest rate of 0.75% per month.
4.31 Fort Bliss, a U.S. Army military base, contributed
$3.3 million of the $87 million capital cost for a
desalting plant constructed and operated by El Paso
Water Utilities (EPWU). In return, EPWU agreed
to sell water to Fort Bliss at $0.85 per 1000 gallons
for 20 years. If the army base uses 200 million gallons of water per month, what is the Army’s cost
per month for water? The $3.3 million capital cost
is amortized at an interest rate of 6% per year,
compounded monthly.
4.32 Beginning in 2011, city hall, administrative offices, and municipal courts in the city of El Paso,
Texas, will go on a 10 hour兾day, 4-day workweek
from the beginning of May through the end of September. The shortened workweek will affect 25%
of the city’s 6100 employees and will save $42,600
per month for those 5 months, because of reduced
fuel, utility, and janitorial expenses. If this work
schedule continues for the next 10 years, what is
the future worth of the savings at the end of that
time (i.e., end of year 2020)? Use an interest rate
of 0.5% per month.
4.33 In October 2009, Wal-Mart started selling caskets
on its website that undercut many funeral homes.
Prices ranged from $999 for steel models such as
Dad Remembered to $3199 for the Sienna Bronze
casket. Part of the business model is to get people
to plan ahead, so the company is allowing people
to pay for the caskets over a 12-month period with
no interest. An individual purchased a Sienna
Bronze casket and made 12 equal monthly payments (in months 1 through 12) at no interest. How
much did this person save each month compared to
another person who paid an interest rate of 6% per
year, compounded monthly?
Nominal and Effective Interest Rates
4.34 NRG Energy plans to construct a giant solar plant
in Santa Teresa, New Mexico, to supply electricity
to West Texas Electric. The plant will have 390,000
heliostats to concentrate sunlight onto 32 water
towers to generate steam. It will provide enough
power for 30,000 homes. If the company spends
$28 million per month for 2 years in constructing
the plant, how much will the company have to
make each month in years 3 through 22 (that is,
20 years) to recover its investment plus 18% per
year, compounded monthly?
4.35 Many college students have Visa credit cards that
carry an interest rate of “simple 24% per year”
(that is, 2% per month). When the balance on
such a card is $5000, the minimum payment is
$110.25.
(a) What is the amount of interest in the first
payment?
(b) How long will it take, in months, to pay off
the balance, if the cardholder continues to
make payments of $110.25 per month and
adds no other charges to the card?
4.36 Bart is an engineering graduate who did not take
the engineering economy elective during his B.S.
degree course work. After working for a year or so,
he found himself in financial trouble, and he borrowed $500 from a friend in the finance department at his office. Bart agreed to repay the loan
principal plus $75 interest 1 month later. The two
got separated doing different jobs, and 1 year went
by. The friend e-mailed Bart after exactly 1 year
and asked for the loan repayment plus the interest
with monthly compounding, since Bart had made
no effort to repay the loan during the year.
(a) What does Bart now owe his friend? (b) What
effective annual interest rate did Bart pay on this
$500 loan?
4.37 AT&T announced that the early termination fee for
smart phones will jump from $175 to $375. The
fee will be reduced by $10 each month of the
2-year contract. Wireless carriers justify the fees
by pointing out that the cost of a new phone is
heavily discounted from what the carrier pays the
manufacturer. Assume AT&T pays $499 for an
iPhone that it sells for $199. How much profit
would the company have to make each month (i.e.,
prior to the termination), if it wanted to make a rate
of return of 1.5% per month on its $300 investment in a customer who terminates the contract
after 12 months?
4.38 What is the future worth of a present cost of
$285,000 to Monsanto, Inc. 5 years from now at an
interest rate of 2% per month?
Problems
4.39 Environmental recovery company RexChem Partners plans to finance a site reclamation project
that will require a 4-year cleanup period. The
company plans to borrow $3.6 million now. How
much will the company have to get in a lump-sum
payment when the project is over in order to earn
24% per year, compounded quarterly, on its investment?
4.40 For the cash flows shown below, determine the
equivalent uniform worth in years 1 through 5 at an
interest rate of 18% per year, compounded monthly.
Year
1
2
3
4
5
Cash Flow, $
200,000
0
350,000
0
400,000
4.41 For the cash flow diagram shown, solve for F,
using an interest rate of 1% per month.
F=?
i = 1% per month
0
1
2
3
4
5
$30
$30
$30
$30
$30
$30
6
7
$50
$50
8 Years
$50
121
Lee antiscalant) for use at its nanofiltration water
conditioning plant. The cost of the pumps was
$950 each. If the chemical cost is $11 per day,
determine the equivalent cost per month at an interest rate of 12% per year, compounded monthly.
Assume 30 days per month and a 3-year pump
life.
4.46 Income from recycling paper and cardboard at the
U.S. Army's Fort Benning Maneuver Center has
averaged $3000 per month for 2½ years. What is
the future worth of the income (after the 2½ years)
at an interest rate of 6% per year, compounded
quarterly? Assume there is no interperiod compounding.
4.47 The Autocar E3 refuse truck has an energy recovery system developed by Parker Hannifin
LLC that is expected to reduce fuel consumption
by 50%. Pressurized fluid flows from carbon
fiber-reinforced accumulator tanks to two hydrostatic motors that propel the vehicle forward.
(The truck recharges the accumulators when it
brakes.) The fuel cost for a regular refuse truck is
$900 per month. How much can a private wastehauling company afford to spend now on the
recovery system, if it wants to recover its investment in 3 years plus a return of 14% per year,
compounded semiannually? Assume no interperiod compounding.
4.42 According to the Government Accountability Office (GAO), if the U.S. Postal Service does not
change its business model, it will lose $480 million
next month and $500 million the month after that,
and the losses will increase by $20 million per
month for the next 10 years. At an interest rate of
0.25% per month, what is the equivalent uniform
amount per month of the losses through year 10?
Continuous Compounding
4.43 Equipment maintenance costs for manufacturing
explosion-proof pressure switches are projected to
be $100,000 in year 1 and increase by 4% each
year through year 5. What is the present worth of
the maintenance costs at an interest rate of 10%
per year, compounded quarterly?
4.50 What nominal rate per month is equivalent to an
effective 1.3% per month, compounded continuously?
Equivalence When PP ⬍ CP
4.44 If you deposit $1000 per month into an investment
account that pays interest at a rate of 6% per year,
compounded quarterly, how much will be in the
account at the end of 5 years? There is no interperiod compounding.
4.45 Freeport McMoran purchased two model MTVS
peristaltic pumps (to inject sulfuric acid and King
4.48 What effective interest rate per year is equal to
1.2% per month, compounded continuously? Show
hand and spreadsheet solutions.
4.49 What effective interest rate per quarter is equal to
a nominal 1.6% per month, compounded continuously?
4.51 Companies such as GE that have huge amounts of
cash flow every day base their financial calculations
on continuous compounding. If the company wants
to make an effective 25% per year, compounded
continuously, what nominal daily rate of return has
to be realized? Assume 365 days per year.
4.52 U.S. Steel is planning a plant expansion that is expected to cost $13 million. How much money must
the company set aside now in a lump-sum investment to have the money in 2 years? Capital funds
earn interest at a rate of 12% per year, compounded
continuously.
122
Nominal and Effective Interest Rates
Chapter 4
4.53 Periodic outlays for inventory control software at
Baron Chemicals are expected to be $150,000
immediately, $200,000 in 1 year, and $350,000 in
2 years. What is the present worth of the costs at
an interest rate of 10% per year, compounded
continuously?
nanowires can be extruded by blasting the carbon
nanotubes with an electron beam. If Gentech Technologies plans to spend $1.7 million in year 1,
$2.1 million in year 2, and $3.4 million in year 3 to
develop the technology, determine the present
worth of the investments in year 0, if the interest
rate in year 1 is 10% and in years 2 and 3 it is 12%
per year.
Varying Interest Rates
4.54 Many small companies use accounts receivable as
collateral to borrow money for continuing operations and meeting payrolls. If a company borrows
$300,000 now at an interest rate of 1% per month,
but the rate changes to 1.25% per month after
4 months, how much will the company owe at the
end of 1 year?
4.57 Find (a) the present worth P, and (b) the equivalent uniform annual worth A for the cash flows
shown below.
P=?
i = 10%
4.55 The maintenance cost for furnaces at a copper
smelting plant have been constant at $140,000 per
year for the past 5 years. If the interest rate was 8%
per year for the first 3 years and then it increased to
10% in years 4 and 5, what is the equivalent future
worth (in year 5) of the maintenance cost? Show
hand and spreadsheet solutions.
0
i = 14%
1
2
3
4
5
$100
$100
$100
$100
$100
4.56 By filling carbon nanotubes with miniscule wires
made of iron and iron carbide, incredibly thin
6
7
8
$160
$160
$160
Year
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
4.58 The term annual percentage rate is the same as:
(a) Effective rate
(b) Nominal rate
(c) Annual percentage yield
(d) All of the above
4.59 An interest rate is an effective rate under all of the
following conditions, except when:
(a) The compounding period is not stated
(b) The interest period and compounding period
are the same
(c) The interest statement says that the interest
rate is effective
(d) The interest period is shorter than the compounding period
4.60 An interest rate of nominal 12% per year, compounded weekly, is:
(a) An effective rate per year
(b) An effective rate per week
(c) A nominal rate per year
(d) A nominal rate per week
4.61 An interest rate of 1.5% per month, compounded
continuously, is the same as:
(a) An effective 1.5% per month
(b) 4.5% per quarter, compounded continuously
(c)
(d)
6.0% per quarter, compounded continuously
9% per 6 months
4.62 An effective 12.68% per year, compounded monthly,
is the closest to:
(a) 12% per year
(b) 12% per year, compounded annually
(c) 1% per month
(d) 1% per month, compounded annually
4.63 For an interest rate of 2% per month, the effective
semiannual interest rate is:
(a) 2.02%
(b) 12.005%
(c) 12.31%
(d) 12.62%
4.64 If you make quarterly deposits for 3 years into an
account that compounds interest at 1% per month,
the value of n in the F兾A factor that will determine
F at the end of the 3-year period is:
(a) 3
(b) 12
(c) 36
(d) None of these
123
Additional Problems and FE Exam Review Questions
4.65 Identify the following interest rates as nominal or
effective.
Rate 1:
1.5% per quarter
Rate 2:
1.5% per quarter, compounded
monthly
(a) Both are nominal rates.
(b) Rate 1 is nominal and rate 2 is effective.
(c) Rate 1 is effective and rate 2 is nominal.
(d) Both are effective.
4.66 In solving uniform series problems, n in the standard factor notation equation is equal to the
number of arrows (i.e., cash flows) in the original
cash flow diagram when:
(a) The payment period (PP) is longer than the
compounding period (CP).
(b) The compounding period is equal to the payment period.
(c) Both (a) and (b) are correct.
(d) The compounding period is longer than the
payment period.
4.67 An engineer who is saving for her retirement plans
to deposit $500 every quarter, starting one quarter
from now, into an investment account. If the account pays interest at 6% per year, compounded
semiannually, the total she will have at the end of
25 years is closest to:
(a) $50,000
(b) $56,400
(c) $79,700
(d) $112,800
4.68 A company that makes flange-mount, motorized rotary potentiometers expects to spend $50,000 for a
certain machine 4 years from now. At an interest rate
of 12% per year, compounded quarterly, the present
worth of the machine’s cost is represented by the following equation:
(a) P ⫽ 50,000(P兾F, 3%, 16)
(b) P ⫽ 50,000(P兾F, effective i兾6 months, 8)
(c) P ⫽ 50,000(P兾F, effective i兾year, 4)
(d) Any of the above
4.69 For the cash flow diagram shown, the unit of the
payment period (PP) is:
(a) Months
(b) Quarters
(c) Semiannual (d) Years
F=?
i = 1% per month
0
1
2
3
4
5
6
7
8
1
9
10
11
2
12 Quarter
3
Year
$400 $400
$420
$440
$460
$480
$500
4.70 A small company plans to spend $10,000 in year 2
and $10,000 in year 5. At an interest rate of effective
10% per year, compounded semiannually, the equation that represents the equivalent annual worth A in
years 1 through 5 is:
(a) A ⫽ 10,000(P兾F,10%,2)(A兾P,10%,5)
⫹ 10,000(A兾F,10%,5)
(b) A ⫽ 10,000(A兾P,10%,4) ⫹ 10,000(A兾F,10%,5)
(c) A ⫽ 10,000(P兾F,5%,2)(A兾P,5%,10)
⫹ 10,000(A兾F,5%,10)
(d) A ⫽ [10,000(F兾P,10%,5)
⫹ 10,000](A兾F,10%,5)
4.71 Assume you make monthly deposits of $200 starting one month from now into an account that pays
6% per year, compounded semiannually. If you
want to know the total after 4 years, the value of n
you should use in the F兾A factor is:
(a) 2
(b) 4
(c) 8
(d) 12
4.72 The cost of replacing part of a cell phone videochip production line in 6 years is estimated to be
$500,000. At an interest rate of 14% per year,
compounded semiannually, the uniform amount
that must be deposited into a sinking fund every
6 months is closest to:
(a) $21,335
(b) $24,825
(c) $27,950
(d) $97,995
124
Nominal and Effective Interest Rates
Chapter 4
CASE STUDY
IS OWNING A HOME A NET GAIN OR NET LOSS OVER TIME?
Background
The Carroltons are deliberating whether to purchase a house
or continue to rent for the next 10 years. They are assured by
both of their employers that no transfers to new locations will
occur for at least this number of years. Plus, the high school
that their children attend is very good for their college prep
education, and they all like the neighborhood where they
live now.
They have a total of $40,000 available now and estimate
that they can afford up to $2850 per month for the total house
payment.
If the Carroltons do not buy a house, they will continue
to rent the house they currently occupy for $2700 per
month. They will also place the $40,000 into an investment
instrument that is expected to earn at the rate of 6% per
year. Additionally, they will add to this investment at the
end of each year the same amount as the monthly 15-year
mortgage payments. This alternative is called the rent–
don’t buy plan.
Information
Two financing plans using fixed-rate mortgages are currently
available. The details are as follows.
Plan
Description
A
30-year fixed rate of 5.25% per year interest;
10% down payment
15-year fixed rate of 5.0% per year interest;
10% down payment
B
Other information:
• Price of the house is $330,000.
• Taxes and insurance (T&I) are $500 per month.
• Up-front fees (origination fee, survey fee, attorney’s fee,
etc.) are $3000.
Any money not spent on the down payment or monthly payment will be invested and return at a rate of 6% per year
(0.5% per month).
The Carroltons anticipate selling the house after 10 years
and plan for a 10% increase in price, that is, $363,000 (after
all selling expenses are paid)
Case Study Exercises
1. The 30-year fixed-rate mortgage (plan A) is analyzed
below. No taxes are considered on proceeds from the
savings or investments.
Perform a similar analysis for the 15-year loan
(plan B) and the rent–don’t buy plan. The Carroltons
decided to use the largest future worth after 10 years to
select the best of the plans. Do the analysis for them and
select the best plan.
Plan A analysis: 30-year fixed-rate loan
Amount of money required for closing costs:
Down payment (10% of $330,000)
Up-front fees (origination fee, attorney’s
fee, survey, filing fee, etc.)
Total
$33,000
3,000
$36,000
The amount of the loan is $297,000, and equivalent
monthly principal and interest (P&I) is determined at
5.25%兾12 ⫽ 0.4375% per month for 30(12) ⫽ 360 months.
A ⫽ 297,000(A兾P,0.4375%,360) ⫽ 297,000(0.005522)
⫽ $1640
Add the T&I of $500 for a total monthly payment of
PaymentA ⫽ $2140 per month
The future worth of plan A is the sum of three future
worth components: remainder of the $40,000 available
for the closing costs (F1A); left-over money from that
available for monthly payments (F2A); and increase in
the house value when it is sold after 10 years (F3A).
These are calculated here.
F1A ⫽ (40,000 ⫺ 36,000)(F兾P,0.5%,120)
⫽ $7278
Money available each month to invest after the mortgage payment, and the future worth after 10 years is
2850 ⫺ 2140 ⫽ $710
F2A ⫽ 710(F兾A,0.5%,120)
⫽ $116,354
Case Study
Net money from the sale in 10 years (F3A) is the difference between the net selling price ($363,000) and the
remaining balance on the loan.
Loan balance
⫽ 297,000(F兾P,0.4375%,120)
⫺ 1640(F兾A,0.4375%,120)
⫽ 297,000(1.6885) ⫺ 1640(157.3770)
⫽ $243,386
F3A ⫽ 363,000 ⫺ 243,386 ⫽ $119,614
125
Total future worth of plan A is
F A ⴝ F 1A ⴙ F 2A ⴙ F 3A
ⴝ 7278 ⴙ 116,354 ⴙ 119,614
ⴝ $243,246
2. Perform this analysis if all estimates remain the same,
except that when the house sells 10 years after purchase,
the bottom has fallen out of the housing market and the
net selling price is only 70% of the purchase price, that
is, $231,000.
L E A R N I N G S TA G E 2
Basic Analysis Tools
LEARNING STAGE 2
Basic Analysis
Tools
CHAPTER
5
Present Worth
Analysis
CHAPTER
6
Annual Worth
Analysis
CHAPTER
7
Rate of Return
Analysis: One Project
CHAPTER
8
Rate of Return
Analysis: Multiple
Alternatives
CHAPTER
9
Benefit兾Cost Analysis
and Public Sector
Economics
A
n engineering project or alternative is formulated to make or
purchase a product, to develop a process, or to provide a
service with specified results. An engineering economic
analysis evaluates cash flow estimates for parameters such as initial
cost, annual costs and revenues, nonrecurring costs, and possible
salvage value over an estimated useful life of the product; process,
or service. The chapters in this Learning Stage develop and demonstrate the basic tools and techniques to evaluate one or more alternatives using the factors, formulas, and spreadsheet functions
learned in Stage 1.
After completing these chapters, you will be able to evaluate
most engineering project proposals using a well-accepted economic
analysis technique, such as present worth, future worth, capitalized
cost, life-cycle costing, annual worth, rate of return, or benefit /cost
analysis.
The epilogue to this stage provides an approach useful in selecting the engineering economic method that will provide the best
analysis for the estimates and conditions present once the mutually
exclusive alternatives are defined.
Important note: If depreciation and兾or after-tax analysis is to be
considered along with the evaluation methods in Chapters 5 through
9, Chapter 16 and/or Chapter 17 should be covered, preferably after
Chapter 6.
CHAPTER 5
Present Worth
Analysis
L E A R N I N G
O U T C O M E S
Purpose: Utilize different present worth techniques to evaluate and select alternatives.
SECTION
TOPIC
LEARNING OUTCOME
5.1
Formulate alternatives
• Identify mutually exclusive and independent
projects; define revenue and cost alternatives.
5.2
PW of equal-life alternatives
• Select the best of equal-life alternatives using
present worth analysis.
5.3
PW of different-life alternatives
• Select the best of different-life alternatives
using present worth analysis.
5.4
FW analysis
• Select the best alternative using future worth
analysis.
5.5
CC analysis
• Select the best alternative using capitalized cost
(CC) analysis.
A
future amount of money converted to its equivalent value now has a present worth
(PW) that is always less than that of the future cash flow, because all P兾F factors have
a value less than 1.0 for any interest rate greater than zero. For this reason, present
worth values are often referred to as discounted cash flows (DCF), and the interest rate is referred to as the discount rate. Besides PW, two other terms frequently used are present value
(PV ) and net present value (NPV ). Up to this point, present worth computations have been
made for one project or alternative. In this chapter, techniques for comparing two or more
mutually exclusive alternatives by the present worth method are treated. Two additional applications are covered here—future worth and capitalized cost. Capitalized costs are used for
projects with very long expected lives or long planning horizons.
To understand how to organize an economic analysis, this chapter begins with a description of independent and mutually exclusive projects as well as revenue and cost
alternatives.
PE
Water for Semiconductor Manufacturing Case: The worldwide contribution of
semiconductor sales is about $250 billion
per year, or about 10% of the world’s
GDP (gross domestic product). This industry produces the microchips used in many
of the communication, entertainment,
transportation, and computing devices
we use every day. Depending upon the
type and size of fabrication plant (fab),
the need for ultrapure water (UPW) to
manufacture these tiny integrated circuits
is high, ranging from 500 to 2000 gpm
(gallons per minute). Ultrapure water is
obtained by special processes that commonly include reverse osmosis兾deionizing
resin bed technologies. Potable water
obtained from purifying seawater or
brackish groundwater may cost from
$2 to $3 per 1000 gallons, but to obtain
UPW on-site for semiconductor manufacturing may cost an additional $1 to $3 per
1000 gallons.
A fab costs upward of $2.5 billion to
construct, with approximately 1% of this
total, or $25 million, required to provide
the ultrapure water needed, including
the necessary wastewater and recycling
equipment.
A newcomer to the industry, Angular
Enterprises, has estimated the cost profiles for two options to supply its anticipated fab with water. It is fortunate to
have the option of desalinated seawater
or purified groundwater sources in the
location chosen for its new fab. The initial cost estimates for the UPW system are
given below.
Source
Equipment first
cost, $M
AOC, $M per year
Seawater
(S)
⫺20
Groundwater
(G)
⫺22
⫺0.5
⫺0.3
Salvage value, % of
first cost
5
10
Cost of UPW, $ per
1000 gallons
4
5
Angular has made some initial estimates
for the UPW system.
Life of UPW equipment 10 years
UPW needs
1500 gpm
Operating time
16 hours per
day for 250 days
per year
This case is used in the following topics
(Sections) and problems of this chapter:
PW analysis of equal-life alternatives
(Section 5.2)
PW analysis of different-life alternatives (Section 5.3)
Capitalized cost analysis (Section 5.5)
Problems 5.20 and 5.34
5.1 Formulating Alternatives
The evaluation and selection of economic proposals require cash flow estimates over a stated
period of time, mathematical techniques to calculate the measure of worth (review Example 1.2 for possible measures), and a guideline for selecting the best proposal. From all the
Present Worth Analysis
Chapter 5
Mandates
ion
as
at
orm
Id
e
130
Inf
Experience
Estimates
Plans
Proposals
Not viable
1
B
Not viable
2
D
m
C
E
Either
of these
Mutually
exclusive
alternatives
1
2
m
+
DN
Nature
of proposals
Independent
projects
Select
all
justified
DN
1
2
…
…
Select
only
one
…
A
Viable
DN = do nothing
m
Types of cash flow estimates
* Costs only
* Revenues and costs
Revenue alternative
Cost alternative
Perform evaluation and make selection
Figure 5–1
Progression from proposals to economic evaluation to selection.
proposals that may accomplish a stated purpose, the alternatives are formulated. This progression is detailed in Figure 5–1. Up front, some proposals are viable from technological, economic, and兾or legal perspectives; others are not viable. Once the obviously nonviable ideas
are eliminated, the remaining viable proposals are fleshed out to form the alternatives to be
evaluated. Economic evaluation is one of the primary means used to select the best
alternative(s) for implementation.
The nature of the economic proposals is always one of two types:
Mutually exclusive alternatives: Only one of the proposals can be selected. For terminology
purposes, each viable proposal is called an alternative.
Independent projects: More than one proposal can be selected. Each viable proposal is called a
project.
The do-nothing (DN) proposal is usually understood to be an option when the evaluation is
performed.
5.2
131
Present Worth Analysis of Equal-Life Alternatives
The DN alternative or project means that the current approach is maintained; nothing new
is initiated. No new costs, revenues, or savings are generated.
Do nothing
If it is absolutely required that one or more of the defined alternatives be selected, do nothing is
not considered. This may occur when a mandated function must be installed for safety, legal,
government, or other purposes.
Mutually exclusive alternatives and independent projects are selected in completely different ways. A mutually exclusive selection takes place, for example, when an engineer must select the best diesel-powered engine from several available models. Only one is chosen, and the
rest are rejected. If none of the alternatives are economically justified, then all can be rejected
and, by default, the DN alternative is selected. For independent projects one, two or more, in
fact, all of the projects that are economically justified can be accepted, provided capital funds
are available. This leads to the two following fundamentally different evaluation bases:
Mutually exclusive alternatives compete with one another and are compared pairwise.
Independent projects are evaluated one at a time and compete only with the DN project.
Any of the techniques in Chapters 5 through 9 can be used to evaluate either type of proposal—
mutually exclusive or independent. When performed correctly as described in each chapter, any
of the techniques will reach the same conclusion of which alternative or alternatives to select.
This chapter covers the present worth method.
A parallel can be developed between independent and mutually exclusive evaluation. Assume
there are m independent projects. Zero, one, two, or more may be selected. Since each project may
be in or out of the selected group of projects, there are a total of 2m mutually exclusive alternatives. This number includes the DN alternative, as shown in Figure 5–1. For example, if the engineer has three diesel engine models (A, B, and C) and may select any number of them, there are
23 ⫽ 8 alternatives: DN, A, B, C, AB, AC, BC, ABC. Commonly, in real-world applications, there
are restrictions, such as an upper budgetary limit, that eliminate many of the 2m alternatives. Independent project analysis without budget limits is discussed in this chapter and through Chapter 9.
Chapter 12 treats independent projects with a budget limitation; this is called capital budgeting.
Finally, it is important to recognize the nature of the cash flow estimates before starting the
computation of a measure of worth that leads to the final selection. Cash flow estimates determine whether the alternatives are revenue- or cost-based. All the alternatives or projects must be
of the same type when the economic study is performed. Definitions for these types follow:
Revenue: Each alternative generates cost (cash outflow) and revenue (cash inflow) estimates,
and possibly savings, also considered cash inflows. Revenues can vary for each alternative.
Cost: Each alternative has only cost cash flow estimates. Revenues or savings are assumed
equal for all alternatives; thus they are not dependent upon the alternative selected. These are
also referred to as service alternatives.
Revenue or cost
alternative
Although the exact procedures vary slightly for revenue and cost cash flows, all techniques and
guidelines covered through Chapter 9 apply to both. Differences in evaluation methodology are
detailed in each chapter.
5.2 Present Worth Analysis of Equal-Life Alternatives
The PW comparison of alternatives with equal lives is straightforward. The present worth P is
renamed PW of the alternative. The present worth method is quite popular in industry because all
future costs and revenues are transformed to equivalent monetary units NOW; that is, all future
cash flows are converted (discounted) to present amounts (e.g., dollars) at a specific rate of return,
which is the MARR. This makes it very simple to determine which alternative has the best economic advantage. The required conditions and evaluation procedure are as follows:
If the alternatives have the same capacities for the same time period (life), the equal-service
requirement is met. Calculate the PW value at the stated MARR for each alternative.
Equal-service
requirement
132
Present Worth Analysis
Chapter 5
For mutually exclusive (ME) alternatives, whether they are revenue or cost alternatives, the following guidelines are applied to justify a single project or to select one from several alternatives.
Project evaluation
ME alternative
selection
One alternative: If PW ⱖ 0, the requested MARR is met or exceeded and the alternative is
economically justified.
Two or more alternatives: Select the alternative with the PW that is numerically largest,
that is, less negative or more positive. This indicates a lower PW of cost for cost alternatives
or a larger PW of net cash flows for revenue alternatives.
Note that the guideline to select one alternative with the lowest cost or highest revenue uses
the criterion of numerically largest. This is not the absolute value of the PW amount, because
the sign matters. The selections below correctly apply the guideline for two alternatives A and B.
PWA
PWB
Selected Alternative
$⫺2300
⫺500
⫹2500
⫹4800
$⫺1500
⫹1000
⫹2000
⫺400
B
B
A
A
For independent projects, each PW is considered separately, that is, compared with the DN
project, which always has PW ⫽ 0. The selection guideline is as follows:
Independent project
selection
One or more independent projects: Select all projects with PW ⱖ 0 at the MARR.
The independent projects must have positive and negative cash flows to obtain a PW value that
can exceed zero; that is, they must be revenue projects.
All PW analyses require a MARR for use as the i value in the PW relations. The bases used
to establish a realistic MARR were summarized in Chapter 1 and are discussed in detail in
Chapter 10.
EXAMPLE 5.1
A university lab is a research contractor to NASA for in-space fuel cell systems that are hydrogenand methanol-based. During lab research, three equal-service machines need to be evaluated
economically. Perform the present worth analysis with the costs shown below. The MARR is
10% per year.
First cost, $
Annual operating cost (AOC), $/year
Salvage value S, $
Life, years
Electric-Powered
Gas-Powered
Solar-Powered
⫺4500
⫺900
200
8
3500
⫺700
350
8
⫺6000
⫺50
100
8
Solution
These are cost alternatives. The salvage values are considered a “negative” cost, so a ⫹ sign
precedes them. (If it costs money to dispose of an asset, the estimated disposal cost has a ⫺ sign.)
The PW of each machine is calculated at i ⫽ 10% for n ⫽ 8 years. Use subscripts E, G, and S.
PWE ⫽ ⫺4500 ⫺ 900(P兾A,10%,8) ⫹ 200(P兾F,10%,8) ⫽ $⫺9208
PWG ⫽ ⫺3500 ⫺ 700(P兾A,10%,8) ⫹ 350(P兾F,10%,8) ⫽ $⫺7071
PWS ⫽ ⫺6000 ⫺ 50(P兾A,10%,8) ⫹ 100(P兾F,10%,8) ⫽ $⫺6220
The solar-powered machine is selected since the PW of its costs is the lowest; it has the
numerically largest PW value.
133
Present Worth Analysis of Different-Life Alternatives
5.3
EXAMPLE 5.2 Water for Semiconductor Manufacturing Case
PE
As discussed in the introduction to this chapter, ultrapure water (UPW) is an expensive commodity for the semiconductor industry. With the options of seawater or groundwater sources,
it is a good idea to determine if one system is more economical than the other. Use a MARR of
12% per year and the present worth method to select one of the systems.
Solution
An important first calculation is the cost of UPW per year. The general relation and estimated
costs for the two options are as follows:
gallons minutes hours days
$
cost in $
——— ————
——— ——
UPW cost relation: ——
year ⫽ ——————
1000 gallons minute
hour
day year
(
)(
)(
)(
)( )
Seawater: (4兾1000)(1500)(60)(16)(250) ⫽ $1.44 M per year
Groundwater: (5兾1000)(1500)(60)(16)(250) ⫽ $1.80 M per year
Calculate the PW at i ⫽ 12% per year and select the option with the lower cost (larger PW
value). In $1 million units:
PW relation: PW ⫽ first cost ⫺ PW of AOC ⫺ PW of UPW ⫹ PW of salvage value
PWS ⫽ ⫺20 ⫺ 0.5(P兾A,12%,10) ⫺ 1.44(P兾A,12%,10) ⫹ 0.05(20)(P兾F,12%,10)
⫽ ⫺20 ⫺ 0.5(5.6502) ⫺ 1.44(5.6502) ⫹ 1(0.3220)
⫽ $⫺30.64
PWG ⫽ ⫺22 ⫺ 0.3(P兾A,12%,10) ⫺ 1.80(P兾A,12%,10) ⫹ 0.10(22)(P兾F,12%,10)
⫽ ⫺22 ⫺ 0.3(5.6502) ⫺ 1.80(5.6502) ⫹ 2.2(0.3220)
⫽ $⫺33.16
Based on this present worth analysis, the seawater option is cheaper by $2.52 M.
5.3 Present Worth Analysis of Different-Life
Alternatives
When the present worth method is used to compare mutually exclusive alternatives that have
different lives, the equal-service requirement must be met. The procedure of Section 5.2 is followed, with one exception:
The PW of the alternatives must be compared over the same number of years and must end
at the same time to satisfy the equal-service requirement.
Equal-service
requirement
This is necessary, since the present worth comparison involves calculating the equivalent PW of
all future cash flows for each alternative. A fair comparison requires that PW values represent
cash flows associated with equal service. For cost alternatives, failure to compare equal service
will always favor the shorter-lived mutually exclusive alternative, even if it is not the more economical choice, because fewer periods of costs are involved. The equal-service requirement is
satisfied by using either of two approaches:
LCM: Compare the PW of alternatives over a period of time equal to the least common
multiple (LCM) of their estimated lives.
Study period: Compare the PW of alternatives using a specified study period of n years.
This approach does not necessarily consider the useful life of an alternative. The study period
is also called the planning horizon.
For either approach, calculate the PW at the MARR and use the same selection guideline as
that for equal-life alternatives. The LCM approach makes the cash flow estimates extend to the
same period, as required. For example, lives of 3 and 4 years are compared over a 12-year period.
LCM or study
period
134
Present Worth Analysis
Chapter 5
The first cost of an alternative is reinvested at the beginning of each life cycle, and the estimated
salvage value is accounted for at the end of each life cycle when calculating the PW values over
the LCM period. Additionally, the LCM approach requires that some assumptions be made about
subsequent life cycles.
The assumptions when using the LCM approach are that
1. The service provided will be needed over the entire LCM years or more.
2. The selected alternative can be repeated over each life cycle of the LCM in exactly the same
manner.
3. Cash flow estimates are the same for each life cycle.
As will be shown in Chapter 14, the third assumption is valid only when the cash flows are
expected to change by exactly the inflation (or deflation) rate that is applicable through the LCM
time period. If the cash flows are expected to change by any other rate, then the PW analysis must
be conducted using constant-value dollars, which considers inflation (Chapter 14).
A study period analysis is necessary if the first assumption about the length of time the alternatives are needed cannot be made. For the study period approach, a time horizon is chosen over
which the economic analysis is conducted, and only those cash flows which occur during that time
period are considered relevant to the analysis. All cash flows occurring beyond the study period are
ignored. An estimated market value at the end of the study period must be made. The time horizon
chosen might be relatively short, especially when short-term business goals are very important. The
study period approach is often used in replacement analysis (Chapter 11). It is also useful when the
LCM of alternatives yields an unrealistic evaluation period, for example, 5 and 9 years.
EXAMPLE 5.3
National Homebuilders, Inc., plans to purchase new cut-and-finish equipment. Two manufacturers offered the estimates below.
First cost, $
Annual M&O cost, $ per year
Salvage value, $
Life, years
Vendor A
Vendor B
⫺15,000
⫺3,500
1,000
6
⫺18,000
⫺3,100
2,000
9
(a) Determine which vendor should be selected on the basis of a present worth comparison,
if the MARR is 15% per year.
(b) National Homebuilders has a standard practice of evaluating all options over a 5-year
period. If a study period of 5 years is used and the salvage values are not expected to
change, which vendor should be selected?
Solution
(a) Since the equipment has different lives, compare them over the LCM of 18 years. For life
cycles after the first, the first cost is repeated in year 0 of each new cycle, which is the last
year of the previous cycle. These are years 6 and 12 for vendor A and year 9 for B. The
cash flow diagram is shown in Figure 5–2. Calculate PW at 15% over 18 years.
PWA ⫽ ⫺15,000 ⫺ 15,000(P兾F,15%,6) ⫹ 1000(P兾F,15%,6)
⫺15,000(P兾F,15%,12) ⫹ 1000(P兾F,15%,12) ⫹ 1000(P兾F,15%,18)
⫺3,500(P兾A,15%,18)
⫽ $⫺45,036
PWB ⫽ ⫺18,000 ⫺ 18,000(P兾F,15%,9) ⫹ 2000(P兾F,15%,9)
⫹ 2000(P兾F,15%,18) ⫺ 3100(P兾A,15%,18)
⫽ $⫺41,384
Present Worth Analysis of Different-Life Alternatives
5.3
135
PWA = ?
$1000
1
2
$1000
6
12
$1000
16
17 18
Year
$3500
$15,000
$15,000
$15,000
Vendor A
PWB = ?
$2000
1
2
$2000
16
9
17 18
Year
$3100
$18,000
$18,000
Vendor B
Figure 5–2
Cash flow diagram for different-life alternatives, Example 5.3a.
Vendor B is selected, since it costs less in PW terms; that is, the PWB value is numerically larger than PWA.
(b) For a 5-year study period, no cycle repeats are necessary. The PW analysis is
PWA ⫽ ⫺15,000 ⫺ 3500(P兾A,15%,5) ⫹ 1000(P兾F,15%,5)
⫽ $⫺26,236
PWB ⫽ ⫺18,000 ⫺ 3100(P兾A,15%,5) ⫹ 2000(P兾F,15%,5)
⫽ $⫺27,397
Vendor A is now selected based on its smaller PW value. This means that the shortened
study period of 5 years has caused a switch in the economic decision. In situations such as
this, the standard practice of using a fixed study period should be carefully examined to
ensure that the appropriate approach, that is, LCM or fixed study period, is used to satisfy
the equal-service requirement.
EXAMPLE 5.4 Water for Semiconductor Manufacturing Case
When we discussed this case in the introduction, we learned that the initial estimates of equipment life were 10 years for both options of UPW (ultrapure water)—seawater and groundwater.
As you might guess, a little research indicates that seawater is more corrosive and the equipment
life is shorter—5 years rather than 10. However, it is expected that, instead of complete replacement, a total refurbishment of the equipment for $10 M after 5 years will extend the life through
the anticipated 10th year of service.
With all other estimates remaining the same, it is important to determine if this 50%
reduction in expected usable life and the refurbishment expense may alter the decision to go
with the seawater option, as determined in Example 5.2. For a complete analysis, consider
both a 10-year and a 5-year option for the expected use of the equipment, regardless of the
source of UPW.
PE
136
Present Worth Analysis
Chapter 5
Seawater option
⫽ 〈OC ⫹ UPW cost/year
⫽ ⫺0.5 ⫺ 1.44
Groundwater option
⫽ 〈OC ⫹ UPW cost/year
⫽ ⫺0.3 ⫺ 1.80
Both options
Salvage values
included here
Salvage values are the
same after 5 and 10 years
Figure 5–3
PW analyses using LCM and study period approaches for water for semiconductor manufacturing case,
Example 5.4.
Solution
A spreadsheet and the NPV function are a quick and easy way to perform this dual analysis.
The details are presented in Figure 5–3.
LCM of 10 years: In the top part of the spreadsheet, the LCM of 10 years is necessary to satisfy the equal-service requirement; however, the first cost in year 5 is the refurbishment cost of
$⫺10 M, not the $⫺20 M expended in year 0. Each year’s cash flow is entered in consecutive
cells; the $⫺11.94 M in year 5 accounts for the continuing AOC and annual UPW cost of
$⫺1.94 M, plus the $⫺10 M refurbishment cost. The NPV functions shown on the spreadsheet
determine the 12% per year PW values in $1 million units.
PWS ⫽ $⫺36.31
PWG ⫽ $⫺33.16
Now, the groundwater option is cheaper; the economic decision is reversed with this new
estimate of life and year 5 refurbishment expense.
Study period of 5 years: The lower portion of Figure 5–3 details a PW analysis using the
second approach to evaluating different-life alternatives, that is, a specific study period, which
is 5 years in this case study. Therefore, all cash flows after 5 years are neglected.
Again the economic decision is reversed as the 12% per year PW values favor the seawater
option.
PWS ⫽ $⫺26.43
PWG ⫽ $⫺28.32
Comments
The decision switched between the LCM and study period approaches. Both are correct answers given the decision of how the equal-service requirement is met. This analysis demonstrates how important it is to compare mutually exclusive alternatives over time periods that
are believable and to take the time necessary to make the most accurate cost, life, and MARR
estimates when the evaluation is performed.
137
Future Worth Analysis
5.4
If the PW evaluation is incorrectly performed using the respective lives of the two options,
the equal-service requirement is violated, and PW values favor the shorter-lived option, that is,
seawater. The PW values are
Option S:
Option G:
n ⫽ 5 years, PWS ⫽ $⫺26.43 M, from the bottom left calculation in Figure 5–3.
n ⫽ 10 years, PWG ⫽ $⫺33.16 M, from the top right calculation in Figure 5–3.
For independent projects, use of the LCM approach is unnecessary since each project is compared to the do-nothing alternative, not to each other, and satisfying the equal-service requirement is not a problem. Simply use the MARR to determine the PW over the respective life of
each project, and select all projects with a PW ⱖ 0.
5.4 Future Worth Analysis
The future worth (FW) of an alternative may be determined directly from the cash flows, or by
multiplying the PW value by the F兾P factor, at the established MARR. The n value in the F兾P
factor is either the LCM value or a specified study period. Analysis of alternatives using FW
values is especially applicable to large capital investment decisions when a prime goal is to
maximize the future wealth of a corporation’s stockholders.
Future worth analysis over a specified study period is often utilized if the asset (equipment, a
building, etc.) might be sold or traded at some time before the expected life is reached. Suppose
an entrepreneur is planning to buy a company and expects to trade it within 3 years. FW analysis
is the best method to help with the decision to sell or keep it 3 years hence. Example 5.5 illustrates
this use of FW analysis. Another excellent application of FW analysis is for projects that will come
online at the end of a multiyear investment period, such as electric generation facilities, toll roads,
airports, and the like. They are analyzed using the FW value of investment commitments made
during construction.
The selection guidelines for FW analysis are the same as for PW analysis; FW ⱖ 0 means the
MARR is met or exceeded. For two or more mutually exclusive alternatives, select the one
with the numerically largest FW value.
EXAMPLE 5.5
A British food distribution conglomerate purchased a Canadian food store chain for £75 million 3 years ago. There was a net loss of £10 million at the end of year 1 of ownership. Net cash
flow is increasing with an arithmetic gradient of £⫹5 million per year starting the second year,
and this pattern is expected to continue for the foreseeable future. This means that breakeven
net cash flow was achieved this year. Because of the heavy debt financing used to purchase the
Canadian chain, the international board of directors expects a MARR of 25% per year from
any sale.
(a) The British conglomerate has just been offered £159.5 million by a French company
wishing to get a foothold in Canada. Use FW analysis to determine if the MARR will be
realized at this selling price.
(b) If the British conglomerate continues to own the chain, what selling price must be obtained at the end of 5 years of ownership to just make the MARR?
Solution
(a) Set up the future worth relation in year 3 (FW3) at i ⫽ 25% per year and an offer price of
£159.5 million. Figure 5–4a presents the cash flow diagram in million £ units.
FW3 ⫽ ⫺75(F兾P,25%,3) ⫺ 10(F兾P,25%,2) ⫺ 5(F兾P,25%,1) ⫹ 159.5
⫽ ⫺168.36 ⫹ 159.5 ⫽ £⫺8.86 million
No, the MARR of 25% will not be realized if the £159.5 million offer is accepted.
ME alternative
selection
138
Present Worth Analysis
Chapter 5
FW = ?
£159.5
i ⱖ 25%?
i = 25%
£10
£5
0
1
2
3
Year
0
1
2
£5
£10
4
5
Year
£5
£10
£75
3
£75
(a)
(b)
Figure 5–4
Cash flow diagrams for Example 5.5. (a) i ⫽ ?; (b) FW ⫽ ?.
(b) Determine the future worth 5 years from now at 25% per year. Figure 5–4b presents the
cash flow diagram. The A兾G and F兾A factors are applied to the arithmetic gradient.
FW5 ⫽ ⫺75(F兾P,25%,5) ⫺ 10(F兾A,25%,5) ⫹ 5(A兾G,25%,5)(F兾A,25%,5)
⫽ £⫺246.81 million
The offer must be for at least £246.81 million to make the MARR. This is approximately
3.3 times the purchase price only 5 years earlier, in large part based on the required
MARR of 25%.
5.5 Capitalized Cost Analysis
Many public sector projects such as bridges, dams, highways and toll roads, railroads, and hydroelectric and other power generation facilities have very long expected useful lives. A perpetual
or infinite life is the effective planning horizon. Permanent endowments for charitable organizations and universities also have perpetual lives. The economic worth of these types of projects or
endowments is evaluated using the present worth of the cash flows.
Capitalized Cost (CC) is the present worth of a project that has a very long life (more than, say,
35 or 40 years) or when the planning horizon is considered very long or infinite.
The formula to calculate CC is derived from the PW relation P ⫽ A(P兾A,i%,n), where n ⫽ ⬁
time periods. Take the equation for P using the P兾A factor and divide the numerator and denominator by (1 ⫹ i)n to obtain
1
1 ⫺ ————
(1 ⫹ i)n
P ⫽ A —————
i
As n approaches ⬁, the bracketed term becomes 1兾i. We replace the symbols P and PW with
CC as a reminder that this is a capitalized cost equivalence. Since the A value can also be termed
AW for annual worth, the capitalized cost formula is simply
[
A
CC ⴝ —
i
]
or
AW
CC ⴝ ——
i
[5.1]
Solving for A or AW, the amount of new money that is generated each year by a capitalization
of an amount CC is
AW ⴝ CC (i)
[5.2]
This is the same as the calculation A ⫽ P(i) for an infinite number of time periods. Equation [5.2]
can be explained by considering the time value of money. If $20,000 is invested now (this is the
Capitalized Cost Analysis
5.5
capitalization) at 10% per year, the maximum amount of money that can be withdrawn at the end
of every year for eternity is $2000, which is the interest accumulated each year. This leaves the
original $20,000 to earn interest so that another $2000 will be accumulated the next year.
The cash flows (costs, revenues, and savings) in a capitalized cost calculation are usually of
two types: recurring, also called periodic, and nonrecurring. An annual operating cost of $50,000
and a rework cost estimated at $40,000 every 12 years are examples of recurring cash flows.
Examples of nonrecurring cash flows are the initial investment amount in year 0 and one-time
cash flow estimates at future times, for example, $500,000 in fees 2 years hence.
The procedure to determine the CC for an infinite sequence of cash flows is as follows:
1. Draw a cash flow diagram showing all nonrecurring (one-time) cash flows and at least two
cycles of all recurring (periodic) cash flows.
2. Find the present worth of all nonrecurring amounts. This is their CC value.
3. Find the A value through one life cycle of all recurring amounts. (This is the same value in
all succeeding life cycles, as explained in Chapter 6.) Add this to all other uniform
amounts (A) occurring in years 1 through infinity. The result is the total equivalent uniform
annual worth (AW).
4. Divide the AW obtained in step 3 by the interest rate i to obtain a CC value. This is an
application of Equation [5.1].
5. Add the CC values obtained in steps 2 and 4.
Drawing the cash flow diagram (step 1) is more important in CC calculations than elsewhere,
because it helps separate nonrecurring and recurring amounts. In step 5 the present worths of all
component cash flows have been obtained; the total capitalized cost is simply their sum.
EXAMPLE 5.6
The Haverty County Transportation Authority (HCTA) has just installed new software to charge
and track toll fees. The director wants to know the total equivalent cost of all future costs incurred to purchase the software system. If the new system will be used for the indefinite future,
find the equivalent cost (a) now, a CC value, and (b) for each year hereafter, an AW value.
The system has an installed cost of $150,000 and an additional cost of $50,000 after
10 years. The annual software maintenance contract cost is $5000 for the first 4 years and
$8000 thereafter. In addition, there is expected to be a recurring major upgrade cost of $15,000
every 13 years. Assume that i ⫽ 5% per year for county funds.
Solution
(a) The five-step procedure to find CC now is applied.
1. Draw a cash flow diagram for two cycles (Figure 5–5).
2. Find the present worth of the nonrecurring costs of $150,000 now and $50,000 in year
10 at i ⫽ 5%. Label this CC1.
CC1 ⫽ ⫺150,000 ⫺ 50,000(P兾F,5%,10) ⫽ $⫺180,695
0
2
4
6
8
10
12
14
20
26
$5000
$8000
$15,000
i = 5% per year
$50,000
$150,000
Figure 5–5
Cash flows for two cycles of recurring costs and all nonrecurring amounts, Example 5.6.
$15,000
Year
139
140
Present Worth Analysis
Chapter 5
3 and 4. Convert the $15,000 recurring cost to an A value over the first cycle of 13 years, and
find the capitalized cost CC2 at 5% per year using Equation [5.1].
A ⫽ ⫺15,000(A兾F,5%,13) ⫽ $⫺847
CC2 ⫽ ⫺847兾0.05 ⫽ $⫺16,940
There are several ways to convert the annual software maintenance cost series to A
and CC values. A straightforward method is to, first, consider the $−5000 an A series
with a capitalized cost of
CC3 ⫽ ⫺5000兾0.05 ⫽ $⫺100,000
Second, convert the step-up maintenance cost series of $−3000 to a capitalized cost
CC4 in year 4, and find the present worth in year 0. (Refer to Figure 5–5 for cash flow
timings.)
⫺3,000
CC4 ⫽ ———— (P兾F,5%,4) ⫽ $⫺49,362
0.05
5. The total capitalized cost CCT for Haverty County Transportation Authority is the sum
of the four component CC values.
CCT ⫽ ⫺180,695 ⫺ 16,940 ⫺ 100,000 ⫺ 49,362
⫽ $⫺346,997
(b) Equation [5.2] determines the AW value forever.
AW ⫽ Pi ⫽ CCT (i) ⫽ $346,997(0.05) ⫽ $17,350
Correctly interpreted, this means Haverty County officials have committed the equivalent
of $17,350 forever to operate and maintain the toll management software.
For the comparison of two alternatives on the basis of capitalized cost, use the procedure
above to find the A value and CCT for each alternative. Since the capitalized cost represents the
total present worth of financing and maintaining a given alternative forever, the alternatives will
automatically be compared for the same number of years (i.e., infinity). The alternative with the
smaller capitalized cost will represent the more economical one. This evaluation is illustrated in
Example 5.7 using the progressive example for this chapter.
EXAMPLE 5.7 Water for Semiconductor Manufacturing Case
PE
Our case study has progressed (in Example 5.4) to the point that the life of the seawater option
can be extended to 10 years with a major refurbishment cost after 5 years. This extension is
possible only one time, after which a new life cycle would commence. In $1 million units, the
estimates and PW values (from Figure 5–3) are as follows:
Seawater: PS ⫽ $⫺20; AOCS ⫽ $⫺1.94; nS ⫽ 10 years; refurbishment, year 5 ⫽ $⫺10;
SS ⫽ 0.05(20) ⫽ $1.00; PWS ⫽ $⫺36.31
Groundwater: PG ⫽ $⫺22; AOCG ⫽ $⫺2.10; nG ⫽ 10 years; SG ⫽ 0.10(22) ⫽ $2.2;
PWG ⫽ $⫺33.16
If we assume that the UPW (ultrapure water) requirement will continue for the foreseeable
future, a good number to know is the present worth of the long-term options at the selected
MARR of 12% per year. What are these capitalized costs for the two options using the estimates made thus far?
Solution
Find the equivalent A value for each option over its respective life, then determine the CC
value using the relation CC ⫽ A兾i. Select the option with the lower CC. This approach satisfies
the equal-service requirement because the time horizon is infinity when the CC is determined.
Capitalized Cost Analysis
5.5
Seawater: AS ⫽ PWS(A兾P,12%,10) ⫽ ⫺36.31(0.17698) ⫽ $⫺6.43
CCS ⫽ ⫺6.43兾0.12 ⫽ $⫺53.58
Groundwater: AG ⫽ PWG(A兾P,12%,10) ⫽ ⫺33.16(0.17698) ⫽ $⫺5.87
CCG ⫽ ⫺5.87兾0.12 ⫽ $⫺48.91
In terms of capitalized cost, the groundwater alternative is cheaper.
Comment
If the seawater-life extension is not considered a viable option, the original alternative of
5 years could be used in this analysis. In this case, the equivalent A value and CC computations
in $1 million units are as follows.
AS,5 years ⫽ ⫺20(A兾P,12%,5) ⫺ 1.94 ⫹ 0.05(20)(A兾F,12%,5)
⫽ $⫺7.33
CCS, 5 years ⫽ ⫺7.33/0.12 ⫽ $⫺61.08
Now, the economic advantage of the groundwater option is even larger.
If a finite-life alternative (for example, 5 years) is compared to one with an indefinite or very
long life, capitalized costs can be used. To determine capitalized cost for the finite life alternative,
calculate the equivalent A value for one life cycle and divide by the interest rate (Equation [5.1]).
This procedure is illustrated in Example 5.8 using a spreadsheet.
EXAMPLE 5.8
The State Legislature has mandated a statewide recycling program to include all types of
plastic, paper, metal, and glass refuse. The goal is zero landfill by 2020. Two options for the
materials separation equipment are outlined below. The interest rate for state-mandated projects
is 5% per year.
Contractor option (C): $8 million now and $25,000 per year will provide separation services
at a maximum of 15 sites. No contract period is stated; thus the contract and services are offered for as long as the State needs them.
Purchase option (P): Purchase equipment at each site for $275,000 per site and expend an
estimated $12,000 in annual operating costs (AOC). Expected life of the equipment is 5 years
with no salvage value.
(a) Perform a capitalized cost analysis for a total of 10 recycling sites.
(b) Determine the maximum number of sites at which the equipment can be purchased and
still have a capitalized cost less than that of the contractor option.
Solution
(a) Figure 5–6, column B, details the solution. The contract, as proposed, has a long life.
Therefore, the $8 million is already a capitalized cost. The annual charge of A ⫽ $25,000
is divided by i ⫽ 0.05 to determine its CC value. Summing the two values results in
CCC ⫽ $⫺8.5 million.
For the finite, 5-year purchase alternative, column B shows the first cost ($⫺275,000 per
site), AOC ($⫺12,000), and equivalent A value of $−755,181, which is determined via
the PMT function (cell tag). Divide A by the interest rate of 5% to determine CCP ⫽
$⫺15.1 million.
The contractor option is by far more economical for the anticipated 10 sites.
(b) A quick way to find the maximum number of sites for which CCP ⬍ CCC is to use Excel’s Goal Seek tool, introduced in Chapter 2, Example 2.10. (See Appendix A for details
on how to use this tool.) The template is set up in Figure 5–6 to make the two CC values
equal as the number of sites is altered (decreased). The result, shown in column C, indicates that 5.63 sites make the options economically equivalent. Since the number of sites
141
142
Present Worth Analysis
Chapter 5
⫽ PMT($B$1,B13,⫺B12*B4) ⫹ B14*B4
Figure 5–6
Spreadsheet solution of Example 5.8 using capitalized cost (a) for 10 recycling sites and (b) to determine the
number of sites to make the alternatives economically equal.
must be an integer, 5 or fewer sites will favor purchasing the equipment and 6 or more
sites will favor contracting the separation services.
This approach to problem solution will be called breakeven analysis in later chapters
of the text. By the way, another way to determine the number of sites is by trial
and error. Enter different values in cell B4 until the CC values favor the purchase
alternative.
CHAPTER SUMMARY
The present worth method of comparing alternatives involves converting all cash flows to present
dollars at the MARR. The alternative with the numerically larger (or largest) PW value is selected. When the alternatives have different lives, the comparison must be made for equal-service
periods. This is done by performing the comparison over either the LCM of lives or a specific
study period. Both approaches compare alternatives in accordance with the equal-service requirement. When a study period is used, any remaining value in an alternative is recognized
through the estimated future market value.
If the life of the alternatives is considered to be very long or infinite, capitalized cost is the
comparison method. The CC value is calculated as A兾i, because the P兾A factor reduces to 1兾i in
the limit of n ⫽ ⬁.
PROBLEMS
Types of Projects
5.1 What is the difference between mutually exclusive
alternatives and independent projects?
5.2 (a)
(b)
What is meant by the do-nothing alternative?
When is the do-nothing alternative not an
option?
5.3 (a)
(b)
How many alternatives are possible from
four independent projects identified as W, X,
Y, and Z?
List all of the possibilities.
5.4 What is the difference between a revenue and a
cost alternative?
Problems
5.11
5.5 What is meant by the term equal service?
5.6 What two approaches can be used to satisfy the
equal-service requirement?
Alternative Comparison—Equal Lives
5.7 A company that manufactures magnetic membrane
switches is investigating two production options
that have the estimated cash flows shown ($1 million units). Which one should be selected on the
basis of a present worth analysis at 10% per year?
First cost, $
Annual cost, $ per year
Annual income, $ per year
Salvage value, $
Life, years
In-house
Contract
⫺30
⫺5
14
2
5
0
⫺2
3.1
—
5
5.8 The manager of a canned food processing plant
must decide between two different labeling machines. Machine A will have a first cost of $42,000,
an annual operating cost of $28,000, and a service
life of 4 years. Machine B will cost $51,000 to buy
and will have an annual operating cost of $17,000
during its 4-year life. At an interest rate of 10% per
year, which should be selected on the basis of a
present worth analysis?
5.9 A metallurgical engineer is considering two materials for use in a space vehicle. All estimates are
made. (a) Which should be selected on the basis of
a present worth comparison at an interest rate of
12% per year? (b) At what first cost for the material not selected above will it become the more
economic alternative?
First cost, $
Maintenance cost, $ per year
Salvage value, $
Life, years
Material X
Material Y
⫺15,000
⫺9,000
2,000
5
⫺35,000
⫺7,000
20,000
5
5.10 To retain high-performing engineers, a large semiconductor company provides corporate stock as
part of the compensation package. In one particular year, the company offered 1000 shares of either
class A or class B stock. The class A stock was selling for $30 per share at the time, and stock market
analysts predicted that it would increase at a rate of
6% per year for the next 5 years. Class B stock was
selling for $20 per share, but its price was expected
to increase by 12% per year. At an interest rate of
8% per year, which stock should the engineers select on the basis of a present worth analysis and a
5-year planning horizon?
143
The Murphy County Fire Department is considering two options for upgrading its aging physical facilities. Plan A involves remodeling the fire stations
on Alameda Avenue and Trowbridge Boulevard that
are 57 and 61 years old, respectively. (The industry
standard is about 50 years of use for a station.) The
cost for remodeling the Alameda station is estimated at $952,000 while the cost of redoing the
Trowbridge station is $1.3 million. Plan B calls for
buying 5 acres of land somewhere between the two
stations, building a new fire station, and selling the
land and structures at the previous sites. The cost of
land in that area is estimated to be $366,000 per
acre. The size of the new fire station would be 9000
square feet with a construction cost of $151.18 per
square foot. Contractor fees for overhead, profit,
etc. are expected to be $340,000, and architect fees
will be $81,500. (Assume all of the costs for plan B
occur at time 0.) If plan A is adopted, the extra cost
for personnel and equipment will be $126,000 per
year. Under plan B, the sale of the old sites is anticipated to net a positive $500,000 five years in the
future. Use an interest rate of 6% per year and a 50year useful life for the remodeled and new stations
to determine which plan is better on the basis of a
present worth analysis.
5.12 Delcon Properties is a commercial developer of
shopping centers and malls in various places around
the country. The company needs to analyze the economic feasibility of rainwater drains in a 60-acre
area that it plans to develop. Since the development
won’t be started for 3 years, this large open space
will be subject to damage from heavy thunderstorms that cause soil erosion and heavy rutting. If
no drains are installed, the cost of refilling and
grading the washed out area is expected to be
$1500 per thunderstorm. Alternatively, a temporary
corrugated steel drainage pipe could be installed
that will prevent the soil erosion. The cost of the
pipe will be $3 per foot for the total length of
7000 feet required. Some of the pipe will be salvageable for $4000 at the end of the 3-year period
between now and when the construction begins.
Assuming that thunderstorms occur regularly at
3-month intervals, starting 3 months from now,
which alternative should be selected on the basis of
a present worth comparison using an interest rate of
4% per quarter?
5.13 A public water utility is trying to decide between
two different sizes of pipe for a new water main. A
250-mm line will have an initial cost of $155,000,
whereas a 300-mm line will cost $210,000. Since
there is more head loss through the 250-mm pipe,
the pumping cost is expected to be $3000 more per
year than for the 250-mm line. If the lines are
144
Present Worth Analysis
Chapter 5
expected to last for 30 years, which size should be
selected on the basis of a present worth analysis
using an interest rate of 10% per year?
5.14 The supervisor of a community swimming pool
has developed two methods for chlorinating the
pool. If gaseous chlorine is added, a chlorinator
will be required that has an initial cost of $8000
and a useful life of 5 years. The chlorine will cost
$650 per year, and the labor cost will be $800 per
year. Alternatively, dry chlorine can be added manually at a cost of $1000 per year for chlorine and
$1900 per year for labor. Which method should be
used on the basis of a present worth analysis if the
interest rate is 10% per year?
5.15 Anion, an environmental engineering consulting
firm, is trying to be eco-friendly in acquiring an automobile for general office use. It is considering a
gasoline-electric hybrid and a gasoline-free allelectric hatchback. The hybrid under consideration
is GM’s Volt, which will cost $35,000 and have a
range of 40 miles on the electric battery and several
hundred more miles when the gasoline engine
kicks in. Nissan’s Leaf, on the other hand, is a pure
electric that will have a range of only 100 miles,
after which its lithium-ion battery will have to be
recharged. The Leaf’s relatively limited range creates a psychological effect known as range anxiety.
This fact alone has caused the company to lean toward purchasing the Volt, which is assumed to have
a salvage value of $15,000 in 5 years. The Leaf
could be leased for $349 per month (end-of-month
payments) for 5 years after an initial $1500 down
payment for “account activation.” If the consulting
company plans to ignore the range anxiety effect in
making its decision, which automobile is the better
option on the basis of a present worth analysis at an
interest rate of 0.75% per month? Assume the operating cost will be the same for both vehicles.
5.16 A pipeline engineer working in Kuwait for the oil
giant BP wants to perform a present worth analysis on
alternative pipeline routings—the first predominately
by land and the second primarily undersea. The undersea route is more expensive initially due to extra
corrosion protection and installation costs, but cheaper
security and maintenance reduces annual costs. Perform the analysis for the engineer at 15% per year.
Land
Installation cost, $ million
⫺215
Pumping, operating, security,
⫺22
$ million per year
Replacement of valves and
⫺30
appurtenances in year 25, $ million
Expected life, years
50
Undersea
⫺350
⫺2
⫺70
50
Alternative Comparison—Different Lives
5.17 An electric switch manufacturing company has to
choose one of three different assembly methods.
Method A will have a first cost of $40,000, an annual operating cost of $9000, and a service life of
2 years. Method B will cost $80,000 to buy and
will have an annual operating cost of $6000 over
its 4-year service life. Method C will cost $130,000
initially with an annual operating cost of $4000
over its 8-year life. Methods A and B will have no
salvage value, but method C will have some equipment worth an estimated $12,000. Which method
should be selected? Use present worth analysis at
an interest rate of 10% per year.
5.18 Midwest Power and Light operates 14 coal-fired
power plants in several states around the United
States. The company recently settled a lawsuit by
agreeing to pay $60 million in mitigation costs related to acid rain. The settlement included $21 million to reduce emissions from barges and trucks in
the Ohio River Valley, $24 million for projects to
conserve energy and produce alternative energy,
$3 million for Chesapeake Bay, $2 million for
Shenandoah National Park, and $10 million to acquire ecologically sensitive lands in Appalachia.
The question of how to distribute the money over
time has been posed. Plan A involves spending
$5 million now and the remaining $55 million
equally over a 10-year period (that is, $5.5 million
in each of years 1 through 10). Plan B requires expenditures of $5 million now, $25 million 2 years
from now, and $30 million 7 years from now. Determine which plan is more economical on the
basis of a present worth analysis over a 10-year
period at an interest rate of 10% per year.
5.19 Machines that have the following costs are under
consideration for a robotized welding process.
Using an interest rate of 10% per year, determine
which alternative should be selected on the basis
of a present worth analysis. Show (a) hand and
(b) spreadsheet solutions.
Machine X Machine Y
First cost, $
Annual operating cost, $ per year
Salvage value, $
Life, years
⫺250,000
⫺60,000
70,000
3
⫺430,000
⫺40,000
95,000
6
5.20 Water for Semiconductor Manufacturing Case
PE
Throughout the present worth analyses, the decision between seawater and groundwater switched
multiple times in Examples 5.2 and 5.4. A summary is given here in $1 million units.
145
Problems
Seawater (S)
Life n,
years
10
5
5
(study
period)
Groundwater (G)
PW at
12%, $
Selected
Life n,
years
⫺20
⫺30.64
Yes
⫺20, plus
⫺10 after
5 years
⫺36.31
⫺20
⫺26.43
First
cost, $
First
cost, $
PW at
12%, $
Selected
10
⫺22
⫺33.16
No
No
10
⫺22
⫺33.16
Yes
Yes
5
(study
period)
⫺22
⫺28.32
No
The confusion about the recommended source for
UPW has not gone unnoticed by the general manager. Yesterday, you were asked to settle the issue
by determining the first cost XS of the seawater option to ensure that it is the economic choice over
groundwater. The study period is set by the manager
as 10 years, simply because that is the time period on
the lease agreement for the building where the fab
will be located. Since the seawater equipment must
be refurbished or replaced after 5 years, the general
manager told you to assume that the equipment will
be purchased anew after 5 years of use. What is the
maximum first cost that Angular Enterprises should
pay for the seawater option?
5.21 Accurate airflow measurement requires straight
unobstructed pipe for a minimum of 10 diameters
upstream and 5 diameters downstream of the measuring device. In a field application, physical constraints compromise the pipe layout, so the engineer
is considering installing the airflow probes in an
elbow, knowing that flow measurement will be less
accurate but good enough for process control. This
is plan 1, which will be in place for only 3 years,
after which a more accurate flow measurement system with the same costs as plan 1 will be available.
This plan will have a first cost of $26,000 with an
annual maintenance cost estimated at $5000.
Plan 2 involves installation of a recently designed submersible airflow probe. The stainless
steel probe can be installed in a drop pipe with the
transmitter located in a waterproof enclosure on the
handrail. The first cost of this system is $83,000,
but because it is accurate and more durable, it will
not have to be replaced for at least 6 years. Its maintenance cost is estimated to be $1400 per year plus
$2500 in year 3 for replacement of signal processing software. Neither system will have a salvage
value. At an interest rate of 10% per year, which
one should be selected on the basis of a present
worth comparison?
5.22
An engineer is considering two different liners for an
evaporation pond that will receive salty concentrate
from a brackish water desalting plant. A plastic liner
will cost $0.90 per square foot initially and will
require replacement in 15 years when precipitated
solids will have to be removed from the pond using
heavy equipment. This removal will cost $500,000.
A rubberized elastomeric liner is tougher and, therefore, is expected to last 30 years, but it will cost
$2.20 per square foot. If the size of the pond is
110 acres (1 acre ⫽ 43,560 square feet), which liner
is more cost effective on the basis of a present worth
comparison at an interest rate of 8% per year?
5.23
A sports mortgage is the brainchild of Stadium Capital Financing Group, a company headquartered in
Chicago, Illinois. It is an innovative way to finance
cash-strapped sports programs by allowing fans to
sign up to pay a “mortgage” over a certain number of
years for the right to buy good seats at football games
for several decades with season ticket prices locked
in at current prices. In California, the locked-in price
period is 50 years. Assume UCLA fan X purchases a
$130,000 mortgage and pays for it now to get season
tickets for $290 each for 50 years, while fan Y buys
season tickets at $290 in year 1, with prices increasing by $20 per year for 50 years. (a) Which fan made
the better deal if the interest rate is 8% per year?
(b) What should fan X be willing to pay up front for
the mortgage to make the two plans exactly equivalent economically? (Assume he has no reason to
give extra money to UCLA at this point.)
5.24 A chemical processing corporation is considering
three methods to dispose of a non-hazardous chemical sludge: land application, fluidized-bed incineration, and private disposal contract. The estimates for
each method are shown. Determine which has the
least cost on the basis of a present worth comparison at 10% per year for the following scenarios:
(a) The estimates as shown
(b) The contract award cost increases by 20%
every 2-year renewal
Land Application Incineration
First cost, $
Annual operating
cost, $ per year
Salvage value, $
Life, years
Contract
⫺130,000
⫺95,000
⫺900,000
⫺60,000
0
⫺120,000
25,000
3
300,000
6
0
2
146
Present Worth Analysis
Chapter 5
5.25 An assistant to Stacy gave her the PW values for
four alternatives they are comparing for the development of a remote control vibration control system for offshore platform application. The results
in the table use a MARR of 14% per year. Determine which alternative(s) should be selected (a) if
the alternatives are exclusive, and (b) if the projects
are independent.
I
Life n, years
PW over n years, $
PW over 6 years, $
PW over 12 years, $
3
16.08
26.94
39.21
J
4
31.12
15.78
60.45
K
L
12
6
⫺257.46 140.46
⫺653.29 140.46
⫺257.46 204.46
Future Worth Comparison
5.26 An industrial engineer is considering two robots
for purchase by a fiber-optic manufacturing company. Robot X will have a first cost of $80,000, an
annual maintenance and operation (M&O) cost of
$30,000, and a $40,000 salvage value. Robot Y
will have a first cost of $97,000, an annual M&O
cost of $27,000, and a $50,000 salvage value.
Which should be selected on the basis of a future
worth comparison at an interest rate of 15% per
year? Use a 3-year study period.
5.27 Two processes can be used for producing a polymer that reduces friction loss in engines. Process T
will have a first cost of $750,000, an operating cost
of $60,000 per year, and a salvage value of $80,000
after its 2-year life. Process W will have a first cost
of $1,350,000, an operating cost of $25,000 per
year, and a $120,000 salvage value after its 4-year
life. Process W will also require updating at the
end of year 2 at a cost of $90,000. Which process
should be selected on the basis of a future worth
analysis at an interest rate of 12% per year?
5.28 Compare the alternatives shown below on the
basis of a future worth analysis, using an interest
rate of 8% per year.
P
First cost, $
Annual operating cost, $ per year
Salvage value, $
Life, years
Q
⫺23,000 ⫺30,000
⫺4,000 ⫺2,500
3,000
1,000
3
6
5.29 Two manufacturers supply MRI systems for medical imaging. St. Jude’s Hospital wishes to replace its
current MRI equipment that was purchased 8 years
ago with the newer technology and clarity of a stateof-the-art system. System K will have a first cost of
$1,600,000, an operating cost of $70,000 per year,
and a salvage value of $400,000 after its 4-year life.
System L will have a first cost of $2,100,000, an
operating cost of $50,000 the first year with an expected increase of $3000 per year thereafter, and no
salvage value after its 8-year life. Which system
should be selected on the basis of a future worth
analysis at an interest rate of 12% per year?
5.30 A retail shopping center developer signed a contract
to build a $100 million high-end shopping center in
City Center, because the city and county governments agreed to sales and tax rebates totaling
$18.7 million over 10 years. The contract called for
the developer to raze existing buildings 2 years
from the date the contract was signed and to have
the shopping center built by the end of year 3. However, due to a real estate–induced recession in the
United States, the developer sought and was granted
a new contract. The new contract required the developer to raze the existing buildings at the end of
year 1, but the shopping center would not have to be
completed for 7 years from the date the contract was
signed. Assume that the cost for razing the existing
buildings is $1.3 million and the developer does not
build the shopping center until 7 years from now (at
a cost of $100 million). Determine the difference in
the future worth cost in year 7 of the two contracts
at an interest rate of 10% per year.
Capitalized Cost
5.31 A wealthy businessman wants to start a permanent
fund for supporting research directed toward sustainability. The donor plans to give equal amounts
of money for each of the next 5 years, plus one now
(i.e., six donations) so that $100,000 per year can
be withdrawn each year forever, beginning in
year 6. If the fund earns interest at a rate of 8% per
year, how much money must be donated each time?
5.32 Bob, a philanthropist, is not sure what rate of return
his gifts may realize once donated to his favorite
charity. Determine the capitalized cost of $10,000
every 5 years forever, starting 5 years from now
at an interest rate of (a) 3% and (b) 8% per year.
(c) Explain the significant difference between the
two capitalized costs.
5.33 Find the capitalized cost of a present cost of
$300,000, annual costs of $35,000, and periodic
costs every 5 years of $75,000. Use an interest rate
of 12% per year.
5.34 Water for Semiconductor Manufacturing Case
PE
It is anticipated that the needs for UPW (ultrapure
water) at the new Angular Enterprises site will
continue for a long time, as long as 50 years. This
is the rationale for using capitalized cost as a basis
for the economic decision between desalinated
147
Additional Problems and FE Exam Review Questions
seawater (S) and purified groundwater (G). These
costs were determined (Example 5.7) to be CCS ⫽
$⫺53.58 million and CCG ⫽ $⫺48.91 million.
Groundwater is the clear economic choice.
Yesterday, the general manager had lunch with
the president of Brissa Water, who offered to supply the needed UPW at a cost of $5 million per year
for the indefinite future. It would mean a dependence upon a contractor to supply the water, but the
equipment, treatment, and other costly activities to
obtain UPW on-site would be eliminated. The
manager asks you to make a recommendation
about this seemingly attractive alternative under
the following conditions at the same MARR of
12% per year as used for the other analyses:
(a) The annual cost of $5 million remains constant throughout the time it is needed.
(b) The annual cost starts at $5 million for the first
year only, and then it increases 2% per year.
(This increase is above the cost of providing
UPW by either of the other two methods.)
5.36 The cost of maintaining a certain permanent monument in Washington, DC occurs as periodic outlays of $1000 every year and $5000 every 4 years.
Calculate the capitalized cost of the maintenance
using an interest rate of 10% per year.
5.35 Compare the alternatives shown on the basis of
their capitalized costs using an interest rate of
10% per year.
5.38 A patriotic group of firefighters is raising money
to erect a permanent (i.e., infinite life) monument
in New York City to honor those killed in the line
of duty. The initial cost of the monument will be
$150,000, and the annual maintenance will cost
$5000. There will be an additional one-time cost
of $20,000 in 2 years to add names of those who
were missed initially. At an interest rate of 6% per
year, how much money must they raise now in
order to construct and maintain the monument
forever?
Alternative M
First cost, $
Annual operating
cost, $ per year
Salvage value, $
Life, years
Alternative N
⫺150,000
⫺50,000
⫺800,000
⫺12,000
8,000
5
1,000,000
⬁
5.37 Because you are thankful for what you learned in
engineering economy, you plan to start a permanent scholarship fund in the name of the professor
who taught the course. You plan to deposit money
now with the stipulation that the scholarships be
awarded beginning 12 years from now (which
happens to be the exact time that your daughter
plans to begin college). The interest that is accumulated between now and year 12 is to be added to
the principal of the endowment. After that, the interest that is earned each year will be awarded as
scholarship money. If you want the amount of the
scholarships to be $40,000 per year, how much
must you donate now if the fund earns interest at a
rate of 8% per year?
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
5.39 One assumption inherent in the present worth
method of analysis is that:
(a) The alternatives will be used only through
the life of the shortest-lived alternative.
(b) The alternatives will be used only through
the life of the longest-lived alternative.
(c) The cash flows of each alternative will
change only by the inflation or deflation rate
in succeeding life cycles.
(d) At least one of the alternatives will have a
finite life.
5.40 When only one alternative can be selected from
two or more, the alternatives are said to be:
(a) Mutually exclusive
(b) Independent alternatives
(c) Cost alternatives
(d) Revenue alternatives
5.41 For the mutually exclusive alternatives shown, the
one(s) that should be selected are:
(a)
(b)
(c)
(d)
Alternative
PW, $
A
B
C
D
⫺25,000
⫺12,000
10,000
15,000
Only C
Only A
C and D
Only D
5.42 The alternatives shown are to be compared on the
basis of their present worth values. At an interest
rate of 10% per year, the values of n that you
should use in the uniform series factors to make a
148
Present Worth Analysis
Chapter 5
correct comparison by the present worth method
are:
Alternative A
Alternative B
⫺50,000
⫺10,000
⫺90,000
⫺4000
13,000
3
15,000
6
First cost, $
Annual operating cost,
$ per year
Salvage value, $
Life, years
(a)
(b)
(c)
(d)
n ⫽ 3 years for A and n ⫽ 3 years for B
n ⫽ 3 years for A and n ⫽ 6 years for B
n ⫽ 6 years for A and n ⫽ 6 years for B
None of the above
5.43 The value of the future worth for alternative P at an
interest rate of 8% per year is closest to:
First cost, $
Annual operating cost,
$ per year
Salvage value, $
Life, years
(a)
(b)
(c)
(d)
P
Q
⫺23,000
⫺4,000
⫺30,000
⫺2,500
3,000
3
1,000
6
FWP ⫽ $⫺88,036
FWP ⫽ $⫺86,026
FWP ⫽ $⫺81,274
FWP ⫽ $⫺70,178
5.44 The present worth of $50,000 now, $10,000 per year
in years 1 through 15, and $20,000 per year in years
16 through infinity at 10% per year is closest to:
(a) Less than $⫺169,000
(b) $⫺169,580
(c) $⫺173,940
(d) $⫺195,730
5.45 A donor (you) wishes to start an endowment that
will provide scholarship money of $40,000 per year
beginning in year 5 and continuing indefinitely. If
the university earns 10% per year on the endowment, the amount you must donate now is closest to:
(a) $⫺225,470
(b) $⫺248,360
(c) $⫺273,200
(d) $⫺293,820
Problems 5.46 through 5.48 are based on the following
information.
Initial cost, $
Annual income, $ per year
Annual expenses, $ per year
Salvage value, $
Life, years
Alternative I
Alternative J
⫺150,000
20,000
⫺9,000
25,000
3
⫺250,000
40,000
⫺14,000
35,000
6
The interest rate is 15% per year.
5.46 In comparing alternatives I and J by the present
worth method, the value of n that must be used in
11,000(P兾A,i,n) for alternative I is:
(a) 3
(b) 6
(c) 18
(d) 36
5.47 In comparing alternatives I and J by the present
worth method, the equation that yields the present
worth of alternative J is:
(a) PWJ ⫽ ⫺250,000 ⫹ 40,000(P兾A,15%,6) ⫹
35,000(P兾F,15%,6)
(b) PWJ ⫽ ⫺250,000 ⫹ 26,000(P兾A,15%,6) ⫹
35,000(P兾F,15%,6)
(c) PWJ ⫽ ⫺250,000 ⫺ 26,000(P兾A,15%,6) ⫹
35,000(P兾F,15%,6)
(d) PWJ ⫽ ⫺250,000 ⫺ 26,000(P兾A,15%,6) ⫺
35,000(P兾F,15%,6)
5.48 In comparing alternatives I and J by the present
worth method, the equation that yields the present
worth of alternative I is:
(a) PWI ⫽ ⫺150,000 ⫹ 11,000(P兾A,15%,3) ⫹
25,000(P兾F,15%,3)
(b) PWI ⫽ ⫺150,000 ⫹ 11,000(P兾A,15%,6) ⫹
25,000(P兾F,15%,6)
(c) PWI ⫽ ⫺150,000 ⫹ 11,000(P兾A,15%,6) ⫹
175,000(P兾F,15%,3) ⫹ 25,000(P兾F,15%,6)
(d) PWI ⫽ ⫺150,000 ⫹ 11,000(P兾A,15%,6) ⫺
125,000(P兾F,15%,3) ⫹ 25,000(P兾F,15%,6)
Problems 5.49 and 5.50 are based on the following
information.
Initial cost, $
Annual operating cost, $ per year
Salvage value, $
Life, years
Machine X
Machine Y
⫺80,000
⫺20,000
10,000
2
⫺95,000
⫺15,000
30,000
4
The interest rate is 10% per year.
5.49 The equation that will calculate the present worth
of machine X is:
(a) PWX ⫽ ⫺80,000 ⫺ 15,000(P兾A,10%,4) ⫹
30,000(P兾F,10%,4)
(b) PWX ⫽ ⫺80,000 ⫺ 20,000(P兾A,10%,4) ⫺
80,000(P兾F,10%,2) ⫹ 10,000(P兾F,10%,4)
(c) PWX ⫽ ⫺80,000 ⫺ 20,000(P兾A,10%,2) ⫹
10,000(P兾F,10%,2)
(d) PWX ⫽ ⫺80,000 ⫺ 20,000(P兾A,10%,4) ⫺
70,000(P兾F,10%,2) ⫹ 10,000(P兾F,10%,4)
5.50 In comparing the machines on a present worth
basis, the present worth of machine Y is closest to:
(a) $⫺112,320
(b) $⫺122,060
(c) $⫺163,040
(d) $⫺175,980
Case Study
5.51 The capitalized cost of $10,000 every 5 years
forever, starting now at an interest rate of 10% per
year, is closest to:
(a) $⫺13,520
(b) $⫺16,380
(c) $⫺26,380
(d) $⫺32,590
149
5.52 At an interest rate of 10% per year, the capitalized
cost of $10,000 in year 0, $5000 per year in years 1
through 5, and $1000 per year thereafter forever is
closest to:
(a) ⫺$29,652
(b) ⫺$35,163
(c) ⫺$38,954
(d) ⫺$43,221
CASE STUDY
COMPARING SOCIAL SECURITY BENEFITS
Background
When Sheryl graduated from Northeastern University in
2000 and went to work for BAE Systems, she did not pay
much attention to the monthly payroll deduction for social
security. It was a “necessary evil” that may be helpful in retirement years. However, this was so far in the future that she
fully expected this government retirement benefit system to
be broke and gone by the time she could reap any benefits
from her years of contributions.
This year, Sheryl and Brad, another engineer at BAE,
got married. Recently, they both received notices from the
Social Security Administration of their potential retirement
amounts, were they to retire and start social security benefits at preset ages. Since both of them hope to retire a few
years early, they decided to pay closer attention to the predicted amount of retirement benefits and to do some analysis on the numbers.
Information
They found that their projected benefits are substantially the
same, which makes sense since their salaries are very close to
each other. Although the numbers were slightly different in
their two mailings, the similar messages to Brad and Sheryl
can be summarized as follows:
If you stop working and start receiving benefits . . .
At age 62, your payment would
be about
At you full retirement age (67
years), your payment would be
about
At age 70, your payment would
be about
$1400 per month
$2000 per month
$2480 per month
These numbers represent a reduction of 30% for early retirement (age 62) and an increase of 24% for delayed retirement (age 70).
This couple also learned that it is possible for a spouse to
take spousal benefits at the time that one of them is at full
retirement age. In other words, if Sheryl starts her $2000 benefit at age 67, Brad can receive a benefit equal to 50% of hers.
Then, when Brad reaches 70 years of age, he can discontinue
spousal benefits and start his own. In the meantime, his benefits will have increased by 24%. Of course, this strategy
could be switched with Brad taking his benefits and Sheryl
receiving spousal benefits until age 70.
All these options led them to define four alternative plans.
A: Each takes early benefits at age 62 with a 30% reduction to $1400 per month.
B: Each takes full benefits at full retirement age of 67
and receives $2000 per month.
C: Each delays benefits until age 70 with a 24% increase
to $2480 per month.
D: One person takes full benefits of $2000 per month at
age 67, and the other person receives spousal benefits
($1000 per month at age 67) and switches to delayed
benefits of $2480 at age 70.
They realize, of course, that the numbers will change over
time, based on their respective salaries and number of years
of contribution to the social security system by them and by
their employers.
Case Study Exercises
Brad and Sheryl are the same age. Brad determined that most
of their investments make an average of 6% per year. With
this as the interest rate, the analysis for the four alternatives is
possible. Sheryl and Brad plan to answer the following questions, but don’t have time this week. Can you please help
them? (Do the analysis for one person at a time, not the couple, and stop at the age of 85.)
1. How much in total (without the time value of money considered) will each plan A through D pay through age 85?
2. What is the future worth at 6% per year of each plan at
age 85?
3. Plot the future worth values for all four plans on one
spreadsheet graph.
4. Economically, what is the best combination of plans for
Brad and Sheryl, assuming they both live to be 85 years
old?
5. Develop at least one additional question that you think
Sheryl and Brad may have. Answer the question.
CHAPTER 6
Annual Worth
Analysis
L E A R N I N G
O U T C O M E S
Purpose: Utilize different annual worth techniques to evaluate and select alternatives.
SECTION
TOPIC
LEARNING OUTCOME
6.1
Advantages of AW
• Demonstrate that the AW value is the same for
each life cycle.
6.2
CR and AW values
• Calculate and interpret the capital recovery (CR)
and AW amounts.
6.3
AW analysis
• Select the best alternative using an annual worth
analysis.
6.4
Perpetual life
• Evaluate alternatives with very long lives using
AW analysis.
6.5
LCC analysis
• Perform a life-cycle cost (LCC) analysis using AW
methods.
I
n this chapter, we add to our repertoire of alternative comparison tools. In Chapter 5 we learned the PW method. Here we learn the equivalent annual worth, or
AW, method. AW analysis is commonly considered the more desirable of the two
methods because the AW value is easy to calculate; the measure of worth—AW in monetary
units per year—is understood by most individuals; and its assumptions are essentially identical to those of the PW method.
Annual worth is also known by other titles. Some are equivalent annual worth (EAW),
equivalent annual cost (EAC), annual equivalent (AE), and equivalent uniform annual cost
(EUAC). The alternative selected by the AW method will always be the same as that selected
by the PW method, and all other alternative evaluation methods, provided they are performed correctly.
An additional application of AW analysis treated here is life-cycle cost (LCC) analysis. This
method considers all costs of a product, process, or system from concept to phaseout.
6.1 Advantages and Uses of Annual Worth Analysis
For many engineering economic studies, the AW method is the best to use, when compared to
PW, FW, and rate of return (Chapters 7 and 8). Since the AW value is the equivalent uniform annual worth of all estimated receipts and disbursements during the life cycle of the project or alternative, AW is easy to understand by any individual acquainted with annual amounts, for example, dollars per year. The AW value, which has the same interpretation as A used thus far, is
the economic equivalent of the PW and FW values at the MARR for n years. All three can be
easily determined from each other by the relation
AW ⴝ PW(A兾P,i,n) ⴝ FW(A兾F,i,n)
[6.1]
The n in the factors is the number of years for equal-service comparison. This is the LCM or the
stated study period of the PW or FW analysis.
When all cash flow estimates are converted to an AW value, this value applies for every year
of the life cycle and for each additional life cycle.
The annual worth method offers a prime computational and interpretation advantage because
the AW value needs to be calculated for only one life cycle. The AW value determined over
one life cycle is the AW for all future life cycles. Therefore, it is not necessary to use the
LCM of lives to satisfy the equal-service requirement.
As with the PW method, there are three fundamental assumptions of the AW method that should
be understood. When alternatives being compared have different lives, the AW method makes the
assumptions that
1. The services provided are needed for at least the LCM of the lives of the alternatives.
2. The selected alternative will be repeated for succeeding life cycles in exactly the same manner as for the first life cycle.
3. All cash flows will have the same estimated values in every life cycle.
In practice, no assumption is precisely correct. If, in a particular evaluation, the first two assumptions are not reasonable, a study period must be established for the analysis. Note that for assumption 1, the length of time may be the indefinite future (forever). In the third assumption, all
cash flows are expected to change exactly with the inflation (or deflation) rate. If this is not a
reasonable assumption, new cash flow estimates must be made for each life cycle, and again a
study period must be used. AW analysis for a stated study period is discussed in Section 6.3.
EXAMPLE 6.1
In Example 5.3, National Homebuilders, Inc. evaluated cut-and-finish equipment from vendor A
(6-year life) and vendor B (9-year life). The PW analysis used the LCM of 18 years. Consider
only the vendor A option now. The diagram in Figure 6–1 shows the cash flows for all three life
cycles (first cost $⫺15,000; annual M&O costs $⫺3500; salvage value $1000). Demonstrate
the equivalence at i ⫽ 15% of PW over three life cycles and AW over one cycle. In Example 5.3, present worth for vendor A was calculated as PW ⫽ $⫺45,036.
Equal-service
requirement and LCM
152
Figure 6–1
Annual Worth Analysis
Chapter 6
PW = $45,036
PW and AW values
for three life cycles,
Example 6.1.
3 life cycles
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
$1000
0
1
2
3
4
5
6
Life cycle 1
i = 15%
$1000
$3500
0
1
2
3
4
5
6
$15,000
Life cycle 2
$1000
$3500
0
1
2
3
4
5
6
$15,000
Life cycle 3
$3500
$15,000
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
•••
continues
AW = $7349
Solution
Calculate the equivalent uniform annual worth value for all cash flows in the first life cycle.
AW ⫽ ⫺15,000(A兾P,15%,6) ⫹ 1000(A兾F,15%,6) ⫺ 3500 ⫽ $⫺7349
When the same computation is performed on each succeeding life cycle, the AW value is
$⫺7349. Now Equation [6.1] is applied to the PW value for 18 years.
AW ⫽ ⫺45,036(A兾P,15%,18) ⫽ $⫺7349
The one-life-cycle AW value and the AW value based on 18 years are equal.
Not only is annual worth an excellent method for performing engineering economy studies,
but also it is applicable in any situation where PW (and FW and benefit/cost) analysis can be
utilized. The AW method is especially useful in certain types of studies: asset replacement and
retention time studies to minimize overall annual costs (both covered in Chapter 11), breakeven
studies and make-or-buy decisions (Chapter 13), and all studies dealing with production or manufacturing costs where a cost /unit or profit /unit measure is the focus.
6.2
Calculation of Capital Recovery and AW Values
153
If income taxes are considered, a slightly different approach to the AW method is used by some
large corporations and financial institutions. It is termed economic value added, or EVA. This approach,
covered in Chapter 17, concentrates upon the wealth-increasing potential that an alternative offers a
corporation. The resulting EVA values are the equivalent of an AW analysis of after-tax cash flows.
6.2 Calculation of Capital Recovery and AW Values
An alternative should have the following cash flow estimates:
Initial investment P. This is the total first cost of all assets and services required to initiate
the alternative. When portions of these investments take place over several years, their present
worth is an equivalent initial investment. Use this amount as P.
Salvage value S. This is the terminal estimated value of assets at the end of their useful life.
The S is zero if no salvage is anticipated; S is negative when it will cost money to dispose of
the assets. For study periods shorter than the useful life, S is the estimated market value or
trade-in value at the end of the study period.
Annual amount A. This is the equivalent annual amount (costs only for cost alternatives;
costs and receipts for revenue alternatives). Often this is the annual operating cost (AOC) or
M&O cost, so the estimate is already an equivalent A value.
Salvage/market value
The annual worth (AW) value for an alternative is comprised of two components: capital
recovery for the initial investment P at a stated interest rate (usually the MARR) and the equivalent
annual amount A. The symbol CR is used for the capital recovery component. In equation form,
AW ⴝ CR ⴙ A
[6.2]
Both CR and A represent costs. The total annual amount A is determined from uniform recurring
costs (and possibly receipts) and nonrecurring amounts. The P兾A and P兾F factors may be necessary to first obtain a present worth amount; then the A兾P factor converts this amount to the A
value in Equation [6.2]. (If the alternative is a revenue project, there will be positive cash flow
estimates present in the calculation of the A value.)
The recovery of an amount of capital P committed to an asset, plus the time value of the
capital at a particular interest rate, is a fundamental principle of economic analysis.
Capital recovery (CR) is the equivalent annual amount that the asset, process, or system must
earn (new revenue) each year to just recover the initial investment plus a stated rate of return
over its expected life. Any expected salvage value is considered in the computation of CR.
The A兾P factor is used to convert P to an equivalent annual cost. If there is some anticipated
positive salvage value S at the end of the asset’s useful life, its equivalent annual value is recovered using the A兾F factor. This action reduces the equivalent annual cost of the asset. Accordingly, CR is calculated as
CR ⴝ ⴚP(A兾P,i,n) ⴙ S(A兾F,i,n)
[6.3]
EXAMPLE 6.2
Lockheed Martin is increasing its booster thrust power in order to win more satellite launch contracts from European companies interested in opening up new global communications markets. A
piece of earth-based tracking equipment is expected to require an investment of $13 million, with
$8 million committed now and the remaining $5 million expended at the end of year 1 of the
project. Annual operating costs for the system are expected to start the first year and continue at
$0.9 million per year. The useful life of the tracker is 8 years with a salvage value of $0.5 million.
Calculate the CR and AW values for the system, if the corporate MARR is 12% per year.
Solution
Capital recovery: Determine P in year 0 of the two initial investment amounts, followed by the
use of Equation [6.3] to calculate the capital recovery. In $1 million units,
P ⫽ 8 ⫹ 5(P兾F,12%,1) ⫽ $12.46
Capital recovery
154
Annual Worth Analysis
Chapter 6
CR ⫽ ⫺12.46(A兾P,12%,8) ⫹ 0.5(A兾F,12%,8)
⫽ ⫺12.46(0.20130) ⫹ 0.5(0.08130)
⫽ $⫺2.47
The correct interpretation of this result is very important to Lockheed Martin. It means that
each and every year for 8 years, the equivalent total net revenue from the tracker must be at
least $2,470,000 just to recover the initial present worth investment plus the required return of
12% per year. This does not include the AOC of $0.9 million each year.
Annual worth: To determine AW, the cash flows in Figure 6–2a must be converted to an
equivalent AW series over 8 years (Figure 6–2b). Since CR ⫽ $⫺2.47 million is an equivalent
annual cost, as indicated by the minus sign, total AW is determined.
AW ⫽ ⫺2.47 ⫺ 0.9 ⫽ $⫺3.37 million per year
This is the AW for all future life cycles of 8 years, provided the costs rise at the same rate as
inflation, and the same costs and services are expected to apply for each succeeding life cycle.
Figure 6–2
$0.5
0
1
2
3
4
5
$0.9
6
7 8
0
1
2
3
4
5
6
AW = ?
7
8
(a) Cash flow diagram
for satellite tracker
costs, and (b) conversion to an equivalent
AW (in $1 million),
Example 6.2.
$5.0
$8.0
(a)
(b)
There is a second, equally correct way to determine CR. Either method results in the same
value. There is a relation between the A兾P and A兾F factors.
(A兾F,i,n) ⫽ (A兾P,i,n) ⫺ i
Both factors are present in the CR Equation [6.3]. Substitute for the A兾F factor to obtain
CR ⫽ ⫺P(A兾P,i,n) ⫹ S[(A兾P,i,n) ⫺ i]
⫽ ⫺[(P ⫺ S)(A兾P,i,n) ⫹ S(i)]
There is a basic logic to this formula. Subtracting S from the initial investment P before applying
the A兾P factor recognizes that the salvage value will be recovered. This reduces CR, the annual
cost of asset ownership. However, the fact that S is not recovered until year n of ownership is
compensated for by charging the annual interest S(i) against the CR. Although either CR relation
results in the same amount, it is better to consistently use the same method. The first method,
Equation [6.3], will be used in this text.
For solution by spreadsheet, use the PMT function to determine CR only in a single spreadsheet cell. The general function ⫽ PMT(i%,n,P,F ) is rewritten using the initial investment as
P and ⫺S for the salvage value. The format is
ⴝ PMT(i%,n,P,ⴚS)
[6.4]
As an illustration, determine only the CR in Example 6.2. The equivalent initial investment
in year 0 is $12.46 million. The complete function for the CR amount (in $1 million units) is
⫽ PMT(12%,8,12.46,⫺0.5). The answer of $⫺2.47 (million) will be displayed in the spreadsheet cell.
As we learned in Section 3.1, one spreadsheet function can be embedded in another function.
In the case of Example 6.2, the initial investment is distributed over a 2-year period. The one-cell
PMT function, with the PV function embedded (in bold), can be written as ⫽ PMT(12%,8,
8ⴙPV(12%,1,ⴚ5),⫺0.5) to display the same value CR ⫽ $⫺2.47.
155
Evaluating Alternatives by Annual Worth Analysis
6.3
6.3 Evaluating Alternatives by Annual
Worth Analysis
The annual worth method is typically the easiest to apply of the evaluation techniques when the
MARR is specified. The AW is calculated over the respective life of each alternative, and the selection guidelines are the same as those used for the PW method. For mutually exclusive alternatives,
whether cost- or revenue-based, the guidelines are as follows:
One alternative: If AW ⱖ 0, the requested MARR is met or exceeded and the alternative is
economically justified.
Two or more alternatives: Select the alternative with the AW that is numerically largest,
that is, less negative or more positive. This indicates a lower AW of cost for cost alternatives
or a larger AW of net cash flows for revenue alternatives.
If any of the three assumptions in Section 6.1 is not acceptable for an alternative, a study period
analysis must be used. Then the cash flow estimates over the study period are converted to AW
amounts. The following two examples illustrate the AW method for one project and two alternatives.
EXAMPLE 6.3
Heavenly Pizza, which is located in Toronto, fares very well with its competition in offering
fast delivery. Many students at the area universities and community colleges work part-time
delivering orders made via the web. The owner, Jerry, a software engineering graduate, plans
to purchase and install five portable, in-car systems to increase delivery speed and accuracy.
The systems provide a link between the web order-placement software and the On-Star system
for satellite-generated directions to any address in the area. The expected result is faster, friendlier service to customers and larger income.
Each system costs $4600, has a 5-year useful life, and may be salvaged for an estimated
$300. Total operating cost for all systems is $1000 for the first year, increasing by $100 per
year thereafter. The MARR is 10%. Perform an annual worth evaluation for the owner that
answers the following questions. Perform the solution by hand and by spreadsheet.
(a) How much new annual net income is necessary to recover the investment at the MARR
of 10% per year?
(b) Jerry estimates increased net income of $6000 per year for all five systems. Is this project
financially viable at the MARR?
(c) Based on the answer in part (b), determine how much new net income Heavenly Pizza
must have to economically justify the project. Operating costs remain as estimated.
Solution by Hand
(a) The capital recovery amount calculated by Equation [6.3] answers the first question.
CR ⫽ ⫺5[4600(A兾P,10%,5)] ⫹ 5[300(A兾F,10%,5)]
⫽ ⫺5[4600(0.26380)] ⫹ 5[300(0.16380)]
⫽ $⫺5822
The five systems must generate an equivalent annual new revenue of $5822 to recover the
initial investment plus a 10% per year return.
(b) Figure 6–3 presents the cash flows over 5 years. The annual operating cost series,
combined with the estimated $6000 annual income, forms an arithmetic gradient series
with a base amount of $5000 and G ⫽ $⫺100. The project is financially viable if
AW ⱖ 0 at i ⫽ 10% per year. Apply Equation [6.2], where A is the equivalent annual
net income series.
AW ⫽ CR ⫹ A ⫽ ⫺5822 ⫹ 5000 ⫺ 100(A兾G,10%,5)
⫽ $⫺1003
The system is not financially justified at the net income level of $6000 per year.
Project evaluation
ME alternative
selection
156
Annual Worth Analysis
Chapter 6
Figure 6–3
$1500
Cash flow diagram used to
compute AW, Example 6.3.
$6000
0
1
2
3
4
5
$1000
$1100
$1200
$1300
$1400
$23,000
(c) Let the required income equal R, and set the AW relation equal to zero to find the minimum
income to justify the system.
0 ⫽ ⫺5822 ⫹ (R ⫺ 1000) ⫺ 100(A兾G,10%,5)
R ⫽ ⫺5822 ⫺ 1000 ⫺ 100(1.8101)
⫽ $7003 per year
Solution by Spreadsheet
The spreadsheet in Figure 6–4 summarizes the estimates and answers the questions posed
for Heavenly Pizza with the same values that were determined in the hand solution.
Figure 6–4
Spreadsheet solution
of Example 6.3.
(a) Capital recovery
in cell B16, (b) AW in
cell E17, and (c) Goal
Seek template and
outcome in cell B5.
⫽ ⫺PMT($B$2, 5, NPV($B$2,B10:B14)+B9)
⫽ ⫺PMT($B$2, 5, NPV($B$2,E10:E14)+E9)
(a) and (b)
(c) Required income to
justify system
(c)
AW of a Permanent Investment
6.4
157
Cell references are used in the spreadsheet functions to accommodate changes in estimated values.
(a) The capital recovery CR ⫽ $–5822 is displayed in column B using the PMT function
with an embedded NPV function, as shown in the cell tag.
(b) The annual worth AW ⫽ $–1003 is displayed in column E using the PMT function
shown. The arithmetic gradient series of costs and estimated income of $6000 in columns C and D, respectively, are added to obtain the net income necessary for the PMT
function.
(c) The minimum required income is determined in the lower part of Figure 6–4. This is
easily accomplished by setting AW ⫽ 0 (column E) in the Goal Seek tool and letting it
find the income per year of $7003 to balance the AW equation.
EXAMPLE 6.4
Luby’s Cafeterias is in the process of forming a separate business unit that provides meals to
facilities for the elderly, such as assisted care and long-term care centers. Since the meals are
prepared in one central location and distributed by trucks throughout the city, the equipment
that keeps food and drink cold and hot is very important. Michele is the general manager of this
unit, and she wishes to choose between two manufacturers of temperature retention units that
are mobile and easy to sterilize after each use. Use the cost estimates below to select the more
economic unit at a MARR of 8% per year.
Hamilton (H)
Initial cost P, $
Annual M&O, $兾year
Refurbishment cost, $
Trade-in value S, % of P
Life, years
⫺15,000
⫺6,000
0
20
4
Infinity Care (IC)
⫺20,000
⫺9,000
⫺2,000 every 4 years
40
12
Solution
The best evaluation technique for these different-life alternatives is the annual worth method,
where AW is taken at 8% per year over the respective lives of 4 and 12 years.
AWH ⫽ annual equivalent of P ⫺ annual M&O ⫹ annual equivalent of S
⫽ ⫺15,000(A兾P,8%,4) ⫺ 6000 ⫹ 0.2(15,000)(A兾F,8%,4)
⫽ ⫺15,000(0.30192) ⫺ 6000 ⫹ 3000(0.22192)
⫽ $⫺9,863
AWIC ⫽ annual equivalent of P ⫺ annual M&O ⫺ annual equivalent of refurbishment
⫹ annual equivalent of S
⫽ ⫺20,000(A兾P,8%,12) ⫺ 9000 ⫺ 2000[(P兾F,8%,4) ⫹ (P兾F,8%,8)](A兾P,8%,12)
⫹ 0.4(20,000)(A兾F,8%,12)
⫽ ⫺20,000(0.13270) ⫺ 9000 ⫺ 2000[0.7350 ⫹ 0.5403](0.13270) ⫹ 8000(0.05270)
⫽ $⫺11,571
The Hamilton unit is considerably less costly on an annual equivalent basis.
If the projects are independent, the AW at the MARR is calculated. All projects with AW ⱖ 0
are acceptable.
Independent project
selection
6.4 AW of a Permanent Investment
This section discusses the annual worth equivalent of the capitalized cost introduced in Section 5.5. Evaluation of public sector projects, such as flood control dams, irrigation canals,
bridges, or other large-scale projects, requires the comparison of alternatives that have such long
158
Annual Worth Analysis
Chapter 6
lives that they may be considered infinite in economic analysis terms. For this type of analysis,
the annual worth (and capital recovery amount) of the initial investment is the perpetual annual
interest on the initial investment, that is, A ⫽ Pi ⫽ (CC) i. This is Equation [5.2].
Cash flows recurring at regular or irregular intervals are handled exactly as in conventional
AW computations; convert them to equivalent uniform annual amounts A for one cycle. This
automatically annualizes them for each succeeding life cycle. Add all the A values to the CR
amount to find total AW, as in Equation [6.2].
EXAMPLE 6.5
The U.S. Bureau of Reclamation is considering three proposals for increasing the capacity of the
main drainage canal in an agricultural region of Nebraska. Proposal A requires dredging the
canal to remove sediment and weeds that have accumulated during previous years’ operation.
The capacity of the canal will have to be maintained in the future near its design peak flow because of increased water demand. The Bureau is planning to purchase the dredging equipment
and accessories for $650,000. The equipment is expected to have a 10-year life with a $17,000
salvage value. The annual operating costs are estimated to total $50,000. To control weeds in the
canal itself and along the banks, environmentally safe herbicides will be sprayed during the irrigation season. The yearly cost of the weed control program is expected to be $120,000.
Proposal B is to line the canal with concrete at an initial cost of $4 million. The lining is
assumed to be permanent, but minor maintenance will be required every year at a cost of
$5000. In addition, lining repairs will have to be made every 5 years at a cost of $30,000.
Proposal C is to construct a new pipeline along a different route. Estimates are an initial cost
of $6 million, annual maintenance of $3000 for right-of-way, and a life of 50 years.
Compare the alternatives on the basis of annual worth, using an interest rate of 5% per
year.
Solution
Since this is an investment for a permanent project, compute the AW for one cycle of all recurring costs. For proposals A and C, the CR values are found using Equation [6.3], with nA ⫽ 10
and nC ⫽ 50, respectively. For proposal B, the CR is simply P(i).
Proposal A
CR of dredging equipment:
⫺650,000(A兾P,5%,10) ⫹ 17,000(A兾F,5%,10)
Annual cost of dredging
Annual cost of weed control
Proposal B
CR of initial investment: ⫺4,000,000(0.05)
Annual maintenance cost
Lining repair cost: ⫺30,000(A兾F,5%,5)
Proposal C
CR of pipeline: ⫺6,000,000(A兾P,5%,50)
Annual maintenance cost
$ ⫺82,824
⫺50,000
⫺120,000
$⫺252,824
$⫺200,000
⫺5,000
⫺5,429
$⫺210,429
$⫺328,680
⫺3,000
$⫺331,680
Proposal B, which is a permanent solution, is selected due to its lowest AW of costs.
Comment
Note the use of the A兾F factor for the lining repair cost in proposal B. The A兾F factor is used
instead of A兾P, because the lining repair cost begins in year 5, not year 0, and continues
indefinitely at 5-year intervals.
If the 50-year life of proposal C is considered infinite, CR ⫽ P(i) ⫽ $⫺300,000, instead of
$⫺328,680 for n ⫽ 50. This is a small economic difference. How long lives of 40 or more
years are treated economically is a matter of “local” practice.
159
AW of a Permanent Investment
6.4
EXAMPLE 6.6
At the end of each year, all owners and employees at Bell County Utility Cooperative are given
a bonus check based on the net profit of the Coop for the previous year. Bart just received his
bonus in the amount of $8530. He plans to invest it in an annuity program that returns
7% per year. Bart’s long-term plans are to quit the Coop job some years in the future when he is
still young enough to start his own business. Part of his future living expenses will be paid from
the proceeds that this year’s bonus accumulates over his remaining years at the Coop.
(a) Use a spreadsheet to determine the amount of annual year-end withdrawal that he can anticipate (starting 1 year after he quits) that will continue forever. He is thinking of working 15 or 20 more years.
(b) Determine the amount Bart must accumulate after 15 and 20 years to generate $3000 per
year forever.
Solution by Spreadsheet
(a) Figure 6–5 presents the cash flow diagram for n ⫽ 15 years of accumulation at 7% per
year on the $8530 deposited now. The accumulated amount after n ⫽ 15 years is indicated
as Fafter 15 ⫽ ? and the withdrawal series starts at the end of year 16. The diagram for
n ⫽ 20 would appear the same, except there is a 20-year accumulation period.
The spreadsheet in Figure 6–6 shows the functions and answers for n ⫽ 15 years in
columns C and D. The FV function displays the total accumulated at 7% after 15 years
as $23,535. The perpetual withdrawal is determined by viewing this accumulated amount
as a P value and by applying the formula
A ⫽ P(i) ⫽ 23,535(0.07) ⫽ $1647 per year
The spreadsheet function ⫽ D9*B7 performs this same computation in cell reference format in column D.
Answers for n ⫽ 20 years are displayed in column E. At a consistent 7% per year return,
Bart’s perpetual income is estimated as $1647 after 15 years, or $2311 per year if he
waits for 20 years.
A=?
i = 7%
0
1
2
3
4
13
14
15
∞
16
17
18
19
20
21
22 …
Figure 6–5
P = $8530
Fafter 15 = ?
Diagram for a perpetual
series starting after 15
years of accumulation,
Example 6.6.
Figure 6–6
Spreadsheet solution,
Example 6.6.
160
Chapter 6
Annual Worth Analysis
(b) To obtain a perpetual annual withdrawal of $3000, it is necessary to determine how much
must be accumulated 1 year before the first withdrawal of $3000. This is an application of
the relation A ⫽ P(i) solved for P, or
3000 ⫽ $42,857
A ⫽ ———
P⫽—
i
0.07
This P value is independent of how long Bart works at the Coop, because he must accumulate
this amount to achieve his goal. Figure 6–6, row 13, shows the function and result. Note that
the number of years n does not enter into the function ⫽ 3000兾B13.
Comment
The NPER function can be used to determine how many years it will take for the current
amount of $8530 to accumulate the required $42,857 at 7% per year. The row 15 function indicates that Bart will have to work at the Coop for just under 24 additional years.
6.5 Life-Cycle Cost Analysis
The PW and AW analysis techniques discussed thus far have concentrated on estimates for first
cost P, annual operating and maintenance costs (AOC or M&O), salvage value S, and predictable
periodic repair and upgrade costs, plus any revenue estimates that may favor one alternative over
another. There are usually a host of additional costs involved when the complete project life costs
are evaluated. A life-cycle cost analysis includes these additional estimates to the extent that they
can be reliably determined.
Life-cycle cost (LCC) analysis utilizes AW or PW methods to evaluate cost estimates for the entire life cycle of one or more projects. Estimates will cover the entire life span from the early
conceptual stage, through the design and development stages, throughout the operating stage, and
even the phaseout and disposal stages. Both direct and indirect costs are included to the extent
possible, and differences in revenue and savings projections between alternatives are included.
Some typical LCC applications are life-span analysis for military and commercial aircraft,
new manufacturing plants, new automobile models, new and expanded product lines, and government systems at federal and state levels. For example, the U.S. Department of Defense
requires that a government contractor include an LCC budget and analysis in the originating
proposal for most defense systems.
Most commonly the LCC analysis includes costs, and the AW method is used for the analysis,
especially if only one alternative is evaluated. If there are expected revenue or other benefit differences between alternatives, a PW analysis is recommended. Public sector projects are usually
evaluated using a benefit/cost analysis (Chapter 9), rather than LCC analysis, because estimates
to the citizenry are difficult to make with much accuracy. The direct costs mentioned above include material, human labor, equipment, supplies, and other costs directly related to a product,
process, or system. Some examples of indirect cost components are taxes, management, legal,
warranty, quality, human resources, insurance, software, purchasing, etc. Direct and indirect
costs are discussed further in Chapter 15.
LCC analysis is most effectively applied when a substantial percentage of the life span (postpurchase) costs, relative to the initial investment, will be expended in direct and indirect operating, maintenance, and similar costs once the system is operational. For example, the evaluation
of two equipment purchase alternatives with expected useful lives of 5 years and M&O costs of
5% to 10% of the initial investment does not require an LCC analysis. However, let’s assume that
Exxon-Mobil wants to evaluate the design, construction, operation, and support of a new type
and style of tanker that can transport oil over long distances of ocean. If the initial costs are in the
$100 millions with support and operating costs ranging from 25% to 35% of this amount over a
25-year life, the logic of an LCC analysis will offer a better understanding of the economic viability of the project.
To understand how a LCC analysis works, first we must understand the phases and stages of
systems engineering or systems development. Many books and manuals are available on systems
development and analysis. Generally, the LCC estimates may be categorized into a simplified
Life-Cycle Cost Analysis
6.5
format for the major phases of acquisition, operation, and phaseout/disposal, and their respective
stages.
Acquisition phase: all activities prior to the delivery of products and services.
• Requirements definition stage—Includes determination of user/customer needs, assessing them relative to the anticipated system, and preparation of the system requirements
documentation.
• Preliminary design stage—Includes feasibility study, conceptual, and early-stage plans;
final go–no go decision is probably made here.
• Detailed design stage—Includes detailed plans for resources—capital, human, facilities, information systems, marketing, etc.; there is some acquisition of assets, if economically justifiable.
Operation phase: all activities are functioning, products and services are available.
• Construction and implementation stage—Includes purchases, construction, and implementation of system components; testing; preparation, etc.
• Usage stage—Uses the system to generate products and services; the largest portion of
the life cycle.
Phaseout and disposal phase: covers all activities to transition to a new system; removal/
recycling/disposal of old system.
EXAMPLE 6.7
In the 1860s, General Mills Inc. and Pillsbury Inc. both started in the flour business in the Twin
Cities of Minneapolis–St. Paul, Minnesota. In the decade of 2000 to 2010, General Mills purchased Pillsbury for a combination cash and stock deal worth more than $10 billion and integrated the product lines. Food engineers, food designers, and food safety experts made many
cost estimates as they determined the needs of consumers and the combined company’s ability
to technologically and safely produce and market new food products. At this point only cost
estimates have been addressed—no revenues or profits.
Assume that the major cost estimates below have been made based on a 6-month study
about two new products that could have a 10-year life span for the company. Use LCC analysis at the industry MARR of 18% to determine the size of the commitment in AW terms.
(Time is indicated in product-years. Since all estimates are for costs, they are not preceded by
a minus sign.)
Consumer habits study (year 0)
Preliminary food product design (year 1)
Preliminary equipment/plant design (year 1)
Detail product designs and test marketing (years 1, 2)
Detail equipment/plant design (year 2)
$0.5 million
0.9 million
0.5 million
1.5 million each year
1.0 million
Equipment acquisition (years 1 and 2)
Current equipment upgrades (year 2)
New equipment purchases (years 4 and 8)
$2.0 million each year
1.75 million
2.0 million (year 4) ⫹
10% per purchase
thereafter
200,000 (year 3) ⫹
4% per year thereafter
Annual equipment operating cost (AOC) (years 3–10)
Marketing, year 2
years 3–10
year 5 only
Human resources, 100 new employees for 2000 hours
per year (years 3–10)
Phaseout and disposal (years 9 and 10)
$8.0 million
5.0 million (year 3)
and ⫺0.2 million
per year thereafter
3.0 million extra
$20 per hour (year 3) ⫹
5% per year
$1.0 million each year
161
162
Annual Worth Analysis
Chapter 6
Solution
LCC analysis can get complicated rapidly due to the number of elements involved. Calculate
the PW by phase and stage, add all PW values, then find the AW over 10 years. Values are in
$1 million units.
Acquisition phase:
Requirements definition: consumer study
PW ⫽ $0.5
Preliminary design: product and equipment
PW ⫽ 1.4(P兾F,18%,1) ⫽ $1.187
Detailed design: product and test marketing, and equipment
PW ⫽ 1.5(P兾A,18%,2) ⫹ 1.0(P兾F,18%,2) ⫽ $3.067
Operation phase:
Construction and implementation: equipment and AOC
PW ⫽ 2.0(P兾A,18%,2) ⫹ 1.75(P兾F,18%,2) ⫹ 2.0(P兾F,18%,4) ⫹ 2.2(P兾F,18%,8)
(
冤
)
冥
1.04 8
1 ⫺ ——
1.18
⫹ 0.2 ————— (P兾F,18%,2) ⫽ $6.512
0.14
Use: marketing
PW ⫽ 8.0(P兾F,18%,2) ⫹ [5.0(P兾A,18%,8) ⫺ 0.2(P兾G,18%,8)](P兾F,18%,2)
⫹ 3.0(P兾F,18%,5)
⫽ $20.144
Use: human resources: (100 employees)(2000 h兾yr)($20兾h) ⫽ $4.0 million in year 3
冤
(
)
冥
1.05 8
1 ⫺ ——
1.18
PW ⫽ 4.0 ————— (P兾F,18%,2) ⫽ $13.412
0.13
Phaseout phase:
PW ⫽ 1.0(P兾A,18%,2)(P兾F,18%,8) ⫽ $0.416
The sum of all PW of costs is PW ⫽ $45.238 million. Finally, determine the AW over the
expected 10-year life span.
AW ⴝ 45.238 million(A兾P,18%,10) ⴝ $10.066 million per year
This is the LCC estimate of the equivalent annual commitment to the two proposed
products.
Often the alternatives compared by LCC do not have the same level of output or amount of
usage. For example, if one alternative will produce 20 million units per year and a second alternative will operate at 35 million per year, the AW values should be compared on a currency unit/
unit produced basis, such as dollar/unit or euro/hour operated.
Figure 6–7 presents an overview of how costs may be distributed over an entire life cycle. For
some systems, typically defense systems, operating and maintenance costs rise fast after acquisition and remain high until phaseout occurs.
The total LCC for a system is established or locked in early in the life cycle. It is not unusual
to have 75% to 85% of the entire life span LCC committed during the preliminary and detail
Life-Cycle Cost Analysis
Costs
6.5
Requirements Acquisition and
installation
and
design
Operating and
maintenance
Start
Time
Phaseout
and disposal
End of
life cycle
Figure 6–7
Typical distribution of life-cycle costs of the phases for one
life cycle.
D
%
%
B
B
Reduced
total LCC
Committed
Cumulative LCC
Cumulative LCC
Committed
Actual
#1
#1
C
E
F
#2
A
A
Time
Acquisition
phase
Operation
phase
Time
Acquisition
phase
(a)
Operation
phase
(b)
Figure 6–8
LCC envelopes for committed and actual costs: (a) design 1, (b) improved design 2.
design stages. As shown in Figure 6–8a, the actual or observed LCC (bottom curve AB) will trail
the committed LCC throughout the life span (unless some major design flaw increases the total
LCC of design #1 above point B).
The potential for significantly reducing total LCC occurs primarily during the early stages. A
more effective design and more efficient equipment can reposition the envelope to design #2 in
Figure 6–8b. Now the committed LCC curve AEC is below AB at all points, as is the actual LCC
curve AFC. It is this lower envelope #2 we seek. The shaded area represents the reduction in
actual LCC.
Even though an effective LCC envelope may be established early in the acquisition phase,
it is not uncommon that unplanned cost-saving measures are introduced during the acquisition
phase and early operation phase. These apparent “savings” may actually increase the total
LCC, as shown by curve AFD. This style of ad hoc cost savings, often imposed by management early in the design stage and/or construction stage, can substantially increase costs later,
especially in the after-sale portion of the use stage. For example, the use of inferior-strength
concrete and steel has been the cause of structural failures many times, thus increasing the
overall life span LCC.
163
164
Annual Worth Analysis
Chapter 6
CHAPTER SUMMARY
The annual worth method of comparing alternatives is often preferred to the present worth method,
because the AW comparison is performed for only one life cycle. This is a distinct advantage when
comparing different-life alternatives. AW for the first life cycle is the AW for the second, third, and
all succeeding life cycles, under certain assumptions. When a study period is specified, the AW
calculation is determined for that time period, regardless of the lives of the alternatives.
For infinite-life (perpetual) alternatives, the initial cost is annualized simply by multiplying P
by i. For finite-life alternatives, the AW through one life cycle is equal to the perpetual equivalent
annual worth.
Life-cycle cost analysis is appropriate for systems that have a large percentage of costs in
operating and maintenance. LCC analysis helps in the analysis of all costs from design to
operation to disposal phases.
PROBLEMS
after its 4-year life. If the project will be needed for
6 years, what would the market (salvage) value of
the 2-year-old asset have to be for the annual worth
to be the same as it is for one life cycle of the asset?
Use an interest rate of 10% per year.
Annual Worth Calculations
6.1 If you are asked to provide an annual worth (AW)
comparison of alternatives after a present worth
(PW) comparison has already been done, what factor multiplied by the PW values provide the correct
AW values?
6.6 A sports mortgage is an innovative way to finance
cash-strapped sports programs by allowing fans to
sign up to pay a “mortgage” for the right to buy
good seats at football games for several decades
with season tickets locked in at current prices. At
Notre Dame, the locked-in price period is 50 years.
If a fan pays a $130,000 “mortgage” fee now (i.e.,
in year 0) when season tickets are selling for $290
each, what is the equivalent annual cost of the
football tickets over the 50-year period at an interest rate of 8% per year?
6.2 List three assumptions that are inherent in the
annual worth method of comparing alternatives.
6.3 In the annual worth method of comparing alternatives that have different lives, why do you calculate the AW of the alternatives over their respective life cycles instead of over the least common
multiple of their lives?
6.4 James developed the two cash flow diagrams shown
at the bottom of this page. The cash flows for alternative B represent two life cycles of A. Calculate the annual worth value of each over the
respective life cycles to demonstrate that they are
the same. Use an interest rate of 10% per year.
6.7 If a fan buys a sports mortgage to USC football
games by paying $130,000 in 10 equal payments
starting now, and then pays a fixed price of $290
per year for 50 years (starting 1 year from now)
for season tickets, what is the AW in years 1
through 50 of the season tickets at 8% per year
interest?
6.5 An asset with a first cost of $20,000 has an annual
operating cost of $12,000 and a $4000 salvage value
Alternative A
Alternative B
i = 10% per year
$1000
0
1
2
$25
$25
3
Year
0
$25
$1000
1
2
3
4
5
$25
$25
$25
$25
$25
$4000
$5000
$5000
6
$25
Year
Problems
6.8 Eight years ago, Ohio Valley Trucking purchased a
large-capacity dump truck for $115,000 to provide
short-haul earthmoving services. The company
sold it today for $45,000. Operating and maintenance costs averaged $10,500 per year. A complete
overhaul at the end of year 4 cost an extra $3600.
Calculate the annual cost of the truck at 8% per
year interest.
6.9 A major repair on the suspension system of a
5-year-old car cost $2000 because the warranty
expired after 3 years of ownership. The cost of
periodic maintenance has been $800 every 2 years.
If the owner donates the car to charity after 8 years
of ownership, what is the equivalent annual cost of
the repair and maintenance in the 8-year period
of ownership? Use an interest rate of 8% per year,
and assume that the owner paid the $800 maintenance cost immediately before donating the car in
year 8.
Capital Recovery
6.10 Ten years ago, Jacobson Recovery purchased a
wrecker for $285,000 to move disabled 18-wheelers.
He anticipated a salvage value of $50,000 after 10
years. During this time his average annual revenue
totaled $52,000. (a) Did he recover his investment
and a 12% per year return? (b) If the annual M&O
cost was $10,000 the first year and increased by a
constant $1000 per year, was the AW positive or
negative at 12% per year? Assume the $50,000 salvage was realized.
6.11
Sylvia has received a $500,000 inheritance from
her favorite, recently deceased aunt in Hawaii.
Sylvia is planning to purchase a condo in Hawaii
in the same area where her aunt lived all her life
and to rent it to vacationers. She hopes to make 8%
per year on this purchase over an ownership period
of 20 years. The condo’s total first cost is $500,000,
and she conservatively expects to sell it for 90% of
the purchase price. No annual M&O costs are considered in the analysis. (a) What is the capital recovery amount? (b) If there is a real boom in rental
real estate 10 years in the future, what sales price
(as a percentage of original purchase price) is necessary at that time (year 10) to realize the same
amount as the 8% return expected over the 20-year
ownership period?
6.12 Humana Hospital Corporation installed a new MRI
machine at a cost of $750,000 this year in its medical professional clinic in Cedar Park. This state-ofthe-art system is expected to be used for 5 years
and then sold for $75,000. Humana uses a return
requirement of 24% per year for all of its medical
diagnostic equipment. As a bioengineering student
165
currently serving a coop semester on the management staff of Humana Corporation in Louisville,
Kentucky, you are asked to determine the minimum revenue required each year to realize the expected recovery and return. Also, you are asked to
draw two cash flow diagrams, one showing the
MRI purchase and sale cash flow and a second depicting the required capital recovery each year.
Alternative Comparison
6.13 Polypropylene wall caps, used for covering exterior vents for kitchen cooktops, bathroom fans,
dryers, and other building air exhausts, can be
made by two different methods. Method X will
have a first cost of $75,000, an operating cost of
$32,000 per year, and a $9000 salvage value after
4 years. Method Y will have a first cost of $140,000,
an operating cost of $24,000 per year, and a
$19,000 salvage value after its 4-year life. At an
interest rate of 10% per year, which method should
be used on the basis of an annual worth analysis?
6.14 Nissan’s all-electric car, the Leaf, has a base price
of $32,780 in the United States, but it is eligible
for a $7500 federal tax credit. A consulting engineering company wants to evaluate the purchase
or lease of one of the vehicles for use by its employees traveling to job sites in the local area. The
cost for leasing the vehicle will be $4200 per year
(payable at the end of each year) after an initialization charge of $2500 paid now. If the company
purchases the vehicle, it will also purchase a home
charging station for $2200 that will be partially
offset by a 50% tax credit. If the company expects
to be able to sell the car and charging station for
40% of the base price of the car alone at the end of
3 years, should the company purchase or lease the
car? Use an interest rate of 10% per year and annual worth analysis.
6.15 A new structural design software package is available for analyzing and designing three-sided guyed
towers and three- and four-sided self-supporting
towers. A single-user license will cost $6000 per
year. A site license has a one-time cost of $22,000.
A structural engineering consulting company is
trying to decide between two alternatives: buy a
single-user license now and one each year for the
next 3 years (which will provide 4 years of service), or buy a site license now. Determine which
strategy should be adopted at an interest rate of
10% per year for a 4-year planning period using
the annual worth method of evaluation.
6.16 The city council in a certain southwestern city is
considering whether to construct permanent restrooms in 22 of its smaller parks (i.e., parks of less
166
Annual Worth Analysis
Chapter 6
than 12 acres) or pay for portable toilets on a yearround basis. The cost of constructing the 22 permanent restrooms will be $3.8 million. The 22
portable restrooms can be rented for $7500 each
for 1 year. The service life of a permanent restroom is 20 years. Using an interest rate of 6% per
year and an annual worth analysis, determine if the
city should build the permanent restrooms or lease
the portable ones.
6.17
A remotely located air sampling station can be powered by solar cells or by running an above ground
electric line to the site and using conventional
power. Solar cells will cost $16,600 to install and
will have a useful life of 5 years with no salvage
value. Annual costs for inspection, cleaning, etc.,
are expected to be $2400. A new power line will
cost $31,000 to install, with power costs expected
to be $1000 per year. Since the air sampling project
will end in 5 years, the salvage value of the line is
considered to be zero. At an interest rate of 10% per
year, (a) which alternative should be selected on the
basis of an annual worth analysis and (b) what must
be the first cost of the above ground line to make the
two alternatives equally attractive economically?
6.18 The cash flows for two small raw water treatment
systems are shown. Determine which should be
selected on the basis of an annual worth analysis at
10% per year interest.
First cost, $
Annual cost, $ per year
Salvage value, $
Life, years
MF
UF
–33,000
–8,000
4,000
3
–51,000
–3,500
11,000
6
6.19 PGM Consulting is under contract to Montgomery
County for evaluating alternatives that use a
robotic, liquid-propelled “pig” to periodically inspect the interior of buried potable water pipes for
leakage, corrosion, weld strength, movement over
time, and a variety of other parameters. Two
equivalent robot instruments are available. Robot
Joeboy will have a first cost of $85,000, annual
M&O costs of $30,000, and a $40,000 salvage
value after 3 years. Robot Watcheye will have a
first cost of $125,000, annual M&O costs of
$27,000, and a $33,000 salvage value after its
5-year life. Assume an interest rate of 8% per year.
(a) Which robot is the better economic option?
(b) Using the spreadsheet Goal Seek tool, determine the first cost of the robot not selected in
(a) so that it will be the economic selection.
6.20 TT Racing and Performance Motor Corporation
wishes to evaluate two alternative CNC machines
for NHRA engine building. Use the AW method at
10% per year to select the better alternative.
First cost, $
Annual operating cost,
$ per year
Salvage value, $
Life, years
Machine R
Machine S
⫺250,000
⫺40,000
⫺370,500
–50,000
20,000
3
30,000
5
6.21 The maintenance and operation (M&O) cost of
front-end loaders working under harsh environmental conditions tends to increase by a constant
$1200 per year for the first 5 years of operation.
For a loader that has a first cost of $39,000 and
first-year M&O cost of $17,000, compare the
equivalent annual worth of a loader kept for 4 years
with one kept for 5 years at an interest rate of
12% per year. The salvage value of a used loader is
$23,000 after 4 years and $18,000 after 5 years.
6.22 You work for Midstates Solar Power. A manager
asked you to determine which of the following
two machines will have the lower (a) capital recovery and (b) equivalent annual total cost. Machine Semi2 has a first cost of $80,000 and an
operating cost of $21,000 in year 1, increasing
by $500 per year through year 5, after which
time it will have a salvage value of $13,000. Machine Auto1 has a first cost of $62,000 and an
operating cost of $21,000 in year 1, increasing
by 8% per year through year 5, after which time
it will have a scavenge value of $2000. Utilize an
interest rate of 10% per year to determine both
estimates.
Permanent Investments
6.23 The State of Chiapas, Mexico, decided to fund a
program for literacy. The first cost is $200,000
now, and an update budget of $100,000 every
7 years forever is requested. Determine the perpetual equivalent annual cost at an interest rate of
10% per year.
6.24 Calculate the perpetual equivalent annual cost
(years 1 through infinity) of $5 million in year 0,
$2 million in year 10, and $100,000 in years 11
through infinity. Use an interest rate of 10% per
year.
6.25 A Pennsylvania coal mining operation has installed an in-shaft monitoring system for oxygen
tank and gear readiness for emergencies. Based on
maintenance patterns for previous systems, there
are no maintenance costs for the first 2 years, they
167
Problems
increase for a time period, and then they level off.
Maintenance costs are expected to be $150,000 in
year 3, $175,000 in year 4, and amounts increasing by $25,000 per year through year 6 and remain constant thereafter for the expected 10-year
life of the system. If similar systems with similar
costs will replace the current one, determine the
perpetual equivalent annual maintenance cost at
i ⫽ 10% per year.
6.26 Compare two alternatives for a security system
surrounding a power distribution substation using
annual worth analysis and an interest rate of 10%
per year.
First cost, $
Annual cost, $ per year
Salvage value, $
Life, years
Condi
Torro
⫺25,000
⫺9,000
3,000
3
⫺130,000
⫺2,500
150,000
⬁
in usage for crop irrigation, is considering the
purchase of one of the Blanton systems. There are
three options with two levels of automation for
the first two options. The estimated costs and associated cash flow diagrams over a 10-year period are summarized below and on the next page,
respectively, for each of the five alternatives.
Costs are categorized as design (Des), development (Dev), and operation (Oper). For alternatives A and B, there is an extra cost of $15,000 per
installation year to maintain the manual system in
place now. Level 2 development costs are distributed equally over a 2-year period. Alternative C
is a retrofit of the current manual system with no
design or development costs, and there is no level
1 option. At an interest rate of 10% per year and a
10-year study period, determine which alternative and level has the lowest LCC.
Alternative
6.27 A new bridge across the Allegheny River in
Pittsburgh is expected to be permanent and will
have an initial cost of $30 million. This bridge must
be resurfaced every 5 years at a cost of $1 million.
The annual inspection and operating costs are estimated to be $50,000. Determine its equivalent annual worth at an interest rate of 10% per year.
6.28 For the cash flows shown, use an annual worth
comparison and an interest rate of 10% per year.
(a) Determine the alternative that is economically best.
(b) Determine the first cost required for each of
the two alternatives not selected in (a) so that
all alternatives are equally acceptable. Use a
spreadsheet to answer this question.
First cost, $
Annual cost,
$ per year
Overhaul every
10 years, $
Salvage value, $
Life, years
X
Y
Z
⫺90,000
⫺40,000
⫺400,000
⫺20,000
⫺650,000
⫺13,000
—
—
⫺80,000
7,000
3
25,000
10
200,000
⬁
Life-Cycle Cost
6.29 Blanton Agriculture of Santa Monica, California,
offers different types and levels of irrigation
water conservation systems for use in areas where
groundwater depletion is a serious concern. A
large farming corporation in India, where depletion is occurring at an alarming rate of 1.6 inches
(4 centimeters) per year due to exponential growth
Cost
Component
Level
1
Level
2
A
Des
Dev
Oper
Install time
$100,000
175,000
60,000
1 year
$200,000
350,000
55,000
2 years
B
Des
Dev
Oper
Install time
$ 50,000
200,000
45,000
1 year
$100,000
500,000
30,000
2 years
C
Oper
$100,000
6.30 The Pentagon asked a defense contractor to estimate the life-cycle cost for a proposed light-duty
support vehicle. The list of items included the following categories: R&D costs (R&D), nonrecurring
investment costs (NRI), recurring upgrade costs
(RU), scheduled and unscheduled maintenance
costs (Maint), equipment usage costs (Equip), and
phaseout兾disposal costs (Po兾D). Use the cost estimates (shown in $1 million) for the 20-year life
cycle to calculate the annual LCC at an interest rate
of 7% per year.
Year
R&D
NRI
0
1
2
3
4
5
6–10
11 on
18–20
5.5
3.5
2.5
0.5
1.1
5.2
10.5
10.5
RU
Maint
Equip
1.3
3.1
4.2
6.5
2.2
0.6
1.4
1.6
2.7
3.5
1.5
3.6
5.3
7.8
8.5
Po兾D
2.7
168
Annual Worth Analysis
Chapter 6
0
1
2
3
4
5
6
7
8
9
10
6
7
8
9
10
7
8
9
10
7
8
9
10
7
8
9
10
Alternative A
Level 1
$60,000
$100,000
$175,000
⫹ 15,000
0
1
2
3
4
5
Alternative A
Level 2
$55,000
$200,000 $175,000
⫹ 15,000
0
1
2
3
4
Alternative B
Level 1
$50,000
5
6
$45,000
$200,000
⫹ 15,000
0
1
2
3
4
5
Alternative B
Level 2
6
$30,000
$100,000
$250,000
⫹ 15,000
0
1
2
3
4
5
6
Alternative C
Level 2
$100,000
Cash flow diagrams for Problem 6.29
6.31 A manufacturing software engineer at a major
aerospace corporation has been assigned the
management responsibility of a project to design, build, test, and implement AREMSS, a
new-generation automated scheduling system
for routine and expedited maintenance. Reports
on the disposition of each service will also be
entered by field personnel, then filed and archived by the system. The initial application will
be on existing Air Force in-flight refueling aircraft. The system is expected to be widely used
over time for other aircraft maintenance scheduling. Once it is fully implemented, enhancements
will have to be made, but the system is expected
to serve as a worldwide scheduler for up to
15,000 separate aircraft. The engineer, who must
make a presentation next week of the best estimates of costs over a 20-year life period, has decided to use the life-cycle cost approach of cost
estimations. Use the following information to
determine the current annual LCC at 6% per year
for the AREMSS scheduling system.
169
Additional Problems and FE Exam Review Questions
Cost in Year ($ millions)
Cost Category
1
2
3
4
5
6 on 10 18
Field study
0.5
Design of system
2.1 1.2 0.5
Software design
0.6 0.9
Hardware purchases
5.1
Beta testing
0.1 0.2
Users manual
0.1 0.1 0.2 0.2 0.06
development
System implementation
1.3 0.7
Field hardware
0.4 6.0 2.9
Training trainers
0.3 2.5 2.5 0.7
Software upgrades
0.6 3.0 3.7
6.32
The U.S. Army received two proposals for a turnkey
design-build project for barracks for infantry unit soldiers in training. Proposal A involves an off-the-shelf
“bare-bones” design and standard grade construction
of walls, windows, doors, and other features. With
this option, heating and cooling costs will be greater,
maintenance costs will be higher, and replacement
will be sooner than for proposal B. The initial cost for
A will be $750,000. Heating and cooling costs will
average $72,000 per year, with maintenance costs
averaging $24,000 per year. Minor remodeling will
be required in years 5, 10, and 15 at a cost of $150,000
each time in order to render the units usable for
20 years. They will have no salvage value.
Proposal B will include tailored design and
construction costs of $1.1 million initially, with
estimated heating and cooling costs of $36,000 per
year and maintenance costs of $12,000 per year.
There will be no salvage value at the end of the
20-year life. Which proposal should be accepted
on the basis of an annual life-cycle cost analysis, if
the interest rate is 6% per year?
6.33 A medium-size municipality plans to develop a
software system to assist in project selection during the next 10 years. A life-cycle cost approach
has been used to categorize costs into development, programming, operating, and support costs
for each alternative. There are three alternatives
under consideration, identified as M, N, and O.
The costs are summarized below. Use an annual
life-cycle cost approach to identify the best alternative at 8% per year.
Alternative
Component
Estimated Cost
M
Development
$250,000 now,
$150,000 years 1–4
$45,000 now,
$35,000 years 1, 2
$50,000 years 1–10
$30,000 years 1–5
Programming
Operation
Support
N
O
Development
Programming
Operation
Support
$10,000 now
$45,000 year 0,
$30,000 years 1–3
$80,000 years 1–10
$40,000 years 1–10
Operation
$175,000 years 1–10
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
6.34 All of the following are fundamental assumptions
for the annual worth method of analysis except:
(a) The alternatives will be needed for only one
life cycle.
(b) The services provided are needed for at least
the LCM of the lives of the alternatives.
(c) The selected alternative will be repeated for
the succeeding life cycles in exactly the same
manner as for the first life cycle.
(d) All cash flows will have the same estimated
values in every life cycle.
6.35 When comparing five alternatives that have different lives by the AW method, you must:
(a) Find the AW of each over the life of the
longest-lived alternative.
(b) Find the AW of each over the life of the
shortest-lived alternative.
(c) Find the AW of each over the LCM of all of
the alternatives.
(d) Find the AW of each alternative over its life
without considering the life of the other
alternatives.
6.36 The annual worth of an alternative can be calculated from the alternative’s:
(a) Present worth by multiplying by (A/P,i,n)
(b) Future worth by multiplying by (F/A,i,n)
(c) Either (a) or (b)
(d) Neither (a) nor (b)
6.37 The alternatives shown are to be compared on the
basis of annual worth. At an interest rate of 10%
per year, the values of n that you could use in the
(A兾P,i,n) factors to make a correct comparison by
the annual worth method are:
First cost, $
Annual cost, $ per year
Salvage value, $
Life, years
(a)
(b)
(c)
(d)
A
B
⫺50,000
⫺10,000
13,000
3
⫺90,000
⫺4,000
15,000
6
n ⫽ 3 years for A and 3 years for B
n ⫽ 3 years for A and 6 years for B
Either (a) or (b)
Neither (a) nor (b)
170
Annual Worth Analysis
Chapter 6
6.38 The alternatives shown are to be compared on the
basis of a perpetual (i.e., forever) equivalent annual worth. At an interest rate of 10% per year, the
equation that represents the perpetual AW of X1 is:
First cost, $
Annual cost, $ per year
Salvage value, $
Life, years
(a)
(b)
(c)
(d)
X1
Y1
⫺50,000
⫺10,000
13,000
3
⫺90,000
⫺4,000
15,000
6
AWX1 ⫽ ⫺50,000(0.10) ⫺ 10,000
⫹ 13,000(0.10)
AWX1 ⫽ ⫺50,000(0.10) ⫺ 10,000
⫹ 13,000(A/F,10%,3)
AWX1 ⫽ ⫺50,000(0.10) ⫺ 10,000
⫺ 37,000(P/F,10%,3)(0.10)
⫹ 13,000(0.10)
AWX1 ⫽ ⫺50,000(A/P,10%,3) ⫺ 10,000
⫹ 13,000(A/F,10%,3)
6.39 To get the AW of a cash flow of $10,000 that
occurs every 10 years forever, with the first one
occurring 10 years from now, you should:
(a) Multiply $10,000 by (A/P,i,10).
(b) Multiply $10,000 by (A/F,i,10).
(c) Multiply $10,000 by i.
(d) Multiply $10,000 by (A/F,i,n) and then
multiply by i.
Problems 6.40 through 6.43 refer to the following
estimates.
The alternatives are mutually exclusive and the MARR is
6% per year.
Vendor 1
Vendor 2
Vendor 3
First cost, $
⫺200,000
Annual cost, $ per year ⫺50,000
Revenue, $ per year
120,000
Salvage value, $
25,000
Life, years
10
⫺550,000
⫺20,000
120,000
0
15
⫺1,000,000
⫺10,000
110,000
500,000
⬁
6.40 The annual worth of vendor 2 cash flow estimates
is closest to:
(a) $⫺63,370
(b) $43,370
(c) $⫺43,370
(d) $63,370
6.41 Of the following three relations, the correct one or ones
to calculate the annual worth of vendor 1 cash flow estimates is (note: all dollar values are in thousands):
Relation 1: AW1 ⫽ ⫺200(A兾P,6%,10) ⫹ 70 ⫹
25(A兾F,6%,10)
Relation 2: AW1 ⫽ [⫺200 ⫺ 50(P兾A,6%,10) ⫹
120(P兾A,6%,10)
⫹ 25(P兾F,6%,10)](A兾P,6%,10)
Relation 3: AW1 ⫽ ⫺200(F兾P,6%,10) ⫹ 25 ⫹
(⫺50 ⫹ 120)(A兾P,6%,10)
(a)
(b)
(c)
(d)
1 and 3
Only 1
1 and 2
Only 3
6.42 The AW values for the alternatives are listed
below. The vendor or vendors that should be recommended is:
AW1 ⫽ $44,723
AW2 ⫽ $43,370
AW3 ⫽ $40,000
(a)
(b)
(c)
(d)
1 and 2
3
2
1
6.43 The capital recovery amount for vendor 3 is:
(a) $40,000 per year
(b) $60,000 per year
(c) $43,370 per year
(d) $100,000 per year
6.44 If a revenue alternative has a negative AW value and
it was correctly calculated, it means the following:
(a) The equivalent annual worth of revenues
does not exceed that of the costs.
(b) The estimates are wrong somewhere.
(c) A minus or plus sign of a cash flow was entered incorrectly into the PMT spreadsheet
function.
(d) The alternative should have a longer life so
revenues will exceed costs.
6.45 Estimates for one of two process upgrades are as
follows: first cost of $40,000, annual cost of $5000
per year, market value that decreases by $2000 per
year to the salvage value of $20,000 after the expected life of 10 years. If a 4-year study period is
used for AW analysis at 15% per year, the correct
AW value is closest to:
(a) $⫺15,000
(b) $⫺11,900
(c) $⫺7600
(d) $⫺12,600
6.46 The perpetual annual worth of investing $50,000
now and $20,000 per year starting in year 16 and
continuing forever at 12% per year is closest to:
(a) $⫺4200
(b) $⫺8650
(c) $⫺9655
(d) $⫺10,655
Case Study
6.47 All the following statements about the capital recovery amount for an alternative are false except:
(a) Annual revenue can be no more than this
amount, if the alternative is selected.
(b) A monetary estimate of new capital funds required each year for the life of the alternative.
(c)
(d)
171
An amount of revenue required to recover
the first cost plus a stated return over the life
of the alternative.
Does not consider the salvage value, since it
is returned at the end of the alternative’s life.
CASE STUDY
THE CHANGING SCENE OF AN ANNUAL WORTH ANALYSIS
Background and Information
Harry, owner of an automobile battery distributorship in
Atlanta, Georgia, performed an economic analysis 3 years ago
when he decided to place surge protectors in-line for all his
major pieces of testing equipment. The estimates used and the
annual worth analysis at MARR ⫽ 15% are summarized
below. Two different manufacturers’ protectors were compared.
Cost and installation, $
Annual maintenance cost,
$ per year
Salvage value, $
Equipment repair savings, $
Useful life, years
PowrUp
Lloyd’s
⫺26,000
⫺800
⫺36,000
⫺300
2,000
25,000
6
3,000
35,000
10
The spreadsheet in Figure 6–9 is the one Harry used to make
the decision. Lloyd’s was the clear choice due to its substantially larger AW value. The Lloyd’s protectors were installed.
During a quick review this last year (year 3 of operation),
it was obvious that the maintenance costs and repair savings
have not followed (and will not follow) the estimates made
3 years ago. In fact, the maintenance contract cost (which includes quarterly inspection) is going from $300 to $1200 per
year next year and will then increase 10% per year for the next
10 years. Also, the repair savings for the last 3 years were
$35,000, $32,000, and $28,000, as best as Harry can determine. He believes savings will decrease by $2000 per year
hereafter. Finally, these 3-year-old protectors are worth nothing on the market now, so the salvage in 7 years is zero, not
$3000.
Case Study Exercises
1. Plot a graph of the newly estimated maintenance costs
and repair savings projections, assuming the protectors
last for 7 more years.
2. With these new estimates, what is the recalculated AW
for the Lloyd’s protectors? Use the old first cost and
maintenance cost estimates for the first 3 years. If these
estimates had been made 3 years ago, would Lloyd’s
still have been the economic choice?
3. How has the capital recovery amount changed for the
Lloyd’s protectors with these new estimates?
Figure 6–9
Annual worth analysis of surge protector alternatives, case study.
CHAPTER 7
Rate of Return
Analysis: One
Project
L E A R N I N G
O U T C O M E S
Purpose: Understand the meaning of rate of return and perform an ROR evaluation of a single project.
SECTION
TOPIC
LEARNING OUTCOME
7.1
Definition
• State and understand the meaning of rate of
return.
7.2
Calculate ROR
• Use a PW or AW relation to determine the
ROR of a series of cash flows.
7.3
Cautions
• State the difficulties of using the ROR method,
relative to the PW and AW methods.
7.4
Multiple RORs
• Determine the maximum number of possible
ROR values and their values for a specific cash
flow series.
7.5
Calculate EROR
• Determine the external rate of return using
the techniques of modified ROR and return on
invested capital.
7.6
Bonds
• Calculate the nominal and effective rate of
return for a bond investment.
T
he most commonly quoted measure of economic worth for a project or alternative is its rate of return (ROR). Whether it is an engineering project with cash
flow estimates or an investment in a stock or bond, the rate of return is a wellaccepted way of determining if the project or investment is economically acceptable. Compared to the PW or AW value, the ROR is a generically different type of measure of worth,
as is discussed in this chapter. Correct procedures to calculate a rate of return using a PW or
AW relation are explained here, as are some cautions necessary when the ROR technique is
applied to a single project’s cash flows.
The ROR is known by other names such as the internal rate of return (IROR), which is the
technically correct term, and return on investment (ROI). We will discuss the computation of
ROI in the latter part of this chapter.
In some cases, more than one ROR value may satisfy the PW or AW equation. This chapter describes how to recognize this possibility and an approach to find the multiple values.
Alternatively, one reliable ROR value can be obtained by using additional information established separately from the project cash flows. Two of the techniques are covered: the modified ROR technique and the ROIC (return on invested capital) technique.
Only one alternative is considered here; Chapter 8 applies these same principles to multiple alternatives. Finally, the rate of return for a bond investment is discussed.
7.1 Interpretation of a Rate of Return Value
From the perspective of someone who has borrowed money, the interest rate is applied to the
unpaid balance so that the total loan amount and interest are paid in full exactly with the last loan
payment. From the perspective of a lender of money, there is an unrecovered balance at each time
period. The interest rate is the return on this unrecovered balance so that the total amount lent and
the interest are recovered exactly with the last receipt. Rate of return describes both of these
perspectives.
Rate of return (ROR) is the rate paid on the unpaid balance of borrowed money, or the rate
earned on the unrecovered balance of an investment, so that the final payment or receipt
brings the balance to exactly zero with interest considered.
The rate of return is expressed as a percent per period, for example, i ⫽ 10% per year. It is
stated as a positive percentage; the fact that interest paid on a loan is actually a negative rate of
return from the borrower’s perspective is not considered. The numerical value of i can range from
⫺100% to infinity, that is, ⫺100% ⱕ i ⬍ ⬁. In terms of an investment, a return of i ⫽ ⫺100%
means the entire amount is lost.
The definition above does not state that the rate of return is on the initial amount of the investment; rather it is on the unrecovered balance, which changes each time period. Example 7.1
illustrates this difference.
EXAMPLE 7.1
To get started in a new telecommuting position with AB Hammond Engineers, Jane took out
a $1000 loan at i ⫽ 10% per year for 4 years to buy home office equipment. From the lender’s
perspective, the investment in this young engineer is expected to produce an equivalent net
cash flow of $315.47 for each of 4 years.
A ⫽ $1000(A兾P,10%,4) ⫽ $315.47
This represents a 10% per year rate of return on the unrecovered balance. Compute the amount
of the unrecovered investment for each of the 4 years using (a) the rate of return on the unrecovered balance (the correct basis) and (b) the return on the initial $1000 investment. (c) Explain why all of the initial $1000 amount is not recovered by the final payment in part (b).
Solution
(a) Table 7–1 shows the unrecovered balance at the end of each year in column 6 using the
10% rate on the unrecovered balance at the beginning of the year. After 4 years the total
$1000 is recovered, and the balance in column 6 is exactly zero.
Rate of return
Rate of Return Analysis: One Project
Chapter 7
TABLE 7–1
Unrecovered Balances Using a Rate of Return of 10% on the Unrecovered
Balance
(1)
(2)
(3) ⴝ 0.10 ⴛ (2)
(4)
(5) ⴝ (4) ⴚ (3)
(6) ⴝ (2) ⴙ (5)
Year
Beginning
Unrecovered
Balance
Interest on
Unrecovered
Balance
Cash
Flow
Recovered
Amount
Ending
Unrecovered
Balance
0
1
2
3
4
—
$⫺1000.00
⫺784.53
⫺547.51
⫺286.79
—
$100.00
78.45
54.75
28.68
$261.88
$⫺1000.00
⫹315.47
⫹315.47
⫹315.47
⫹315.47
—
$215.47
237.02
260.72
286.79
$1000.00
TABLE 7–2
$⫺1000.00
⫺784.53
⫺547.51
⫺286.79
0
Unrecovered Balances Using a 10% Return on the Initial Amount
(1)
(2)
(3) ⴝ 0.10 ⴛ (2)
(4)
(5) ⴝ (4) ⴚ (3)
(6) ⴝ (2) ⴙ (5)
Year
Beginning
Unrecovered
Balance
Interest on
Initial Amount
Cash
Flow
Recovered
Amount
Ending
Unrecovered
Balance
0
1
2
3
4
—
$⫺1000.00
⫺784.53
⫺569.06
⫺353.59
—
$100
100
100
100
$400
$⫺1000.00
⫹315.47
⫹315.47
⫹315.47
⫹315.47
—
$215.47
215.47
215.47
215.47
$861.88
$⫺1000.00
⫺784.53
⫺569.06
⫺353.59
⫺138.12
(b) Table 7–2 shows the unrecovered balance if the 10% return is always figured on the initial $1000. Column 6 in year 4 shows a remaining unrecovered amount of $138.12,
because only $861.88 is recovered in the 4 years (column 5).
(c) As shown in column 3, a total of $400 in interest must be earned if the 10% return each year is
based on the initial amount of $1000. However, only $261.88 in interest must be earned if a
10% return on the unrecovered balance is used. There is more of the annual cash flow available
to reduce the remaining loan when the rate is applied to the unrecovered balance as in part (a)
and Table 7–1. Figure 7–1 illustrates the correct interpretation of rate of return in Table 7–1.
Loan balance
of $1000
Interest
$100.00
1000.00
Loan balance
reduction
$78.45
$215.47
784.53
Loan balance, $
174
$237.02
$54.75
547.51
$260.72
$28.68
286.79
$286.79
0
1
2
3
4
Loan balance
of $0
Figure 7–1
Plot of unrecovered balances and 10% per year rate of return on a $1000
amount, Table 7–1.
Year
7.2
175
Rate of Return Calculation Using a PW or AW Relation
Each year the $315.47 receipt represents 10% interest on the unrecovered balance in column 2
plus the recovered amount in column 5.
Because rate of return is the interest rate on the unrecovered balance, the computations in
Table 7–1 for part (a) present a correct interpretation of a 10% rate of return. Clearly, an interest rate applied only to the principal represents a higher rate than is stated. In practice, a socalled add-on interest rate is frequently based on principal only, as in part (b). This is sometimes referred to as the installment financing problem.
Installment financing can be discovered in many forms in everyday finances. One popular
example is a “no-interest program” offered by retail stores on the sale of major appliances, audio
and video equipment, furniture, and other consumer items. Many variations are possible, but in
most cases, if the purchase is not paid for in full by the time the promotion is over, usually
6 months to 1 year later, finance charges are assessed from the original date of purchase. Further,
the program’s fine print may stipulate that the purchaser use a credit card issued by the retail
company, which often has a higher interest rate than that of a regular credit card, for example,
24% per year compared to 15% per year. In all these types of programs, the one common theme
is more interest paid over time by the consumer. Usually, the correct definition of i as interest on
the unpaid balance does not apply directly; i has often been manipulated to the financial disadvantage of the purchaser. This was demonstrated by Example 4.4 using the Credit Card Case in
Chapter 4.
7.2 Rate of Return Calculation Using a PW
or AW Relation
The ROR value is determined in a generically different way compared to the PW or AW value for
a series of cash flows. For a moment, consider only the present worth relation for a cash flow
series. Using the MARR, which is established independent of any particular project’s cash flows,
a mathematical relation determines the PW value in actual monetary units, say, dollars or euros.
For the ROR values calculated in this and later sections, only the cash flows themselves are used
to determine an interest rate that balances the present worth relation. Therefore, ROR may be
considered a relative measure, while PW and AW are absolute measures. Since the resulting interest rate depends only on the cash flows themselves, the correct term is internal rate of return
(IROR); however, the term ROR is used interchangeably. Another definition of rate of return is
based on our previous interpretations of PW and AW.
The rate of return is the interest rate that makes the present worth or annual worth of a cash
flow series exactly equal to 0.
To determine the rate of return, develop the ROR equation using either a PW or AW relation, set
it equal to 0, and solve for the interest rate. Alternatively, the present worth of cash outflows
(costs and disbursements) PWO may be equated to the present worth of cash inflows (revenues
and savings) PWI. That is, solve for i using either of the relations
0 ⴝ PW
or
[7.1]
PWO ⴝ PWI
The annual worth approach utilizes the AW values in the same fashion to solve for i.
0 ⴝ AW
or
[7.2]
AWO ⴝ AWI
The i value that makes these equations numerically correct is called i*. It is the root of the ROR
relation. To determine if the investment project’s cash flow series is viable, compare i* with the
established MARR.
Rate of return
176
Rate of Return Analysis: One Project
Chapter 7
Figure 7–2
$1500
Cash flow for which
a value of i is to be
determined.
$500
0
1
2
3
4
5
i* = ?
$1000
The guideline is as follows:
Project evaluation
If i* ⱖ MARR, accept the project as economically viable.
If i* ⬍ MARR, the project is not economically viable.
The purpose of engineering economy calculations is equivalence in PW or AW terms for a
stated i ⱖ 0%. In rate of return calculations, the objective is to find the interest rate i* at which
the cash flows are equivalent. The calculations are the reverse of those made in previous chapters,
where the interest rate was known. For example, if you deposit $1000 now and are promised payments of $500 three years from now and $1500 five years from now, the rate of return relation
using PW factors and Equation [7.1] is
1000 ⫽ 500(P兾F,i*,3) ⫹ 1500(P兾F,i*,5)
[7.3]
The value of i* that makes the equality correct is to be determined (see Figure 7–2). If the $1000
is moved to the right side of Equation [7.3], we have the form 0 ⫽ PW.
0 ⫽ ⫺1000 ⫹ 500(P兾F,i*,3) ⫹ 1500(P兾F,i*,5)
The equation is solved for i* ⫽ 16.9% by hand using trial and error or using a spreadsheet function. The rate of return will always be greater than zero if the total amount of cash inflow is greater
than the total amount of outflow, when the time value of money is considered. Using i* ⫽ 16.9%,
a graph similar to Figure 7–1 can be constructed. It will show that the unrecovered balances each
year, starting with $⫺1000 in year 1, are exactly recovered by the $500 and $1500 receipts in
years 3 and 5.
It should be evident that rate of return relations are merely a rearrangement of a present worth
equation. That is, if the above interest rate is known to be 16.9%, and it is used to find the present
worth of $500 three years from now and $1500 five years from now, the PW relation is
PW ⫽ 500(P兾F,16.9%,3) ⫹ 1500(P兾F,16.9%,5) ⫽ $1000
This illustrates that rate of return and present worth equations are set up in exactly the same fashion. The only differences are what is given and what is sought.
There are several ways to determine i* once the PW relation is established: solution via trial
and error by hand, using a programmable calculator, and solution by spreadsheet function. The
spreadsheet is faster; the first helps in understanding how ROR computations work. We summarize two methods here and in Example 7.2. Refer to Appendix D for the discussion about solution
by calculator.
i* Using Trial and Error
The general procedure of using a PW-based equation is as follows:
1. Draw a cash flow diagram.
2. Set up the rate of return equation in the form of Equation [7.1].
3. Select values of i by trial and error until the equation is balanced.
When the trial-and-error method is applied to determine i*, it is advantageous in step 3 to get
fairly close to the correct answer on the first trial. If the cash flows are combined in such a manner
that the income and disbursements can be represented by a single factor such as P兾F or P兾A, it is
Rate of Return Calculation Using a PW or AW Relation
7.2
possible to look up the interest rate (in the tables) corresponding to the value of that factor for n
years. The problem, then, is to combine the cash flows into the format of only one of the factors.
This may be done through the following procedure:
1. Convert all disbursements into either single amounts (P or F) or uniform amounts (A) by
neglecting the time value of money. For example, if it is desired to convert an A to an F
value, simply multiply the A by the number of years n. The scheme selected for movement
of cash flows should be the one that minimizes the error caused by neglecting the time value
of money. That is, if most of the cash flow is an A and a small amount is an F, convert the F
to an A rather than the other way around.
2. Convert all receipts to either single or uniform values.
3. Having combined the disbursements and receipts so that a P兾F, P兾A, or A兾F format applies,
use the interest tables to find the approximate interest rate at which the P兾F, P兾A, or A兾F
value is satisfied. The rate obtained is a good estimate for the first trial.
It is important to recognize that this first-trial rate is only an estimate of the actual rate of return,
because the time value of money is neglected. The procedure is illustrated in Example 7.2.
i* by Spreadsheet The fastest way to determine an i* value when there is a series of equal
cash flows (A series) is to apply the RATE function. This is a powerful one-cell function, where it
is acceptable to have a separate P value in year 0 and a separate F value in year n. The format is
ⴝ RATE(n,A,P,F)
[7.4]
When cash flows vary from year to year (period to period), the best way to find i* is to enter
the net cash flows into contiguous cells (including any $0 amounts) and apply the IRR function
in any cell. The format is
ⴝ IRR(first_cell:last_cell,guess)
[7.5]
where “guess” is the i value at which the function starts searching for i*.
The PW-based procedure for sensitivity analysis and a graphical estimation of the i* value is
as follows:
1.
2.
3.
4.
5.
Draw the cash flow diagram.
Set up the ROR relation in the form of Equation [7.1], PW ⫽ 0.
Enter the cash flows onto the spreadsheet in contiguous cells.
Develop the IRR function to display i*.
Use the NPV function to develop a PW graph (PW versus i values). This graphically shows
the i* value at which PW ⫽ 0.
EXAMPLE 7.2
Applications of green, lean manufacturing techniques coupled with value stream mapping can
make large financial differences over future years while placing greater emphasis on environmental factors. Engineers with Monarch Paints have recommended to management an investment of $200,000 now in novel methods that will reduce the amount of wastewater, packaging
materials, and other solid waste in their consumer paint manufacturing facility. Estimated savings are $15,000 per year for each of the next 10 years and an additional savings of $300,000
at the end of 10 years in facility and equipment upgrade costs. Determine the rate of return
using hand and spreadsheet solutions.
Solution by Hand
Use the trial-and-error procedure based on a PW equation.
1. Figure 7–3 shows the cash flow diagram.
2. Use Equation [7.1] format for the ROR equation.
0 ⫽ ⫺200,000 ⫹ 15,000(P兾A,i*,10) ⫹ 300,000(P兾F,i*,10)
[7.6]
177
178
Rate of Return Analysis: One Project
Chapter 7
$300,000
i* = ?
$15,000
0
1
2
3
4
5
6
7
8
9
10
$200,000
Figure 7–3
Cash flow diagram, Example 7.2.
3. Use the estimation procedure to determine i for the first trial. All income will be regarded
as a single F in year 10 so that the P兾F factor can be used. The P兾F factor is selected because most of the cash flow ($300,000) already fits this factor and errors created by
neglecting the time value of the remaining money will be minimized. Only for the first
estimate of i, define P ⫽ $200,000, n ⫽ 10, and F ⫽ 10(15,000) ⫹ 300,000 ⫽ $450,000.
Now we can state that
200,000 ⫽ 450,000(P兾F,i,10)
(P兾F,i,10) ⫽ 0.444
The roughly estimated i is between 8% and 9%. Use 9% as the first trial because this approximate rate for the P兾F factor will be lower than the true value when the time value of money is
considered.
0 ⫽ ⫺200,000 ⫹ 15,000(P兾A,9%,10) ⫹ 300,000(P兾F,9%,10)
0 ⬍ $22,986
The result is positive, indicating that the return is more than 9%. Try i ⫽ 11%.
0 ⫽ ⫺200,000 ⫹ 15,000(P兾A,11%,10) ⫹ 300,000(P兾F,11%,10)
0 ⬎ $⫺6002
Since the interest rate of 11% is too high, linearly interpolate between 9% and 11%.
22,986 ⫺ 0
i* ⫽ 9.00 ⫹ ———————— (2.0)
22,986 ⫺ (⫺6002)
⫽ 9.00 ⫹ 1.58 ⫽ 10.58%
Solution by Spreadsheet
The fastest way to find i* is to use the RATE function (Equation [7.4]). The entry
⫽ RATE(10,15000,⫺200000,300000) displays i* ⫽ 10.55% per year. It is equally correct to
use the IRR function. Figure 7–4, column B, shows the cash flows and ⫽ IRR(B2:B12)
function to obtain i*.
For a complete spreadsheet analysis, use the procedure outlined above.
1.
2.
3.
4.
5.
Figure 7–3 shows cash flows.
Equation [7.6] is the ROR relation.
Figure 7–4 shows the net cash flows in column B.
The IRR function in cell B14 displays i* ⫽ 10.55%.
To graphically observe i * ⫽ 10.55%, column D displays the PW graph for different
i values. The NPV function is used repeatedly to calculate PW for the xy scatter
chart.
Special Considerations When Using the ROR Method
7.3
⫽ NPV(C4,$B$3:$B$12) + $B$2
⫽ IRR(B2:B12)
Figure 7–4
Spreadsheet to determine i* and develop a PW graph, Example 7.2.
Just as i* can be found using a PW equation, it may equivalently be determined using an AW
relation. This method is preferred when uniform annual cash flows are involved. Solution by
hand is the same as the procedure for a PW-based relation, except Equation [7.2] is used. In the
case of Example 7.2, i* ⫽ 10.55% is determined using the AW-based relation.
0 ⫽ ⫺200,000(A兾P,i*,10) ⫹ 15,000 ⫹ 300,000(A兾F,i*,10)
The procedure for solution by spreadsheet is exactly the same as outlined above using the IRR
function. Internally, IRR calculates the NPV function at different i values until NPV ⫽ 0. (There is no
equivalent way to utilize the PMT function, since it requires a fixed value of i to calculate an A value.)
7.3 Special Considerations When Using
the ROR Method
The rate of return method is commonly used in engineering and business settings to evaluate
one project, as discussed in this chapter, and to select one alternative from two or more, as
explained in the next chapter. As mentioned earlier, an ROR analysis is performed using a different basis than PW and AW analyses. The cash flows themselves determine the (internal) rate
of return. As a result, there are some assumptions and special considerations with ROR analysis that must be made when calculating i* and in interpreting its real-world meaning. A summary is provided below.
• Multiple i* values. Depending upon the sequence of net cash inflows and outflows, there may
be more than one real-number root to the ROR equation, resulting in more than one i* value.
This possibility is discussed in Section 7.4.
• Reinvestment at i*. Both the PW and AW methods assume that any net positive investment
(i.e., net positive cash flows once the time value of money is considered) is reinvested at the
MARR. However, the ROR method assumes reinvestment at the i* rate. When i* is not close
to the MARR (e.g., if i* is substantially larger than MARR), this is an unrealistic assumption.
In such cases, the i* value is not a good basis for decision making. This situation is discussed
in Section 7.5.
• Different procedure for multiple alternative evaluations. To correctly use the ROR method
to choose from two or more mutually exclusive alternatives requires an incremental analysis
procedure that is significantly more involved than PW and AW analysis. Chapter 8 explains
this procedure.
If possible, from an engineering economic study perspective, the AW or PW method at a
stated MARR should be used in lieu of the ROR method. However, there is a strong appeal
179
180
Rate of Return Analysis: One Project
Chapter 7
for the ROR method because rate of return values are very commonly quoted. And it is easy to
compare a proposed project’s return with that of in-place projects.
When it is important to know the exact value of i*, a good approach is to determine PW or
AW at the MARR, then determine the specific i* for the selected alternative.
As an illustration, if a project is evaluated at MARR ⫽ 15% and has PW ⬍ 0, there is no need to
calculate i*, because i* ⬍ 15%. However, if PW is positive, but close to 0, calculate the exact i*
and report it along with the conclusion that the project is financially justified.
7.4 Multiple Rate of Return Values
In Section 7.2 a unique rate of return i* was determined. In the cash flow series presented thus
far, the algebraic signs on the net cash flows changed only once, usually from minus in year 0 to
plus at some time during the series. This is called a conventional (or simple) cash flow series.
However, for some series the net cash flows switch between positive and negative from one year
to another, so there is more than one sign change. Such a series is called nonconventional (nonsimple). As shown in the examples of Table 7–3, each series of positive or negative signs may be
one or more in length. Relatively large net cash flow (NCF) changes in amount and sign can
occur in projects that require significant spending at the end of the expected life. Nuclear plants,
open-pit mines, petroleum well sites, refineries, and the like often require environmental restoration, waste disposal, and other expensive phaseout costs. The cash flow diagram will appear
similar to Figure 7–5a. Plants and systems that have anticipated major refurbishment costs or
upgrade investments in future years may have considerable swings in cash flow and sign changes
over the years, as shown by the pattern in Figure 7–5b.
TABLE 7–3 Examples of Conventional and Nonconventional Net Cash Flow for
a 6-year Project
Sign on Net Cash Flow by Year
Type of Series
0
1
2
3
4
5
6
Conventional
Conventional
Conventional
Nonconventional
Nonconventional
Nonconventional
⫺
⫺
⫹
⫺
⫹
⫺
⫹
⫺
⫹
⫹
⫹
⫹
⫹
⫺
⫹
⫹
⫺
⫺
⫹
⫹
⫹
⫹
⫺
⫺
⫹
⫹
⫹
⫺
⫺
⫹
⫹
⫹
⫺
⫺
⫹
⫹
⫹
⫹
⫺
⫺
⫹
⫹
Positive
NCF
0
1
Year
0
1
1
1
1
2
2
3
Positive
NCF
Positive
NCF
n⫺1 n
2
Number of
Sign Changes
n⫺1
2
n
i* = ?
Phaseout
costs
Initial
investment
Initial
investment
(a)
Figure 7–5
Midlife
investments
i* = ?
(b)
Typical cash flow diagrams for projects with (a) large restoration or remediation costs, and (b) upgrade or refurbishment costs.
Year
Multiple Rate of Return Values
7.4
When there is more than one sign change in the net cash flows, it is possible that there will be
multiple i* values in the ⫺100% to plus infinity range. There are two tests to perform in sequence
on the nonconventional series to determine if there is one unique value or possibly multiple i*
values that are real numbers.
Test 1: (Descartes’) rule of signs states that the total number of real-number roots is always less
than or equal to the number of sign changes in the series.
This rule is derived from the fact that the relation set up by Equation [7.1] or [7.2] to find i*
is an nth-order polynomial. (It is possible that imaginary values or infinity may also satisfy the
equation.)
Test #2: Cumulative cash flow sign test, also known as Norstrom’s criterion, states that only
one sign change in a series of cumulative cash flows which starts negatively indicates that there
is one positive root to the polynomial relation.
Zero values in the series are neglected when applying Norstrom’s criterion. This is a more discriminating test that determines if there is one, real-number, positive i* value. There may be
negative roots that satisfy the ROR relation, but these are not useful i* values. To perform the
test, determine the series
St ⫽ cumulative cash flows through period t
Observe the sign of S0 and count the sign changes in the series S0, S1, . . . , Sn. Only if S0 ⬍ 0 and
signs change one time in the series is there a single, real-number, positive i*.
With the results of these two tests, the ROR relation is solved for either the unique i* or the
multiple i* values, using trial and error by hand, using a programmable calculator, or by spreadsheet using an IRR function that incorporates the “guess” option. Development of the PW graph
is recommended, especially when using a spreadsheet. Examples 7.3 and 7.4 illustrate the tests
and solution for i*.
EXAMPLE 7.3
Sept-Îles Aluminum Company operates a bauxite mine to supply its aluminum smelter located about 2 km from the current open pit. A new branch for the pit is proposed that will
supply an additional 10% of the bauxite currently available over the next 10-year period. The
lease for the land will cost $400,000 immediately. The contract calls for the restoration of the
land and development as part of a state park and wildlife area at the end of the 10 years. This
is expected to cost $300,000. The increased production capacity is estimated to net an additional $75,000 per year for the company. Perform an ROR analysis that will provide the following information:
(a) Type of cash flow series and possible number of ROR values
(b) PW graph showing all i* values
(c) Actual i* values determined using the ROR relation and spreadsheet function
(d) Conclusions that can be drawn about the correct rate of return from this analysis
Solution
(a) The net cash flows will appear like those in Figure 7–5a with an initial investment of
$⫺400,000, annual net cash flow (NCF) of $75,000 for years 1 through 10, and a phaseout cost of $⫺300,000 in year 10. Figure 7–6 details the NCF series (column B) and cumulative NCF (column C) for use in the two tests for unique and multiple i* values. The
series is nonconventional based on the sign changes throughout the series.
Test #1: There are two sign changes in the NCF series, which indicates a possible
maximum of two roots to the polynomial equation or i* values for the ROR
equation.
Test #2: There is one sign change in the cumulative NCF series, which indicates a unique
positive root or one positive i* value.
181
182
Rate of Return Analysis: One Project
Chapter 7
⫽ ⌵PV(D3,$B$4:$B$13)⫹$B$3
⫽ IRR(B3:B13,guess)
Figure 7–6
Spreadsheet determination of multiple i* values and PW graph, Example 7.3.
(b) Columns D and E of the spreadsheet in Figure 7–6 use i values ranging from ⫺20% to
⫹20% per year to plot the PW vs i curve via the NPV function. There are two times that
the parabolic-shaped curve crosses the PW ⫽ 0 line; these are approximately i1* ⫽ ⫺18%
and i2*⫽ 5%.
(c) The ROR equation based on PW computations is
0 ⫽ ⫺400,000 ⫹ 75,000(P兾A,i*%,10) ⫺ 300,000(P兾F,i*%,10)
[7.7]
i* values by hand If hand solution is chosen, the same procedure used in Example 7.2
can be applied here. However, the technique to estimate the initial i value will not work
as well in this case since the majority of the cash flows do not fit either the P兾F or the
F兾P factor. In fact, using the P兾F factor, the initial i value is indicated to be 1.25%.
Trial-and-error solution of Equation [7.7] with various i values will approximate the
correct answer of about 4.5% per year. This complies with the test results of one positive i* value.
i* by spreadsheet function Use the ⫽ IRR(B3:B13,guess) function to determine the i*
value for the NCF series in column B, Figure 7–6. Entering different values in the
optional “guess” field will force the function to find multiple i* values, if they exist.
As shown in row 17, two are found.
i1* ⫽ ⫺18.70%
i2*⫽ ⫹4.53%
This result does not conflict with test results, as there is one positive value, but a negative
value also balances the ROR equation.
(d) The positive i* ⫽ 4.53% is accepted as the correct internal rate of return (IROR) for the
project. The negative value is not useful in economic conclusions about the project.
EXAMPLE 7.4
The engineering design and testing group for Honda Motor Corp. does contract-based work for
automobile manufacturers throughout the world. During the last 3 years, the net cash flows for
contract payments have varied widely, as shown below, primarily due to a large manufacturer’s
inability to pay its contract fee.
Year
0
1
2
3
Cash Flow ($1000)
⫹2000
⫺500
⫺8100
⫹6800
Multiple Rate of Return Values
7.4
(a) Determine the maximum number of i* values that may satisfy the ROR equation.
(b) Write the PW equation and approximate the i* value(s) by plotting PW vs i.
(c) What do the i* values mean?
Solution
(a) Table 7–4 shows the annual cash flows and cumulative cash flows. Since there are two
sign changes in the cash flow sequence, the rule of signs indicates a maximum of two i*
values. The cumulative cash flow sequence starts with a positive number S0 ⫽ ⫹2000,
indicating that test #2 is inconclusive. As many as two i* values can be found.
TABLE 7–4 Cash Flow and Cumulative Cash Flow Sequences, Example 7.4
Year
Cash Flow
($1000)
Sequence
Number
0
1
2
3
⫹2000
⫺500
⫺8100
⫹6800
S0
S1
S2
S3
Cumulative Cash Flow
($1000)
⫹2000
⫹1500
⫺6600
⫹200
(b) The PW relation is
PW ⫽ 2000 ⫺ 500(P兾F,i,1) ⫺ 8100(P兾F,i,2) ⫹ 6800(P兾F,i,3)
The PW values are shown below and plotted in Figure 7–7 for several i values. The characteristic parabolic shape for a second-degree polynomial is obtained, with PW crossing
the i axis at approximately i1* ⫽ 8% and i2* ⫽ 41%.
i%
PW ($1000)
5
10
20
30
40
50
⫹51.44
⫺39.55
⫺106.13
⫺82.01
⫺11.83
⫹81.85
Figure 7–7
100
Present worth of cash
flows at several interest
rates, Example 7.4.
75
PW (⫻ $1000)
50
25
i%
0
10
20
30 40
50
⫺25
⫺50
⫺75
⫺100
⫺125
Figure 7–8 presents the spreadsheet PW graph with the PW curve crossing the x axis at
PW ⫽ 0 two times. Also, the solution for two positive i* values using the IRR function
with different guess values is displayed. The values are
i1* ⫽ 7.47%
i2* ⫽ 41.35%
(c) Since both i* values are positive, they are not of much value, because neither can be
considered the true ROR of the cash flow series. This result indicates that additional
183
184
Rate of Return Analysis: One Project
Chapter 7
IRR function format
⫽ IRR($B4:$B7,guess)
Figure 7–8
Spreadsheet solution, Example 7.4.
information is needed to calculate a more useful project ROR, namely, some information
about the anticipated return on funds invested external to the project and the cost of capital to borrow money to continue the project. This problem is a good example of when an
approach discussed in the next section should be taken.
If the guess option is not used in the IRR function, the starting point is 10% per year. The function will find the one ROR closest to 10% that satisfies the PW relation. Entering various guess
values will allow IRR to find multiple i* values in the range of ⫺100% to ⬁, if they exist. Often,
the results are unbelievable or unacceptable values that are rejected. Some helpful guidelines can
be developed. Assume there are two i* values for a particular cash flow series.
If the Results Are
What to Do
Both i* ⬍ 0
Discard both values.
Both i* ⬎ 0
Discard both values.
One i* ⬎ 0; one i* ⬍ 0
Use i* ⬎ 0 as ROR.
If both i* values are discarded, proceed to the approach discussed in the next section to determine
one rate of return value for the project. However, remember the prime recommendation.
Always determine the PW or AW at the MARR first for a reliable measure of economic justification. If the PW or AW is greater than zero and the ROR is needed, then find the actual i* of the
project cash flows.
This recommendation is not to dissuade you from using the ROR method. Rather it is a recommendation that the use of the ROR technique be reserved for times when the actual i* value is
essential to the decision-making process.
7.5 Techniques to Remove Multiple Rates of Return
The techniques developed here are used under the following conditions:
• The PW or AW value at the MARR is determined and could be used to make the decision, but information on the ROR is deemed necessary to finalize the economic decision, and
• The two tests of cash flow sign changes (Descartes’ and Norstrom’s) indicate multiple roots
(i* values) are possible, and
7.5
Techniques to Remove Multiple Rates of Return
• More than one positive i* value or all negative i* values are obtained when the PW graph and
IRR function are developed, and
• A single, reliable rate of return value is required by management or engineers to make a clear
economic decision.
We will present a couple of ways to remove multiple i* values. The selected approach depends
upon what estimates are the most reliable for the project being evaluated. An important fact to
remember is the following.
The result of follow-up analysis to obtain a single ROR value when multiple, nonuseful i* values
are present does not determine the internal rate of return (IROR) for nonconventional net
cash flow series. The resulting rate is a function of the additional information provided to make
the selected technique work, and the accuracy is further dependent upon the reliability of this
information.
We will refer to the resulting value as the external rate of return (EROR) as a reminder that
it is different from the IROR obtained in all previous sections. First, it is necessary to identify
the perspective about the annual net cash flows of a project. Take the following view: You are
the project manager and the project generates cash flows each year. Some years produce positive
NCF, and you want to invest the excess money at a good rate of return. We will call this
the investment rate ii. This can also be called the reinvestment rate. Other years, the net cash flow
will be negative and you must borrow funds from some source to continue. The interest rate you
pay should be as small as possible; we will call this the borrowing rate ib, also referred to as the
finance rate. Each year, you must consider the time value of money, which must utilize either the
investment rate or the borrowing rate, depending upon the sign on the NCF of the preceding year.
With this perspective, it is now possible to outline two approaches that rectify the multiple
i* situation. The resulting ROR value will not be the same for each method, because slightly
different additional information is necessary and the cash flows are treated in slightly different
fashions from the time value of money viewpoint.
Modified ROR (MIRR) Approach This is the easier approach to apply, and it has a spreadsheet function that can find the single EROR value quickly. However, the investment and borrowing rates must be reliably estimated, since the results may be quite sensitive to them. The
symbol i⬘ will identify the result.
Return on Invested Capital (ROIC) Approach Though more mathematically rigorous,
this technique provides a more reliable estimate of the EROR and it requires only the investment
rate ii. The symbol i⬙ is used to indentify the result.
Before covering the techniques, it would be good to review the material in Section 7.1, including Example 7.1. Though the i⬘ or i⬙ value determined here is not the ROR defined earlier in the
chapter, the concepts used to make the ending cash flow balance equal to zero are used.
Modified ROR Approach
The technique requires that two rates external to the project net cash flows be estimated.
• Investment rate ii is the rate at which extra funds are invested in some source external to the
project. This applies to all positive annual NCF. It is reasonable that the MARR is used for this
rate.
• Borrowing rate ib is the rate at which funds are borrowed from an external source to provide
funds to the project. This applies to all negative annual NCF. The weighted average cost of
capital (WACC) can be used for this rate.
It is possible to make all rates the same, that is, ii ⫽ ib ⫽ MARR ⫽ WACC. However, this is not
a good idea as it implies that the company is willing to borrow funds and invest in projects at the
same rate. This implies no profit margin over time, so the company can’t survive for long using
this strategy. Commonly MARR ⬎ WACC, so usually ii ⬎ ib. (See Section 1.9 for a quick review
of MARR and WACC and Chapter 10 for a more detailed discussion of WACC.)
185
186
Rate of Return Analysis: One Project
Chapter 7
Find FWn of all
NCF > 0 at investment rate ii
PW0 at ib
NCF1
0
NCF4
NCF2
NCFn⫺1
1
2
3
4
n⫺1
NCF3
NCF0
Year
n
NCFn
Find PW0 of all
NCF < 0 at borrowing rate ib
FWn at ii
Figure 7–9
Typical cash flow diagram to determine modified rate of return i⬘.
Figure 7–9 is a reference diagram that has multiple i* values, since the net cash flows change
sign multiple times. The modified ROR method uses the following procedure to determine a
single external rate of return i⬘ and to evaluate the economic viability of the project.
1. Determine the PW value in year 0 of all negative NCF at the borrowing rate ib (lightly
shaded area and resulting PW0 value in Figure 7–9).
2. Determine the FW value in year n of all positive NCF at the investment rate ii (darker shaded
area and resulting FWn value in Figure 7–9).
3. Calculate the modified rate of return i⬘ at which the PW and FW values are equivalent over
the n years using the following relation, where i⬘ is to be determined.
FWn ⫽ PW0 (F兾P,i⬘%,n)
[7.8]
If using a spreadsheet rather than hand computation, the MIRR function displays i⬘ directly with
the format
⫽ MIRR(first_cell:last_cell, ib, ii)
[7.9]
4. The guideline for economic decision making compares the EROR or i⬘ to MARR.
If i⬘ ⬎ MARR, the project is economically justified.
If i⬘ ⬍ MARR, the project is not economically justified.
As in other situations, on the rare occasion that i⬘ ⫽ MARR, there is indifference to the project's economic acceptability; however, acceptance is the usual decision.
EXAMPLE 7.5
The cash flows experienced by Honda Motors in Example 7.4 are repeated below. There are
two positive i* values that satisfy the PW relation, 7.47% and 41.35% per year. Use the modified ROR method to determine the EROR value. Studies indicate that Honda has a WACC of
8.5% per year and that projects with an estimated return of less than 9% per year are routinely
rejected. Due to the nature of this contract business, any excess funds generated are expected
to earn at a rate of 12% per year.
Year
Net Cash Flow ($1000)
0
1
2
3
⫹2000
⫺500
⫺8100
⫹6800
Techniques to Remove Multiple Rates of Return
7.5
Solution by Spreadsheet
Using the information in the problem statement, the rate estimates are as follows:
MARR:
Investment rate, ii:
Borrowing rate, ib:
9% per year
12% per year
8.5% per year
The fast way to find i⬘ is with the MIRR function. Figure 7–10 shows the result of i' ⫽ 9.39%
per year. Since 9.39% ⬎ MARR, the project is economically justified.
⫽ MIRR(B2:B5,E2,E3)
Figure 7–10
Spreadsheet application of MIRR function, Example 7.5.
It is vital that the interpretation be correct. The 9.39% is not the internal rate of return
(IROR); it is the external ROR (EROR) based on the two external rates for investing and borrowing money.
Solution by Hand
Figure 7–9 can serve as a reference as the procedure to find i' manually is applied.
Step 1. Find PW0 of all negative NCF at ib ⫽ 8.5%.
PW0 ⫽ ⫺500(P兾F,8.5%,1) ⫺ 8100(P兾F,8.5%,2)
⫽ $⫺7342
Step 2. Find FW3 of all positive NCF at ii ⫽ 12%.
FW3 ⫽ 2000(F兾P,12%,3) ⫹ 6800
⫽ $9610
Step 3. Find the rate i' at which the PW and FW are equivalent.
PW0(F兾P,i⬘,3) ⫹ FW3 ⫽ 0
⫺7342(1 ⫹ i⬘)3 ⫹ 9610 ⫽ 0
(
)
9610 1兾3 ⫺ 1
i' ⫽ ———
7342
⫽ 0.939 (9.39%)
Step 4. Since i' ⬎ MARR of 9%, the project is economically justified using this EROR
approach.
Return on Invested Capital Approach
The definition of ROIC should be understood before we discuss the approach.
Return on invested capital (ROIC) is a rate-of-return measure of how effectively a project utilizes
the funds invested in it, that is, funds that remain internal to the project. For a corporation, ROIC
is a measure of how effectively it utilizes the funds invested in its operations, including facilities,
equipment, people, systems, processes, and all other assets used to conduct business.
187
188
Rate of Return Analysis: One Project
Chapter 7
The technique requires that the investment rate ii be estimated for excess funds generated in any
year that they are not needed by the project. The ROIC rate, which has the symbol i⬙, is determined
using an approach called the net-investment procedure. It involves developing a series of future
worth (F) relations moving forward 1 year at a time. In those years that the net balance of the
project cash flows is positive (extra funds generated by the project), the funds are invested at the ii
rate. Usually, ii is set equal to the MARR. When the net balance is negative, the ROIC rate is used,
since the project keeps all of its funds internal to itself. The ROIC method uses the following procedure to determine a single external rate of return i⬙ and to evaluate the economic viability of the
project. Remember that the perspective is that you are the project manager and when the project
generates extra cash flows, they are invested external to the project at the investment rate ii.
1. Develop a series of future worth relations by setting up the following relation for each year t
(t ⫽ 1, 2, . . . , n years).
Ft ⴝ Ftⴚ1(1 ⴙ k) ⴙ NCFt
[7.10]
where Ft ⴝ future worth in year t based on previous year and time value of money
NCFt ⴝ net cash flow in year t
i
if Ftⴚ1 ⬎ 0
(extra funds available)
kⴝ i
i⬙
if Ftⴚ1 ⬍ 0
(project uses all available funds)
{
2. Set the future worth relation for the last year n equal to 0, that is, Fn ⫽ 0, and solve for i⬙ to
balance the equation. The i⬙ value is the ROIC for the specified investment rate ii.
The Ft series and solution for i⬙ in the Fn ⫽ 0 relation can become involved mathematically. Fortunately, the Goal Seek spreadsheet tool can assist in the determination of i⬙ because there is only one unknown in the Fn relation and the target value is zero. (Both the hand
and spreadsheet solutions are demonstrated in Example 7.6.)
3. The guideline for economic decision making is the same as above, namely,
If ROIC ⱖ MARR, the project is economically justified.
If ROIC ⬍ MARR, the project is not economically justified.
It is important to remember that the ROIC is an external rate of return dependent upon
the investment rate choice. It is not the same as the internal rate of return discussed at the beginning of this chapter, nor is it the multiple rates, nor is it the MIRR rate found by the previous
method. This is a separate technique to find a single rate for the project.
EXAMPLE 7.6
Once again, we will use the cash flows experienced by Honda Motors in Example 7.4 (repeated
below). Use the ROIC method to determine the EROR value. The MARR is 9% per year, and
any excess funds generated by the project can earn at a rate of 12% per year.
Year
0
Net Cash Flow ($1000) ⫹2000
1
⫺500
2
⫺8100
3
⫹6800
Solution by Hand
The hand solution is presented first to provide the logic of the ROIC method. Use MARR ⫽
9% and ii ⫽ 12% per year in the procedure to determine i⬙, which is the ROIC. Figure 7–11
details the cash flows and tracks the progress as each Ft is developed. Equation [7.10] is applied to develop each Ft.
Step 1. Year 0:
Year 1:
F0 ⫽ $⫹2000
Since F0 ⬎ 0, externally invest in year 1 at ii ⫽ 12%.
F1 ⫽ 2000(1.12) ⫺ 500 ⫽ $⫹1740
Since F1 ⬎ 0, use ii ⫽ 12% for year 2.
189
Techniques to Remove Multiple Rates of Return
7.5
$6800
$6800
F0 ⴝ $2000
$6800
F1 ⴝ $1740
F3 = 0
at iⴖ = ?
0
1
$500
2
3
0
1
2
3
0
1
2
3
0
1
2
3 Year
F2 ⴝ $ⴚ6151
$8100
$8100
(a)
(b)
(c)
(d)
Figure 7–11
Application of ROIC method at ii ⫽ 12% per year: (a) original cash flow; equivalent form in (b) year 1, (c) year 2, and (d) year 3.
Year 2:
Year 3:
F2 ⫽ 1740(1.12) ⫺ 8100 ⫽ $⫺6151
Now F2 ⬍ 0, use i⬙ for year 3, according to Equation [7.10].
F3 ⫽ ⫺6151(1 ⫹ i⬙) ⫹ 6800
This is the last year. See Figure 7–11 for equivalent net cash flow diagrams.
Go to step 2.
Step 2. Solve for i⬙ ⫽ ROIC from F3 ⫽ 0.
⫺6151(1 ⫹ i⬙) ⫹ 6800 ⫽ 0
i⬙ ⫽ 6800兾6151 ⫺ 1
⫽ 0.1055 (10.55%)
Step 3. Since ROIC > MARR ⫽ 9%, the project is economically justified.
Solution by Spreadsheet
Figure 7–12 provides a spreadsheet solution. The future worth values F1 through F3 are determined by the conditional IF statements in rows 3 through 5. The functions are shown in column
D. In each year, Equation [7.10] is applied. If there are surplus funds generated by the project,
Ft–1 ⬎ 0 and the investment rate ii (in cell E7) is used to find Ft. For example, because the F1
value (in cell C3) of $1740 ⬎ 0, the time value of money for the next year is calculated at the
investment rate of 12% per year, as shown in the hand solution above for year 2.
Figure 7–12
Spreadsheet application of ROIC method using Goal Seek, Example 7.6.
190
Rate of Return Analysis: One Project
Chapter 7
The Goal Seek template sets the F3 value to zero by changing the ROIC value (cell E8). The
result is i⬙ ⫽ ROIC ⫽ 10.55% per year. As before, since 10.55% ⬎ 9%, the MARR, the project
is economically justified.
Comment
Note that the rate by the ROIC method (10.55%) is different than the MIRR rate (9.39%). Also
these are both different than the multiple rates determined earlier (7.47% and 41.35%). This
shows how dependent the different methods are upon the additional information provided
when multiple i* rates are indicated by the two sign tests.
Now that we have learned two techniques to remove multiple i* values, there are some connections between the multiple i* values, the external rate estimates, and the resulting external
rates (i⬘ and i⬙) obtained by the two methods.
Modified ROR technique When both the borrowing rate ib and the investment rate ii are
exactly equal to any one of the multiple i* values, the rate i' found by the MIRR function, or by
hand solution, will equal the i* value. That is, all four parameters have the same value.
If any i* ⫽ ib ⫽ ii,
then i' ⫽ i*
ROIC technique Similarly, if the investment rate ii is exactly equal to any one of the multiple
i* values, the rate found by the Goal Seek tool, or by hand when the equation Fn ⫽ 0 is solved,
will be i⬙ ⫽ i* value.
Finally, it is very important to remember the following fact.
None of the details of the modified ROR (MIRR) technique or the return on invested capital
(ROIC) technique are necessary if the PW or AW method of project evaluation is applied at a
specific MARR. When the MARR is established, this is, in effect, fixing the i* value. Therefore,
a definitive economic decision can be made directly from the PW or AW value.
7.6 Rate of Return of a Bond Investment
A time-tested method of raising capital funds is through the issuance of an IOU, which is financing through debt, not equity (see Chapter 1). One very common form of IOU is a bond—a longterm note issued by a corporation or a government entity (the borrower) to finance major projects. The borrower receives money now in return for a promise to pay the face value V of the
bond on a stated maturity date. Bonds are usually issued in face value amounts of $1000, $5000,
or $10,000. Bond dividend I, also called bond interest, is paid periodically between the time the
money is borrowed and the time the face value is repaid. The bond dividend is paid c times per
year. Expected payment periods are usually semiannually or quarterly. The amount of interest is
determined using the stated dividend or interest rate, called the bond coupon rate b.
(face value) (bond coupon rate)
I ⴝ ————————————————
number of payment periods per year
Vb
I ⴝ ——
c
[7.11]
There are many types or classifications of bonds. Four general classifications are summarized
in Table 7–5 according to their issuing entity, fundamental characteristics, and example names or
purposes. For example, Treasury securities are issued in different monetary amounts ($1000 and
up) with varying periods of time to the maturity date (Bills up to 1 year; Notes for 2 to 10 years).
In the United States, Treasury securities are considered a very safe bond purchase because they are
backed with the “full faith and credit of the U.S. government.” The safe investment rate indicated
in Figure 1⫺6 as the lowest level for establishing a MARR is the coupon rate on a U.S. Treasury
security. Funds obtained through corporate bond issues are used for new product development,
facilities upgrade, expansion into international markets, and similar business ventures.
Rate of Return of a Bond Investment
7.6
TABLE 7–5
Classification and Characteristics of Bonds
Classification
Issued by
Characteristics
Examples
Treasury securities
Federal government
Backed by faith and credit of the federal
government
Bills (ⱕ 1 year)
Notes (2–10 years)
Bonds (10–30 years)
Municipal
Local governments
Federal tax-exempt
Issued against taxes received
General obligation
Revenue
Zero coupon
Put
Mortgage
Corporation
Backed by specified assets or mortgage
Low rate兾low risk on first mortgage
Foreclosure, if not repaid
First mortgage
Second mortgage
Equipment trust
Debenture
Corporation
Not backed by collateral, but by
reputation of corporation
Bond rate may “float”
Higher interest rates and higher risks
Convertible
Subordinated
Junk or high yield
EXAMPLE 7.7
General Electric just released $10 million worth of $10,000 ten-year bonds. Each bond pays dividends semiannually at a rate of 6% per year. (a) Determine the amount a purchaser will receive each
6 months and after 10 years. (b) Suppose a bond is purchased at a time when it is discounted by 2%
to $9800. What are the dividend amounts and the final payment amount at the maturity date?
Solution
(a) Use Equation [7.11] for the dividend amount.
10,000 (0.06)
I ⫽ —————— ⫽ $300 per 6 months
2
The face value of $10,000 is repaid after 10 years.
(b) Purchasing the bond at a discount from face value does not change the dividend or final repayment amounts. Therefore, $300 per 6 months and $10,000 after 10 years remain the amounts.
The cash flow series for a bond investment is conventional and has one unique i*, which is
best determined by solving a PW-based rate of return equation in the form of Equation [7.1], that
is, 0 ⫽ PW.
EXAMPLE 7.8
Allied Materials needs $3 million in debt capital for expanded composites manufacturing.
It is offering small-denomination bonds at a discount price of $800 for a 4% $1000 bond
that matures in 20 years with a dividend payable semiannually. What nominal and effective
interest rates per year, compounded semiannually, will Allied Materials pay an investor?
Solution
The income that a purchaser will receive from the bond purchase is the bond dividend I ⫽ $20 every
6 months plus the face value in 20 years. The PW-based equation for calculating the rate of return is
0 ⫽ ⫺800 ⫹ 20(P兾A,i*,40) ⫹ 1000(P兾F,i*,40)
Solve by the IRR function or by hand to obtain i* ⫽ 2.8435% semiannually. The nominal interest rate per year is computed by multiplying i* by 2.
Nominal i ⫽ (2.8435)(2) ⫽ 5.6870% per year, compounded semiannually
Using Equation [4.5], the effective annual rate is
ia ⫽ (1.028435)2 ⫺ 1 ⫽ 5.7678%
191
192
Rate of Return Analysis: One Project
Chapter 7
EXAMPLE 7.9
Gerry is a project engineer. He took a financial risk and bought a bond from a corporation that
had defaulted on its interest payments. He paid $4240 for an 8% $10,000 bond with dividends
payable quarterly. The bond paid no interest for the first 3 years after Gerry bought it. If interest
was paid for the next 7 years and then Gerry was able to resell the bond for $11,000, what rate
of return did he make on the investment? Assume the bond is scheduled to mature 18 years
after he bought it. Perform hand and spreadsheet analysis.
Solution by Hand
The bond interest received in years 4 through 10 was
(10,000)(0.08)
I ⫽ —————— ⫽ $200 per quarter
4
The effective rate of return per quarter can be determined by solving the PW equation developed on a per quarter basis.
0 ⫽ ⫺4240 ⫹ 200(P兾A,i* per quarter,28)(P兾F,i* per quarter,12)
⫹ 11,000(P兾F,i* per quarter,40)
The equation is correct for i* ⫽ 4.1% per quarter, which is a nominal 16.4% per year, compounded quarterly.
Solution by Spreadsheet
Once all the cash flows are entered into contiguous cells, the function ⫽ IRR(B2:B42) is used
in Figure 7–13, row 43, to display the answer of a nominal rate of return of 4.10% per quarter.
(Note that many of the row entries have been hidden to conserve space.) This is the same as the
nominal annual rate of
i* ⫽ 4.10%(4) ⫽ 16.4% per year, compounded quarterly
Gerry did well on his bond investment.
Figure 7–13
Spreadsheet solution for
a bond investment,
Example 7.9.
If a bond investment is being considered and a required rate of return is stated, the same
PW-based relation used to find i* can be used to determine the maximum amount to pay for
the bond now to ensure that the rate is realized. The stated rate is the MARR, and the PW
computations are performed exactly as they were in Chapter 5. As an illustration, in the last
example, if 12% per year, compounded quarterly, is the target MARR, the PW relation is
used to find the maximum that Gerry should pay now; P is determined to be $6004. The
quarterly MARR is 12%兾4 ⫽ 3%.
0 ⫽ ⫺P ⫹ 200(P兾A,3%,28)(P兾F,3%,12) ⫹ 11,000(P兾F,3%,40)
P ⫽ $6004
Problems
193
CHAPTER SUMMARY
The rate of return of a cash flow series is determined by setting a PW-based or AW-based relation
equal to zero and solving for the value of i*. The ROR is a term used and understood by almost
everybody. Most people, however, can have considerable difficulty in calculating a rate of return
correctly for anything other than a conventional cash flow series. For some types of series, more
than one ROR possibility exists. The maximum number of i* values is equal to the number of
changes in the sign of the net cash flow series (Descartes’ rule of signs). Also, a single positive
rate can be found if the cumulative net cash flow series starts negatively and has only one sign
change (Norstrom’s criterion).
When multiple i* values are indicated, either of the two techniques covered in this chapter can
be applied to find a single, reliable rate for the nonconventional net cash flow series. In the case
of the ROIC technique, additional information is necessary about the investment rate that excess
project funds will realize, while the modified ROR technique requires this same information, plus
the borrowing rate for the organization considering the project. Usually, the investment rate is set
equal to the MARR, and the borrowing rate takes on the historical WACC rate. Each technique
will result in slightly different rates, but they are reliable for making the economic decision,
whereas the multiple rates are often not useful to decision making.
If an exact ROR is not necessary, it is strongly recommended that the PW or AW method at
the MARR be used to decide upon economic justification.
PROBLEMS
Understanding ROR
7.1 Under what circumstances would the rate of return
be (a) ⫺100%, and (b) infinite?
7.2 A shrewd investor loaned $1,000,000 to a start-up
company at 10% per year interest for 3 years, but
the terms of the agreement were such that interest
would be charged on the principal rather than on
the unpaid balance. How much extra interest did
the company pay?
7.3 What is the nominal rate of return per year on an
investment that increases in value by 8% every
3 months?
Determination of ROR
7.6 In 2010, the city of Houston, Texas, collected
$24,112,054 in fines from motorists because of
traffic violations caught by red-light cameras. The
cost of operating the system was $8,432,372. The
net profit, that is, profit after operating costs, is
split equally (that is, 50% each) between the city
and the operator of the camera system. What will
be the rate of return over a 3-year period to the
contractor that paid for, installed, and operates the
system, if its initial cost was $9,000,000 and the
profit for each of the 3 years is the same as it was
in 2010?
7.4 Assume you borrow $50,000 at 10% per year interest and you agree to repay the loan in five equal
annual payments. What is the amount of the unrecovered balance immediately after you make the
third payment?
7.7 P&G sold its prescription drug business to
Warner-Chilcott, Ltd. for $3.1 billion. If income
from product sales is $2 billion per year and net
profit is 20% of sales, what rate of return will
the company make over a 10-year planning
horizon?
7.5 International Potash got a $50 million loan amortized over a 10-year period at 10% per year interest. The loan agreement stipulates that the loan
will be repaid in 10 equal annual payments with
interest charged on the principal amount of the
loan (not on the unrecovered balance).
(a) What is the amount of each payment?
(b) What is the total amount of interest paid?
How does the total interest paid compare
with the principal of the loan?
7.8 Water damage from a major flood in a Midwestern city resulted in damages estimated at
$108 million. As a result of the claimant payouts,
insurance companies raised homeowners' insurance rates by an average of $59 per year for each
of the 160,000 households in the affected city. If
a 20-year study period is considered, what was
the rate of return on the $108 million paid by the
insurance companies?
194
Rate of Return Analysis: One Project
Chapter 7
7.9 Determine the rate of return for the cash flows shown
in the diagram. (If requested by your instructor,
show both hand and spreadsheet solutions.)
$7000
i=?
0
1
2
3
4
$200
$200
$200
5
6
7
$90
$90
8
Year
$90
$3000
to 1,694,247 in 2015. If the increase were to occur
uniformly, what rate of increase would be required
each year to meet the goal?
7.14 U.S. Census Bureau statistics show that the annual
earnings for persons with a high school diploma
are $35,220 versus $57,925 for someone with a
bachelor’s degree. If the cost of attending college
is assumed to be $30,000 per year for 4 years and
the forgone earnings during those years are assumed to be $35,220 per year, what rate of return
does earning a bachelor’s degree represent? Use a
35-year study period. (Hint: The investment in
years 1 through 4 is the cost of college plus the
foregone earnings, and the income in years 5
through 35 is the difference in income between a
high school diploma and a bachelor’s degree.)
7.10 The Office of Naval Research sponsors a contest
for college students to build underwater robots that
can perform a series of tasks without human intervention. The University of Florida, with its SubjuGator robot, won the $7000 first prize (and serious
bragging rights) over 21 other universities. If the
team spent $2000 for parts (at time 0) and the project took 2 years, what annual rate of return did the
team make?
7.15 The Ester Municipal Water Utility issued 20-year
bonds in the amount of $53 million for several
high-priority flood control improvement projects.
The bonds carried a 5.38% dividend rate with the
dividend payable annually. The U.S. economy was
in a recession at that time, so as part of the federal
stimulus program, the Utility gets a 35% reimbursement on the dividend it pays.
(a) What is the effective dividend rate that the
Utility is paying on the bonds?
7.11 For the cash flows shown, determine the rate of
(b) What is the total dollar amount the Utility will
return.
save in dividends over the life of the bonds?
(c) What is the future worth in year 20 of the
Year
0
1
2
3
4
5
dividend savings, if the interest rate is
Expense, $ ⫺17,000 ⫺2,500 ⫺2,500 ⫺2,500 ⫺2,500 ⫺2,500
6% per year?
Revenue, $
0
5,000
6,000
7,000
8,000
12,000
7.12 In an effort to avoid foreclosure proceedings on
struggling mortgage customers, Bank of America
proposed an allowance that a jobless customer
make no payment on their mortgage for up to
9 months. If the customer did not find a job within
that time period, they would have to sign over their
house to the bank. The bank would give them
$2000 for moving expenses.
Assume John and his family had a mortgage
payment of $2900 per month and he was not able
to find a job within the 9-month period. If the bank
saved $40,000 in foreclosure costs, what rate of
return per month did the bank make on the allowance? Assume the first payment that was skipped
was due at the end of month 1 and the $40,000
foreclosure savings and $2000 moving expense
occurred at the end of the 9-month forbearance
period.
7.13 The Closing the Gaps initiative by the Texas
Higher Education Coordinating Board established
the goal of increasing the number of students in
higher education in Texas from 1,064,247 in 2000
7.16 A contract between BF Goodrich and the Steelworkers Union of America called for the company
to spend $100 million in capital investment to keep
the facilities competitive. The contract also required the company to provide buyout packages
for 400 workers. If the average buyout package is
$100,000 and the company is able to reduce costs
by $20 million per year, what rate of return will the
company make over a 10-year period? Assume all
of the company’s expenditures occur at time 0 and
the savings begin 1 year later.
7.17 Rubber sidewalks made from ground up tires are
said to be environmentally friendly and easier on
people’s knees. Rubbersidewalks, Inc. of Gardena,
California, manufactures the small rubberized
squares that are being installed where tree roots,
freezing weather, and snow removal have required
sidewalk replacement or major repairs every
3 years. The District of Columbia spent $60,000 for
a rubber sidewalk to replace broken concrete in a
residential neighborhood lined with towering willow oaks. If a concrete sidewalk costs $28,000 and
195
Problems
lasts only 3 years versus a 9-year life for the rubber
sidewalks, what rate of return does this represent?
7.18 Efficient light jets (ELJs) are smaller aircraft that
may revolutionize the way people travel by plane.
They cost between $1.5 and $3 million, seat 5 to 7
people, and can fly up to 1100 miles at cruising
speeds approaching 425 mph. Eclipse Aerospace
was founded in 2009, and its sole business is making ELJs. The company invested $500 million
(time 0) and began taking orders 2 years later. If
the company accepted orders for 2500 planes and
received 10% down (in year 2) on planes having
an average cost of $1.8 million, what rate of return
will the company make over a 10-year planning
period? Assume 500 of the planes are delivered
each year in years 6 through 10 and that the company’s M&O costs average $10 million per year in
years 1 through 10. (If requested by your instructor, show both hand and spreadsheet solutions.)
7.19 Betson Enterprises distributes and markets the Big
Buck video game which allows players to “hunt”
for elk, antelope, moose, and bucks without shivering outside in the cold. E-sports entertainment in
New York City purchased five machines for $6000
each and took in an average of $600 total per week
in sales. What rate of return does this represent
(a) per week and (b) per year (nominal)? Use a
3-year study period with 52 weeks per year.
7.20 A 473-foot, 7000-ton World War II troop carrier
(once commissioned as the USS Excambion) was
sunk in the Gulf of Mexico to serve as an underwater habitat and diving destination. The project
took 10 years of planning and cost $4 million.
Assume the $4 million was expended equally in
years 1 through 10. What rate of return does the
venture represent, if increased fishing and recreation activities are valued at $270,000 per year
beginning in year 11 and they continue in perpetuity? (If assigned by your instructor, show both
hand and spreadsheet solutions.)
7.24 According to Descartes’ rule of signs, what is the
maximum number of real-number values that will
balance a rate of return equation?
7.25 According to Descartes’ rule of signs, how many
possible i* values are there for net cash flows that
have the following signs?
(a) ⫹⫺⫹⫹⫹⫺⫹
(b) ⫺ ⫺ ⫺ ⫹⫹⫹⫹⫹
(c) ⫹⫹⫹⫹ ⫺ ⫺ ⫺ ⫹⫺ ⫺ ⫺⫹⫺⫹⫺ ⫺ ⫺
7.26 According to Norstrom’s criterion, there are two
requirements regarding the cumulative cash flows
that must be satisfied to ensure that there is only
one positive root in a rate of return equation. What
are they?
7.27 According to Descartes’ rule of signs, how many
possible i* values are there for the cash flows
shown?
Year
1
2
3
4
5
6
Net Cash ⫹4100 ⫺2000 ⫺7000 ⫹12,000 ⫺700 ⫹800
Flow, $
7.28 According to Descartes’ rule of signs, how many
i* values are possible for the cash flows shown?
Year
Revenue, $
Costs, $
1
2
25,000
⫺30,000
13,000
⫺7,000
3
4
4,000
70,000
⫺6,000 ⫺12,000
7.29 According to Descartes’ rule and Norstrom’s criterion, how many i* values are possible for the cash
flow (CF ) sequence shown?
Year
1
2
3
4
5
Net Cash
⫹16,000 ⫺32,000 ⫺25,000 ⫹50,000 ⫺8,000
Flow, $
Cumulative ⫹16,000 ⫺16,000 ⫺41,000 ⫹9,000 ⫹1,000
CF, $
7.30 For the cash flows shown, determine the sum of
the cumulative cash flows.
Multiple ROR Values
Year
7.21 What is meant by a nonconventional cash flow
series?
Revenue, $
Costs, $
⫺6,000
7.22 Explain at least three types of projects in which
large net cash flow changes may cause sign changes
during the life of the project, thus indicating the
possible presence of multiple ROR values.
7.31 Stan-Rite Corp of Manitowoc, Wisconsin, is a B to
B company that manufactures many types of industrial products, including portable measuring
arms with absolute encoders, designed to perform
3D inspections of industrial parts. If the company’s
cash flow (in millions) for one of its product divisions is as shown on the next page, determine
(a) the number of possible i* values and (b) all rate
of return values between 0% and 100%.
7.23 Explain a situation with which you are personally familiar for which the net cash flows have
changed signs in a fashion similar to those in
Figure 7–5.
0
1
2
3
4
25,000 15,000
4,000
18,000
⫺30,000 ⫺7,000 ⫺6,000 ⫺12,000
196
Rate of Return Analysis: One Project
Chapter 7
Year
Expenses
Revenues
0
1
2
3
4
5
6
$⫺30
⫺20
⫺25
⫺15
⫺22
⫺20
⫺30
$0
18
19
36
52
38
70
7.32 Julie received a $50 bill for her birthday at the end
of January. At the end of February, she spent this
$50 and an additional $150 to buy clothes. Her
parents then gave her $50 and $125 at the end of
March and April, respectively, as she prepared to
go to summer school and needed the clothes. Her
conclusion was that over the 4 months, she had received $25 more than she spent. Determine if Julie
has a multiple rate of return situation for these cash
flows. If so, determine the multiple rates and comment on their validity. The cash flow values are as
follows:
Month
Jan (0)
Feb (1)
Mar (2)
Apr (3)
Cash Flow, $
50
⫺200
50
125
7.33 Veggie Burger Boy sells franchises to individuals
who want to start small in the sandwiches-forvegeterians business and grow in net cash flow
over the years. A franchisee in Mississippi provided the $5000 up-front money, but did very
poorly the first year. He was allowed to borrow at
the end of his first year from the corporation’s capital incentive fund with a promise to repay the loan
in addition to the annual share that the corporation
contractually receives from annual sales. The net
cash flows from the corporation’s perspective are
shown below.
Year
0
1
NCF, $ 5000 ⫺10,100
2
500
3
4
5
6
2000 2000 2000 2000
The corporate chief financial officer (CFO) has
some questions concerning this NCF series. Help
her by doing the following, using a spreadsheet.
(a) Plot the PW versus i graph and estimate the
rate of return for this franchise.
(b) Use the IRR function on a spreadsheet to find
the corresponding return.
(c) Basing your conclusions on Descartes’ and
Norstrom’s rules, provide the CFO with
some advice on what ROR value is the most
reliable for this franchise over the 6-year period. The normal corporate MARR used for
franchisee evaluation is 30% per year.
based on an older technology to produce meat
products. It had positive NCF until 2010 and discontinued operation in 2011 due to labor and
safety problems. In 2012, prior to the sale of the
facility and property, Vaught spent $1 million to
make the site environmentally acceptable to a potential buyer. The net cash flows in $100,000 over
the years are listed below. Use a spreadsheet to do
the following.
(a) Check for multiple rates of return.
(b) Find all rates that are real numbers between
⫺25% and ⫹50%, and calculate the PW
value for interest rates in this range.
(c) Indicate which is the best and correct i* value
to use in a PW analysis.
Year
NCF, $ ⫺25 ⫹10 ⫹10 ⫹15 ⫹15
⫺5
⫺6 ⫺10
7.35 Five years ago, VistaCare spent $5 million to develop and introduce a new service in home health
care for people who require frequent blood dialysis treatments. The service was not well received
after the first year and was removed from the market. When reintroduced 4 years after its initial
launch, it was much more profitable. Now, in year
5, VistaCare has spent a large sum on research to
broaden the application of this service. Use the
NCF series below to plot the PW versus i graph
and estimate the ROR over the 5 years. NCF values
are in $1 million units.
Year
2
3
Net Cash Flow, $ ⫺5000 5000 0
0
0
1
4
5
15,000 ⫺15,000
Removing Multiple i* Values
7.36 In calculating the external rate of return by the
modified rate of return approach, it is necessary to
use two different rates of return, the investment rate
ii and the borrowing rate ib. When is each used?
7.37 In the modified rate of return approach for determining a single interest rate from net cash flows,
state which interest rate is usually higher, the investment rate ii or the borrowing rate ib. State why.
7.38 Use the modified rate of return approach with an
investment rate of 18% per year and a borrowing
rate of 10% to find the external rate of return for
the following cash flows.
Year
0
1
2
3
Net Cash Flow, $ ⫹16,000 ⫺32,000 ⫺25,000 ⫹70,000
7.39
7.34 In 2011, Vaught Industries closed its plant in
Marionsville following labor, environmental, and
safety problems. The plant was built in 2005
2005 2006 2007 2008 2009 2010 2011 2012
Harley worked for many years to save enough money
to start his own residential landscape design business.
The cash flows shown are those he recorded for the
first 6 years as his own boss. Find the external rate of
197
Problems
return using the modified rate of return approach, an
investment rate of 15% per year, and a borrowing rate
of 8%. (After using the procedure, use the MIRR
function to confirm your answer.)
Year
0
1
2
3
4
5
7.40 Samara, an engineer working for GE, invested her
bonus money each year in company stock. Her
bonus has been $8000 each year for the past
6 years (i.e., at the end of years 1 to 6). At the end
of year 7, she sold the stock for $52,000 to buy a
condo; she purchased no stock that year. In years 8
to 10, she again invested the $8000 bonus. Samara
sold all of the remaining stock for $28,000 immediately after the investment at the end of year 10.
(a) Determine the number of possible rate of return values in the net cash flow series using
the two sign tests.
(b) Determine the external rate of return by
hand, using the modified rate of return approach with an investment rate of 12% per
year and a borrowing rate of 8%.
(c) Find the external rate of return by spreadsheet using the ROIC approach with an investment rate of 12% per year.
(d) Enter the cash flows into a spreadsheet, and use
the IRR function to find the i* value. You should
get the same value as the ROIC in part (c). Explain why this is so, given that the investment
rate is 12% per year. (Hint: Look carefully at the
column labeled “Future worth, F, $” when you
solved part (c) using the spreadsheet.)
7.41 Swagelok Co. of Solon, Ohio, makes variable area
flowmeters (VAFs) that measure liquid and gas flow
rates by means of a tapered tube and float. If tooling
and setup costs were $400,000 in year 0 and an additional $190,000 in year 3, determine the external
rate of return using the modified rate of return approach. The revenue was $160,000 per year in
years 1 through 10. Assume the company’s MARR
is 20% per year and its cost of capital is 9% per year.
7.42 A company that makes clutch disks for race cars
has the cash flows shown for one department.
(a)
(b)
Cash Flow, $1000
0
1
2
3
4
⫺65
30
84
⫺10
⫺12
Calculate the external rate of return using the
return on invested capital (ROIC) approach
with an investment rate of 15% per year. (As
assigned by your instructor, solve by hand
and兾or spreadsheet.)
6
NCF, $ ⫺9000 ⫹4100 ⫺2000 ⫺7000 ⫹12,000 ⫹700 ⫹800
Year
(c)
Determine the number of positive roots to the
rate of return relation.
Calculate the internal rate of return.
7.43 For the cash flow series below, calculate the external rate of return, using the return on invested capital approach with an investment rate of 14% per
year.
Year
Cash Flow, $
0
1
2
3
4
3000
⫺2000
1000
⫺6000
3800
7.44 Five years ago, a company made a $500,000 investment in a new high-temperature material. The
product did poorly after only 1 year on the market.
However, with a new name and advertising campaign 4 years later it did much better. New development funds have been expended this year
(year 5) at a cost of $1.5 million. Determine the
external rate of return using the ROIC approach
and an investment rate of 15% per year. The i* rate
is 44.1% per year.
Year
Cash Flow, $
0
1
2
3
4
5
⫺500,000
400,000
0
0
2,000,000
⫺1,500,000
Bonds
7.45 What is the bond coupon rate on a $25,000 mortgage bond that has semiannual interest payments
of $1250 and a 20-year maturity date?
7.46 An equipment trust bond with a face value of
$10,000 has a bond coupon rate of 8% per year,
payable quarterly. What are the amount and frequency of the dividend payments?
7.47 What is the face value of a municipal bond that
matures in 20 years and has a bond coupon rate
of 6% per year with semiannual payments of
$900?
7.48 What is the present worth of a $50,000 debenture
bond that has a bond coupon rate of 8% per year,
payable quarterly? The bond matures in 15 years.
The interest rate in the marketplace is 6% per year,
compounded quarterly.
198
Chapter 7
Rate of Return Analysis: One Project
7.49 Best Buy issued collateral bonds 4 years ago that
have a face value of $20,000 each and a coupon
rate of 8% per year, payable semiannually. If the
bond maturity date is 20 years from the date they
were issued and the interest rate in the marketplace
is now 12% per year, compounded semiannually,
what is the present worth (now) of one bond?
issued 30-year bonds with a face value of $25 million. The bond coupon rate was set at 5% per year,
payable semiannually. Because the market interest
rate increased immediately before the bonds were
sold, the city received only $23.5 million from the
bond sale. What was the semiannual interest rate
when the bonds were sold?
7.50 In 2011, El Paso Water Utilities (EPWU) issued
bonds worth $9.125 million to improve the Van
Buren dam in central El Paso and to finance three
other drainage projects. The bonds were purchased
by the Texas Water Development Board under the
federal stimulus program wherein EPWU did not
have to pay any dividend on the bonds. If the bond
dividend rate would have been 4% per year, payable quarterly, with a bond maturity date 18 years
after issuance, what is the present worth of the
dividend savings to EPWU rate payers? Assume
the market interest rate is 6% per year.
7.55 An investor who purchased a $10,000 mortgage
bond today paid only $6000 for it. The bond coupon rate is 8% per year, payable quarterly, and the
maturity date is 18 years from the year of issuance.
Because the bond is in default, it will pay no dividend for the next 2 years. If the bond dividend is in
fact paid for the following 5 years (after the
2 years) and the investor then sells the bond for
$7000, what rate of return will be realized (a) per
quarter and (b) per year (nominal)?
7.51 A recently issued industrial bond with a face value
of $10,000 has a coupon rate of 8% per year, payable
annually. The bond matures 20 years from now. Jeremy is interested in buying one bond. If he pays
$10,000 for the bond and plans to hold it to maturity, what rate of return per year will he realize?
7.52 Due to a significant troop buildup at the local military base, a school district issued $10,000,000 in
bonds to build new schools. The bond coupon rate
is 6% per year, payable semiannually, with a maturity date of 20 years. If an investor is able to purchase one of the bonds that has a face value of
$5000 for $4800, what rate of return per 6 months
will the investor realize? Assume the bond is kept
to maturity.
7.53 As the name implies, a zero-coupon bond pays no
dividend, only the face value when it matures. If a
zero coupon bond that has a face value of $10,000
and a maturity date of 15 years is for sale for
$2000, what rate of return will the purchaser make,
provided the bond is held to maturity?
7.54 To provide infrastructure in the outlying areas of
Morgantown, West Virginia, the city council
7.56 Five years ago, GSI, an oil services company
headquartered in Texas, issued $10 million worth
of 12% 30-year bonds with the dividend payable
quarterly. The bonds have a call date of this year if
GSI decides to take advantage of it. The interest
rate in the marketplace decreased enough that the
company is considering calling the bonds since the
coupon rate is relatively high. If the company buys
the bonds back now for $11 million, determine the
rate of return that the company will make (a) per
quarter and (b) per year (nominal). (Hint: The
“call” option means the following: By spending
$11 million now, the company will not have to
make the quarterly bond dividend payments or pay
the face value of the bonds when they come due
25 years from now.)
7.57 Four years ago, Chevron issued $5 million worth
of debenture bonds with a coupon rate of 10% per
year, payable semiannually. Market interest rates
dropped, and the company called the bonds (i.e.,
paid them off in advance) at a 10% premium on the
face value. Therefore, it cost the corporation
$5.5 million to retire the bonds. What semiannual
rate of return did an investor make who purchased
a $5000 bond 4 years ago and held it until it was
called 4 years later?
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
7.58 All of the following mean the same as rate of
return except:
(a) Internal rate of return
(b) Time for return of capital
(c) Interest rate
(d) Return on investment
7.59 The numerical value of i in a rate of return equation can range from:
(a) 0% to 100%
(b) 0 to ⬁
(c) ⫺100% to ⫹100%
(d) ⫺100% to ⬁
199
Additional Problems and FE Exam Review Questions
7.60 The internal rate of return on an investment refers
to the interest rate earned on the:
(a) Initial investment
(b) Unrecovered balance of the investment
(c) Money recovered from an investment
(d) Income from an investment
7.61 A conventional (or simple) cash flow series is one
wherein:
(a) The algebraic signs on the net cash flows
change only once.
(b) The interest rate you get is a simple interest
rate.
(c) The total of the net cash flows is equal to 0.
(d) The total of the cumulative cash flows is
equal to 0.
7.62 According to Descartes’ rule of signs, for a net
cash flow sequence of ⫺ ⫺⫹⫹⫺⫹, the number of
possible i values is:
(a) 2
(b) 3
(c) 4
(d) 5
7.63 According to Norstrom’s criterion, the one statement below that is correct is:
(a) The cumulative cash flow must start out
positively.
(b) The cumulative cash flow must start out
negatively.
(c) The cumulative cash flow must equal 0.
(d) The net cash flow must start out positively.
7.64 According to Descartes’ rule and Norstrom’s criterion, the number of positive i* values for the
following cash flow sequence is:
Year
Revenue, $
Cost, $
(a)
(b)
(c)
(d)
1
2
3
4
25,000
30,000
15,000
7,000
4,000
6,000
18,000
12,000
1
2
3
4
7.65 For the net cash flows and cumulative cash flows
shown, the value of x is:
Year
1
NCF, $
Cumulative
NCF, $
(a)
(b)
(c)
(d)
2
3
4
5
⫹13,000 ⫺29,000 ⫺25,000
⫺8000
x
⫹13,000 ⫺16,000 ⫺41,000 ⫹9000 ⫹1000
$7000
$16,000
$36,000
$50,000
7.66 A company that uses a minimum attractive rate of
return of 10% per year is evaluating new processes
to improve operational efficiency. The estimates
associated with candidate processes are shown.
Alternative I
Alternative J
⫺40,000
⫺15,000
5,000
3
First cost, $
Annual cost, $ per year
Salvage value, $
Life, years
⫺50,000
⫺12,000
5,000
6
The statement that is most correct is:
(a) The alternatives are revenue alternatives.
(b) The alternatives are cost alternatives.
(c) The alternatives are revenue alternatives and
DN is an option.
(d) The alternatives are cost alternatives and DN
is an option.
7.67 Scientific Instruments, Inc. uses a MARR of 8% per
year. The company is evaluating a new process to
reduce water effluents from its manufacturing processes. The estimate associated with the process follows. In evaluating the process on the basis of a rate
of return analysis, the correct equation to use is:
New Process
⫺40,000
13,000
5,000
3
First cost, $
NCF, $ per year
Salvage value, $
Life, years
(a)
(b)
(c)
(d)
0 ⫽ ⫺40,000 ⫹ 13,000(P兾A,i,3) ⫹
5000(P兾F,i,3)
0 ⫽ ⫺40,000(A兾P,i,3) ⫹ 13,000 ⫹
5000(A兾F,i,3)
0 ⫽ ⫺40,000(F兾P,i,3) ⫹ 13,000(F兾A,i,3) ⫹
5000
Any of the above
7.68 When one is using the modified ROR method to remove multiple ROR values, an additional estimate
needed besides the cash flows and their timings is:
(a) The ROIC value
(b) External rate of return
(c) Investment rate
(d) Internal rate of return
7.69 For the following cash flows, the modified rate of
return method uses a borrowing rate of 10%, and
an investment rate is 12% per year. The correct
computation for the present worth in year 0 is:
Year
NCF, $
(a)
(b)
1
2
3
⫺10,000
0
0
4
5
⫺19,000 ⫹25,000
⫺10,000 ⫺ 19,000(P兾F,12%,4)
⫺10,000 ⫺ 19,000(P兾F,12%,4) ⫹
25,000(P兾F,10%,5)
200
Rate of Return Analysis: One Project
Chapter 7
(c)
(d)
25,000(P兾F,10%,5)
⫺10,000 ⫺ 19,000(P兾F,10%,4)
7.70 The return on invested capital (ROIC) method removes multiple ROR values from a cash flow sequence. If the future worth computation in year t is
Ft ⬍ 0, the ROIC rate i⬙ is used. The interpretation
of Ft ⬍ 0 most closely means:
(a) The net balance of project cash flows in year
t is negative.
(b) The resulting external rate of return will be
positive.
(c) The net balance of project cash flows is positive in year t.
(d) The sequence has nonremovable negative
ROR values.
7.71 The meaning of return on invested capital for a
corporation is best stated as:
(a) A rate-of-return measure that equates the internal and external ROR
(b) A measure of how effectively the corporation
uses capital funds invested in it
(c) The value at which borrowing ROR and investing ROR are equal
(d) The external rate of return value is based on
total capital invested
7.72 A corporate $10,000 bond has a coupon rate of 8%
per year, payable semiannually. The bond matures
20 years from now. At an interest rate of 6% per
year, compounded semiannually, the amount and
frequency of the bond dividend payments are:
(a) $600 every 6 months
(b) $800 every 6 months
(c) $300 every 6 months
(d) $400 every 6 months
7.73 A $20,000 mortgage bond that is due in 1 year
pays interest of $500 every 3 months. The bond’s
coupon rate is:
(a) 2.5% per year, payable quarterly
(b) 5% per year, payable quarterly
(c) 5% per year, payable semiannually
(d) 10% per year, payable quarterly
7.74 A $10,000 municipal bond due in 10 years pays
interest of $400 every 6 months. If an investor purchases the bond now for $9000 and holds it to maturity, the rate of return received can be determined
by the following equation:
(a) 0 ⫽ ⫺9000 ⫹ 400(P兾A,i,10)
⫹ 10,000(P兾F,i,10)
(b) 0 ⫽ ⫺9000 ⫹ 400(P兾A,i,20)
⫹ 10,000(P兾F,i,20)
(c) 0 ⫽ ⫺10,000 ⫹ 400(P兾A,i,20)
⫹ 10,000(P兾F,i,20)
(d) 0 ⫽ ⫺9000 ⫹ 800(P兾A,i,10)
⫹ 10,000(P兾F,i,10)
7.75 A debenture bond issued 3 years ago has a face
value of $5000, a coupon rate of 4% per year, payable semiannually, and a maturity date of 20 years
from the date it was issued. The bond is for sale
now for $4500. If the interest rate in the marketplace is compounded quarterly, the value of n that
must be used in the P兾A factor to calculate the rate
of return for the bond is:
(a) 34
(b) 40
(c) 68
(d) 80
CASE STUDY
DEVELOPING AND SELLING AN INNOVATIVE IDEA
Background
Three engineers who worked for Mitchell Engineering, a
company specializing in public housing development, went
to lunch together several times a week. Over time they decided to work on solar energy production ideas. After a lot of
weekend time over several years, they had designed and developed a prototype of a low-cost, scalable solar energy plant
for use in multifamily dwellings on the low end and mediumsized manufacturing facilities on the upper end. For residential applications, the collector could be mounted along side a
TV dish and be programmed to track the sun. The generator
and additional equipment are installed in a closet-sized area
in an apartment or on a floor for multiple-apartment supply.
The system serves as a supplement to the electricity provided
by the local power company. After some 6 months of testing,
it was agreed that the system was ready to market and reliably
state that an electricity bill in high-rises could be reduced by
approximately 40% per month. This was great news for lowincome dwellers on government subsidy that are required to
pay their own utility bills.
Information
With a hefty bank loan and $200,000 of their own capital,
they were able to install demonstration sites in three cities in
the sunbelt. Net cash flow after all expenses, loan repayment,
and taxes for the first 4 years was acceptable; $55,000 at the
Case Study
end of the first year, increasing by 5% each year thereafter. A
business acquaintance introduced them to a potential buyer of
the patent rights and current subscriber base with an estimated $500,000 net cashout after only these 4 years of ownership. However, after serious discussion replaced the initial
excitement of the sales offer, the trio decided to not sell at this
time. They wanted to stay in the business for a while longer
to develop some enhancement ideas and to see how much
revenue may increase over the next few years.
During the next year, the fifth year of the partnership, the
engineer who had received the patents upon which the collector and generator designs were based became very displeased
with the partnering arrangements and left the trio to go into
partnership with an international firm in the energy business.
With new research and development funds and the patent
rights, a competing design was soon on the market and took
much of the business away from the original two developers.
Net cash flow dropped to $40,000 in year 5 and continued to
decrease by $5000 per year. Another offer to sell in year 8
was presented, but it was only for $100,000 net cash. This
was considered too much of a loss, so the two owners did not
accept. Instead, they decided to put $200,000 more of their
own savings into the company to develop additional applications in the housing market.
It is now 12 years since the system was publicly launched.
With increased advertising and development, net cash flow
has been positive the last 4 years, starting at $5000 in year 9
and increasing by $5000 each year until now.
201
Case Study Exercises
It is now 12 years after the products were developed, and the
engineers invested most of their savings in an innovative
idea. However, the question of “When do we sell?” is always
present in these situations. To help with the analysis, determine the following:
1. The rate of return at the end of year 4 for two situations:
(a) The business is sold for the net cash amount of
$500,000 and (b) no sale.
2. The rate of return at the end of year 8 for two situations:
(a) The business is sold for the net cash amount of
$100,000 and (b) no sale.
3. The rate of return now at the end of year 12.
4. Consider the cash flow series over the 12 years. Is there
any indication that multiple rates of return may be present? If so, use the spreadsheet already developed to
search for ROR values in the range ⫾100% other than
the one determined in exercise 3 above.
5. Assume you are an investor with a large amount of
ready cash, looking for an innovative solar energy product. What amount would you be willing to offer for the
business at this point (end of year 12) if you require a
12% per year return on all your investments and, if purchased, you plan to own the business for 12 additional
years? To help make the decision, assume the current
NCF series continues increasing at $5000 per year for
the years you would own it. Explain your logic for
offering this amount.
CHAPTER 8
Rate of Return
Analysis:
Multiple
Alternatives
L E A R N I N G
O U T C O M E S
Purpose: Select the best alternative on the basis of incremental rate of return analysis.
SECTION
TOPIC
LEARNING OUTCOME
8.1
Incremental analysis
• State why the ROR method of comparing
alternatives requires an incremental cash flow
analysis.
8.2
Incremental cash flows
• Calculate the incremental cash flow series for two
alternatives.
8.3
Meaning of ⌬i*
• Interpret the meaning of the incremental ROR (⌬i*)
determined from the incremental cash flow series.
8.4
⌬i* from PW relation
• Based on a PW relation, select the better of two
alternatives using incremental ROR analysis or a
breakeven ROR value.
8.5
⌬i* from AW relation
• Select the better of two alternatives using
incremental ROR analysis based on an AW
relation.
8.6
More than two alternatives
• Select the best from several alternatives using
incremental ROR analysis.
8.7
All-in-one spreadsheet
• Use a single spreadsheet to perform PW, AW, ROR,
and incremental ROR analyses for mutually
exclusive and independent alternatives.
T
his chapter presents the methods by which two or more alternatives can be evaluated using a rate of return (ROR) comparison based on the methods of the previous
chapter. The ROR evaluation, correctly performed, will result in the same selection
as the PW and AW analyses, but the computational procedure is considerably different for ROR
evaluations. The ROR analysis evaluates the increments between two alternatives in pairwise
comparisons. As the cash flow series becomes more complex, spreadsheet functions help speed
computations.
8.1 Why Incremental Analysis Is Necessary
When two or more mutually exclusive alternatives are evaluated, engineering economy can identify the one alternative that is the best economically. As we have learned, the PW and AW techniques can be used to do so, and are the recommended methods. Now the procedure using ROR
to identify the best is presented.
Let’s assume that a company uses a MARR of 16% per year, that the company has $90,000
available for investment, and that two alternatives (A and B) are being evaluated. Alternative
A requires an investment of $50,000 and has an internal rate of return i*
A of 35% per year. Alternative B requires $85,000 and has an i*
B of 29% per year. Intuitively we may conclude that
the better alternative is the one that has the larger return, A in this case. However, this is not
necessarily so. While A has the higher projected return, its initial investment ($50,000) is much
less than the total money available ($90,000). What happens to the investment capital that is
left over? It is generally assumed that excess funds will be invested at the company’s MARR,
as we learned in previous chapters. Using this assumption, it is possible to determine the consequences of the two alternative investments. If alternative A is selected, $50,000 will return
35% per year. The $40,000 left over will be invested at the MARR of 16% per year. The rate
of return on the total capital available, then, will be the weighted average. Thus, if alternative
A is selected,
50,000(0.35) ⫹ 40,000(0.16)
Overall RORA ⫽ ———————————— ⫽ 26.6%
90,000
If alternative B is selected, $85,000 will be invested at 29% per year, and the remaining $5000
will earn 16% per year. Now the weighted average is
85,000(0.29) ⫹ 5000(0.16)
Overall RORB ⫽ ——————————— ⫽ 28.3%
90,000
These calculations show that even though the i* for alternative A is higher, alternative B presents
the better overall ROR for the $90,000. If either a PW or AW comparison is conducted using the
MARR of 16% per year as i, alternative B will be chosen.
This simple example illustrates a major fact about the rate of return method for ranking and
comparing alternatives:
Under some circumstances, project ROR values do not provide the same ranking of alternatives
as do PW and AW analyses. This situation does not occur if we conduct an incremental ROR
analysis (discussed below).
When independent projects are evaluated, no incremental analysis is necessary between
projects. Each project is evaluated separately from others, and more than one can be selected.
Therefore, the only comparison is with the do-nothing alternative for each project. The project
ROR can be used to accept or reject each one.
8.2 Calculation of Incremental Cash Flows for
ROR Analysis
To conduct an incremental ROR analysis, it is necessary to calculate the incremental cash flow
series over the lives of the alternatives. Based upon the equivalence relations (PW and AW), ROR
evaluation makes the equal-service assumption.
Independent project
selection
204
Rate of Return Analysis: Multiple Alternatives
Chapter 8
TABLE 8–1
Format for Incremental Cash Flow Tabulation
Cash Flow
Year
Alternative A
(1)
Alternative B
(2)
Incremental
Cash Flow
(3) ⴝ (2) ⴚ (1)
0
1
Equal-service requirement
The incremental ROR method requires that the equal-service requirement be met. Therefore,
the LCM (least common multiple) of lives for each pairwise comparison must be used. All
the assumptions of equal service present for PW analysis are necessary for the incremental
ROR method.
A format for hand or spreadsheet solutions is helpful (Table 8–1). Equal-life alternatives have
n years of incremental cash flows, while unequal-life alternatives require the LCM of lives for
analysis. At the end of each life cycle, the salvage value and initial investment for the next cycle
must be included for the LCM case.
When a study period is established, only this number of years is used for the evaluation. All
incremental cash flows outside the period are neglected. As we learned earlier, using a study period, especially one shorter than the life of either alternative, can change the economic decision
from that rendered when the full lives are considered.
Only for the purpose of simplification, use the convention that between two alternatives, the
one with the larger initial investment will be regarded as alternative B. Then, for each year in
Table 8–1,
Incremental cash flow ⴝ cash flowB ⴚ cash flowA
[8.1]
The initial investment and annual cash flows for each alternative (excluding the salvage value)
are one of the types identified in Chapter 5:
Revenue alternative, where there are both negative and positive cash flows
Cost alternative, where all cash flow estimates are negative
Revenue or cost
alternative
In either case, Equation [8.1] is used to determine the incremental cash flow series with the sign
of each cash flow carefully determined.
EXAMPLE 8.1
A tool and die company in Hanover is considering the purchase of a drill press with fuzzy-logic
software to improve accuracy and reduce tool wear. The company has the opportunity to buy a
slightly used machine for $15,000 or a new one for $21,000. Because the new machine is a
more sophisticated model, its operating cost is expected to be $7000 per year, while the used
machine is expected to require $8200 per year. Each machine is expected to have a 25-year life
with a 5% salvage value. Tabulate the incremental cash flow.
Solution
Incremental cash flow is tabulated in Table 8–2. The subtraction performed is (new – used)
since the new machine has a larger initial cost. The salvage values in year 25 are separated
from ordinary cash flow for clarity. When disbursements are the same for a number of consecutive years, for hand solution only, it saves time to make a single cash flow listing, as is
done for years 1 to 25. However, remember that several years were combined when performing
the analysis. This approach cannot be used for spreadsheets, when the IRR or NPV function is
used, as each year must be entered separately.
Calculation of Incremental Cash Flows for ROR Analysis
8.2
TABLE 8–2
Cash Flow Tabulation for Example 8.1
Year
Used Press
New Press
Incremental
Cash Flow
(New – Used)
0
1–25
25
$⫺15,000
⫺8,200
⫹750
$⫺21,000
⫺7,000
⫹1,050
$⫺6,000
⫹1,200
⫹300
Cash Flow
EXAMPLE 8.2
A sole-source vendor can supply a new industrial park with large transformers suitable for
underground utilities and vault-type installation. Type A has an initial cost of $70,000 and
a life of 8 years. Type B has an initial cost of $95,000 and a life expectancy of 12 years.
The annual operating cost for type A is expected to be $9000, while the AOC for type B is
expected to be $7000. If the salvage values are $5000 and $10,000 for type A and type B,
respectively, tabulate the incremental cash flow using their LCM for hand and spreadsheet
solutions.
Solution by Hand
The LCM of 8 and 12 is 24 years. In the incremental cash flow tabulation for 24 years
(Table 8–3), note that the reinvestment and salvage values are shown in years 8 and 16 for type
A and in year 12 for type B.
Solution by Spreadsheet
Figure 8–1 shows the incremental cash flows for the LCM of 24 years. As in the hand tabulation, reinvestment is made in the last year of each intermediate life cycle. The incremental
values in column D are the result of subtractions of column B from C.
Note that the final row includes a summation check. The total incremental cash flow should
agree in both the column D total and the subtraction C29 ⫺ B29. Also note that the incremental values change signs three times, indicating the possibility of multiple i* values, per Descartes’ rule of signs. This possible dilemma is discussed later in the chapter.
TABLE 8–3
Incremental Cash Flow Tabulation, Example 8.2
Type A
Type B
Incremental
Cash Flow
(B ⴚ A)
$ ⫺70,000
⫺9,000
⫺70,000
⫺9,000
⫹5,000
⫺9,000
$ ⫺95,000
–7,000
$⫺25,000
⫹2,000
–7,000
⫹67,000
–7,000
–95,000
–7,000
⫹10,000
–7,000
⫹2,000
⫺83,000
–7,000
⫹67,000
–7,000
–7,000
⫹10,000
$⫺338,000
⫹2,000
Cash Flow
Year
0
1–7
8
9–11
12
13–15
16
17–23
24
⫺9,000
⫺9,000
⫺70,000
⫺9,000
⫹5,000
⫺9,000
⫺9,000
⫹5,000
$⫺411,000
⫹2,000
⫹7,000
$⫹73,000
205
206
Chapter 8
Rate of Return Analysis: Multiple Alternatives
Figure 8–1
Spreadsheet computation of incremental
cash flows for
unequal-life alternatives, Example 8.2.
Starting new life cycle for A
⫽ initial cost ⫹ AOC ⫹ salvage
ⴝ ⴚ 70,000 ⴚ 9,000 ⴙ 5,000
Check on summations
Incremental column should
equal difference of columns
8.3 Interpretation of Rate of Return
on the Extra Investment
The incremental cash flows in year 0 of Tables 8–2 and 8–3 reflect the extra investment or cost
required if the alternative with the larger first cost is selected. This is important in an incremental
ROR analysis in order to determine the ROR earned on the extra funds expended for the largerinvestment alternative. If the incremental cash flows of the larger investment don’t justify it, we
must select the cheaper one. In Example 8.1 the new drill press requires an extra investment of
$6000 (Table 8–2). If the new machine is purchased, there will be a “savings” of $1200 per year
for 25 years, plus an extra $300 in year 25. The decision to buy the used or new machine can
be made on the basis of the profitability of investing the extra $6000 in the new machine. If the
equivalent worth of the savings is greater than the equivalent worth of the extra investment at the
MARR, the extra investment should be made (i.e., the larger first-cost proposal should
be accepted). On the other hand, if the extra investment is not justified by the savings, select the
lower-investment proposal.
It is important to recognize that the rationale for making the selection decision is the same as
if only one alternative were under consideration, that alternative being the one represented by the
incremental cash flow series. When viewed in this manner, it is obvious that unless this investment yields a rate of return equal to or greater than the MARR, the extra investment should not
be made. As further clarification of this extra investment rationale, consider the following: The
rate of return attainable through the incremental cash flow is an alternative to investing at the
MARR. Section 8.1 states that any excess funds not invested in the alternative are assumed to
be invested at the MARR. The conclusion is clear:
ME alternative selection
If the rate of return available through the incremental cash flow equals or exceeds the MARR,
the alternative associated with the extra investment should be selected.
8.4
Rate of Return Evaluation Using PW: Incremental and Breakeven
207
Not only must the return on the extra investment meet or exceed the MARR, but also the return
on the investment that is common to both alternatives must meet or exceed the MARR. Accordingly,
prior to performing an incremental ROR analysis, it is advisable to determine the internal rate of
return i* for each alternative. This can be done only for revenue alternatives, because cost alternatives
have only cost (negative) cash flows and no i* can be determined. The guideline is as follows:
For multiple revenue alternatives, calculate the internal rate of return i* for each alternative,
and eliminate all alternatives that have an i* ⬍ MARR. Compare the remaining alternatives
incrementally.
As an illustration, if the MARR ⫽ 15% and two alternatives have i* values of 12% and 21%,
the 12% alternative can be eliminated from further consideration. With only two alternatives, it
is obvious that the second one is selected. If both alternatives have i* ⬍ MARR, no alternative is
justified and the do-nothing alternative is the best economically. When three or more alternatives
are evaluated, it is usually worthwhile, but not required, to calculate i* for each alternative for
preliminary screening. Alternatives that cannot meet the MARR may be eliminated from further
evaluation using this option. This option is especially useful when performing the analysis by
spreadsheet. The IRR function applied to each alternative’s cash flow estimates can quickly indicate unacceptable alternatives, as demonstrated in Section 8.6.
When independent projects are evaluated, there is no comparison on the extra investment.
The ROR value is used to accept all projects with i* ⱖ MARR, assuming there is no budget
limitation. For example, assume MARR ⫽ 10%, and three independent projects are available
with ROR values of
iA* ⫽ 12%
i*B ⫽ 9%
iC* ⫽ 23%
Projects A and C are selected, but B is not because i*B ⬍ MARR.
8.4 Rate of Return Evaluation Using PW:
Incremental and Breakeven
In this section we discuss the primary approach to making mutually exclusive alternative selections by the incremental ROR method. A PW-based relation is developed for the incremental cash
flows and set equal to zero. Use hand solution or spreadsheet functions to find ⌬i*B–A, the internal
ROR for the series. Placing ⌬ (delta) before i*B–A distinguishes it from the overall ROR values iA*
and iB*. (⌬i* may replace ⌬i*B–A when only two alternatives are present.)
Since incremental ROR requires equal-service comparison, the LCM of lives must be used in
the PW formulation. Because of the reinvestment requirement for PW analysis for different-life
assets, the incremental cash flow series may contain several sign changes, indicating multiple ⌬i*
values. Though incorrect, this indication is usually neglected in actual practice. The correct
approach is to follow one of the techniques of Section 7.5. This means that the single external
ROR (⌬i⬘ or ⌬i⬘⬘) for the incremental cash flow series is determined.
These three elements—incremental cash flow series, LCM, and multiple roots—are the primary reasons that the ROR method is often applied incorrectly in engineering economy analyses of
multiple alternatives. As stated earlier, it is always possible, and generally advisable, to use a PW or
AW analysis at an established MARR in lieu of the ROR method when multiple rates are indicated.
The complete procedure for hand or spreadsheet solution for an incremental ROR analysis
for two alternatives is as follows:
1. Order the alternatives by initial investment or cost, starting with the smaller one, called
A. The one with the larger initial investment is in the column labeled B in Table 8–1.
2. Develop the cash flow and incremental cash flow series using the LCM of years, assuming reinvestment in alternatives.
3. Draw an incremental cash flow diagram, if needed.
Independent project
selection
208
Rate of Return Analysis: Multiple Alternatives
Chapter 8
4. Count the number of sign changes in the incremental cash flow series to determine if
multiple rates of return may be present. If necessary, use Norstrom’s criterion to determine if a single positive root exists.
5. Set up the PW ⫽ 0 equation and determine ⌬i*B–A.
6. Select the economically better alternative as follows:
ME alternative
selection
If ⌬i*B⫺A ⬍ MARR, select alternative A.
If ⌬i*B⫺A ⱖ MARR, the extra investment is justified; select alternative B.
If ⌬i* is exactly equal to or very near the MARR, noneconomic considerations help in
the selection of the “better” alternative.
In step 5, if trial and error by hand is used, time may be saved if the ⌬i*B⫺A value is bracketed,
rather than approximated by a point value using linear interpolation, provided that a single ROR
value is not needed. For example, if the MARR is 15% per year and you have established that
⌬i*B⫺A is in the 15% to 20% range, an exact value is not necessary to accept B since you already
know that ⌬i*B⫺A ⱖ MARR.
The IRR function on a spreadsheet will normally determine one ⌬i* value. Multiple guess
values can be input to find multiple roots in the range ⫺100% to ⬁ for a nonconventional series,
as illustrated in Example 7.4. If this is not the case, to be correct, the indication of multiple roots
in step 4 requires that one of the techniques of Section 7.5 be applied to find an EROR.
EXAMPLE 8.3
As Ford Motor Company retools an old truck assembly plant in Michigan to produce a fuelefficient economy model, the Ford Focus. Ford and its suppliers are seeking additional sources for
light, long-life transmissions. Automatic transmission component manufacturers use highly finished dies for precision forming of internal gears and other moving parts. Two United States–
based vendors make the required dies. Use the per unit estimates below and a MARR of 12% per
year to select the more economical vendor bid. Show both hand and spreadsheet solutions.
Initial cost, $
Annual costs, $ per year
Salvage value, $
Life, years
A
B
⫺8,000
⫺3,500
0
10
⫺13,000
⫺1,600
2,000
5
Solution by Hand
These are cost alternatives, since all cash flows are costs. Use the procedure described above to
determine ⌬i*B–A.
1. Alternatives A and B are correctly ordered with the higher first-cost alternative in column 2
of Table 8–4.
2. The cash flows for the LCM of 10 years are tabulated.
TABLE 8–4 Incremental Cash Flow Tabulation, Example 8.3
Year
0
1–5
Cash Flow A
(1)
$ ⫺8,000
⫺3,500
5
—
6–10
10
⫺3,500
—
$⫺43,000
Cash Flow B
(2)
$⫺13,000
⫺1,600
⫹2,000
⫺13,000
⫺1,600
⫹2,000
$⫺38,000
Incremental
Cash Flow
(3) ⴝ (2) ⴚ (1)
$ ⫺5,000
⫹1,900
⫺11,000
⫹1,900
⫹2,000
$ ⫹5,000
Rate of Return Evaluation Using PW: Incremental and Breakeven
8.4
$2000
Diagram of incremental cash
flows, Example 8.3.
$1900
1
0
2
3
4
5
Figure 8–2
6
7
8
9
10
$5000
$11,000
3. The incremental cash flow diagram is shown in Figure 8–2.
4. There are three sign changes in the incremental cash flow series, indicating as many as
three roots. There are also three sign changes in the cumulative incremental series, which
starts negatively at S0 ⫽ $⫺5000 and continues to S10 ⫽ $⫹5000, indicating that more
than one positive root may exist.
5. The rate of return equation based on the PW of incremental cash flows is
0 ⫽ ⫺5000 ⫹ 1900(P兾A,⌬i*,10) ⫺ 11,000(P兾F,⌬i*,5) ⫹ 2000(P兾F,⌬i*,10)
[8.2]
In order to resolve any multiple-root problem, we can assume that the investment rate ii in the
ROIC technique will equal the ⌬i* found by trial and error. Solution of Equation [8.2] for the
first root discovered results in ⌬i* between 12% and 15%. By interpolation ⌬i* ⫽ 12.65%.
6. Since the rate of return of 12.65% on the extra investment is greater than the 12% MARR,
the higher-cost vendor B is selected.
Comment
In step 4, the presence of up to three i* values is indicated. The preceding analysis finds one of
the roots at 12.65%. When we state that the incremental ROR is 12.65%, we assume that any
positive net cash flows are reinvested at 12.65%. If this is not a reasonable assumption, the
ROIC or modified ROR technique (Section 7.5) must be applied to find a different single ⌬i'
or ⌬i'' to compare with MARR ⫽ 12%.
The other two roots are very large positive and negative numbers, as the IRR function of
Excel reveals. So they are not useful to the analysis.
Solution by Spreadsheet
Steps 1 through 4 are the same as above.
5. Figure 8–3 includes the same incremental net cash flows from Table 8–4 calculated in
column D. Cell D15 displays the ⌬i* value of 12.65% using the IRR function.
Figure 8–3
Spreadsheet solution using LCM of
lives and IRR function, Example 8.3.
⫽ NPV(12%,D5:D14) + D4
209
210
Rate of Return Analysis: Multiple Alternatives
Chapter 8
6. Since the rate of return on the extra investment is greater than the 12% MARR, the highercost vendor B is selected.
Comment
Once the spreadsheet is set up, there are a wide variety of analyses that can be performed. For
example, row 17 uses the NPV function to verify that the present worth is positive at MARR⫽12%.
Charts such as PW versus ⌬i and PW versus i help graphically interpret the situation.
The rate of return determined for the incremental cash flow series or the actual cash flows can
be interpreted as a breakeven rate of return value.
The breakeven rate of return is the incremental i* value, ⌬i*, at which the PW (or AW)
value of the incremental cash flows is exactly zero. Equivalently, the breakeven ROR is the i
value, i*, at which the PW (or AW) values of two alternatives’ actual cash flows are exactly
equal to each other.
If the incremental cash flow ROR (⌬i*) is greater than the MARR, the larger-investment
alternative is selected. For example, if the PW versus ⌬i graph for the incremental cash flows in
Table 8–4 (and spreadsheet Figure 8–3) is plotted for various interest rates, the graph shown in
Figure 8–4 is obtained. It shows the ⌬i* breakeven at 12.65%. The conclusions are that
• For MARR ⬍ 12.65%, the extra investment for B is justified.
• For MARR ⬎ 12.65%, the opposite is true—the extra investment in B should not be made,
and vendor A is selected.
• If MARR is exactly 12.65%, the alternatives are equally attractive.
Figure 8–5, which is a breakeven graph of PW versus i for the cash flows (not incremental) of
each alternative in Example 8.3, provides the same results. Since all net cash flows are negative
Breakeven ⌬i
is 12.65%
1800
For MARR
in this range,
select B
1600
For MARR
in this range,
select A
1400
1200
PW of incremental cash flows, $
Breakeven ROR
1000
800
600
400
200
0
6
7
8
9
10
11
12
14
15
16 ⌬i%
– 200
– 400
– 600
Vendor B
Vendor A
– 800
Figure 8–4
Plot of present worth of incremental cash flows for Example 8.3 at various
⌬i values.
211
Rate of Return Evaluation Using PW: Incremental and Breakeven
8.4
Figure 8–5
Interest rate, %
PW of alternative cash flows, $
10
11
12
13
⫺ 25,000
14
15
16
A
⫺ 26,000
⫺ 27,000
B
B
⫺ 28,000
⫺ 29,000
A
⫺ 30,000
(cost alternatives), the PW values are negative. Now, the same conclusions are reached using the
following logic:
• If MARR ⬍ 12.65%, select B since its PW of cost cash flows is smaller (numerically larger).
• If MARR ⬎ 12.65%, select A since its PW of costs is smaller.
• If MARR is exactly 12.65%, either alternative is equally attractive.
Example 8.4 illustrates incremental ROR evaluation and breakeven rate of return graphs for
revenue alternatives. More of breakeven analysis is covered in Chapter 13.
EXAMPLE 8.4
New filtration systems for commercial airliners are available that use an electric field to remove up
to 99.9% of infectious diseases and pollutants from aircraft air. This is vitally important, as many
of the flu germs, viruses, and other contagious diseases are transmitted through the systems that
recirculate aircraft air many times per hour. Investments in the new filtration equipment can cost
from $100,000 to $150,000 per aircraft, but savings in fuel, customer complaints, legal actions,
etc., can also be sizable. Use the estimates below (in $100 units) from two suppliers provided to an
international carrier to do the following, using a spreadsheet and an MARR of 15% per year.
• Plot two graphs: PW versus i values for both alternatives’ cash flows and PW versus ⌬i
values for incremental cash flows.
• Estimate the breakeven ROR values from both graphs, and use this estimate to select one
alternative.
Initial cost per aircraft, $
Estimated savings, $ per year
Estimated life, years
Air Cleanser
(Filter 1)
Purely Heaven
(Filter 2)
⫺1000
375
⫺1500
700 in year 1, decreasing by
100 per year thereafter
5
5
Solution by Spreadsheet
Refer to Figure 8–6 as the solution is explained. For information, row 10 shows the ROR values
calculated using the IRR function for filter 1 and filter 2 cash flows and the incremental cash
flow series (filter 2 ⫺ filter 1).
The cash flow sign tests for each filter indicate no multiple rates. The incremental cash flow
sign test does not indicate the presence of a unique positive root; however, the second rate is an
extremely large and useless value. The PW values of filter 1 and filter 2 cash flows are plotted
on the right side for i values ranging from 0% to 60%. Since the PW curves cross each other at
approximately i* ⫽ 17%, the rate does exceed the MARR of 15%. The higher-cost filter 2
(Purely Heaven) is selected.
The PW curve for the incremental cash flows series (column G) is plotted at the bottom left
of Figure 8–6. As expected, the curve crosses the PW ⫽ 0 line at approximately 17%, indicating the same economic conclusion of filter 2.
Breakeven graph of
Example 8.3 cash flows
(not incremental).
212
MARR
Filter 1 ROR ⬇ 25%
Filter 2 ROR ⬇ 23%
MARR
Breakeven ROR ⬇ 17%
Breakeven
Incremental ROR ⬇ 17%
Figure 8–6
PW versus i graph and PW versus incremental i graph, Example 8.4.
213
Rate of Return Evaluation Using AW
8.5
Figure 8–6 provides an excellent opportunity to see why the ROR method can result in selecting the wrong alternative when only i* values are used to select between two alternatives. This is
sometimes called the ranking inconsistency problem of the ROR method. The inconsistency
occurs when the MARR is set less than the breakeven rate between two revenue alternatives.
Since the MARR is established based on conditions of the economy and market, MARR is established external to any particular alternative evaluation. In Figure 8–6 the incremental breakeven
rate is 16.89%, and the MARR is 15%. The MARR is lower than breakeven; therefore, the incremental ROR analysis results in correctly selecting filter 2. But if only the i* values were used,
filter 1 would be wrongly chosen, because its i* exceeds that of filter 2 (25.41% > 23.57%). This
error occurs because the rate of return method assumes reinvestment at the alternative’s ROR
value, while PW and AW analyses use the MARR as the reinvestment rate. The conclusion is
simple:
If the ROR method is used to evaluate two or more alternatives, use the incremental cash flows
and ⌬i* to make the decision between alternatives.
8.5 Rate of Return Evaluation Using AW
Comparing alternatives by the ROR method (correctly performed) always leads to the same selection as PW and AW analyses, whether the ROR is determined using a PW-based or an AW-based
relation. However, for the AW-based technique, there are two equivalent ways to perform the
evaluation: (1) using the incremental cash flows over the LCM of alternative lives, just as for the
PW-based relation (Section 8.4), or (2) finding the AW for each alternative’s actual cash flows and
setting the difference of the two equal to zero to find the ⌬i* value. There is no difference between
the two approaches if the alternative lives are equal. Both methods are summarized here.
Since the ROR method requires comparison for equal service, the incremental cash flows
must be evaluated over the LCM of lives. There may be no real computational advantage to
using AW, as was found in Chapter 6. The same six-step procedure of the previous section (for
PW-based calculation) is used, except in step 5 the AW-based relation is developed.
The second AW-based method mentioned above takes advantage of the AW technique’s assumption that the equivalent AW value is the same for each year of the first and all succeeding life
cycles. Whether the lives are equal or unequal, set up the AW relation for the cash flows of each
alternative, form the relation below, and solve for i*.
0 ⴝ AWB ⴚ AWA
[8.3]
For both methods, all equivalent values are on an AW basis, so the i* that results from Equation [8.3] is the same as the ⌬i* found using the first approach. Example 8.5 illustrates ROR
analysis using AW-based relations for unequal lives.
EXAMPLE 8.5
Compare the alternatives of vendors A and B for Ford in Example 8.3, using an AW-based
incremental ROR method and the same MARR of 12% per year.
Solution
For reference, the PW-based ROR relation, Equation [8.2], for the incremental cash flow in
Example 8.3 shows that vendor B should be selected with ⌬i* ⫽ 12.65%.
For the AW relation, there are two equivalent solution approaches. Write an AW-based
relation on the incremental cash flow series over the LCM of 10 years, or write Equation [8.3]
for the two actual cash flow series over one life cycle of each alternative.
For the incremental method, the AW equation is
0 ⫽ ⫺5000(A兾P,⌬i*,10) ⫺ 11,000(P兾F,⌬i*,5)(A兾P,⌬i*,10) ⫹ 2000(A兾F,⌬i*,10) ⫹ 1900
Equal-service
requirement
214
Rate of Return Analysis: Multiple Alternatives
Chapter 8
It is easy to enter the incremental cash flows onto a spreadsheet, as in Figure 8–3, column D,
and use the ⫽ IRR(D4:D14) function to display ⌬i* ⫽ 12.65%.
For the second method, the ROR is found using the actual cash flows and the respective
lives of 10 years for A and 5 years for B.
AWA ⫽ ⫺8000(A兾P,i,10) ⫺ 3500
AWB ⫽ ⫺13,000(A兾P,i,5) ⫹ 2000(A兾F,i,5) ⫺ 1600
Now develop 0 ⫽ AWB ⫺ AWA.
0 ⫽ ⫺13,000(A兾P,i*,5) ⫹ 2000(A兾F,i*,5) ⫹ 8000(A兾P,i*,10) ⫹ 1900
Solution again yields i* ⫽ 12.65%.
Comment
It is very important to remember that when an incremental ROR analysis using an AW-based
equation is made on the incremental cash flows, the LCM must be used.
8.6 Incremental ROR Analysis of Multiple Alternatives
This section treats selection from multiple alternatives that are mutually exclusive, using the incremental ROR method. Acceptance of one alternative automatically precludes acceptance of
any others. The analysis is based upon PW (or AW) relations for incremental cash flows between
two alternatives at a time.
When the incremental ROR method is applied, the entire investment must return at least the
MARR. When the i* values on several alternatives exceed the MARR, incremental ROR evaluation
is required. (For revenue alternatives, if not even one i* ⱖ MARR, the do-nothing alternative is selected.) For all alternatives (revenue or cost), the incremental investment must be separately justified.
If the return on the extra investment equals or exceeds the MARR, then the extra investment should
be made in order to maximize the total return on the money available, as discussed in Section 8.1.
For ROR analysis of multiple, mutually exclusive alternatives, the following criteria are used.
ME alternative
selection
Select the one alternative
That requires the largest investment, and
Indicates that the extra investment over another acceptable alternative is justified.
An important rule to apply when evaluating multiple alternatives by the incremental ROR method
is that an alternative should never be compared with one for which the incremental investment is
not justified.
The incremental ROR evaluation procedure for multiple, equal-life alternatives is summarized below. Step 2 applies only to revenue alternatives, because the first alternative is compared
to DN only when revenue cash flows are estimated. The terms defender and challenger are dynamic in that they refer, respectively, to the alternative that is currently selected (the defender)
and the one that is challenging it for acceptance based on ⌬i*. In every pairwise evaluation, there
is one of each. The steps for solution by hand or by spreadsheet are as follows:
1. Order the alternatives from smallest to largest initial investment. Record the annual cash
flow estimates for each equal-life alternative.
2. Revenue alternatives only: Calculate i* for the first alternative. In effect, this makes DN the
defender and the first alternative the challenger. If i* ⬍ MARR, eliminate the alternative and
go to the next one. Repeat this until i* ⱖ MARR for the first time, and define that alternative
as the defender. The next alternative is now the challenger. Go to step 3. (Note: This is where
solution by spreadsheet can be a quick assist. Calculate the i* for all alternatives first using
the IRR function, and select as the defender the first one for which i* ⱖ MARR. Label it the
defender and go to step 3.)
3. Determine the incremental cash flow between the challenger and defender, using the relation
Incremental cash flow ⫽ challenger cash flow ⫺ defender cash flow
Set up the ROR relation.
Incremental ROR Analysis of Multiple Alternatives
8.6
4. Calculate ⌬i* for the incremental cash flow series using a PW- or AW-based equation. (PW
is most commonly used.)
5. If ⌬i* ⱖ MARR, the challenger becomes the defender and the previous defender is eliminated. Conversely, if ⌬i* ⬍ MARR, the challenger is removed, and the defender remains
against the next challenger.
6. Repeat steps 3 to 5 until only one alternative remains. It is the selected one.
Note that only two alternatives are compared at any one time. It is vital that the correct alternatives be compared, or the wrong alternative may be selected.
EXAMPLE 8.6
Caterpillar Corporation wants to build a spare parts storage facility in the Phoenix, Arizona,
vicinity. A plant engineer has identified four different location options. The initial cost of earthwork and prefab building and the annual net cash flow estimates are detailed in Table 8–5. The
annual net cash flow series vary due to differences in maintenance, labor costs, transportation
charges, etc. If the MARR is 10%, use incremental ROR analysis to select the one economically best location.
TABLE 8–5
Estimates for Four Alternative Building Locations, Example 8.6
A
Initial cost, $
Annual cash flow, $ per year
Life, years
⫺200,000
⫹22,000
30
B
C
⫺275,000
⫹35,000
30
⫺190,000
⫹19,500
30
D
⫺350,000
⫹42,000
30
Solution
All sites have a 30-year life, and they are revenue alternatives. The procedure outlined above
is applied.
1. The alternatives are ordered by increasing initial cost in Table 8–6.
2. Compare location C with the do-nothing alternative. The ROR relation includes only the
P兾A factor.
0 ⫽ ⫺190,000 ⫹ 19,500(P兾A,i*,30)
Table 8–6, column 1, presents the calculated (P兾A,⌬i*,30) factor value of 9.7436 and
⌬ic* ⫽ 9.63%. Since 9.63% ⬍ 10%, location C is eliminated. Now the comparison is A to
DN, and column 2 shows that ⌬iA* ⫽ 10.49%. This eliminates the do-nothing alternative;
the defender is now A and the challenger is B.
TABLE 8–6
Computation of Incremental Rate of Return for Four Alternatives,
Example 8.6
C
(1)
Initial cost, $
Cash flow, $ per year
Alternatives compared
Incremental cost, $
Incremental cash flow, $
Calculated (P兾A,⌬i*,30)
⌬i*,%
Increment justified?
Alternative selected
⫺190,000
⫹19,500
C to DN
⫺190,000
⫹19,500
9.7436
9.63
No
DN
A
(2)
⫺200,000
⫹22,000
A to DN
⫺200,000
⫹22,000
9.0909
10.49
Yes
A
B
(3)
⫺275,000
⫹35,000
B to A
⫺75,000
⫹13,000
5.7692
17.28
Yes
B
D
(4)
⫺350,000
⫹42,000
D to B
⫺75,000
⫹7,000
10.7143
8.55
No
B
215
216
Rate of Return Analysis: Multiple Alternatives
Chapter 8
3. The incremental cash flow series, column 3, and ⌬i* for B-to-A comparison are determined from
0 ⫽ ⫺275,000 ⫺ (⫺200,000) ⫹ (35,000 ⫺ 22,000)(P兾A,⌬i*,30)
⫽ ⫺75,000 ⫹ 13,000(P兾A,⌬i*,30)
4. From the interest tables, look up the P兾A factor at the MARR, which is (P兾A,10%,30) ⫽
9.4269. Now, any P兾A value greater than 9.4269 indicates that the ⌬i* will be less than
10% and is unacceptable. The P兾A factor is 5.7692, so B is acceptable. For reference
purposes, ⌬i* ⫽ 17.28%.
5. Alternative B is justified incrementally (new defender), thereby eliminating A.
6. Comparison D-to-B (steps 3 and 4) results in the PW relation 0 ⫽ ⫺75,000 ⫹
7000(P兾A,⌬i*,30) and a P兾A value of 10.7143 (⌬i* ⫽ 8.55%). Location D is eliminated,
and only alternative B remains; it is selected.
Comment
An alternative must always be incrementally compared with an acceptable alternative, and the
do-nothing alternative can end up being the only acceptable one. Since C was not justified in
this example, location A was not compared with C. Thus, if the B-to-A comparison had not
indicated that B was incrementally justified, then the D-to-A comparison would be correct
instead of D-to-B.
To demonstrate how important it is to apply the ROR method correctly, consider the following. If the i* of each alternative is computed initially, the results by ordered alternatives are
Location
i*, %
C
A
B
D
9.63
10.49
12.35
11.56
Now apply only the first criterion stated earlier; that is, make the largest investment that has a
MARR of 10% or more. Location D is selected. But, as shown above, this is the wrong selection, because the extra investment of $75,000 over location B will not earn the MARR. In fact,
it will earn only 8.55%. This is another example of the ranking inconsistency problem of the
ROR method mentioned in Section 8.4.
For cost alternatives, the incremental cash flow is the difference between costs for two alternatives. There is no do-nothing alternative and no step 2 in the solution procedure. Therefore, the
lowest-investment alternative is the initial defender against the next-lowest investment (challenger).
This procedure is illustrated in Example 8.7 using a spreadsheet solution.
EXAMPLE 8.7
The complete failure of an offshore platform and the resulting spillage of up to 800,000 to
1,000,000 gallons per day into the Gulf of Mexico in the spring of 2010 have made major oil producers and transporters very conscious of the harm done to people’s livelihood and all forms of
aquatic life by spills of this magnitude. To address the specific danger to birds that are shoreline
feeders and dwellers, environmental engineers from several international petroleum corporations
and transport companies—Exxon-Mobil, BP, Shell, and some transporters for OPEC producers—
have developed a plan to strategically locate throughout the world newly developed equipment
that is substantially more effective than manual procedures in cleaning crude oil residue from bird
feathers. The Sierra Club, Greenpeace, and other international environmental interest groups are
in favor of the initiative. Alternative machines from manufacturers in Asia, America, Europe, and
Africa are available with the cost estimates in Table 8–7. Annual cost estimates are expected to be
high to ensure readiness at any time. The company representatives have agreed to use the average
of the corporate MARR values, which results in MARR ⫽ 13.5%. Use a spreadsheet and incremental ROR analysis to determine which manufacturer offers the best economic choice.
Incremental ROR Analysis of Multiple Alternatives
8.6
TABLE 8–7
Costs for Four Alternative Machines, Example 8.7
First cost, $
Annual operating cost, $
Salvage value, $
Life, years
Machine 1
Machine 2
Machine 3
Machine 4
⫺5,000
⫺3,500
⫹500
8
⫺6,500
⫺3,200
⫹900
8
⫺10,000
⫺3,000
⫹700
8
⫺15,000
⫺1,400
⫹1,000
8
Solution by Spreadsheet
Follow the procedure for incremental ROR analysis. The spreadsheet in Figure 8–7 contains
the complete solution.
1. The alternatives are already ordered by increasing first costs.
2. These are cost alternatives, so there is no comparison to DN, since i* values cannot be
calculated.
3. Machine 2 is the first challenger to machine 1; the incremental cash flows for the 2-to-1
comparison are in column D.
4. The 2-to-1 comparison results in ⌬i* ⫽ 14.57% by applying the IRR function.
5. This return exceeds MARR ⫽ 13.5%, so machine 2 is the new defender (cell D17).
The comparison continues for 3-to-2 in column E, where the return is negative at ⌬i* ⫽
⫺18.77%; machine 2 is retained as the defender. Finally the 4-to-2 comparison has an incremental ROR of 13.60%, which is slightly larger than MARR ⫽ 13.5%. The conclusion is to
purchase machine 4 because the extra investment is (marginally) justified.
⫽ IRR(D6:D14)
Figure 8–7
Spreadsheet solution to select from multiple cost alternatives, Example 8.7.
Comment
As mentioned earlier, it is not possible to generate a PW versus i graph for each cost alternative
because all cash flows are negative. However, it is possible to generate PW versus ⌬i graphs
for the incremental series in the same fashion as we have done previously. The curves will
cross the PW ⫽ 0 line at the ⌬i* values determined by the IRR functions.
Selection from multiple, mutually exclusive alternatives with unequal lives using ⌬i* values
requires that the incremental cash flows be evaluated over the LCM of the two alternatives being
compared. This is another application of the principle of equal-service comparison. The spreadsheet application in the next section illustrates the computations.
It is always possible to rely on PW or AW analysis of the incremental cash flows at the MARR
to make the selection. In other words, don’t find ⌬i* for each pairwise comparison; find PW or
217
218
Chapter 8
Rate of Return Analysis: Multiple Alternatives
AW at the MARR instead. However, it is still necessary to make the comparison over the LCM
number of years for an incremental analysis to be performed correctly.
8.7 All-in-One Spreadsheet Analysis (Optional)
For professors and students who like to pack a spreadsheet, Example 8.8 combines many of the
economic analysis techniques we have learned so far—(internal) ROR analysis, incremental
ROR analysis, PW analysis, and AW analysis. Now that the IRR, NPV, and PV functions are
mastered, it is possible to perform a wide variety of evaluations for multiple alternatives on a
single spreadsheet. No cell tags are provided in this example. A nonconventional cash flow series
for which multiple ROR values may be found, and selection from both mutually exclusive alternatives and independent projects, are included in this example.
EXAMPLE 8.8
In-flight texting, phone, and Internet connections provided at airline passenger seats are an
expected service by many customers. Singapore Airlines knows it will have to replace 15,000
to 24,000 units in the next few years on its Boeing 757, 777, and A300 aircraft. Four optional
data handling features that build upon one another are available from the manufacturer, but at
an added cost per unit. Besides costing more, the higher-end options (e.g., satellite-based plugin video service) are estimated to have longer lives before the next replacement is forced by
new, advanced features expected by flyers. All four options are expected to boost annual revenues by varying amounts. Figure 8–8 spreadsheet rows 2 through 6 include all the estimates
for the four options.
(a) Using MARR ⫽ 15%, perform ROR, PW, and AW evaluations to select the one level of
options that is the most promising economically.
(b) If more than one level of options can be selected, consider the four that are described as
independent projects. If no budget limitations are considered at this time, which options
are acceptable if the MARR is increased to 20% when more than one option may be
implemented?
Figure 8–8
Spreadsheet analysis using ROR, PW, and AW methods for unequal-life, revenue alternatives, Example 8.8.
Chapter Summary
Solution by Spreadsheet
(a) The spreadsheet (Figure 8–8) is divided into six sections:
Section 1 (rows 1, 2): MARR value and the alternative names (A through D) are in increasing order of initial cost.
Section 2 (rows 3 to 6): Per-unit net cash flow estimates for each alternative. These are
revenue alternatives with unequal lives.
Section 3 (rows 7 to 20): Actual and incremental cash flows are displayed here.
Section 4 (rows 21, 22): Because these are all revenue alternatives, i* values are determined by the IRR function. If an alternative passes the MARR test (i* ⬎ 15%), it is
retained and a column is added to the right of its actual cash flows so the incremental
cash flows can be determined. Columns F and H were inserted to make space for the
incremental evaluations. Alternative A does not pass the i* test.
Section 5 (rows 23 to 25): The IRR functions display the ⌬i* values in columns F and H.
Comparison of C to B takes place over the LCM of 12 years. Since ⌬i*C⫺B ⫽ 19.42% ⬎
15%, eliminate B; alternative C is the new defender and D is the next challenger. The
final comparison of D to C over 12 years results in ⌬i*D⫺C ⫽ 11.23% ⬍ 15%, so D is
eliminated. Alternative C is the chosen one.
Section 6 (rows 26 to 28): These include the AW and PW analyses. The AW value over the life
of each alternative is calculated using the PMT function at the MARR with an embedded
NPV function. Also, the PW value is determined from the AW value for 12 years using the
PV function. For both measures, alternative C has the numerically largest value, as expected.
Conclusion: All methods result in the same, correct choice of alternative C.
(b) Since each option is independent of the others, and there is no budget limitation at this
time, each i* value in row 21 of Figure 8–8 is compared to MARR ⫽ 20%. This is a comparison of each option with the do-nothing alternative. Of the four, options B and C have
i* ⬎ 20%. They are acceptable; the other two are not.
Comment
In part (a), we should have applied the two multiple-root sign tests to the incremental cash flow
series for the C-to-B comparison. The series itself has three sign changes, and the cumulative
cash flow series starts negatively and also has three sign changes. Therefore, up to three realnumber roots may exist. The IRR function is applied in cell F23 to obtain ⌬i*C⫺B ⫽ 19.42%
without using a supplemental (Section 7.5) procedure. This means that the investment assumption of 19.42% for positive cash flows is a reasonable one. If the MARR ⫽ 15%, or some other
earning rate were more appropriate, the ROIC procedure could be applied to determine a single
rate, which would be different from 19.42%. Depending upon the investment rate chosen,
alternative C may or may not be incrementally justified against B. Here, the assumption is
made that the ⌬i* value is reasonable, so C is justified.
CHAPTER SUMMARY
Just as present worth and annual worth methods find the best alternative from among several,
incremental rate of return calculations can be used for the same purpose. In using the ROR technique, it is necessary to consider the incremental cash flows when selecting between mutually
exclusive alternatives. The incremental investment evaluation is conducted between only two
alternatives at a time, beginning with the lowest initial investment alternative. Once an alternative has been eliminated, it is not considered further.
Rate of return values have a natural appeal to management, but the ROR analysis is often
more difficult to set up and complete than the PW or AW analysis using an established MARR.
Care must be taken to perform a ROR analysis correctly on the incremental cash flows; otherwise
it may give incorrect results.
If there is no budget limitation when independent projects are evaluated, the ROR value of
each project is compared to the MARR. Any number, or none, of the projects can be accepted.
219
220
Chapter 8
Rate of Return Analysis: Multiple Alternatives
PROBLEMS
Understanding Incremental ROR
8.1 In a tabulation of cash flow, the column entitled
“Rate of return on the incremental cash flow” represents the rate of return on what?
8.2 If the rate of return on the incremental cash flow
between two alternatives is less than the minimum
attractive rate of return, which alternative should
be selected, if any?
8.3 An engineer is comparing three projects by the incremental ROR method. There are revenue and
cost cash flow estimates. He used the IRR function of Excel to determine the ROR values for
each project. The first one is 3.5% above the
MARR, the second is 1.2% below the MARR, and
the third is 2.4% above the MARR. Which alternatives, if any, must he include in the incremental
ROR analysis?
8.4 The rates of return on alternatives X and Y are
15% and 12%, respectively. Alternative Y requires
a larger investment than alternative X.
(a) What is known about the rate of return on the
increment of investment between the two alternatives?
(b) If the MARR is 12%, which alternative
should be selected and why?
8.5 Victoria is comparing two mutually exclusive
alternatives, A and B. The overall ROR on alternative A is greater than the MARR, and the overall
ROR on alternative B, which requires the larger
investment, is exactly equal to the MARR.
(a) What is known about the ROR on the increment between A and B?
(b) Which alternative should be selected?
8.6 A food processing company is considering two
types of moisture analyzers. Only one can be selected. The company expects an infrared model to
yield a rate of return of 27% per year. A more expensive microwave model will yield a rate of return of 22% per year. If the company’s MARR is
19% per year, can you determine which model
should be purchased solely on the basis of the
overall rate of return information provided? Why
or why not?
8.7 If $80,000 is invested at 30% and another $50,000
is invested at 20% per year, what is the overall rate
of return on the entire $130,000?
8.8 A total of $100,000 was invested in two different
projects identified as Z1 and Z2. If the overall rate
of return on the $100,000 was 30% and the rate of
return on the $30,000 invested in Z1 was 15%,
what was the rate of return on Z2?
8.9 Tuggle, Inc., which manufactures rigid shaft couplings, has $600,000 to invest. The company is
considering three different projects that will yield
the following rates of return.
Project X
Project Y
Project Z
iX ⫽ 24%
iY ⫽ 18%
iZ ⫽ 30%
The initial investment required for each project is
$100,000, $300,000, and $200,000, respectively.
If Tuggle’s MARR is 15% per year and it invests
in all three projects, what rate of return will the
company make?
8.10 Two options are available for setting up a wireless
meter scanner and controller. A simple setup is
good for 2 years and has an initial cost of $12,000,
no salvage value, and an operating cost of $27,000
per year. A more permanent system has a higher
first cost of $73,000, but it has an estimated life of
6 years and a salvage value of $15,000. It costs
only $14,000 per year to operate and maintain. If
the two options are compared using an incremental
rate of return, what are the incremental cash flows
in (a) year 0 and (b) year 2?
8.11 Prepare a tabulation of incremental cash flows for
the two machine alternatives below.
First cost, $
Annual operating cost, $ per year
Salvage value, $
Life, years
Machine X
Machine Y
⫺35,000
⫺31,600
0
2
⫺90,000
⫺19,400
8,000
4
8.12 For the alternatives shown, determine the sum of
the incremental cash flows for Q – P.
Alternative P Alternative Q
First cost, $
Annual operating cost, $ per year
Annual revenue, $ per year
Salvage value, $
Life, years
⫺50,000
⫺8,600
22,000
3,000
3
⫺85,000
⫺2,000
45,000
8,000
6
8.13 The tabulation of the incremental cash flows between alternatives A and B is shown on the next
page. Alternative A has a 3-year life and alternative B a 6-year life. If neither alternative has a salvage value, what is (a) the first cost of alternative
A and (b) the first cost of alternative B?
221
Problems
Year
Incremental
Cash Flow (B ⴚ A), $
0
1
2
3
4
5
6
–20,000
5,000
5,000
12,000
5,000
5,000
5,000
8.14 Standby power for pumps at water distribution
booster stations can be provided by either gasolineor diesel-powered engines. The costs for the gasoline engines are as follows:
Gasoline
First cost, $
Annual M&O, $ per year
Salvage value, $
Life, years
⫺150,000
⫺41,000
23,000
15
The incremental PW cash flow equation associated
with (diesel – gasoline) is
0 ⫽ ⫺40,000 ⫹ 11,000(P兾A,i,15) ⫹ 16,000(P兾F,i,15)
Determine the following:
(a) First cost of the diesel engines
(b) Annual M&O cost of the diesel engines
(c) Salvage value of the diesel engines
8.15 Several high-value parts for NASA’s reusable
space exploration vehicle can be either anodized
or powder-coated. Some of the costs for each
process are shown below:
First cost, $
Annual cost, $ per year
Resale value, $
Life, years
Anodize
Powder Coat
?
⫺21,000
?
3
⫺65,000
?
6,000
3
The incremental AW cash flow equation associated with (powder coat – anodize) is
0 ⫽ –14,000(A兾P,i,3) ⫹ 5000 ⫹ 2000(A兾F,i,3)
What is (a) the first cost for anodizing, (b) the annual cost for powder coating, and (c) the resale
(salvage) value of the anodized parts?
Incremental ROR Comparison (Two Alternatives)
8.16 Specialty Gases & Chemicals manufactures nitrogen trifluoride, a highly specialized gas used as an
industrial cleansing agent for flat panels installed in
laptop computers, televisions, and desktop monitors.
The incremental cash flow associated with two alternatives for chemical storage and handling systems
(identified as P3 and X3) has been calculated in
$1000 units. Determine (a) the rate of return on the
incremental cash flows and (b) which one should be
selected if the company’s MARR is 25% per year.
Alternative X3 requires the larger initial investment.
Year
Incremental Cash Flow
(X3 ⴚ P3), $1000
0
1–9
10
$⫺4600
1100
2000
8.17 As groundwater wells age, they sometimes begin to
pump sand (and they become known as “sanders”),
and this can cause damage to downstream desalting
equipment. This situation can be dealt with by drilling a new well at a cost of $1,000,000 or by installing a tank and self-cleaning screen ahead of the
desalting equipment. The tank and screen will cost
$230,000 to install and $61,000 per year to operate
and maintain. A new well will have a pump that is
more efficient than the old one, and it will require
almost no maintenance, so its operating cost will be
only $18,000 per year. If the salvage values are estimated at 10% of the first cost, use a present worth
relation to (a) calculate the incremental rate of return and (b) determine which alternative is better at
a MARR of 6% per year over a 20-year study
period.
8.18 Konica Minolta plans to sell a copier that prints
documents on both sides simultaneously, cutting in
half the time it takes to complete big commercial jobs.
The costs associated with producing chemicallytreated vinyl rollers and fiber-impregnated rubber
rollers are shown below. Determine which of
the two types should be selected by calculating the
rate of return on the incremental investment.
Assume the company’s MARR is 21% per year.
(Solve by hand and/or spreadsheet, as instructed.)
First cost, $
Annual cost, $ per year
Salvage value, $
Life, years
Treated
Impregnated
⫺50,000
⫺100,000
5,000
3
⫺95,000
⫺85,000
11,000
6
8.19 The Texas Department of Transportation (TxDOT)
is considering two designs for crash barriers along
a reconstructed portion of I-10. Design 2B will
cost $3 million to install and $135,000 per year to
maintain. Design 4R will cost $3.7 million to install and $70,000 per year to maintain. Calculate
the rate of return and determine which design is
preferred if TxDOT uses a MARR of 6% per year
and a 20-year project period.
222
Rate of Return Analysis: Multiple Alternatives
Chapter 8
8.20 Chem-Tex Chemical is considering two additives
for improving the dry-weather stability of its lowcost acrylic paint. Additive A has a first cost of
$110,000 and an annual operating cost of $60,000.
Additive B has a first cost of $175,000 and an annual operating cost of $35,000. If the company
uses a 3-year recovery period for paint products
and a MARR of 20% per year, which process is
economically favored? Use an incremental ROR
analysis.
8.21 The manager of Liquid Sleeve, Inc., a company
that makes a sealing solution for machine shaft surfaces that have been compromised by abrasion,
high pressures, or inadequate lubrication, is considering adding metal-based nanoparticles of either
type Al or Fe to its solution to increase the product’s performance at high temperatures. The costs
associated with each are shown below. The company’s MARR is 20% per year. Do the following
using a PW-based rate of return analysis and a
spreadsheet:
(a) Determine which nanoparticle type the company should select using the ⌬i* value.
(b) On the same graph, plot the PW versus different i values for each alternative. Indicate
the breakeven i* value and the MARR value
on the plot.
(c) Use the plot of PW versus ⌬i values to select
the better alternative with MARR ⫽ 20% per
year. Is the answer the same as in part (a)?
Type Fe
First cost, $
Annual operating cost, $ per year
Salvage value, $
Life, years
Type Al
⫺150,000 ⫺280,000
⫺92,000 ⫺74,000
30,000
70,000
2
4
8.22 A chemical company is considering two processes
for isolating DNA material. The incremental cash
flows between the two alternatives, J and S, have an
incremental rate of return that is less than 40%,
which is the MARR of the company. However, the
company CEO prefers the more expensive process
S. She believes the company can implement cost
controls to reduce the annual cost of the more
expensive process. By how much would she have to
reduce the annual operating cost of alternative S (in
$ per year) for it to have an incremental rate of
return of exactly 40%?
Year
Incremental Cash Flow
(S ⴚ J), $
0
1
2
3
⫺900,000
400,000
400,000
400,000
8.23 The incremental cash flows for two alternative
electrode setups are shown. The MARR is 12%
per year, and alternative Dryloc requires a larger
initial investment compared to NPT.
(a) Determine which should be selected using an
AW-based rate of return analysis.
(b) Use a graph of incremental values to determine the largest MARR value that will justify
the NPT alternative.
Year
Incremental Cash Flow
(Dryloc ⴚ NPT), $
0
1⫺8
9
⫺56,000
⫹8,900
⫹12,000
8.24 Hewett Electronics manufactures amplified pressure transducers. It must decide between two machines for a finishing operation. Select one for them
on the basis of AW-based rate of return analysis.
The company’s MARR is 18% per year.
First cost, $
Annual operating cost, $ per year
Salvage value, $
Life, years
Variable
Speed
Dual
Speed
⫺270,000
⫺135,000
75,000
6
⫺245,000
⫺139,000
35,000
6
8.25 A manufacturer of hydraulic equipment is trying to
determine whether it should use monoflange double block and bleed (DBB) valves or a multi-valve
system (MVS) for chemical injection. The costs
are shown below. Use an AW-based rate of return
analysis and a MARR of 18% per year to determine the better of the two options.
DBB
First cost, $
Annual cost, $ per year
Salvage value, $
Life, years
⫺40,000
⫺60,000
0
2
MVS
⫺71,000
⫺65,000
18,000
4
8.26 Poly-Chem Plastics is considering two types of injection molding machines—hydraulic and electric.
The hydraulic press (HP) will have a first cost of
$600,000, annual costs of $200,000, and a salvage
value of $70,000 after 5 years. Electric machine
technology (EMT) will have a first cost of $800,000,
annual costs of $150,000, and a salvage value of
$130,000 after 5 years.
(a) Use an AW-based rate of return equation to
determine the ROR on the increment of
investment between the two.
223
Problems
(b)
(c)
Determine which machine the company
should select, if the MARR ⫽ 16% per year.
Plot the AW versus i graph for each alternative’s cash flows, and utilize it to determine
the largest MARR that will justify the EMT
extra investment of $200,000.
8.27 Last week Eduardo calculated the overall project
ROR values for two alternatives A and B using the
estimates below. He calculated iA* ⫽ 34.2% and
iB* ⫽ 31.2% and recommended acceptance of A
since its rate of return exceeded the established
MARR of 30% by a greater amount than project B.
Yesterday, the general manager of the company
announced a major capital investment program,
which includes a large drop in the MARR from
30% to 20% per year. Do the following to help
Eduardo better understand the rate of return method
and what this reduction in MARR means.
(a) Explain the error that Eduardo made in performing the rate of return analysis.
(b) Perform the correct analysis using each
MARR value.
(c) Illustrate the ranking inconsistency problem
using the two MARR values, and determine
the maximum MARR that will justify alternative B.
Alternative A
First cost, $
Annual operating cost,
$ per year
Annual revenue, $ per year
Salvage value, $
Life, years
i*, %
a MARR of 10% per year to determine which
alternative is best using an incremental rate of
return analysis.
Bandwidth, First Cost,
Mbps
$1000
44
55
88
⫺85,000
⫺15,000
22,000
0
6
34.2
45,000
20,000
6
29.2
Multiple Alternative Comparison
8.28 Four mutually exclusive revenue alternatives
are under consideration to automate a baking
and packaging process at Able Bakery Products. The alternatives are ranked in order of increasing initial investment and compared by
incremental rate of return analysis. The rate of
return on each increment of investment was less
than the MARR. Which alternative should be
selected?
8.29 A WiMAX wireless network integrated with a
satellite network can provide connectivity to
any location within 10 km of the base station.
The number of sectors per base station can be
varied to increase the bandwidth. An independent cable operator is considering three bandwidth alternatives. Assume a life of 20 years and
⫹5000
⫹5000
⫹8000
–2000
–1000
–500
8.30 Xerox’s iGenX high-speed commercial printers
cost $1.5 billion to develop. The machines cost
$500,000 to $750,000 depending on what options
the client selects. Spectrum Imaging Systems is
considering the purchase of a new printer based
on recent contracts it received for printing weekly
magazine and mail-out advertising materials. The
operating costs and revenues generated are related
to a large extent to the speed and other capabilities
of the copier. Spectrum is considering the four
machines shown below. The company uses a
3-year planning period and a MARR of 15% per
year. Determine which copier the company should
acquire on the basis of an incremental rate of return analysis.
Copier
Initial
Investment,
$
Operating
Cost,
$ per Year
Annual
Revenue,
$ per Year
Salvage
Value, $
iGen-1
iGen-2
iGen-3
iGen-4
–500,000
–600,000
–650,000
–750,000
–350,000
–300,000
–275,000
–200,000
⫹450,000
⫹460,000
⫹480,000
⫹510,000
⫹70,000
⫹85,000
⫹95,000
⫹120,000
Alternative B
⫺40,000
⫺5,500
–40,000
–46,000
–61,000
Operating
Annual
Cost,
Income,
$1000 per Year $1000 per Year
8.31 Ashley Foods, Inc. has determined that any one of
five machines can be used in one phase of its chili
canning operation. The costs of the machines are
estimated below, and all machines are estimated to
have a 4-year useful life. If the minimum attractive
rate of return is 20% per year, determine which
machine should be selected on the basis of a rate of
return analysis.
Machine
First Cost, $
Annual
Operating
Cost, $ per Year
1
2
3
4
5
⫺31,000
⫺29,000
⫺34,500
⫺49,000
⫺41,000
⫺16,000
⫺19,300
⫺17,000
⫺12,200
⫺15,500
8.32 Five revenue projects are under consideration by
General Dynamics for improving material flow
through an assembly line. The initial cost in
224
Rate of Return Analysis: Multiple Alternatives
Chapter 8
Alternative
$1000 and the life of each project are as follows
(revenue estimates are not shown):
Project
Initial cost, $1000
Life, years
A
B
C
D
E
⫺700
5
⫺2300
8
⫺900
5
⫺300
5
⫺1600
6
An engineer made the comparisons shown below.
From the calculations, determine which project, if
any, should be undertaken if the company’s MARR
is (a) 11.5% per year and (b) 13.5% per year. If
other calculations are necessary to make a decision,
state which ones.
Comparison
Incremental Rate of Return, %
B vs DN
A vs B
D vs DN
E vs B
E vs D
E vs A
C vs DN
C vs A
E vs DN
A vs DN
E vs C
D vs C
D vs B
13%
19%
11%
15%
24%
21%
7%
19%
12%
10%
33%
33%
29%
8.33 The five alternatives shown here are being evaluated by the rate of return method.
Initial
ROR
Investment, versus
Alternative
$
DN, %
A
B
C
D
E
–25,000
–35,000
–40,000
–60,000
–75,000
(a)
(b)
(c)
9.6
15.1
13.4
25.4
20.2
A
B
C
D
— 27.3 9.4 35.3 25.0
— 0
38.5 24.4
— 46.5 27.3
— 6.8
—
If the alternatives are mutually exclusive and
the MARR is 26% per year, which alternative should be selected?
If the alternatives are mutually exclusive and
the MARR is 15% per year, which alternative should be selected?
If the alternatives are independent and the
MARR is 15% per year, which alternative(s)
should be selected?
8.34 Five mutually exclusive revenue alternatives that
have infinite lives are under consideration for increasing productivity in a manufacturing operation. The initial costs and cash flows of each
project are shown. If the MARR is 14.9% per year,
which alternative should be selected?
D
E
8.35 The plant manager at Automaton Robotics is looking at the summarized incremental rate of return
information shown below for five mutually exclusive alternatives, one of which must be chosen.
The table includes the overall ROR and the incremental comparison of alternatives. Which alternative is best if the minimum attractive rate of return
is (a) 15% per year and (b) 12% per year?
Alternative
First
Cost, $
A
B
C
D
E
⫺80,000
⫺60,000
⫺40,000
⫺30,000
⫺20,000
Overall Incremental Rate of Return, %
ROR, % A
B
C
D
E
14
16
17
12
8
—
12
11
17
24
12
—
14
23
21
11
14
—
35
29
17
23
35
—
17
24
21
29
17
—
8.36 Only one of four different machines can be purchased for Glass Act Products. An engineer performed the following analysis to select the best
machine, all of which have a 10-year life. Which
machine, if any, should the company select at a
MARR of 22% per year?
Machine
1
E
C
Initial cost, $
⫺7,000 ⫺23,000 ⫺9,000 ⫺3,000 ⫺16,000
Cash flow, $ per
1,000
3,500 1,400
500
2,200
year
Rate of return
14.3
15.2
15.6
16.7
13.8
(vs DN), %
Incremental
Rate of Return, %
A
B
2
3
4
Initial cost, $
⫺44,000 ⫺60,000 ⫺72,000 ⫺98,000
Annual cost, $ per year ⫺70,000 ⫺64,000 ⫺61,000 ⫺58,000
Annual savings,
⫹80,000 ⫹80,000 ⫹80,000 ⫹82,000
$ per year
Overall ROR, %
18.6
23.4
23.1
20.8
Machines compared
Incremental
investment, $
Incremental cash flow,
$ per year
ROR on increment, %
8.37
2 to 1
3 to 2
4 to 3
⫺16,000 ⫺12,000 ⫺26,000
⫹6,000
⫹3,000
⫹5,000
35.7
21.4
14.1
The U.S. Bureau of Reclamation is considering five
national park projects shown below, all of which can
be considered to last indefinitely. At a MARR of
7.5% per year, determine which should be selected, if
they are (a) independent and (b) mutually exclusive.
Project ID
First
Cost, $1000
Annual
Income, $1000
Rate of
Return, %
A
B
C
D
E
⫺20,000
⫺10,000
⫺15,000
⫺70,000
⫺50,000
⫹2000
⫹1300
⫹1000
⫹4000
⫹2600
10.0
13.0
6.6
5.7
5.2
225
Additional Problems and FE Exam Review Questions
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
8.38 When conducting a rate of return (ROR) analysis
involving multiple mutually exclusive alternatives, the first step is to:
(a) Rank the alternatives according to decreasing initial investment cost
(b) Rank the alternatives according to increasing
initial investment cost
(c) Calculate the present worth of each alternative using the MARR
(d) Find the LCM between all of the alternatives
8.39 In comparing mutually exclusive alternatives by
the ROR method, you should:
(a) Find the ROR of each alternative and pick
the one with the highest ROR
(b) Select the alternative whose incremental
ROR is the highest
(c) Select the alternative with ROR ⱖ MARR
that has the lowest initial investment cost
(d) Select the alternative with the largest initial investment that has been incrementally justified
8.40 When comparing independent projects by the
ROR method, you should:
(a) Find the ROR of each project and pick the
ones with the highest ROR
(b) Select all projects that have an overall ROR ⱖ
MARR
(c) Select the project with an overall ROR ⱖ
MARR that involves the lowest initial
investment cost
(d) Select the project with the largest initial investment that has been incrementally justified
8.41 Of the following scenarios, alternative Y requires a
higher initial investment than alternative X, and
the MARR is 20% per year. The only scenario that
requires an incremental investment analysis to
select an alternative is that:
(a) X has an overall ROR of 22% per year, and Y
has an overall ROR of 24% per year
(b) X has an overall ROR of 19% per year, and Y
has an overall ROR of 23% per year
(c) X has an overall ROR of 18% per year, and Y
has an overall ROR of 19% per year
(d) X has an overall ROR of 28% per year, and Y
has an overall ROR of 26% per year
8.42 Alternatives whose cash flows (excluding the
salvage value) are all negative are called:
(a) Revenue alternatives
(b) Nonconventional alternatives
(c) Cost alternatives
(d) Independent alternatives
8.43 For these alternatives, the sum of the incremental
cash flows is:
Year
A
B
0
1
2
3
4
5
⫺10,000
⫹2,500
⫹2,500
⫹2,500
⫹2,500
⫹2,500
⫺14,000
⫹4,000
⫹4,000
⫹4,000
⫹4,000
⫹4,000
(a)
(b)
(c)
(d)
$2500
$3500
$6000
$8000
8.44 Helical Systems, Inc. uses a minimum attractive
rate of return of 8% per year, compounded annually. The company is evaluating two new processes for improving the efficiency of its manufacturing operation. The cash flow estimates
associated with each process are shown below. A
correct equation to use for an incremental rate of
return analysis is:
First cost, $
Annual cost, $ per year
Salvage value, $
Life, years
(a)
(b)
(c)
(d)
Process I
Process J
⫺40,000
⫺15,000
5,000
3
⫺50,000
⫺12,000
6,000
3
0 ⫽ ⫺10,000 ⫹ 3000(P兾A,⌬i*,3) ⫹
1000(P兾F,⌬i*,3)
0 ⫽ ⫺40,000(A兾P,⌬i*,3) ⫺ 15,000 ⫹
5000(A兾F,⌬i*,3)
0 ⫽ ⫺50,000(A兾P,⌬i*,3) ⫺ 12,000 ⫹
6000(A兾F,⌬i*,3)
0 ⫽ ⫺10,000 ⫺ 3000(P兾A,⌬i*,3) ⫹
1000(P兾F,⌬i*,3)
8.45 For the four independent projects shown, the one or
ones to select using a MARR of 14% per year are:
(a)
(b)
(c)
(d)
Project
Rate of Return,
% per Year
A
B
C
D
14
12
15
10
Only C
Only A and C
Only A
Can’t tell; need to conduct incremental
analysis
226
Rate of Return Analysis: Multiple Alternatives
Chapter 8
Problems 8.46 through 8.48 are based on the following
information.
Five alternatives are being evaluated by the incremental
rate of return method.
Initial
Overall ROR
Alternative Investment, $ versus DN, %
A
B
C
D
E
⫺25,000
⫺35,000
⫺40,000
⫺60,000
⫺75,000
9.6
15.1
13.4
25.4
20.2
Incremental
Rate of Return, %
A
B
C
D
E
— 27.3 9.4 35.3 25.0
— 0 38.5 24.4
— 46.5 27.3
—
6.8
—
8.47 If the projects are mutually exclusive and the
MARR is 20% per year, the best alternative is:
(a) B
(b) C
(c) D
(d) E
8.48 If the projects are independent, instead of mutually
exclusive, the one or ones to select at an MARR of
18% per year are:
(a) B and C
(b) B, D, and E
(c) D and E
(d) B, C, and E
8.46 If the projects are mutually exclusive and the minimum attractive rate of return is 14% per year, the
best alternative is:
(a) B
(b) C
(c) D
(d) E
CASE STUDY
ROR ANALYSIS WITH ESTIMATED LIVES THAT VARY
Background
Make-to-Specs is a software system under development by
ABC Corporation. It will be able to translate digital versions of
three-dimensional computer models, containing a wide variety
of part shapes with machined and highly finished (ultrasmooth) surfaces. The product of the system is the numerically
controlled (NC) machine code for the part’s manufacturing.
Additionally, Make-to-Specs will build the code for superfine
finishing of surfaces with continuous control of the finishing
machines.
Information
There are two alternative computers that can provide the
server function for the software interfaces and shared database updates on the manufacturing floor while Make-toSpecs is operating in parallel mode. The server first cost
and estimated contribution to annual net cash flow are
summarized below.
First cost, $
Net cash flow, $/year
Life, years
Server 1
Server 2
100,000
35,000
200,000
50,000 year 1, plus 5000 per
year for years 2, 3, and 4
(gradient)
70,000 maximum for years
5 on, even if the server is
replaced
5 or 8
3 or 4
The life estimates were developed by two different individuals:
a design engineer and a manufacturing manager. They have
asked that, at this stage of the project, all analyses be performed
using both life estimates for each system.
Case Study Exercises
Use spreadsheet analysis to determine the following:
1. If the MARR ⫽ 12%, which server should be selected?
Use the PW or AW method to make the selection.
2. Use incremental ROR analysis to decide between the
servers at MARR ⫽ 12%.
3. Use any method of economic analysis to display on the
spreadsheet the value of the incremental ROR between
server 2 with a life estimate of 5 years and a life estimate
of 8 years.
Case Study
227
CASE STUDY
HOW A NEW ENGINEERING GRADUATE CAN HELP HIS FATHER1
Background
“I don’t know whether to sell it, expand it, lease it, or what. But
I don’t think we can keep doing the same thing for many more
years. What I really want to do is to keep it for 5 more years,
then sell it for a bundle,” Elmer Kettler said to his wife, Janise,
their son, John Kettler, and new daughter-in-law, Suzanne
Gestory, as they were gathered around the dinner table. Elmer
was sharing thoughts on Gulf Coast Wholesale Auto Parts, a
company he has owned and operated for 25 years on the southern outskirts of Houston, Texas. The business has excellent contracts for parts supply with several national retailers operating
in the area—NAPA, AutoZone, O’Reilly, and Advance. Additionally, Gulf Coast operates a rebuild shop serving these same
retailers for major automobile components, such as carburetors,
transmissions, and air conditioning compressors.
At his home after dinner, John decided to help his father
with an important and difficult decision: What to do with his
business? John graduated just last year with an engineering
degree from a major state university in Texas, where he completed a course in engineering economy. Part of his job at
Energcon Industries is to perform basic rate of return and
present worth analyses on energy management proposals.
Information
Over the next few weeks, John outlined five options, including his dad’s favorite of selling in 5 years. John summarized
all the estimates over a 10-year horizon. The options and
estimates were given to Elmer, and he agreed with them.
Option 1: Remove rebuild. Stop operating the rebuild shop and concentrate on selling wholesale
parts. The removal of the rebuild operations and the
switch to an “all-parts house” are expected to cost
$750,000 in the first year. Overall revenues will
drop to $1 million the first year with an expected 4%
increase per year thereafter. Expenses are projected
at $0.8 million the first year, increasing 6% per year
thereafter.
Option 2: Contract rebuild operations. To get the
rebuild shop ready for an operations contractor to
take over will cost $400,000 immediately. If expenses stay the same for 5 years, they will average
$1.4 million per year, but they can be expected to
rise to $2 million per year in year 6 and thereafter.
Elmer thinks revenues under a contract arrangement
can be $1.4 million the first year and can rise 5% per
year for the duration of a 10-year contract.
Option 3: Maintain status quo and sell out after 5 years
(Elmer’s personal favorite). There is no cost now, but
1
the current trend of negative net profit will probably
continue. Projections are $1.25 million per year for expenses and $1.15 million per year in revenue. Elmer
had an appraisal last year, and the report indicated Gulf
Coast Wholesale Auto Parts is worth a net $2 million.
Elmer’s wish is to sell out completely after 5 more
years at this price, and to make a deal that the new
owner pay $500,000 per year at the end of year 5 (sale
time) and the same amount for the next 3 years.
Option 4: Trade-out. Elmer has a close friend in the
antique auto parts business who is making a “killing,”
so he says, with e-commerce. Although the possibility
is risky, it is enticing to Elmer to consider a whole new
line of parts, but still in the basic business that he already understands. The trade-out would cost an estimated $1 million for Elmer immediately. The 10-year
horizon of annual expenses and revenues is considerably higher than for his current business. Expenses are
estimated at $3 million per year and revenues at
$3.5 million each year.
Option 5: Lease arrangement. Gulf Coast could be
leased to some turnkey company with Elmer remaining
the owner and bearing part of the expenses for building,
delivery trucks, insurance, etc. The first-cut estimates
for this option are $1.5 million to get the business ready
now, with annual expenses at $500,000 per year and
revenues at $1 million per year for a 10-year contract.
Case Study Exercises
Help John with the analysis by doing the following:
1. Develop the actual cash flow series and incremental
cash flow series (in $1000 units) for all five options in
preparation for an incremental ROR analysis.
2. Discuss the possibility of multiple rate of return values
for all the actual and incremental cash flow series. Find
any multiple rates in the range of 0% to 100%.
3. If John’s father insists that he make 25% per year or more
on the selected option over the next 10 years, what should
he do? Use all the methods of economic analysis you
have learned so far (PW, AW, ROR) so John’s father can
understand the recommendation in one way or another.
4. Prepare plots of the PW versus i for each of the five options.
Estimate the breakeven rate of return between options.
5. What is the minimum amount that must be received in
each of years 5 through 8 for option 3 (the one Elmer
wants) to be best economically? Given this amount,
what does the sale price have to be, assuming the same
payment arrangement as presented above?
Based upon a study by Alan C. Stewart, Consultant, Communications and High Tech Solutions Engineering, Accenture LLP.
CHAPTER 9
Benefit/Cost
Analysis and
Public Sector
Economics
L E A R N I N G
O U T C O M E S
Purpose: Understand public sector projects and select the best alternative on the basis of incremental benefit /cost analysis.
SECTION
TOPIC
LEARNING OUTCOME
9.1
Public sector
• Explain some of the fundamental differences
between private and public sector projects.
9.2
B/C for single project
• Calculate the benefit/cost ratio and use it to
evaluate a single project.
9.3
Incremental B/C
• Select the better of two alternatives using the
incremental B/C ratio method.
9.4
More than two alternatives
• Based on the incremental B/C ratios, select the
best of multiple alternatives.
9.5
Service projects and CEA
• Explain service sector projects and use costeffectiveness analysis (CEA) to evaluate projects.
9.6
Ethical considerations
• Explain the major aspects of public project
activities, and describe how ethical compromise
may enter public sector project analysis.
T
he evaluation methods of previous chapters are usually applied to alternatives
in the private sector, that is, for-profit and not-for-profit corporations and businesses. This chapter introduces public sector and service sector alternatives and
their economic consideration. In the case of public projects, the owners and users (beneficiaries) are the citizens and residents of a government unit—city, county, state, province,
or nation. Government units provide the mechanisms to raise capital and operating funds.
Public-private partnerships have become increasingly common, especially for large infrastructure projects such as major highways, power generation plants, water resource developments, and the like.
The benefit/cost (B/C) ratio introduces objectivity into the economic analysis of public sector evaluation, thus reducing the effects of politics and special interests. The different formats of B/C analysis, and associated disbenefits of an alternative, are discussed here. The B/C
analysis can use equivalency computations based on PW, AW, or FW values. Performed correctly, the benefit/cost method will always select the same alternative as PW, AW, and ROR
analyses.
This chapter also introduces service sector projects and discusses how their economic evaluation is different from that for other projects. Finally, there is a discussion on professional
ethics and ethical dilemmas in the public sector.
PE
Water Treatment Facility #3 Case: Allen
Water Utilities has planned for the last
25 years to construct a new drinking
water treatment facility that will supply
the rapidly growing north and northwest areas of the city. An expectation of
over 100,000 new residents in the next
several years and 500,000 by 2040
prompted the development of the plant
starting in 2012. The supply is from a
large surface lake currently used to provide water to all of Allen and the surrounding communities. The project is
termed WTF3, and its initial capital investment is $540 million for the treatment plant and two large steel-pipe
transmission mains (84- and 48-inch)
that will be installed via tunneling approximately 100 to 120 feet under suburban areas of the city to reach current
reservoirs.
Tunneling was selected after geotechnical borings indicated that open trenching was not supportable by the soil and
based upon a large public outcry against
trenching in the living areas along the
selected transmission routes. Besides the
treatment plant construction on the 95acre site, there must be at least three
large vertical shafts (25 to 50 feet in diameter) bored along each transmission
main to gain underground access for
equipment and debris removal during
the tunneling operations.
The stated criteria used to make decisions for WTF3 and the transmission
mains were economics, environment,
community impact, and constructability.
There are major long-term benefits for
the new facility. These are some mentioned by city engineers:
• It will meet projected water needs
of the city for the next 50 years.
• The new treatment plant is at a
higher elevation than the current
two plants, allowing gravity flow to
replenish reservoirs, thereby using
little or no electric pumping.
• There will be an increase in the
diversity and reliability of supply as
other plants age.
• It will provide a water quality that
is more consistent due to the location of the raw water intakes.
• The facility uses water supplies
already purchased; therefore, there
is no need to negotiate additional
allowances.
The disbenefits are mostly short-term
during the construction of WTF3 and
transmission mains. Some of these are
mentioned by citizen groups and one retired city engineer:
• There will be disruption of habitat for some endangered species
of birds, lizards, and trees not
230
Chapter 9
Benefit/Cost Analysis and Public Sector Economics
found in any other parts of the
country.
• Large amounts of dust and smoke
will enter the atmosphere in a
residential area during the 3½ years
of construction, tunneling, and
transmission main completion.
• Noise pollution and traffic congestion will result during an estimated
26,000 truck trips to remove debris
from the plant site and tunnel
shafts, in addition to the problems
from regular construction traffic.
• Natural landscape in plant and tunnel shaft sites will be destroyed.
• Safety will be compromised for children in a school where large trucks
will pass about every 5 minutes for
approximately 12 hours per day,
6 days per week for 2½ years.
• There may be delays in fire and
ambulance services in emergencies,
since many neighborhood streets are
country-road width and offer only
single ingress/egress streets for neighborhoods along the indicated routes.
• The need for the facility has not
been proved, as the water will be
sold to developers outside the city
limits, not provided to residences
within Allen.
• Newly generated revenues will be
used to pay off the capital funding
bonds approved for the plant’s construction.
Last year, the city engineers did a benefit/cost analysis for this massive public
sector project; none of the results were
publicized. Public and elected official intervention has now caused some of the
conclusions using the criteria mentioned
above to be questioned by the general
manager of Allen Water Utilities.
This case is used in the following topics
of this chapter:
Public sector projects (Section 9.1)
Incremental B/C analysis, two alternatives (Section 9.3)
Incremental B/C analysis, more than
two alternatives (Section 9.4)
9.1 Public Sector Projects
Virtually all the examples and problems of previous chapters have involved the private sector,
where products, systems, and services are developed and offered by corporations and businesses
for use by individual customers and clients, the government, or other companies. (Notable exceptions are the long-life alternatives discussed in Chapters 5 (PW) and 6 (AW) where capitalized
cost analysis was applied.) Now we will explore projects that concentrate on government units
and the citizens they serve. These are called public sector projects.
A public sector project is a product, service, or system used, financed, and owned by the citizens
of any government level. The primary purpose is to provide service to the citizenry for the
public good at no profit. Areas such as public health, criminal justice, safety, transportation,
welfare, and utilities are publically owned and require economic evaluation.
Upon reflection, it is surprising how much of what we use on a daily or as-needed basis is
publicly owned and financed to serve us—the citizenry. These are some public sector
examples:
Hospitals and clinics
Parks and recreation
Utilities: water, electricity, gas,
sewer, sanitation
Schools: primary, secondary, community
colleges, universities
Economic development projects
Convention centers
Sports arenas
Transportation: highways, bridges,
waterways
Public Sector Projects
9.1
Police and fire protection
Courts and prisons
Food stamp and rent relief programs
Job training
Public housing
Emergency relief
Codes and standards
There are significant differences in the characteristics of private and public sector alternatives.
They are summarized here.
Characteristic
Public sector
Private sector
Size of investment
Large
Some large; more medium to small
Often alternatives developed to serve public needs require large initial investments, possibly
distributed over several years. Modern highways, public transportation systems, universities,
airports, and flood control systems are examples.
Characteristic
Public sector
Private sector
Life estimates
Longer (30–50ⴙ years)
Shorter (2–25 years)
The long lives of public projects often prompt the use of the capitalized cost method, where
infinity is used for n and annual costs are calculated as A ⫽ P(i). As n gets larger, especially
over 30 years, the differences in calculated A values become small. For example, at i ⫽ 7%,
there will be a very small difference in 30 and 50 years, because (A/P,7%,30) ⫽ 0.08059 and
(A/P,7%,50) ⫽ 0.07246.
Characteristic
Annual cash flow
estimates
Public sector
Private sector
No profit; costs, benefits, and
disbenefits are estimated
Revenues contribute to
profits; costs are estimated
Public sector projects (also called publicly owned) do not have profits; they do have costs that are
paid by the appropriate government unit; and they benefit the citizenry. Public sector projects
often have undesirable consequences, as interpreted by some sectors of the public. It is these
consequences that can cause public controversy about the projects. The economic analysis should
consider these consequences in monetary terms to the degree estimable. (Often in private sector
analysis, undesirable consequences are not considered, or they may be directly addressed as
costs.) To perform a benefit/cost economic analysis of public alternatives, the costs (initial and
annual), the benefits, and the disbenefits, if considered, must be estimated as accurately as
possible in monetary units.
Costs—estimated expenditures to the government entity for construction, operation, and maintenance of the project, less any expected salvage value.
Benefits—advantages to be experienced by the owners, the public.
Disbenefits—expected undesirable or negative consequences to the owners if the alternative is
implemented. Disbenefits may be indirect economic disadvantages of the alternative.
It is difficult to estimate and agree upon the economic impact of benefits and disbenefits for a
public sector alternative. For example, assume a short bypass around a congested area in town is
recommended. How much will it benefit a driver in dollars per driving minute to be able to bypass five traffic lights while averaging 35 miles per hour, as compared to currently driving
through the lights averaging 20 miles per hour and stopping at an average of two lights for an
average of 45 seconds each? The bases and standards for benefits estimation are always difficult
to establish and verify. Relative to revenue cash flow estimates in the private sector, benefit estimates are much harder to make, and vary more widely around uncertain averages. (The inability
to make economic estimates for benefits may be overcome by using the evaluation technique
discussed in Section 9.5.) And the disbenefits that accrue from an alternative are even harder to
estimate. In fact, the disbenefit itself may not be known at the time the evaluation is performed.
231
232
Benefit/Cost Analysis and Public Sector Economics
Chapter 9
Characteristic
Funding
Public sector
Private sector
Taxes, fees, bonds,
private funds
Stocks, bonds, loans,
individual owners
The capital used to finance public sector projects is commonly acquired from taxes, bonds,
and fees. Taxes are collected from those who are the owners—the citizens (e.g., federal gasoline
taxes for highways are paid by all gasoline users, and health care costs are covered by insurance
premiums). This is also the case for fees, such as toll road fees for drivers. Bonds are often issued:
U.S. Treasury bonds, municipal bond issues, and special-purpose bonds, such as utility district
bonds. Private lenders can provide up-front financing. Also, private donors may provide funding
for museums, memorials, parks, and garden areas through gifts.
Characteristic
Interest rate
Public sector
Private sector
Lower
Higher, based on cost of capital
Because many of the financing methods for public sector projects are classified as low-interest,
the interest rate is virtually always lower than for private sector alternatives. Government
agencies are exempt from taxes levied by higher-level units. For example, municipal projects
do not have to pay state taxes. (Private corporations and individual citizens do pay taxes.)
Many loans are very low-interest, and grants with no repayment requirement from federal
programs may share project costs. This results in interest rates in the 4% to 8% range. It is
common that a government agency will direct that all projects be evaluated at a specific rate.
As a matter of standardization, directives to use a specific interest rate are beneficial because
different government agencies are able to obtain varying types of funding at different rates.
This can result in projects of the same type being rejected in one state or city but accepted in
another. Standardized rates tend to increase the consistency of economic decisions and to reduce gamesmanship.
The determination of the interest rate for public sector evaluation is as important as the determination of the MARR for a private sector analysis. The public sector interest rate is identified
as i; however, it is referred to by other names to distinguish it from the private sector rate. The
most common terms are discount rate and social discount rate.
Characteristic
Alternative selection
criteria
Public sector
Private sector
Multiple criteria
Primarily based on rate
of return
Multiple categories of users, economic as well as noneconomic interests, and special-interest
political and citizen groups make the selection of one alternative over another much more difficult in public sector economics. Seldom is it possible to select an alternative on the sole basis of
a criterion such as PW or ROR. It is important to describe and itemize the criteria and selection
method prior to the analysis. This helps determine the perspective or viewpoint when the evaluation is performed. Viewpoint is discussed below.
Characteristic
Environment of the evaluation
Public sector
Private sector
Politically inclined
Primarily economic
There are often public meetings and debates associated with public sector projects to accommodate the various interests of citizens (owners). Elected officials commonly assist with the selection, especially when pressure is brought to bear by voters, developers, environmentalists, and
others. The selection process is not as “clean” as in private sector evaluation.
The viewpoint of the public sector analysis must be determined before cost, benefit, and disbenefit estimates are made and before the evaluation is formulated and performed. There are
several viewpoints for any situation, and the different perspectives may alter how a cash flow
estimate is classified.
Public Sector Projects
9.1
233
Some example perspectives are the citizen; the city tax base; number of students in the school district; creation and retention of jobs; economic development potential; a particular industry interest
(agriculture, banking, electronics manufacturing); even the reelection of a public officeholder (often
termed pork projects). In general, the viewpoint of the analysis should be as broadly defined as those
who will bear the costs of the project and reap its benefits. Once established, the viewpoint assists in
categorizing the costs, benefits, and disbenefits of each alternative, as illustrated in Example 9.1.
EXAMPLE 9.1 Water Treatment Facility #3 Case
The situation with the location and construction of the new WTF3 and associated transmission
mains described in the chapter’s introduction has reached a serious level because of recent
questions posed by some city council members and citizen groups. Before going public to the
city council with the analysis performed last year, the director of Allen Water Utilities has
asked an engineering management consultant to review it and determine if it was an acceptable
analysis and correct economic decision, then and now. The lead consultant, Joel Whiterson,
took engineering economy as a part of his B.S. education and has previously worked on
economic studies in the government sector, but never as the lead person.
Within the first hour of checking background notes, Joel found several initial estimates
(shown below) from last year for expected consequences if WTF3 were built. He realized that
no viewpoint of the study was defined, and, in fact, the estimates were never classified as costs,
benefits, or disbenefits. He did determine that disbenefits were considered at some point in the
analysis, though the estimates for them are very sketchy.
Joel defined two viewpoints: a citizen of Allen and the Allen Water Utilities budget. He
wants to identify each of the estimates as a cost, benefit, or disbenefit from each viewpoint.
Please help with this classification.
Economic Dimension
Monetary Estimate
1. Cost of water: 10% annual increase to Allen
households
Average of $29.7 million (years 1–5, steady
thereafter)
2. Bonds: Annual debt service at 3% per year on
$540 million
$16.2 million (years 1–19); $516.2 million (year 20)
3. Use of land: Payment to Parks and Recreation
for shaft sites and construction areas
$300,000 (years 1–4)
4. Property values: Loss in value, sales price,
and property taxes
$4 million (years 1–5)
5. Water sales: Increases in sales to surrounding
communities
$5 million (year 4) plus 5% per year (years 5–20)
6. M&O: Annual maintenance and operations
costs
$300,000 plus 4% per year increase (years 1–20)
7. Peak load purchases: Savings in purchases of
treated water from secondary sources
$500,000 (years 5–20)
Solution
The perspective of each viewpoint is identified and estimates are classified. (How this classification is done will vary depending upon who does the analysis. This solution offers only one
logical answer.)
Viewpoint 1: Citizen of the city of Allen. Goal: Maximize the quality of life and wellness of
citizens with family and neighborhood as prime concerns.
Costs: 1, 2, 4, 6
Benefits: 5, 7
Disbenefits: 3
Viewpoint 2: Allen Water Utilities budget. Goal: Ensure the budget is balanced and of sufficient size to fund rapidly growing city service demands.
Costs: 2, 3, 6
Benefits: 1, 5, 7
Disbenefits: 4
Citizens view costs in a different light than a city budget employee does. For example, the loss
of property values (item 4) is considered a real cost to a citizen, but is an unfortunate disbenefit
PE
234
Chapter 9
Benefit/Cost Analysis and Public Sector Economics
from the city budget perspective. Similarly, the Allen Water Utilities budget interprets estimate
3 (payment for use of land to Parks and Recreation) as a real cost; but a citizen might interpret
this as merely a movement of funds between two municipal budgets—therefore, it is a disbenefit, not a real cost.
Comment
The inclusion of disbenefits can easily change the economic decision. However, agreement on
the disbenefits and their monetary estimates is difficult (to impossible) to develop, often resulting in the exclusion of any disbenefits from the economic analysis. Unfortunately, this usually
transfers the consideration of disbenefits to the noneconomic (i.e., political) realm of public
project decision making.
Most of the large public sector projects are developed through public-private partnerships
(PPPs). A partnership is advantageous in part because of the greater efficiency of the private sector and in part because of the sizable cost to design, construct, and operate such projects. Full
funding by the government unit may not be possible using traditional means—fees, taxes, and
bonds. Some examples of the projects are as follows:
Project
Some Purposes of the Project
Mass transportation
Reduce transit time; reduce congestion; improve environment; decrease road
accidents
Speed traffic flows; reduce congestion; improve safety
Increase cargo capacity; support industrial development; increase tourism
Increase capacity; improve passenger safety; support development
Desalination for drinking water; meet irrigation and industrial needs; improve
wastewater treatment
Bridges and tunnels
Ports and harbors
Airports
Water resources
In these joint ventures, the public sector (government) is responsible for the funding and service
to the citizenry, and the private sector partner (corporation) is responsible for varying aspects of
the projects as detailed below. The government unit cannot make a profit, but the corporation(s)
involved can realize a reasonable profit; in fact, the profit margin is usually written into the contract that governs the design, construction, and operation of the project.
Traditional methods of contracting were fixed-price (historically called lump-sum) and cost
reimbursable (also called cost-plus). In these formats, a government unit took responsibility for
funding and possibly some of the design elements, and later all operation activities, while the
contractor did not share in the risks involved—liability, natural disasters, funding shortfalls, etc.
More recently, the PPP has become the arrangement of choice for most large public projects.
Commonly these are called design-build contracts, under which contractors take on more and
more of the functions from design to operation. Details are explained in publications such as
Design-Build Contracting Handbook (Cushman and Loulakis). The most reliance is placed upon
a contractor or contractors with a DBOMF contract, as described below.
The Design-Build-Operate-Maintain-Finance (DBOMF) contract is considered a turnkey approach to a project. It requires the contractor(s) to perform all the DBOMF activities with collaboration and approval of the owner (the government unit). The activity of financing is the
management of cash flow to support project implementation by a contracting firm. Although a
contractor may assist in some instances, the funding (obtaining the capital funds) remains the
government’s responsibility through bonding, commercial loans, taxation, grants, and gifts.
When the financing activity is not managed by a contractor, the contract is a DBOM; it is also
common to develop a design-build contract. In virtually all cases, some forms of design-build
arrangements for public projects are made because they offer several advantages to the government and citizens served:
• Cost and time savings in the design, build, and operate phases
• Earlier and more reliable (less variable) cost estimates
Benefit/Cost Analysis of a Single Project
9.2
235
• Reduced administrative responsibilities for the owner
• Better efficiency of resource allocation of private enterprise
• Environmental, liability, and safety issues addressed by the private sector, where there usually
is greater expertise
Many of the projects in international settings and in developing countries utilize the publicprivate partnership. There are, of course, disadvantages to this arrangement. One risk is that the
amount of funding committed to the project may not cover the actual build cost because it is
considerably higher than estimated. Another risk is that a reasonable profit may not be realized
by the private corporation due to low usage of the facility during the operate phase. To prevent
such problems, the original contract may provide for special subsidies and loans guaranteed by
the government unit. The subsidy may cover costs plus (contractually agreed-to) profit if usage
is lower than a specified level. The level used may be the breakeven point with the agreed-to
profit margin considered.
9.2 Benefit/Cost Analysis of a Single Project
The benefit/cost ratio is relied upon as a fundamental analysis method for public sector projects.
The B/C analysis was developed to introduce greater objectivity into public sector economics,
and as one response to the U.S. Congress approving the Flood Control Act of 1936. There are
several variations of the B/C ratio; however, the fundamental approach is the same. All cost and
benefit estimates must be converted to a common equivalent monetary unit (PW, AW, or FW) at
the discount rate (interest rate). The B/C ratio is then calculated using one of these relations:
PW of benefits ⴝ ———————
AW of benefits ⴝ ———————
FW of benefits
B/C ⴝ ———————
PW of costs
AW of costs
FW of costs
[9.1]
Present worth and annual worth equivalencies are preferred to future worth values. The sign
convention for B/C analysis is positive signs; costs are preceded by a ⴙ sign. Salvage values
and additional revenues to the government, when they are estimated, are subtracted from costs in
the denominator. Disbenefits are considered in different ways depending upon the model used.
Most commonly, disbenefits are subtracted from benefits and placed in the numerator. The
different formats are discussed below.
The decision guideline is simple:
If B/C ⱖ 1.0, accept the project as economically justified for the estimates and discount rate
applied.
If B/C ⬍ 1.0, the project is not economically acceptable.
If the B/C value is exactly or very near 1.0, noneconomic factors will help make the decision.
The conventional B/C ratio, probably the most widely used, is calculated as follows:
benefits ⴚ disbenefits ⴝ ———
BⴚD
B/C ⴝ ——————————
costs
C
[9.2]
In Equation [9.2] disbenefits are subtracted from benefits, not added to costs. The B/C value
could change considerably if disbenefits are regarded as costs. For example, if the numbers
10, 8, and 5 are used to represent the PW of benefits, disbenefits, and costs, respectively, the
correct procedure results in B/C ⫽ (10 ⫺ 8)兾5 ⫽ 0.40. The incorrect placement of disbenefits
in the denominator results in B/C ⫽ 10兾(8 ⫹ 5) ⫽ 0.77, which is approximately twice the correct B/C value of 0.40. Clearly, then, the method by which disbenefits are handled affects the
magnitude of the B/C ratio. However, regardless of whether disbenefits are (correctly) subtracted from the numerator or (incorrectly) added to costs in the denominator, a B/C ratio of
less than 1.0 by the first method will always yield a B/C ratio less than 1.0 by the second
method, and vice versa.
The modified B/C ratio includes all the estimates associated with the project, once operational.
Maintenance and operation (M&O) costs are placed in the numerator and treated in a manner
Project evaluation
236
Benefit/Cost Analysis and Public Sector Economics
Chapter 9
similar to disbenefits. The denominator includes only the initial investment. Once all amounts are
expressed in PW, AW, or FW terms, the modified B/C ratio is calculated as
benefits ⴚ disbenefits ⴚ M&O costs
Modified B/C ⴝ ————————————————
initial investment
[9.3]
Salvage value is usually included in the denominator as a negative cost. The modified B/C ratio
will obviously yield a different value than the conventional B/C method. However, as with disbenefits, the modified procedure can change the magnitude of the ratio but not the decision to
accept or reject the project.
The benefit and cost difference measure of worth, which does not involve a ratio, is based on
the difference between the PW, AW, or FW of benefits and costs, that is, B − C. If (B − C) ⱖ 0,
the project is acceptable. This method has the advantage of eliminating the discrepancies noted
above when disbenefits are regarded as costs, because B represents net benefits. Thus, for the
numbers 10, 8, and 5, the same result is obtained regardless of how disbenefits are treated.
Subtracting disbenefits from benefits:
Adding disbenefits to costs:
B ⫺ C ⫽ (10 ⫺ 8) ⫺ 5 ⫽ ⫺3
B ⫺ C ⫽ 10 ⫺ (8 ⫹ 5) ⫽ ⫺3
Before calculating the B/C ratio by any formula, check whether the alternative with the larger
AW or PW of costs also yields a larger AW or PW of benefits. It is possible for one alternative
with larger costs to generate lower benefits than other alternatives, thus making it unnecessary to
further consider the larger-cost alternative.
By the very nature of benefits and especially disbenefits, monetary estimates are difficult to
make and will vary over a wide range. The extensive use of sensitivity analysis on the more
questionable parameters helps determine how sensitive the economic decision is to estimate
variation. This approach assists in determining the economic and public acceptance risk associated with a defined project. Also, the use of sensitivity analysis can alleviate some of the public’s concerns commonly expressed that people (managers, engineers, consultants, contractors,
and elected officials) designing (and promoting) the public project are narrowly receptive to different approaches to serving the public’s interest.
EXAMPLE 9.2
In the past, the Afram Foundation has awarded many grants to improve the living and medical
conditions of people in war-torn and poverty-stricken countries throughout the world. In a
proposal for the foundation’s board of directors to construct a new hospital and medical clinic
complex in a deprived central African country, the project manager has developed some estimates. These are developed, so she states, in a manner that does not have a major negative
effect on prime agricultural land or living areas for citizens.
Award amount:
Annual costs:
Benefits:
Disbenefits:
$20 million (end of) first year, decreasing by $5 million per year for 3
additional years; local government will fund during the first year only
$2 million per year for 10 years, as proposed
Reduction of $8 million per year in health-related expenses for citizens
$0.1 to $0.6 million per year for removal of arable land and commercial
districts
Use the conventional and modified B/C methods to determine if this grant proposal is economically justified over a 10-year study period. The foundation’s discount rate is 6% per year.
Solution
Initially, determine the AW for each parameter over 10 years. In $1 million units,
Award:
Annual costs:
Benefits:
Disbenefits:
20 − 5(A/G,6%,4) ⫽ $12.864 per year
$2 per year
$8 per year
Use $0.6 for the first analysis
Benefit/Cost Analysis of a Single Project
9.2
237
The conventional B/C analysis applies Equation [9.2].
8.0 ⫺ 0.6 ⫽ 0.50
B/C ⫽ ——————
12.864 ⫹ 2.0
The modified B/C analysis uses Equation [9.3].
8.0 ⫺ 0.6 ⫺ 2.0 ⫽ 0.42
Modified B/C ⫽ ———————
12.864
The proposal is not justified economically since both measures are less than 1.0. If the low
disbenefits estimate of $0.1 million per year is used, the measures increase slightly, but not
enough to justify the proposal.
It is possible to develop a direct formula connection between the B/C of a public sector and
B/C of a private sector project that is a revenue alternative; that is, both revenues and costs are
estimated. Further, we can identify a direct correspondence between the modified B/C relation in
Equation [9.3] and the PW method we have used repeatedly. (The following development also
applies to AW or FW values.) Let’s neglect the initial investment in year 0 for a moment, and
concentrate on the cash flows of the project for year 1 through its expected life. For the private
sector, the PW for project cash flows is
PW of project ⫽ PW of revenue ⫺ PW of costs
Since private sector revenues are approximately the same as public sector benefits minus disbenefits (B ⫺ D), the modified B/C relation in Equation [9.3] may be written as
PW of (B ⫺ D) ⫺ PW of C
Modified B/C ⫽ ————————————
PW of initial investment
This relation can be slightly rewritten to form the profitability index (PI), which can be used to
evaluate revenue projects in the public or private sector. For years t ⫽ 1, 2, . . . , n,
PW of NCFt
PI ⴝ ———————————
PW of initial investment
[9.4]
Note that the denominator includes only first cost (initial investment) items, while the numerator
has only cash flows that result from the project for years 1 through its life. The PI measure of
worth provides a sense of getting the most for the investment dollar (euro, yen, etc.). That is, the
result is in PW units per PW of money invested at the beginning. This is a “bang for the buck”
measure. When used solely for a private sector project, the disbenefits are usually omitted,
whereas they should be estimated and included in the modified B/C version of this measure for a
public project.
The evaluation guideline for a single project using the PI is the same as for the conventional
B/C or modified B/C.
If PI ⱖ 1.0, the project is economically acceptable at the discount rate.
If PI ⬍ 1.0, the project is not economically acceptable at the discount rate.
Remember, the computations for PI and modified B/C are essentially the same, except the PI is
usually applied without disbenefits estimated. The PI has another name: the present worth index
(PWI). It is often used to rank and assist in the selection of independent projects when the capital
budget is limited. This application is discussed in Chapter 12, Section 12.5.
EXAMPLE 9.3
The Georgia Transportation Directorate is considering a public-private partnership with Young
Construction as the prime contractor using a DBOMF contract for a new 22.51-mile toll road on
the outskirts of Atlanta’s suburban area. The design includes three 4-mile-long commercial/
retail corridors on both sides of the toll road. Highway construction is expected to require
Project evaluation
238
Chapter 9
Benefit/Cost Analysis and Public Sector Economics
5 years at an average cost of $3.91 million per mile. The discount rate is 4% per year, and the
study period is 30 years. Evaluate the economics of the proposal using (a) the modified B/C
analysis from the State of Georgia perspective and (b) the profitability index from the Young
corporate viewpoint in which disbenefits are not included.
Initial investment: $88 million distributed over 5 years; $4 million now and in year 5 and
$20 million in each of years 1 through 4.
Annual M&O cost: $1 million per year, plus an additional $3 million each fifth year, including year 30.
Annual revenue/benefits: Include tolls and retail/commercial growth; start at $2 million in
year 1, increasing by a constant $0.5 million annually through year 10, and then increasing by
a constant $1 million per year through year 20 and remaining constant thereafter.
Estimable disbenefits: Include loss of business income, taxes, and property value in surrounding areas; start at $10 million in year 1, decrease by $0.5 million per year through year
21, and remain at zero thereafter.
Solution
The PW values in year 0 for all estimates must be developed initially usually by hand, calculator, or spreadsheet computations. If the 30 years of estimates are entered into a spreadsheet and
NPV functions at 4% are applied, the results in $1 million units are obtained. All values are
positive because of the sign convention for B/C and PI measures.
PW of investment ⫽ $71.89
PW of costs ⫽ $26.87
PW of benefits ⫽ $167.41
PW of disbenefits ⫽ $80.12
(a) From the public project perspective, the State will apply Equation [9.3].
167.41 ⫺ 80.12 ⫺ 26.87 ⫽ 0.84
Modified B/C ⫽ ——————————
71.89
The toll road proposal is not economically acceptable, since B/C ⬍ 1.0.
(b) From the private corporation viewpoint, Young Construction will apply Equation [9.4].
167.41 ⫺ 26.87 ⫽ 1.95
PI ⫽ ———————
71.89
The proposal is clearly justified without the disbenefits, since PI > 1.0. The private project
perspective predicts that every investment dollar will return an equivalent of $1.95 over
30 years at a 4% per year discount rate.
Comment
The obvious question that arises concerns the correct measure to use. When PI is used in the private project setting, there is no problem, since disbenefits are virtually never considered in the
economic analysis. The public project setting will commonly use some form of the B/C ratio with
the disbenefit considered. When a public-private partnership is initiated, there should be some
agreement beforehand that establishes the economic measure acceptable for analysis and decision
making throughout the project. Then the numerical dilemma presented above should not occur.
9.3 Alternative Selection Using Incremental
B/C Analysis
The technique to compare two mutually exclusive alternatives using benefit/cost analysis is virtually the same as that for incremental ROR in Chapter 8. The incremental (conventional) B/C ratio,
which is identified as ⌬B/C, is determined using PW, AW, or FW calculations. The higher-cost
alternative is justified if ⌬B/C is equal to or larger than 1.0. The selection rule is as follows:
ME alternative
selection
If ⌬B/C ⱖ 1.0, choose the higher-cost alternative, because its extra cost is economically justified.
If ⌬B/C ⬍ 1.0, choose the lower-cost alternative.
Alternative Selection Using Incremental B/C Analysis
9.3
To perform a correct incremental B/C analysis, it is required that each alternative be compared
only with another alternative for which the incremental cost is already justified. This same rule
was used for incremental ROR analysis.
There are two dimensions of an incremental B/C analysis that differ from the incremental
ROR method in Chapter 8. We already know the first, all costs have a positive sign in the B/C
ratio. The second, and significantly more important, concerns the ordering of alternatives prior to
incremental analysis.
Alternatives are ordered by increasing equivalent total costs, that is, PW or AW of all cost
estimates that will be utilized in the denominator of the B/C ratio. When not done correctly, the
incremental B/C analysis may reject a justified higher-cost alternative.
If two alternatives, A and B, have equal initial investments and lives, but B has a larger equivalent
annual cost, then B must be incrementally justified against A. (This is illustrated in Example 9.4
below.) If this convention is not correctly followed, it is possible to get a negative cost value in
the denominator, which can incorrectly make B/C ⬍ 1 and reject a higher-cost alternative that is
actually justified.
Follow these steps to correctly perform a conventional B/C ratio analysis of two alternatives.
Equivalent values can be expressed in PW, AW, or FW terms.
1. Determine the equivalent total costs for both alternatives.
2. Order the alternatives by equivalent total cost: first smaller, then larger. Calculate the incremental cost (⌬C) for the larger-cost alternative. This is the denominator in ⌬B/C.
3. Calculate the equivalent total benefits and any disbenefits estimated for both alternatives.
Calculate the incremental benefits (⌬B) for the larger-cost alternative. This is ⌬(B ⫺ D) if
disbenefits are considered.
4. Calculate the ⌬B/C ratio using Equation [9.2], (B ⫺ D)/C.
5. Use the selection guideline to select the higher-cost alternative if ⌬B/C ⱖ 1.0.
When the B/C ratio is determined for the lower-cost alternative, it is a comparison with the donothing (DN) alternative. If B/C ⬍ 1.0, then DN should be selected and compared to the second
alternative. If neither alternative has an acceptable B/C value and one of the alternatives does not
have to be selected, the DN alternative must be selected. In public sector analysis, the DN alternative is usually the current condition.
EXAMPLE 9.4
The city of Garden Ridge, Florida, has received designs for a new patient room wing to the
municipal hospital from two architectural consultants. One of the two designs must be accepted in order to announce it for construction bids. The costs and benefits are the same in most
categories, but the city financial manager decided that the estimates below should be considered to determine which design to recommend at the city council meeting next week and to
present to the citizenry in preparation for an upcoming bond referendum next month.
Construction cost, $
Building maintenance cost, $/year
Patient usage copay, $/year
Design A
Design B
10,000,000
35,000
450,000
15,000,000
55,000
200,000
The patient usage copay is an estimate of the amount paid by patients over the insurance coverage
generally allowed for a hospital room. The discount rate is 5%, and the life of the building is
estimated at 30 years.
(a) Use incremental B/C analysis to select design A or B.
(b) Once the two designs were publicized, the privately owned hospital in the directly adjacent
city of Forest Glen lodged a complaint that design A will reduce its own municipal hospital’s income by an estimated $500,000 per year because some of the day-surgery features
of design A duplicate its services. Subsequently, the Garden Ridge merchants’ association
argued that design B could reduce its annual revenue by an estimated $400,000, because it
239
240
Benefit/Cost Analysis and Public Sector Economics
Chapter 9
will eliminate an entire parking lot used by their patrons for short-term parking. The city
financial manager stated that these concerns would be entered into the evaluation as disbenefits of the respective designs. Redo the B/C analysis to determine if the economic decision is still the same as when disbenefits were not considered.
Solution
(a) Since most of the cash flows are already annualized, the incremental B/C ratio will use
AW values. No disbenefit estimates are considered. Follow the steps of the procedure
above:
1. The AW of costs is the sum of construction and maintenance costs.
AWA ⫽ 10,000,000(A/P,5%,30) ⫹ 35,000 ⫽ $685,500
AWB ⫽ 15,000,000(A/P,5%,30) ⫹ 55,000 ⫽ $1,030,750
2. Design B has the larger AW of costs, so it is the alternative to be incrementally justified. The incremental cost is
⌬C ⫽ AWB ⫺ AWA ⫽ $345,250 per year
3. The AW of benefits is derived from the patient usage copays, since these are consequences to the public. The benefits for the ⌬B/C analysis are not the estimates themselves, but the difference if design B is selected. The lower usage copay is a positive
benefit for design B.
⌬B ⫽ usageA ⫺ usageB ⫽ $450,000 ⫺ $200,000 ⫽ $250,000 per year
4. The incremental B/C ratio is calculated by Equation [9.2].
$250,000
⌬B兾C ⫽ ———— ⫽ 0.72
$345,250
5. The B/C ratio is less than 1.0, indicating that the extra costs associated with design B
are not justified. Therefore, design A is selected for the construction bid.
(b) The revenue loss estimates are considered disbenefits. Since the disbenefits of design B are
$100,000 less than those of A, this positive difference is added to the $250,000 benefits of
B to give it a total benefit of $350,000. Now
$350,000
⌬B兾C ⫽ ———— ⫽ 1.01
$345,250
Design B is slightly favored. In this case the inclusion of disbenefits has reversed the previous economic decision. This has probably made the situation more difficult politically.
New disbenefits will surely be claimed in the near future by other special-interest groups.
Like other methods, incremental B/C analysis requires equal-service comparison of alternatives. Usually, the expected useful life of a public project is long (25 or 30 or more years), so
alternatives generally have equal lives. However, when alternatives do have unequal lives, the
use of PW or AW to determine the equivalent costs and benefits requires that the LCM of lives be
used to calculate ⌬B/C. As with ROR analysis of two alternatives, this is an excellent opportunity
to use the AW equivalency of estimated (not incremental) costs and benefits, if the implied assumption that the project could be repeated is reasonable. Therefore, use AW-based analysis of
actual costs and benefits for B/C ratios when different-life alternatives are compared.
EXAMPLE 9.5
Water Treatment Facility #3 Case
PE
As our case unfolds, the consultant, Joel Whiterson, has pieced together some of the B/C
analysis estimates for the 84-inch Jolleyville transmission main study completed last year. The
two options for constructing this main were open trench (OT) for the entire 6.8-mile distance
or a combination of trenching and bore tunneling (TT) for a shorter route of 6.3 miles. One of
the two options had to be selected to transport approximately 300 million gallons per day (gpd)
of treated water from the new WTF3 to an existing aboveground reservoir.
Alternative Selection Using Incremental B/C Analysis
9.3
The general manager of Allen Water Utilities has stated publicly several times that the
trench-tunnel combination option was selected over the open-trench alternative based on analysis of both quantitative and nonquantitative data. He stated the equivalent annual costs in an
internal e-mail some months ago, based on the expected construction periods of 24 and
36 months, respectively, as equivalent to
AWOT ⫽ $1.20 million per year
AWTT ⫽ $2.37 million per year
This analysis indicated that the open-trench option was economically better, at that time. The
planning horizon for the transmission mains is 50 years; this is a reasonable study period, Joel
concluded. Use the estimates below that Joel has unearthed to perform a correct incremental
B/C analysis and comment on the results. The interest (discount) rate is 3% per year, compounded annually, and 1 mile is 5280 feet.
Open trench (OT)
Trench-tunnel (TT)
6.8
700
24
250,000
6.3
Trench for 2.0 miles: 700
Tunnel for 4.3 miles: 2100
36
175,000
150,000
140,000
20,000
20,000
60,000
5,000
Distance, miles
First cost, $ per foot
Time to complete, months
Construction support costs, $ per month
Ancillary expenses, $ per month:
Environmental
Safety
Community interface
Solution
One of the alternatives must be selected, and the construction lives are unequal. Since it is not
reasonable to assume that this construction project will be repeated many cycles in the future,
it is incorrect to conduct an AW analysis over the respective completion periods of 24 and
36 months, or the LCM of these time periods. However, the study period of 50 years is a reasonable evaluation time frame, since the mains are considered permanent installations. We can
assume that the construction first costs are a present worth value in year 0, but the equivalent
PW and 50-year AW of other monthly costs must be determined.
PWOT ⫽ PW of construction ⫹ PW of construction support costs
⫽ 700(6.8)(5280) ⫹ 250,000(12)(P兾A,3%,2)
⫽ $30,873,300
AWOT ⫽ 30,873,300(A兾P,3%,50)
⫽ $1.20 million per year
PWTT ⫽ [700(2.0) ⫹ 2100(4.3)](5280) ⫹ 175,000(12)(P兾A,3%,3)
⫽ $61,010,460
AWTT ⫽ 61,010,460(A兾P,3%,50)
⫽ $2.37 million per year
The trench-tunnel (TT) alternative has a larger equivalent cost; it must be justified against the
OT alternative. The incremental cost is
⌬C ⫽ AWTT ⫺ AWOT ⫽ 2.37 − 1.20 ⫽ $1.17 million per year
The difference between ancillary expenses defines the incremental benefit for TT.
PWOT-anc ⫽ 310,000(12)(P兾A,3%,2)
⫽ $7,118,220
AWOT-anc ⫽ 7,118,220(A兾P,3%,50)
⫽ $276,685 per year
241
242
Benefit/Cost Analysis and Public Sector Economics
Chapter 9
PWTT-anc ⫽ 85,000(12)(P兾A,3%,3)
⫽ $2,885,172
AWTT-anc ⫽ 2,885,172(A兾P,3%,50)
⫽ $112,147 per year
⌬B ⫽ AWOT-anc ⫺ AWTT-anc ⫽ 276,685 ⫺ 112,147 ⫽ $164,538 per year
($0.16 million)
Calculate the incremental B/C ratio.
⌬B/C ⫽ 0.16/1.17 ⫽ 0.14
Since ⌬B/C ⬍⬍ 1.0, the trench-tunnel option is not economically justified. Joel can now conclude that the general manager’s earlier comment that the TT option was selected based on
quantitative and nonquantitative data must have had heavy dependence on nonquantitative
information not yet discovered.
9.4 Incremental B/C Analysis of Multiple, Mutually
Exclusive Alternatives
The procedure necessary to select one from three or more mutually exclusive alternatives using
incremental B/C analysis is essentially the same as that of Section 9.3. The procedure also parallels that for incremental ROR analysis in Section 8.6. The selection guideline is as follows:
Choose the largest-cost alternative that is justified with an incremental B/C ⱖ 1.0 when this selected alternative has been compared with another justified alternative.
There are two types of benefit estimates—estimation of direct benefits, and implied benefits
based on usage cost estimates. The previous two examples (9.4 and 9.5) are good illustrations of
the second type of implied benefit estimation. When direct benefits are estimated, the B/C ratio
for each alternative may be calculated first as an initial screening mechanism to eliminate unacceptable alternatives. At least one alternative must have B/C ⱖ 1.0 to perform the incremental
B/C analysis. If all alternatives are unacceptable, the DN alternative is the choice. (This is the
same approach as that of step 2 for “revenue alternatives only” in the ROR procedure of Section
8.6. However, the term revenue alternative is not applicable to public sector projects.)
As in the previous section when comparing two alternatives, selection from multiple alternatives by incremental B/C ratio utilizes equivalent total costs to initially order alternatives from
smallest to largest. Pairwise comparison is then undertaken. Also, remember that all costs are
considered positive in B/C calculations. The terms defender and challenger alternative are used
in this procedure, as in a ROR-based analysis. The procedure for incremental B/C analysis of
multiple alternatives is as follows:
Determine the equivalent total cost for all alternatives. Use AW, PW, or FW equivalencies.
Order the alternatives by equivalent total cost, smallest first.
Determine the equivalent total benefits (and any disbenefits estimated) for each alternative.
Direct benefits estimation only: Calculate the B/C for the first ordered alternative. If B/C
⬍ 1.0, eliminate it. By comparing each alternative to DN in order, we eliminate all that have
B/C ⬍ 1.0. The lowest-cost alternative with B/C ⱖ 1.0 becomes the defender and the next
higher-cost alternative is the challenger in the next step. (For analysis by spreadsheet, determine the B/C for all alternatives initially and retain only acceptable ones.)
5. Calculate incremental costs (⌬C) and benefits (⌬B) using the relations
1.
2.
3.
4.
⌬C ⫽ challenger cost ⫺ defender cost
⌬B ⫽ challenger benefits ⫺ defender benefits
If relative usage costs are estimated for each alternative, rather than direct benefits, ⌬B may
be found using the relation
⌬B ⴝ defender usage costs ⴚ challenger usage costs
[9.5]
Incremental B/C Analysis of Multiple, Mutually Exclusive Alternatives
9.4
6. Calculate the ⌬B/C for the first challenger compared to the defender.
⌬B/C ⫽ ⌬B兾⌬C
[9.6]
If ⌬B/C ⱖ 1.0 in Equation [9.6], the challenger becomes the defender and the previous defender is eliminated. Conversely, if ⌬B/C ⬍ 1.0, remove the challenger and the defender
remains against the next challenger.
7. Repeat steps 5 and 6 until only one alternative remains. It is the selected one.
In all the steps above, incremental disbenefits may be considered by replacing ⌬B with ⌬(B ⫺ D).
EXAMPLE 9.6
Schlitterbahn Waterparks of Texas, a very popular water and entertainment park headquartered in New Braunfels, has been asked by four different cities outside of Texas to consider building a park in their area. All the offers include some version of the following
incentives:
•
•
•
•
Immediate cash incentive (year 0)
A 10% of first-year incentive as a direct property tax reduction for 8 years
Sales tax rebate sharing plan for 8 years
Reduced entrance (usage) fees for area residents for 8 years
Table 9–1 (top section) summarizes the estimates for each proposal, including the present
worth of the initial construction cost and anticipated annual revenue. The annual M&O costs
are expected to be the same for all locations. Use incremental B/C analysis at 7% per year and
an 8-year study period to advise the board of directors if they should consider any of the offers
to be economically attractive.
Solution
The viewpoint is that of Schlitterbahn, and the benefits are direct estimates. Develop the AW
equivalents over 8 years, and use the procedure detailed above. The results are presented in
Table 9–1.
1. AW of total costs and an example for city 1 are determined in $1 million units.
AW of costs ⫽ first cost(A兾P,7%,8) ⫹ entrance fee reduction to residents
⫽ 38.5(0.16747) ⫹ 0.5
⫽ $6.948 ($6,948,000 per year)
2. The four alternatives are correctly ordered by increasing equivalent total cost in
Table 9–1.
TABLE 9–1
Incremental B/C Analysis of Water Park Proposals, Example 9.6
City 1
City 2
City 3
City 4
First cost, $ million
Entrance fee costs, $/year
Annual revenue, $ million/year
Initial cash incentive, $
Property tax reduction, $/year
Sales tax sharing, $/year
38.5
500,000
7.0
250,000
25,000
310,000
40.1
450,000
6.2
350,000
35,000
320,000
45.9
425,000
10.0
500,000
50,000
320,000
60.3
250,000
10.4
800,000
80,000
340,000
AW of total costs, $ million/year
AW of total benefits, $ million/year
Overall B/C
Alternatives compared
Incremental costs ⌬C, $/year
Incremental benefits ⌬B, $/year
⌬B/C
Increment justified?
City selected
6.948
7.377
1.06
1-to-DN
6.948
7.377
1.06
Yes
1
7.166
6.614
0.92
B/C ⬍ 1.0
8.112
10.454
1.29
3-to-1
1.164
3.077
2.64
Yes
3
10.348
10.954
1.06
4-to-3
2.236
0.50
0.22
No
3
Eliminated
243
244
Chapter 9
Benefit/Cost Analysis and Public Sector Economics
3. AW of total benefits and an example for city 1 are also determined in $1 million units.
AW of benefits ⫽ revenue ⫹ initial incentive(A兾P,7%,8)
⫹ property tax reduction ⫹ sales tax sharing
⫽ 7.0 ⫹ 0.25(0.16747) ⫹ 0.025 ⫹ 0.31
⫽ $7.377 ($7,377,000 per year)
4. Since benefits are directly estimated (and no disbenefits are included), determine the overall B/C for each alternative using Equation [9.1]. In the case of city 1,
B兾C1 ⫽ 7.377兾6.948 ⫽ 1.06
City 2 is eliminated with B/C2 ⫽ 0.92; the rest are initially acceptable.
5. The ⌬C and ⌬B values are the actual estimates for the 1-to-DN comparison.
6. The overall B/C is the same as ⌬B/C ⫽ 1.06, using Equation [9.6]. City 1 is economically
justified and becomes the defender.
7. Repeat steps 5 and 6. Since city 2 is eliminated, the 3-to-1 comparison results in
⌬C ⫽ 8.112 ⫺ 6.948 ⫽ 1.164
⌬B ⫽ 10.454 ⫺ 7.377 ⫽ 3.077
⌬B兾C ⫽ 3.077兾1.164 ⫽ 2.64
City 3 is well justified and becomes the defender against city 4. From Table 9–1, ⌬B/C ⫽ 0.22
for the 4-to-3 comparison. City 4 falls out easily, and city 3 is the one to recommend to the
board. Note that the DN alternative could have been selected had no proposal met the B/C or
⌬B/C requirements.
Independent project
selection
When two or more independent projects are evaluated using B/C analysis and there is no
budget limitation, no incremental comparison is necessary. The only comparison is between
each project separately with the do-nothing alternative. The project B/C values are calculated,
and those with B/C ⱖ 1.0 are accepted.
This is the same procedure as that used to select from independent projects using the ROR method
(Chapter 8). When a budget limitation is imposed, the capital budgeting procedure discussed in
Chapter 12 must be applied.
When the lives of mutually exclusive alternatives are so long that they can be considered
infinite, the capitalized cost is used to calculate the equivalent PW or AW values for costs
and benefits. Equation [5.3], A ⫽ P(i), is used to determine the equivalent AW values in the
incremental B/C analysis. Example 9.7 illustrates this using the progressive example and a
spreadsheet.
EXAMPLE 9.7 Water Treatment Facility #3 Case
PE
Land for Water Treatment Facility #3 was initially purchased in the year 2010 for $19.3 million; however, when it was publicized, influential people around Allen spoke strongly against
the location. We will call this location 1. Some of the plant design had already been completed
when the general manager announced that this site was not the best choice anyway, and that it
would be sold and a different, better site (location 2) would be purchased for $28.5 million.
This was well over the budget amount of $22.0 million previously set for land acquisition. As
it turns out, there was a third site (location 3) available for $35.0 million that was never seriously considered.
In his review and after much resistance from Allen Water Utilities staff, the consultant, Joel,
received a copy of the estimated costs and benefits for the three plant location options. The
revenues, savings, and sale of bulk water rights to other communities are estimated as increments from a base amount for all three locations. Using the assumption of a very long life for
the WTF3 facility and the established discount rate of 3% per year, determine what Joel discovered when he did the B/C analysis. Was the general manager correct in concluding that
location 2 was the best, all said and done?
Incremental B/C Analysis of Multiple, Mutually Exclusive Alternatives
9.4
Land cost, $ million
Facility first cost, $ million
Benefits, $ per year:
Pumping cost savings
Sales to area communities
Added revenue from Allen
Total benefits, $ per year
Location 1
Location 2
Location 3
19.3
460.0
28.5
446.0
35.0
446.0
5
12
6
3
10
6
0
8
6
23
19
14
Solution
A spreadsheet can be very useful when performing an incremental B/C analysis of three or
more alternatives. Figure 9–1a presents the analysis with the preliminary input of AW values
for costs using the relation A ⫽ P(i) and annual benefits. Figure 9–1b details all the functions
used in the analysis. Logical IF statements indicate alternative elimination and selection decisions. In $1 million units,
AW of costs ⫽ A of land cost ⫹ A of facility first cost
⫽ (19.3 ⫹ 460.0)(0.03)
⫽ $14.379 per year
AW of benefits ⫽ $23
Location 2:
AW of costs ⫽ $14.235
AW of benefits ⫽ $19
Location 3:
AW of costs ⫽ $14.430
AW of benefits ⫽ $14
Location 1:
Though the AW of cost values are close to one another, the increasing order is locations 2, 1,
and 3 to determine ⌬B/C values. The benefits are direct estimates; therefore, the overall B/C
ratios indicate that location 3 (row 5; B/C3 ⫽ 0.97) is not economically justified at the outset.
It is eliminated, and one of the remaining locations must be selected. Location 2 is justified
against the DN alternative (B/C2 ⫽ 1.33); the only remaining comparison is 1-to-2 as detailed
in column C of Figure 9–1. Location 1 is a clear winner with ⌬B/C ⫽ 27.78.
In conclusion, Joel has learned that location 1 is indeed the best and that, from the economic
perspective, the general manager was incorrect in stating that location 2 was better. However,
given the original evaluation criteria listed in the introduction—economics, environment, community impact, and constructability—location 2 is likely a good compromise selection.
(a)
(b)
Figure 9–1
Incremental B/C analysis for WTF3 case: (a) numerical results and (b) functions developed for the analysis.
245
246
Chapter 9
Benefit/Cost Analysis and Public Sector Economics
Comment
This is an actual situation with changed names and values. Location 1 was initially purchased
and planned for WTF3. However, the presence of political, community, and environmental
stress factors changed the decision to location 2, when all was said and done.
9.5 Service Sector Projects and Cost-Effectiveness
Analysis
Much of the GDP of the United States and some countries in Europe and Asia is generated by
what has become known as the service sector of the economy. A large percentage of service sector projects are generated by and dependent upon the private sector (corporations, businesses,
and other for-profit institutions). However, many projects in the public sector are also service
sector projects.
A service sector project is a process or system that provides services to individuals, businesses,
or government units. The economic value is developed primarily by the intangibles of the process or system, not the physical entities, such as buildings, machines, and equipment. Manufacturing and construction activities are commonly not considered a service sector project, though
they may support the theme of the service provided.
Service projects have a tremendous range of variety and purpose; to name a few: health care
systems, health and life insurance, airline reservation systems, credit card services, police and
court systems, security programs, safety training programs, and all types of consulting projects.
The intangible and intellectual work done by engineers and other professionals is often a part of
a service sector project.
The economic evaluation of a service project is difficult to a great degree because the cost and
benefit estimates are not accurate and often not within an acceptable degree of error. In other
words, undue risk may be introduced into the decision because of poor monetary estimates. For
example, consider the decision to place red-light cameras at stop lights to ticket drivers who run
the red light. This is a public and a service project, but its (economic) benefits are quite difficult
to estimate. Depending upon the viewpoint, benefits could be in terms of accidents averted,
deaths prevented, police personnel released from patrolling the intersection, or, from a more
mercenary viewpoint, amount of fines collected. In all but the last case, benefits in monetary
terms will be poor estimates. These are examples where B/C analysis does not work well and a
different form of analysis is needed.
In service and public sector projects, as expected, it is the benefits that are the more difficult
to estimate. An evaluation method that combines monetary cost estimates with nonmonetary
benefit estimates is cost-effectiveness analysis (CEA). The CEA approach utilizes a costeffectiveness measure or the cost-effectiveness ratio (CER) as a basis of ranking projects and
selecting the best of independent projects or mutually exclusive alternatives. The CER ratio is
defined as
equivalent total costs
C
CER ⴝ ————————————— ⴝ —
total of effectiveness measure E
[9.7]
In the red-light camera example, the effectiveness measure (the benefit) may be one of the samples mentioned earlier, accidents averted or deaths prevented. Different from the B/C ratio of
costs to benefits, CER places the PW or AW of total costs in the numerator and the effectiveness
measure in the denominator. (The reciprocal of Equation [9.7] can also be used as the measure of
worth, but we will use CER as defined above.) With costs in the numerator, smaller ratio values
are more desirable for the same value of the denominator, since smaller ratio values indicate a
lower cost for the same level of effectiveness.
Like ROR and B/C analysis, cost-effectiveness analysis requires the ordering (ranking) of
alternatives prior to selection and the use of incremental analysis for mutually exclusive
Service Sector Projects and Cost-Effectiveness Analysis
9.5
247
alternative selection. Cost-effectiveness analysis utilizes a different ranking criterion than ROR or
BC analysis. The ordering criteria are as follows:
Independent projects: Initially rank projects by CER value.
Mutually exclusive alternatives: Initially rank alternatives by effectiveness measure, then
perform an incremental CER analysis.
Return again to the public/service project of red-light cameras. If the CER is defined as “cost per
total accidents averted ” and the projects are independent, increasing CER value is the ranking
basis. If the projects are mutually exclusive, “total accidents averted ” is the correct ranking basis
and an incremental analysis is necessary.
There are significantly different analysis procedures for independent and mutually exclusive
proposals. To select some from several (independent) projects, a budget limit, termed b, is inherently necessary once ordering is complete. However, for selecting one from several (mutually
exclusive) alternatives, a pairwise incremental analysis is necessary and selection is made on the
basis of ⌬C/E ratios. The procedures and examples follow.
For independent projects, the procedure is as follows:
1. Determine the equivalent total cost C and effectiveness measure E, and calculate the CER
measure for each project.
2. Order projects from the smallest to the largest CER value.
3. Determine cumulative cost for each project and compare with the budget limit b.
4. The selection criterion is to fund all projects such that b is not exceeded.
EXAMPLE 9.8
Recent research indicates that corporations throughout the world need employees who demonstrate creativity and innovation for new processes and products. One measure of these talents
is the number of patents approved each year through the R&D efforts of a company. Rollings
Foundation for Innovative Thinking has allocated $1 million in grant funds to award to corporations that enroll their top R&D personnel in a 1- to 2-month professional training program in
their home state that has a historically proven track record over the last 5 years in helping
individuals earn patents.
Table 9–2 summarizes data for six corporations that submitted proposals. Columns 2 and 3
give the proposed number of attendees and cost per person, respectively, and column 4 provides the historical track record of program graduates in patents per year. Use cost-effectiveness
analysis to select the corporations and programs to fund.
Solution
We assume that across all programs and all patent awards there is equal quality. Use the procedure for independent projects and b ⫽ $1 million to select from the proposals.
TABLE 9–2
Data for Programs to Increase Patents Used for CEA
Program
(1)
Total Personnel
(2)
Cost/Person, $
(3)
5-Year History,
Patents/Graduate/Year
(4)
1
2
3
4
5
6
50
35
57
24
12
87
5000
4500
8000
2500
5500
3800
0.5
3.1
1.9
2.1
2.9
0.6
Independent project
selection
248
Benefit/Cost Analysis and Public Sector Economics
Chapter 9
1. Using Equation [9.7], the effectiveness measure E is patents per year, and the CER is
program cost per person C
CER ⫽ —————————— ⫽ —
E
patents per graduate
The program cost C is a PW value, and the E values are obtained from the proposals.
2. The CER values are shown in Table 9–3 in increasing order, column 5.
3. Cost per course, column 6, and cumulative costs, column 7, are determined.
4. Programs 4, 2, 5, 3, and 6 (68 of the 87 people) are selected to not exceed $1 million.
TABLE 9–3 Programs Ordered by CER Value, Example 9.8
Program
(1)
4
2
5
3
6
1
Total
Cost/Person
Personnel
C, $
(2)
(3)
24
35
12
57
87
50
2,500
4,500
5,500
8,000
3,800
5,000
Patents
per Year E
(4)
CER, $ per
Patent
(5) ⴝ (3)/(4)
Program
Cost, $
(6) ⴝ (2)(3)
Cumulative
Cost, $
(7) ⴝ ⌺(6)
2.1
3.1
2.9
1.9
0.6
0.5
1,190
1,452
1,897
4,211
6,333
10,000
60,000
157,500
66,000
456,000
330,600
250,000
60,000
217,500
283,500
739,500
1,070,100
1,320,100
Comment
This is the first time that a budget limit has been imposed for the selection among independent
projects. This is often referred to as capital budgeting, which is discussed further in Chapter 12.
For mutually exclusive alternatives and no budget limit, the alternative with the highest effectiveness measure E is selected without further analysis. Otherwise, an incremental CER analysis is necessary and the budget limit is applied to the selected alternative(s). The analysis is
based on the incremental ratio ⌬C/E, and the procedure is similar to that we have applied for
incremental ROR and B/C, except now the concept of dominance is utilized.
Dominance occurs when the incremental analysis indicates that the challenger alternative offers an improved incremental CER measure compared to the defender’s CER, that is,
ME alternative
selection
(⌬C/E)challenger ⬍ (C/E)defender
Otherwise, no dominance is present, and both alternatives remain in the analysis.
For mutually exclusive alternatives, the selection procedure is as follows:
1. Order the alternatives from smallest to largest effectiveness measure E. Record the cost for
each alternative.
2. Calculate the CER measure for the first alternative. This, in effect, makes DN the defender
and the first alternative the challenger. This CER is a baseline for the next incremental comparison, and the first alternative becomes the new defender.
3. Calculate incremental costs (⌬C) and effectiveness (⌬E) and the incremental measure ⌬C/E
for the new challenger using the relation
cost of challenger ⫺ cost of defender
⌬C
⌬C/E ⫽ ——————————————————————— ⫽ ——
effectiveness of challenger ⫺ effectiveness of defender ⌬E
4. If ⌬C/E ⬍ C/Edefender, the challenger dominates the defender and it becomes the new defender; the previous defender is eliminated. Otherwise, no dominance is present and both
alternatives are retained for the next incremental evaluation.
5. Dominance present: Repeat steps 3 and 4 to compare the next ordered alternative (challenger) and new defender. Determine if dominance is present.
Service Sector Projects and Cost-Effectiveness Analysis
9.5
Dominance not present: The current challenger becomes the new defender, and the next alternative is the new challenger. Repeat steps 3 and 4 to compare the new challenger and new
defender. Determine if dominance is present.
6. Continue steps 3 through 5 until only one alternative or only nondominated alternatives remain.
7. Apply the budget limit (or other criteria) to determine which of the remaining alternative(s)
can be funded.
EXAMPLE 9.9
One of the corporations not selected for funding in Example 9.8 decided to fund its 50 R&D
personnel to attend one of the innovation and creativity programs at its own expense. One
criterion is that the program must have a historical average for a graduate of at least 2.0 patents
per year. Use the data in Table 9–2 to select the best program.
TABLE 9–4
Mutually Exclusive Alternatives Evaluated by Cost-Effectiveness
Analysis, Example 9.9
Total
Program Personnel
(2)
(1)
4
5
2
50
50
50
Cost/Person
C, $
(3)
Patents per
Year E
(4)
CER, $ per
Patent
(5) ⴝ (3)/(4)
Program Cost,
$
(6) ⴝ (2)(3)
2,500
5,500
4,500
2.1
2.9
3.1
1190
1897
1452
125,000
275,000
225,000
Solution
From Table 9–2, three programs—2, 4, and 5—have a historical record of at least two patents
per graduate per year. Since only one program will be selected, these are now mutually exclusive alternatives. Use the procedure to perform the incremental analysis.
1. The alternatives are ranked by increasing patents per year in Table 9–4, column 4.
2. The CER measure for program 4 is compared to the DN alternative.
program cost per person 2500
⫽ 1190
C/E4 ⫽ —————————— ⫽ ———
2.1
patents per graduate
3. Program 5 is now the challenger.
⌬C ⫽ ——————
5500 ⫺ 2500 ⫽ 3750
5-to-4 comparison: ⌬C/E ⫽ ——
2.9 ⫺ 2.1
⌬E
4. In comparison to C/E4 ⫽ 1190, it costs $3750 per additional patent if program 5 is chosen
over 4. Program 5 is more expensive for more patents; however, clear dominance is not
present; both programs are retained for further evaluation.
5. Dominance not present: Program 5 becomes the new defender, and program 2 is the new
challenger. Perform the 2-to-5 comparison.
⌬C ⫽ ——————
4500 ⫺ 5500 ⫽ ⫺5000
2-to-5 comparison: ⌬C/E ⫽ ——
3.1 ⫺ 2.9
⌬E
Compared to C/E5 ⫽ 1897, this increment is much cheaper—more patents for less money
per person. Dominance is present; eliminate program 5 and compare 2 to 4.
6. Repeat steps 3 through 5 and compare ⌬C/E to C/E4 ⫽ 1190.
⌬C ⫽ ——————
4500 ⫺ 2500 ⫽ 2000
2-to-4 comparison: ⌬C/E ⫽ ——
3.1 ⫺ 2.1
⌬E
This does not represent dominance of program 2 over 4. The conclusion is that both programs are eligible for funding, that is, CEA in this case does not indicate only one program. This occurs when there is not lower cost and higher effectiveness of one alternative
over another; that is, one alternative does not dominate all the others.
7. Now the budget and other considerations (probably noneconomic) are brought to bear to
make the final decision. The fact that program 4 costs $125,000—significantly less than
program 2 at $225,000—will likely enter into the decision.
249
250
Chapter 9
Benefit/Cost Analysis and Public Sector Economics
Cost-effectiveness analysis is a form of multiattribute decision-making in which economic
and noneconomic dimensions are integrated to evaluate alternatives from several perspectives by
different decision makers. See Chapter 10 for further discussion of alternative analysis using
multiple attributes.
9.6 Ethical Considerations in the Public Sector
Usual expectations by citizens of their elected officials—locally, nationally, and internationally—
are that they make decisions for the good of the public, ensuring safety and minimizing risk and cost
to the public. Above these is the long-standing expectation that public servants have integrity.
Similarly, the expectations of engineers employed by government departments, and those
serving as consultants to government agencies, are held to high standards. Impartiality, consideration of a wide range of circumstances, and the use of realistic assumptions are but three of
the foundation elements upon which engineers should base their recommendations to decision
makers. This implies that engineers in public service avoid
• Self-serving, often greedy individuals and clients with goals of excessive profits and future
contract awards
• Using a politically favorable perspective that compromises the results of a study
• Narrowly-defined assumptions that serve special interest groups and subcommunities potentially affected by the findings
Many people are disappointed and discouraged with government when elected officials and
public employees (engineers and others, alike) do not have real commitment to integrity and
unbiasedness in their work.
Engineers are routinely involved in two of the major aspects of public sector activities:
Public policy making—the development of strategy for public service, behavior, fairness,
and justice. This may involve literature study, background discovery, data collection, opinion
giving, and hypothesis testing. An example is transportation management. Engineers make
virtually all the recommendations based on data and long-standing decision algorithms for
policy items such as capacity of roads, expansion of highways, planning and zoning rules,
traffic signal usage, speed limit corridors, and many related topics in transportation policy.
Public officials use these findings to establish public transportation policy.
Public planning—the development of projects that implement strategy and affect people,
the environment, and financial resources in a variety of ways. Consider traffic control,
where the use and placement of traffic control signs, signals, speed limits, parking restrictions, etc. are detailed based upon established policy and current data. (In effect, this is
systems engineering, that is, an application of the life-cycle phases and stages explained in
Section 6.5.)
Whether in the arena of policy making or public planning, engineers can find ethical compromise
a possibility when working with the public sector. A few circumstances are summarized here.
• Use of technology Many public projects involve the use of new technology. The public risks
and safety factors are not always known for these new advances. It is common and expected
that engineers make every attempt to apply the latest technology while ensuring that the public
is not exposed to undue risk.
• Scope of study A client may pressure the engineers to limit the range of options, the assumption base, or the breadth of alternative solutions. These restrictions may be based on financial
reasons, politically-charged topics, client-favored options, or a wide variety of other reasons.
To remain impartial, it is the responsibility of the engineer to submit a fully unbiased analysis,
report, and recommendation, even though it may jeopardize future contract possibilities, promote public disfavor, or generate other negative consequences.
• Negative community impact It is inevitable that public projects will adversely affect some
groups of people, or the environment, or businesses. The intentional silencing of these projected effects is often the cause of strong public outcries against what may be a project that is
in the best interest of the community at large. Engineers who find (stumble onto) such negative impacts may be pressured by clients, managers, or public figures to overlook them, though
Chapter Summary
TABLE 9–5 Some Ethical Considerations When Performing B/C and CEA Analysis
What the Study Includes Ethical Dimension
Example
Audience for study
Is it ethical to select a specific
group of people affected by the
project and neglect possible
effects on other groups?
Construct children’s health care clinics
for city dwellers, but neglect rural families with poor transportation means.
Impact time of decision
Is it ethical to decide now for future generations who may be adversely and economically affected
by the current project decision?
Accomplish financial bailouts of corporations when future generations’ taxes
will be significantly higher to recover the
costs, plus interest and inflation effects.
Greater good for community as a whole
Vulnerable minority groups, especially economically deprived
ones, may be disproportionally
affected. Is this ethical if the impact is predictable?
Allow a chemical plant that is vital to
the community’s employment to pollute
a waterway when a minority group is
known to eat fish from the water that is
predictably contaminated.
Reliance on economic
measures only
Is it acceptable to reduce all
costs and benefits to monetary
estimates for a decision, then
subjectively impute nonquantified factors in the final decision?
Softening of building codes can improve the financial outlook for home
builders; however, increased risks of
fire loss, storm and water damage to
structures, and reduced future resale
values are considered only in passing as
a new subdivision is approved by the
planning and zoning committee.
Scope of disbenefits esti- Is it ethical to disregard any dismated and evaluated
benefits in the B/C study or use
indirect effectiveness measures
in a CEA study based on the difficulty to estimate some of them?
Noise and air pollution caused by a
planned open-pit quarry will have a negative effect on area ranchers, residents,
wildlife, and plant life; but the effectiveness measure considers only suburban
residents due to estimation difficulty of
effects on other constituencies.
the Code of Ethics for Engineers dictates a full and fair analysis and report. For example, a
planned rerouting of a city street may effectively cut off a section of international citizens’
businesses, thus resulting in a clearly predictable economic downturn. Considering this outcome in the recommendation to the transportation department should be a goal of the analyzing engineers, yet pressure to bias the results may be quite high.
The results of a B/C or CEA analysis are routinely depended upon by public officials and
government staff members to assist in making public planning decisions. As discussed earlier,
estimations for benefits, disbenefits, effectiveness measures, and costs can be difficult and inaccurate, but these analysis tools are often the best available to structure a study. Some example
ethically-oriented challenges that may be confronted during B/C and CEA analyses are summarized in Table 9–5.
CHAPTER SUMMARY
The benefit/cost method is used primarily to evaluate alternatives in the public sector. When one
is comparing mutually exclusive alternatives, the incremental B/C ratio must be greater than or
equal to 1.0 for the incremental equivalent total cost to be economically justified. The PW, AW,
or FW of the initial costs and estimated benefits can be used to perform an incremental B/C
analysis. For independent projects, no incremental B/C analysis is necessary. All projects with
B/C ⱖ 1.0 are selected provided there is no budget limitation. It is usually quite difficult to make
accurate estimates of benefits for public sector projects. The characteristics of public sector projects are substantially different from those of the private sector: initial costs are larger; expected
life is longer; additional sources of capital funds include taxation, user fees, and government
grants; and interest (discount) rates are lower.
251
252
Chapter 9
Benefit/Cost Analysis and Public Sector Economics
Service projects develop economic value largely based on the intangibles of the services provided to users, not the physical items associated with the process or system. Evaluation by B/C
analysis can be difficult with no good way to make monetary estimates of benefits. Cost-effectiveness analysis (CEA) combines cost estimates and a nonmonetary effectiveness measure (the
benefit) to evaluate independent or mutually exclusive projects using procedures that are similar
to incremental ROR and B/C analysis. The concept of dominance is incorporated into the procedure for comparing mutually exclusive alternatives.
As a complement to the discussion on professional ethics in Chapter 1, some potential ethical
challenges in the public sector for engineers, elected officials, and government consultants are
discussed here. Examples are included.
PROBLEMS
Public Projects and B/C Concepts
9.1 What is the difference between disbenefits and
costs?
9.2 Identify the following as primarily public or private sector undertakings: eBay, farmer’s market,
state police department, car racing facility, social
security, EMS, ATM, travel agency, amusement
park, gambling casino, swap meet, football
stadium.
9.3 State whether the following characteristics are primarily associated with public or private sector
projects: large initial investment, park user fees,
short-life projects, profit, disbenefits, tax-free
bonds, subsidized loans, low interest rate, income
tax, water quality regulations.
9.4 Identify the following cash flows as a benefit, disbenefit, or cost.
(a) Loss of income to local businesses because
of a new freeway
(b) Less travel time because of a loop bypass
(c) $400,000 annual income to local businesses
because of tourism created by a national park
(d) Cost of fish from a hatchery to stock a lake at
the state park
(e) Less tire wear because of smoother road
surfaces
(f) Decrease in property values due to the closure of a government research lab
(g) School overcrowding because of a military
base expansion
(h) Revenue to local motels because of an extended weekend holiday
public-private partnership between the sheriff’s
office and a private security company. In preparing
for a B/C analysis of the proposal, estimates must
be categorized as a benefit, cost, or disbenefit. The
categories chosen will vary depending upon a person’s viewpoint. Assume that possible viewpoints
include sales revenues, politics, service to the
public, the environment, contract obligations, future generations of residents, legal matters, revenue and budget (in general), and customers.
(a) Select the top two viewpoints (in your opinion) for each of the following individuals as
they would categorize estimates as a cost,
benefit, or disbenefit.
1. An industrial plant manager in the county
2. County sheriff’s deputy (appointed office)
3. County commissioner (elected office)
4. Security company president
(b) Explain your answers by writing a short description of why you selected as you did.
Project B/C Value
9.7 If an alternative has a salvage value, how is it handled in the calculation of a B/C ratio relative to
benefits, disbenefits, costs, or savings?
9.5 What is a fundamental difference between DBOM
and DBOMF contracts?
9.8 The cost of grading and spreading gravel on a
short rural road is expected to be $300,000. The
road will have to be maintained at a cost of $25,000
per year. Even though the new road is not very
smooth, it allows access to an area that previously
could only be reached with off-road vehicles. The
improved accessibility has led to a 150% increase
in the property values along the road. If the previous market value of a property was $900,000,
calculate the B/C ratio using an interest rate of 6%
per year and a 20-year study period.
9.6 Buster County has proposed a strict water conservation policy for all industrial plants within the
county limits. Enforcement is proposed to be a
9.9 Arsenic enters drinking water supplies from natural deposits in the earth or from agricultural and
industrial practices. Since it has been linked to
253
Problems
cancer of the bladder, kidney, and other internal
organs, the EPA has lowered the arsenic standard
for drinking water from 0.050 parts per million to
0.010 parts per million (10 parts per billion). The
annual cost to public water utilities to meet the
new standard is estimated to be $200 per household. If it is estimated that there are 90 million
households in the United States and that the lower
standard can save 50 lives per year valued at
$4,000,000 per life, what is the benefit/cost ratio of
the regulation?
9.10 From the following estimates, determine the B/C
ratio for a project that has a 20-year life. Use an
interest rate of 8% per year.
Consequences to
the People
Consequences to
the Government
Annual
$90,000
benefits
per year
Annual
$10,000
disbenefits
per year
First
$750,000
cost
Annual
$50,000
cost
per year
Annual
$30,000
savings
per year
9.11 A project to extend irrigation canals into an area
that was recently cleared of mesquite trees (a nuisance tree in Texas) and large weeds is projected to
have a capital cost of $2,000,000. Annual maintenance and operation costs will be $100,000 per
year. Annual favorable consequences to the general public of $820,000 per year will be offset to
some extent by annual adverse consequences of
$400,000 to a portion of the general public. If the
project is assumed to have a 20-year life, what is
the B/C ratio at an interest rate of 8% per year?
9.12 Calculate the B/C ratio for the following cash flow
estimates at a discount rate of 7% per year.
Item
Cash Flow
FW of benefits, $
AW of disbenefits, $ per year
First cost, $
M&O costs, $ per year
Life of project, years
30,800,000
105,000
1,200,000
400,000
20
9.13 The benefits associated with a nuclear power plant
cooling water filtration project located on the Ohio
River are $10,000 per year forever, starting in
year 1. The costs are $50,000 in year 0 and $50,000
at the end of year 2. Calculate the B/C ratio at i ⫽
10% per year.
9.14 A privately funded wind-based electric power
generation company in the southern part of the
country has developed the following estimates
(in $1000) for a new turbine farm. The MARR is
10% per year, and the project life is 25 years. Calculate (a) the profitability index and (b) the modified B/C ratio.
Benefits:
Savings:
Cost:
Disbenefits:
$20,000 in year 0 and $30,000 in
year 5
$2000 in years 1–20
$50,000 in year 0
$3000 in years 1–10
9.15 For the values shown, calculate the conventional
B/C ratio at i ⫽ 10% per year.
First cost
M&O cost
Benefits
Disbenefits
PW, $
AW, $/Year
FW, $
100,000
61,446
—
30,723
—
10,000
40,000
5,000
259,370
159,374
637,496
—
9.16 A proposal to reduce traffic congestion on I-5 has
a B/C ratio of 1.4. The annual worth of benefits
minus disbenefits is $560,000. What is the first
cost of the project if the interest rate is 6% per year
and the project is expected to have a 20-year life?
9.17 Oil spills in the Gulf of Mexico have been known
to cause extensive damage to both public and private oyster grounds along the Louisiana and Mississippi shores. One way to protect shellfish along
the shoreline is to release large volumes of freshwater from the Mississippi River to flush oil out to
sea. This procedure inevitably results in death to
some of the saltwater shellfish while preventing
more widespread destruction to public reefs. Oil
containment booms and other temporary structures can also be used to intercept floating oil before it damages sensitive fishing grounds. If the
Fish and Wildlife Service spent $110 million in
year 0 and $50 million in years 1 and 2 to minimize environmental damage from one particular
oil spill, what is the benefit-to-cost ratio provided
the efforts resulted in saving 3000 jobs valued at a
total of $175 million per year? Assume disbenefits
associated with oyster deaths amounted to $30 million in year 0. Use a 5-year study period and an
interest rate of 8% per year.
9.18 On-site granular ferric hydroxide (GFH) systems
can be used to remove arsenic from water when
daily flow rates are relatively low. If the operating
cost is $600,000 per year and the public health
benefits are assumed to be $800,000 per year, what
initial investment in the GFH system is necessary
to guarantee a modified B/C ratio of at least 1.0?
Assume that the equipment life is 10 years and the
interest rate is 6% per year.
254
Benefit/Cost Analysis and Public Sector Economics
Chapter 9
9.19 The Parks and Recreation Department of Burkett
County has estimated that the initial cost of a
“bare-bones” permanent river park will be
$2.3 million. Annual upkeep costs are estimated at
$120,000. Benefits of $340,000 per year and disbenefits of $40,000 per year have also been identified. Using a discount rate of 6% per year, calculate (a) the conventional B/C ratio and (b) the
modified B/C ratio.
9.20 From the following data, calculate the (a) conventional and (b) modified benefit/cost ratios using an
interest rate of 6% per year and an infinite project
period.
To the
People
Benefits:
$300,000 now and
$100,000 per
year thereafter
Disbenefits: $40,000 per year
To the
Government
Costs:
$1.5 million now
and $200,000
three years
from now
Savings: $70,000 per year
9.21 In 2010, Brazil began construction of the Belo
Monte hydroelectric dam on the Xingu River
(which feeds the Amazon River). The project is
funded by a consortium of investors and is expected to cost $11 billion. It will begin producing
electricity in 2015. Even though the dam will provide clean energy for millions of people, environmentalists are sharply opposed. They say it will
devastate wildlife and the livelihoods of 40,000
people who live in the area to be flooded.
Assume that the funding will occur evenly over
the 5-year period from 2010 through 2014 at
$2.2 billion per year. The disbenefits are estimated
to be $100,000 for each displaced person and $1
billion for wildlife destruction. Assume that the
disbenefits will occur evenly through the 5-year
construction period and anticipated benefits will
begin at the end of 2015 and continue indefinitely.
Use an interest rate of 8% per year to determine
what the equivalent annual benefits must be to ensure a B/C ratio of at least 1.0.
9.22 In the United States, the average number of airplanes in the sky on an average morning is 4000.
There are another 16,000 planes on the ground.
Aerospace company Rockwell Collins developed
what it calls a digital parachute—a panic-button
technology that will land any plane in a pinch at
the closest airport, no matter what the weather or
geography and without the help of a pilot. The
technology can be applied if a pilot is no longer
capable of flying the plane or is panicked and confused about what to do in an emergency. Assume
that the cost of retrofitting 20,000 commercial air-
planes is $100,000 each and the plane stays in service for 15 years. If the technology saves an average of 30 lives per year, with the value of a human
life placed at $4,000,000, what is the B/C ratio?
Use an interest rate of 10% per year.
9.23 Although the lower Rio Grande is regulated by
the Elephant Butte Dam and the Caballo Reservoir, serious flooding has occurred in El Paso and
other cities located along the river. This has required homeowners living in valley areas near the
river to purchase flood insurance costing between
$145 and $2766 per year. To alleviate the possibility of flooding, the International Boundary and
Water Commission undertook a project costing
$220 million to raise the levees along flood-prone
portions of the river. As a result, 13,000 properties
were freed of the federal mandate to purchase
flood insurance. In addition, historical records indicate that damage to infrastructure will be
avoided, which amounts to an average benefit of
$8,200,000 per year. If the average cost of flood
insurance is $460 per household per year, calculate the benefit-to-cost ratio of the levee-raising
project. Use an interest rate of 6% per year and a
30-year study period.
9.24 For the data shown, calculate the conventional B/C
ratio at i ⫽ 6% per year.
Benefits:
$20,000 in year 0 and
$30,000 in year 5
Disbenefits:
$7000 in year 3
Savings (to government): $25,000 in years 1–4
Cost:
$100,000 in year 0
Project life:
5 years
9.25 Explain a fundamental difference between the
modified B/C ratio and the profitability index.
9.26 Gerald Corporation entered a public-private partnership using a DBOM contract with the state of
Massachusetts 10 years ago for railroad system
upgrades. Determine the profitability index for the
financial results listed below using a MARR of
8% per year.
Investments: Year 0
$⫺4.2 million
Year 5
$⫺3.5 million
Net savings: Years 1⫺5
Years 6⫺10
$1.2 million per year
$2.5 million per year
9.27 A project had a staged investment distributed over
the 6-year contract period. For the cash flows
shown (next page) and an interest rate of 10% per
year, determine the profitability index and determine if the project was economically justified.
255
Problems
Year
Investment, $1000
NCF, $1000 per year
0
1
2
3
⫺25
0
0
5
⫺10
7
0
9
4
5
6
selected at an interest rate of 8% per year and a
5-year study period.
⫺5 0
0
11 13 15
Initial cost, $
Annual cost, $ per year
Two Alternative Comparison
9.28 In comparing two alternatives by the B/C method,
if the overall B/C ratio for both alternatives is calculated to be exactly 1.0, which alternative should
you select?
9.29 In comparing alternatives X and Y by the B/C
method, if B/CX ⫽ 1.6 and B/CY ⫽ 1.8, what is
known about the B/C ratio on the increment of investment between X and Y?
9.30
Logan Well Services Group is considering two sites
for storage and recovery of reclaimed water. The
mountain site (MS) will use injection wells that cost
$4.2 million to develop and $280,000 per year for
M&O costs. This site will be able to accommodate
150 million gallons per year. The valley site (VS) will
involve recharge basins that cost $11 million to construct and $400,000 to operate and maintain. At this
site, 890 million gallons can be injected each year. If
the value of the injected water is $3.00 per 1000 gallons, which alternative, if either, should be selected
according to the B/C ratio method? Use an interest
rate of 8% per year and a 20-year study period.
9.31 The estimates shown are for a bridge under consideration for a river crossing in Wheeling, West
Virginia. Use the B/C ratio method at an interest
rate of 6% per year to determine which bridge, if
either, should be built.
East Location
Initial cost, $
Annual M&O, $/year
Benefits, $/year
Disbenefits, $/year
Life, years
6
11 ⫻ 10
100,000
990,000
120,000
⬁
West Location
27 ⫻ 106
90,000
2,400,000
100,000
⬁
9.32 Select the better of two proposals to improve street
safety and lighting in a colonia in south central
New Mexico. Use a B/C analysis and an interest
rate of 8% per year.
Initial cost, $
Annual M&O cost, $/year
Annual benefits, $/year
Annual disbenefits, $/year
Life, years
Proposal 1
Proposal 2
900,000
120,000
530,000
300,000
10
1,700,000
60,000
650,000
195,000
20
9.33 Conventional and solar alternatives are available
for providing energy at a remote radar site. Use the
B/C ratio to determine which method should be
Conventional
Solar
300,000
700,000
2,500,000
5,000
9.34 The two alternatives shown are under consideration
for improving security at a county jail in Travis
County, New York. Determine which one should be
selected, based on a B/C analysis, an interest rate of
7% per year and a 10-year study period.
Extra Cameras
(EC)
New Sensors
(NS)
38,000
49,000
110,000
26,000
87,000
64,000
160,000
—
First cost, $
Annual M&O, $/year
Benefits, $/year
Disbenefits, $/year
9.35
The U.S. government recently released an RFP to
construct a second-story floor on an existing building
at the Pentagon Complex. Separate contractors proposed two methods. Method 1 will use lightweight
expanded shale on a metal deck with open web joists
and steel beams. For this method, the costs will be
$14,100 for concrete, $6000 for metal decking,
$4300 for joists, and $2600 for beams. Method 2 will
construct a reinforced concrete slab costing $5200
for concrete, $1400 for rebar, $2600 for equipment
rental, and $1200 for expendable supplies. Special
additives will be included in the lightweight concrete
that will improve the heat-transfer properties of the
floor. If the energy costs for method 1 will be $600
per year lower than for method 2, which one is more
attractive? Use an interest rate of 7% per year, a
20-year study period, and the B/C method.
9.36 A project to control flooding from rare, but sometimes heavy rainfalls in the arid southwest will
have the cash flows shown below. Determine
which project should be selected on the basis of a
B/C analysis at i ⫽ 8% per year and a 20-year
study period.
First cost, $
M&O cost, $ per year
Homeowner cleanup costs,
$ per year
Sanitary
Sewers
Open
Channels
26 million
400,000
60,000
53 million
30,000
0
9.37 Two routes are under consideration for a new interstate highway. The long intervalley route would be
25 kilometers in length and would have an initial
cost of $25 million. The short transmountain route
256
Benefit/Cost Analysis and Public Sector Economics
Chapter 9
would be 10 kilometers long and would have an initial cost of $45 million. Maintenance costs are estimated at $150,000 per year for the long route and
$35,000 per year for the short route. Regardless of
which route is selected, the volume of traffic is expected to be 400,000 vehicles per year. If the vehicle
operating expenses are assumed to be $0.30 per kilometer, determine which route should be selected on
the basis of (a) conventional B/C analysis and
(b) modified B/C analysis. Assume an infinite life for
each road, and use an interest rate of 8% per year.
9.38 The Idaho Department of Fish and Wildlife (IDFW)
is considering two locations for a new state park.
Location E would require an investment of $3 million and $50,000 per year to maintain. Location W
would cost $7 million to construct, but the IDFW
would receive an additional $25,000 per year in
park fees. The operating cost of location W will be
$65,000 per year. The revenue to park concessionaires will be $500,000 per year at location E and
$700,000 at location W. The disbenefits associated
with each location are $30,000 per year for location
E and $40,000 per year for location W. Assume the
park will be maintained indefinitely. Use an interest
rate of 12% per year to determine which location, if
either, should be selected on the basis of (a) the
B/C method and (b) the modified B/C method.
9.39 Three engineers made the estimates shown below
for two optional methods by which new construction
technology would be implemented at a site for public housing. Either one of the two options or the current method may be selected. Set up a spreadsheet
for B/C sensitivity analysis and determine if option
1, option 2 or the do-nothing option is selected by
each of the three engineers. Use a life of 5 years and
a discount rate of 10% per year for all analyses.
Engineer Bob
Engineer Judy
Engineer Chen
Alternative
A
B
C
D
E
F
PW of capital, $
PW of benefits, $
80
70
50
55
72
76
43
52
89
85
81
84
9.41 Comparison of five mutually exclusive alternatives is shown. One must be accepted. According
to the B/C ratio, which alternative should be selected (costs increase from A to E).
Comparison
⌬B/C
Ratio
A versus B
B versus C
C versus D
A versus C
A versus D
B versus D
C versus E
D versus E
0.75
1.4
1.3
1.1
0.2
1.9
1.2
0.9
9.42 A consulting engineer is currently evaluating four
different projects for the Department of Housing
and Urban Development. The future worth of
costs, benefits, disbenefits, and cost savings is
shown. The interest rate is 10% per year, compounded continuously. Determine which of the
projects, if any, should be selected, if the projects
are (a) independent and (b) mutually exclusive.
Project ID
FW of first costs, $
FW of benefits, $
FW of disbenefits, $
FW of cost savings, $
Good
Better
Best
Best of All
10,000
15,000
6,000
1,500
8,000
11,000
1,000
2,000
20,000
25,000
20,000
16,000
14,000
42,000
32,000
3,000
9.43 From the data shown below for six mutually
exclusive projects, determine which project, if
any, should be selected.
Project ID
Option 1 Option 2 Option 1 Option 2 Option 1 Option 2
Initial cost, $
50,000
90,000
75,000
90,000
60,000
70,000
Cost, $/year
3,000
4,000
3,800
3,000
6,000
3,000
20,000
29,000
30,000
35,000
30,000
35,000
500
1,500
1,000
0
5,000
1,000
Benefits, $/year
Disbenefits, $/year
Multiple Alternatives
9.40 A group of engineers responsible for developing
advanced missile detection and tracking technologies, such as shortwave infrared, thermal infrared
detection, target tracking radar, etc., recently came
up with six proposals for consideration. The present worth (in $ billions) of the capital requirements
and benefits is shown for each alternative in the
table. Determine which one(s) should be undertaken, if they are (a) independent and (b) mutually
exclusive.
A
Annual cost, $
per year
Annual benefits,
$ per year
B/C ratio (alternative vs. DN)
B
C
D
8000 25,000 15,000 32,000
E
F
17,000 20,000
?
?
?
?
?
?
1.23
1.12
0.87
0.97
0.71
1.10
Selected Incremental B/C Ratios
A versus B ⫽ 1.07
A versus C ⫽ 0.46
A versus F ⫽ 1.02
B versus D ⫽ 0.43
B versus E ⫽ 2.00
B versus F ⫽ 1.20
C versus D ⫽ 1.06
C versus F ⫽ 1.80
257
Problems
9.44 Four mutually exclusive revenue alternatives are
being compared using the B/C method. Which alternative, if any, should be selected?
Strategy
Cost/Employee, $
Measurement Score
A
B
C
D
E
F
5.20
23.40
3.75
10.80
8.65
15.10
50
182
40
75
53
96
Incremental B/C
When Compared
with Alternative
Initial
B/C Ratio
Alternative Cost, $ Millions vs. DN A
A
B
C
D
30
38
52
81
0.87
1.18
1.04
1.16
B
C
D
— 2.38 1.30 1.38
— 0.58 1.13
— 1.45
—
9.47 There are a number of techniques to help people
stop smoking, but their cost and effectiveness vary
widely. One accepted measure of effectiveness of
a program is the percentage of enrollees quitting.
The table below shows several techniques touted
as effective stop-smoking methods, some historical data on the approximate cost of each program
per person, and the percentage of people smokefree 3 months after the program ended.
9.45 The city of St. Louis, Missouri, is considering various proposals regarding the disposal of used tires.
All of the proposals involve shredding, but the
charges for the service and the handling of the tire
shreds differ in each plan. An incremental B/C
analysis was initiated but never completed.
(a) Fill in all the missing blanks in the table.
(b) Determine which alternative should be selected.
PW of
PW of
PW of
Alternative Costs,$ Benefits, $ Disbenefits, $
J
K
L
M
20
23
28
?
?
28
35
51
1
?
3
4
B/C
Ratio
1.05
1.13
?
1.34
Incremental
B/C When
Compared with
Alternative
J
K
L
M
— ? ? ?
— ? ?
— ?
—
The Cancer Society provides annual cost-offset
funding to cancer patients so more people can afford
these programs. A large clinic in St. Louis has the
capacity to treat each year the number of people
shown. If the clinic plans to place a proposal with the
Cancer Society to treat a specified number of people
annually, estimate the amount of money the clinic
should ask for in its proposal to do the following:
(a) Conduct programs at the capacity level
for the technique with the lowest costeffectiveness ratio.
(b) Offer programs using as many techniques as
possible to treat up to 1300 people per year
using the most cost-effective techniques.
Cost-Effectiveness Analysis
9.46 In an effort to improve productivity in a large
semiconductor manufacturing plant, the plant
manager decided to undertake on a trial basis a series of actions directed toward improving employee morale. Six different strategies were implemented, such as increased employee autonomy,
flexible work schedules, improved training, company picnics, electronic suggestion box, and better
work environment. Periodically, the company surveyed the employees to measure the change in morale. The measure of effectiveness is the difference
between the number of employees who rate their
job satisfaction as very high and those who rate it
very low. The per-employee cost of each strategy
(identified as A through F) and the resultant measurement score are shown in the next column.
The manager has a maximum of $50 per employee
to spend on the permanent implementation of as
many of the strategies as are justified from both the
effectiveness and economic viewpoints. Determine which strategies are the best to implement.
(Hand or spreadsheet solution is acceptable, as you
are instructed.)
Technique
Acupuncture
Subliminal message
Aversion therapy
Outpatient clinic
In-patient clinic
Nicotine replacement
therapy (NRT)
Cost, $/
Enrollee
% Quitting
Treatment
Capacity per Year
700
150
1700
2500
1800
1300
9
1
10
39
41
20
250
500
200
400
550
100
9.48 The cost in $ per year and effectiveness measure
in items salvaged per year for four mutually
exclusive service sector alternatives have been
collected. (a) Calculate the cost effectiveness ratio
for each alternative, and (b) use the CER to identify the best alternative.
Alternative
Cost C, $/Year
Salvaged Items/Year E
W
X
Y
Z
355
208
660
102
20
17
41
7
9.49 An engineering student has only 30 minutes before
the final exam in statics and dynamics. He wants to
258
Benefit/Cost Analysis and Public Sector Economics
Chapter 9
get help solving a type of problem that he knows
will be on the test from the professor’s review during the previous class. There is time for using only
one method of assistance before the exam; he must
select well. In a rapid process of estimation, he determines how many minutes it would take for each
method of assistance and how many points it might
gain for him on the final. The method and estimates follow. Where should he seek help to be
most effective?
Assistance from
Minutes for Assistance Points Gained
Teaching assistant
Course slides on Web
Friend in class
Professor
15
20
10
20
15
10
5
15
Public Sector Ethics
9.50 During the design and specifications development
stages of a remote meter reading system for residential electricity use (a system that allows monthly
usage to be transmitted via phone lines with no
need to physically view meters), the two engineers
working on the project for the city of Forest Ridge
noted something different from what they expected.
The first, an electrical/software engineer, noted that
the city liaison staff member provided all the information on the software options, but only one option, the one from Lorier Software, was ever discussed and detailed. The second designer, an
industrial/systems engineer, further noted that all
the hardware specifications provided to them by
this same liaison came from the same distributor,
namely, Delsey Enterprises. Coincidently, at a
weekend family picnic for city employees, to which
the engineers had been invited, they met a couple
named Don Delsey and Susan Lorier. Upon review,
they learned that Don is the son-in-law of the city
liaison and Susan is his stepdaughter. Based on
these observations and before they complete the
system design and specifications, what should the
two engineers do, if anything, about their suspicions that the city liaison person is trying to bias the
design to favor of the use of his relatives’ software
and hardware businesses?
9.51 Explain the difference between public policy making and public planning.
9.52 Since transportation via automobile was introduced,
drivers throughout the country of Yalturia in eastern
Europe have driven on the left side of the road. Recently the Yalturian National Congress passed a law
that within 3 years, a right-hand driving convention
will be adopted and implemented throughout the
country. This is a major policy change for the country and will require significant public planning and
project development to implement successfully and
safely. Assume you are the lead engineering consultant from Halcrow Engineers, responsible for developing and describing many of these major projects.
Identify six of the projects you deem necessary.
State a name and provide a one- or two-sentence
description of each project.
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
9.53 All of the following are examples of public sector
projects, except:
(a) Bridges
(b) Emergency relief
(c) Prisons
(d) Oil wells
9.54 All of the following are usually associated with
public sector projects except:
(a) Funding from taxes
(b) Profit
(c) Disbenefits
(d) Infinite life
9.55 All of the following would be examples of public
projects except:
(a) Air traffic control system
(b) Establishing a dot.com company
(c) Dam with irrigation canals
(d) Mass transit system
9.56 In a conventional B/C ratio, revenue received by
the government from admission to national parks
should be:
(a) Added to benefits in the numerator
(b) Subtracted from benefits in the numerator
(c)
(d)
Subtracted from costs in the denominator
Added to costs in the denominator
9.57 In a conventional B/C ratio:
(a) Disbenefits and M&O costs are subtracted
from benefits.
(b) Disbenefits are subtracted from benefits and
M&O costs are included in costs.
(c) Disbenefits and M&O costs are added to
costs.
(d) Disbenefits are added to costs and M&O
costs are subtracted from benefits.
9.58 In a modified B/C ratio:
(a) Disbenefits are put in the denominator.
(b) Benefits are subtracted from costs.
(c) M&O costs are put in the denominator.
(d) M&O costs are put in the numerator.
9.59 If two mutually exclusive alternatives have B/C
ratios of 1.4 and 1.5 for the lower- and higher-cost
alternatives, respectively, the following is correct:
Case Study
(a)
(b)
(c)
(d)
9.60
The B/C ratio on the increment between
them is equal to 1.5.
The B/C ratio on the increment between
them is between 1.4 and 1.5.
The B/C ratio on the increment between
them is less than 1.5.
The higher-cost alternative is the better one
economically.
In evaluating three independent alternatives by the B/C
method, the alternatives were ranked A, B, and C,
respectively, in terms of increasing cost, and the following results were obtained for overall B/C ratios: 1.1,
0.9, and 1.3. On the basis of these results, you should:
(a) Select only alternative A
(b) Select only alternative C
(c) Compare A and C incrementally
(d) Select alternatives A and C
9.61 An alternative has the following cash flows:
Benefits of $50,000 per year
Disbenefits of $27,000 per year
Initial cost of $250,000
M&O costs of $10,000 per year
If the alternative has an infinite life and the interest
rate is 10% per year, the B/C ratio is closest to:
(a) 0.52
(b) 0.66
(c) 0.91
(d) 1.16
9.62 At the interest rate of 10% per year, an alternative
with the following estimates has a modified B/C
ratio that is closest to:
Benefits of $60,000 per year
Disbenefits of $29,000 per year
Amortized first cost of $20,000 per year
M&O costs of $15,000 per year
(a) 0.65
(b) 0.72
(c) 0.80
(d) 1.04
9.63 In evaluating three mutually exclusive alternatives
by the B/C method, the alternatives are ranked A,
B, and C, respectively, in terms of increasing cost,
and the following results are obtained for the
overall B/C ratios: 1.1, 0.9, and 1.06. On the basis
of these results, you should:
(a) Select A
(b) Select C
(c)
(d)
259
Select A and C
Compare A and C incrementally
9.64 An alternative with an infinite life has a B/C ratio
of 1.5. The alternative has benefits of $50,000 per
year and annual maintenance costs of $10,000 per
year. The first cost of the alternative at an interest
rate of 10% per year is closest to:
(a) $23,300
(b) $85,400
(c) $146,100
(d) $233,000
9.65 Cost-effectiveness analysis (CEA) differs from
cost-benefit (B/C) analysis in that:
(a) CEA cannot handle multiple alternatives.
(b) CEA expresses outcomes in natural units
rather than in currency units.
(c) CEA cannot handle independent alternatives.
(d) CEA is more time-consuming and resourceintensive.
9.66 Several private colleges claim to have programs
that are very effective at teaching enrollees how
to become entrepreneurs. Two programs, identified as program X and program Y, have produced
4 and 6 persons per year, respectively, who were
recognized as entrepreneurs. If the total cost of
the programs is $25,000 and $33,000, respectively, the incremental cost-effectiveness ratio is
closest to:
(a) 6250
(b) 5500
(c) 4000
(d) 1333
9.67 The statements contained in a code of ethics are
variously known as all of the following except:
(a) Canons
(b) Laws
(c) Standards
(d) Norms
9.68 Of the following, the word not related to ethics is:
(a) Virtuous
(b) Honest
(c) Lucrative
(d) Proper
9.69 All of the following are examples of unethical
behavior except:
(a) Offering services at prices lower than the
competition
(b) Price fixing
(c) Bait and switch
(d) Selling on the black market
CASE STUDY
COMPARING B/C ANALYSIS AND CEA OF TRAFFIC ACCIDENT REDUCTION
Background
This case study compares benefit/cost analysis and costeffectiveness analysis on the same information about highway
lighting and its role in accident reduction.
Poor highway lighting may be one reason that proportionately more traffic accidents occur at night. Traffic accidents are categorized into six types by severity and value.
For example, an accident with a fatality is valued at
260
Benefit/Cost Analysis and Public Sector Economics
Chapter 9
approximately $4 million, while an accident in which there
is property damage (to the car and contents) is valued at
$6000. One method by which the impact of lighting is measured compares day and night accident rates for lighted and
unlighted highway sections with similar characteristics. Observed reductions in accidents seemingly caused by too low
lighting can be translated into either monetary estimates of
the benefits B of lighting or used as the effectiveness
measure E of lighting.
Information
Freeway accident data were collected in a 5-year study. The
property damage category is commonly the largest based on
the accident rate. The number of accidents recorded on a section of highway is presented here.
Number of Accidents Recorded1
Unlighted
Lighted
Accident Type
Day
Night
Day
Night
Property damage
379
199
2069
839
The ratios of night to day accidents involving property damage for the unlighted and lighted freeway sections are 199/379 ⫽
0.525 and 839/2069 ⫽ 0.406, respectively. These results indicate that the lighting was beneficial. To quantify the benefit, the
accident rate ratio from the unlighted section will be applied to
the lighted section. This will yield the number of accidents that
were prevented. Thus, there would have been (2069)(0.525) ⫽
1086 accidents instead of 839 if there had not been lights on the
freeway. This is a difference of 247 accidents. At a cost of $6000
per accident, this results in a net annual benefit of
B ⫽ (247)($6000) ⫽ $1,482,000
For an effectiveness measure of number of accidents prevented,
this results in E ⫽ 247.
To determine the cost of the lighting, it will be assumed that
the light poles are center poles 67 meters apart with 2 bulbs
each. The bulb size is 400 watts, and the installation cost is
$3500 per pole. Since these data were collected over 87.8 kilometers of lighted freeway, the installed cost of the lighting is
(with number of poles rounded off):
(
87.8
Installation cost ⫽ $3500 ———
0.067
⫽ 3500(1310)
⫽ $4,585,000
)
There are a total of 87.8/0.067⫽1310 poles, and electricity
costs $0.10 per kWh. Therefore, the annual power cost is
Annual power cost
⫽ 1310 poles(2 bulbs/pole)(0.4 kilowatt/bulb)
⫻ (12 hours/day)(365 days/year)
⫻ ($0.10/kilowatt-hour)
⫽ $459,024 per year
1
The data were collected over a 5-year period. Therefore, the
annualized cost C at i ⫽ 6% per year is
Total annual cost ⫽ $4,585,000(A/P,6%,5)
⫹ 459,024
⫽ $1,547,503
If a benefit/cost analysis is the basis for a decision on additional lighting, the B/C ratio is
1,482,000
B/C ⫽ ————— ⫽ 0.96
1,547,503
Since B/C ⬍ 1.0, the lighting is not justified. Consideration
of other categories of accidents is necessary to obtain a better
basis for decisions. If a cost-effectiveness analysis (CEA) is
applied, due to a judgment that the monetary estimates for
lighting’s benefit is not accurate, the C/E ratio is
1,547,503
C/E ⫽ ————— ⫽ 6265
247
This can serve as a base ratio for comparison when an incremental CEA is performed for additional accident reduction proposals.
These preliminary B/C and C/E analyses prompted the
development of four lighting options:
W)
X)
Y)
Z)
Implement the plan as detailed above; light poles
every 67 meters at a cost of $3500 per pole.
Install poles at twice the distance apart (134 meters).
This is estimated to cause the accident prevention
benefit to decrease by 40%.
Install cheaper poles and surrounding safety guards,
plus slightly lowered lumen bulbs (350 watts) at a
cost of $2500 per pole; place the poles 67 meters
apart. This is estimated to reduce the benefit by 25%.
Install cheaper equipment for $2500 per pole with
350-watt lightbulbs and place them 134 meters apart.
This plan is estimated to reduce the accident prevention measure by 50% from 247 to 124.
Case Study Exercises
Determine if a definitive decision on lighting can be determined by doing the following:
1. Use a benefit/cost analysis to compare the four alternatives to determine if any are economically justified.
2. Use a cost-effectiveness analysis to compare the four
alternatives.
From an understanding viewpoint, consider the following:
3. How many property-damage accidents could be prevented on the unlighted portion if it were lighted?
4. What would the lighted, night-to-day accident ratio
have to be to make alternative Z economically justified
by the B/C ratio?
5. Discuss the analysis approaches of B/C and C/E. Does
one seem more appropriate in this type of situation than
the other? Why? Can you think of other bases that might
be better for decisions for public projects such as this one?
Portion of data reported in Michael Griffin, “Comparison of the Safety of Lighting on Urban Freeways,” Public Roads, vol. 58, pp. 8–15, 1994.
LEARNING STAGE 2: EPILOGUE
Selecting the Basic Analysis Tool
I
n the previous five chapters, several equivalent evaluation techniques have been discussed. Any
method—PW, AW, FW, ROR, or B/C—can be used to select one alternative from two or more and
obtain the same, correct answer. Only one method is needed to perform the engineering economy
analysis, because any method, correctly performed, will select the same alternative. Yet different information about an alternative is available with each different method. The selection of a method and
its correct application can be confusing.
Table LS2–1 gives a recommended evaluation method for different situations, if it is not specified
by the instructor in a course or by corporate practice in professional work. The primary criteria for
selecting a method are speed and ease of performing the analysis. Interpretation of the entries in each
column follows.
Evaluation period: Most private sector alternatives (revenue and cost) are compared over their
equal or unequal estimated lives, or over a specific period of time. Public sector projects are
commonly evaluated using the B/C ratio and usually have long lives that may be considered infinite for economic computation purposes.
Type of alternatives: Private sector alternatives have cash flow estimates that are revenuebased (includes income and cost estimates) or cost-based (cost estimates only). For cost alternatives, the revenue cash flow series is assumed to be equal for all alternatives. For public sector projects, the difference between costs and timing is used to select one alternative over
another. Service sector projects for which benefits are estimated using a nonmonetary effectiveness measure are usually evaluated with a method such as cost-effectiveness analysis. This
applies to all evaluation periods.
Recommended method: Whether an analysis is performed by hand, calculator, or spreadsheet, the
method(s) recommended in Table LS2–1 will correctly select one alternative from two or more as
TABLE LS2–1
Recommended Method to Compare Mutually Exclusive Alternatives,
Provided the Method Is Not Preselected
Evaluation Period
Type of Alternatives
Recommended Method
Series to
Evaluate
Equal lives of
alternatives
Revenue or cost
AW or PW
Cash flows
Public sector
B/C, based on AW or PW
Incremental
cash flows
Revenue or cost
AW
Cash flows
Public sector
B/C, based on AW
Incremental
cash flows
Revenue or cost
AW or PW
Public sector
B/C, based on AW or PW
Updated cash
flows
Updated incremental
cash flows
Revenue or cost
Public sector
AW or PW
B/C, based on AW
Unequal lives of
alternatives
Study period
Long to infinite
Cash flows
Incremental
cash flows
262
Benefit/Cost Analysis and Public Sector Economics
Chapter 9
rapidly as possible. Any other method can be applied subsequently to obtain additional information and, if needed, verification of the selection. For example, if lives are unequal and the
rate of return is needed, it is best to first apply the AW method at the MARR and then determine the selected alternative’s i* using the same AW relation with i as the unknown.
Series to evaluate: The estimated cash flow series for one alternative and the incremental
series between two alternatives are the only two options for present worth or annual worth
evaluation. For spreadsheet analyses, this means that the NPV or PV functions (for present
worth) or the PMT function (for annual worth) is applied. The word updated is added as a
reminder that a study period analysis requires that cash flow estimates (especially salvage/
market values) be reexamined and updated before the analysis is performed.
Once the evaluation method is selected, a specific procedure must be followed. These procedures were the primary topics of the last five chapters. Table LS2–2 summarizes the important
TABLE LS2–2
Evaluation
Method
Present worth
Future worth
Characteristics of an Economic Analysis of Mutually Exclusive Alternatives Once the
Evaluation Method Is Determined
Lives of
Alternatives
Time
Period for
Analysis
Series to
Evaluate
Rate of
Return;
Interest Rate
Decision
Guideline:
Select1
PW
Equal
Lives
Cash flows
MARR
PW
Unequal
LCM
Cash flows
MARR
PW
Study period
Study period
MARR
CC
Long to
infinite
Infinity
Updated
cash flows
Cash flows
Numerically
largest PW
Numerically
largest PW
Numerically
largest PW
Numerically
largest CC
Equivalence
Relation
FW
AW
Annual worth
Rate of return
Benefit/cost
AW
Same as present worth for equal lives, unequal lives,
and study period
Equal or
unequal
Study period
Lives
Cash flows
MARR
Study period
Updated
cash flows
Cash flows
MARR
AW
Long to
infinite
Infinity
PW or AW
Equal
Lives
PW or AW
Unequal
LCM of pair
AW
Unequal
Lives
PW or AW
Study period
Study period
Updated
incremental
cash flows
Find ⌬i*
PW
Equal or
unequal
Equal or
unequal
Long to
infinite
LCM of pairs
Incremental
cash flows
Incremental
cash flows
Incremental
cash flows
Discount rate
AW
AW or PW
1
MARR
Lowest equivalent cost or largest equivalent income.
Lives
Infinity
Incremental
cash flows
Incremental
cash flows
Cash flows
MARR
Find ⌬i*
Find ⌬i*
Find ⌬i*
Discount rate
Discount rate
Numerically
largest FW
Numerically
largest AW
Numerically
largest AW
Numerically
largest AW
Last ⌬i* ⱖ
MARR
Last ⌬i* ⱖ
MARR
Last ⌬i* ⱖ
MARR
Last ⌬i* ⱖ
MARR
Last ⌬B/C ⱖ
1.0
Last ⌬B/C ⱖ
1.0
Last ⌬B/C ⱖ
1.0
Learning Stage 2: Epilogue
elements of the procedure for each method—PW, AW, ROR, and B/C. FW is included as an extension of PW. The meaning of the entries in Table LS2–2 follows.
Equivalence relation The basic equation written to perform any analysis is either a PW
or an AW relation. The capitalized cost (CC) relation is a PW relation for infinite life, and
the FW relation is likely determined from the PW equivalent value. Additionally, as we
learned in Chapter 6, AW is simply PW times the A/P factor over the LCM or study
period.
Lives of alternatives and time period for analysis The length of time for an evaluation (the
n value) will always be one of the following: equal lives of the alternatives, LCM of unequal
lives, specified study period, or infinity because the lives are very long.
• PW analysis always requires the LCM of compared alternatives.
• Incremental ROR and B/C methods require the LCM of the two alternatives being compared.
• The AW method allows analysis over the respective alternative lives.
• CC analysis has an infinite time line and uses the relation P ⫽ A兾i.
The one exception is for the incremental ROR method for unequal-life alternatives using an AW
relation for incremental cash flows. The LCM of the two alternatives compared must be used.
This is equivalent to using an AW relation for the actual cash flows over the respective lives. Both
approaches find the incremental rate of return ⌬i*.
Series to evaluate Either the estimated cash flow series or the incremental series is used to
determine the PW value, the AW value, the i* value, or the B/C ratio.
Rate of return (interest rate) The MARR value must be stated to complete the PW, FW,
or AW method. This is also correct for the discount rate for public sector alternatives analyzed by the B/C ratio. The ROR method requires that the incremental rate be found in order
to select one alternative. It is here that the dilemma of multiple rates appears, if the sign tests
indicate that a unique, real number root does not necessarily exist for a nonconventional
series.
Decision guideline The selection of one alternative is accomplished using the general guideline in the rightmost column. Always select the alternative with the numerically largest PW,
FW, or AW value. This is correct for both revenue and cost alternatives. The incremental cash
flow methods—ROR and B/C—require that the largest initial cost and incrementally justified
alternative be selected, provided it is justified against an alternative that is itself justified. This
means that the ⌬i* exceeds MARR, or the ⌬B/C exceeds 1.0.
EXAMPLE LS2–1
Read through the problem statement of the following examples, neglecting the evaluation
method used in the example. Determine which evaluation method is probably the fastest and
easiest to apply. Is this the method used in the example? (a) 8.6, (b) 6.5, (c) 5.8, (d) 5.4.
Solution
Referring to the contents of Table LS2–1, the following methods should be applied first.
(a) Example 8.6 involves four revenue alternatives with equal lives. Use the AW or PW value
at the MARR of 10%. The incremental ROR method was applied in the example.
(b) Example 6.5 requires selection between three public sector alternatives with unequal lives,
one of which is 50 years and another is infinite. The B/C ratio of AW values is the best
choice. This is how the problem was solved.
263
264
Chapter 9
Benefit/Cost Analysis and Public Sector Economics
(c) Since Example 5.8 involves two cost alternatives with one having a long life, either AW or
PW can be used. Since one life is long, capitalized cost, based on P ⫽ A/i, is best in this
case. This is the method applied in the example.
(d) Example 5.4 is in the series of progressive examples. It involves 5-year and 10-year cost
alternatives. The AW method is the best to apply in this case. The PW method for the LCM
of 10 years and a study period of 5 years were both presented in the example.
L E A R N I N G S TA G E 3
Making Better Decisions
LEARNING STAGE 3
Making Better
Decisions
CHAPTER
10
Project Financing and
Noneconomic
Attributes
CHAPTER
11
Replacement and
Retention Decisions
CHAPTER
12
Independent
Projects with
Budget Limitation
CHAPTER
13
Breakeven and
Payback Analysis
M
ost of the evaluations in the real world involve more than
a simple economic selection of new assets or projects. The
chapters in this stage introduce information-gathering and
techniques that make decisions better. For example, noneconomic
parameters can be introduced into the project analysis study
through multiple attribute evaluation, and the appropriate MARR
for a corporation or type of alternative can tailor and improve the
economic decision.
The future is certainly not exact. However, techniques such as
replacement/retention studies, breakeven analysis, and payback analysis help make informed decisions about future uses of
existing assets and systems.
After completing these chapters, you will be able to go beyond
the basic alternative analysis tools of the previous chapters. The
techniques covered in this learning stage take into consideration the
moving targets of change over time.
Important note: If asset depreciation and taxes are to be considered by an after-tax analysis, Chapters 16 and 17 should be covered
before or in conjunction with these chapters.
CHAPTER 10
Project
Financing and
Noneconomic
Attributes
L E A R N I N G
O U T C O M E S
Purpose: Explain debt and equity financing, select the appropriate MARR, and consider multiple attributes when
comparing alternatives.
SECTION
TOPIC
LEARNING OUTCOME
10.1
COC and MARR
• Explain the relation between cost of capital and
the MARR; explain why MARR values vary.
10.2
D-E mix and WACC
• Understand debt-to-equity mix and calculate the
weighted average cost of capital.
10.3
Cost of debt capital
• Estimate the cost of debt capital, considering tax
advantages.
10.4
Cost of equity capital
• Estimate the cost of equity capital and describe its
relation to MARR and WACC.
10.5
High D-E mixes
• Demonstrate the connection between high D-E
mixes and corporate (or personal) financial risk.
10.6
Multiple attributes
• Develop weights for multiple attributes used in
alternative evaluation and selection.
10.7
Additive weights
• Apply the weighted attribute method to
alternative evaluations that include noneconomic
attributes.
T
his chapter discusses the different ways to finance a project through debt and
equity sources and explains how the MARR is established. The descriptions here
complement the introductory material of Chapter 1 on the same topics. Some of
the parameters specified earlier are unspecified here, and in future chapters. As a result,
some of the textbook aspects apparent in previous chapters are removed, thus coming closer
to treating the more complex, real-world situations in which professional practice and decision making occur.
Until now, only one dimension—the economic one—has been the basis for judging the
economic viability of one project, or the selection basis from two or more alternatives. In this
chapter, guidelines and techniques explain the determination and use of multiple (noneconomic) attributes helpful in selecting between alternatives.
10.1 MARR Relative to the Cost of Capital
The MARR value used in alternative evaluation is one of the most important parameters of
a study. In Chapter 1 the MARR was described relative to the weighted costs of debt and
equity capital. This and the next four sections explain how to establish a MARR under varying conditions.
To form the basis for a realistic MARR, the types and cost of each source of project financing
should be understood and estimated. There is a strong connection between the costs of debt and
equity capital and the MARR used to evaluate one or more alternatives, whether they are mutually exclusive or independent. There are several terms and relationships important to the understanding of project financing and the MARR that is specified to evaluate projects using PW, AW,
FW, or B/C methods. (Reference to Section 1.9 will complement the following material.)
The cost of capital is the weighted average interest rate paid based on the proportion of
investment capital from debt and equity sources.
The MARR is then set relative to the cost of capital. The MARR can be set for one project, a
series of projects, a division of a corporation, or the entire company. MARR values change
over time due to changing circumstances.
When no specific MARR is established, the estimated net cash flows and available capital
establish an inherent MARR. This rate is determined by finding the ROR (i*) value of the
project cash flows. This rate is utilized as the opportunity cost, which is the ROR of the first
project not funded due to the lack of capital funds.
Before we discuss cost of capital, we review the two primary sources of capital.
Debt capital represents borrowing from outside the company, with the principal repaid at a
stated interest rate following a specified time schedule. Debt financing includes borrowing via
bonds, loans, and mortgages. The lender does not share in the profits made using the debt
funds, but there is risk in that the borrower could default on part of or all the borrowed funds.
The amount of outstanding debt financing is indicated in the liabilities section of the corporate
balance sheet.
Equity capital is corporate money comprised of the funds of owners and retained earnings.
Owners’ funds are further classified as common and preferred stock proceeds or owners’ capital for a private (non-stock-issuing) company. Retained earnings are funds previously retained
in the corporation for capital investment. The amount of equity is indicated in the net worth
section of the corporate balance sheet.
To illustrate the relation between cost of capital and MARR, assume a new greenhouse gas
emission control system will be completely financed by a $25,000,000 bond issue (100% debt
financing), and assume the dividend rate on the bonds is 8%. Therefore, the cost of debt capital
is 8% as shown in Figure 10–1. This 8% is the minimum for MARR. Management may increase
this MARR in increments that reflect its desire for added return and its perception of risk. For
example, management may add an amount for all capital commitments in this area. Suppose this
amount is 2%. This increases the expected return to 10% (Figure 10–1). Also, if the risk associated with the investment is considered substantial enough to warrant an additional 1% return requirement, the final MARR is 11%.
Cost of capital
MARR
Opportunity cost
268
Project Financing and Noneconomic Attributes
Chapter 10
Established
MARR
11%
Risk factor added
10%
Expected return
requirement added
1%
2%
Minimum
MARR
8%
Cost of capital
Figure 10–1
A fundamental relation between cost of capital
and MARR used in practice.
The recommended approach does not follow the logic presented above. Rather, the cost of
capital (8% here) should be the established MARR. Then the i* value is determined from the
estimated net cash flows. Using this approach, suppose the control system is estimated to return
11%. Now, additional return requirements and risk factors are considered to determine if 3%
above the MARR of 8% is sufficient to justify the capital investment. After these considerations,
if the project is not funded, the effective MARR is now 11%. This is the opportunity cost discussed previously—the unfunded project i* has established the effective MARR for emission
control system alternatives at 11%, not 8%.
The setting of the MARR for an economy study is not an exact process. The debt and equity
capital mix changes over time and between projects. Also, the MARR is not a fixed value established corporatewide. It is altered for different opportunities and types of projects. For example,
a corporation may use a MARR of 10% for evaluating the purchase of assets (equipment, cars)
and a MARR of 20% for expansion investments, such as acquiring smaller companies.
The effective MARR varies from one project to another and through time because of factors
such as the following:
Project risk. Where there is greater risk (perceived or actual) associated with proposed
projects, the tendency is to set a higher MARR. This is encouraged by the higher cost of debt
capital for projects considered risky. This usually means that there is some concern that the
project will not realize its projected revenue requirements.
Investment opportunity. If management is determined to expand in a certain area, the
MARR may be lowered to encourage investment with the hope of recovering lost revenue in
other areas. This common reaction to investment opportunity can create havoc when the
guidelines for setting a MARR are too strictly applied. Flexibility becomes very important.
Government intervention. Depending upon the state of the economy, international relations, and a host of other factors, the federal government (and possibly lower levels) can dictate the forces and direction of the free market. This may occur through price limits, subsidies,
import tariffs, and limitation on availability. Both short-term and long-term government interventions are commonly present in different areas of the economy. Examples are steel imports,
foreign capital investment, car imports, and agricultural product exports. During the time that
such government actions are in force, there is a strong impact to increase or decrease taxes,
prices, etc., thus tending to move the MARR up or down.
Tax structure. If corporate taxes are rising (due to increased profits, capital gains, local
taxes, etc.), pressure to increase the MARR is present. Use of after-tax analysis may assist in
eliminating this reason for a fluctuating MARR, since accompanying business expenses will
tend to decrease taxes and after-tax costs.
Limited capital. As debt and equity capital become limited, the MARR is increased. If the
demand for limited capital exceeds supply, the MARR may tend to be set even higher. The
opportunity cost has a large role in determining the MARR actually used.
Debt-Equity Mix and Weighted Average Cost of Capital
10.2
Market rates at other corporations. If the MARR increases at other corporations, especially competitors, a company may alter its MARR upward in response. These variations are
often based on changes in interest rates for loans, which directly impact the cost of capital.
If the details of after-tax analysis are not of interest, but the effects of income taxes are important, the MARR may be increased by incorporating an effective tax rate using the formula
after-tax MARR
Before-tax MARR ⴝ ————————
1 ⴚ tax rate
[10.1]
The total or effective tax rate, including federal, state, and local taxes, for most corporations is in
the range of 30% to 50%. If an after-tax rate of return of 10% is required and the effective tax rate
is 35%, the MARR for the before-tax economic analysis is 10%/(1 ⫺ 0.35) ⫽ 15.4%.
EXAMPLE 10.1
Twin brother and sister, Carl and Christy graduated several years ago from college. Carl, an
architect, has worked in home design with Bulte Homes since graduation. Christy, a civil engineer, works with Butler Industries in structural components and analysis. They both reside in
Richmond, Virginia. They have started a creative e-commerce network through which
Virginia-based builders can buy their “spec home” plans and construction materials much
more cheaply. Carl and Christy want to expand into a regional e-business corporation. They
have gone to the Bank of America (BA) in Richmond for a business development loan. Identify
some factors that might cause the loan rate to vary when BA provides the quote. Also, indicate
any impact on the established MARR when Carl and Christy make economic decisions for
their business.
Solution
In all cases the direction of the loan rate and the MARR will be the same. Using the six factors
mentioned above, some loan rate considerations are as follows:
Project risk: The loan rate may increase if there has been a noticeable downturn in housing
starts, thus reducing the need for the e-commerce connection.
Investment opportunity: The rate could increase if other companies offering similar services have already applied for a loan at other BA branches regionally or nationwide.
Government intervention: The loan rate may decrease if the federal government has
recently offered Federal Reserve loan money at low rates to banks. The intervention may
be designed to boost the housing economic sector in an effort to offset a significant
slowdown in new home construction.
Taxes: If the state recently removed house construction materials from the list of items
subject to sales tax, the rate might be lowered slightly.
Capital limitation: Assume the computer equipment and software rights held by Carl and
Christy were bought with their own funds and there are no outstanding loans. If additional equity capital is not available for this expansion, the rate for the loan (debt capital)
should be lowered.
Market loan rates: The local BA branch probably obtains its development loan money
from a large national pool. If market loan rates to this BA branch have increased, the rate
for this loan will likely increase, because money is becoming “tighter.”
10.2 Debt-Equity Mix and Weighted
Average Cost of Capital
The debt-to-equity (D-E) mix identifies the percentages of debt and equity financing for a corporation. A company with a 40–60 D-E mix has 40% of its capital originating from debt capital
sources (bonds, loans, and mortgages) and 60% derived from equity sources (stocks and retained
269
270
Project Financing and Noneconomic Attributes
Chapter 10
Figure 10–2
General shape of different
cost of capital curves.
Cost of capital for each source
Equity
WACC
Debt
Normal
operating
range
Minimum
WACC
0% debt
45%
100% debt
Percent debt capital (D-E mix)
earnings). Most projects are funded with a combination of debt and equity capital made available
specifically for the project or taken from a corporate pool of capital. The weighted average cost
of capital (WACC) of the pool is estimated by the relative fractions from debt and equity sources.
If known exactly, these fractions are used to estimate WACC; otherwise the historical fractions
for each source are used in the relation
WACC ⴝ (equity fraction)(cost of equity capital)
ⴙ (debt fraction)(cost of debt capital)
[10.2]
The two cost terms are expressed as percentage interest rates.
Since virtually all corporations have a mixture of capital sources, the WACC is a value between the debt and equity costs of capital. If the fraction of each type of equity financing—
common stock, preferred stock, and retained earnings—is known, Equation [10.2] is expanded.
WACC ⴝ (common stock fraction)(cost of common stock capital)
ⴙ (preferred stock fraction)(cost of preferred stock capital)
ⴙ (retained earnings fraction)(cost of retained earnings capital)
ⴙ (debt fraction)(cost of debt capital)
[10.3]
Figure 10–2 indicates the usual shape of cost of capital curves. If 100% of the capital is derived
from equity or 100% is from debt sources, the WACC equals the cost of capital of that source of
funds. There is virtually always a mixture of capital sources involved for any capitalization program. As an illustration only, Figure 10–2 indicates a minimum WACC at about 45% debt capital.
Most firms operate over a range of D-E mixes. For example, a range of 30% to 50% debt financing
for some companies may be very acceptable to lenders, with no increases in risk or MARR. However, another company may be considered “risky” with only 20% debt capital. It takes knowledge
about management ability, current projects, and the economic health of the specific industry to
determine a reasonable operating range of the D-E mix for a particular company.
EXAMPLE 10.2
Historically, Hong Kong has imported over 95% of its fresh vegetables each day. In an effort
to develop sustainable and renewable vegetable sources, a new commercial vertical crop technology is being installed through a public-private partnership with Valcent Products.1 For
1
“Valcent Announces Agreement to Supply Verticrop™ Vertical Farming Technology to Hong Kong’s
VF Innovations Ltd.,” www.valcent.net, June 16, 2010 news release.
Determination of the Cost of Debt Capital
10.3
illustration purposes, assume that the present worth of the total system cost is $20 million with
financing sources and costs as follows.
Commercial loan for debt financing
Retained earnings from partnering corporations
Sale of stock (common and preferred)
$10 million at 6.8% per year
$4 million at 5.2% per year
$6 million at 5.9% per year
There are three existing international vertical farming projects with capitalization and WACC
values as follows:
Project 1:
Project 2:
Project 3:
$5 million with WACC1 ⫽ 7.9%
$30 million with WACC2 ⫽ 10.2%
$7 million with WACC3 ⫽ 4.8%
Compare the WACC for the Hong Kong (HK) project with the WACC of the existing projects.
Solution
To apply Equation [10.3] to this new project, the fraction of equity (stock and retained earnings) and debt financing is needed. These are 0.3 for stock ($6 out of $20 million), 0.2 for retained earnings, and 0.5 for debt ($10 out of $20 million).
WACCHK ⫽ 0.3(5.9%) ⫹ 0.2(5.2%) ⫹ 0.5(6.8%) ⫽ 6.210%
To correctly weight the other three project WACCs by size, determine the fraction in each one
of the $42 million in total capital: project 1 has $5 million/$42 million ⫽ 0.119; project 2 has
0.714; project 3 has 0.167. The WACC weighted by project size is WACCW.
WACCW ⫽ 0.119(7.9%) ⫹ 0.714(10.2%) ⫹ 0.167(4.8%) ⫽ 9.025%
The Hong Kong project has a considerably lower cost of capital than the weighted average of
other projects, considering all sources of funding.
The WACC value can be computed using before-tax or after-tax values for cost of capital. The
after-tax method is the correct one since debt financing has a distinct tax advantage, as discussed
in Section 10.3 below. Approximations of after-tax or before-tax cost of capital are made using
the effective tax rate Te in the relation
After-tax cost of debt capital ⴝ (before-tax cost)(1 ⴚ Te)
[10.4]
The effective tax rate is a combination of federal, state, and local tax rates. They are reduced to a
single number Te to simplify computations. Equation [10.4] may be used to approximate the cost
of debt capital separately or inserted into Equation [10.2] for an after-tax WACC rate. Chapter 17
treats taxes and after-tax economic analysis in detail.
10.3 Determination of the Cost of Debt Capital
Debt financing includes borrowing, primarily via bonds and loans. (We learned about bonds in
Section 7.6.) In most industrialized countries, bond dividends and loan interest payments are taxdeductible as a corporate expense. This reduces the taxable income base upon which taxes are
calculated, with the end result of less taxes paid. The cost of debt capital is, therefore, reduced
because there is an annual tax savings of the expense cash flow times the effective tax rate Te.
This tax savings is subtracted from the debt-capital expense in order to calculate the cost of debt
capital. In formula form,
Tax savings ⴝ (expenses) (effective tax rate) ⴝ expenses (Te)
Net cash flow ⴝ expenses ⴚ tax savings ⴝ expenses (1 ⴚ Te)
[10.5]
[10.6]
To find the cost of debt capital, develop a PW- or AW-based relation of the net cash flow
(NCF) series with i* as the unknown. Find i* by trial and error, by calculator, or by the RATE
271
272
Project Financing and Noneconomic Attributes
Chapter 10
or IRR function on a spreadsheet. This is the cost of debt capital used in the WACC computation, Equation [10.2].
EXAMPLE 10.3
AT&T will generate $5 million in debt capital by issuing five thousand $1000 8% per year
10-year bonds. If the effective tax rate of the company is 30% and the bonds are discounted
2%, compute the cost of debt capital (a) before taxes and (b) after taxes from the company
perspective. Obtain the answers by hand and spreadsheet.
Solution by Hand
(a) The annual bond dividend is $1000(0.08) ⫽ $80, and the 2% discounted sales price is $980
now. Using the company perspective, find the i* in the PW relation
0 ⫽ 980 ⫺ 80(P/A, i*,10) ⫺ 1000(P/F, i*,10)
i* ⫽ 8.3%
The before-tax cost of debt capital is i* ⫽ 8.3%, which is slightly higher than the
8% bond interest rate, because of the 2% sales discount.
(b) With the allowance to reduce taxes by deducting the bond dividend, Equation [10.5]
shows a tax savings of $80(0.3) ⫽ $24 per year. The bond dividend amount for the PW
relation is now $80 ⫺ 24 ⫽ $56. Solving for i* after taxes reduces the cost of debt
capital to 5.87%.
Solution by Spreadsheet
Figure 10–3 is a spreadsheet image for both before-tax (column B) and after-tax (column C)
analysis using the IRR function. The after-tax net cash flow is calculated using Equation [10.6]
with Te ⫽ 0.3.
Bond dividend before taxes
⫽ ⫺1000*0.08
Bond dividend after taxes
⫽ (⫺1000*0.08)*(1 ⫺ 0.3)
⫽ IRR(C3:C13)
Figure 10–3
Use of IRR function to determine cost of debt capital before taxes and after taxes, Example10.3.
EXAMPLE 10.4
LST Trading Company will purchase a $20,000 ten-year-life asset. Company managers
have decided to put $10,000 down now from retained earnings and borrow $10,000 at an
interest rate of 6%. The simplified loan repayment plan is $600 in interest each year, with
the entire $10,000 principal paid in year 10. (a) What is the after-tax cost of debt capital if
the effective tax rate is 42%? (b) How are the interest rate and cost of debt capital used to
calculate WACC?
Determination of the Cost of Equity Capital and the MARR
10.4
Solution
(a) The after-tax net cash flow for interest on the $10,000 loan is an annual amount of
600(1 ⫺ 0.42) ⫽ $348 by Equation [10.6]. The loan repayment is $10,000 in year 10.
PW is used to estimate a cost of debt capital of 3.48%.
0 ⫽ 10,000 ⫺ 348(P/A, i*,10) ⫺ 10,000(P/F, i*,10)
(b) The 6% annual interest on the $10,000 loan is not the WACC because 6% is paid only on
the borrowed funds. Nor is 3.48% the WACC, since it is only the cost of debt capital. The
cost of the $10,000 equity capital is needed to determine the WACC.
10.4 Determination of the Cost of Equity
Capital and the MARR
Equity capital is usually obtained from the following sources:
Sale of preferred stock
Sale of common stock
Use of retained earnings
Use of owner’s private capital
The cost of each type of financing is estimated separately and entered into the WACC computation. A summary of one commonly accepted way to estimate each source’s cost of capital is
presented here. One additional method for estimating the cost of equity capital via common stock
is presented. There are no tax savings for equity capital, because dividends paid to stockholders
and owners are not tax-deductible.
Issuance of preferred stock carries with it a commitment to pay a stated dividend annually.
The cost of capital is the stated dividend percentage, for example, 10%, or the dividend amount
divided by the price of the stock. Preferred stock may be sold at a discount to speed the sale, in
which case the actual proceeds from the stock should be used as the denominator. For example,
if a 10% dividend preferred stock with a value of $200 is sold at a 5% discount for $190 per
share, there is a cost of equity capital of ($20/$190) ⫻ 100% ⫽ 10.53%.
Estimating the cost of equity capital for common stock is more involved. The dividends paid
are not a true indication of what the stock issue will actually cost in the future. Usually a valuation of the common stock is used to estimate the cost. If Re is the cost of equity capital (in
decimal form),
first-year dividend
Re ⴝ ———————— ⴙ expected dividend growth rate
price of stock
DV
ⴝ ——1 ⴙ g
P
[10.7]
The growth rate g is an estimate of the annual increase in returns that the shareholders receive.
Stated another way, it is the compound growth rate on dividends that the company believes is
required to attract stockholders. For example, assume a multinational corporation plans to raise
capital through its U.S. subsidiary for a new plant in South America by selling $2,500,000 worth
of common stock valued at $20 each. If a 5% or $1 dividend is planned for the first year and an
appreciation of 4% per year is anticipated for future dividends, the cost of capital for this common stock issue from Equation [10.7] is 9%.
1 ⫹ 0.04 ⫽ 0.09
Re ⫽ ——
20
The retained earnings and owner’s funds cost of equity capital is usually set equal to the
common stock cost, since it is the shareholders and owners who will realize any returns from
projects in which these funds are invested.
273
274
Project Financing and Noneconomic Attributes
Chapter 10
Figure 10–4
Expected return on common stock issue using
CAPM.
Market
security
line
Re
⬎1
Premium
increases
for more
risky securities
Rm – Rf
Premium
Rm
Rf
Selected
market
portfolio
0

1.0
Once the cost of capital for all planned equity sources is estimated, the WACC is calculated
using Equation [10.3].
A second method used to estimate the cost of common stock capital is the capital asset pricing model (CAPM). Because of the fluctuations in stock prices and the higher return demanded
by some corporations’ stocks compared to others, this valuation technique is commonly applied.
The cost of equity capital from common stock Re, using CAPM, is
Re ⴝ risk-free return ⴙ premium above risk-free return
ⴝ Rf ⴙ (Rm ⴚ Rf)
[10.8]
where  ⫽ volatility of a company’s stock relative to other stocks in the market ( ⫽ 1.0 is
the norm)
Rm ⫽ return on stocks in a defined market portfolio measured by a prescribed index
Rf ⫽ return on a “safe investment,” which is usually the U.S. Treasury bill rate
The term (Rm ⫺ Rf) is the premium paid above the safe or risk-free rate. The coefficient 
(beta) indicates how the stock is expected to vary compared to a selected portfolio of stocks in
the same general market area, often the Standard and Poor’s 500 stock index. If  ⬍ 1.0, the
stock is less volatile, so the resulting premium can be smaller; when  ⬎ 1.0, larger price movements are expected, so the premium is increased.
Security is a word that identifies a stock, bond, or any other instrument used to develop capital. To better understand how CAPM works, consider Figure 10–4. This is a plot of a market security line, which is a linear fit by regression analysis to indicate the expected return for different
 values. When  ⫽ 0, the risk-free return Rf is acceptable (no premium). As  increases, the
premium return requirement grows. Beta values are published periodically for most stock-issuing
corporations. Once complete, this estimated cost of common stock equity capital can be included
in the WACC computation in Equation [10.3].
EXAMPLE 10.5
The lead software engineer at SafeSoft, a food industry service corporation, has convinced the
president to develop new software technology for the meat and food safety industry. It is envisioned that processes for prepared meats can be completed more safely and faster using this
automated control software. A common stock issue is a possibility to raise capital if the cost of
equity capital is below 9%. SafeSoft, which has a historical beta value of 1.09, uses CAPM to
determine the premium of its stock compared to other software corporations. The security
market line indicates that a 5% premium above the risk-free rate is desirable. If U.S. Treasury
bills are paying 2%, estimate the cost of common stock capital.
10.5
Effect of Debt-Equity Mix on Investment Risk
Solution
The premium of 5% represents the term Rm ⫺ Rf in Equation [10.8].
Re ⫽ 2.0 ⫹ 1.09(5.0) ⫽ 7.45%
Since this cost is lower than 9%, SafeSoft should issue common stock to finance this new
venture.
In theory, a correctly performed engineering economy study uses a MARR equal to the cost
of the capital committed to the specific alternatives in the study. Of course, such detail is not
known. For a combination of debt and equity capital, the calculated WACC sets the minimum for
the MARR. The most rational approach is to set MARR between the cost of equity capital and
the corporation’s WACC. The risks associated with an alternative should be treated separately
from the MARR determination, as stated earlier. This supports the guideline that the MARR
should not be arbitrarily increased to account for the various types of risk associated with the
cash flow estimates. Unfortunately, the MARR is often set above the WACC because management does want to account for risk by increasing the MARR.
EXAMPLE 10.6
The Engineering Products Division of 4M Corporation has two mutually exclusive alternatives
A and B with ROR values of i*A ⫽ 9.2% and i*B ⫽ 5.9%. The financing scenario is yet unsettled,
but it will be one of the following: plan 1—use all equity funds, which are currently earning
8% for the corporation; plan 2—use funds from the corporate capital pool which is 25% debt
capital costing 14.5% and the remainder from the same equity funds mentioned above. The
cost of debt capital is currently high because the company has narrowly missed its projected
revenue on common stock for the last two quarters, and banks have increased the borrowing
rate for 4M. Make the economic decision on alternative A versus B under each financing
scenario. The MARR is set by the calculated WACC.
Solution
The capital is available for one of the two mutually exclusive alternatives. For plan 1, 100%
equity, the financing is specifically known, so the cost of equity capital is the MARR, that is,
8%. Only alternative A is acceptable; alternative B is not since the estimated return of 5.9%
does not exceed this MARR.
Under financing plan 2, with a D-E mix of 25–75,
WACC ⫽ 0.25(14.5) ⫹ 0.75(8.0) ⫽ 9.625%
Now, neither alternative is acceptable since both ROR values are less than MARR ⫽
WACC ⫽ 9.625%. The selected alternative should be to do nothing, unless one alternative
absolutely must be selected, in which case noneconomic attributes must be considered.
10.5 Effect of Debt-Equity Mix on Investment Risk
The D-E mix was introduced in Section 10.2. As the proportion of debt capital increases, the
calculated cost of capital decreases due to the tax advantages of debt capital.
The leverage offered by larger debt capital percentages increases the riskiness of projects undertaken by the company. When large debts are already present, additional financing using debt (or
equity) sources gets more difficult to justify, and the corporation can be placed in a situation
where it owns a smaller and smaller portion of itself. This is sometimes referred to as a highly
leveraged corporation.
Inability to obtain operating and investment capital means increased difficulty for the company
and its projects. Thus, a reasonable balance between debt and equity financing is important for
275
276
Project Financing and Noneconomic Attributes
Chapter 10
the financial health of a corporation. Example 10.7 illustrates the disadvantages of unbalanced
D-E mixes.
EXAMPLE 10.7
Three auto parts manufacturing companies have the following debt and equity capital amounts
and D-E mixes. Assume all equity capital is in the form of common stock.
Amount of Capital
Company
Debt
($ in Millions)
Equity
($ in Millions)
D-E Mix (%–%)
A
B
C
10
20
40
40
20
10
20−80
50−50
80−20
Assume the annual revenue is $15 million for each one and that after interest on debt is considered, the net incomes are $14.4, $13.4, and $10.0 million, respectively. Compute the return on
common stock for each company, and comment on the return relative to the D-E mixes.
Solution
Divide the net income by the stock (equity) amount to compute the common stock return. In
million dollars,
14.4 ⫽ 0.36 (36%)
ReturnA ⫽ ——
40
13.4
ReturnB ⫽ —— ⫽ 0.67 (67%)
20
10.0
ReturnC ⫽ —— ⫽ 1.00 (100%)
10
As expected, the return is by far the largest for highly leveraged C, where only 20% of the
company is in the hands of the ownership. The return is excellent, but the risk associated with
this firm is high compared to A, where the D-E mix is only 20% debt.
The use of large percentages of debt financing greatly increases the risk taken by lenders and
stock owners. Long-term confidence in the corporation diminishes, no matter how large the
short-term return on stock.
The leverage of large D-E mixes does increase the return on equity capital, as shown in previous examples; but it can also work against the owners and investors. A decrease in asset value
will more negatively affect a highly debt-leveraged company compared to one with small leveraging. Example 10.8 illustrates this fact.
EXAMPLE 10.8
During the last several years, the U.S. airline industry has had financial problems, in part due
to high fuel costs, fewer customers, security problems, government regulations, aging aircraft,
and union dissatisfaction. As a consequence, the D-E mixes of the so-called traditional companies (American, United, Delta, and others) have become larger on the debt side than is historically acceptable. Meanwhile, the D-E mixes of so-called low-cost airlines (Southwest, JetBlue,
and others) have suffered, but not to the same degree. In an effort to reduce costs, assume that
three airlines joined forces to cooperate on a range of services (baggage handling, onboard
food preparation, ticket services, and software development) by forming a new company called
FullServe, Inc. This required $5 billion ($5 B) up-front funding from each airline.
Table 10–1 summarizes the D-E mixes and the total equity capitalization for each airline
after its share of $5 B was removed from available equity funds. The percentage of the $5 B
obtained as debt capital was the same proportion as the debt in the company’s D-E mix. For
Effect of Debt-Equity Mix on Investment Risk
10.5
TABLE 10–1 Debt and Equity Statistics, Example 10.8
Airline
Company
Corporate D-E
Mix, %
Amount Borrowed,
$B
Equity Capital
Available, $ B
National
Global
PanAm
30–70
65–35
91–9
1.50
3.25
4.55
5.0
3.7
6.7
example, National had 30% of its capitalization in debt capital; therefore, 30% of $5 B was
borrowed, and 70% was provided from National’s equity fund.
Unfortunately, after a short time, it was clear that the three-way collaborative effort was a
complete failure, and FullServe was dissolved and its assets were distributed or sold for a total
of $3.0 billion, only 20% of its original value. A total of $1.0 billion in equity capital was returned to each airline. The commercial banks that provided the original loans then required that
the airlines each pay back the entire borrowed amount now, since FullServe was dissolved and
no profit from the venture could be realized. Assuming the loan and equity amounts are the
same as shown in Table 10–1, determine the resulting equity capital situation for each airline
after it pays off the loan from its own equity funds. Also, describe one impact on each company
as a result of this expensive and risky venture that failed.
Solution
Determine the level of post-FullServe equity capital using the following relation, in $ billions.
Equity capital ⫽ pre-FullServe level ⫹ returned capital ⫺ loan repayment
National:
Global:
PanAm:
Equity capital ⫽ 5.0 ⫹ 1.0 ⫺ 1.50 ⫽ $4.50
Equity capital ⫽ 3.7 ⫹ 1.0 ⫺ 3.25 ⫽ $1.45
Equity capital ⫽ 6.7 ⫹ 1.0 ⫺ 4.55 ⫽ $3.15
Comparing the equity capital levels (Table 10–1) with the levels above indicates that the
FullServe effort reduced equity amounts by 10% for National, 60% for Global, and 53%
for PanAm. The debt capital to fund the failed FullServe effort has affected National
airlines the least, in large part due to its low D-E mix of 30%–70%. However, Global
and PanAm are in much worse shape financially, and they now must maintain business
with a significantly lower ownership level and much reduced ability to obtain future
capital—debt or equity.
The same principles discussed above for corporations are applicable to individuals. The person who is highly leveraged has large debts in terms of credit card balances, personal loans, and
house mortgages. As an example, assume two engineers each have a take-home amount of
$40,000 after all income tax, social security, and insurance premiums are deducted from their
annual salaries. Further, assume that the cost of the debt (money borrowed via credit cards and
loans) averages 15% per year and that the total debt is being repaid in equal amounts over
20 years. If Jamal has a total debt of $25,000 and Barry owes $100,000, the remaining amount of
the annual take-home pay may be calculated as follows:
Amount Paid, $ per Year
Person
Total
Debt, $
Cost of
Debt at 15%, $
Repayment of Debt, $
Amount Remaining
from $40,000, $
Jamal
Barry
25,000
100,000
3,750
15,000
1,250
5,000
35,000
20,000
Jamal has 87.5% of his base available while Barry has only 50% available.
277
278
Chapter 10
Project Financing and Noneconomic Attributes
10.6 Multiple Attribute Analysis: Identification
and Importance of Each Attribute
In Chapter 1 the role and scope of engineering economy in decision making were outlined. The
decision-making process explained in that chapter (Figure 1–1) included the seven steps listed on
the right side of Figure 10–5. Step 4 is to identify the one or multiple attributes (criteria) upon
which the selection will be based. In all prior evaluations, only one attribute—the economic
one—has been identified and used to select the best alternative. The criterion has been the maximization of the equivalent value of PW, AW, FW, ROR, B/C ratio, or the CER for service projects. As we are all aware, most evaluations do and should take into account multiple attributes in
decision making. These are the factors labeled as noneconomic in step 5 of Figure 1–1. However,
these noneconomic dimensions tend to be intangible and often difficult, if not impossible, to
quantify with economic and other scales. Nonetheless, among the many attributes that can be
identified, there are key ones that must be considered in earnest before the alternative selection
process is complete. This section and the next describe some of the techniques that accommodate
multiple attributes in an engineering study.
Multiple attributes enter into the decision-making process in many studies. Public and service sector projects are excellent examples of multiple-attribute problem solving. For example, the proposal
to construct a dam to form a lake in a low-lying area or to widen the catch basin of a river usually has
several purposes, such as flood control; drinking water; industrial use; commercial development;
recreation; nature conservation for fish, plants, and birds; and possibly other less obvious purposes.
High levels of complexity are introduced into the selection process by the multiple attributes thought
to be important in selecting an alternative for the dam’s location, design, environmental impact, etc.
The left side of Figure 10–5 expands steps 4 and 5 to consider multiple attributes. The discussion below concentrates on the expanded step 4 and the next section focuses on the evaluation
measure and alternative selection of step 5.
4-1 Attribute Identification Attributes to be considered in the evaluation methodology can
be identified and defined by several methods, some much better than others depending upon the
situation surrounding the study itself. To seek input from individuals other than the analyst is
important; it helps focus the study on key attributes. The following is an incomplete listing of
ways in which key attributes are identified.
•
•
•
•
Comparison with similar studies that include multiple attributes
Input from experts with relevant past experience
Surveys of constituencies (customers, employees, managers) impacted by the alternatives
Small group discussions using approaches such as focus groups, brainstorming, or nominal
group technique
• Delphi method, which is a progressive procedure to develop reasoned consensus from different
perspectives and opinions
Figure 10–5
Consider multiple attributes
Expansion of the decisionmaking process to include
multiple attributes.
Emphasis on one attribute
1. Understand the problem; define the objective.
2. Collect relevant information; define alternatives.
3. Make estimates.
4-1. Identify the attributes for decision
making.
4-2. Determine the relative
importance (weights) of attributes.
4-3. For each alternative, determine
each attribute’s value rating.
5. Evaluate each alternative using
a multiple-attribute technique.
Use sensitivity analysis for key
attributes.
4. Identify the selection criteria (one or more
attributes).
5. Evaluate each alternative; use sensitivity and
risk analysis.
6. Select the best alternative.
7. Implement the solution and monitor results.
10.6
279
Multiple Attribute Analysis: Identification and Importance of Each Attribute
As an illustration, assume that Delta Airlines has decided to purchase five new Boeing 787s
for overseas flights, primarily between the North American west coast and Asian cities, principally Hong Kong, Tokyo, and Singapore. There are approximately 8000 options for each plane
that must be decided upon by Delta’s engineering, purchasing, maintenance, and marketing personnel before the order to Boeing is placed. Options range in scope from the material and color
of the plane’s interior to the type of latching devices used on the engine cowlings, and in function
from maximum engine thrust to pilot instrument design. An economic study based on the equivalent AW of the estimated passenger income per trip has determined that 150 of these options are
clearly advantageous. But other noneconomic attributes are to be considered before some of the
more expensive options are specified. A Delphi study was performed using input from 25 individuals. Concurrently, option choices for another, unidentified airline’s recent order were shared with
Delta personnel. From these two studies it was determined that there are 10 key attributes for options selection. Four of the most important attributes are
• Repair time: mean time to repair or replace (MTTR) if the option is or affects a flight-critical
component.
• Safety: mean time to failure (MTTF) of flight-critical components.
• Economic: estimated extra revenue for the option. (Basically, this is the attribute evaluated by
the economic study already performed.)
• Crewmember needs: some measure of the necessity and/or benefits of the option as judged by
representative crewmembers—pilots and attendants.
The economic attribute of extra revenue may be considered an indirect measure of customer satisfaction, one that is more quantitative than customer opinion/satisfaction survey results. Of course, there
are many other attributes that can be, and are, used. However, the point is that the economic study
may directly address only one or a few of the key attributes vital to alternative decision making.
An attribute routinely identified by individuals and groups is risk.
Risk is a possible variation in a parameter from an expected, desired, or predicted value that
may be detrimental to observing the intended outcome(s) of the product, process, or system.
It represents the absence of or deviation from certainty. Risk is present when there are two or
more observable values of a parameter and it is possible to assume or estimate the chance that
each value may occur.
Actually, risk is not a stand-alone attribute, because it is a part of every attribute in one form or
another. Considerations of variation, probabilistic estimates, etc., in the decision-making process
are treated in Chapters 18 and 19. Formalized sensitivity analysis, expected values, simulation,
and decision trees are some of the techniques useful in handling risk.
4-2 Importance (Weights) for the Attributes Determination of the extent of importance
for each attribute i results in a weight Wi that is incorporated into the final evaluation measure. The
weight, a number between 0 and 1, is based upon the experienced opinion of one individual or a
group of persons familiar with the attributes, and possibly the alternatives. If a group is utilized to
determine the weights, there must be consensus among the members for each weight. Otherwise,
some averaging technique must be applied to arrive at one weight value for each attribute.
Table 10–2 is a tabular layout of attributes and alternatives used to perform a multiple attribute
evaluation. Weights Wi for each attribute are entered on the left side. The remainder of the table
is discussed as we proceed through steps 4 and 5 of the expanded decision-making process.
Attribute weights are usually normalized such that their sum over all the alternatives is 1.0. This
normalizing implies that each attribute’s importance score is divided by the sum S over all attributes.
Expressed in formula form, these two properties of weights for attribute i (i ⫽ 1, 2, . . . , m) are
m
Normalized weights:
兺 W ⫽ 1.0
i
[10.9]
i⫽1
importance scorei
importance score
ⴝ ————————i [10.10]
Weight calculation: Wi ⴝ —————————
m
S
importance scorei
兺
i ⴝ1
Risk
280
Project Financing and Noneconomic Attributes
Chapter 10
TABLE 10–2
Tabular Layout of Attributes and Alternatives
Used for Multiple Attribute Evaluation
Alternatives
Attributes
Weights
1
2
3
W1
W2
W3
m
Wm
1
2
3
...
n
Value ratings Vij
Of the many procedures developed to assign weights to an attribute, an analyst is likely to rely
upon one that is relatively simple, such as equal weighting, rank order, or weighted rank order.
Pairwise comparison is another technique. Each is briefly presented below.
Equal Weighting All attributes are considered to be of approximately the same importance, or there is no rationale to distinguish the more important from the less important attribute. This is the default approach. Each weight in Table 10–2 will be 1兾m, according to Equation [10.10]. Alternatively, the normalizing can be omitted, in which case each weight is 1 and
their sum is m. In this case, the final evaluation measure for an alternative will be the sum over
all attributes.
Rank Order The m attributes are ordered (ranked) by increasing importance with a score of 1
assigned to the least important and m assigned to the most important. By Equation [10.10], the
weights follow the pattern 1兾S, 2兾S, . . . , m兾S. With this method, the difference in weights between attributes of increasing importance is constant.
Weighted Rank Order The m attributes are again placed in the order of increasing importance. However, now differentiation between attributes is possible. The most important attribute is assigned a score, usually 100, and all other attributes are scored relative to it between 100
and 0. Now, define the score for each attribute as si, and Equation [10.10] takes the form
si
Wi ⴝ ——
m
兺
[10.11]
si
iⴝ1
This is a very practical method to determine weights because one or more attributes can
be heavily weighted if they are significantly more important than the remaining ones, and
Equation [10.11] automatically normalizes the weights. For example, suppose the four key
attributes in the previous aircraft purchase example are ordered: safety, repair time, crewmember needs, and economic. If repair time is only one-half as important as safety, and the
last two attributes are each one-half as important as repair time, the scores and weights are
as follows.
Attribute
Score
Weights
Safety
Repair time
Crewmember needs
Economic
100
50
25
25
100/200 ⫽ 0.50
50/200 ⫽ 0.25
25/200 ⫽ 0.125
25/200 ⫽ 0.125
Sum of scores and weights
200
1.000
Pairwise Comparison Each attribute is compared to each other attribute in a pairwise fashion using a rating scale that indicates the importance of one attribute over the other. Assume the
10.6
TABLE 10–3
Attribute i
Cost
Constructability
Environment
Sum of scores
Weight Wi
Multiple Attribute Analysis: Identification and Importance of Each Attribute
Pairwise Comparison of Three Attributes to Determine Weights
1 ⴝ Cost
2 ⴝ Constructability
3 ⴝ Environment
—
2
1
0
—
1
1
1
—
3
0.500
1
0.167
2
0.333
three criteria (attributes) upon which a public works project decision is based are cost, constructability, and environmental impact. Define the importance comparison scale as follows:
0 if attribute is less important than one compared to
1 if attribute is equally important as one compared to
2 if attribute is more important than one compared to
Set up a table listing attributes across the top and down the side, and perform the pairwise comparison for each column attribute with each row attribute. Table 10–3 presents a comparison with
importance scores included. The arrow to the right of the table indicates the direction of comparison, i.e., column with row attribute. For example, cost is judged more important than constructability, thus a score of 2. The complement score of 0 is placed in the reverse comparison of
constructability with cost. The weights are determined by normalizing the scores using Equation [10.11], where the sum for each column is si. For the first attribute, cost i ⫽ 1.
s1 ⫽ 3
兺s ⫽ 3 ⫹ 1 ⫹ 2 ⫽ 6
i
Cost weight W1 ⫽ 3兾6 ⫽ 0.500
Similarly, the other weights are W2 ⫽ 1兾6 ⫽ 0.167 and W3 ⫽ 2兾6 ⫽ 0.333.
There are other attribute weighting techniques, especially for group processes, such as utility
functions, and the Dunn-Rankin procedure. These become increasingly sophisticated, but they are
able to provide an advantage that these simple methods do not afford the analyst: consistency of
ranks and scores between attributes and between individuals. If this consistency is important in that
several decision makers with diverse opinions about attribute importance are involved in a study, a
more sophisticated technique may be warranted. There is substantial literature on this topic.
4-3 Value Rating of Each Alternative by Attribute This is the final step prior to calculating the evaluation measure. Each alternative j is awarded a value rating Vij for each attribute i.
These are the entries within the cells in Table 10–2. The ratings are appraisals by decision makers
of how well an alternative will perform as each attribute is considered.
The scale for the value rating can vary depending upon what is easiest to understand for those
who do the valuation. A scale of 0 to 100 can be used for attribute importance scoring. However,
the most popular is a scale of 4 or 5 gradations about the perceived ability of an alternative to
accomplish the intent of the attribute. This is called a Likert scale, which can have descriptions
for the gradations (e.g., very poor, poor, good, very good), or numbers assigned between 0 and
10, or ⫺1 to ⫹1, or ⫺2 to ⫹2. The last two scales can give a negative impact to the evaluation
measure for poor alternatives. An example numerical scale of 0 to 10 is as follows:
If You Value the
Alternative as
Very poor
Poor
Good
Very good
Give It a Rating
between the Numbers
0–2
3–5
6–8
9–10
It is preferable to have a Likert scale with four choices (an even number) so that the central tendency of “fair” is not overrated.
281
282
Project Financing and Noneconomic Attributes
Chapter 10
TABLE 10–4
Completed Layout for Four Attributes and Three
Alternatives for Multiple Attribute Evaluation
Alternatives
Attributes
Weights
1
2
3
Safety
Repair
Crew needs
Economic
0.50
0.25
0.125
0.125
6
9
5
5
4
3
6
9
8
1
6
7
If we now build upon the aircraft purchase illustration to include value ratings, the cells are
filled with ratings awarded by a decision maker. Table 10–4 includes example ratings Vij and the
weights Wi determined above. Initially, there will be one such table for each decision maker. Prior
to calculating the final evaluation measure Rj, the ratings can be combined in some fashion; or a
different Rj can be calculated using each decision maker’s ratings. Determination of this evaluation measure is discussed below.
10.7 Evaluation Measure for Multiple Attributes
The need for an evaluation measure that accommodates multiple attributes is indicated in step 5
of Figure 10–5. The measure can be one that attempts to retain all of the ratings, values, and
complexity of previous assessments by multiple decision makers, or the measure can reduce
these inputs to a single-dimension measure. This section introduces a single-dimension measure
that is widely accepted.
A single-dimension measure effectively combines the different aspects addressed by the attribute importance weights Wi and the alternative value ratings Vij. The resulting evaluation measure is a formula that calculates an aggregated measure for use in selecting from two or more
alternatives. The approach applied in this process is called the rank-and-rate method.
This reduction process removes much of the complexity of trying to balance the different attributes;
however, it also eliminates much of the robust information captured by the process of ranking attributes for their importance and rating each alternative’s performance against each attribute.
There are additive, multiplicative, and exponential measures, but by far the most commonly
applied is the additive model. The most used additive model is the weighted attribute method,
also called the additive weight technique. The evaluation measure, symbolized by Rj for each
alternative j, is defined as
Rj ⴝ sum of (weight ⴛ value rating)
m
ⴝ
兺W ⴛ V
i
ij
[10.12]
iⴝ1
The Wi numbers are the attribute importance weights, and Vij is the value rating by attribute i for
each alternative j. If the attributes are of equal weight (also called unweighted ), all Wi ⫽ 1兾m, as
determined by Equation [10.10]. This means that Wi can be moved outside of the summation in
the formula for Rj. (If an equal weight of Wi ⫽ 1.0 is used for all attributes, in lieu of 1兾m, then
the Rj value is simply the sum of all ratings for the alternative.)
The selection guideline is as follows:
ME alternative selection
Choose the alternative with the largest Rj value. This measure assumes that increasing
weights Wi mean more important attributes and increasing ratings Vij mean better performance
of an alternative.
Sensitivity analysis for any score, weight, or value rating is used to determine sensitivity of the decision to it. The Chapter 18 case study includes an example of multiple attribute sensitivity analysis.
Chapter Summary
EXAMPLE 10.9
The Island of Niue in the South Pacific Ocean (www.niueisland.com) released a request for proposal (RFP)1 for a new or reconditioned workboat as it upgrades the infrastructure of its port services. The spreadsheet in Figure 10–6, left two columns, presents the attributes and normalized
weights Wi published in the RFP for use in selecting one of the tenders presenting proposals. Four
acceptable proposals were received. The next four columns (C through F ) include value ratings
between 0 and 100 developed by a group of decision makers when the details of each proposal were
evaluated against each attribute. For example, proposal 2 received a perfect score of 100 on delivery
time, but lifetime costs were considered too high (rating of 20) and the price was considered relatively high (rating of 55). Use these weights and ratings to determine which proposal to pursue first.
Solution
Assume an additive weighting model is appropriate and apply the weighted attribute method. Equation [10.12] determines the Rj measure for the four alternatives. As an illustration, for proposal 3,
R3 ⫽ 0.30(95) ⫹ 0.20(60) ⫹ 0.05(90) ⫹ 0.35(85) ⫹ 0.10(100)
⫽ 28.5 ⫹ 12.0 ⫹ 4.5 ⫹ 29.8 ⫹ 10.0
⫽ 84.8
The four totals in Figure 10–6 (columns G through J, row 8) indicate that proposal 3 is the
overall best choice for the attributes and weights published in the RFP.
Comment
Any economic measure can be incorporated into a multiple attribute evaluation using this method.
All measures of worth—PW, AW, ROR, B/C, and C/E—can be included; however, their impact
on the final selection will vary relative to the importance placed on the noneconomic attributes.
Figure 10–6
Attributes, weights, ratings, and evaluation measure for Niue workboat proposals, Example 10.9.
1
Used with permission of Government of Niue, Infrastructure Department, “Request for Proposal: Supply
of Workboat,” released March 30, 2010, www.gov.nu/Documents/workboattender3022.pdf.
CHAPTER SUMMARY
The interest rate at which the MARR is established depends principally upon the cost of capital
and the mix between debt and equity financing. The MARR is strongly influenced by the weighted
average cost of capital (WACC). Risk, profit, and other factors can be considered after the AW,
PW, or ROR analysis is completed and prior to final alternative selection. A high debt-to-equity
mix can significantly increase the riskiness of a project and make further debt financing difficult
to acquire for the corporation.
If multiple attributes, which include more than the economic dimension of a study, are to be
considered in making the alternative decision, first the attributes must be identified and their
relative importance assessed. Then each alternative can be value-rated for each attribute. The
evaluation measure is determined using a model such as the weighted attribute method, where the
measure is calculated by Equation [10.12]. The largest value indicates the best alternative.
283
284
Project Financing and Noneconomic Attributes
Chapter 10
PROBLEMS
Working with MARR
funded, with the total initial investment not to
exceed $18 million. Use the results below to determine the opportunity cost for each measure.
10.1 List at least three factors that affect the MARR,
and discuss how each one affects it.
Ranking by ROR
10.2 State whether each of the following involves debt
financing or equity financing.
(a) $10,000 taken from one partner’s savings account to pay for equipment repair
(b) Issuance of preferred stock worth $1.3 million
(c) Short-term loan of $75,000 from a local bank
(d ) Issuance of $3 million worth of 20-year
bonds
(e) Del Engineering buyback of $8 million of its
own stock using internal funds
10.3 Helical Products, Inc. uses an after-tax MARR of
12% per year. If the company’s effective tax rate
(federal, state, and local taxes) is 40%, determine
the company’s before-tax MARR.
10.4 The owner of a small pipeline construction company is trying to figure out how much he should
bid in his attempt to win his first “big” contract. He
estimates that his cost to complete the project will
be $7.2 million. He wants to bid an amount that
will give him an after-tax rate of return of 15% per
year if he gets the job, but he doesn’t know how
much he should bid on a before-tax basis. He told
you that his effective state tax rate is 7% and his
effective federal tax rate is 22%.
(a) The expression for determining the overall
effective tax rate is
state rate ⫹ (1 ⫺ state rate)(federal rate)
What should his before-tax MARR be in
order for him to make an after-tax MARR
of 15% per year?
(b) How much should he bid on the job?
10.5 A group of private equity investors provided
$16 million to a start-up company involved in
making high-technology detection systems for
drugs and other types of contraband. Immediately
after the investment was made, another investment opportunity came up for which the investors
didn’t have enough capital. That project would
have yielded an estimated rate of return of 29%
per year before taxes. If the group’s effective tax
rate is 32%, what after-tax rate of return would
the forgone project have yielded?
10.6 Five projects were ranked in decreasing order by
two measures—rate of return (ROR) and present
worth (PW)—to determine which ones should be
Project
A
B
C
D
E
Ranking by PW
PW at
Initial
Cumulative
Cumulative
ROR,
15%,
Investment,
Investment,
Investment,
%
$1000
$1000
Project
$1000
Project
$1000
44.5
7,138
12.8 ⫺1,162
20.4
1,051
9.6
⫺863
26.0
936
8,000
15,000
8,000
8,000
5,000
A
E
C
B
D
8,000
13,000
21,000
36,000
44,000
A
C
E
D
B
8,000
16,000
21,000
29,000
44,000
10.7 Tom, the owner of Burger Palace, determined that
his weighted average cost of capital is 8%. He expects a return of 4% per year on all of his investments. A proposal presented by the owner of the
Dairy Choice next door seems quite risky to Tom,
but is an intriguing partnership opportunity. Tom
has determined that the proposal’s “risk factor”
will require an additional 3% per year return for
him to accept it.
(a) Use the recommended approach to determine
the MARR that Tom should use, and explain
how the 3% risk factor is compensated for in
this MARR.
(b) Determine the effective MARR for his business if Tom turns down the proposal.
D-E Mix and WACC
10.8 Electrical generators produce not only electricity,
but also heat from conductor resistance and from
friction losses in bearings. A company that manufactures generator coolers for nuclear and gas turbine power plants undertook a plant expansion
through financing that had a debt-equity mix of
45–55. If $18 million came from mortgages and
bond sales, what was the total amount of the
financing?
10.9 Determine the debt-to-equity mix when Applied
Technology bought out Southwest Semiconductor
using financing as follows: $12 million from mortgages, $5 million from retained earnings, $10 million from cash on hand, and $20 million from
bonds.
10.10 Business and engineering seniors are comparing
methods of financing their college education during their senior year. The business student has
$30,000 in student loans that come due at graduation. Interest is an effective 4% per year. The
engineering senior owes $50,000, 50% from his
parents with no interest due and 50% from a
285
Problems
credit union loan. This latter amount is also due at
graduation with an effective rate of 7% per year.
(a) What is the D-E mix for each student?
(b) If their grandparents pay the loans in full at
graduation, what are the amounts on the
checks they write for each graduate?
(c) When grandparents pay the full amount at
graduation, what percent of the principal
does the interest represent?
dividends at a rate of 5% per year, and the remaining 60% from retained earnings, which currently
earn 9% per year.
Debt capital: 40%, or $15 million, obtained
through two sources—bank loans for $10 million
borrowed at 8% per year, and the remainder in
convertible bonds at an estimated 10% per year
bond dividend rate.
10.11 Two public corporations, First Engineering and
Midwest Development, each show capitalization
of $175 million in their annual reports. The balance sheet for First Engineering indicates a total
debt of $87 million, and that of Midwest Development indicates a net worth of $62 million. Determine the D-E mix for each company.
10.16 A public corporation in which you own common
stock reported a WACC of 11.1% for the year in its
report to stockholders. The common stock that you
own has averaged a total return of 7% per year
over the last 3 years. The annual report also mentions that projects within the corporation are 75%
funded by its own capital. Estimate the company’s
cost of debt capital.
10.12 Forest Products, Inc. invested $50 million. The
company’s overall D-E mix is 60–40.What is the
return on the company’s equity, if the net income is
$5 million on a revenue base of $6 million?
10.13 Determine the weighted average cost of capital for
a company that manufactures miniature triaxial accelerometers for space-restricted applications. The
financing profile, with interest rates, is as follows:
$3 million in stock sales at 15% per year, $4 million in bonds at 9%, and $6 million in retained
earnings at 7% per year.
10.14 Growth Transgenics Enterprises (GTE) is contemplating the purchase of its rival. One of GTE’s genetics engineers got interested in the financing
strategy of the buyout. He learned there are two
plans being considered. Plan 1 requires 50% equity funds from GTE’s retained earnings that currently earn 9% per year, with the balance borrowed
externally at 6%, based on the company’s excellent stock rating. Plan 2 requires only 20% equity
funds with the balance borrowed at a higher rate of
8% per year.
(a) Which plan has the lower average cost of
capital?
(b) If GTE’s owners decide that the current corporate WACC of 8.2% will not be exceeded,
what is the maximum cost of debt capital allowed for each plan? Are these rates higher
or lower than the current estimates?
10.15 Midac Corporation wants to arrange for $50 million in capital to finance the manufacturing of a
new consumer product. The current plan is 60%
equity capital and 40% debt financing. Calculate
the WACC for the following scenario:
Equity capital: 60%, or $35 million, via common
stock sales for 40% of this amount that will pay
10.17 BASF will invest $14 million this year to upgrade
its ethylene glycol processes. This chemical is
used to produce polyester resins to manufacture
products varying from construction materials to
aircraft, and from luggage to home appliances. Equity capital costs 14.5% per year and will supply
65% of the capital funds. Debt capital costs 10%
per year before taxes. The effective tax rate for
BASF is 36%.
(a) Determine the amount of annual revenue
after taxes that is consumed in covering the
interest on the project’s initial cost.
(b) If the corporation does not want to use 65%
of its own funds, the financing plan may include 75% debt capital. Determine the
amount of annual revenue needed to cover
the interest with this plan, and explain the effect it may have on the corporation’s ability
to borrow in the future.
10.18 A couple planning for their child’s college education can fund part of or all the expected $100,000
tuition cost from their own funds (through an education IRA) or borrow all or part of it. The average
return for their own funds is 7% per year, but the
loan is expected to have a higher interest rate as
the loan amount increases. Use a spreadsheet to
generate a plot of the WACC curve with the estimated loan interest rates below and determine the
best D-E mix for the couple.
Loan Amount, $
Interest Rate, % per year
10,000
30,000
50,000
60,000
75,000
100,000
5.0
6.0
8.0
9.0
11.0
13.0
286
Project Financing and Noneconomic Attributes
Chapter 10
10.19 Over the last few years, Carol’s Fashion Store, a
statewide franchise, has experienced the D-E
mixes and costs of debt and equity capital on several projects summarized below.
(a) Plot debt, equity, and weighted average cost
of capital.
(b) Determine what mix of debt and equity capital provided the lowest WACC.
Debt capital
Project
Percent
A
B
C
D
E
F
G
100%
70
65
50
35
20
Rate
14.5%
13.0
12.0
11.5
9.9
12.4
Equity capital
Percent
30%
35
50
65
80
100
Rate
7.8%
7.8
7.9
9.8
12.5
12.5
10.20 For Problem 10.19, use a spreadsheet to (a) determine the best D-E mix and (b) determine the best
D-E mix if the cost of debt capital increases by
10% per year, for example, 13.0% increases
to 14.3%.
Cost of Debt Capital
10.21 The cash flow plan associated with a debt financing transaction allowed a company to receive
$2,800,000 now in lieu of future interest payments
of $196,000 per year for 10 years plus a lump sum
of $2,800,000 in year 10. If the company’s effective tax rate is 33%, determine the company’s cost
of debt capital (a) before taxes and (b) after taxes.
10.22 A company that makes several different types of
skateboards, Jennings Outdoors, incurred interest
expenses of $1,200,000 per year from various
types of debt financing. The company borrowed
$19,000,000 in year 0 and repaid the principal of
the loans in year 15 in a lump-sum payment of
$20,000,000. If the company’s effective tax rate is
29%, what was the company’s cost of debt capital
(a) before taxes and (b) after taxes?
10.23 Molex Inc., a manufacturer of cable assemblies for
polycrystalline photovoltaic solar modules, requires $3.1 million in debt capital. The company
plans to sell 15-year bonds that carry a dividend of
6% per year, payable semiannually. Molex has an
effective tax rate of 32% per year. Determine
(a) the nominal annual after-tax cost of debt capital and (b) the effective annual after-tax cost of
debt capital.
10.24 Tri-States Gas Producers expects to borrow
$800,000 for field engineering improvements. Two
methods of debt financing are possible—borrow it
all from a bank or issue debenture bonds. The company will pay an effective 8% per year to the bank
for 8 years. The principal on the loan will be reduced
uniformly over the 8 years, with the remainder of
each annual payment going toward interest. The
bond issue will be for 800 10-year bonds of $1000
each that require a 6% per year dividend payment.
(a) Which method of financing is cheaper after
an effective tax rate of 40% is considered?
(b) Which is the cheaper method using a beforetax analysis?
10.25 The Sullivan Family Partnership plans to purchase
a refurbished condo in their hometown for investment purposes. The negotiated $200,000 purchase
price will be financed with 20% of savings (retained earnings) which consistently makes 6.5%
per year after all relevant income taxes are paid.
Eighty percent will be borrowed at a before-tax
rate of 9% per year for 15 years with the principal
repaid in equal annual installments. If the effective
tax rate is 22% per year, based only on these data,
answer the following.
(a) What is the partnership’s annual loan payment for each of the 15 years?
(b) What is the net present worth difference between the $200,000 now and the PW of the
cost of the 80–20 D-E mix series of cash
flows necessary to finance the purchase?
What does this PW value mean?
(c) What is the after-tax WACC for this
purchase?
Cost of Equity Capital
10.26 Determine the cost of equity capital to Hy-Lok
USA if the company sells 500,000 shares of its
preferred stock at a 5% discount from its price of
$130. The stock carries a $10 per year dividend.
10.27 The initial public offering price for the common
stock of SW Refining is $23 per share, and it will
pay a first-year dividend of $0.92 per share. If the
appreciation rate in dividends is anticipated to be
3.2% per year, determine the cost of equity capital
for the stock offering.
10.28 The cost of debt capital is lower after taxes than
before taxes. The cost of equity capital is more difficult to estimate using the dividend method or the
CAPM model, for example, yet the after-tax and
before-tax cost of equity capital is the same. Why
are the after-tax rates not the same for both types
of financing?
10.29 H2W Technologies is considering raising capital
to expand its offerings of 2-phase and 4-phase
Problems
linear stepper motors. The beta value for its stock
is 1.41. Use the capital asset pricing model and a
3.8% premium above the risk-free return to determine the cost of equity capital if the risk-free return is 3.2%.
10.30 Management at Hirschman Engineering has asked
you to determine the cost of equity capital based
on the company’s common stock. The company
wants you to use two methods: the dividend
method and the CAPM. Last year, the first year for
dividends, the stock paid $0.75 per share on the
average of $11.50 on the New York Stock Exchange. Management hopes to grow the dividend
rate at 3% per year. Hirschman Engineering stock
has a volatility that is higher than the norm at 1.3.
If safe investments are returning 5.5% and the 3%
growth on common stocks is also the premium
above the risk-free investments that Hirschman
Engineering plans to pay, calculate the cost of equity capital using the two methods.
10.31 Common stocks issued by Meggitt Sensing Systems paid stockholders $0.93 per share on an average price of $18.80 last year. The company expects
to grow the dividend rate at a maximum of 1.5%
per year. The stock volatility is 1.19, and other
stocks in the same industry are paying an average
of 4.95% per year dividend. U.S Treasury bills are
returning 4.5%. Determine the company’s cost of
equity capital last year using (a) the dividend
method and (b) the CAPM.
10.32 Last year a Japanese engineering materials corporation, Yamachi Inc., purchased some U.S. Treasury bonds that return an average of 4% per year.
Now, Euro bonds are being purchased with a realized average return of 3.9% per year. The volatility
factor of Yamachi stock last year was 1.10 and has
increased this year to 1.18. Other publicly traded
stocks in this same business are paying an average
of 5.1% dividends per year. Determine the cost of
equity capital for each year, and explain why the
increase or decrease seems to have occurred.
10.33 The engineering manager at FXO Plastics wants to
complete an alternative evaluation study. She
asked the finance manager for the corporate
MARR. The finance manager gave her some data
on the project and stated that all projects must
clear their average (pooled) cost by at least 4%.
Funds Source
Amount, $ Average Cost, %
Retained earnings
Stock sales
Long-term loans
4,000,000
6,000,000
5,000,000
7.4
4.8
9.8
(a)
(b)
287
Use the data to determine the minimum
MARR.
The study is after taxes and part (a) provided
the before-tax MARR. Determine the correct
MARR to use if the tax rate was 32% last year
and the finance manager meant that the 4%
above the cost is for after-tax evaluations.
Different D-E Mixes
10.34 Why is it financially unhealthy for an individual to
maintain a large percentage of debt financing over
a long time, that is, to be highly leveraged?
10.35 In a leveraged buyout of one company by another,
the purchasing company usually obtains borrowed
money and inserts as little of its own equity as possible into the purchase. Explain some circumstances under which such a buyout may put the
purchasing company at economic risk.
10.36 Grainger and Company has an opportunity to invest $500,000 in a new line of direct-drive rotary
screw compressors. Financing will be equally split
between common stock ($250,000) and a loan
with an 8% after-tax interest rate. The estimated
annual NCF after taxes is $48,000 for the next
7 years. The effective tax rate is 50%. Grainger
uses the capital asset pricing model for evaluation
of its common stock. Recent analysis shows that it
has a volatility rating of 0.95 and is paying a premium of 5% above a safe return on its common
stock. Nationally, the safest investment is currently paying 3% per year. Is the investment financially attractive if Grainger uses as the MARR its
(a) equity cost of capital and (b) WACC?
10.37 Fairmont Industries primarily relies on 100% equity financing to fund projects. A good opportunity
is available that will require $250,000 in capital.
The Fairmont owner can supply the money from
personal investments that currently earn an average of 7.5% per year. The annual net cash flow
from the project is estimated at $30,000 for the
next 15 years. Alternatively, 60% of the required
amount can be borrowed for 15 years at 7% per
year. If the MARR is the WACC, determine which
plan, if either, should be undertaken. This is a
before-tax analysis.
10.38 Omega Engineering Inc. has an opportunity to invest $10,000,000 in a new engineering remote
control system for offshore drilling platforms. Financing will be split between common stock sales
($5,000,000) and a loan with an 8% per year interest rate. Omega’s share of the annual net cash flow
is estimated to be $1.35 million for each of the
next 5 years. Omega is about to initiate CAPM as
288
Project Financing and Noneconomic Attributes
Chapter 10
its common stock evaluation model. Recent analysis shows that it has a volatility rating of 1.22 and
is paying a premium of 5% on its common stock
dividend. The U.S. Treasury bills are currently
paying 4% per year. Is the venture financially attractive if the MARR equals (a) the cost of equity
capital and (b) the WACC?
10.39 A new annular die process is to be installed for extruding pipes, tubes, and tubular films. The phase I
installed price for the dies and machinery is
$2,000,000. The manufacturer has not decided
how to finance the system. The WACC over the
last 5 years has averaged 10% per year.
(a) Two financing alternatives have been defined. The first requires an investment of
40% equity funds at 9% and a loan for the
balance at an interest rate of 10% per year.
The second alternative requires only 25% equity funds and the balance borrowed at
10.5% per year. Which approach will result
in the smaller average cost of capital?
(b) Yesterday, the corporate finance committee
decided that the WACC for all new projects
must not exceed the 5-year historical average
of 10% per year. With this restriction, what is
the maximum loan interest rate that can be incurred for each of the financing alternatives?
10.40 Shadowland, a manufacturer of air-freightable pet
crates, has identified two projects that, though having a relatively high risk, are expected to move the
company into new revenue markets. Utilize a
spreadsheet solution to (a) select any combination
of the projects if the MARR is equal to the aftertax WACC and (b) determine if the same projects
should be selected if the risk factors are enough to
require an additional 2% per year for the investment to be made.
Project
Wildlife (W)
Reptiles (R)
Initial
After-Tax Cash
Investment, $ Flow, $/Year Life, Years
⫺250,000
⫺125,000
48,000
30,000
10
5
Financing will be developed using a D-E mix of
60–40 with equity funds costing 7.5% per year.
Debt financing will be developed from $10,000,
5% per year, paid quarterly, 10-year bonds. The effective tax rate is 30% per year.
10.41 Two friends each invested $20,000 of their own
(equity) funds. Stan, being more conservative, purchased utility and manufacturing corporation
stocks. Theresa, being a risk taker, leveraged the
$20,000 and purchased a $100,000 condo for rental
property. Considering no taxes, dividends, or revenues, analyze these two purchases by doing the following for one year after the funds were invested.
(a) Determine the year-end values of their equity
funds if there was a 10% increase in the
value of the stocks and the condo.
(b) Determine the year-end values of their equity
funds if there was a 10% decrease in the
value of the stocks and the condo.
(c) Use your results to explain why leverage can
be financially risky.
Multiple, Noneconomic Attributes
10.42 In multiple attribute analysis, if three different alternatives are to be evaluated on the basis of eight
attributes that are considered of equal importance,
what is the weight of each attribute?
10.43 A consulting engineer asked a company manager to
assign importance values (0 to 100) to five attributes that will be included in an alternative evaluation process. Determine the weight of each attribute
using the importance scores.
Attribute
1. Safety
2. Cost
3. Impact
4. Environmental
5. Acceptability
Importance Score
60
40
80
30
20
10.44 Ten attributes were rank-ordered in terms of increasing importance and were identified as A, B,
C, . . . , and J. Determine the weight of (a) attribute C and (b) attribute J.
10.45 A team of three people submitted the following
statements about the attributes to be used in a
weighted attribute evaluation. Use the statements
to determine the normalized weights if assigned
scores are between 0 and 100.
Attribute
1. Flexibility (F)
2. Safety (S)
3. Uptime (U)
4. Rate of return (R)
Comment
The most important factor
70% as important as uptime
One-half as important as flexibility
Twice as important as safety
10.46 Different types and capacities of crawler hoes are
being considered for use in a major excavation on
a pipe-laying project. Several supervisors who
served on similar projects in the past have identified some of the attributes and their view of relative importance. For the information that follows,
determine the weighted rank order, using a 0-to-10
scale and the normalized weights.
289
Additional Problems and FE Exam Review Questions
Attribute
Comment
1. Truck versus hoe height 90% as important as trenching speed
2. Type of topsoil
Only 10% of most important
attribute
3. Type of subsoil
30% as important as trenching speed
4. Hoe cycle time
Twice as important as type of
subsoil
5. Hoe trenching speed
Most important attribute
6. Pipe laying speed
80% as important as hoe cycle time
10.47 John, who works at Swatch, has decided to use the
weighted attribute method to compare three systems for manufacturing a watchband. The vice
president and her assistant VP have evaluated each
of three attributes in terms of importance to them,
and John has placed an evaluation from 0 to 100
on each alternative for the three attributes. John’s
ratings for each alternative are as follows:
Alternative
Attribute
Economic return ⬎ MARR
High throughput
Low scrap rate
1
2
3
50
100
100
70
60
40
100
30
50
Use the weights below to evaluate the alternatives.
Are the results the same for both persons’ weights?
Why?
Importance Score
VP
Assistant VP
Economic return ⬎ MARR
High throughput
Low scrap rate
20
80
100
100
80
20
10.48 The Athlete’s Shop has evaluated two proposals
for weight lifting and exercise equipment. A present worth analysis at i ⫽ 15% per year of estimated revenues and costs resulted in PWA ⫽
$440,000 and PWB ⫽ $390,000. In addition to
this economic measure, three more attributes
were independently assigned a relative importance score from 0 to 100 by the shop manager
and the lead trainer.
Importance Score
Attribute
Economics
Durability
Flexibility
Maintainability
Manager
Trainer
80
35
30
20
80
80
100
50
Separately, you have used the four attributes to
rate the two equipment proposals on a scale of 0 to
1.0 as shown in the following table. The economic
attribute was rated using the PW values.
Attribute
Economics
Durability
Flexibility
Maintainability
Proposal A
Proposal B
1.00
0.35
1.00
0.25
0.90
1.00
0.90
1.00
Select the better proposal using each of the following methods.
(a) Present worth
(b) Weighted evaluations of the shop manager
(c) Weighted evaluations of the lead trainer
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
10.49 The term opportunity cost refers to:
(a) The first cost of an alternative that has been
accepted for funding
(b) The total cost of an alternative that has been
accepted for funding
(c) The rate of return or profit available on the
next-best alternative that had to be forgone
due to lack of capital funds
(d) The cost of an alternative that was not recognized as an alternative that actually represented a good opportunity
10.50 The cost of capital is established on the basis of:
(a) The cost of debt financing
(b) The weighted average of debt and equity
financing
(c)
(d)
The cost of equity financing
The cost of debt financing plus the expected
inflation rate
10.51 All of the following are examples of debt capital
except:
(a) Retained earnings
(b) Long-term bonds
(c) Loan from a local bank
(d) Purchase of equipment using a credit card
10.52 All of the following are examples of equity capital
except:
(a) Sale of preferred stock
(b) Long-term bonds
(c) Company cash on hand
(d) Use of retained earnings
290
Chapter 10
Project Financing and Noneconomic Attributes
10.53 If a public utility expands its capacity to generate
electricity by obtaining $41 million from retained
earnings and $30 million from municipal bond
sales, the utilities’ debt-to-equity mix is closest to:
(a) 58% debt and 42% equity
(b) 73% debt and 27% equity
(c) 27% debt and 73% equity
(d) 42% debt and 58% equity
If Medzyme does not want to exceed its historical
weighted average cost of capital (WACC), and it is
forced to go to a D-E mix of 75–25, the maximum
acceptable cost of equity capital is closest to:
(a) 7.6%
(b) 9.2%
(c) 9.8%
(d) 10.9%
10.54 Gentech, Inc. financed a new product as follows:
$5 million in stock sales at 13.7% per year, $2 million in retained earnings at 8.9% per year, and
$3 million through convertible bonds at 7.8% per
year. The company’s WACC is closest to:
(a) 9% per year
(b) 10% per year
(c) 11% per year
(d) 12% per year
10.57 The importance values (0 to 100) for five attributes
are shown below. The weight to assign to attribute
1 is:
(a) 0.16
(b) 0.20
(c) 0.22
(d) 0.25
10.55 If the after-tax rate of return for a cash flow series
is 11.2% and the corporate effective tax rate is
39%, the approximated before-tax rate of return is
closest to:
(a) 6.8%
(b) 5.4%
(c) 18.4%
(d) 28.7%
10.56 Medzyme Pharmaceuticals has maintained a 50-50
D-E mix for capital investments. Equity capital
has cost 11%; however, debt capital that historically cost 9% has now increased by 20% per year.
Attribute
1
2
3
4
5
Importance Score
55
45
85
30
60
10.58 For eight attributes rank-ordered in terms of increasing importance, the weighting of the sixth attribute is closest to:
(a) 0.17
(b) 0.14
(c) 0.08
(d) 0.03
CASE STUDY
WHICH IS BETTER—DEBT OR EQUITY FINANCING?
Background
Information
Pizza Hut Corporation has decided to enter the catering business in three states within its Southeastern U.S. Division,
using the name Pizza Hut At-Your-Place. To deliver the meals
and serving personnel, it is about to purchase 200 vans with
custom interiors for a total of $1.5 million. Each van is expected to be used for 10 years and have a $1000 salvage
value.
A feasibility study completed last year indicated that the
At-Your-Place business venture could realize an estimated
annual net cash flow of $300,000 before taxes in the three
states. After-tax considerations would have to take into account the effective tax rate of 35% paid by Pizza Hut.
An engineer with Pizza Hut’s Distribution Division has
worked with the corporate finance office to determine how to
best develop the $1.5 million capital needed for the purchase
of vans. There are two viable financing plans.
Plan A is debt financing for 50% of the capital ($750,000)
with the 8% per year compound interest loan repaid over
10 years with uniform year-end payments. (A simplifying assumption that $75,000 of the principal is repaid with each
annual payment can be made.)
Plan B is 100% equity capital raised from the sale of $15
per share common stock. The financial manager informed the
engineer that stock is paying $0.50 per share in dividends and
that this dividend rate has been increasing at an average of
5% each year. This dividend pattern is expected to continue,
based on the current financial environment.
Case Study Exercises
1. What values of MARR should the engineer use to determine the better financing plan?
Case Study
2. The engineer must make a recommendation on the financing plan by the end of the day. He does not know
how to consider all the tax angles for the debt financing
in plan A. However, he does have a handbook that gives
these relations for equity and debt capital about taxes
and cash flows:
Equity capital: no income tax advantages
After-tax net cash flow
⫽ (before-tax net cash flow)(1 ⫺ tax rate)
Debt capital: income tax advantage comes from interest
paid on loans
After-tax net cash flow ⫽ before-tax net cash flow
⫺ loan principal
⫺ loan interest ⫺ taxes
291
Taxes ⫽ (taxable income)(tax rate)
Taxable income ⫽ net cash flow
⫺ loan interest
He decides to forget any other tax consequences and use
this information to prepare a recommendation. Is A or B
the better plan?
3. The division manager would like to know how much
the WACC varies for different D-E mixes, especially
about 15% to 20% on either side of the 50% debt financing option in plan A. Plot the WACC curve and compare
its shape with that of Figure 10–2.
CHAPTER 11
Replacement
and Retention
Decisions
L E A R N I N G
O U T C O M E S
Purpose: Perform a replacement兾retention study between an in-place asset, process, or system and one that could
replace it.
SECTION
TOPIC
LEARNING OUTCOME
11.1
Replacement study basics
•
Explain the fundamental approach and
terminology of replacement analysis.
11.2
Economic service life
•
Determine the ESL that minimizes the total AW
for estimated costs and salvage value.
11.3
Replacement analysis
•
Perform a replacement兾retention study between
a defender and the best challenger.
11.4
Additional considerations
•
Understand the approach to special situations in
a replacement study.
11.5
Study period analysis
•
Perform a replacement兾retention study over a
specified number of years.
11.6
Replacement value
•
Calculate the minimum market (trade-in) value
required to make the challenger economically
attractive.
O
ne of the most common and important issues in industrial practice is that of replacement or retention of an asset, process, or system that is currently installed.
This differs from previous situations where all the alternatives were new. The fundamental question answered by a replacement study (also called a replacement兾retention
study) about a currently installed system is, Should it be replaced now or later? When an
asset is currently in use and its function is needed in the future, it will be replaced at some
time. In reality, a replacement study answers the question of when, not if, to replace.
A replacement study is usually designed to first make the economic decision to retain or
replace now. If the decision is to replace, the study is complete. If the decision is to retain,
the cost estimates and decision can be revisited each year to ensure that the decision to
retain is still economically correct. This chapter explains how to perform the initial-year and
follow-on year replacement studies.
A replacement study is an application of the AW method of comparing unequal-life alternatives, first introduced in Chapter 6. In a replacement study with no specified study period,
the AW values are determined by a technique called the economic service life (ESL) analysis.
If a study period is specified, the replacement study procedure is different from that used
when no study period is set.
If asset depreciation and taxes are to be considered in an after-tax replacement analysis,
Chapters 16 and 17 should be covered before or in conjunction with this chapter. After-tax
replacement analysis is included in Chapter 17.
PE
Keep or Replace the Kiln Case: B&T
Enterprises manufactures and sells highmelting-temperature ceramics and highperformance metals to other corporations.
The products are sold to a wide range of
industries from the nuclear and solar power
industry to sports equipment manufacturers
of specialty golf and tennis gear, where kiln
temperatures up to approximately 1700°C
are needed. For years, B&T has owned and
been very satisfied with Harper International pusher-plate tunnel kilns. Two are in use
currently at plant locations on each coast
of the country; one kiln is 10 years old, and
the second was purchased only 2 years ago
and serves, primarily, the ceramics industry
needs on the west coast. This newer kiln
can reach temperatures of 2800°C.
During the last two or three quarterly
maintenance visits, the Harper team
leader and the head of B&T quality have
discussed the ceramic and metal industry
needs for higher temperatures. In some
cases the temperatures are as high as
3000°C for emerging nitride, boride, and
carbide transition metals that form very
high-melting-temperature oxides. These
may find use in hypersonic vehicles,
engines, plasma arc electrodes, cutting
tools, and high-temperature shielding.
A looming question on the mind of the
senior management and financial officers
of B&T revolves around the need to seriously consider a new graphite hearth kiln,
which can meet higher temperature and
other needs of the current and projected
customer base. This unit will have lower
operating costs and significantly greater
furnace efficiency in heat time, transit,
and other crucial parameters. Since virtually all of this business is on the west coast,
the graphite hearth kiln would replace the
newer of the two kilns currently in use.
For identification, let
PT identify the currently installed
pusher-plate tunnel kiln (defender)
GH identify the proposed new graphite hearth kiln (challenger)
Relevant estimates follow in $ millions
for monetary units.
PT
GH
First cost,
$M
$25; 2 years
ago
AOC, $ M
per year
year 1: $5.2; starts at $3.4,
year 2: $6.4
increases
10%兾year
Life, years
6 (remaining)
Heating
element,
$M
—
$38; with no
trade-in
12 (estimated)
$2.0 every
6 years
This case is used in the following topics
of this chapter:
Economic service life (Section 11.2)
Replacement study (Section 11.3)
Replacement study with study period
(Section 11.5)
Replacement value (Section 11.6)
Problems 11.18 and 11.41
294
Chapter 11
Replacement and Retention Decisions
11.1 Basics of a Replacement Study
The need for a replacement study can develop from several sources:
Reduced performance. Because of physical deterioration, the ability to perform at an
expected level of reliability (being available and performing correctly when needed) or productivity (performing at a given level of quality and quantity) is not present. This usually
results in increased costs of operation, higher scrap and rework costs, lost sales, reduced quality, diminished safety, and larger maintenance expenses.
Altered requirements. New requirements of accuracy, speed, or other specifications cannot
be met by the existing equipment or system. Often the choice is between complete replacement or enhancement through retrofitting or augmentation.
Obsolescence. International competition and rapidly changing technology make currentlyused systems and assets perform acceptably but less productively than equipment coming
available. The ever-decreasing development cycle time to bring new products to market is
often the reason for premature replacement studies, that is, studies performed before the
estimated useful or economic life is reached.
Replacement studies use some terminology that is closely related to terms in previous
chapters.
Defender and challenger are the names for two mutually exclusive alternatives. The defender is the currently installed asset, and the challenger is the potential replacement. A
replacement study compares these two alternatives. The challenger is the “best” challenger
because it has been selected as the best one to possibly replace the defender. (This is the
same terminology used earlier for incremental ROR and B兾C analysis, but both alternatives
were new).
Salvage兾market value
Market value is the current value of the installed asset if it were sold or traded on the
open market. Also called trade-in value, this estimate is obtained from professional appraisers, resellers, or liquidators familiar with the industry. As in previous chapters, salvage value is the estimated value at the end of the expected life. In replacement analysis,
the salvage value at the end of one year is used as the market value at the beginning of the
next year.
AW values are used as the primary economic measure of comparison between the defender
and challenger. The term equivalent uniform annual cost (EUAC) may be used in lieu of
AW, because often only costs are included in the evaluation; revenues generated by the
defender or challenger are assumed to be equal. (Since EUAC calculations are exactly the
same as for AW, we use the term AW.) Therefore, all values will be negative when only
costs are involved. Salvage or market value is an exception; it is a cash inflow and carries
a plus sign.
Economic service life
Economic service life (ESL) for an alternative is the number of years at which the lowest AW
of cost occurs. The equivalency calculations to determine ESL establish the life n for the best
challenger and the lowest cost life for the defender in a replacement study. The next section
explains how to find the ESL.
Defender first cost is the initial investment amount P used for the defender. The current
market value (MV) is the correct estimate to use for P for the defender in a replacement
study. The estimated salvage value at the end of one year becomes the market value at the
beginning of the next year, provided the estimates remain correct as the years pass. It is
incorrect to use the following as MV for the defender first cost: trade-in value that does
not represent a fair market value, or the depreciated book value taken from accounting
records. If the defender must be upgraded or augmented to make it equivalent to the challenger (in speed, capacity, etc.), this cost is added to the MV to obtain the estimated
defender first cost.
Challenger first cost is the amount of capital that must be recovered (amortized) when
replacing a defender with a challenger. This amount is almost always equal to P, the first cost
of the challenger.
11.1
295
Basics of a Replacement Study
If an unrealistically high trade-in value is offered for the defender compared to its fair
market value, the net cash flow required for the challenger is reduced, and this fact should be
considered in the analysis. The correct amount to recover and use in the economic analysis for
the challenger is its first cost minus the difference between the trade-in value (TIV) and market value (MV) of the defender. In equation form, this is P ⫺ (TIV ⫺ MV). This amount
represents the actual cost to the company because it includes both the opportunity cost (i.e.,
market value of the defender) and the out-of-pocket cost (i.e., first cost ⫺ trade-in) to acquire
the challenger. Of course, when the trade-in and market values are the same, the challenger P
value is used in all computations.
The challenger first cost is the estimated initial investment necessary to acquire and install it.
Sometimes, an analyst or manager will attempt to increase this first cost by an amount equal to the
unrecovered capital remaining in the defender, as shown on the accounting records for the asset.
This incorrect treatment of capital recovery is observed most often when the defender is working
well and in the early stages of its life, but technological obsolescence, or some other reason, has
forced consideration of a replacement. This leads us to identify two additional characteristics of
replacement analysis, in fact, of any economic analysis: sunk costs and nonowner’s viewpoint.
A sunk cost is a prior expenditure or loss of capital (money) that cannot be recovered by a
decision about the future. The replacement alternative for an asset, system, or process that has
incurred a nonrecoverable cost should not include this cost in any direct fashion; sunk costs
should be handled in a realistic way using tax laws and write-off allowances.
A sunk cost should never be added to the challenger’s first cost, because it will make the challenger appear to be more costly than it actually is. For example, assume an asset costing $100,000
two years ago has a depreciated value of $80,000 on the corporate books. It must be replaced
prematurely due to rapidly advancing technology. If the replacement alternative (challenger) has
a first cost of $150,000, the $80,000 from the current asset is a sunk cost were the challenger
purchased. For the purposes of an economic analysis, it is incorrect to increase the challenger’s
first cost to $230,000 or any number between this and $150,000.
The second characteristic is the perspective taken when conducting a replacement study. You,
the analyst, are a consultant from outside the company.
The nonowner’s viewpoint, also called the outsider’s viewpoint or consultant’s viewpoint, provides the greatest objectivity in a replacement study. This viewpoint performs the analysis without bias; it means the analyst owns neither the defender nor the challenger. Additionally, it assumes the services provided by the defender can be purchased now by making an “initial
investment” equal to the market value of the defender.
Besides being unbiased, this perspective is correct because the defender’s market value is a forgone opportunity of cash inflow were the replacement not selected, and the defender chosen.
As mentioned in the introduction, a replacement study is an application of the annual worth
method. As such, the fundamental assumptions for a replacement study parallel those of an AW
analysis. If the planning horizon is unlimited, that is, a study period is not specified, the assumptions are as follows:
1. The services provided are needed for the indefinite future.
2. The challenger is the best challenger available now and in the future to replace the defender.
When this challenger replaces the defender (now or later), it will be repeated for succeeding
life cycles.
3. Cost estimates for every life cycle of the defender and challenger will be the same as in their
first cycle.
As expected, none of these assumptions is precisely correct. We discussed this previously for the
AW method (and the PW method). When the intent of one or more of the assumptions becomes
incorrect, the estimates for the alternatives must be updated and a new replacement study conducted. The replacement procedure discussed in Section 11.3 explains how to do this. When the
planning horizon is limited to a specified study period, the assumptions above do not hold. The
procedure of Section 11.5 discusses replacement analysis over a fixed study period.
Sunk cost
296
Replacement and Retention Decisions
Chapter 11
EXAMPLE 11.1
Only 2 years ago, Techtron purchased for $275,000 a fully loaded SCADA (supervisory control and data acquisition) system including hardware and software for a processing plant operating on the Houston ship channel. When it was purchased, a life of 5 years and salvage of 20%
of first cost were estimated. Actual M&O costs have been $25,000 per year, and the book value
is $187,000. There has been a series of insidious malware infections targeting Techtron’s command and control software, plus next-generation hardware marketed only recently could
greatly reduce the competitiveness of the company in several of its product lines. Given these
factors, the system is likely worth nothing if kept in use for the final 3 years of its anticipated
useful life.
Model K2-A1, a new replacement turnkey system, can be purchased for $300,000 net cash,
that is, $400,000 first cost and a $100,000 trade-in for the current system. A 5-year life, salvage
value of 15% of stated first cost or $60,000, and an M&O cost of $15,000 per year are good
estimates for the new system. The current system was appraised this morning, and a market
value of $100,000 was confirmed for today; however, with the current virus discovery, the appraiser anticipates that the market value will fall rapidly to the $80,000 range once the virus
problem and new model are publicized.
Using the above values as the best possible today, state the correct defender and challenger
estimates for P, M&O, S, and n in a replacement study to be performed today.
Solution
Defender: Use the current market value of $100,000 as the first cost for the defender. All
others—original cost of $275,000, book value of $187,000, and trade-in value of
$100,000—are irrelevant to a replacement study conducted today. The estimates are as
follows:
First cost
M&O cost
Expected life
Salvage value
P ⫽ $⫺100,000
A ⫽ $⫺25,000 per year
n ⫽ 3 years
S⫽0
Challenger: The $400,000 stated first cost is the correct one to use for P, because the
trade-in and market values are equal.
First cost
P ⫽ $⫺400,000
M&O cost
A ⫽ $⫺15,000 per year
Expected life
n ⫽ 5 years
Salvage value
S ⫽ $60,000
Comment
If the replacement study is conducted next week when estimates will have changed, the defender’s first cost will be $80,000, the new market value according to the appraiser. The challenger’s first cost will be $380,000, that is, P ⫺ (TIV ⫺ MV) ⫽ 400,000 ⫺ (100,000 – 80,000).
11.2 Economic Service Life
Until now the estimated life n of an alternative or asset has been stated. In reality, the best life
estimate to use in the economic analysis is not known initially. When a replacement study or an
analysis between new alternatives is performed, the best value for n should be determined using
current cost estimates. The best life estimate is called the economic service life.
Economic service life
The economic service life (ESL) is the number of years n at which the equivalent uniform
annual worth (AW) of costs is the minimum, considering the most current cost estimates over
all possible years that the asset may provide a needed service.
297
Economic Service Life
11.2
Figure 11–1
Larger
costs
AW of costs, $/year
Annual worth curves of
cost elements that determine the economic
service life.
T
AW
otal
AW
of c
o sts
OC
of A
C a p it a l r
e c o v ery
0
Years
Economic
service life
The ESL is also referred to as the economic life or minimum cost life. Once determined, the ESL
should be the estimated life for the asset used in an engineering economy study, if only economics are considered. When n years have passed, the ESL indicates that the asset should be replaced
to minimize overall costs. To perform a replacement study correctly, it is important that the ESL
of the challenger and the ESL of the defender be determined, since their n values are usually not
preestablished.
The ESL is determined by calculating the total AW of costs if the asset is in service 1 year,
2 years, 3 years, and so on, up to the last year the asset is considered useful. Total AW of costs is
the sum of capital recovery (CR), which is the AW of the initial investment and any salvage
value, and the AW of the estimated annual operating cost (AOC), that is,
Total AW ⴝ capital recovery ⴚ AW of annual operating costs
ⴝ CR ⴚ AW of AOC
[11.1]
The ESL is the n value for the smallest total AW of costs. (Remember: These AW values are cost
estimates, so the AW values are negative numbers. Therefore, $–200 is a lower cost than $⫺500.)
Figure 11–1 shows the characteristic shape of a total AW of cost curve. The CR component of total
AW decreases, while the AOC component increases, thus forming the concave shape. The two AW
components are calculated as follows.
Decreasing cost of capital recovery. The capital recovery is the AW of investment; it decreases with each year of ownership. Capital recovery is calculated by Equation [6.3], which
is repeated here. The salvage value S, which usually decreases with time, is the estimated
market value (MV) in that year.
Capital recovery ⫽ ⫺P(A兾P,i,n) ⫹ S(A兾F,i,n)
[11.2]
Increasing cost of AW of AOC. Since the AOC (or M&O) estimates usually increase over the
years, the AW of AOC increases. To calculate the AW of the AOC series for 1, 2, 3, . . . years,
determine the present worth of each AOC value with the P兾F factor, then redistribute this
P value over the years of ownership, using the A兾P factor.
The complete equation for total AW of costs over k years (k ⫽ 1, 2, 3, . . . ) is
关兺
jⴝk
Total AWk ⴝ ⴚP(A兾P,i,k) ⴙ Sk(A兾F,i,k) ⴚ
where
兴
AOCj(P兾F,i,j) (A兾P,i,k) [11.3]
jⴝ1
P ⫽ initial investment or current market value
Sk ⫽ salvage value or market value after k years
AOCj ⫽ annual operating cost for year j ( j ⫽ 1 to k)
Capital recovery
298
Replacement and Retention Decisions
Chapter 11
The current MV is used for P when the asset is the defender, and the estimated future MV values
are substituted for the S values in years 1, 2, 3, . . . . Plotting the AWk series as in Figure 11–1
clearly indicates where the ESL is located and the trend of the AWk curve on each side of the ESL.
To determine ESL by spreadsheet, the PMT function (with embedded NPV functions as
needed) is used repeatedly for each year to calculate capital recovery and the AW of AOC. Their
sum is the total AW for k years of ownership. The PMT function formats for the capital recovery
and AOC components for each year k (k ⫽ 1, 2, 3, . . .) are as follows:
Capital recovery for the challenger:
Capital recovery for the defender:
AW of AOC:
PMT(i%,years,P,−MV_in_year_k)
PMT(i%,years,current_MV,ⴚMV_in_
year_k)
[11.4]
ⴚPMT(i%,years,NPV(i%,year_1_AOC:
year_k_AOC)ⴙ0)
When the spreadsheet is developed, it is recommended that the PMT functions in year 1 be developed using cell-reference format; then drag down the function through each column. A final
column summing the two PMT results displays total AW. Augmenting the table with an Excel xy
scatter chart graphically displays the cost curves in the general form of Figure 11–1, and the ESL
is easily identified. Example 11.2 illustrates ESL determination by hand and by spreadsheet.
EXAMPLE 11.2
A 3-year-old backup power system is being considered for early replacement. Its current market value is $20,000. Estimated future market values and annual operating costs for the next
5 years are given in Table 11–1, columns 2 and 3. What is the economic service life of this
defender if the interest rate is 10% per year? Solve by hand and by spreadsheet.
Solution by Hand
Equation [11.3] is used to calculate total AWk for k ⫽ 1, 2, . . . , 5. Table 11–1, column 4, shows
the capital recovery for the $20,000 current market value ( j ⫽ 0) plus 10% return. Column 5
gives the equivalent AW of AOC for k years. As an illustration, the computation of total AW
for k ⫽ 3 from Equation [11.3] is
Total AW3 ⫽ ⫺P(A兾P,i,3) ⫹ MV3(A兾F,i,3) ⫺ [PW of AOC1,AOC2, and AOC3](A兾P,i,3)
⫽ ⫺20,000(A兾P,10%,3) ⫹ 6000(A兾F,10%,3) ⫺ [5000(P兾F,10%,1)
⫹ 6500(P兾F,10%,2) ⫹ 8000(P兾F,10%,3)](A兾P,10%,3)
⫽ ⫺6230 ⫺ 6405 ⫽ $⫺12,635
A similar computation is performed for each year 1 through 5. The lowest equivalent cost
(numerically largest AW value) occurs at k ⫽ 3. Therefore, the defender ESL is n ⫽ 3 years,
and the AW value is $⫺12,635. In the replacement study, this AW will be compared with the
best challenger AW determined by a similar ESL analysis.
Solution by Spreadsheet
See Figure 11–2 for the spreadsheet screen shot and chart that shows the ESL is n ⫽ 3 years
and AW ⫽ $⫺12,634. (This format is a template for any ESL analysis; simply change the
TABLE 11–1
Computation of Economic Service Life
Year j
(1)
MVj, $
(2)
AOCj, $
(3)
Capital
Recovery, $
(4)
AW of
AOC, $
(5)
Total
AWk, $
(6) ⴝ (4) ⴙ (5)
1
2
3
4
5
10,000
8,000
6,000
2,000
0
⫺5,000
⫺6,500
⫺8,000
⫺9,500
⫺12,500
⫺12,000
⫺7,714
⫺6,230
⫺5,878
⫺5,276
⫺5,000
⫺5,714
⫺6,405
⫺7,072
⫺7,961
⫺17,000
⫺13,428
⫺12,635
⫺12,950
⫺13,237
Economic Service Life
11.2
Current market value
⫽ PMT($B$1,$A9,$B$2,⫺$B9)
⫽ ⫺PMT($B$1,$A9,NPV($B$1,$C$5:$C9)⫹0)
ESL of defender
Total AW is minimum
at n ⫽ 3 years
Total AW
curve
Capital
recovery
curve
Figure 11–2
Determination of ESL and plot of curves, Example 11.2
estimates and add rows for more years.) Contents of columns D and E are described below. The
PMT functions apply the formats as described in Equation [11.4]. Cell tags show detailed cellreference format for year 5. The $ symbols are included for absolute cell referencing, needed
when the entry is dragged down through the column.
Column D: Capital recovery is the AW of the $20,000 investment in year 0 for each year 1
through 5 with the estimated MV in that year. For example, in actual numbers, the
cell-reference PMT function in year 5 shown on the spreadsheet reads ⫽ PMT (10%,5,20000,
⫺0), resulting in $⫺5276. This series is plotted in Figure 11–2.
Column E: The NPV function embedded in the PMT function obtains the present worth in
year 0 of all AOC estimates through year k. Then PMT calculates the AW of AOC over
k years. For example, in year 5, the PMT in numbers is ⫽ ⫺PMT(10%,5,NPV
(10%,C5:C9)⫹0). The 0 is the AOC in year 0; it is optional. The graph plots the AW of
AOC curve, which constantly increases in cost because the AOC estimates increase
each year.
Comment
The capital recovery curve in Figure 11–2 (middle curve) is not the expected shape (see year 4)
because the estimated market value changes each year. If the same MV were estimated for each
year, the curve would appear like Figure 11–1. When several total AW values are approximately equal, the curve will be flat over several periods. This indicates that the ESL is relatively insensitive to costs.
299
300
Replacement and Retention Decisions
Chapter 11
It is reasonable to ask about the difference between the ESL analysis above and the AW
analyses performed in previous chapters. Previously we had a specific life estimated to be n years
with associated other estimates: first cost in year 0, possibly a salvage value in year n, and an
AOC that remained constant or varied each year. For all previous analyses, the calculation of AW
using these estimates determined the AW over n years. This is the economic service life when n
is fixed. Also, in all previous cases, there were no year-by-year market value estimates. Therefore,
we can conclude the following:
When the expected life n is known and specified for the challenger or defender, no ESL computations are necessary. Determine the AW over n years, using the first cost or current market
value, estimated salvage value after n years, and AOC estimates. This AW value is the correct one
to use in the replacement study.
However, when n is not fixed, the following is useful. First the market/salvage series is needed.
It is not difficult to estimate this series for a new or current asset. For example, an asset with a
first cost of P can lose market value of, say, 20% per year, so the market value series for years
0, 1, 2, . . . is P, 0.8P, 0.64P, . . . , respectively. If it is reasonable to predict the MV series on a
year-by-year basis, it can be combined with the AOC estimates to produce what is called the
marginal costs for the asset.
Marginal costs (MC) are year-by-year estimates of the costs to own and operate an asset for that
year. Three components are added to determine the marginal cost:
• Cost of ownership (loss in market value is the best estimate of this cost)
• Forgone interest on the market value at the beginning of the year
• AOC for each year
Once the marginal costs are estimated for each year, their equivalent AW value is calculated. The
sum of the AW values of the first two of these components is the capital recovery amount. Now, it
should be clear that the total AW of all three marginal cost components over k years is the same
value as the total annual worth for k years calculated in Equation [11.3]. That is, the following
relation is correct.
AW of marginal costs ⴝ total AW of costs
[11.5]
Therefore, there is no need to perform a separate, detailed marginal cost analysis when yearly
market values are estimated. The ESL analysis presented in Example 11.2 is sufficient in that
it results in the same numerical values. This is demonstrated in Example 11.3 using the progressive example.
EXAMPLE 11.3 Keep or Replace the Kiln Case
PE
In our progressive example, B&T Enterprises is considering the replacement of a 2-year-old
kiln with a new one to meet emerging market needs. When the current tunnel kiln was purchased 2 years ago for $25 million, an ESL study indicated that the minimum cost life was between 3 and 5 years of the expected 8-year life. The analysis was not very conclusive because
the total AW cost curve was flat for most years between 2 and 6, indicating insensitivity of the
ESL to changing costs. Now, the same type of question arises for the proposed graphite hearth
model that costs $38 million new: What are the ESL and the estimated total AW of costs? The
Manager of Critical Equipment at B&T estimates that the market value after only 1 year will
drop to $25 million and then retain 75% of the previous year’s value over the 12-year expected
life. Use this market value series and i ⫽ 15% per year to illustrate that an ESL analysis and
marginal cost analysis result in exactly the same total AW of cost series.
Solution
Figure 11–3 is a spreadsheet screen shot of the two analyses in $ million units. The market
value series is detailed in column B starting at $25 (million) and decreasing by 25% per year.
A brief description of each analysis follows.
Economic Service Life
11.2
ESL analysis
Two AW series
are identical
Marginal cost analysis
Figure 11–3
Comparison of annual worth series resulting from ESL analysis and marginal cost analysis, Example 11.3.
ESL analysis: Equation [11.4] is applied repeatedly for k ⫽ 1, 2, . . . , 12 years (columns C,
D, and E) in the top of Figure 11–3. Row 16 details the spreadsheet functions for year 12.
The result in column F is the total AW series that is of interest now.
Marginal cost (MC): The functions in the bottom of Figure 11–3 (columns C, D, and E)
develop the three components added to obtain the MC series. Row 33 details the functions
for year 12. The resulting AW of marginal costs (column G) is the series to compare with
the corresponding ESL series above (column F).
The two AW series are identical, thus demonstrating that Equation [11.5] is correct. Therefore, either an ESL or a marginal cost analysis will provide the same information for a replacement study. In this case, the results show that the new kiln will have a minimum AW of costs
of $–12.32 million at its full 12-year life.
We can draw two important conclusions about the n and AW values to be used in a replacement study. These conclusions are based on the extent to which detailed annual estimates are
made for the market value.
1. Year-by-year market value estimates are made. Use them to perform an ESL analysis,
and determine the n value with the lowest total AW of costs. These are the best n and AW
values for the replacement study.
2. Yearly market value estimates are not available. The only estimate available is market value (salvage value) in year n. Use it to calculate the AW over n years. These are
the n and AW values to use; however, they may not be the “best” values in that they may
not represent the best equivalent total AW of cost value found if an ESL analysis were
performed.
301
302
Replacement and Retention Decisions
Chapter 11
Upon completion of the ESL analysis (item 1 above), the replacement study procedure in Section 11.3 is applied using the values
Challenger alternative (C):
Defender alternative (D):
AWC for nC years
AWD for nD years
11.3 Performing a Replacement Study
Replacement studies are performed in one of two ways: without a study period specified or with
one defined. Figure 11–4 gives an overview of the approach taken for each situation. The procedure discussed in this section applies when no study period (planning horizon) is specified. If a
specific number of years is identified for the replacement study, for example, over the next
5 years, with no continuation considered after this time period in the economic analysis, the procedure in Section 11.5 is applied.
A replacement study determines when a challenger replaces the in-place defender. The complete study is finished if the challenger (C) is selected to replace the defender (D) now. However,
if the defender is retained now, the study may extend over a number of years equal to the life of
the defender nD, after which a challenger replaces the defender. Use the annual worth and life
values for C and D determined in the ESL analysis in the following procedure. Assume the services provided by the defender could be obtained at the AWD amount.
The replacement study procedure is:
New replacement study:
1. On the basis of the better AWC or AWD value, select the challenger C or defender D. When the
challenger is selected, replace the defender now, and expect to keep the challenger for nC
years. This replacement study is complete. If the defender is selected, plan to retain it for up
to nD more years. (This is the leftmost branch of Figure 11–4.) Next year, perform the
following steps.
One-year-later analysis:
2. Determine if all estimates are still current for both alternatives, especially first cost, market
value, and AOC. If not, proceed to step 3. If yes and this is year nD, replace the defender. If
this is not year nD, retain the defender for another year and repeat this same step. This step
may be repeated several times.
3. Whenever the estimates have changed, update them and determine new AWC and AWD values. Initiate a new replacement study (step 1).
Figure 11–4
Replacement study
Overview of replacement
study approaches.
No study period
specified
Study period
specified
Develop succession
options for D and C
using AW of
respective cash flows
Perform ESL
analysis
AWD
AWC
PW or AW for
each option
Select better AW
Select best option
Performing a Replacement Study
11.3
If the defender is selected initially (step 1), estimates may need updating after 1 year of retention (step 2). Possibly there is a new best challenger to compare with D. Either significant
changes in defender estimates or availability of a new challenger indicates that a new replacement study is to be performed. In actuality, a replacement study can be performed each year or
more frequantly to determine the advisability of replacing or retaining any defender, provided
a competitive challenger is available.
Example 11.4 illustrates the application of ESL analysis for a challenger and defender, followed by the use of the replacement study procedure. The planning horizon is unspecified in this
example.
EXAMPLE 11.4
Two years ago, Toshiba Electronics made a $15 million investment in new assembly line
machinery. It purchased approximately 200 units at $70,000 each and placed them in plants
in 10 different countries. The equipment sorts, tests, and performs insertion-order kitting on
electronic components in preparation for special-purpose circuit boards. This year, new international industry standards will require a $16,000 retrofit on each unit, in addition to the expected operating cost. Due to the new standards, coupled with rapidly changing technology, a
new system is challenging the retention of these 2-year-old machines. The chief engineer at
Toshiba USA realizes that the economics must be considered, so he has asked that a replacement study be performed this year and each year in the future, if need be. The i is 10% and the
estimates are below.
Challenger:
First cost: $50,000
Future market values: decreasing by 20% per year
Estimated retention period: no more than 10 years
AOC estimates: $5000 in year 1 with increases of $2000 per year
thereafter
Defender:
Current international market value: $15,000
Future market values: decreasing by 20% per year
Estimated retention period: no more than 3 more years
AOC estimates: $4000 next year, increasing by $4000 per year
thereafter, plus the $16,000 retrofit next year
(a) Determine the AW values and economic service lives necessary to perform the replacement
study.
(b) Perform the replacement study now.
(c) After 1 year, it is time to perform the follow-up analysis. The challenger is making large
inroads to the market for electronic components assembly equipment, especially with the
new international standards features built in. The expected market value for the defender is
still $12,000 this year, but it is expected to drop to virtually nothing in the future—$2000
next year on the worldwide market and zero after that. Also, this prematurely outdated
equipment is more costly to keep serviced, so the estimated AOC next year has been increased from $8000 to $12,000 and to $16,000 two years out. Perform the follow-up replacement study analysis.
Solution
(a) The results of the ESL analysis, shown in Figure 11–5, include all the MV and AOC estimates in columns B and C. For the challenger, note that P ⫽ $50,000 is also the MV in
year 0. The total AW of costs is for each year, should the challenger be placed into service
for that number of years. As an example, the year k ⫽ 4 amount of $⫺19,123 is determined using Equation [11.3], where the A兾G factor accommodates the arithmetic gradient
series in the AOC.
Total AW4 ⫽ ⫺50,000(A兾P,10%,4) ⫹ 20,480(A兾F,10%,4)
⫺ [5000 ⫹ 2000(A兾G,10%,4)]
⫽ $⫺19,123
303
304
Replacement and Retention Decisions
Chapter 11
ESL Analysis of Challenger
ESL results
ESL Analysis of Defender
Figure 11–5
ESL analysis of challenger and defender, Example 11.4.
For spreadsheet-based ESL analysis, this same result is achieved in cell F8 using Equation [11.4]. The functions are
Total AW4 ⫽ PMT(10%,4,50000,⫺20480) ⫺ PMT(10%,4,NPV(10%,C5:C8)⫹0)
⫽ ⫺11,361 ⫺ 7,762
⫽ ⫺19,123
The defender costs are analyzed in the same way up to the maximum retention period of
3 years.
The lowest AW cost (numerically largest) values for the replacement study are as follows:
Challenger:
Defender:
AWC ⫽ $⫺19,123
AWD ⫽ $⫺17,307
for nC ⫽ 4 years
for nD ⫽ 3 years
The challenger total AW of cost curve (Figure 11–5) is classically shaped and relatively flat
between years 3 and 6; there is virtually no difference in the total AW for years 4 and 5. For
the defender, note that the estimated AOC values change substantially over 3 years, and
they do not constantly increase or decrease.
(b) To perform the replacement study now, apply only the first step of the procedure. Select the
defender because it has the better AW of costs ($⫺17,307), and expect to retain it for
3 more years. Prepare to perform the one-year-later analysis 1 year from now.
(c) One year later, the situation has changed significantly for the equipment Toshiba retained
last year. Apply the steps for the one-year-later analysis:
2. After 1 year of defender retention, the challenger estimates are still reasonable, but the
defender market value and AOC estimates are substantially different. Go to step 3 to
perform a new ESL analysis for the defender.
3. The defender estimates in Figure 11–5 are updated below for the ESL analysis. New
AW values are calculated using Equation [11.3]. There is now a maximum of 2 more
years of retention, 1 year less than the 3 years determined last year.
Year k
Market
Value, $
AOC, $
Total AW
If Retained k More Years, $
0
1
2
12,000
2,000
0
—
⫺12,000
⫺16,000
—
⫺23,200
⫺20,819
Performing a Replacement Study
11.3
305
The AW and n values for the new replacement study are as follows:
Challenger:
Defender:
unchanged at AWC ⫽ $⫺19,123 for nC ⫽ 4 years
new AWD ⫽ $⫺20,819 for nD ⫽ 2 more years
Now select the challenger based on its favorable AW value. Therefore, replace the defender
now, not 2 years from now. Expect to keep the challenger for 4 years, or until a better challenger appears on the scene.
PE
EXAMPLE 11.5 Keep or Replace the Kiln Case
We continue with the progressive example of possibly replacing a kiln at B&T Enterprises. A
marketing study revealed that the improving business activity on the west coast implies that
the revenue profile between the installed kiln (PT) and the proposed new one (GH) would be
the same, with the new kiln possibly bringing in new revenue within the next couple of years.
The president of B&T decided it was time to do a replacement study. Assume you are the lead
engineer and that you previously completed the ESL analysis on the challenger (Example 11.3). It indicates that for the GH system the ESL is its expected useful life.
Challenger: ESL nGH ⴝ 12 years with total equivalent annual cost AWGH ⴝ $ⴚ12.32 million
The president asked you to complete the replacement study, stipulating that, due to the rapidly
rising annual operating costs (AOC), the defender would be retained a maximum of 6 years.
You are expected to make the necessary estimates for the defender (PT) and perform the study
at a 15% per year return.
Solution
After some data collection, you have good evidence that the market value for the PT system
will stay high, but that the increasing AOC is expected to continue rising about $1.2 million per
year. The best estimates for the next 6 years in $ million units are these:
Year
1
2
3
4
5
6
Market Value, $ M
AOC, $ M per year
22.0
⫺5.2
22.0
⫺6.4
22.0
⫺7.6
20.0
⫺8.8
18.0
⫺10.0
18.0
⫺11.2
You developed a spreadsheet and performed the analysis in Figure 11–6. As an illustration,
total AW computation for 3 years of retention, in $ million units, is
Total AW3 ⫽ ⫺22.0(A兾P,15%,3) ⫹ 22.0(A兾F,15%,3) ⫺ [5.2(P兾F,15%,1) ⫹ 6.4(P兾F,15%,2)
⫹7.6(P兾F,15%,3)](A兾P,15%,3)
⫽ –9.63 ⫹ 6.34 ⫺[14.36] (0.43798)
⫽ $⫺9.59 per year
Though the system could be retained up to 6 years, the ESL is much shorter at 1 year.
Defender: ESL nPT ⴝ 1 year with total equivalent annual cost AWPT ⴝ $ⴚ8.50 million
Figure 11–6
ESL analysis of defender kiln PT for progressive example, Example 11.5.
306
Chapter 11
Replacement and Retention Decisions
To make the replacement兾retention decision, apply step 1 of the procedure. Since AWPT ⫽
$⫺8.50 million per year is considerably less than AW⫽ $⫺12.32 million, you should recommend keeping the current kiln only 1 more year and doing another study during the year
to determine if the current estimates are still reliable.
Comment
An observation of the trends of the two final AW series in this problem is important. A comparison of Figure 11–3 (top), column F, and Figure 11–6, column F, shows us that the largest
total AW of the current system ($–11.47 M for 6 years) is still below the smallest total AW of
the proposed system ($⫺12.32 M for 12 years). This indicates that the graphite hearth system
(the challenger) will not be chosen on an economic basis, if the decision to consider the installed kiln as a defender in the future were made. It would take some significant estimate
changes to justify the challenger.
11.4 Additional Considerations in a
Replacement Study
There are several additional aspects of a replacement study that may be introduced. Three of
these are identified and discussed in turn.
• Future-year replacement decisions at the time of the initial replacement study
• Opportunity cost versus cash flow approaches to alternative comparison
• Anticipation of improved future challengers
In most cases when management initiates a replacement study, the question is best framed as,
“Replace now, 1 year from now, 2 years from now, etc.?” The procedure above does answer this
question provided the estimates for C and D do not change as each year passes. In other words,
at the time it is performed, step 1 of the procedure does answer the replacement question for
multiple years. It is only when estimates change over time that the decision to retain the defender
may be prematurely reversed (prior to nD years) in favor of the then-best challenger.
The first costs (P values) for the challenger and defender have been correctly taken as the
initial investment for the challenger C and current market value for the defender D. This is called
the opportunity cost approach because it recognizes that a cash inflow of funds equal to the
market value is forgone if the defender is selected. This approach, also called the conventional
approach, is correct for every replacement study. A second approach, called the cash flow approach, recognizes that when C is selected, the market value cash inflow for the defender is received and, in effect, immediately reduces the capital needed to invest in the challenger. Use of
the cash flow approach is strongly discouraged for at least two reasons: possible violation of the
equal-service requirement and incorrect capital recovery value for C. As we are aware, all economic evaluations must compare alternatives with equal service. Therefore, the cash flow approach can work only when challenger and defender lives are exactly equal. This is commonly
not the case; in fact, the ESL analysis and the replacement study procedure are designed to compare two mutually exclusive, unequal-life alternatives via the annual worth method. If this equalservice comparison reason is not enough to avoid the cash flow approach, consider what happens
to the challenger’s capital recovery amount when its first cost is decreased by the market value of
the defender. The capital recovery (CR) terms in Equation [11.3] will decrease, resulting in a
falsely low value of CR for the challenger, were it selected. From the vantage point of the economic study itself, the decision for C or D will not change; but when C is selected and implemented, this CR value is not reliable. The conclusion is simple:
Use the initial investment of C and the market value of D as the first costs in the ESL analysis and
in the replacement study.
A basic premise of a replacement study is that some challenger will replace the defender at a future time, provided the service continues to be needed and a worthy challenger is available. The expectation of ever-improving challengers can offer strong encouragement to retain the defender until
some situational elements—technology, costs, market fluctuations, contract negotiations, etc.—
11.5
307
Replacement Study over a Specified Study Period
stabilize. This was the case in the previous two examples. A large expenditure on equipment when
the standards changed soon after purchase forced an early replacement consideration and a large loss
of invested capital. The replacement study is no substitute for forecasting challenger availability. It is
important to understand trends, new advances, and competitive pressures that can complement the
economic outcome of a good replacement study. It is often better to compare a challenger with an
augmented defender in the replacement study. Adding needed features to a currently installed defender may prolong its useful life and productivity until challenger choices are more appealing.
It is possible that a significant tax impact may occur when a defender is traded early in its
expected life. If taxes should be considered, proceed now, or after the next section, to Chapter 17
and the after-tax replacement analysis in Section 17.7.
11.5 Replacement Study over a
Specified Study Period
When the time period for the replacement study is limited to a specified study period or planning
horizon, for example, 6 years, the ESL analysis is not performed.
The AW values for the challenger and for the remaining life of the defender are not based on the
economic service life; the AW is calculated over the study period only. What happens to the alternatives after the study period is not considered in the replacement analysis.
This means that the defender or challenger is not needed beyond the study period. In fact,
a study period of fixed duration does not comply with the three assumptions stated in
Section 11.1—service needed for indefinite future, best challenger available now, and estimates will be identical for future life cycles.
When performing a replacement study over a fixed study period, it is crucial that the estimates
used to determine the AW values be accurate and used in the study. This is especially important for
the defender. Failure to do the following violates the requirement of equal-service comparison.
When the defender’s remaining life is shorter than the study period, the cost of providing
the defender’s services from the end of its expected remaining life to the end of the study
period must be estimated as accurately as possible and included in the replacement study.
The right branch of Figure 11–4 presents an overview of the replacement study procedure for a
stated study period.
1. Succession options and AW values. Develop all the viable ways to use the defender and challenger during the study period. There may be only one option or many options; the longer the
study period, the more complex this analysis becomes. The AW values for the challenger and
defender cash flows are used to build the equivalent cash flow values for each option.
2. Selection of the best option. The PW or AW for each option is calculated over the study
period. Select the option with the lowest cost, or highest income if revenues are estimated.
(As before, the best option will have the numerically largest PW or AW value.)
The following examples use this procedure and illustrate the importance of making cost estimates for the defender alternative when its remaining life is less than the study period.
EXAMPLE 11.6
Claudia works with Lockheed-Martin (LMCO) in the aircraft maintenance division. She is
preparing for what she and her boss, the division chief, hope to be a new 10-year defense contract with the U.S. Air Force on C-5A cargo aircraft. A key piece of equipment for maintenance
operations is an avionics circuit diagnostics system. The current system was purchased 7 years
ago on an earlier contract. It has no capital recovery costs remaining, and the following are
reliable estimates: current market value ⫽ $70,000, remaining life of 3 more years, no salvage
value, and AOC ⫽ $30,000 per year. The only options for this system are to replace it now or
retain it for the full 3 additional years.
Claudia has found that there is only one good challenger system. Its cost estimates are: first
cost ⫽ $750,000, life ⫽ 10 years, S ⫽ 0, and AOC ⫽ $50,000 per year.
Study period
308
Replacement and Retention Decisions
Chapter 11
Realizing the importance of accurate defender alternative cost estimates, Claudia asked the
division chief what system would be a logical follow-on to the current one 3 years hence, if
LMCO wins the contract. The chief predicted LMCO would purchase the very system she had
identified as the challenger, because it is the best on the market. The company would keep it
for the entire 10 additional years for use on an extension of this contract or some other application that could recover the remaining 3 years of invested capital. Claudia interpreted the response to mean that the last 3 years would also be capital recovery years, but on some project
other than this one. Claudia’s estimate of the first cost of this same system 3 years from now is
$900,000. Additionally, the $50,000 per year AOC is the best estimate at this time.
The division chief mentioned any study had to be conducted using the interest rate of 10%,
as mandated by the U.S. Office of Management and Budget (OMB). Perform a replacement
study for the fixed contract period of 10 years.
Solution
The study period is fixed at 10 years, so the intent of the replacement study assumptions is not
present. This means the defender follow-on estimates are very important to the analysis.
Further, any analyses to determine the ESL values are unnecessary since alternative lives are
already set and no projected annual market values are available. The first step of the replacement study procedure is to define the options. Since the defender will be replaced now or in
3 years, there are only two options:
1. Challenger for all 10 years.
2. Defender for 3 years, followed by the challenger for 7 years.
Cash flows are diagrammed in Figure 11–7. For option 1, the challenger is used for all 10 years.
Equation [11.3] is applied to calculate AW using the following estimates:
P ⫽ $⫺750,000
n ⫽ 10 years
Challenger:
AOC ⫽ $⫺50,000
S⫽0
AWC ⫽ ⫺750,000(A兾P,10%,10) ⫺ 50,000 ⫽ $⫺172,063
0
1
2
3
4
5
6
7
8
9
10 Year
AOC = $50,000
End of
study period
P = $750,000
(a) Challenger (option 1)
0
1
2
3
4
5
6
7
8
9
10
11
12
13 Year
AOC = $30,000
AOC = $50,000
MV = $70,000
Defendercurrent
CR = $146,475
Defender-follow-on
CR for another
project
(b) Defender alternative (option 2)
Figure 11–7
Cash flow diagrams for a 10-year study period replacement study, Example 11.6.
Replacement Study over a Specified Study Period
11.5
The second option has more complex cost estimates. The AW for the in-place system is calculated over the first 3 years. Added to this is the capital recovery for the defender follow-on for the
next 7 years. However in this case, the CR amount is determined over its full 10-year life. (It is not
unusual for the recovery of invested capital to be moved between projects, especially for contract
work.) Refer to the AW components as AWDC (subscript DC for defender current) and AWDF (subscript DF for defender follow-on). The final cash flows are shown in Figure 11–7b.
market value ⫽ $⫺70,000
n ⫽ 3 years
Defender current:
AOC ⫽ $⫺30,000
S⫽0
AWDC ⫽ [⫺70,000 ⫺ 30,000(P兾A,10%,3)](A兾P,10%,10) ⫽ $⫺23,534
Defender follow-on: P ⫽ $⫺900,000, n ⫽ 10 years for capital recovery calculation
only, AOC ⫽ $–50,000 for years 4 through 10, S ⫽ 0.
The CR and AW for all 10 years are
CRDF ⫽ ⫺900,000(A兾P,10%,10) ⫽ $⫺146,475
AWDF ⫽ (⫺146,475 ⫺ 50,000)(F兾A,10%,7)(A兾F,10%,10) ⫽ $⫺116,966
[11.6]
Total AWD for the defender is the sum of the two annual worth values above. This is the AW
for option 2.
AWD ⫽ AWDC ⫹ AWDF ⫽ ⫺23,534 ⫺ 116,966 ⫽ $⫺140,500
Option 2 has a lower cost ($⫺140,500 versus $⫺172,063). Retain the defender now and expect
to purchase the follow-on system 3 years hence.
Comment
The capital recovery cost for the defender follow-on will be borne by some yet-to-be-identified
project for years 11 through 13. If this assumption were not made, its capital recovery cost would
be calculated over 7 years, not 10, in Equation [11.6], increasing CR to $⫺184,869. This raises the
annual worth to AWD ⫽ $⫺163,357. The defender alternative (option 2) is still selected.
EXAMPLE 11.7
Three years ago Chicago’s O’Hare Airport purchased a new fire truck. Because of flight increases, new fire-fighting capacity is needed once again. An additional truck of the same capacity can be purchased now, or a double-capacity truck can replace the current fire truck. Estimates are presented below. Compare the options at 12% per year using (a) a 12-year study
period and (b) a 9-year study period.
First cost P, $
AOC, $
Market value, $
Salvage value, $
Life, years
Presently Owned
New Purchase
Double Capacity
⫺151,000 (3 years ago)
⫺1,500
70,000
10% of P
12
⫺175,000
⫺1,500
—
12% of P
12
⫺190,000
⫺2,500
—
10% of P
12
Solution
Identify option 1 as retention of the presently owned truck and augmentation with a new samecapacity vehicle. Define option 2 as replacement with the double-capacity truck.
Option 1
P, $
AOC, $
S, $
n, years
Option 2
Presently Owned
Augmentation
Double Capacity
⫺70,000
⫺1,500
15,100
9
⫺175,000
⫺1,500
21,000
12
⫺190,000
⫺2,500
19,000
12
309
310
Replacement and Retention Decisions
Chapter 11
(a) For a full-life 12-year study period of option 1,
AW1 ⫽ (AW of presently owned) ⫹ (AW of augmentation)
⫽ [⫺70,000(A兾P,12%,9) ⫹ 15,100(A兾F,12%,9) ⫺ 1500]
⫹ [⫺175,000(A兾P,12%,12) ⫹ 21,000(A兾F,12%,12) ⫺ 1500]
⫽ ⫺13,616 ⫺ 28,882
⫽ $⫺42,498
This computation assumes the equivalent services provided by the current fire truck can be
purchased at $−13,616 per year for years 10 through 12.
AW2 ⫽ ⫺190,000(A兾P,12%,12) ⫹ 19,000(A兾F,12%,12) ⫺ 2500
⫽ $⫺32,386
Replace now with the double-capacity truck (option 2) at an advantage of $10,112 per year.
(b) The analysis for an abbreviated 9-year study period is identical, except that n ⫽ 9 in each
factor; that is, 3 fewer years are allowed for the augmentation and double-capacity trucks
to recover the capital investment plus a 12% per year return. The salvage values remain the
same since they are quoted as a percentage of P for all years.
AW1 ⫽ $⫺46,539
AW2 ⫽ $⫺36,873
Option 2 is again selected.
The previous two examples indicate an important consideration for setting the length of the
study period for a replacement analysis. It involves the capital recovery amount for the challenger, when the strict definition of a study period is applied.
Study period
Capital recovery
When a study period shorter than the life of the challenger is defined, the challenger’s capital
recovery amount increases in order to recover the initial investment plus a return in this
shortened time period. Highly abbreviated study periods tend to disadvantage the challenger
because no consideration of time beyond the end of the study period is made in calculating the
challenger’s capital recovery amount.
If there are several options for the number of years that the defender may be retained before
replacement with the challenger, the first step of the replacement study—succession options and
AW values—must include all the viable options. For example, if the study period is 5 years and
the defender will remain in service 1 year, or 2 years, or 3 years, cost estimates must be made to
determine AW values for each defender retention period. In this case, there are four options; call
them W, X, Y, and Z.
Option
Defender
Retained, Years
Challenger
Serves, Years
W
X
Y
Z
3
2
1
0
2
3
4
5
The respective AW values for defender retention and challenger use define the cash flows for
each option. Example 11.8 illustrates the procedure using the progressive example.
EXAMPLE 11.8 Keep or Replace the Kiln Case
PE
We have progressed to the point that the replacement study between the defender PT and
challenger GH was completed (Example 11.5). The defender was the clear choice with a
much smaller AW value ($⫺8.50 M) than that of the challenger ($⫺12.32 M). Now the
management of B&T is in a dilemma. They know the current tunnel kiln is much cheaper
than the new graphite hearth, but the prospect of future new business should not be dismissed.
Replacement Study over a Specified Study Period
11.5
TABLE 11–2
Replacement Study Options and Total AW Values, Example 11.8
Defender PT
Challenger GH
Option
Years Retained
AW, $ M兾Year
Years Retained
AW, $ M兾Year
A
B
C
D
E
F
1
2
3
4
5
6
⫺8.50
⫺9.06
⫺9.59
⫺10.49
⫺11.16
⫺11.47
5
4
3
2
1
0
⫺14.21
⫺15.08
⫺16.31
⫺18.21
⫺22.10
—
The president asked, “Is it possible to determine when it is economically the cheapest to
purchase the new kiln, provided the current one is kept at least 1 year, but no more than 6
years, its remaining expected life?” The chief financial officer answered, yes, of course.
(a) Determine the answer for the president. (b) Discuss the next step in the analysis based on
the conclusion reached here.
Solution
(a) Actually, this is a quite easy question to answer, because all the information has been determined previously. We know the MARR is 15% per year, the study period has been established at 6 years, and the defender PT will stay in place between 1 and 6 years. Therefore, the challenger GH will be considered for 0 to 5 years of service. The total AW values
were determined for the defender in Example 11.5 (Figure 11–6) and for the challenger in
Example 11.3 (Figure 11–3). They are repeated in Table 11–2 for convenience. Use the
procedure for a replacement study with a fixed study period.
Step 1: Succession options and AW values. There are six options in this case; the defender
is retained from 1 to 6 years while the challenger is installed from 0 to 5 years. We will
label them A through F. Figure 11–8 presents the options and the AW series for each option
from Table 11–2. No consideration of the fact that the challenger has an expected life of
12 years is made since the study period is fixed at 6 years.
Step 2: Selection of the best option. The PW value for each option is determined over the
6-year study period in column J of Figure 11–8. The conclusion is clearly to keep the
defender in place for 6 more years.
(b) Every replacement analysis has indicated that the defender should be retained for the near
future. If the analysis is to be carried further, the possibility of increased revenue based on
services of the challenger’s high-temperature and operating efficiency should be considered next. In the introductory material, new business opportunities were mentioned. A revenue increase for the challenger will reduce its AW of costs and possibly make it more
economically viable.
Conclusion: Keep defender all 6 years
Figure 11–8
PW values for 6-year study period replacement analysis, Example 11.8.
311
312
Replacement and Retention Decisions
Chapter 11
11.6 Replacement Value
Often it is helpful to know the minimum market value of the defender necessary to make the
challenger economically attractive. If a realizable market value or trade-in of at least this amount
can be obtained, from an economic perspective the challenger should be selected immediately.
This is a breakeven value between AWC and AWD; it is referred to as the replacement value
(RV). Set up the relation AWC ⫽ AWD with the market value for the defender identified as RV,
which is the unknown. The AWC is known, so RV can be determined. The selection guideline is
as follows:
If the actual market trade-in exceeds the breakeven replacement value, the challenger is the better alternative and should replace the defender now.
Determination of the RV for a defender is an excellent opportunity to utilize the Goal Seek tool
in Excel. The target cell is the current market value, and the AWD value is forced to equal the
AWC amount. Example 11.9 discusses a replacement value of the progressive example.
EXAMPLE 11.9 Keep or Replace the Kiln Case
PE
As one final consideration of the challenger kiln, you decide to determine what the trade-in
amount would have to be so that the challenger is the economic choice next year. This is based
on the ESL analysis that concluded the following (Examples 11.3 and 11.5):
Defender:
Challenger:
ESL nPT ⫽ 1 year with AWPT ⫽ $⫺8.50 million
ESL nGH ⫽ 12 years with AWGH ⫽ $⫺12.32 million
The original defender price was $25 million, and a current market value of $22 million was
estimated earlier (Figure 11–6). Since the installed kiln is known for retention of its market
value (MV), you are hopeful the difference between RV and estimated MV may not be so significant. What will you discover RV to be? The MARR is 15% per year.
Solution
Set the AW relation for the defender for the ESL time of 1 year equal to AWGH ⫽ $⫺12.32
and solve for RV. The estimates for AOC and MV next year are in Figure 11–6; they are, in
$ million,
Year 1:
AOC ⫽ $−5.20
MV ⫽ $22.0
⫺12.32 ⫽ ⫺RV(A兾P,15%,1) ⫹ 22.00(A兾F,15%,1) ⫺ 5.20
1.15RV ⫽ 12.32 ⫹ 22.00 ⫺ 5.20
RV ⫽ $25.32
Though the RV is larger than the defender’s estimated MV of $22 million, some flexibility in
the trade-in offer or the challenger’s first cost may cause the challenger to be economically
justifiable.
Comment
To find RV using a spreadsheet, return to Figure 11–6. In the Goal Seek template, the “set” cell
is the AW for 1 year (currently $⫺8.50), and the required value is $⫺12.32, the AWGH for its
ESL of 12 years. The “changing” cell is the current market value (cell F2), currently $22.00.
When “OK” is touched, $25.32 is displayed as the breakeven market value. This is the RV.
CHAPTER SUMMARY
It is important in a replacement study to compare the best challenger with the defender. Best
(economic) challenger is described as the one with the lowest annual worth (AW) of costs for
some number of years. If the expected remaining life of the defender and the estimated life of the
challenger are specified, the AW values over these years are determined and the replacement
study proceeds. However, if reasonable estimates of the expected market value (MV) and AOC
Problems
313
for each year of ownership can be made, these year-by-year (marginal) costs help determine the
best challenger.
The economic service life (ESL) analysis is designed to determine the best challenger’s years
of service and the resulting lowest total AW of costs. The resulting nC and AWC values are used
in the replacement study. The same analysis can be performed for the ESL of the defender.
Replacement studies in which no study period (planning horizon) is specified utilize the annual worth method of comparing two unequal-life alternatives. The better AW value determines
how long the defender is retained before replacement.
When a study period is specified for the replacement study, it is vital that the market value and
cost estimates for the defender be as accurate as possible. When the defender’s remaining life is
shorter than the study period, it is critical that the cost for continuing service be estimated carefully. All the viable options for using the defender and challenger are enumerated, and their AW
equivalent cash flows are determined. For each option, the PW or AW value is used to select the
best option. This option determines how long the defender is retained before replacement.
PROBLEMS
Foundations of Replacement
11.1 In a replacement study, what is meant by “taking
the nonowner’s viewpoint”?
11.2 An asset that was purchased 3 years ago for
$100,000 is becoming obsolete faster than expected. The company thought the asset would last
5 years and that its book value would decrease by
$20,000 each year and, therefore, be worthless at
the end of year 5. In considering a more versatile,
more reliable high-tech replacement, the company
discovered that the presently owned asset has a
market value of only $15,000. If the replacement is
purchased immediately at a first cost of $75,000
and if it will have a lower annual worth, what is the
amount of the sunk cost? Assume the company’s
MARR is 15% per year.
11.3 As a muscle car aficionado, a friend of yours likes
to restore cars of the 60s and 70s and sell them for
a profit. He started his latest project (a 1965 Shelby
GT350) four months ago and has a total of
$126,000 invested so far. Another opportunity has
come up (a 1969 Dodge Charger) that he is thinking of buying because he believes he could sell it
for a profit of $60,000 after it is completely restored. To do so, however, he would have to sell
the unfinished Shelby first. He thought that the
completely restored Shelby would be worth
$195,000, resulting in a tidy profit of $22,000, but
in its half-restored condition, the most he could get
now is $115,000. In discussing the situation with
you, he stated that if he could sell the Shelby now
and buy the Charger at a reduced price, he would
make up for the money he will lose in selling the
Shelby at a lower-than-desired price.
(a) What is wrong with this thinking?
(b) What is his sunk cost in the Shelby?
11.4 In conducting a replacement study wherein the
planning horizon is unspecified, list three assumptions that are inherent in an annual worth analysis
of the defender and challenger.
11.5 A civil engineer who owns his own design兾build兾
operate company purchased a small crane 3 years
ago at a cost of $60,000. At that time, it was expected to be used for 10 years and then traded in
for its salvage value of $10,000. Due to increased
construction activities, the company would prefer
to trade for a new, larger crane now that will cost
$80,000. The company estimates that the old crane
can be used, if necessary, for another 3 years, at
which time it would have a $23,000 estimated
market value. Its current market value is estimated
to be $39,000, and if it is used for another 3 years,
it will have M&O costs (exclusive of operator
costs) of $17,000 per year. Determine the values of
P, n, S, and AOC that should be used for the existing crane in a replacement analysis.
11.6 Equipment that was purchased by Newport Corporation for making pneumatic vibration isolators
cost $90,000 two years ago. It has a market value
that can be described by the relation $90,000 ⫺
8000k, where k is the years from time of purchase.
Experience with this type of equipment has shown
that the operating cost for the first 4 years is
$65,000 per year, after which it increases by $6300
per year. The asset’s salvage value was originally
estimated to be $7000 after a predicted 10-year
useful life. Determine the values of P, S, and AOC
if a replacement study is done (a) now and
(b) 1 year from now.
11.7 A piece of equipment that was purchased 2 years
ago by Toshiba Imaging for $50,000 was expected
to have a useful life of 5 years with a $5000 salvage
314
Replacement and Retention Decisions
Chapter 11
value. Its performance was less than expected, and
it was upgraded for $20,000 one year ago. Increased demand now requires that the equipment
be upgraded again for another $17,000 so that it
can be used for 3 more years. If upgraded, its annual operating cost will be $27,000 and it will
have a $12,000 salvage after 3 years. Alternatively,
it can be replaced with new equipment priced at
$65,000 with operating costs of $14,000 per year
and a salvage value of $23,000 after 6 years. If replaced now, the existing equipment will be sold for
$7000. Determine the values of P, S, AOC, and n
for the defender in a replacement study.
Economic Service Life
11.8 For equipment that has a first cost of $10,000 and
the estimated operating costs and year-end salvage
values shown below, determine the economic service life at i ⫽ 10% per year.
Year
Operating Cost,
$ per Year
Salvage
Value, $
1
2
3
4
5
⫺1000
⫺1200
⫺1300
⫺2000
⫺3000
7000
5000
4500
3000
2000
11.9 To improve package tracking at a UPS transfer facility, conveyor equipment was upgraded with
RFID sensors at a cost of $345,000. The operating
cost is expected to be $148,000 per year for the
first 3 years and $210,000 for the next 3 years. The
salvage value of the equipment is expected to be
$140,000 for the first 3 years, but due to obsolescence, it won’t have a significant value after that.
At an interest rate of 10% per year, determine
(a) The economic service life of the equipment
and associated annual worth
(b) The percentage increase in the AW of cost if
the equipment is retained 2 years longer than
the ESL
11.10 Economic service life calculations for an asset are
shown below. If an interest rate of 10% per year
was used in making the calculations, determine the
values of P and S that were used in calculating the
AW for year 3.
Years
Retained
AW of First
Cost, $
AW of Operating
Cost, $ per Year
AW of
Salvage Value, $
1
2
3
4
5
⫺51,700
⫺27,091
⫺18,899
⫺14,827
⫺12,398
⫺15,000
⫺17,000
⫺19,000
⫺21,000
⫺23,000
35,000
13,810
6,648
4,309
2,457
11.11 The initial cost of a bridge that is expected to be in
place forever is $70 million. Maintenance can be
done at 1-, 2-, 3-, or 4-year intervals, but the longer
the interval between servicing, the higher the cost.
The costs of servicing are estimated at $83,000,
$91,000, $125,000, and $183,000 for intervals of 1
through 4 years, respectively. What interval should
be scheduled for maintenance to minimize the
overall equivalent annual cost? The interest rate is
8% per year.
11.12 A construction company bought a 180,000 metric
ton earth sifter at a cost of $65,000. The company
expects to keep the equipment a maximum of
7 years. The operating cost is expected to follow the
series described by 40,000 ⫹ 10,000k, where k is
the number of years since it was purchased (k ⫽ 1,
2, . . . , 7). The salvage value is estimated to be
$30,000 for years 1 and 2 and $20,000 for years 3
through 7. At an interest rate of 10% per year, determine the economic service life and the associated
equivalent annual cost of the sifter.
11.13 An engineer determined the ESL of a new
$80,000 piece of equipment and recorded the
calculations shown below. [Note that the numbers are annual worth values associated with
various years of retention; that is, if the equipment is kept for, say, 3 years, the AW (years 1
through 3) of the first cost is $32,169, the AW of
the operating cost is $51,000, and the AW of the
salvage value is $6042.] The engineer forgot to
enter the AW of the salvage value for 2 years of
retention. From the information available, determine the following:
(a) The interest rate used in the ESL calculations.
(b) The salvage value after 2 years, if the total
AW of the equipment in year 2 was
$78,762. Use the interest rate determined
in part (a).
Years
AW of
AW of
AW of
Retained First Cost, $ Operating Cost, $ Salvage Value, $
1
2
3
⫺88,000
⫺46,095
⫺32,169
⫺45,000
⫺46,000
⫺51,000
50,000
?
6,042
11.14 A large, standby electricity generator in a hospital
operating room has a first cost of $70,000 and may
be used for a maximum of 6 years. Its salvage
value, which decreases by 15% per year, is described by the equation S ⫽ 70,000(1 – 0.15)n,
where n is the number of years after purchase. The
operating cost of the generator will be constant at
$75,000 per year. At an interest rate of 12% per
year, what are the economic service life and the
associated AW value?
315
Problems
11.15 A piece of equipment has a first cost of $150,000, a
maximum useful life of 7 years, and a market (salvage) value described by the relation S ⫽ 120,000 ⫺
20,000k, where k is the number of years since it was
purchased. The salvage value cannot go below zero.
The AOC series is estimated using AOC ⫽ 60,000 ⫹
10,000k. The interest rate is 15% per year. Determine the economic service life (a) by hand solution,
using regular AW computations, and (b) by spreadsheet, using annual marginal cost estimates.
11.16 Determine the economic service life and corresponding AW value for a machine that has the following cash flow estimates. Use an interest rate of
14% per year and hand solution.
Year
Salvage
Value, $
Operating
Cost, $ per Year
0
1
2
3
4
5
6
7
100,000
75,000
60,000
50,000
40,000
25,000
15,000
0
—
–28,000
–31,000
–34,000
–34,000
–34,000
–45,000
–49,000
11.17 Use the annual marginal costs to find the economic
service life for Problem 11.16 on a spreadsheet. Assume the salvage values are the best estimates of future market value. Develop an Excel chart of annual
marginal costs (MC) and AW of MC over 7 years.
11.18 Keep or Replace the Kiln Case
PE
In Example 11.3, the market value (salvage value)
series of the proposed $38 million replacement kiln
(GH) dropped to $25 million in only 1 year and then
retained 75% of the previous year’s market value
through the remainder of its 12-year expected life.
Based on the experience with the current kiln, and
the higher-temperature capability of the replacement, the new kiln is actually expected to retain
only 50% of the previous year’s value starting in
year 5. Additionally, the heating element replacement in year 6 will probably cost $4 million, not
$2 million. And finally, the maintenance costs will
be considerably higher as the kiln ages. Starting in
year 5, the AOC is expected to increase by 25% per
year, not 10% as predicted earlier. The Manager of
Critical Equipment is now very concerned that the
ESL will be significantly decreased from the
12 years calculated earlier (Example 11.3).
(a) Determine the new ESL and associated AW
value.
(b) In percentage changes, estimate how much
these new cost estimates may affect the
minimum-cost life and AW of cost estimate.
Replacement Study
11.19 In a one-year-later analysis, what action should be
taken if (a) all estimates are still current and the
year is nD, (b) all estimates are still current and the
year is not nD, and (c) the estimates have changed?
11.20 Based on company records of similar equipment, a
consulting aerospace engineer at Aerospatiale estimated AW values for a presently owned, highly
accurate steel rivet inserter.
If Retained This
Number of Years
AW Value,
$ per Year
1
2
3
4
5
⫺62,000
⫺51,000
⫺49,000
⫺53,000
⫺70,000
A challenger has ESL ⫽ 2 years and AWC ⫽
$⫺48,000 per year. (a) If the consultant must recommend a replace兾retain decision today, should
the company keep the defender or purchase the
challenger? Why? The MARR is 15% per year.
(b) When should the next replacement evaluation
take place?
11.21 In planning a plant expansion, MedImmune has an
economic decision to make—upgrade the existing
controlled-environment rooms or purchase new
ones. The presently owned ones were purchased 4
years ago for $250,000. They have a current “quick
sale” value of $20,000, but for an investment of
$100,000 now, they would be adequate for another
4 years, after which they would be sold for $40,000.
Alternatively, new controlled-environment rooms
could be purchased at a cost of $270,000. They are
expected to have a 10-year life with a $50,000 salvage value at that time. Determine whether the
company should upgrade or replace. Use a MARR
of 20% per year.
11.22 Three years ago, Witt Gas Controls purchased
equipment for $80,000 that was expected to have a
useful life of 5 years with a $9000 salvage value.
Increased demand necessitated an upgrade costing
$30,000 one year ago. Technology changes now require that the equipment be upgraded again for another $25,000 so that it can be used for 3 more
years. Its annual operating cost will be $47,000, and
it will have a $22,000 salvage after 3 years. Alternatively, it can be replaced with new equipment that
will cost $68,000 with operating costs of $35,000
per year and a salvage value of $21,000 after
3 years. If replaced now, the existing equipment will
be sold for $9000. Calculate the annual worth of the
defender at an interest rate of 10% per year.
316
Replacement and Retention Decisions
Chapter 11
11.23 A recent environmental engineering graduate is trying to decide whether he should keep his presently
owned car or purchase a more environmentally
friendly hybrid. A new car will cost $26,000 and have
annual operation and maintenance costs of $1200 per
year with an $8000 salvage value in 5 years (which is
its estimated economic service life).
The presently owned car has a resale value now
of $5000; one year from now it will be $3000, two
years from now $2500, and 3 years from now
$2200. Its operating cost is expected to be $1900
this year, with costs increasing by $200 per year.
The presently owned car will definitely not be kept
longer than 3 more years. Assuming used cars like
the one presently owned will always be available,
should the presently owned car be sold now, 1 year
from now, 2 years from now, or 3 years from now?
Use annual worth calculations at i ⫽ 10% per year
and show your work.
11.24 A pulp and paper company is evaluating whether it
should retain the current bleaching process that
uses chlorine dioxide or replace it with a proprietary “oxypure” process. The relevant information
for each process is shown. Use an interest rate of
15% per year to perform the replacement study.
Original cost 6
years ago, $
Investment cost
now, $
Current market
value, $
Annual operating
cost, $兾year
Salvage value, $
Remaining life,
years
Current Process
Oxypure Process
⫺450,000
—
—
⫺700,000
25,000
⫺180,000
0
5
—
⫺70,000
50,000
10
11.25 A machine that is critical to the Phelps-Dodge copper refining operation was purchased 7 years ago
for $160,000. Last year a replacement study was
performed with the decision to retain it for 3 more
years. The situation has changed. The equipment is
estimated to have a value of $8000 if “scavenged”
for parts now or anytime in the future. If kept in
service, it can be minimally upgraded at a cost of
$43,000, which will make it usable for up to
2 more years. Its operating cost is expected to be
$22,000 the first year and $25,000 the second year.
Alternatively, the company can purchase a new
system that will have an equivalent annual worth
of $⫺47,063 per year over its ESL. The company
uses a MARR of 10% per year. Calculate the relevant annual worth values, and determine when the
company should replace the machine.
11.26 A crushing machine that is a basic component of a
metal recycling operation is wearing out faster than
expected. The machine was purchased 2 years ago
for $400,000. At that time, the buyer thought the
machine would serve its needs for at least 5 years,
at which time the machine would be sold to a
smaller independent recycler for $80,000. Now,
however, the company thinks the market value of
the diminished machine is only $50,000. If it is
kept, the machine’s operating cost will be $37,000
per year for the next 2 years, after which it will be
scrapped for $1000. If it is kept for only 1 year, the
market value is estimated to be $10,000. Alternatively, the company can outsource the process now
for a fixed cost of $56,000 per year. At an interest
rate of 10% per year, should the company continue
with the current machine or outsource the process?
11.27 The data associated with operating and maintaining an asset are shown below. The company manager has already decided to keep the machine for 1
more year (i.e., until the end of year 1), but you
have been asked to determine the cost of keeping it
1 more year after that. At an interest rate of 10%
per year, estimate the AW of keeping the machine
from year 1 to year 2.
Year
Market Value, $
Operating Cost,
$ per Year
0
1
2
3
30,000
25,000
14,000
10,000
⫺15,000
⫺15,000
⫺15,000
⫺15,000
11.28 A machine that cost $120,000 three years ago can
be sold now for $54,000. Its market value for the
next 2 years is expected to be $40,000 and $20,000
one year and 2 years from now, respectively. Its
operating cost was $18,000 for the first 3 years of
its life, but the M&O cost is expected to be
$23,000 for the next 2 years. A new improved machine that can be purchased for $138,000 will
have an economic life of 5 years, an operating
cost of $9000 per year, and a salvage value of
$32,000 after 5 years. At an interest rate of 10%
per year, determine if the presently owned machine should be replaced now, 1 year from now, or
2 years from now.
11.29 The projected market value and M&O costs associated with a presently owned machine are shown
(next page). An outside vendor of services has offered to provide the service of the existing machine at a fixed price per year. If the presently
owned machine is replaced now, the cost of the
fixed-price contract will be $33,000 for each of the
next 3 years. If the presently owned machine is
317
Problems
replaced next year or the year after that, the contract price will be $35,000 per year. Determine if
and when the defender should be replaced with the
outside vendor using an interest rate of 10% per
year. Assume used equipment similar to the defender will always be available.
Year
Market Value, $
M&O Cost,
$ per Year
0
1
2
3
4
30,000
25,000
14,000
10,000
8,000
—
⫺24,000
⫺25,000
⫺26,000
—
11.30 BioHealth, a biodevice systems leasing company,
is considering a new equipment purchase to replace a currently owned asset that was purchased
2 years ago for $250,000. It is appraised at a current market value of only $50,000. An upgrade is
possible for $200,000 now that would be adequate
for another 3 years of lease rights, after which the
entire system could be sold on the international
circuit for an estimated $40,000. The challenger,
which can be purchased for $300,000, has an expected life of 10 years and a $50,000 salvage
value. Determine whether the company should upgrade or replace at a MARR of 12% per year. Assume the AOC estimates are the same for both
alternatives.
11.31 For the estimates in Problem 11.30, use a
spreadsheet-based analysis to determine the first
cost for the augmentation of the current system
that will make the defender and challenger break
even. Is this a maximum or minimum for the upgrade, if the current system is to be retained?
11.32 Herald Richter and Associates, 5 years ago, purchased for $45,000 a microwave signal graphical
plotter for corrosion detection in concrete structures. It is expected to have the market values and
annual operating costs shown below for its remaining useful life of up to 3 years. It could be
traded now at an appraised market value of $8000.
Richter should retain the present plotter. Solve
(a) by hand and (b) by using a spreadsheet.
11.33 In the opportunity cost approach to replacement
analysis, what does the opportunity refer to?
11.34 State what is meant by the cash flow approach to
replacement analysis, and list two reasons why it is
not a good idea to use this method.
Replacement Study over a Specified Study Period
11.35 ABB Communications is considering replacing
equipment that had a first cost of $300,000 five
years ago. The company CEO wants to know if the
equipment should be replaced now or at any other
time over the next 3 years to minimize the cost of
producing miniature background suppression sensors. Since the present equipment or the proposed
equipment can be used for any or all of the 3-year
period, one of the company’s industrial engineers
produced AW cost information for the defender
and challenger as shown below. The values represent the annual costs of the respective equipment if
used for the indicated number of years. Determine
when the defender should be replaced to minimize
the cost to ABB for the 3-year study period using
an interest rate of 10% per year.
AW If Kept Stated
Number of Years, $ per Year
Number of
Years Kept
Defender
Challenger
1
2
3
⫺22,000
⫺24,000
⫺27,000
⫺29,000
⫺26,000
⫺25,000
11.36 The table below shows present worth calculations
of the costs associated with using a presently
owned machine (defender) and a possible replacement (challenger) for different numbers of years.
Determine when the defender should be replaced
using an interest rate of 10% per year and a 5-year
study period. Show solutions (a) by hand and
(b) by spreadsheet.
PW If Kept兾Used Stated
Number of Years, $
Year
Market Value
at End of Year, $
AOC,
$ per Year
Number of Years
Kept兾Used
Defender
Challenger
1
2
3
6000
4000
1000
–50,000
–53,000
–60,000
1
2
3
4
5
⫺36,000
⫺75,000
⫺125,000
⫺166,000
⫺217,000
⫺89,000
⫺96,000
⫺102,000
⫺113,000
⫺149,000
A replacement plotter with new Internet-based,
digital technology costing $125,000 has an estimated $10,000 salvage value after its 5-year life
and an AOC of $31,000 per year. At an interest rate
of 15% per year, determine how many more years
11.37 Nano Technologies intends to use the newest and
finest equipment in its labs. Accordingly, a senior
engineer has recommended that a 2-year-old piece
318
Replacement and Retention Decisions
Chapter 11
of precision measurement equipment be replaced
immediately. This engineer believes it can be
shown that the proposed equipment is economically advantageous at a 10% per year return and a
planning horizon of 3 years.
(a) Perform the replacement analysis using the
annual worth method for a specified 3-year
study period.
(b) Determine the challenger’s capital recovery
amount for the 3-year study period and the
expected full life. Comment on the effect
made by the 3-year study period.
Original purchase price, $
Current market value, $
Remaining life, years
Estimated value in 3 years, $
Salvage value after 15 years, $
Annual operating cost, $ per year
Current
Proposed
⫺30,000
17,000
5
9,000
—
⫺8,000
⫺40,000
—
15
20,000
5,000
⫺3,000
11.38 An industrial engineer at a fiber-optic manufacturing company is considering two robots to reduce
costs in a production line. Robot X will have a first
cost of $82,000, an annual maintenance and operation (M&O) cost of $30,000, and salvage values of
$50,000, $42,000, and $35,000 after 1, 2, and
3 years, respectively. Robot Y will have a first cost
of $97,000, an annual M&O cost of $27,000, and
salvage values of $60,000, $51,000, and $42,000
after 1, 2, and 3 years, respectively. Which robot
should be selected if a 2-year study period is specified at an interest rate of 15% per year and replacement after 1 year is not an option?
11.39 A 3-year-old machine purchased for $140,000 is not
able to meet today’s market demands. The machine
can be upgraded now for $70,000 or sold to a subcontracting company for $40,000. The current machine will have an annual operating cost of $85,000
per year and a $30,000 salvage value in 3 years. If
upgraded, the presently owned machine will be retained for only 3 more years, then replaced with a
machine to be used in the manufacture of several
other product lines. The replacement machine,
which will serve the company now and for at least
8 years, will cost $220,000. Its salvage value will be
$50,000 for years 1 through 4; $20,000 after 5 years;
and $10,000 thereafter. It will have an estimated operating cost of $65,000 per year. You want to perform an economic analysis at 15% per year using a
3-year planning horizon.
(a) Should the company replace the presently
owned machine now, or do it 3 years from now?
(b) Compare the capital recovery requirements for
the replacement machine (challenger) over the
study period and an expected life of 8 years.
11.40 Two processes can be used for producing a polymer that reduces friction loss in engines. Process
K, which is currently in place, has a market value
of $165,000 now, an operating cost of $69,000 per
year, and a salvage value of $50,000 after 1 more
year and $40,000 after its maximum 2-year remaining life. Process L, the challenger, will have a
first cost of $230,000, an operating cost of $65,000
per year, and salvage values of $100,000 after 1
year, $70,000 after 2 years, $45,000 after 3 years,
and $26,000 after its maximum expected 4-year
life. The company’s MARR is 12% per year. You
have been asked to determine which process to select when (a) a 2-year study period is used and
(b) a 3-year study period is used.
PE
11.41 Keep or Replace the Kiln Case
In Example 11.8, the in-place kiln and replacement
kiln (GH) were evaluated using a fixed study period of 6 years. This is a significantly shortened
period compared to the expected 12-year life of
the challenger. Use the best estimates available
throughout this case to determine the impact on the
capital recovery amount for the GH kiln of shortening the evaluation time from 12 to 6 years.
11.42 Nabisco Bakers currently employs staff to operate
the equipment used to sterilize much of the mixing,
baking, and packaging facilities in a large cookie
and cracker manufacturing plant in Iowa. The plant
manager, who is dedicated to cutting costs but not
sacrificing quality and hygiene, has the projected
data shown in the table below if the current system
were retained for up to its maximum expected life
of 5 years. A contract company has proposed a
turnkey sanitation system for $5.0 million per year
if Nabisco signs on for 4 to 10 years, and $5.5 million per year for a shorter number of years.
Years Retained
AW, $ per Year
Close-Down Expense, $
0
1
2
3
4
5
—
⫺2,300,000
⫺2,300,000
⫺3,000,000
⫺3,000,000
⫺3,500,000
⫺3,000,000
⫺2,500,000
⫺2,000,000
⫺1,000,000
⫺1,000,000
⫺500,000
(a)
At a MARR ⫽ 8% per year, perform a replacement study for the plant manager with a
fixed study period of 5 years, when it is anticipated that the plant will be shut down due
to the age of the facility and projected technological obsolescence. As you perform the
study, take into account that regardless of the
number of years that the current sanitation
system is retained, a one-time close-down
Additional Problems and FE Exam Review Questions
(b)
cost will be incurred for personnel and equipment during the last year of operation. (Hint:
Calculate AW values for all combinations of
defender兾challenger options.)
What is the percentage change in the AW
amount each year for the 5-year period? If
the decision to retain the current sanitation
system for all 5 years is made, what is the
economic disadvantage in AW compared to
that of the most economic retention period?
Replacement Value
11.43 In 2008, Amphenol Industrial purchased a new
quality inspection system for $550,000. The estimated salvage value was $50,000 after 8 years.
Currently the expected remaining life is 3 years
with an AOC of $27,000 per year and an estimated
salvage value of $30,000. The new president has
recommended early replacement of the system
with one that costs $400,000 and has a 5-year
economic service life, a $45,000 salvage value,
and an estimated AOC of $50,000 per year. If the
MARR for the corporation is 12% per year, find
the minimum trade-in value now necessary to
make the president’s replacement economically
advantageous.
11.44 A CNC milling machine purchased by Proto Tool
and Die 10 years ago for $75,000 can be used for 3
more years. Estimates are an annual operating cost
of $63,000 and a salvage value of $25,000. A challenger will cost $130,000 with an economic life of
6 years and an operating cost of $32,000 per year.
Its salvage value will be $45,000. On the basis of
319
these estimates, what market value for the existing
asset will render the challenger equally attractive?
Use an interest rate of 12% per year.
11.45 Hydrochloric acid, which fumes at room temperatures, creates a very corrosive work environment,
causing steel tools to rust and equipment to fail
prematurely. A distillation system purchased 4
years ago for $80,000 can be used for only 2 more
years, at which time it will be scrapped with no
value. Its operating cost is $75,000 per year. A
more corrosion-resistant challenger will cost
$220,000 with an operating cost of $49,000 per
year. It is expected to have a $30,000 salvage
value after its 6-year ESL. At an interest rate of
10% per year, what minimum replacement value
now for the present system will render the challenger attractive?
11.46 Engine oil purifier machines can effectively remove acid, pitch, particles, water, and gas from
used oil. Purifier A was purchased 5 years ago for
$90,000. Its operating cost is higher than expected,
so if it is not replaced now, it will likely be used for
only 3 more years. Its operating cost this year will
be $140,000, increasing by $2000 per year through
the end of its useful life, at which time it will be
donated for its recyclable scrap value. A more efficient challenger, purifier B, will cost $150,000
with a $50,000 salvage value after its 8-year ESL.
Its operating cost is expected to be $82,000 for
year 1, increasing by $500 per year thereafter.
What is the market value for A that will make the
two purifiers equally attractive at an interest rate of
12% per year?
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
11.47 In conducting a replacement study, all of the following are correct viewpoints for the analyst except:
(a) Consultant’s
(b) Owner’s
(c) Outsider’s
(d) Nonowner’s
11.48 A sunk cost is the difference between:
(a) The first cost and the salvage value
(b) The market value and the salvage value
(c) The first cost and the market value
(d) The book value and the market value
11.49 A truck was purchased 3 years ago for $45,000 and
can be sold today for $24,000. The operating costs
are $9000 per year, and it is expected to last 4 more
years with a $5000 salvage value. A new truck,
which will perform that same service, can be purchased for $50,000, and it will have a life of
10 years with operating costs of $28,000 per year
and a $10,000 salvage value. The value that should
be used as P for the presently owned vehicle in a
replacement study is:
(a) $45,000
(b) $5000
(c) $50,000
(d) $24,000
11.50 The economic service life of an asset with the PW
and AW values on the next page is:
(a) 1 year
(b) 2 years
(c) 3 years
(d) 4 years
320
Replacement and Retention Decisions
Chapter 11
Years n
Present Worth If
Kept n Years, $
Annual Worth If
Kept n Years,
$兾Year
1
2
3
4
5
⫺15,455
⫺28,646
⫺45,019
⫺57,655
⫺72,867
⫺17,000
⫺16,286
⫺18,102
⫺18,188
⫺19,222
11.51 In looking for ways to cut costs and increase
profit (to make the company’s stock go up), one
of the company’s industrial engineers (IEs) determined that the equivalent annual worth of an
existing machine over its remaining useful life of
3 years will be $–70,000 per year. The IE also
determined that a replacement with more advanced features will have an AW of $–80,000 per
year if it is kept for 2 years or less, $–75,000 if it
is kept between 3 and 4 years, and $–65,000 if it
is kept for 5 to 10 years. If the engineer uses a
3-year study period and an interest rate of 15%
per year, she should recommend that the existing
machine be:
(a) Replaced now
(b) Replaced 1 year from now
(c) Replaced 2 years from now
(d) Not replaced
11.52 The equivalent annual worth of an existing machine at American Semiconductor is estimated to
be $–70,000 per year over its remaining useful life
of 3 years. It can be replaced now or later with a
machine that will have an AW of $–90,000 per
year if it is kept for 2 years or less, –$65,000 if it
is kept between 3 and 5 years, and $–110,000 if it
is kept for 6 to 8 years. The company wants an
analysis of what it should do for a 3-year study
period at an interest rate of 15% per year. The replacement must be made now or 3 years from
now, according to the department supervisor. You
should recommend that the existing machine be
replaced:
(a) Now
(b) 1 year from now
(c) 2 years from now
(d) Can’t tell; must find AW of different retention combinations for the two machines
11.53 The cost characteristics of a CO testing machine at
Dytran Instruments are shown below. The cost of a
new tester is $100,000. The equation for determining the AW of keeping the tester for 2 years is:
(a) AW ⫽ ⫺100,000(A兾P,i,8) ⫺ [42,000(P兾F,i,1)
⫹ 47,000(P兾F,i,2)](A兾P,i,8) ⫹ 40,000(A兾F,i,8)
(b) AW ⫽ ⫺100,000(A兾P,i,2) ⫺ [42,000(P兾F,i,1)
⫹ 47,000(P兾F,i,2)](A兾P,i,2) ⫹ 40,000(A兾F,i,2)
(c)
(d)
AW ⫽ ⫺100,000(A兾P,i,2) ⫺ 47,000
⫹ 40,000(A兾F,i,2)
AW ⫽ ⫺100,000(A兾P,i,2) ⫺ 42,000
⫹ 40,000(A兾F,i,2)
Machine Age,
Years
M&O Costs,
$ per Year
Salvage Value
at End of Year, $
1
2
3
4
5
6
7
8
⫺42,000
⫺47,000
⫺49,000
⫺50,000
⫺52,000
⫺54,000
⫺63,000
⫺67,000
60,000
40,000
31,000
24,000
15,000
10,000
10,000
10,000
11.54 An engineer with Haliburton calculated the AW
values shown for a presently owned machine by
using estimates he obtained from the vendor and
company records.
Retention
Period, Years
AW Value,
$ per Year
1
2
3
4
5
⫺92,000
⫺81,000
⫺87,000
⫺89,000
⫺95,000
A challenger has an economic service life of
7 years with an AW of $–86,000 per year. Assume
that used machines like the one presently owned
will always be available and that the MARR is
12% per year. If all future costs remain as estimated for the analysis, the company should purchase the challenger:
(a) Now
(b) After 2 years
(c) After 3 years
(d) Never
11.55 The annual worth values for a defender, which can
be replaced with a similar used asset, and a challenger are estimated. The defender should be replaced:
(a) Now
(b) 1 year from now
(c) 2 years from now
(d) 3 years from now
AW Value, $ per Year
Number of
Years Retained
Defender
Challenger
1
2
3
4
5
⫺14,000
⫺13,700
⫺16,900
⫺17,000
⫺18,000
⫺21,000
⫺18,000
⫺13,100
⫺15,600
⫺17,500
Case Study
321
CASE STUDY
WILL THE CORRECT ESL PLEASE STAND?
Background
New pumper system equipment is under consideration by a
Gulf Coast chemical processing plant. One crucial pump
moves highly corrosive liquids from specially lined tanks on
intercoastal barges into storage and preliminary refining facilities dockside. Because of the variable quality of the raw
chemical and the high pressures imposed on the pump chassis
and impellers, a close log is maintained on the number of
hours per year that the pump operates. Safety records and
pump component deterioration are considered critical control
points for this system. As currently planned, rebuild and
M&O cost estimates are increased accordingly when cumulative operating time reaches the 6000-hour mark.
Information
You are the safety engineer at the plant. Estimates made for
this pump are as follows:
First cost:
Rebuild cost:
M&O costs:
MARR:
$⫺800,000
$⫺150,000 whenever 6000 cumulative hours are logged. Each
rework will cost 20% more than
the previous one. A maximum of
3 rebuilds is allowed.
$⫺25,000 for each year 1 through 4
$⫺40,000 per year starting the
year after the first rebuild, plus
15% per year thereafter
10% per year
Based on previous logbook data, the current estimates for
number of operating hours per year are as follows:
Year
Hours per Year
1
2
3 on
500
1500
2000
Case Study Questions
1. Determine the economic service life of the pump. How
does the ESL compare with the maximum allowed
rebuilds?
2. The plant superintendent told you, the safety engineer,
that only one rebuild should be planned for, because
these types of pumps usually have their minimum-cost
life before the second rebuild. Determine a market
value for this pump that will force the ESL to be 6 years.
Comment on the practicality of ESL ⫽ 6 years, given
the MV calculated.
3. In a separate conversation, the line manager told you
to not plan for a rebuild after 6000 hours, because
the pump will be replaced after a total of 10,000
hours of operation. The line manager wants to know
what the base AOC in year 1 can be to make the ESL
6 years. He also told you to assume now that the 15%
growth rate applies from year 1 forward. How does
this base AOC value compare with the rebuild cost
after 6000 hours?
4. What do you think of these suggestions from the plant
superintendent and the line manager?
CHAPTER 12
Independent
Projects with
Budget
Limitation
L E A R N I N G
O U T C O M E S
Purpose: Select independent projects for funding when there is a limitation on the amount of capital available for investment.
SECTION
TOPIC
LEARNING OUTCOME
12.1
Capital rationing
• Explain how a capital budgeting problem is
approached.
12.2
Equal-life projects
• Use PW-based capital budgeting to select from
several equal-life independent projects.
12.3
Unequal-life projects
• Use PW-based capital budgeting to select from
several unequal-life independent projects.
12.4
Linear programming
• Set up the linear programming model and use the
Solver spreadsheet tool to select projects.
12.5
Ranking options
• Use the internal rate of return (IROR) and
profitability index (PI) to rank and select from
independent projects.
I
n most of the previous economic comparisons, the alternatives have been mutually
exclusive; only one could be selected. If the projects are not mutually exclusive,
they are categorized as independent of one another, as discussed at the beginning of Chapter 5. Now we learn techniques to select from several independent projects. It
is possible to select any number of projects from none (do nothing) to all viable projects.
There is virtually always some upper limit on the amount of capital available for investment in new projects. This limit is considered as each independent project is economically
evaluated. The techniques applied are called capital budgeting methods, also referred to
as capital rationing. They determine the economically best rationing of initial investment
capital among independent projects based upon different measures, such as PW, ROR, and
the profitability index. These three are discussed here.
12.1 An Overview of Capital Rationing
among Projects
Investment capital is a scarce resource for all corporations; thus there is virtually always a limited
amount to be distributed among competing investment opportunities. When a corporation has
several options for placing investment capital, a “reject or accept” decision must be made for
each project. Effectively, each option is independent of other options, so the evaluation is performed on a project-by-project basis. Selection of one project does not impact the selection decision for any other project. This is the fundamental difference between mutually exclusive alternatives and independent projects.
The term project is used to identify each independent option. We use the term bundle to
identify a collection of independent projects. The term mutually exclusive alternative continues
to identify a project when only one may be selected from several.
There are two exceptions to purely independent projects: A contingent project is one that has
a condition placed upon its acceptance or rejection. Two examples of contingent projects A and
B are as follows: A cannot be accepted unless B is accepted; and A can be accepted in lieu of B,
but both are not needed. A dependent project is one that must be accepted or rejected based on
the decision about another project(s). For example, B must be accepted if both A and C are accepted. In practice, these complicating conditions can be bypassed by forming packages of related projects that are economically evaluated themselves as independent projects along with the
remaining, unconditioned projects.
A capital budgeting study has the following characteristics:
• Several independent projects are identified, and net cash flow estimates are available.
• Each project is either selected entirely or not selected; that is, partial investment in a project
is not possible.
• A stated budgetary constraint restricts the total amount available for investment. Budget
constraints may be present for the first year only or for several years. This investment limit
is identified by the symbol b.
• The objective is to maximize the return on the investments using a measure of worth, such
as the PW value.
By nature, independent projects are usually quite different from one another. For example, in the
public sector, a city government may develop several projects to choose from: drainage, city
park, metro rail, and an upgraded public bus system. In the private sector, sample projects may
be a new warehousing facility, expanded product base, improved quality program, upgraded information system, automation, and acquisition of another firm. The size of the study can range
from only four or five projects to a complex study involving 50 to 100 projects. The typical
capital budgeting problem is illustrated in Figure 12–1. For each independent project there is an
initial investment, project life, and estimated net cash flows that can include a salvage value.
Present worth analysis using the capital budgeting process is the recommended method to
select projects. The general selection guideline is as follows:
Accept projects with the best PW values determined at the MARR over the project life, provided
the investment capital limit is not exceeded.
Independent project
selection
Limited budget
324
Independent Projects with Budget Limitation
Chapter 12
Figure 12–1
Independent
projects
Basic characteristics of a
capital budgeting study.
Estimates
Initial
investment
Estimated
net cash flows
A.
Life
Capital
investment
limit
Investment
B.
Life
(can’t invest
more than
this much)
C.
Life
Select 0 to all 3 projects
Objective: Maximize PW
value of selection
within capital limit
Equal-service requirement
This guideline is not different from that used for selection in previous chapters for independent
projects. As before, each project is compared with the do-nothing project; that is, incremental
analysis between projects is not necessary. The primary difference now is that the amount of
money available to invest is limited, thus the title capital budgeting or rationing. A specific solution procedure that incorporates this constraint is needed.
Previously, PW analysis had the requirement of equal service between alternatives. This assumption is not necessary for capital rationing, because there is no life cycle of a project beyond
its estimated life. Rather, the selection guideline has the following implied assumption.
When the present worth at the MARR over the respective project life is used to select projects,
the reinvestment assumption is that all positive net cash flows are reinvested at the MARR from
the time they are realized until the end of the longest-lived project.
Opportunity cost
This fundamental assumption is demonstrated to be correct at the end of Section 12.3, which
treats PW-based capital rationing for unequal-life projects.
Another dilemma of capital rationing among independent projects concerns the flexibility of the
capital investment limit b. The limit may marginally disallow an acceptable project that is next in
line for acceptance. For example, assume project A has a positive PW value at the MARR. If A will
cause the capital limit of $5,000,000 to be exceeded by only $1000, should A be included in the PW
analysis? Commonly, a capital investment limit is somewhat flexible, so project A would usually be
included. However, in the examples here, we will not exceed a stated investment limit.
As we learned earlier (Sections 1.9 and 10.1), the rate of return on the first unfunded project
is an opportunity cost. The lack of capital to fund the next project defines the ROR level that
is forgone. The opportunity cost will vary with each set of independent projects evaluated, but
over time it provides information to fine-tune the MARR and other measures used by the
company in future evaluations.
Capital Rationing Using PW Analysis of Equal-Life Projects
12.2
It is common to rank independent projects and select from them based on measures other than
PW at the MARR. Two are the internal rate of return (IROR), discussed in Chapter 7, and the
profitability index (PI), also called the present worth index (PWI), introduced in Chapter 9. Neither of these measures guarantees an optimal PW-based selection. The capital budgeting process,
covered in the next 3 sections, does find the optimal solution for PW values. We recommend use
of this PW-based technique; however, it should be recognized that both PI and IROR usually
provide excellent, near-optimal selections, and they both work very well when the number of
projects is large. Application of these two measures is presented in Section 12.5.
12.2 Capital Rationing Using PW Analysis
of Equal-Life Projects
To select from projects that have the same expected life while investing no more than the limit b,
first formulate all mutually exclusive bundles—one project at a time, two at a time, etc. Each
feasible bundle must have a total investment that does not exceed b. One of these bundles is the
do-nothing (DN) project. The total number of bundles for m projects is 2m. The number increases
rapidly with m. For m ⫽ 4, there are 24 ⫽ 16 bundles, and for m ⫽ 16, 216 ⫽ 65,536 bundles.
Then the PW of each bundle is determined at the MARR. The bundle with the largest PW value
is selected.
To illustrate the development of mutually exclusive bundles, consider these four projects with
equal lives.
Project
Initial Investment, $
A
B
C
D
⫺10,000
⫺5,000
⫺8,000
⫺15,000
If the investment limit is b ⫽ $25,000, of the 16 bundles, there are 12 feasible ones to evaluate.
The bundles ABD, ACD, BCD, and ABCD have investment totals that exceed $25,000. The
viable bundles are shown below.
Projects
Total Initial
Investment, $
Projects
Total Initial
Investment, $
A
B
C
D
AB
AC
⫺10,000
⫺5,000
⫺8,000
⫺15,000
⫺15,000
⫺18,000
AD
BC
BD
CD
ABC
Do nothing
⫺25,000
⫺13,000
⫺20,000
⫺23,000
⫺23,000
0
The procedure to conduct a capital budgeting study using PW analysis is as follows:
1. Develop all mutually exclusive bundles with a total initial investment that does not exceed
the capital limit b.
2. Sum the net cash flows NCFjt for all projects in each bundle j ( j ⫽ 1, 2, . . . , 2m) and each
year t (t ⫽ 1, 2, . . . , nj). Refer to the initial investment of bundle j at time t ⫽ 0 as NCFj0.
3. Compute the present worth value PWj for each bundle at the MARR.
PWj ⴝ PW of bundle net cash flows ⴚ initial investment
tⴝnj
ⴝ
兺 NCF (P兾F,i,t) ⴚ NCF
jt
j0
[12.1]
tⴝ1
4. Select the bundle with the (numerically) largest PWj value.
Selecting the maximum PWj means that this bundle has a PW value larger than any other bundle.
Any bundle with PWj ⬍ 0 is discarded, because it does not produce a return of at least the MARR.
325
326
Independent Projects with Budget Limitation
Chapter 12
EXAMPLE 12.1
The projects review committee of Microsoft has $20 million to allocate next year to new software product development. Any or all of five projects in Table 12–1 may be accepted. All
amounts are in $1000 units. Each project has an expected life of 9 years. Select the project(s)
if a 15% return is expected.
TABLE 12–1 Five Equal-Life Independent Projects ($1000 Units)
Project
Initial
Investment, $
Annual
Net Cash Flow, $
Project
Life, Years
A
B
C
D
E
⫺10,000
⫺15,000
⫺8,000
⫺6,000
⫺21,000
2870
2930
2680
2540
9500
9
9
9
9
9
Solution
Use the procedure above with b ⫽ $20,000 to select one bundle that maximizes present worth.
Remember the units are in $1000.
1. There are 25 ⫽ 32 possible bundles. The eight bundles that require no more than $20,000
in initial investments are described in columns 2 and 3 of Table 12–2. The $21,000 investment for E eliminates it from all bundles.
TABLE 12–2
Summary of Present Worth Analysis of Equal-Life Independent
Projects ($1000 Units)
Bundle
j
(1)
Projects
Included
(2)
Initial
Investment
NCFj0, $
(3)
Annual Net
Cash Flow
NCFj, $
(4)
Present
Worth
PWj, $
(5)
1
2
3
4
5
6
7
8
A
B
C
D
AC
AD
CD
Do nothing
⫺10,000
⫺15,000
⫺8,000
⫺6,000
⫺18,000
⫺16,000
⫺14,000
0
2,870
2,930
2,680
2,540
5,550
5,410
5,220
0
⫹3,694
⫺1,019
⫹4,788
⫹6,120
⫹8,482
⫹9,814
⫹10,908
0
2. The bundle net cash flows, column 4, are the sum of individual project net cash
flows.
3. Use Equation [12.1] to compute the present worth for each bundle. Since the annual NCF
and life estimates are the same for a bundle, PWj reduces to
PWj ⫽ NCFj(P兾A,15%,9) ⫺ NCFj0
4. Column 5 of Table 12–2 summarizes the PWj values at i ⫽ 15%. Bundle 2 does not return
15%, since PW2 ⬍ 0. The largest is PW7 ⫽ $10,908; therefore, invest $14 million in C and
D. This leaves $6 million uncommitted.
This analysis assumes that the $6 million not used in this initial investment will return the
MARR by placing it in some other, unspecified investment opportunity.
Capital Rationing Using PW Analysis of Unequal-Life Projects
12.3
12.3 Capital Rationing Using PW Analysis
of Unequal-Life Projects
Usually independent projects do not have the same expected life. As stated in Section 12.1, the
PW method for solution of the capital budgeting problem assumes that each project will last for
the period of the longest-lived project nL. Additionally, reinvestment of any positive net cash
flows is assumed to be at the MARR from the time they are realized until the end of the longestlived project, that is, from year nj through year nL. Therefore, use of the LCM of lives is not
necessary, and it is correct to use Equation [12.1] to select bundles of unequal-life projects by
PW analysis using the procedure of the previous section.
EXAMPLE 12.2
For MARR ⫽ 15% per year and b ⫽ $20,000, select from the following independent projects.
Solve by hand and by spreadsheet.
Project
Initial
Investment, $
Annual Net
Cash Flow, $
Project
Life, Years
A
B
C
D
⫺8,000
⫺15,000
⫺8,000
⫺8,000
3870
2930
2680
2540
6
9
5
4
Solution by Hand
The unequal-life values make the net cash flows vary over a bundle’s life, but the selection
procedure is the same as above. Of the 24 ⫽ 16 bundles, 8 are economically feasible. Their PW
values by Equation [12.1] are summarized in Table 12–3. As an illustration, for bundle 7:
PW7 ⫽ ⫺16,000 ⫹ 5220(P兾A,15%,4) ⫹ 2680(P兾F,15%,5) ⫽ $235
Select bundle 5 (projects A and C) for a $16,000 investment.
TABLE 12–3
Present Worth Analysis for Unequal-Life Independent Projects,
Example 12.2
Initial
Investment,
NCFj0, $
(3)
Bundle J
(1)
Project
(2)
1
2
3
4
5
A
B
C
D
AC
⫺8,000
⫺15,000
⫺8,000
⫺8,000
⫺16,000
6
AD
⫺16,000
7
CD
⫺16,000
8
Do nothing
0
Net Cash Flows
Year t
(4)
NCFjt, $
(5)
1–6
1–9
1–5
1–4
1–5
6
1–4
5–6
1–4
5
3,870
2,930
2,680
2,540
6,550
3,870
6,410
3,870
5,220
2,680
0
Present
Worth
PWj, $
(6)
⫹6,646
⫺1,019
⫹984
⫺748
⫹7,630
⫹5,898
⫹235
0
Solution by Spreadsheet
Figure 12–2 presents a spreadsheet with the same information as in Table 12–3. It is necessary
to initially develop the mutually exclusive bundles manually and total net cash flows each year
using each project’s NCF. The NPV function is used to determine PW for each bundle j over
its respective life. Bundle 5 (projects A and C) has the largest PW value (row 16).
327
328
Independent Projects with Budget Limitation
Chapter 12
⫽ D7⫹E7
⫽ NPV($B$1,F7:F15) ⫹ F6
Figure 12–2
Computation of PW values for independent project selection, Example 12.2.
The rest of this section will help you understand why solution of the capital budgeting problem
by PW evaluation using Equation [12.1] is correct. The following logic verifies the assumption of
reinvestment at the MARR for all net positive cash flows when project lives are unequal. Refer to
Figure 12–3, which uses the general layout of a two-project bundle. Assume each project has the
same net cash flow each year. The P兾A factor is used for PW computation. Define nL as the life of
the longer-lived project. At the end of the shorter-lived project, the bundle has a total future worth
of NCFj(F兾A,MARR,nj) as determined for each project. Now, assume reinvestment at the MARR
from year nj⫹1 through year nL (a total of nL − nj years). The assumption of the return at the MARR
is important; this PW approach does not necessarily select the correct projects if the return is not
at the MARR. The results are the two future worth arrows in year nL in Figure 12–3. Finally,
Figure 12–3
Representative cash flows
used to compute PW for a
bundle of two independent unequal-life projects
by Equation [12.1].
FWB
PWB
nB = nL
Project B
Investment
for B
FWA
PWA
Future worth
Period of
reinvestment
at MARR
nA
Project A
Investment
for A
Bundle PW = PWA + PWB
nL
12.4
329
Capital Budgeting Problem Formulation Using Linear Programming
FW = $57,111
Initial investment and
cash flows for bundle 7,
projects C and D,
Example 17.2.
$5220
PW = $235
0
$2680
1
2
3
4
5
6
7
8
9
$16,000
compute the bundle PW value in the initial year. This is the bundle PW ⫽ PWA ⫹ PWB. In general
form, the bundle j present worth is
PWj ⫽ NCFj(F兾A,MARR,nj)(F兾P,MARR,nL⫺nj)(P兾F,MARR,nL)
[12.2]
Substitute the symbol i for the MARR, and use the factor formulas to simplify.
(1 ⫹ i)nj ⫺ 1
1
PWj ⫽ NCFj ——————(1 ⫹ i)nL⫺nj ————
i
(1 ⫹ i)nL
[
(1 ⫹ i)nj ⫺ 1
⫽ NCFj ——————
i(1 ⫹ i)nj
Figure 12–4
]
[12.3]
⫽ NCFj(P兾A,i,nj)
Since the bracketed expression in Equation [12.3] is the (P兾A,i,nj) factor, computation of PWj for
nj years assumes reinvestment at the MARR of all positive net cash flows until the longest-lived
project is completed in year nL.
To demonstrate numerically, consider bundle j ⫽ 7 in Example 12.2. The evaluation is in
Table 12–3, and the net cash flow is pictured in Figure 12–4. Calculate the future worth in year 9,
which is the life of the longest-lived project (B).
FW ⫽ 5220(F兾A,15%,4)(F兾P,15%,5) ⫹ 2680(F兾P,15%,4) ⫽ $57,111
The present worth at the initial investment time is
PW ⫽ ⫺16,000 ⫹ 57,111(P兾F,15%,9) ⫽ $235
The PW value is the same as PW7 in Table 12–3 and Figure 12–2. This demonstrates the reinvestment assumption for positive net cash flows. If this assumption is not realistic, the PW analysis
must be conducted using the LCM of all project lives.
12.4 Capital Budgeting Problem Formulation
Using Linear Programming
The procedure discussed above requires the development of mutually exclusive bundles one
project at a time, two projects at a time, etc., until all 2m bundles are developed and each one is
compared with the capital limit b. As the number of independent projects increases, this process
becomes prohibitively cumbersome and unworkable. Fortunately, the capital budgeting problem
can be stated in the form of a linear programming model. The problem is formulated using the
integer linear programming (ILP) model, which means simply that all relations are linear and that
the variable x can take on only integer values. In this case, the variables can only take on the
values 0 or 1, which makes it a special case called the 0-or-1 ILP model. The formulation in
words follows.
330
Independent Projects with Budget Limitation
Chapter 12
Maximize: Sum of PW of net cash flows of independent projects.
Constraints:
• Capital investment constraint is that the sum of initial investments must not exceed a specified limit.
• Each project is completely selected or not selected.
For the math formulation, define b as the capital investment limit, and let xk (k ⫽ 1 to m projects)
be the variables to be determined. If xk ⫽ 1, project k is completely selected; if xk ⫽ 0, project k
is not selected. Note that the subscript k represents each independent project, not a mutually
exclusive bundle.
If the sum of PW of the net cash flows is Z, the math programming formulation is as follows:
k⫽m
Maximize:
兺 PW x
⫽Z
k k
k⫽1
k⫽m
Constraints:
兺 NCF
k 0 xk
[12.4]
ⱕb
k⫽1
xk ⫽ 0 or 1
for k ⫽ 1, 2, . . . , m
The PWk of each project is calculated using Equation [12.1] at MARR ⫽ i.
PWk ⴝ PW of project net cash flows for nk years
tⴝnk
ⴝ
兺 NCF
kt (P兾F,i,t)
ⴚ NCFk0
[12.5]
tⴝ1
Computer solution is accomplished by a linear programming software package which treats the ILP
model. Also, Excel and its optimizing tool Solver can be used to develop the formulation and select
the projects. The Solver tool is similar to Goal Seek with significantly more capabilities. For example,
Solver allows the target cell to be maximized, minimized, or set to a specific value. This means that the
function Z in Equation [12.4] can be maximized. Also, multiple changing cells can be identified, so
the 0 or 1 value of the unknowns can be determined. Additionally, with the added capability to include
constraints, the investment limit b and 0-or-1 requirement on the unknowns in Equation [12.4] can be
accommodated. Solver is explained in detail in Appendix A, and Example 12.3 illustrates its use.
EXAMPLE 12.3
Review Example 12.2. (a) Formulate the capital budgeting problem using the math programming model presented in Equation [12.4]. (b) Select the projects using Solver.
Solution
(a) Define the subscript k ⫽ 1 through 4 for the four projects, which are relabeled as 1, 2, 3,
and 4. The capital investment limit is b ⫽ $20,000 in Equation [12.4].
k⫽4
兺 PW x
Maximize:
k k
⫽Z
k⫽1
k⫽4
兺 NCF
Constraints:
k0 xk
ⱕ 20,000
k⫽1
xk ⫽ 0 or 1
for k ⫽ 1, 2, 3, 4
Now, substitute the PWk and NCFk0 values from Table 12–3 into the model. Plus signs are
used for all values in the budget constraint. We have the complete 0-or-1 ILP formulation.
Maximize:
6646x1 ⫺ 1019x2 ⫹ 984x3 ⫺ 748x4 ⫽ Z
Constraints:
8000x1 ⫹ 15,000x2 ⫹ 8000x3 ⫹ 8000x4 ⱕ 20,000
x1, x2, x3, and x4 ⫽ 0 or 1
The maximum PW is $7630, and the solution from Example 12.2 is written as
x1 ⫽ 1
x2 ⫽ 0
x3 ⫽ 1
x4 ⫽ 0
Capital Budgeting Problem Formulation Using Linear Programming
12.4
⫽ NPV($B$1,E7:E18) ⫹ E6
Figure 12–5
Spreadsheet and Solver template configured to solve a capital budgeting problem, Example 12.3.
(b) Figure 12–5 presents a spreadsheet template developed to select from six or fewer independent projects with 12 years or less of net cash flow estimates per project. The spreadsheet
template can be expanded in either direction if needed. Figure 12–5 (inset) shows the
Solver parameters set to solve this example for four projects and an investment limit of
$20,000. The descriptions below and the cell tag identify the contents of the rows and cells
in Figure 12–5, and their linkage to Solver parameters.
Rows 4 and 5: Projects are identified by numbers to distinguish them from spreadsheet column letters. Cell I5 is the expression for Z, the sum of the PW values for
the projects. This is the target cell for Solver to maximize.
Rows 6 to 18: These are initial investments and net cash flow estimates for each
project. Zero values that occur after the life of a project need not be entered;
however, any $0 estimates that occur during a project’s life must be entered.
Row 19: The entry in each cell is 1 for a selected project and 0 if not selected. These
are the changing cells for Solver. Since each entry must be 0 or 1, a binary constraint is placed on all row 19 cells in Solver, as shown in Figure 12–5. When a
problem is to be solved, it is best to initialize the spreadsheet with 0s for all projects. Solver will find the solution to maximize Z.
Row 20: The NPV function is used to find the PW for each net cash flow series. The
NPV functions are developed for any project with a life up to 12 years at the
MARR entered in cell B1.
Row 21: When a project is selected, the contribution to the Z function is shown by
multiplying rows 19 and 20.
Row 22: This row shows the initial investment for the selected projects. Cell I22 is
the total investment. This cell has the budget limitation placed on it by the constraint in Solver. In this example, the constraint is I22 ⬍⫽ $20,000.
To solve the example, set all values in row 19 to 0, set up the Solver parameters as described
above, and click on Solve. (Since this is a linear model, the Solver options choice “Assume Linear Model” may be checked, if desired.) If needed, further directions on saving the solution,
making changes, etc., are available in Appendix A, Section A.5, and on the Excel help function.
For this problem, the selection is projects 1 and 3 with Z ⫽ $7630, the same as determined
previously, and $16,000 of the $20,000 limit is invested.
331
332
Independent Projects with Budget Limitation
Chapter 12
12.5 Additional Project Ranking Measures
The PW-based method of solving a capital budgeting problem covered in Sections 2.2 to 2.4
provides an optimal solution that maximizes the PW of the competing projects. However, it is
very common in industrial, professional, and government settings to learn that the rate of return
is the basis for ranking projects. The internal rate of return (IROR), as we learned in Section 7.2, is determined by setting a PW or AW relation equal to zero and solving for i* — the
IROR. Using a PW basis and the estimated net cash flow (NCF) series for each project j, solve
for i* in the relation
t⫽nj
0⫽
兺 NCF (P兾F,i*,t) ⫺ NCF
jt
j0
[12.6]
t⫽1
This is the same as Equation [12.1] set equal to zero with i ⫽ i* as the unknown. The spreadsheet
function RATE or IRR will provide the same answer. The selection guideline is as follows:
Independent project
selection
Once the project ranking by IROR is complete, select all projects in order without exceeding
the investment limit b.
If there is no budget limit, select all projects that have IROR ⱖ MARR.
The ordering of projects using IROR ranking may differ from the PW-based ranking we used in
previous sections. This can occur because IROR ranking maximizes the overall rate of return, not
necessarily the PW value. The use of IROR ranking is illustrated in the next example.
Another common ranking method is the profitability index (PI) that we learned in
Section 9.2. This is a “bang for the buck” measure that provides a sense of getting the most for
the investment dollar over the life of the project. (Refer to Section 9.2 for more details.) When
utilized for project ranking, it is often called the present worth index (PWI); however, we will
use the PI term for consistency, and because there are other ways to mathematically define measures also referred to as a PW index. The PI measure is defined as
tⴝnj
兺 NCF (P兾F, i, t)
jt
PW of net cash flows ⴝ ————————
tⴝ1
———————————
PW of initial investment
円 NCFj0 円
[12.7]
Note that the denominator includes only the initial investment, and it is its absolute value that is
used. The numerator has only cash flows that result from the project for years 1 through its life
nj. Salvage value, if there is one estimated, is incorporated into the numerator. Similar to the
previous case, the selection guideline is as follows:
Independent project
selection
Once the project ranking by PI is complete, select all projects in order without exceeding the
investment limit b.
If there is no budget limit, select all projects that have PI ⱖ 1.0.
Depending upon the project NCF estimates, the PI ranking can differ from the IROR ranking. Example 12.4 compares results using the different ranking methods—IROR, PI, and PW values. None
of these results are incorrect; they simply maximize different measures, as you will see. The use of
IROR, PI, or other measures is common when there are a large number of projects, because the PW
basis (when solved by hand, Solver, or ILP software) becomes increasingly cumbersome as the number of possible mutually exclusive bundles grows using the formula 2m. Additionally, greater complexity is introduced when dependent and contingent projects are involved.
EXAMPLE 12.4
Georgia works as a financial analyst in the Management Science Group of General Electronics.
She has been asked to recommend which of the five projects detailed in Table 12–4 should be
funded if the MARR is 15% per year and the investment budget limit for next year is a firm
333
Additional Project Ranking Measures
12.5
IROR, PI, and PW Values for Five Projects, Example 12.4
TABLE 12–4
Projects
1
2
3
4
5
Investment, NCF0, $1000
NCF, $1000 per year
Life nj, years
IROR, %
PI
PW at 15%, $1000
⫺8,000
4,000
6
44.5
1.89
7,138
⫺15,000
2,900
9
12.8
0.92
⫺1,162
⫺8,000
2,700
5
20.4
1.13
1,051
⫺8,000
2,500
4
9.6
0.89
⫺863
⫺5,000
2,600
3
26.0
1.19
936
$18 million. She has confirmed the computations and is ready to do the ranking and make the
selection. Help her by doing the following:
(a)
(b)
(c)
(d)
Use the IROR measure to rank and select projects.
Use the PI measure to rank and select projects.
Use the PW measure to rank and select projects.
Compare the selected projects by the three methods and determine which one will maximize the overall ROR value of the $18 million budget.
Solution
Refer to Table 12–5 for the ranking, cumulative investment for each project, and selection
based on the ranking and budget limit b ⫽ $18 million.
(a) Ranking by overall IROR values indicates that projects 1 and 5 should be selected with
$13 million of the $18 million budget expended. The remaining $5 million is assumed to
be invested at the MARR of 15% per year.
(b) As an example, the PI for project 1 is calculated using Equation [12.7].
4000(P兾A,15%,6)
PI1 ⫽ ————————
|⫺8000|
⫽ 15,138兾8,000
⫽ 1.89
Ranking and projects selected are the same as in the IROR-based analysis. Again, the remaining $5 million is assumed to generate a return of MARR ⫽ 15% per year.
(c) Ranking by PW value results in a different selection from that of IROR and PI ranking.
Projects 1 and 3, rather than 1 and 5, are selected for a total PW ⫽ $8.189 million and an
investment of $16 million. The remaining $2 million is assumed to earn 15% per year.
(d ) IROR and PI rankings result in projects 1 and 5 being selected. PW ranking results in the
selection of projects 1 and 3. With MARR ⫽ 15%, the $18 million will earn at the
following ROR values. In $1000 units,
Projects 1 and 5
NCF, year 0:
NCF, years 1–3:
NCF, years 4–6:
$⫺13,000
$6,600
$4,000
0 ⫽ ⫺13,000 ⫹ 6600(P兾A,i,3) ⫹ 4000(P兾A,i,3)(P兾F,i,3)
Ranking of Projects by Different Measures, Example 12.4 (Monetary units in $1000)
TABLE 12–5
Ranking by IROR
Ranking by PI
Ranking by PW
IROR, %
(1)
Project
(2)
Cumulative
Investment, $
(3)
PI
(4)
Project
(5)
Cumulative
Investment, $
(6)
44.5
26.0
1
5
8,000
13,000
1.89
1.19
1
5
8,000
13,000
20.4
12.8
9.6
3
2
4
21,000
1.13
0.92
0.89
3
2
4
21,000
Project
(8)
Cumulative
Investment, $
(9)
7,138
1,051
1
3
8,000
16,000
936
⫺863
⫺1,162
5
4
2
21,000
PW, $
(7)
334
Independent Projects with Budget Limitation
Chapter 12
By IRR function, the rate of return is 39.1%. The overall return on the entire budget is
ROR ⫽ [39.1(13,000) ⫹ 15.0(5000)]兾18,000
⫽ 32.4%
Projects 1 and 3
NCF, year 0:
NCF, years 1–5:
NCF, year 6:
$⫺16,000
$6,700
$4,000
0 ⫽ ⫺16,000 ⫹ 6700(P兾A,i,5) ⫹ 4000(P兾F,i,6)
By IRR function, the rate of return is 33.5%. The overall return on the entire budget is
ROR ⫽ [33.5(16,000) ⫹ 15.0(2000)]兾18,000
⫽ 31.4%
In conclusion, the IROR and PI selections maximize the overall rate of return at 32.4%. The
PW selection maximizes the PW value at $8.189 (i.e., 7.138 ⫹ 1.051) million, as determined
from Table 12–5, column 7.
CHAPTER SUMMARY
Investment capital is always a scarce resource, so it must be rationed among competing projects
using economic and noneconomic criteria. Capital budgeting involves proposed projects, each
with an initial investment and net cash flows estimated over the life of the project. The fundamental capital budgeting problem has specific characteristics (Figure 12–1).
•
•
•
•
Selection is made from among independent projects.
Each project must be accepted or rejected as a whole.
Maximizing the present worth of the net cash flows is the objective.
The total initial investment is limited to a specified maximum.
The present worth method is used for evaluation. To start the procedure, formulate all mutually exclusive bundles that do not exceed the investment limit, including the do-nothing bundle.
There are a maximum of 2m bundles for m projects. Calculate the PW at MARR for each bundle,
and select the bundle with the largest PW value. Reinvestment of net positive cash flows at the
MARR is assumed for all projects with lives shorter than that of the longest-lived project.
The capital budgeting problem may be formulated as a linear programming problem to select
projects directly in order to maximize the total PW. Excel’s Solver tool solves this problem by
spreadsheet.
Projects can be ranked and selected on bases other than PW values. Two measures are the
internal rate of return (IROR) and the profitability index (PI), also called the PW index. The
ordering of projects may differ between the various ranking bases since different measures are
optimized. When there are a large number of projects, the IROR basis is commonly applied in
industrial settings.
PROBLEMS
Understanding Capital Rationing
12.1 Define the following terms: bundle, contingent
project, dependent project.
12.2 State two assumptions made when doing capital rationing using a PW analysis for unequal-life projects.
12.3 For independent projects identified as A, B, C, D,
E, F, and G, how many mutually exclusive bundles
can be formed?
12.4 For independent projects identified as W, X, Y, and
Z, develop all of the mutually exclusive bundles.
Projects X and Y perform the same function with
different processes; both should not be selected.
12.5 Five projects have been identified for possible
implementation by a company that makes dry ice
blasters—machines that propel tiny dry ice pellets at supersonic speeds so they flash-freeze and
then lift grime, paint, rust, mold, asphalt, and
335
Problems
other contaminants off in-place machines and a
wide range of surfaces. The total present worth
of the initial investment for each project is
shown. Determine which bundles are possible,
provided the budget limitation is (a) $34,000 and
(b) $45,000.
Project
PW of Investment
at 15%, $
L
M
N
O
P
28,000
11,000
43,000
38,000
6,000
Project
12.7 Develop all acceptable mutually exclusive bundles for the four independent projects described
below if the investment limit is $400 and the following project selection restriction applies: Project 1 can be selected only if both projects 3 and 4
are selected.
Investment, $
1
2
3
4
⫺100
⫺150
⫺75
⫺235
Initial
Investment, $
PW at 18%,
$
⫺15,000
⫺25,000
⫺20,000
⫺40,000
⫺400
8500
500
9600
1
2
3
4
12.6 Four independent projects (1, 2, 3, and 4) are proposed for investment by Perfect Manufacturing,
Inc. List all the acceptable mutually exclusive
bundles based on the following selection restrictions developed by the department of engineering
production:
Project 2 can be selected only if project 3 is
selected.
Projects 1 and 4 should not both be selected;
they are essentially duplicates.
Project
improve surface durability on stainless steel
products. The project costs and 18% per year PW
values are as shown. What projects should be accepted if the budget limit is (a) no limit and
(b) $55,000?
Selection from Independent Projects
12.8 Listed below are bundles, each comprised of three
independent proposals for which the PW has been
estimated. Select the best bundle if the capital budget limit is $45,000 and the MARR is the cost of
capital, which is 9% per year.
Proposal
Bundle
Initial Investment,
For Bundle, $
PW at 9%,
$
1
2
3
4
⫺18,000
⫺26,000
⫺34,000
⫺41,000
⫺1,400
8,500
7,100
10,500
12.9 The general manager for Woodslome Appliance
Company Plant #A14 in Mexico City has four
independent projects that she can fund this year to
12.10 The capital fund for research project investment at
SummaCorp is limited to $100,000 for next year.
Select any or all of the following proposals if a
MARR of 15% per year is established by the board
of directors.
Initial
Project Investment, $
I
II
III
⫺25,000
⫺30,000
⫺50,000
Annual Net Cash
Flow, $兾Year
6,000
9,000
15,000
Life, Salvage
Years Value, $
4
4
4
4,000
⫺1,000
20,000
12.11 An electrical engineer at GE is assigned the responsibility to determine how to invest up to $100,000
in none, some, or all of the following independent
proposals. Use (a) hand and (b) spreadsheet-based
PW analysis and a 15% per year return requirement
to help this engineer make the best decision from a
purely economic perspective.
Project
Initial
Investment, $
Annual Net Cash
Flow, $兾Year
Life,
Years
Salvage
Value, $
A
B
C
⫺25,000
⫺20,000
⫺50,000
6,000
9,000
15,000
4
4
4
4,000
0
20,000
12.12 Dwayne has four independent vendor proposals to
contract the nationwide oil recycling services for
Ford Corporation manufacturing plants. All combinations are acceptable, except that vendors B
and C cannot both be chosen. Revenue sharing of
recycled oil sales with Ford is a part of the requirement. Develop all possible mutually exclusive
bundles under the additional following restrictions
and select the best projects. The corporate MARR
is 10% per year.
(a) A maximum of $4 million can be spent.
(b) A larger budget of $5.5 million is allowed,
but no more than two vendors can be
selected.
(c) There is no limit on spending.
336
Independent Projects with Budget Limitation
Chapter 12
Vendor
Initial
Investment, $
Life,
Years
Annual Net
Revenue, $ per Year
A
B
C
D
⫺1.5 million
⫺3.0 million
⫺1.8 million
⫺2.0 million
8
10
5
4
360,000
600,000
620,000
630,000
12.13 Use the PW method at 8% per year to select up to
three projects from the four available ones if no
more than $20,000 can be invested. Estimated
lives and annual net cash flows vary.
Initial
Project Investment, $
W
X
Y
Z
⫺5,000
⫺8,000
⫺8,000
⫺10,000
Net Cash Flow, $ per Year
1
2
3
4
5
6
1,000 1,700 1,800 2,500 2,000
900 950 1,000 1,050 10,500
4,000 3,000 1,000
500
500 2,000
0
0
0 17,000
12.14 Charlie’s Garage has $70,000 to spend on new
equipment that may increase revenue for his car
repair shop. Use the PW method to determine
which of these independent investments are financially acceptable at 6% per year, compounded
monthly. All are expected to last 3 years.
Feature
Installed
Cost, $
Estimated Added
Revenue, $ per Month
Diagnostics system
Exhaust analyzer
Hybrid engine tester
⫺45,000
⫺30,000
⫺22,000
2200
2000
1500
12.15 (a)
Initial
Investment, $
Life, Years
PW at 12%
per Year, $
S
A
M
E
H
⫺15,000
⫺25,000
⫺10,000
⫺25,000
⫺40,000
6
8
6
4
12
8,540
12,325
3,000
10
15,350
12.17 The independent project estimates below have
been developed by the engineering and finance
managers. The corporate MARR is 8% per year,
and the capital investment limit is $4 million. Select the economically best projects using the PW
method and (a) hand solution and (b) spreadsheet
solution.
Project
Project
Cost, $ M
Life, Years
NCF,
$ per Year
1
2
3
4
⫺1.5
⫺3.0
⫺1.8
⫺2.0
8
10
5
4
360,000
600,000
520,000
820,000
12.18 Use the PW method to evaluate four independent
projects. Select as many as three of the four projects. The MARR is 12% per year, and up to
$16,000 in capital investment funds are available.
Project
Investment, $
Life, years
Determine which of the following independent projects should be selected for investment if $315,000 is available and the MARR
is 10% per year. Use the PW method to
evaluate mutually exclusive bundles to
make the selection. (Solve by hand or
spreadsheet as instructed.)
1
2
3
4
⫺5000
5
⫺8,000
5
⫺9,000
3
⫺10,000
4
Year
1
2
3
4
5
NCF Estimates, $ per Year
1000
1700
2400
3000
3800
500
500
500
500
10,500
5000
5000
2000
0
0
0
17,000
12.19 Work Problem 12.18 using a spreadsheet.
Project
Initial
Investment, $
NCF,
$ per Year
Life, Years
A
B
C
D
E
⫺100,000
⫺125,000
⫺120,000
⫺220,000
⫺200,000
50,000
24,000
75,000
39,000
82,000
8
8
8
8
8
(b)
Project
If the five projects are mutually exclusive
alternatives, perform the present worth analysis and select the best alternative.
12.16 Use the following analysis of five independent
projects to select the best, if the capital limitation
is (a) $30,000, (b) $52,000, and (c) unlimited.
12.20 A capital rationing problem is defined for you as
follows: Three projects are to be evaluated at a
MARR of 12.5% per year. No more than $3.0
million can be invested.
(a) Use a spreadsheet to select from the independent projects.
(b) If the life of project 3 can be increased from
5 to 10 years for the same $1 million investment, use Goal Seek to determine the NCF in
year 1 for project 3 alone to have the same
PW as the best bundle in part (a). All other
estimates remain the same. With these new
NCF and life estimates, what are the best
projects for investment?
337
Problems
Estimated NCF, $ per Year
Project
Investment,
$M
Life,
Years
Year 1
Gradient
after Year 1
1
2
3
⫺0.9
⫺2.1
⫺1.0
6
10
5
250,000
485,000
200,000
⫺5000
⫹5000
⫹20%
Linear Programming and Capital Budgeting
12.21 Formulate the linear programming model, develop
a spreadsheet, and solve the capital rationing problem in Example 12.1 (a) as presented and (b) using
an investment limit of $13 million.
12.22 Use linear programming and a spreadsheet-based
solution technique to select from the independent
unequal-life projects in Problem 12.17.
12.25 Using the estimates in Problem 12.18 and repeated
spreadsheet solution of the capital budgeting problem for budget limits ranging from b ⫽ $5000 to b
⫽ $25,000, develop a spreadsheet chart that plots b
versus the value of Z.
Other Ranking Measures
12.26 A newly proposed project has a first cost of
$325,000 and estimated annual income of $60,000
per year for 8 years.
(a) Determine the IROR, PI, and PW values if
the MARR is 15% per year.
(b) Is the project economically justified?
12.27 An engineer at Delphi Systems is considering the
projects below, all of which can be considered to
last indefinitely. The company’s MARR is 13%
per year.
(a) Determine which projects should be selected
on the basis of IROR if the budget limitation
is $39,000.
(b) What is the overall rate of return if the money
not invested in projects is assumed to earn a
rate of return equal to the MARR?
First Cost, $
Net Income,
$ per Year
Rate of
Return, %
A
B
C
D
E
⫺20,000
⫺10,000
⫺15,000
⫺70,000
⫺50,000
4,000
1,900
2,600
10,000
6,000
20.0
19.0
17.3
14.3
12.0
12.28 The five independent projects shown below are under
consideration for implementation by KNF Neuberger, Inc. The company’s MARR is 15% per year.
(a) Determine which projects should be undertaken on the basis of IROR if the budget
limitation is $97,000. (Solve by hand or
spreadsheet as instructed.)
(b) Determine the overall rate of return if the
money not invested in projects is assumed to
earn a rate of return equal to the MARR.
12.23 Solve the capital budgeting problem in Problem
12.20(a), using the linear programming model and
a spreadsheet.
12.24 Johnson and Johnson is expanding its first-aid
products line for individuals allergic to latex. The
research team comprised of doctors, engineers,
and chemists has proposed the four projects estimated in Problem 12.18. Develop the linear
programming model and use a spreadsheet to select the projects to be funded. The MARR is 12%
per year, and the budget limit is $16 million. All
monetary values are in $1000 units.
Project
Annual
Project
Project First Cost, $ Income, $ per Year Life, Years
L
A
N
D
T
⫺30,000
⫺15,000
⫺45,000
⫺70,000
⫺40,000
9,000
4,900
11,100
9,000
10,000
10
10
10
10
10
12.29 An estimated 6 billion gallons of clean drinking
water disappear each day across the United States
due to aging, leaky pipes and water mains before
it gets to the consumer or industrial user. The American Society of Civil Engineers (ASCE) has teamed
with municipalities, counties, and several excavation
companies to develop robots that can travel through
mains, detect leaks, and repair many of them immediately. Four proposals have been received for funding. There is a $100 million limit on capital funding,
and the MARR is established at 12% per year.
(a) Use the IROR method to rank and determine
which of the four independent projects should
be funded. Solve by spreadsheet, unless instructed to use hand solution.
(b) Determine the rate of return for the combination of projects selected in part (a). Is it economically justified?
(c) Determine the overall rate of return for the
$100 million with the projects selected in
part (a). Assume that excess funds are invested at the MARR.
Project
First
Cost, $ M
Estimated Annual
Savings, $M per Year
Project
Life, Years
W
X
Y
Z
⫺12
⫺25
⫺45
⫺60
5.0
7.3
12.1
9.0
3
4
6
8
338
Independent Projects with Budget Limitation
Chapter 12
12.30 Determine the profitability index at 10% per year
interest for a project that has a first cost of
$200,000 in year 0 and $200,000 in year 2, an annual operating cost of $80,000 per year, revenue
of $170,000 per year, and a salvage value of
$60,000 after its 5-year life.
12.31 The budget limit is $120,000, and the interest rate
is 10% per year. All projects have a 10-year life.
Use (a) the PI method and (b) the IROR method to
rank and select from the independent projects.
(c) Are different projects selected using the two
methods?
Project
First Cost, $
Net Income,
$ per Year
A
B
C
D
E
⫺18,000
⫺15,000
⫺35,000
⫺60,000
⫺50,000
4,000
2,800
12,600
13,000
8,000
IROR, %
18.0
13.3
34.1
17.3
9.6
12.32 The six independent projects shown below are
under consideration by Peyton Packing under
budget-constrained conditions. The company always has more projects to engage in than it has
capital to fund projects. Therefore, it uses a relatively high MARR of 25% per year. Since all projects are considered long-term ventures, the company uses an infinite period for their life. Determine
which projects the company should fund and the
total investment for a capital budget of $700,000 if
the capital budgeting method used is (a) the IROR
method, (b) the PI method, and (c) the PW method.
Project
First Cost, $
Estimated Annual
Income, $ per Year
F
G
H
I
J
K
⫺200,000
⫺120,000
⫺250,000
⫺370,000
⫺50,000
⫺9,000
54,000
21,000
115,000
205,000
26,000
2,100
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
12.33 A project that has a condition associated with its
acceptance or rejection is known as:
(a) A mutually exclusive alternative
(b) A contingent project
(c) A dependent project
(d) Both (b) and (c)
12.34 An assumption inherent in capital budgeting is that
all positive net cash flows are reinvested at the
MARR from the time they are realized until:
(a) The end of the shortest-lived project
(b) The end of the longest-lived project
(c) The end of the respective project that generated the cash flow
(d) The average of the lives of the projects included in the bundle
12.35 All of the following are correct when a capital
budgeting problem is solved using the 0-1 integer
linear programming model except:
(a) Partial investment in a project is acceptable.
(b) The objective is to maximize the present
worth of the investments.
(c) Budget constraints may be present for the
first year only or for several years.
(d) Contingent and dependent project restrictions may be considered.
12.36 All of the following are true when formulating
mutually exclusive bundles of independent projects except:
(a)
(b)
(c)
(d)
One of the bundles is the do-nothing project.
A bundle may consist of only one project.
The capital limit may be exceeded as long as
it is exceeded by less than 3%.
A bundle may include contingent and dependent projects.
12.37 When there are 5 projects involved in a capital
budgeting study, the maximum number of bundles
that can be formulated is:
(a) 6
(b) 10
(c) 31
(d) 32
12.38 For a capital limit of $25,000, the selected independent projects are:
(a) P only
(b) Q only
(c) R only
(d) P and R
Project
Initial
Investment, $
Life, Years
PW at
12%, $
P
Q
R
S
T
⫺15,000
⫺25,000
⫺10,000
⫺25,000
⫺40,000
6
8
6
4
12
8,540
12,325
3,000
10
15,350
Additional Problems and FE Exam Review Questions
12.39 The independent projects shown below are under
consideration for possible implementation by
Renishaw Inc. of Hoffman Estates, Rhode Island.
If the company’s MARR is 14% per year and it
uses the IROR method of capital budgeting, the
projects it should select under a budget limitation
of $105,000 are:
(a) A, B, and C
(b) A, B, and D
(c) B, C, and D
(d) A, C, and D
Project
First Cost, $
Annual
Income, $ per Year
Rate of
Return, %
A
B
C
D
E
⫺20,000
⫺10,000
⫺15,000
⫺70,000
⫺50,000
4,000
1,900
2,600
10,000
6,000
20.0
19.0
17.3
14.3
12.0
339
12.40 For a project that requires an initial investment
of $26,000 and yields $10,000 per year for
4 years, the PI at an interest rate of 10% per year
is closest to:
(a) 1.03
(b) 1.22
(c) 1.38
(d) 1.56
CHAPTER 13
Breakeven and
Payback
Analysis
L E A R N I N G
O U T C O M E S
Purpose: Determine the breakeven for one or two alternatives and calculate the payback period with and without a
return required.
SECTION
TOPIC
LEARNING OUTCOME
13.1
Breakeven point
• Determine the breakeven point for one
parameter.
13.2
Two alternatives
• Calculate the breakeven point of a parameter
and use it to select between two alternatives.
13.3
Payback period
• Determine the payback period of a project at
i ⫽ 0% and i ⬎ 0%. Illustrate the cautions when
using payback analysis.
13.4
Spreadsheet
• Answer breakeven and payback questions that
are best resolved by spreadsheet and Goal Seek
tools.
B
reakeven analysis is performed to determine the value of a variable or parameter of a project or alternative that makes two elements equal, for example, the
sales volume that will equate revenues and costs. A breakeven study is performed
for two alternatives to determine when either alternative is equally acceptable. Breakeven
analysis is commonly applied in make-or-buy decisions when a decision is needed about the
source for manufactured components, services, etc.
Payback analysis determines the required minimum life of an asset, process, or system to
recover the initial investment. There are two types of payback: return (i ⬎ 0%) and no return
(i ⫽ 0%). Payback analysis should not be considered the final decision maker; it is used as
a screening tool or to provide supplemental information for a PW, AW, or other analysis.
These aspects are discussed in depth in this chapter.
Breakeven and payback studies use estimates that are considered to be certain; that is, if
the estimated values are expected to vary enough to possibly change the outcome, another
study is necessary using different estimates. If the variable of interest is allowed to vary, the
approaches of sensitivity analysis (Chapter 18) should be used. Additionally, if probability
and risk assessment are considered, the tools of simulation (Chapter 19) can be used to
supplement the static nature of a breakeven or payback study.
13.1 Breakeven Analysis for a Single Project
When one of the engineering economy symbols—P, F, A, i, or n—is not known or not estimated,
a breakeven quantity can be determined by setting an equivalence relation for PW or AW equal
to zero. This form of breakeven analysis has been used many times so far. For example, we have
solved for the rate of return i*, found the replacement value for a defender, and determined the P,
F, A, or salvage value S at which a series of cash flow estimates return a specific MARR. Methods
used to determine the quantity include
Direct solution by hand if only one factor is present (say, P兾A) or only single amounts are
estimated (for example, P and F)
Trial and error by hand or calculator when multiple factors are present
Spreadsheet when cash flow and other estimates are entered into cells and used in resident
functions (PV, FV, RATE, IRR, NPV, PMT, and NPER) or tools (Goal Seek and Solver).
We now concentrate on the determination of the breakeven quantity QBE for one parameter
or decision variable. For example, the variable may be a design element to minimize cost or the
production level needed to realize revenues that exceed costs by 10%.
Breakeven analysis finds the value of a parameter that makes two elements equal. The breakeven point QBE is determined from mathematical relations, e.g., product revenue and costs or
materials supply and demand or other parameters that involve the parameter Q. Breakeven
analysis is fundamental to evaluations such as make-buy decisions.
The unit of the parameter Q may vary widely: units per year, cost per kilogram, hours per month,
percentage of full plant capacity, etc.
Figure 13–1a presents different shapes of a revenue relation identified as R. A linear revenue
relation is commonly assumed, but a nonlinear relation is often more realistic. It can model an
increasing per unit revenue with larger volumes (curve 1 in Figure 13–1a) or a decreasing per
unit revenue that usually prevails at higher quantities (curve 2).
Costs, which may be linear or nonlinear, usually include two components—fixed and
variable—as indicated in Figure 13–1b.
Fixed costs (FC). These include costs such as buildings, insurance, fixed overhead, some
minimum level of labor, equipment capital recovery, and information systems.
Variable costs (VC). These include costs such as direct labor, materials, indirect costs, contractors, marketing, advertisement, and warranty.
The fixed-cost component is essentially constant for all values of the variable, so it does not vary
for a large range of operating parameters, such as production level or workforce size. Even if no
units are produced, fixed costs are incurred at some threshold level. Of course, this situation
Breakeven
342
Breakeven and Payback Analysis
Chapter 13
Figure 13–1
(1)
Linear and nonlinear
revenue and cost relations.
R, revenue per year
Nonlinear
(2)
Linear
Q, units per year
(a) Revenue relations—(1) increasing and
(2) decreasing revenue per unit
TC ⫽ FC ⫹ VC
Total
cost, TC
Variable, VC
Cost per year
Cost per year
TC ⫽ FC ⫹ VC
TC
VC
FC
Fixed, FC
Q, units per year
(b) Linear cost relations
Q, units per year
(c) Nonlinear cost relations
cannot last long before the plant must shut down to reduce fixed costs. Fixed costs are reduced
through improved equipment, information systems, and workforce utilization; and less costly
fringe benefit packages, subcontracting specific functions, and so on.
Variable costs change with production level, workforce size, and other parameters. It is usually possible to decrease variable costs through better product design, manufacturing efficiency,
improved quality and safety, and higher sales volume.
When FC and VC are added, they form the total cost relation TC. Figure 13–1b illustrates
the TC relation for linear fixed and variable costs. Figure 13–1c shows a general TC curve for a
nonlinear VC in which unit variable costs decrease as the quantity level rises.
At a specific but unknown value Q of the decision variable, the revenue R and total cost TC
relations will intersect to identify the breakeven point QBE (Figure 13–2). If Q ⬎ QBE, there is a
predictable profit; but if Q ⬍ QBE, there is a loss. For linear models of R and VC, the greater the
quantity, the larger the profit. Profit is calculated as
Profit ⴝ revenue ⴚ total cost
ⴝ R ⴚ TC
ⴝ R ⴚ (FC ⴙ VC)
[13.1]
A relation for the breakeven point may be derived when revenue and total cost are linear functions
of quantity Q by setting the relations for R and TC equal to each other, indicating a profit of zero.
R ⫽ TC
rQ ⫽ FC ⫹ vQ
where
r ⫽ revenue per unit
v ⫽ variable cost per unit
Solve for the breakeven quantity Q ⫽ QBE for linear R and TC functions.
FC
QBE ⴝ ———
rⴚv
[13.2]
Breakeven Analysis for a Single Project
13.1
$
R
TC
Breakeven
TC with
lowered VC
Breakeven with
lowered VC
Profit
maximized
Loss
Breakeven
point moves
Profit
QBE
Q, units per year
Figure 13–2
Effect on the breakeven point when the variable cost per unit is reduced.
$
TC
R
Profit
maximized
Profit range
Loss
Loss
QBE
QP
QBE
Q, units per year
Figure 13–3
Breakeven points and maximum profit point for a nonlinear analysis.
The breakeven graph is an important management tool because it is easy to understand and may be used
in decision making in a variety of ways. For example, if the variable cost per unit is reduced, then the
TC line has a smaller slope (Figure13–2) and the breakeven point will decrease. This is an advantage
because the smaller the value of QBE, the greater the profit for a given amount of revenue. A similar
analysis is possible for fixed VC and increased levels of production, as shown in the next example.
If nonlinear R or TC models are used, there may be more than one breakeven point. Figure 13–3
presents this situation for two breakeven points. The maximum profit occurs at QP between the
two breakeven points where the distance between the R and TC relations is greatest.
343
344
Breakeven and Payback Analysis
Chapter 13
Of course, no static R and TC relations—linear or nonlinear—are able to estimate exactly the
revenue and cost amounts over an extended period of time. But the breakeven point is an excellent target for planning purposes.
EXAMPLE 13.1
Indira Industries is a major producer of diverter dampers used in the gas turbine power industry
to divert gas exhausts from the turbine to a side stack, thus reducing the noise to acceptable
levels for human environments. Normal production level is 60 diverter systems per month, but
due to significantly improved economic conditions in Asia, production is at 72 per month. The
following information is available.
Fixed costs
Variable cost per unit
Revenue per unit
FC ⫽ $2.4 million per month
v ⫽ $35,000
r ⫽ $75,000
(a) How does the increased production level of 72 units per month compare with the current
breakeven point?
(b) What is the current profit level per month for the facility?
(c) What is the difference between the revenue and variable cost per damper that is necessary
to break even at a significantly reduced monthly production level of 45 units, if fixed costs
remain constant?
Solution
(a) Use Equation [13.2] to determine the breakeven number of units. All dollar amounts are
in $1000 units.
FC
QBE ⫽ ———
r⫺v
2400 ⫽ 60 units per month
⫽ ————
75 ⫺ 35
Figure 13–4 is a plot of R and TC lines. The breakeven value is 60 damper units. The
increased production level of 72 units is above the breakeven value.
7000
Revenue line
6000
5000
$1000
4500
Profit
4000
Total cost line
3000
2400
2000
1000
QBE = 60
0
15
30
Figure 13–4
Breakeven graph, Example 13.1.
45
60
75
Q, units per month
90
13.2
345
Breakeven Analysis Between Two Alternatives
(b) To estimate profit (in $1000 units) at Q ⫽ 72 units per month, use Equation [13.1].
Profit ⫽ R ⫺ TC ⫽ rQ ⫺ (FC ⫹ vQ)
⫽ (r ⫺ v)Q ⫺ FC
⫽ (75 ⫺ 35)72 ⫺ 2400
⫽ $480
[13.3]
There is a profit of $480,000 per month currently.
(c) To determine the required difference r ⫺ v, use Equation [13.3] with profit ⫽ 0, Q ⫽ 45,
and FC ⫽ $2.4 million. In $1000 units,
0 ⫽ (r ⫺ v) (45) ⫺ 2400
2400 ⫽ $53.33 per unit
r ⫺ v ⫽ ———
45
The spread between r and v must be $53,330. If v stays at $35,000, the revenue per damper
must increase from $75,000 to $88,330 (i.e., 35,000 ⫹ 53,330) just to break even at a production level of Q ⫽ 45 per month.
In some circumstances, breakeven analysis performed on a per unit basis is more meaningful.
The value of QBE is still calculated using Equation [13.2], but the TC relation is divided by Q to
obtain an expression for cost per unit, also termed average cost per unit Cu.
FC ⫹ vQ FC
TC ⫽ ————
⫽ —— ⫹ v
[13.4]
Cu ⫽ ——
Q
Q
Q
At the breakeven quantity Q ⫽ QBE, the revenue per unit is exactly equal to the cost per unit. If
graphed, the FC per unit term in Equation [13.4] takes on the shape of a hyperbola.
The breakeven point for a project with one unknown variable can always be determined by
equating revenue and total cost. This is the same as setting profit equal to zero in Equation [13.1].
It may be necessary to perform some dimensional analysis initially to obtain the correct revenue
and total cost relations in order to use the same dimension for both relations, for example, $ per
unit, miles per month, or units per year.
13.2 Breakeven Analysis Between Two Alternatives
Now we consider breakeven analysis between two mutually exclusive alternatives.
Breakeven analysis determines the value of a common variable or parameter between two
alternatives. Equating the two PW or AW relations determines the breakeven point. Selection
of the alternative is different depending upon two facts: slope of the variable cost curve and
the parameter value relative to the breakeven point.
The parameter can be the interest rate i, first cost P, annual operating cost (AOC), or any parameter. We have already performed breakeven analysis between alternatives on several parameters.
For example, the incremental ROR value (⌬i*) is the breakeven rate between alternatives. If the
MARR is lower than ⌬i*, the extra investment of the larger-investment alternative is justified. In
Section 11.6, the replacement value (RV) of a defender was determined. If the market value is
larger than RV, the decision should favor the challenger.
Often breakeven analysis involves revenue or cost variables common to both alternatives,
such as price per unit, operating cost, cost of materials, or labor cost. Figure 13–5 illustrates
the breakeven concept for two alternatives with linear cost relations. The fixed cost of alternative 2 is greater than that of alternative 1. However, alternative 2 has a smaller variable cost,
as indicated by its lower slope. The intersection of the total cost lines locates the breakeven
point, and the variable cost establishes the slope. Thus, if the number of units of the common
variable is greater than the breakeven amount, alternative 2 is selected, since the total cost
will be lower. Conversely, an anticipated level of operation below the breakeven point favors
alternative 1.
Breakeven
346
Breakeven and Payback Analysis
Chapter 13
Figure 13–5
Alt. 1 TC
Total cost
Breakeven between two
alternatives with linear
cost relations.
Alt. 2 FC
Alt. 1 FC
Alt. 2 TC
Breakeven
point
Common variable, units
Instead of plotting the total costs of each alternative and estimating the breakeven point graphically, it may be easier to calculate the breakeven point numerically using engineering economy
expressions for the PW or AW at the MARR. The AW is preferred when the variable units are
expressed on a yearly basis, and AW calculations are simpler for alternatives with unequal lives.
The following steps determine the breakeven point of the common variable and the slope of a
linear total cost relation.
1. Define the common variable and its dimensional units.
2. Develop the PW or AW relation for each alternative as a function of the common variable.
3. Equate the two relations and solve for the breakeven value of the variable.
Selection between alternatives is based on this guideline:
If the anticipated level of the common variable is below the breakeven value, select the alternative with the higher variable cost (larger slope).
If the level is above the breakeven point, select the alternative with the lower variable cost.
(Refer to Figure 13–5.)
EXAMPLE 13.2
A small aerospace company is evaluating two alternatives: the purchase of an automatic feed
machine and a manual feed machine for a finishing process. The auto feed machine has an
initial cost of $23,000, an estimated salvage value of $4000, and a predicted life of 10 years.
One person will operate the machine at a rate of $12 per hour. The expected output is 8 tons per
hour. Annual maintenance and operating cost is expected to be $3500.
The alternative manual feed machine has a first cost of $8000, no expected salvage value, a
5-year life, and an output of 6 tons per hour. However, three workers will be required at $8 per
hour each. The machine will have an annual maintenance and operation cost of $1500. All
projects are expected to generate a return of 10% per year. How many tons per year must be
finished to justify the higher purchase cost of the auto feed machine?
Solution
Use the steps above to calculate the breakeven point between the two alternatives.
1. Let x represent the number of tons per year.
2. For the auto feed machine, the annual variable cost is
$12 1 hour ———
x tons
Annual VC ⫽ —— ———
hour 8 tons year
⫽ 1.5x
The VC is developed in dollars per year. The AW expression for the auto feed
machine is
AWauto ⫽ ⫺23,000(A兾P,10%,10) ⫹ 4000(A兾F,10%,10) ⫺ 3500 ⫺ 1.5x
⫽ $–6992 ⫺ 1.5x
Breakeven Analysis Between Two Alternatives
13.2
Similarly, the annual variable cost and AW for the manual feed machine are
$8
x tons
1 hour ———
Annual VC ⫽ —— (3 operators) ———
hour
6 tons year
⫽ 4x
AWmanual ⫽ ⫺8000(A兾P,10%,5) ⫺ 1500 ⫺ 4x
⫽ $⫺3610 ⫺ 4x
3. Equate the two cost relations and solve for x.
AWauto ⫽ AWmanual
⫺6992 ⫺ 1.5x ⫽ ⫺3610 – 4x
x ⫽ 1353 tons per year
If the output is expected to exceed 1353 tons per year, purchase the auto feed machine, since
its VC slope of 1.5 is smaller than the manual feed VC slope of 4.
The breakeven analysis approach is commonly used for make-or-buy decisions. This means
the company contracts to buy the product or service from the outside, or makes it within the company. The alternative to buy usually has no fixed cost and a larger variable cost than the option to
make. Where the two cost relations cross is the make-buy decision quantity. Amounts above this
indicate that the item should be made, not purchased outside.
EXAMPLE 13.3
Guardian is a national manufacturing company of home health care appliances. It is faced with
a make-or-buy decision. A newly engineered lift can be installed in a car trunk to raise and
lower a wheelchair. The steel arm of the lift can be purchased internationally for $3.50 per unit
or made in-house. If manufactured on site, two machines will be required. Machine A is estimated to cost $18,000, have a life of 6 years, and have a $2000 salvage value; machine B will
cost $12,000, have a life of 4 years, and have a $⫺500 salvage value (carry-away cost). Machine A will require an overhaul after 3 years costing $3000. The annual operating cost for
machine A is expected to be $6000 per year and for machine B is $5000 per year. A total of four
operators will be required for the two machines at a rate of $12.50 per hour per operator. In a
normal 8-hour period, the operators and two machines can produce parts sufficient to manufacture 1000 units. Use a MARR of 15% per year to determine the following.
(a) Number of units to manufacture each year to justify the in-house (make) option.
(b) The maximum capital expense justifiable to purchase machine A, assuming all other estimates for machines A and B are as stated. The company expects to produce 10,000 units
per year.
Solution
(a) Use steps 1 to 3 stated previously to determine the breakeven point.
1. Define x as the number of lifts produced per year.
2. There are variable costs for the operators and fixed costs for the two machines for the
make option.
Annual VC ⫽ (cost per unit)(units per year)
4 operators $12.50
⫽ ————— ——— (8 hours)x
1000 units hour
⫽ 0.4x
The annual fixed costs for machines A and B are the AW amounts.
AWA ⫽ ⫺18,000(A兾P,15%,6) ⫹ 2000(A兾F,15%,6)
⫺6000 ⫺ 3000(P兾F,15%,3)(A兾P,15%,6)
AWB ⫽ ⫺12,000(A兾P,15%,4) ⫺ 500(A兾F,15%,4) ⫺ 5000
Total cost is the sum of AWA, AWB, and VC.
347
Breakeven and Payback Analysis
Chapter 13
3. Equating the annual costs of the buy option (3.50x) and the make option yields
⫺3.50x ⫽ AWA ⫹ AWB ⫺ VC
⫽ ⫺18,000(A兾P,15%,6) ⫹ 2000(A兾F,15%,6) ⫺ 6000
⫺3000(P兾F,15%,3)(A兾P,15%,6) ⫺ 12,000(A兾P,15%,4)
⫺500(A兾F,15%,4) ⫺ 5000 ⫺ 0.4x
⫺3.10x ⫽ ⫺20,352
x ⫽ 6565 units per year
[13.5]
A minimum of 6565 lifts must be produced each year to justify the make option, which
has the lower variable cost of 0.4x.
(b) Substitute 10,000 for x and PA for the to-be-determined first cost of machine A (currently
$18,000) in Equation [13.5]. Solution yields PA ⫽ $58,295. This is approximately three
times the estimated first cost of $18,000, because the production of 10,000 per year is
considerably larger than the breakeven amount of 6565.
Even though the preceding examples treat only two alternatives, the same type of analysis can
be performed for three or more alternatives. To do so, compare the alternatives in pairs to find
their respective breakeven points. The results are the ranges through which each alternative is
more economical. For example, in Figure 13–6, if the output is less than 40 units per hour, alternative 1 should be selected. Between 40 and 60, alternative 2 is more economical; and above 60,
alternative 3 is favored.
If the variable cost relations are nonlinear, analysis is more complicated. If the costs increase
or decrease uniformly, mathematical expressions that allow direct determination of the breakeven point can be developed.
13.3 Payback Analysis
Payback analysis is another use of the present worth technique. It is used to determine the amount
of time, usually expressed in years, required to recover the first cost of an asset or project. Payback is allied with breakeven analysis; this is illustrated later in the section. The payback period,
also called payback or payout period, has the following definition and types.
Alternative 1
Alternative 2
Alternative 3
Total cost, $/year
348
Breakeven
points
40
60
Output, units/hour
Figure 13–6
Breakeven points for three alternatives.
349
Payback Analysis
13.3
The payback period np is an estimated time for the revenues, savings, and any other monetary benefits to completely recover the initial investment plus a stated rate of return i.
There are two types of payback analysis as determined by the required return.
No return; i⫽0%: Also called simple payback, this is the recovery of only the initial investment.
Discounted payback; i ⬎ 0%: The time value of money is considered in that some return, for
example, 10% per year, must be realized in addition to recovering the initial investment.
An example application of payback may be a corporate senior manager who insists that every
proposal return the initial cost and some stated return within 3 years. Using payback as an initial
screening tool, no proposal with np ⬎ 3 years can become a viable alternative. The payback period should be determined using a required i ⬎ 0%. Unfortunately in practice, no-return payback
is used too often to make economic decisions. After the formulas are presented, a couple of cautions about payback usage are provided.
The equations used to determine np differ for each type of analysis. For both types, the terminology is P for the initial investment in the asset, project, contract, etc., and NCF for the estimated annual net cash flow. Using Equation [1.5], annual NCF is
NCF ⫽ cash inflows ⫺ cash outflows
To calculate the payback period for i ⫽ 0% or i ⬎ 0%, determine the pattern of the NCF series.
Note that np is usually not an integer. For t ⫽ 1, 2, . . . , np,
tⴝnp
No return, i ⴝ 0%; NCFt varies annually:
0 ⴝ ⴚP ⴙ
兺 NCF
t
[13.6]
tⴝ1
No return, i ⴝ 0%; annual uniform NCF:
P
np ⴝ ———
NCF
[13.7]
tⴝnp
Discounted, i ⬎ 0%; NCFt varies annually:
0 ⴝ ⴚP ⴙ
兺 NCF (P兾F, i, t)
t
[13.8]
tⴝ1
Discounted, i ⬎ 0%; annual uniform NCF:
0 ⴝ ⴚP ⴙ NCF(P兾A, i, np)
[13.9]
After np years, the cash flows will recover the investment in year 0 plus the required return of
i%. If the alternative is used more than np years, with the same or similar cash flows, a larger
return results. If the estimated life is less than np years, there is not enough time to recover the
investment and i% return. It is important to understand that payback analysis neglects all cash
flows after the payback period of np years. Consequently, it is preferable to use payback as an
initial screening method or supplemental tool rather than as the primary means to select an
alternative. The reasons for this caution are that
• No-return payback neglects the time value of money, since no return on an investment is required.
• Either type of payback disregards all cash flows occurring after the payback period. These
cash flows may increase the return on the initial investment.
Payback analysis utilizes a significantly different approach to alternative evaluation than the
primary methods of PW, AW, ROR, and B兾C. It is possible for payback analysis to select a different alternative than these techniques. However, the information obtained from discounted payback analysis performed at an appropriate i ⬎ 0% can be very useful in that a sense of the risk
involved in undertaking an alternative is provided. For example, if a company plans to utilize a
machine for only 3 years and payback is 6 years, indication is that the equipment should not be
obtained. Even here, the 6-year payback is considered supplemental information and does not
replace a complete economic analysis.
EXAMPLE 13.4
The board of directors of Halliburton International has just approved an $18 million worldwide engineering construction design contract. The services are expected to generate new
annual net cash flows of $3 million. The contract has a potentially lucrative repayment clause
Payback period
350
Breakeven and Payback Analysis
Chapter 13
to Halliburton of $3 million at any time that the contract is canceled by either party during the
10 years of the contract period. (a) If i ⫽ 15%, compute the payback period. (b) Determine
the no-return payback period and compare it with the answer for i ⫽ 15%. This is an initial
check to determine if the board made a good economic decision. Show both hand and spreadsheet solutions.
Solution by Hand
(a) The net cash flow each year is $3 million. The single $3 million payment (call it CV for
cancellation value) could be received at any time within the 10-year contract period.
Equation [13.9] is altered to include CV.
0 ⫽ ⫺P ⫹ NCF(P兾A,i,n) ⫹ CV(P兾F,i,n)
In $1,000,000 units,
0 ⫽ ⫺18 ⫹ 3(P兾A,15%,n) ⫹ 3(P兾F,15%,n)
The 15% payback period is np ⫽ 15.3 years, found by trail and error. During the period of
10 years, the contract will not deliver the required return.
(b) If Halliburton requires absolutely no return on its $18 million investment, Equation [13.6]
results in np ⫽ 5 years, as follows (in $ million).
0 ⫽ ⫺18 ⫹ 5(3) ⫹ 3
There is a very significant difference in np for 15% and 0%. At 15% this contract would
have to be in force for 15.3 years, while the no-return payback period requires only 5 years.
A longer time is always required for i ⬎ 0% for the obvious reason that the time value of
money is considered.
Solution by Spreadsheet
Enter the function ⫽ NPER(15%,3,⫺18,3) to display 15.3 years. Change the rate from 15% to
0% to display the no-return payback period of 5 years.
If two or more alternatives are evaluated using payback periods to indicate that one may
be better than the other(s), the second shortcoming of payback analysis (neglect of cash
flows after np) may lead to an economically incorrect decision. When cash flows that occur
after np are neglected, it is possible to favor short-lived assets even when longer-lived assets
produce a higher return. In these cases, PW (or AW) analysis should always be the primary
selection method. Comparison of short- and long-lived assets in Example 13.5 illustrates
this situation.
EXAMPLE 13.5
Two equivalent pieces of quality inspection equipment are being considered for purchase by
Square D Electric. Machine 2 is expected to be versatile and technologically advanced enough
to provide net income longer than machine 1.
First cost, $
Annual NCF, $
Maximum life, years
Machine 1
Machine 2
12,000
3,000
8,000
1,000 (years 1–5),
3,000 (years 6–14)
14
7
The quality manager used a return of 15% per year and software that incorporates Equations [13.8] and [13.9] to recommend machine 1 because it has a shorter payback period of
6.57 years at i ⫽ 15%. The computations are summarized here.
Payback Analysis
13.3
$3000 per year
Cash flow neglected
by payback analysis
0
1
2
3
4
5
Machine 1
6
7
np = 6.57
$12,000
Cash flows neglected
by payback analysis
$3000 per year
$1000 per year
0
1
2
3
4
5
6
7
Machine 2
8
9
10
11
12
13
14
np = 9.52
$8000
Figure 13–7
Illustration of payback periods and neglected net cash flows, Example 13.5.
Machine 1: np ⫽ 6.57 years, which is less than the 7-year life.
Equation used:
0 ⫽ ⫺12,000 ⫹ 3000(P兾A,15%,np)
Machine 2: np ⫽ 9.52 years, which is less than the 14-year life.
Equation used:
0 ⫽ ⫺8000 ⫹ 1000(P兾A,15%,5)
⫹ 3000(P兾A,15%,np⫺5)(P兾F,15%,5)
Recommendation: Select machine 1.
Now, use a 15% PW analysis to compare the machines and comment on any difference in the
recommendation.
Solution
For each machine, consider the net cash flows for all years during the estimated (maximum)
life. Compare them over the LCM of 14 years.
PW1 ⫽ ⫺12,000 ⫺ 12,000(P兾F,15%,7) ⫹ 3000(P兾A,15%,14) ⫽ $663
PW2 ⫽ ⫺8000 ⫹ 1000(P兾A,15%,5) ⫹ 3000(P兾A,15%,9)(P兾F,15%,5)
⫽ $2470
Machine 2 is selected since its PW value is numerically larger than that of machine 1 at 15%.
This result is the opposite of the payback period decision. The PW analysis accounts for the
increased cash flows for machine 2 in the later years. As illustrated in Figure 13–7 (for one life
cycle for each machine), payback analysis neglects all cash flow amounts that may occur after
the payback time has been reached.
Comment
This is a good example of why payback analysis is best used for initial screening and supplemental risk assessment. Often a shorter-lived alternative evaluated by payback analysis may
appear to be more attractive, when the longer-lived alternative has cash flows later in its life
that make it more economically attractive.
As mentioned in the introduction to this section, breakeven and payback analyses are allied.
They can be used in conjunction to determine the payback period when a desired level of breakeven is specified. The reverse is also possible; when a desired payback period is established, the
breakeven value with or without a return requirement can be determined. By working together in
this fashion, better economic decisions can be made. Example 13.6 illustrates the second of the
situations mentioned above.
351
352
Breakeven and Payback Analysis
Chapter 13
EXAMPLE 13.6
The president of a local company expects a product to have a profitable life of between 1 and
5 years. Help her determine the breakeven number of units that must be sold annually (without
any return) to realize payback for each of the time periods 1 year, 2 years, and so on up to
5 years. The cost and revenue estimates are as follows:
Fixed costs: Initial investment of $80,000 with $1000 annual operating cost.
Variable cost: $8 per unit.
Revenue: Twice the variable cost for the first 5 years and 50% of the variable cost thereafter.
Solution by Hand
Define XBE as the breakeven quantity and np as the payback period. Since values of XBE are
sought for np ⫽ 1, 2, 3, 4, 5, solve for breakeven by substituting each payback period. First
develop the FC, r, and v terms.
80,000
———
Fixed cost, FC
np ⫹ 1000
Revenue per unit, r
$16
Variable cost per unit, v
$8
(years 1 through 5 only)
The breakeven relation from Equation [13.2] is
80,000兾np ⫹ 1000
XBE ⫽ ————————
8
Insert np values and solve for XBE, the breakeven value.
[13.10]
np, payback years
1
2
3
4
5
XBE, units per year
10,125
5125
3458
2625
2125
Solution by Spreadsheet
Figure 13–8 presents a spreadsheet solution for the breakeven values. Equation [13.10] is encoded to display the answers in column C. The breakeven values are the same as those above,
e.g., sell 5125 units per year to pay back in 2 years. The breakeven curve rapidly flattens out as
shown in the accompanying chart in Figure 13–8.
-
Figure 13–8
Breakeven number of units for different payback periods, Example 13.6.
13.4 More Breakeven and Payback
Analysis on Spreadsheets
The Goal Seek tool that we have used previously is an excellent tool to perform breakeven and payback analysis. Examples 13.7 and 13.8 demonstrate the use of Goal Seek for both types of problems.
More Breakeven and Payback Analysis on Spreadsheets
13.4
EXAMPLE 13.7
The Naruse brake-accelerator pedal (www.autoblog.com兾tag兾Masuyuki⫹Naruse) is designed
to minimize the chances that a driver will accidently step on the accelerator pedal of the car
when the brake pedal is the intended target. The design is based on the fact that a person naturally steps downward on his or her foot when surprised, shocked, or struck with a medical
emergency. In this pedal design, downward motion of the foot will always engage the brake,
never the accelerator. Assume that for the manufacture of pedal components, two equally qualified machines have been identified and estimates made.
First cost, $
Net cash flow, $ per year
Salvage value, $
Life, years
Machine 1
Machine 2
⫺80,000
25,000
2,000
4
⫺110,000
22,000
3,000
6
Using an AW analysis at MARR ⫽ 10%, the spreadsheet screen shot in Figure 13–9 indicates
that machine 1 is the economic choice with a positive AW value of $193. However, the automated controls, safety features, and ergonomic design of machine 2 make it a better choice for
the plant in the opinion of the project engineer. Use breakeven analysis to find the threshold
values for each of several parameters that will make machine 2 equally qualified economically.
The parameters to concentrate on are (a) first cost, (b) net cash flow, and (c) life of machine 2,
if all other estimates remain the same.
⫽ ⫺PMT($B$1,4,NPV($B1,B5:B8)+B4)
Figure 13–9
AW values for two machines, Example 13.7.
Solution
Figure 13–10 shows the spreadsheet and Goal Seek templates that determine breakeven values
for first cost and NCF.
(a) Figure 13–10a: By forcing the AW for machine 2 to equal $193, Goal Seek finds a breakeven of $96,669. If the first cost can be negotiated down to this cost from $110,000, machine 2 will be economically equivalent to machine 1.
(b) Figure 13–10b: (Remember to reset the first cost to $ –110,000 on the spreadsheet.) By setting all NCFs equal to the value in year 1 (using the function ⫽ $C$5), Goal Seek determines a breakeven of $25,061 per year. Therefore, if the NCF estimate can realistically be
increased from $22,000 to $25,061, again machine 2 will be economically equivalent.
(c) Finding an extended life estimate for machine 2 is a payback question; Goal Seek is not
needed. The easiest approach is to use the NPER function to find the payback period.
Entering ⫽ NPER(10%,22000,–110000,3000) displays np ⫽ 7.13 years. Therefore, extending the estimated life from 6 to between 7 and 8 years and retaining the salvage value
of $3000 will select machine 2 over 1.
353
354
Breakeven and Payback Analysis
Chapter 13
(a) First cost
(b) NCF
Figure 13–10
Breakeven values for (a) first cost and (b) annual net cash flow using Goal Seek, Example 13.7.
EXAMPLE 13.8
Chris and her father just purchased a small office building for $160,000 that is in need of a lot
of repairs, but is located in a prime commercial area of the city. The estimated costs each year
for repairs, insurance, etc. are $18,000 the first year, increasing by $1000 per year thereafter.
At an expected 8% per year return, use spreadsheet analysis to determine the payback period
if the building is (a) kept for 2 years and sold for $290,000 sometime beyond year 2 or (b) kept
for 3 years and sold for $370,000 sometime beyond 3 years.
Solution
Figure 13–11 shows the annual costs (column B) and the sales prices if the building is kept 2
or 3 years (columns C and E, respectively). The NPV function is applied (columns D and F) to
determine when the PW changes sign from plus to minus. These results bracket the payback
period for each retention period and sales price. When PW ⬎ 0, the 8% return is exceeded.
(a) The 8% return payback period is between 3 and 4 years (column D). If the building is sold
after exactly 3 years for $290,000, the payback period was not exceeded; but after 4 years
it is exceeded.
(b) At a sales price of $370,000, the 8% return payback period is between 5 and 6 years (column F). If the building is sold after 4 or 5 years, the payback is not exceeded; however, a
sale after 6 years is beyond the 8%-return payback period.
If kept 2 years and
sold, payback is
between 3 and 4
If kept 3 years and
sold, payback is
between 5 and 6
⫽ NPV(8%,$B$4:B7)+$B$3 ⫺ PV(8%,A7,,290000)
Figure 13–11
Payback period analysis, Example 13.8
355
Problems
CHAPTER SUMMARY
The breakeven point for a variable for one project is expressed in terms such as units per year or
hours per month. At the breakeven amount QBE, there is indifference to accept or reject the project. Use the following decision guideline:
Single Project
(Refer to Figure 13–2.)
Estimated quantity is larger than QBE → accept project
Estimated quantity is smaller than QBE → reject project
For two or more alternatives, determine the breakeven value of the common variable. Use the
following guideline to select an alternative:
Two Alternatives
(Refer to Figure 13–5.)
Estimated level is below breakeven → select alternative with higher
variable cost (larger slope)
Estimated level is above breakeven → select alternative with lower
variable cost (smaller slope)
Payback analysis estimates the number of years necessary to recover the initial investment plus
a stated rate of return. This is a supplemental analysis technique used primarily for initial screening prior to a full evaluation by PW or some other method. The technique has some drawbacks,
especially for no-return payback analysis, where i ⫽ 0% is the stated return.
PROBLEMS
Breakeven Analysis for a Project
13.1 A design-to-cost approach to product pricing involves determining the selling price of the product
and then figuring out if it can be made at a cost lower
than that. Banner Engineering’s QT50R radar-based
sensor features frequency-modulated technology to
accurately monitor or detect objects up to 15 miles
away while resisting rain, wind, humidity, and extreme temperatures. It has a list price of $589, and
the variable cost of manufacturing the unit is $340.
(a) What could the company’s fixed cost per
year be in order for Banner to break even
with sales of 9000 units per year?
(b) If Banner’s fixed cost is actually $750,000
per year, what is the profit at a sales level of
7000 units per year?
13.2 Handheld fiber-optic meters with white light polarization interferometry are useful for measuring
temperature, pressure, and strain in electrically
noisy environments. The fixed costs associated
with manufacturing are $800,000 per year. If a
base unit sells for $2950 and its variable cost is
$2075, (a) how many units must be sold each year
for breakeven and (b) what will the profit be for
sales of 3000 units per year?
13.3 A metallurgical engineer has estimated that the
capital investment cost for recovering valuable
metals (nickel, silver, platinum, gold, etc.) from
the copper refinery’s wastewater stream will be
$12 million. The equipment will have a useful life
of 15 years with no salvage value. Its operating
cost is represented by the relation ($2,600,000)E1.9,
where E is the efficiency of the metal recovery operation (in decimal form). The amount of metal
currently discharged is 2880 pounds per year prior
to recovery operations, and the efficiency of
recovery is estimated at 71%. What must the
average selling price per pound be for the precious metals that are recovered and sold in order
for the company to break even at its MARR of
15% per year?
Problems 13.4 through 13.7 are based on the following
information.
Hambry Enterprises produces a component for recycling
uranium used as a nuclear fuel in power plant generators
in France and the United States. Use the following cost
and revenue figures, quoted in U.S. dollars per hundredweight (cwt), recorded for this year to calculate the answers for each plant.
Location
France
United States
Fixed Cost,
$ million
Revenue,
$ per cwt
Cost,
$ per cwt
3.50
2.65
8,500
12,500
3,900
9,900
356
Chapter 13
13.4 Determine the breakeven point for each plant.
13.5 Estimate the minimum revenue per hundredweight
required for next year if breakeven values and
variable costs remain constant, but fixed costs
increase by 10%.
13.6 During this year, the French plant sold 950 units in
Europe, and the U.S. plant sold 850 units. Determine the year’s profit (loss) for each plant.
13.7 Hambry’s president has a goal of $1 million profit
next year at each plant with no revenue or fixed
cost increases. Determine the decreases in dollar
amounts and percentages in variable cost necessary to meet this goal, if the number of units sold is
the same as this year.
13.8 The National Highway Traffic Safety Administration raised the average fuel efficiency standard to
35.5 miles per gallon (mpg) for cars and light
trucks by the year 2016. The rules will cost consumers an average of $926 extra per vehicle in the
2016 model year. Assume a person purchases a
new car in 2016 that gets 35.5 mpg and keeps it
for 5 years. If the person drives an average of
1000 miles per month and gets an extra 10 miles
per gallon of gasoline, how much will the gasoline
have to cost in order for the buyer to recover the
extra investment in 5 years at an interest rate of
0.75% per month?
13.9 A call center in India used by U.S. and U.K. credit
card holders has a capacity of 1,400,000 calls annually. The fixed cost of the center is $775,000
with an average variable cost of $1 and revenue of
$2.50 per call.
(a) Find the percentage of the call capacity that
must be placed each year to break even.
(b) The center manager expects to dedicate the
equivalent of 500,000 of the 1,400,000 capacity to a new product line. This is expected to
increase the center’s fixed cost to $900,000 of
which 50% will be allocated to the new product line. Determine the average revenue per
call necessary to make 500,000 calls the
breakeven point for only the new product.
How does this required revenue compare with
the current center revenue of $2.50 per call?
13.10 The addition of a turbocharger to a small V-6 engine that gets 18 miles per gallon of gasoline can
boost its power to that of a V-8 engine and increase
fuel efficiency at the same time. If Bill will pay
$800 to turbocharge his engine and his fuel efficiency increases by 3 miles per gallon, how many
miles will he have to drive each month for 3 years
Breakeven and Payback Analysis
in order to break even? Assume the cost of gasoline is $3.25 per gallon and the interest rate is 1%
per month.
13.11 Transporting extremely heavy patients (people
who weigh more than 500 pounds) is much more
difficult than transporting normal-weight patients.
Various cities in Colorado, Nebraska, and Kansas
charge $1421 for an extremely obese patient compared to $758 for a typical patient. The extra fees
are justified by the ambulance companies on the
basis of the specialty equipment required and the
extra personnel involved. If it is assumed that
50 extremely obese patients are transported every
year, how much could the ambulance companies
afford to spend on the specialty equipment now
and break even on the initial cost in 5 years just
from the extra charges? Assume the extra costs are
$400 per patient and the company’s MARR is
10% per year.
13.12 High-profile vehicles have poor fuel efficiency because of increased wind resistance from their large
front area. A number of companies make devices
that they claim will significantly increase a vehicle’s fuel efficiency. One company claims that by
spending $560 on its friction-reducing device, the
fuel efficiency of a pickup truck will increase by
25%. Assuming the device works as claimed, for a
vehicle that currently gets 20 miles per gallon
(mpg), how many miles would the owner have to
drive each year to break even in 5 years? Assume
the cost of gasoline is $3.50 per gallon and the
interest rate is 10% per year.
13.13 As the price of gasoline goes up, people are willing to drive farther to fill their tank in order to save
money. Assume you had been buying gasoline for
$2.90 per gallon and that it went up to $2.98 per
gallon at the station where you usually go. If you
drive an F-150 pickup that gets 18 miles per gallon, what is the round-trip distance you can drive
to break even if it will take 20 gallons to fill your
tank? Use an interest rate of 8% per year.
13.14 An automobile company is investigating the advisability of converting a plant that manufactures
economy cars into one that will make retro sports
cars. The initial cost for equipment conversion
will be $200 million with a 20% salvage value
anytime within a 5-year period. The cost of producing a car will be $21,000, but it is expected to
have a selling price of $33,000 to dealers. The
production capacity for the first year will be
4000 units. At an interest rate of 12% per year, by
what uniform amount will production have to increase each year in order for the company to recover its investment in 3 years?
Problems
13.15 For the last 2 years, The Health Company has
experienced a fixed cost of $850,000 per year
and an (r ⫺ v) value of $1.25 per unit for its
multivitamin line of products. International
competition has become severe enough that
some financial changes must be made to keep
market share at the current level.
(a) Perform a spreadsheet-based graphical analysis to estimate the effect on the breakeven
point if the difference between revenue and
variable cost per unit increases somewhere
between 1% and 15% of its current value.
(b) If fixed costs and revenue per unit remain at
their current values, what type of change
must take place to make the breakeven point
go down?
13.16 (This is an extension of Problem 13.15) Expand
the analysis performed in Problem 13.15 by changing the variable cost per unit. The financial manager estimates that fixed costs will fall to $750,000
when the required production rate to break even is
at or below 600,000 units. What happens to the
breakeven points over the (r ⫺ v) range of 1% to
15% increase as evaluated previously?
Breakeven Analysis Between Alternatives
13.17 Providing restrooms at parks, zoos, and other cityowned recreation facilities is a considerable expense for municipal governments. City councils
usually opt for permanent restrooms in larger
parks and portable restrooms in smaller ones. The
cost of renting and servicing a portable restroom
is $7500 per year. In one northeastern municipality, the parks director informed the city council
that the cost of constructing a permanent restroom
is $218,000 and the annual cost of maintaining it
is $12,000. He remarked that the rather high cost
is due to the necessity to use expensive materials
and construction techniques that are tailored to
minimize damage from vandalism that often occurs in unattended public facilities. If the useful
life of a permanent restroom is assumed to be 20
years, how many portable restrooms could the
city afford to rent each year and break even with
the cost of one permanent facility? Let the interest
rate be 6% per year.
13.18 A consulting engineer is considering two methods
for lining ponds used for evaporating concentrate
generated during reverse osmosis treatment of
brackish groundwater for the Clay County Industrial Park. A geosynthetic bentonite clay liner
(GCL) will cost $1.8 million to install, and if it is
renovated after 4 years at a cost of $375,000, its
life can be extended another 2 years. Alternatively,
357
a high-density polyethylene (HDPE) geomembrane can be installed that will have a useful life of
12 years. At an interest rate of 6% per year, how
much money can be spent on the HDPE liner for
the two methods to break even?
13.19 An irrigation canal contractor wants to determine
whether he should purchase a used Caterpillar
mini excavator or a Toro powered rotary tiller for
servicing irrigation ditches in an agricultural area
of California. The initial cost of the excavator is
$26,500 with a $9000 salvage value after 10 years.
Fixed costs for insurance, license, etc. are expected
to be $18,000 per year. The excavator will require
one operator at $15 per hour and maintenance at
$1 per hour. In 1 hour, 0.15 mile of ditch can be
prepared. Alternatively, the contractor can purchase a tiller and hire 2 workers at $11 per hour
each. The tiller costs $1200 and has a useful life of
5 years with no salvage value. Its operating cost is
expected to be $1.20 per hour, and with the tiller,
the two workers can prepare 0.04 mile of ditch in
1 hour. The contractor’s MARR is 10% per year.
Determine the number of miles of ditch per year
the contractor would have to service for the two
options to break even.
13.20 An effective method to recover water used for regeneration of ion exchange resins is to use a reverse osmosis system in a batch treatment mode.
Such a system involves recirculation of the partially treated water back into the feed tank, causing the water to heat up. The water can be cooled
using one of two systems: a single-pass heat exchanger or a closed-loop heat exchange system.
The single-pass system, good for 3 years, requires
a small chiller costing $920 plus stainless steel
tubing, connectors, valves, etc. costing $360. The
cost of water, treatment charges, electricity, etc.
will be $3.10 per hour. The closed-loop system
will cost $3850 to buy, will have a useful life of
5 years, and will cost $1.28 per hour to operate.
What is the minimum number of hours per year
that the cooling system must be used in order to
justify purchase of the closed-loop system? The
MARR is 10% per year, and the salvage values
are negligible.
13.21 Samsung Electronics is trying to reduce supply
chain risk by making more responsible make-buy
decisions through improved cost estimation. A
high-use component (expected usage is 5000 units
per year) can be purchased for $25 per unit with
delivery promised within a week. Alternatively,
Samsung can make the component in-house and
have it readily available at a cost of $5 per unit, if
equipment costing $150,000 is purchased. Labor
358
Chapter 13
and other operating costs are estimated to be
$35,000 per year over the study period of 5 years.
Salvage is estimated at 10% of first cost and
i ⫽12% per year. Neglect the element of availability (a) to determine the breakeven quantity
and (b) to recommend making or buying at the expected usage level.
13.22 A partner in a medium-size A/E (architectural兾engineering) design firm is evaluating two alternatives
for improving the exterior appearance of the building they occupy. The building can be completely
painted at a cost of $6500. The paint is expected to
remain attractive for 4 years, at which time repainting will be necessary. Every time the building is
repainted, the cost will be 20% higher than the previous time. Alternatively, the building can be sandblasted now and every 6 years at a cost 40% greater
than the previous time. If the company’s MARR is
10% per year, what is the maximum amount that
could be spent now on the sandblasting alternative
that would render the two alternatives indifferent
over a study period of 12 years?
13.23 A junior mechanical engineering student is cooping this semester at Regency Aircraft, which
customizes the interiors of private and corporate
jets. Her first assignment is to develop the specifications for a new machine to cut, shape, and sew
leather or vinyl covers and trims. The first cost is
not easy to estimate due to many options, but the
annual revenue and M&O costs should net out at
$⫹15,000 per year over a 10-year life. Salvage is
expected to be 20% of the first cost. Determine the
breakeven first cost of the machine to just recover
its first cost and a return of 8% per year under two
scenarios:
I: No outside revenue will be developed by
the machine.
II: Outside contracting will occur with estimated revenue of $10,000 the first year, increasing by $5000 per year thereafter.
Solve using (a) hand and (b) spreadsheet solutions.
13.24 Ascarate Fishing Club (a nonprofit organization
dedicated to teaching kids how to fish) is considering two options for providing a heavily stocked
pond for kids who have never caught a fish before.
Option 1 is an above-ground swimming pool made
of heavy vinyl plastic that will be assembled and
disassembled for each quarterly event. The purchase price will be $400. Leaks from hooks piercing the fabric will be repaired with a vinyl repair
kit at a cost of $70 per year, but the pool will have
to be replaced when too many repairs have been
made.
Breakeven and Payback Analysis
Option 2 is an in-ground pond that will be excavated by club members at no cost and lined with
fabric that costs $1 per square foot. The pond will
be 15 ft in diameter and 3 ft deep. Assume 300 ft2
of liner will be purchased. A chain link fence at
$10 per lineal foot will be installed around the
pond (100 ft of fence). Maintenance inside the
fence is expected to cost $20 per year. The park
where the pond will be constructed has committed
the land for only 10 years. At an interest rate of 6%
per year, how long would the above-ground pool
have to last to break even?
13.25 A rural subdivision has several miles of access
roads that need a new surface treatment. Alternative 1 is a gravel base and pavement with an initial cost of $500,000 that will last for 15 years
and has an annual upkeep cost of $100 per mile.
Alternative 2 is to enhance the gravel base now at
a cost of $50,000 and immediately coat the surface with a durable hot oil mix, which costs $130
per barrel applied. Annual reapplication of the
mix is required. A barrel covers 0.05 mile. (a) If
the discount rate is 6% per year, determine the
number of miles at which the two alternatives
break even. (b) A drive in a pickup indicates a
total of 12.5 miles of road. Which is the more
economical alternative?
13.26 A waste-holding lagoon situated near the main
plant receives sludge daily. When the lagoon is
full, it is necessary to remove the sludge to a site
located 8.2 kilometers from the main plant. Currently, when the lagoon is full, the sludge is removed by pump into a tank truck and hauled away.
This process requires the use of a portable pump
that initially costs $800 and has an 8-year life. The
company pays a contract individual to operate the
pump and oversee environmental and safety factors at a rate of $100 per day, plus the truck and
driver must be rented for $200 per day. The company has the option to install a pump and pipeline
to the remote site. The pump would have an initial
cost of $1600 and a life of 10 years and will cost
$3 per day to operate. The company’s MARR is
10% per year.
(a) If the pipeline will cost $12 per meter to construct and will have a 10-year life, how many
days per year must the lagoon require pumping to justify construction of the pipeline?
(b) If the company expects to pump the lagoon
once per week every week of the year, how
much money can it afford to spend now on
the 10-year life pipeline to just break even?
13.27 Lorraine can select from two nutrient injection
systems for her cottage industry of hydroponic
Problems
tomato and lettuce greenhouses. (a) Use an AW
relation to determine the minimum number of
hours per year to operate the pumps that will justify the Auto Green system, if the MARR is 10%
per year. (b) Which pump is economically better if
it operates 7 hours per day, 365 days per year?
$1500 per month if your initial investment is
$28,000 and your MARR is (a) 0% and (b) 3% per
month? Solve by formula. (c) Write the spreadsheet functions to display the payback period for
both 0% and 3% per month.
13.31 (a)
Nutra Jet (N) Auto Green (A)
Initial cost, $
Life, years
Rebuild cost, $
Time before rebuild, annually
or minimum hours
Cost to operate, $ per hour
⫺4,000
3
⫺1,000
2,000
⫺10,300
6
⫺2,200
8,000
1.00
0.90
13.28 An engineering practitioner can lease a fully
equipped computer and color printer system for
$800 per month or purchase one for $8500 now and
pay a $75 per month maintenance fee. If the nominal interest rate is 15% per year, determine the
months of use necessary for the two to break even.
Show both (a) hand and (b) spreadsheet solutions.
13.29 The office manager of an environmental engineering consulting firm was instructed to make an ecofriendly decision in acquiring an automobile for
general office use. He is considering a gasolineelectric hybrid or a gasoline-free, all-electric
hatchback. The hybrid under consideration is
GM’s Volt, which will cost $35,000, will have a
salvage value of $15,000 after 5 years, and will
have a range of 40 miles on the electric battery,
plus several hundred more miles when the gasoline engine kicks in. Nissan’s Leaf, on the other
hand, is a pure electric that will have a range of
only 100 miles, after which its lithium ion battery
must be recharged. The Leaf’s relatively limited
range creates a psychological effect known as
range anxiety (RA), which has the company leaning toward purchasing the Volt. The Leaf can be
leased for $349 per month after an initial $500
down payment.
The accountant for the consulting firm told the
office manager that the Leaf is the better economic
option based on an evaluation she performed earlier. If the office manager purchases the Volt anyway (instead of leasing the Leaf), what is the
monthly equivalent (AW value) of the extra
amount of money the company will be paying to
eliminate range anxiety? Assume the operating
costs will be the same for both vehicles and the
MARR is 0.75% per month.
Payback Analysis
13.30 How long will you have to sell a product that has
an income of $5000 per month and expenses of
359
(b)
Determine the payback period at an interest
rate of 8% per year for an asset that initially
cost $28,000, has a scrap value of $1500
whenever it is sold, and generates cash flow
of $2900 per year.
If the asset will be in service for 12 years,
should it be purchased?
13.32 ABB purchased fieldbus communication equipment for a project in South Africa for $3.15 million. The net cash flow is estimated at $500,000
per year, and a salvage value of $400,000 is anticipated regardless of when it is sold. Determine the
number of years the equipment must be used to
obtain payback at MARR values of (a) 0% and 8%
per year and (b) 15% and 16% per year. (c) Use a
spreadsheet to plot the payback years for all four
return values.
13.33 Sundance Detective Agency purchased new surveillance equipment with the following estimates.
The year index is k ⫽ 1, 2, 3, . . . .
First cost
Annual maintenance cost
Extra annual revenue
Salvage value
(a)
(b)
$1050
$70 ⫹ 5k per year
$200 ⫹ 50k per year
$600 for all years
Calculate the payback period to make a return of 10% per year.
For a preliminary conclusion, should the
equipment be purchased if the actual useful
life is 7 years?
13.34 Clarisa, an engineering manager, wants to purchase a resort accommodation to rent to skiers.
She is considering the purchase of a three-bedroom
lodge in upper Montana that will cost $250,000.
The property in the area is rapidly appreciating in
value because people anxious to get away from
urban developments are bidding up the prices. If
Clarisa spends an average of $500 per month for
utilities and the investment increases at a rate of
2% per month, how long would it be before she
could sell the property for $100,000 more than she
has invested in it?
13.35 Laura’s grandparents helped her purchase a small
self-serve laundry business to make extra money
during her 5 college years. When she completed
her engineering management degree, she sold the
business and her grandparents told her to keep the
360
Breakeven and Payback Analysis
Chapter 13
money as a graduation present. For the net cash
flows listed below, determine the following:
(a) The percentage of the investment recovered
during the 5 years
(b) The actual rate of return over the 5-year
period
(c) How long it took to pay back the $75,000 investment in year 0, plus a 7% per year return
Year
0
1
2
3
4
5
NCF, $ per year ⫺75,000 ⫺10,500 18,600 ⫺2000 28,000 105,000
13.36 Buhler Tractor sold a tractor for $45,000 to Tom
Edwards 10 years ago. (a) What is the uniform net
cash flow that Tom must make each year to realize
payback and a return of 5% per year on his investment over a period of 3 years? 5 years? 8 years?
All 10 years? (b) If the net cash flow was actually
$5000 per year, what is the amount Tom should
have paid for the tractor to realize payback plus the
5% per year return over these 10 years?
13.37 National Parcel Service has historically owned and
maintained its own delivery trucks. Leasing is an
option being seriously considered because costs
for maintenance, fuel, insurance, and some liability issues will be transferred to Pacific Leasing, the
truck leasing company. The study period is no
more than 24 months for either alternative. The annual lease cost is paid at the beginning of each year
and is not refundable for partially used years. Use
the first cost and net cash flow estimates to determine the payback in months with a nominal 9%
per year return for the (a) purchase option and
(b) lease option.
Purchase:
Lease:
P ⫽ $–30,000 now
Monthly cost ⫽ $⫺1000
Monthly revenue ⫽ $4500
P ⫽ $⫺10,000 at the beginning
of each year (months 0 and 12)
Monthly cost ⫽ $−2500
Monthly revenue ⫽ $4500
13.38 Julian Browne, owner of Clear Interior Environments, purchased an air scrubber, HEPA vacuum,
and other equipment for mold removal for $15,000
eight months ago. Net cash flows were $−2000 for
each of the first 2 months, followed by $1000 per
month for months 3 and 4. For the last 4 months, a
contract generated a net $6000 per month. Julian
sold the equipment yesterday for $3000 to a friend.
Determine (a) the no-return payback period and
(b) the nominal 18%-per-year payback period.
13.39 Explain why payback analysis may favor an alternative with a shorter payback period when it is not
the better choice economically.
13.40 When comparing two alternatives, why is it best to
use no-return payback analysis as a preliminary
screening tool prior to conducting a complete PW
or AW evaluation?
Spreadsheet Problems
13.41 Benjamin used regression analysis to fit quadratic
relations to monthly revenue and cost data with the
following results:
R ⫽ ⫺0.007Q2 ⫹ 32Q
TC ⫽ 0.004Q2 ⫹ 2.2Q ⫹ 8
(a) Plot R and TC. Estimate the quantity Qp at
which the maximum profit should occur.
Estimate the amount of profit at this quantity.
(b) The profit relation P ⫽ R ⫺ TC and calculus
can be used to determine the quantity Qp at
which the maximum profit will occur and the
amount of the profit. The equations are
Profit ⫽ aQ2 ⫹ bQ ⫹ c
⫺b
Qp ⫽ ——
2a
2
⫺b
Maximum profit ⫽ —— ⫹ c
4a
Use these relations to confirm the graphical
estimates you made in (a). (Your instructor
may ask you to derive the relations above.)
13.42 The National Potato Cooperative purchased a deskinning machine last year for $150,000. Revenue
for the first year was $50,000. Over the total estimated life of 8 years, what must the remaining
equivalent annual revenues (years 2 through 8)
equal to break even by recovering the investment
and a return of 10% per year? Costs are expected
to be constant at $42,000 per year, and a salvage
value of $20,000 is anticipated.
Problems 13.43 and 13.44 are based on the following
information.
Wilson Partners manufactures thermocouples for electronics
applications. The current system has a fixed cost of $300,000
per year and a variable cost of $10 per unit. Wilson sells the
units for $14 each. A newly proposed process will add onboard features that allow the revenue to increase to $16 per
unit, but the fixed cost will now be $500,000 per year. The
variable cost of the new system will be based on a $48 per
hour rate with 0.2 hour required to produce each unit.
13.43 Determine the annual breakeven quantity for
(a) the current system and (b) the new system.
13.44 Plot the two profit relations and estimate graphically the breakeven quantity between the two
alternatives.
Additional Problems and FE Exam Review Questions
361
Problems 13.45 through 13.48 are based on the following information.
13.45 Determine the breakeven number of tests between
the two options.
Mid-Valley Industrial Extension Service, a state-sponsored
agency, provides water quality sampling services to all
business and industrial firms in a 10-county region. Just
last month, the service purchased all necessary lab equipment for full in-house testing and analysis. Now an outsourcing agency has offered to take over this function on a
per sample basis. Data and quotes for the two options have
been collected. The MARR for government projects is 5%
per year, and a study period of 8 years is chosen.
13.46 Use a spreadsheet to graph the AW curves for both
options for test loads between 0 and 4000 per year
in increments of 1000 tests. What is the estimated
breakeven quantity?
Equipment and supplies initially cost
$125,000 for a life of 8 years, an AOC of
$15,000, and annual salaries of $175,000.
Sample costs average $25 each. There is
no significant market value for the equipment and supplies currently owned.
Outsourced: Contractors quote sample cost averages
of $100 for the first 5 years, increasing to
$125 per sample for years 6 through 8.
13.47 The service director has asked the outsource company to reduce the per sample costs by 25% across
the board over the 8-year study period. What will this
do to the breakeven point? (Hint: Look carefully at
your graph from Problem 13.46 before answering.)
In-house:
13.48 Assume the Extension Service can reduce its annual salaries from $175,000 to $100,000 per
year and the per sample cost from $25 to $20.
What will this do to the breakeven point? (Hint:
Again, look carefully at your graph from the
previous problem before answering.) What is
the new annual breakeven test quantity?
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
13.49 In linear breakeven analysis, if a company expects
to operate at a point above the breakeven point, it
should select the alternative:
(a) With the lower fixed cost
(b) With the higher fixed cost
(c) With the lower variable cost
(d) With the higher variable cost
13.52 AW1 ⫽ ⫺23,000(A兾P,10%,10)⫹4000(A兾F,10%,10)
⫺ 3000 ⫺ 3x
AW2 ⫽ ⫺8,000(A兾P,10%,4) ⫺ 2000 − 6x
For these two AW relations, the breakeven point x,
in miles per year, is closest to:
(a) 1130
(b) 1224
(c) 1590
(d) 655
13.50 A company is considering two alternatives to automate the pH of process liquids. Alternative A will
have fixed costs of $42,000 per year and will require 2 workers at $48 per day each. Together,
these workers can generate 100 units of product
per day. Alternative B will have fixed costs of
$56,000 per year, but with this alternative, 3 workers will generate 200 units of product. If x is the
number of units per year, the variable cost (VC) in
$ per year for alternative B is represented by:
(a) [2(48)兾100]x
(b) [3(48)兾200]x
(c) [3(48)兾200]x ⫹ 56,000
(d) [2(48)兾100]x ⫹ 42,000
13.53 To make an item in-house, equipment costing
$250,000 must be purchased. It will have a life of
4 years, an annual cost of $80,000, and each unit
will cost $40 to manufacture. Buying the item externally will cost $100 per unit. At i ⫽ 15% per
year, it is cheaper to make the item in-house if the
number per year needed is:
(a) Above 1047 units
(b) Above 2793 units
(c) Equal to 2793 units
(d) Below 2793 units
13.51 When the variable cost is reduced for linear total
cost and revenue lines, the breakeven point decreases. This is an economic advantage because:
(a) The revenue per unit will increase.
(b) The two lines will now cross at zero.
(c) The profit will increase for the same revenue
per unit.
(d) The total cost line becomes nonlinear.
13.54 A procedure at Mercy Hospital has fixed costs of
$10,000 per year and variable costs of $50 per test.
If the procedure is automated, its fixed cost will be
$21,500 per year, but its variable cost will be only
$10 per test. The number of tests that must be performed each year for the two operations to break
even is closest to:
(a) 290 (b) 455 (c) 750 (d) Over 800
362
Chapter 13
13.55 An assembly process can be completed using
either alternative X or Y. Alternative X has fixed
costs of $10,000 per year with a variable cost of
$50 per unit. If the process is automated per alternative Y, its fixed cost will be $5000 per year
and its variable cost will be only $10 per unit.
The number of units that must be produced each
year in order for alternative Y to be favored is
closest to:
(a) Y will be favored for any level of production
(b) 125
(c) 375
(d) X will be favored for any level of production
13.56 Two different methods are under consideration
for building a bypass road. Material C will cost
$100,000 per mile and last for 10 years. Its annual maintenance cost will be $10,000 per year
per mile. Material D will cost $30,000 per mile
and last for 5 years. At an interest rate of 6% per
year, the annual maintenance cost for material D
that will make the two methods cost the same is
closest to:
(a) Less than $14,000
(b) $14,270
(c) $16,470
(d) $19,510
13.57 A construction company can purchase a piece of
equipment for $50,000 and spend $100 per day in
operating costs. The equipment will have a 5-year
life with no salvage value. Alternatively, the company can lease the equipment for $400 per day.
The number of days per year the company must
require the equipment to justify its purchase at an
interest rate of 8% per year is closest to:
(a) 10 days
(b) 42 days
(c) 51 days
(d) 68 days
13.58 A tractor has a first cost of $40,000, a monthly operating cost of $1500, and a salvage value of
$12,000 in 10 years. The MARR is 12% per year.
An identical tractor can be rented for $3200 per
month (operating cost not included). If n is the
minimum number of months per year the tractor
must be used in order to justify its purchase, the
relation to find n is represented by:
(a) ⫺40,000(A兾P,1%,10) ⫺ 1500n
⫹ 12,000(A兾F,1%,10) ⫽ 3200n
(b) ⫺40,000(A兾P,12%,10) ⫺ 1500n
⫹ 12,000(A兾F,12%,10) ⫽ 3200n
(c) ⫺40,000(A兾P,1%,120) ⫺ 1500n
⫹ 12,000(A兾F,1%,120) ⫽ 3200n
(d) ⫺40,000(A兾P,11.4%,10) ⫺ 1500n
⫹ 12,000(A兾F,11.4%,10) ⫽ 3200n
13.59 An anticorrosive coating for a chemical storage
tank will cost $5000 and last 5 years if touched
up at the end of 3 years at a cost of $1000. If an
Breakeven and Payback Analysis
oil-base enamel coating could be used that will
last 2 years, the amount the enamel coating can
cost for the two to breakeven at i ⫽ 8% per year
is closest to:
(a) $2120
(b) $2390
(c) $2590
(d) $2725
13.60 The price of a car is $50,000 today. Its price is expected to increase by $2400 each year. You now
have $25,000 in an investment that is earning 20%
per year. The number of years before you have
enough money to buy the car, without borrowing
any money, is closest to:
(a) 3 years
(b) 5 years
(c) 7 years
(d) 9 years
13.61 Process A has a fixed cost of $16,000 per year and
a variable cost of $40 per unit. For process B,
5 units can be produced in 1 day at a cost of $125.
If the company’s MARR is 10% per year, the fixed
cost of process B that will make the two alternatives have the same annual cost at a production
rate of 1000 units per year is closest to:
(a) Less than $10,000
(b) $18,000
(c) $27,000
(d) Over $30,000
13.62 The profit relation for the following estimates at a
quantity that is 20% above breakeven is:
Fixed cost ⫽ $500,000 per year
Variable cost per unit ⫽ $200
Revenue per unit ⫽ $250
(a) Profit ⫽ 200(12,000) ⫺ 250(12,000) ⫺ 500,000
(b) Profit ⫽ 250(12,000) ⫺ 500,000 ⫺ 200 (12,000)
(c) Profit ⫽ 250(12,000) ⫺ 200(12,000) ⫹ 500,000
(d) Profit ⫽ 250(10,000) ⫺ 200(10,000) ⫺ 500,000
13.63 Two methods of weed control in an irrigation canal
are under consideration. Method A involves lining
at a cost of $4000. The lining will last 20 years.
The maintenance cost with this method will be
$3 per mile per year. Method B involves spraying
a chemical that costs $40 per gallon. One gallon
will treat 8 miles, but the treatment must be applied
4 times per year. In determining the number of
miles per year that would result in breakeven, the
variable cost for method B is closest to:
(a) $5 per mile
(b) $15 per mile
(c) $20 per mile
(d) $40 per mile
13.64 How long will you have to maintain a business that
has an income of $5000 per year and expenses of
$1500 per year if your initial investment was
$28,000 and your MARR is 10% per year?
(a) Less than 6 years
(b) 8 years
(c) 12 years
(d) 17 years
Case Study
363
CASE STUDY
WATER TREATMENT PLANT PROCESS COSTS
Background
Aeration and sludge recirculation have been practiced for
many years at municipal and industrial water treatment
plants. Aeration is used primarily for the physical removal of
gases or volatile compounds, while sludge recirculation can
be beneficial for turbidity removal and hardness reduction.
When the advantages of aeration and sludge recirculation
in water treatment were first recognized, energy costs were so
low that such considerations were seldom of concern in treatment plant design and operation. With the huge increases in
electricity cost that have occurred in some localities, however, it became necessary to review the cost-effectiveness of
all water treatment processes that consume significant
amounts of energy. This study was conducted at a municipal
water treatment plant for evaluating the cost-effectiveness of
the pre-aeration and sludge recirculation practices.
there was neither aeration nor recirculation. For turbidity, the
reduction was 28% when both recirculation and aeration
were used. The reduction was 18% when neither aeration nor
recirculation was used. The reduction was also 18% when
aeration alone was used, which means that aeration alone was
of no benefit for turbidity reduction. With sludge recirculation alone, the turbidity reduction was only 6%, meaning that
sludge recirculation alone actually resulted in an increase in
turbidity—the difference between 18% and 6%.
Since aeration and sludge recirculation did cause readily
identifiable effects on treated water quality (some good and
others bad), the cost-effectiveness of each process for turbidity and hardness reduction was investigated. The calculations
are based on the following data:
Aerator motor ⫽ 40 hp
Aerator motor efficiency ⫽ 90%
Sludge recirculation motor ⫽ 5 hp
Recirculation pump efficiency ⫽ 90%
Information
This study was conducted at a 106 m3 per minute watertreatment plant where, under normal operating circumstances, sludge from the secondary clarifiers is returned to
the aerator and subsequently removed in the primary clarifiers. Figure 13–12 is a schematic of the process.
To evaluate the effect of sludge recirculation, the sludge
pump was turned off, but aeration was continued. Next, the
sludge pump was turned back on, and aeration was discontinued. Finally, both processes were discontinued. Results obtained during the test periods were averaged and compared to
the values obtained when both processes were operational.
The results obtained from the four operating modes
showed that the hardness decreased by 4.7% when both processes were in operation (i.e., sludge recirculation and aeration). When only sludge was recirculated, the reduction was
3.8%. There was no reduction due to aeration only, or when
Chemical
additions
Electricity cost ⫽ 9 ¢兾kWh (previous analysis)
Lime cost ⫽ 7.9 ¢兾kg
Lime required ⫽ 0.62 mg兾L per mg兾L hardness
Coagulant cost ⫽ 16.5 ¢兾kg
Days兾month ⫽ 30.5
As a first step, the costs associated with aeration and sludge
recirculation were calculated. In each case, costs are independent of flow rate.
Aeration cost:
40 hp ⫻ 0.75 kW兾hp ⫻ 0.09 $兾kWh ⫻ 24 h兾day
⫼ 0.90 ⫽ $72 per day or $2196 per month
Sludge recirculation cost:
5 hp ⫻ 0.75 kW兾hp ⫻ 0.09 $兾kWh ⫻ 24 h兾day
⫼ 0.90 ⫽ $9 per day or $275 per month
Figure 13–12
Flash
mix
Flocculation
Primary
clarifier
Secondary
clarifier
Aerator
Filter
Schematic of water
treatment plant.
To clear
well
Canal
364
Breakeven and Payback Analysis
Chapter 13
TABLE 13–1
Cost Summary in Dollars per Month
Savings from
Discontinuation of
Alt.
I.D.
Alternative
Description
1
Sludge recirculation
and aeration
2
3
Aeration only
Sludge recirculation
only
Neither aeration nor
sludge recirculation
4
Aeration
(1)
Recirculation
(2)
Extra Cost for
Removal of
Total Savings
(3) ⴝ (1) ⴙ (2)
Hardness
(4)
Turbidity
(5)
Total
Extra Cost
(6) ⴝ (4) ⴙ (5)
Net
Savings
(7) ⴝ (3) ⴚ (6)
Normal operating condition
—
2196
275
—
275
2196
1380
262
469
845
1849
1107
⫺1574
⫹1089
2196
275
2471
1380
469
1849
⫹622
The estimates appear in columns 1 and 2 of the cost summary
in Table 13–1.
Costs associated with turbidity and hardness removal are
a function of the chemical dosage required and the water flow
rate. The calculations below are based on a design flow of
53 m3兾minute.
As stated earlier, there was less turbidity reduction
through the primary clarifier without aeration than there was
with it (28% versus 6%). The extra turbidity reaching the
flocculators could require further additions of the coagulating
chemical. If it is assumed that, as a worst case, these chemical
additions would be proportional to the extra turbidity, then
22 percent more coagulant would be required. Since the average dosage before discontinuation of aeration was 10 mg兾L,
the incremental chemical cost incurred because of the increased turbidity in the clarifier effluent would be
(10 ⫻ 0.22) mg兾L ⫻ 10–6 kg兾mg ⫻ 53 m3兾min
⫻ 1000 L兾m3 ⫻ 0.165 $兾kg ⫻ 60 min兾h
⫻ 24 h兾day ⫽ $27.70兾day or $845兾month
Similar calculations for the other operating conditions (i.e.,
aeration only, and neither aeration nor sludge recirculation)
reveal that the additional cost for turbidity removal would be
$469 per month in each case, as shown in column 5 of
Table 13–1.
Changes in hardness affect chemical costs by virtue of the
direct effect on the amount of lime required for water softening. With aeration and sludge recirculation, the average hardness reduction was 12.1 mg兾L (that is, 258 mg兾L ⫻ 4.7%).
However, with sludge recirculation only, the reduction was
9.8 mg兾L, resulting in a difference of 2.3 mg兾L attributed to
aeration. The extra cost of lime incurred because of the discontinuation of aeration, therefore, was
2.3 mg兾L ⫻ 0.62 mg兾L lime ⫻ 10⫺6 kg兾mg
⫻ 53m3兾min ⫻ 1000 L兾m3 ⫻ 0.079 $兾kg
⫻60 min兾h ⫻ 24 h兾day ⫽ $8.60兾day or
$262兾month
When sludge recirculation was discontinued, there was no
hardness reduction through the clarifier, so that the extra lime
cost would be $1380 per month.
The total savings and total costs associated with changes
in plant operating conditions are tabulated in columns 3 and
6 of Table 13–1, respectively, with the net savings shown in
column 7. Obviously, the optimum condition is represented
by “sludge recirculation only.” This condition would result in
a net savings of $1089 per month, compared to a net savings
of $622 per month when both processes are discontinued and
a net cost of $1574 per month for aeration only. Since the
calculations made here represent worst-case conditions, the
actual savings that resulted from modifying the plant operating procedures were greater than those indicated.
In summary, the commonly applied water treatment practices of sludge recirculation and aeration can significantly affect the removal of some compounds in the primary clarifier.
However, increasing energy and chemical costs warrant continued investigations on a case-by-case basis of the costeffectiveness of such practices.
Case Study Exercises
1. What will be the monthly savings in electricity from
discontinuation of aeration if the cost of electricity is
now 12 ¢兾kWh?
2. Does a decrease in the efficiency of the aerator motor
make the selected alternative of sludge recirculation only
more attractive, less attractive, or the same as before?
3. If the cost of lime were to increase by 50%, would the
cost difference between the best alternative and secondbest alternative increase, decrease, or remain the same?
4. If the efficiency of the sludge recirculation pump were
reduced from 90% to 70%, would the net savings difference between alternatives 3 and 4 increase, decrease, or
stay the same?
5. If hardness removal were to be discontinued at the treatment plant, which alternative would be the most costeffective?
6. If the cost of electricity decreased to 8 ¢兾kWh, which
alternative would be the most cost-effective?
7. At what electricity cost would the following alternatives just break even? (a) Alternatives 1 and 2, (b) alternatives 1 and 3, (c) alternatives 1 and 4.
L E A R N I N G S TA G E 4
Rounding Out the Study
LEARNING STAGE 4
Rounding Out
the Study
CHAPTER
14
Effects of Inflation
CHAPTER
15
Cost Estimation and
Indirect Cost
Allocation
CHAPTER
16
Depreciation Methods
CHAPTER
17
After-Tax Economic
Analysis
CHAPTER
18
Sensitivity Analysis
and Staged Decisions
CHAPTER
19
More on Variation
and Decision Making
under Risk
T
his stage includes topics to enhance your ability to perform a
thorough engineering economic study of one project or several alternatives. The effects of inflation, depreciation,
income taxes in all types of studies, and indirect costs are incorporated into the methods of previous chapters. Techniques of cost
estimation to better predict cash flows are treated in order to base
alternative selection on more accurate estimates. The last two chapters include additional material on the use of engineering economics
in decision making. An expanded version of sensitivity analysis is
developed to examine parameters that vary over a predictable range
of values. The use of decision trees and an introduction to real
options are included. Finally, the elements of risk and probability
are explicitly considered using expected values, probabilistic analysis,
and spreadsheet-based Monte Carlo simulation.
Several of these topics can be covered earlier in the text, depending on the objectives of the course. Use the chart in the Preface to
determine appropriate points at which to introduce the material in
Learning Stage 4.
CHAPTER 14
Effects of
Inflation
L E A R N I N G
O U T C O M E S
Purpose: Consider the effects of inflation when performing an engineering economy evaluation.
SECTION
TOPIC
LEARNING OUTCOME
14.1
Inflationary impact
•
Demonstrate the difference that inflation makes
on money now and money in the future; also,
explain deflation.
14.2
PW with inflation
•
Calculate the PW of cash flows with an
adjustment made for inflation.
14.3
FW with inflation
•
Determine the real interest rate and calculate
the inflation-adjusted FW with different
interpretations of future worth values.
14.4
CR with inflation
•
Calculate capital recovery of an investment
using the AW value with inflation considered.
T
his chapter concentrates upon understanding and calculating the effects of inflation in time value of money computations. Inflation is a reality that we deal with
nearly everyday in our professional and personal lives.
The annual inflation rate is closely watched and historically analyzed by government
units, businesses, and industrial corporations. An engineering economy study can have different outcomes in an environment in which inflation is a serious concern compared to one
in which it is of minor consideration. In the first decade of the 21st century, inflation has not
been a major concern in the United States or most industrialized nations. But the inflation
rate is sensitive to real, as well as perceived, factors of the economy. Factors such as the cost
of energy, interest rates, availability and cost of skilled people, scarcity of materials, political stability, and other, less tangible factors have short-term and long-term impacts on the
inflation rate. In some industries, it is vital that the effects of inflation be integrated into an
economic analysis. The basic techniques to do so are covered here.
14.1 Understanding the Impact of Inflation
We are all very well aware that $20 now does not purchase the same amount as $20 did in 2005
and purchases significantly less than in 2000. Why? Primarily this is due to inflation and the
purchasing power of money.
Inflation is an increase in the amount of money necessary to obtain the same amount of
goods or services before the inflated price was present.
Purchasing power, or buying power, measures the value of a currency in terms of the quantity and quality of goods or services that one unit of money will purchase. Inflation decreases
the purchasing ability of money in that less goods or services can be purchased for the same
one unit of money.
Inflation occurs because the value of the currency has changed—it has gone down in value. The
value of money has decreased, and as a result, it takes more money for the same amount of
goods or services. This is a sign of inflation. To make comparisons between monetary amounts
that occur in different time periods, the different-valued money first must be converted to constant-value money in order to represent the same purchasing power over time. This is especially important when future sums of money are considered, as is the case with all alternative
evaluations.
Money in one period of time t1 can be brought to the same value as money in another period
of time t2 by using the equation
amount in period t2
Amount in period t1 ⴝ —————————————
inflation rate between t1 and t2
[14.1]
Using dollars as the currency, dollars in period t1 are called constant-value dollars or today’s
dollars. Dollars in period t2 are called future dollars or then-current dollars and have inflation
taken into account. If f represents the inflation rate per period (year) and n is the number of time
periods (years) between t1 and t2, Equation [14.1] is
future dollars
Constant-value dollars ⴝ ——————
(1 ⴙ f )n
Future dollars ⴝ constant-value dollars(1 ⴙ f )n
[14.2]
[14.3]
We can express future dollars in terms of constant-value dollars, and vice versa, by applying the
last two equations. This is how the Consumer Price Index (CPI) and cost estimation indices
(of Chapter 15) are determined. As an illustration, use the price of a cheese pizza.
$8.99
March 2011
If inflation on food prices averaged 5% during the last year, in constant-value 2010 dollars, this
cost is last year’s equivalent of
$8.99兾1.05 ⫽ $8.56
March 2010
Inflation
368
Effects of Inflation
Chapter 14
A predicted price in 2012, according to Equation [14.3], is
$8.99(1.05) ⫽ $9.44
March 2012
The price of $9.44 in 2012 buys exactly the same cheese pizza as $8.56 did in 2010. If inflation
averages 5% per year over the next 10 years, Equation [14.3] is used to predict a price in 2020
based on 2010.
$8.56(1.05)10 ⫽ $13.94
March 2020
This is a 63% increase over the 2010 price at 5% inflation for prepared food prices, which is generally not considered excessive. In some areas of the world, hyperinflation may average 50% per
year. In such an unfortunate economy, the cheese pizza in 10 years rises from the dollar equivalent of $8.99 to $518.44! This is why countries experiencing hyperinflation must devalue the
currency by factors of 100 and 1000 when unacceptable inflation rates persist.
Placed into an industrial or business context, at a reasonably low inflation rate averaging 4%
per year, equipment or services with a first cost of $209,000 will increase by 48% to $309,000
over a 10-year span. This is before any consideration of the rate of return requirement is placed
upon the equipment’s revenue-generating ability. Make no mistake: Inflation is a formidable
force in our economy.
There are three different rates that are important to understanding inflation: the real interest
rate (i), the market interest rate (if), and the inflation rate (f). Only the first two are interest rates.
Real or inflation-free interest rate i. This is the rate at which interest is earned when the
effects of changes in the value of currency (inflation) have been removed. Thus, the real interest rate presents an actual gain in purchasing power. (The equation used to calculate i, with the
influence of inflation removed, is derived later in Section 14.3.) The real rate of return that
generally applies for individuals is approximately 3.5% per year. This is the “safe investment”
rate. The required real rate for corporations (and many individuals) is set above this safe rate
when a MARR is established without an adjustment for inflation.
Inflation-adjusted or market interest rate if . As its name implies, this is the interest rate that
has been adjusted to take inflation into account. This is the interest rate we hear everyday. It is
a combination of the real interest rate i and the inflation rate f, and, therefore, it changes as the
inflation rate changes. It is also known as the inflated interest rate. A company’s MARR adjusted for inflation is referred to as the inflation-adjusted or market MARR. The determination
of this value is discussed in Section 14.3.
Inflation rate f. As described above, this is a measure of the rate of change in the value of the
currency.
Deflation is the opposite of inflation in that when deflation is present, the purchasing
power of the monetary unit is greater in the future than at present. That is, it will take fewer
dollars in the future to buy the same amount of goods or services as it does today. Inflation
occurs much more commonly than deflation, especially at the national economy level. In
deflationary economic conditions, the market interest rate is always less than the real interest rate.
Temporary price deflation may occur in specific sectors of the economy due to the introduction of improved products, cheaper technology, or imported materials or products that
force current prices down. In normal situations, prices equalize at a competitive level after a
short time. However, deflation over a short time in a specific sector of an economy can be
orchestrated through dumping. An example of dumping may be the importation of materials, such as steel, cement, or cars, into one country from international competitors at very
low prices compared to current market prices in the targeted country. The prices will go
down for the consumer, thus forcing domestic manufacturers to reduce their prices in order
to compete for business. If domestic manufacturers are not in good financial condition, they
may fail, and the imported items replace the domestic supply. Prices may then return to normal levels and, in fact, become inflated over time, if competition has been significantly
reduced.
On the surface, having a moderate rate of deflation sounds good when inflation has been
present in the economy over long periods. However, if deflation occurs at a more general
Present Worth Calculations Adjusted for Inflation
14.2
level, say nationally, it is likely to be accompanied by the lack of money for new capital. Another result is that individuals and families have less money to spend due to fewer jobs, less
credit, and fewer loans available; an overall “tighter” money situation prevails. As money
gets tighter, less is available to be committed to industrial growth and capital investment. In
the extreme case, this can evolve over time into a deflationary spiral that disrupts the entire
economy. This has happened on occasion, notably in the United States during the Great Depression of the 1930s.
Engineering economy computations that consider deflation use the same relations as those for
inflation. For basic equivalence between constant-value dollars and future dollars, Equations
[14.2] and [14.3] are used, except the deflation rate is a −f value. For example, if deflation is estimated to be 2% per year, an asset that costs $10,000 today would have a first cost 5 years from
now determined by Equation [14.3].
10,000(1 ⫺ f )n ⫽ 10,000(0.98)5 ⫽ 10,000(0.9039) ⫽ $9039
14.2 Present Worth Calculations
Adjusted for Inflation
When the dollar amounts in different time periods are to be expressed in constant-value dollars, the equivalent present and future amounts must be determined using the real interest
rate i. The calculations involved in this procedure are illustrated in Table14–1, where the inflation rate is 4% per year. Column 2 shows the inflation-driven increase for each of the next
4 years for an item that has a cost of $5000 today. Column 3 shows the cost in future dollars,
and column 4 verifies the cost in constant-value dollars via Equation [14.2]. When the future
dollars of column 3 are converted to constant-value dollars (column 4), the cost is always
$5000, the same as the cost at the start. This is predictably true when the costs are increasing
by an amount exactly equal to the inflation rate. The actual cost (in inflated dollars) of the
item 4 years from now will be $5849, but in constant-value dollars the cost in 4 years will still
amount to $5000. Column 5 shows the present worth of future amounts of $5000 at a real
interest rate of i ⫽ 10% per year.
Two conclusions can be drawn. At f ⫽ 4%, $5000 today inflates to $5849 in 4 years. And
$5000 four years in the future has a PW of only $3415 now in constant-value dollars at a real
interest rate of 10% per year.
Figure 14–1 graphs the differences over a 4-year period of the constant-value amount of
$5000, the future-dollar costs at 4% inflation, and the present worth at 10% real interest with
inflation considered. The effect of compounded inflation and interest rates can be large, as you
can see by the shaded area.
An alternative, less complicated method of accounting for inflation in a present worth analysis
involves adjusting the interest formulas themselves to account for inflation. Consider the P兾F
formula, where i is the real interest rate.
1
P ⫽ F ————
(1 ⫹ i)n
TABLE 14–1
Inflation Calculations Using Constant-Value Dollars (f ⴝ 4%, i ⴝ 10%)
Year
n
(1)
Cost
Increase
due to 4%
Inflation, $
(2)
Cost in
Future
Dollars, $
(3)
Future Cost in
Constant-Value
Dollars, $
(4) ⴝ (3)兾1.04n
Present Worth
at Real
i ⴝ 10%,$
(5) ⫽ (4)(P兾F,10%,n)
0
1
2
3
4
5000(0.04) ⫽ 200
5200(0.04) ⫽ 208
5408(0.04) ⫽ 216
5624(0.04) ⫽ 225
5000
5200
5408
5624
5849
5000
5200兾(1.04)1 ⫽ 5000
5408兾(1.04)2 ⫽ 5000
5624兾(1.04)3 ⫽ 5000
5849兾(1.04)4 ⫽ 5000
5000
4545
4132
3757
3415
369
370
Effects of Inflation
Chapter 14
Figure 14–1
Comparison of constantvalue dollars, future dollars, and their present
worth values.
6000
Increased
cost
Actual cost
at 4% inflation
$5849
dollars
Future
5000
Constant value
10%
Pres
ent
$5000
Decrease in
PW due
to inflation
and interest
wor
4000
th
$341
5
3000
0
1
2
Time, years
3
4
The F, which is a future-dollar amount with inflation built in, can be converted to constant-value
dollars by using Equation [14.2].
F
1
————
P ⫽ ————
(1 ⫹ f )n (1 ⫹ i)n
1
⫽ F ———————
[14.4]
(1 ⫹ i ⫹ f ⫹ if )n
If the term i ⫹ f ⫹ if is defined as if, the equation becomes
1
ⴝ F(P兾F,if ,n)
P ⴝ F ————
(1 ⴙ if )n
[14.5]
As described earlier, if is the inflation-adjusted or market interest rate and is defined as
if ⴝ i ⴙ f ⴙ if
[14.6]
where i ⫽ real interest rate
f ⫽ inflation rate
For a real interest rate of 10% per year and an inflation rate of 4% per year, Equation [14.6] yields
a market interest rate of 14.4%.
if ⫽ 0.10 ⫹ 0.04 ⫹ 0.10(0.04) ⫽ 0.144
Table 14–2 illustrates the use of if ⫽ 14.4% in PW calculations for $5000 now, which inflates
to $5849 in future dollars 4 years hence. As shown in column 4, the present worth for each year
is the same as column 5 of Table 14–1.
The present worth of any series of cash flows—uniform, arithmetic gradient, or geometric
gradient—can be found similarly. That is, either i or if is introduced into the P兾A, P兾G, or Pg factors, depending upon whether the cash flow is expressed in constant-value (today’s) dollars or
future dollars, respectively.
If a cash flow series is expressed in today’s (constant-value) dollars, then its PW is the discounted
value using the real interest rate i.
If the cash flow is expressed in future dollars, the PW value is obtained using if.
It is always acceptable to first convert all future dollars to constant-value dollars using
Equation [14.2] and then find the PW at the real interest rate i.
Present Worth Calculations Adjusted for Inflation
14.2
TABLE 14–2
Present Worth Calculation Using an Inflated Interest Rate
Year
n
(1)
Cost in
Future Dollars, $
(2)
(P兾F,14.4%,n)
(3)
PW, $
(4) ⴝ (2)(3)
0
1
2
3
4
5000
5200
5408
5624
5849
1
0.8741
0.7641
0.6679
0.5838
5000
4545
4132
3757
3415
EXAMPLE 14.1
Glyphosate is the active ingredient in the herbicide Roundup® marketed by Monsanto Co.
Roundup has been a dependable product used by farmers, municipalities, and suburbanites
alike to control weeds in fields, yards, gardens, streets, and parks. Contributions to Monsanto’s
revenue have been reduced significantly by international dumping of generic glyphosate, as
announced in mid-2010.1 Monsanto’s sales price was decreased from $16 to $12 per gallon to
compete with the highly competitive pricing, and it is expected that the international price
will settle at approximately $10 per gallon. Assume when the price was set at $16 per gallon,
there was a prediction that in 5 years the price would inflate to $19 per gallon. Perform the
following analysis.
(a) Determine the annual rate of inflation over 5 years to increase the price from $16 to $19.
(b) Using the same annual rate determined above as the rate at which the price continues to
decline from the new $12 price, calculate the expected price in 5 years. Compare this result
with $10 per gallon that Monsanto predicted would be the longer-term price.
(c) Provided Monsanto were somehow able to recover the same market share as it had previously, and the same inflation rate was applied to the reduced $12 per gallon price, determine the price 5 years in the future and compare it with the pre-dumping price of $16 per
gallon.
(d) Determine the market interest rate that must be used in economic equivalence
computations, if inflation is considered and an 8% per year real return is expected
by Monsanto.
Solution
The first three parts involve inflation only—no return on investments.
(a) Solve Equation [14.2] for the annual inflation rate f with known constant-value and future
amounts.
19
16 ⫽ 19(P兾F,f,5) ⫽ ————
(1 ⫹ f )5
1 ⫹ f ⫽ (1.1875)0.2
f ⫽ 0.035
(3.5% per year)
(b) If the price deflation rate is 3.5% per year, find the F value in 5 years with P ⫽ $12.
F ⫽ P(F兾P,⫺3.5%,5) ⫽ 12(1 ⫺ 0.035)5
⫽ 12(0.8368)
⫽ $10.04
The price will fall to exactly $10 per gallon after 5 years, as Monsanto predicted.
1
S. Kilman and I. Berry, “Monsanto Cuts Roundup Prices as Knockoffs Flood Farm Belt,” Wall Street
Journal, May 28, 2010.
371
372
Effects of Inflation
Chapter 14
(c) Five years in the future, at 3.5% per year inflation, the price will be
F ⫽ P(F兾P,3.5%,5) ⫽ 12(1.035)5
⫽ 12(1.1877)
⫽ $14.25
After 5 years of recovery at the same level as historically experienced, the price will still
be considerably lower than it was at the pre-dumping point ($14.25 versus $16 per gallon).
(d) With inflation at 3.5% per year and a real return of 8% per year, Equation [14.6] results in
a market rate of 11.78% per year.
if ⫽ 0.08 ⫹ 0.035 ⫹ (0.08)(0.035)
⫽ 0.1178 (11.78% per year)
EXAMPLE 14.2
A 15-year $50,000 bond that has a dividend rate of 10% per year, payable semiannually, is currently for sale. If the expected rate of return of the purchaser is 8% per year, compounded
semiannually, and if the inflation rate is expected to be 2.5% each 6-month period, what is the
bond worth now (a) without an adjustment for inflation and (b) when inflation is considered?
Show both hand and spreadsheet solutions.
Solution by Hand
(a) Without inflation adjustment: The semiannual dividend is I ⫽ [(50,000)(0.10)]兾2 ⫽ $2500.
At a nominal 4% per 6 months for 30 periods,
PW ⫽ 2500(P兾A,4%,30) ⫹ 50,000(P兾F,4%,30) ⫽ $58,645
(b) With inflation: Use the inflated rate if.
if ⫽ 0.04 ⫹ 0.025 ⫹ (0.04)(0.025) ⫽ 0.066 per semiannual period
PW ⫽ 2500(P兾A,6.6%,30) ⫹ 50,000(P兾F,6.6%,30)
⫽ 2500(12.9244) ⫹ 50,000(0.1470)
⫽ $39,660
Solution by Spreadsheet
Both (a) and (b) require simple, single-cell functions on a spreadsheet (Figure 14–2). Without
an inflation adjustment, the PV function is developed at the nominal 4% rate for 30 periods;
with inflation considered the rate is if ⫽ 6.6%, as determined above.
Comment
The $18,985 difference in PW values illustrates the tremendous negative impact made by only
2.5% inflation each 6 months (5.06% per year). Purchasing the $50,000 bond means receiving
$75,000 in dividends over 15 years and the $50,000 principal in year 15. Yet, this is worth only
$39,660 in constant-value dollars.
Figure 14–2
PW computation of a bond purchase (a) without and (b) with an inflation adjustment,
Example 14.2.
Present Worth Calculations Adjusted for Inflation
14.2
EXAMPLE 14.3
A self-employed chemical engineer is on contract with Dow Chemical, currently working in a
relatively high-inflation country in Central America. She wishes to calculate a project’s PW
with estimated costs of $35,000 now and $7000 per year for 5 years beginning 1 year from now
with increases of 12% per year thereafter for the next 8 years. Use a real interest rate of 15%
per year to make the calculations (a) without an adjustment for inflation and (b) considering
inflation at a rate of 11% per year.
Solution
(a) Figure 14–3 presents the cash flows. The PW without an adjustment for inflation is found
using i ⫽ 15% and g ⫽ 12% in Equations [2.34] and [2.35] for the geometric series.
PW ⫽ ⫺35,000 ⫺ 7000(P兾A,15%,4)
1.12 9
7000 1 ⫺ ——
(P兾F,15%,4)
1.15
⫺ ————————
0.15 ⫺ 0.12
{
[ ( ) ]}
⫽ ⫺35,000 ⫺ 19,985 ⫺ 28,247
⫽ $⫺83,232
In the P兾A factor, n ⫽ 4 because the $7000 cost in year 5 is the A1 term in Equation [2.34].
PW = ?
01
PWg = ?
2
3
0
$7000
i = 15% per year
45
6
7
8
9 10 11 12 13
1
2
3
4
5
6
7
8
9
Year
Geometric series year
$7840
$35,000
$17,331
12% increase
per year
Figure 14–3
Cash flow diagram, Example 14.3.
(b) To adjust for inflation, calculate the inflated interest rate by Equation [14.6] and use it to
calculate PW.
if ⫽ 0.15 ⫹ 0.11 ⫹ (0.15)(0.11) ⫽ 0.2765
PW ⫽ ⫺35,000 ⫺ 7000(P兾A,27.65%,4)
1.12 9
7000 1 ⫺ ———
(P兾F,27.65%,4)
1.2765
⫺ —————————
0.2765 ⫺ 0.12
{
[ (
)]
}
⫽ −35,000 ⫺ 7000(2.2545) ⫺ 30,945(0.3766)
⫽ $⫺62,436
Comment
This result demonstrates that in a high-inflation economy, when negotiating the amount of the
payments to repay a loan, it is economically advantageous for the borrower to use future
(inflated) dollars whenever possible to make the payments. The present value of future inflated
dollars is significantly less when the inflation adjustment is included. And the higher the inflation rate, the larger the discounting because the P兾F and P兾A factors decrease in size.
373
374
Effects of Inflation
Chapter 14
Examples 14.2 and 14.3 above add credence to the “buy now, pay later” philosophy. However, at
some point, the debt-ridden company or individual will have to pay off the debts and the accrued
interest with the inflated dollars. If cash is not readily available at that time, the debts cannot be repaid.
This can happen, for example, when a company unsuccessfully launches a new product, when there
is a serious downturn in the economy, or when an individual loses a salary. In the longer term, this
buy now, pay later approach must be tempered with sound financial practices now, and in the future.
14.3 Future Worth Calculations
Adjusted for Inflation
In future worth calculations, a future amount F can have any one of four different interpretations:
Case 1. The actual amount of money that will be accumulated at time n.
Case 2. The purchasing power of the actual amount accumulated at time n, but stated in
today’s (constant-value) dollars.
Case 3. The number of future dollars required at time n to maintain the same purchasing
power as today; that is, inflation is considered, but interest is not.
Case 4. The amount of money required at time n to maintain purchasing power and earn a
stated real interest rate.
Depending upon which interpretation is intended, the F value is calculated differently, as
described below. Each case is illustrated.
Case 1: Actual Amount Accumulated It should be clear that F, the actual amount of money
accumulated, is obtained using the inflation-adjusted (market) interest rate.
F ⴝ P(1 ⴙ if)n ⴝ P(F兾P,if ,n)
[14.7]
For example, when we quote a market rate of 10%, the inflation rate is included. Over a 7-year
period, $1000 invested at 10% per year will accumulate to
F ⫽ 1000(F兾P,10%,7) ⫽ $1948
Case 2: Constant-Value Dollars with Purchasing Power The purchasing power of future dollars is determined by first using the market rate if to calculate F and then deflating the
future amount through division by (1 ⫹ f )n.
P(1 ⴙ if)n P (F兾P, if ,n)
ⴝ ——————
F ⴝ ————
(1 ⴙ f) n
(1 ⴙ f )n
[14.8]
This relation, in effect, recognizes the fact that inflated prices mean $1 in the future purchases
less than $1 now. The percentage loss in purchasing power is a measure of how much less. As an
illustration, consider the same $1000 now, and a 10% per year market rate, which includes an
inflation rate of 4% per year. In 7 years, the purchasing power has risen, but only to $1481.
1000 (F兾P, 10%, 7) $1948
⫽ ——— ⫽ $1481
F ⫽ ————————
1.3159
(1.04)7
This is $467 (or 24%) less than the $1948 actually accumulated at 10% (case 1). Therefore, we
conclude that 4% inflation over 7 years reduces the purchasing power of money by 24%.
Also for case 2, the future amount of money accumulated with today’s buying power could equivalently be determined by calculating the real interest rate and using it in the F兾P factor to compensate
for the decreased purchasing power. This real interest rate is the i in Equation [14.6].
if ⫽ i ⫹ f ⫹ if
⫽ i(1 ⫹ f ) ⫹ f
if ⴚ f
i ⴝ ———
1ⴙf
[14.9]
14.3
Future Worth Calculations Adjusted for Inflation
The real interest rate i represents the rate at which today’s dollars expand with their same purchasing power into equivalent future dollars. An inflation rate larger than the market interest
rate leads to a negative real interest rate. The use of this interest rate is appropriate for calculating the future worth of an investment (such as a savings account or money market fund) when
the effect of inflation must be removed. For the example of $1000 in today’s dollars from
Equation [14.9]
0.10 ⫺ 0.04 ⫽ 0.0577
i ⫽ —————
1 ⫹ 0.04
(5.77%)
F ⫽ 1000(F兾P,5.77%,7) ⫽ $1481
The market interest rate of 10% per year has been reduced to a real rate that is less than 6% per
year because of the erosive effects of 4% per year inflation.
Case 3: Future Amount Required, No Interest This case recognizes that prices increase
when inflation is present. Simply put, future dollars are worth less, so more are needed. No interest rate is considered in this case—only inflation. This is the situation if someone asks, “How
much will a car cost in 5 years if its current cost is $20,000 and its price will increase by the inflation rate of 6% per year?” (The answer is $26,765.) No interest rate—only inflation—is involved.
To find the future cost, substitute f for the interest rate in the F兾P factor.
Fⴝ P(1 ⴙ f )n ⴝ P(F兾P, f,n)
[14.10]
Reconsider the $1000 used previously. If it is escalating at exactly the inflation rate of 4% per
year, the amount 7 years from now will be
F ⫽ 1000(F兾P,4%,7) ⫽ $1316
Case 4: Inflation and Real Interest This is the case applied when a market MARR is established. Maintaining purchasing power and earning interest must account for both increasing
prices (case 3) and the time value of money. If the growth of capital is to keep up, funds must
grow at a rate equal to or above the real interest rate i plus the inflation rate f. Thus, to make a real
rate of return of 5.77% when the inflation rate is 4%, if is the market (inflation-adjusted) rate that
must be used. For the same $1000 amount,
if ⫽ 0.0577 ⫹ 0.04 ⫹ 0.0577(0.04) ⫽ 0.10
F ⫽ 1000(F兾P,10%,7) ⫽ $1948
This calculation shows that $1948 seven years in the future will be equivalent to $1000 now with
a real return of i ⫽ 5.77% per year and inflation of f ⫽ 4% per year.
Table 14–3 summarizes which rate is used in the equivalence formulas for the different interpretations of F. The calculations made in this section explain the following:
•
•
•
•
The amount of $1000 now at a market rate of 10% per year will accumulate to $1948 in 7 years.
The $1948 will have the purchasing power of $1481 of today’s dollars if f ⫽ 4% per year.
An item with a cost of $1000 now will cost $1316 in 7 years at an inflation rate of 4% per year.
It will take $1948 of future dollars to be equivalent to $1000 now at a real interest rate of 5.77%
with inflation considered at 4% per year.
Most corporations evaluate alternatives at a MARR large enough to cover inflation plus some
return greater than their cost of capital, and significantly higher than the safe investment return of
approximately 3.5% mentioned earlier. Therefore, for case 4, the resulting MARR will normally
be higher than the market rate if. Define the symbol MARRf as the inflation-adjusted or market
MARR, which is calculated in a fashion similar to if.
MARRf ⴝ i ⴙ f ⴙ i(f )
[14.11]
375
376
Effects of Inflation
Chapter 14
TABLE 14–3
Calculation Methods for Various Future Worth Interpretations
Future Worth
Desired
Method of
Calculation
Case 1: Actual dollars
accumulated
Use stated market
rate if in equivalence
formulas
Case 2: Purchasing power
of accumulated dollars in
terms of constant-value
dollars
Use market rate if in
equivalence and
divide by (1 ⫹ f )n
or
Use real i
Example for
P ⴝ $1000, n ⴝ 7,
if ⴝ 10%, f ⴝ 4%
F ⫽ 1000(F兾P,10%,7)
1000 (F兾P,10%,7)
F ⫽ ————————
(1.04)7
or
F ⫽ 1000(F兾P,5.77%,7)
Case 3: Dollars required for
same purchasing power
Use f in place of i in
equivalence
formulas
F ⫽ 1000(F兾P,4%,7)
Case 4: Future dollars to
maintain purchasing
power and to earn a return
Calculate if and use in
equivalence
formulas
F ⫽ 1000(F兾P,10%,7)
The real rate of return i used here is the required rate for the corporation relative to its cost of
capital. Now the future worth F, or FW, is calculated as
F ⫽ P(1 ⫹ MARRf)n ⫽ P(F兾P,MARRf , n)
For example, if a company has a WACC (weighted average cost of capital) of 10% per year and
requires that a project return 3% per year above its WACC, the real return is i ⫽ 13%. The
inflation-adjusted MARR is calculated by including the inflation rate of, say, 4% per year. Then
the project PW, AW, or FW will be determined at the rate obtained from Equation [14.11].
MARRf ⫽ 0.13 ⫹ 0.04 ⫹ 0.13(0.04) ⫽ 0.1752
(17.52%)
EXAMPLE 14.4
Abbott Mining Systems wants to determine whether it should upgrade a piece of equipment
used in deep mining operations in one of its international operations now or later. If the company selects plan A, the upgrade will be purchased now for $200,000. However, if the company selects plan I, the purchase will be deferred for 3 years when the cost is expected to rise
to $300,000. Abbott is ambitious; it expects a real MARR of 12% per year. The inflation rate
in the country has averaged 3% per year. From only an economic perspective, determine
whether the company should purchase now or later (a) when inflation is not considered and
(b) when inflation is considered.
Solution
(a) Inflation not considered: The real rate, or MARR, is i ⫽ 12% per year. The cost of plan I is
$300,000 three years hence. Calculate the FW value for plan A three years from now and
select the lower cost.
FWA ⫽ ⫺200,000(F兾P,12%,3) ⫽ $⫺280,986
FWI ⫽ $−300,000
Select plan A (purchase now).
(b) Inflation considered: This is case 4; the real rate (12%), and inflation of 3% must be
accounted for. First, compute the inflation-adjusted MARR by Equation [14.11].
MARRf ⫽ 0.12 ⫹ 0.03 ⫹ 0.12(0.03) ⫽ 0.1536
14.4
Capital Recovery Calculations Adjusted for Inflation
Use MARRf to compute the FW value for plan A in future dollars.
FWA ⫽ ⫺200,000(F兾P,15.36%,3) ⫽ $⫺307,040
FWI ⫽ $−300,000
Purchase later (plan I) is now selected, because it requires fewer equivalent future dollars.
The inflation rate of 3% per year has raised the equivalent future worth of costs by 9.3%
from $280,986 to $307,040. This is the same as an increase of 3% per year, compounded
over 3 years, or (1.03)3 ⫺ 1 ⫽ 9.3%.
Most countries have inflation rates in the range of 2% to 8% per year, but hyperinflation is a
problem in countries where political instability, overspending by the government, weak international trade balances, etc., are present. Hyperinflation rates may be very high—10% to 100% per
month. In these cases, the government may take drastic actions: redefine the currency in terms of
the currency of another country, control banks and corporations, and control the flow of capital
into and out of the country in order to decrease inflation.
In a hyperinflated environment, people usually spend all their money immediately since the
cost of goods and services will be so much higher the next month, week, or day. To appreciate
the disastrous effect of hyperinflation on a company’s ability to keep up, we can rework Example 14.4b using an inflation rate of 10% per month, that is, a nominal 120% per year (not
considering the compounding effect of inflation). The FWA amount skyrockets and plan I is a
clear choice. Of course, in such an environment the $300,000 purchase price for plan I three
years hence would obviously not be guaranteed, so the entire economic analysis is unreliable.
Good economic decisions in a hyperinflated economy are very difficult to make using traditional engineering economy methods, since the estimated future values are totally unreliable
and the future availability of capital is uncertain.
14.4 Capital Recovery Calculations
Adjusted for Inflation
It is particularly important in capital recovery (CR) calculations used for AW analysis to include
inflation because current capital dollars must be recovered with future inflated dollars. Since future dollars have less buying power than today’s dollars, it is obvious that more dollars will be
required to recover the present investment. This suggests the use of the inflated interest rate in the
A兾P formula. For example, if $1000 is invested today at a real interest rate of 10% per year when
the inflation rate is 8% per year, the equivalent amount that must be recovered each year for
5 years in future dollars is
A ⫽ 1000(A兾P,18.8%,5) ⫽ $325.59
On the other hand, the decreased value of dollars through time means that investors can spend
fewer present (higher-value) dollars to accumulate a specified amount of future (inflated) dollars.
This suggests the use of a higher interest rate, that is, the if rate, to produce a lower A value in the
A兾F formula. The annual equivalent (with adjustment for inflation) of F ⴝ $1000 five years
from now in future dollars is
A ⫽ 1000(A兾F,18.8%,5) ⫽ $137.59
For comparison, the equivalent annual amount to accumulate F ⫽ $1000 at a real i ⫽
10% (without adjustment for inflation) is 1000(A兾F,10%,5) ⫽ $163.80. When F is a future
known cost, uniformly distributed payments should be spread over as long a time period as
possible so that the leveraging effect of inflation will reduce the effective annual payment
($137.59 versus $163.80 here).
377
378
Effects of Inflation
Chapter 14
EXAMPLE 14.5
What annual deposit is required for 5 years to accumulate an amount of money with the same
purchasing power as $680.58 today, if the market interest rate is 10% per year and inflation is
8% per year?
Solution
First, find the actual number of inflated dollars required 5 years in the future that is equivalent
to $680.58 today. This is case 3; Equation [14.10] applies.
F ⫽ (present purchasing power)(1 ⫹ f )5 ⫽ 680.58(1.08)5 ⫽ $1000
The actual amount of the annual deposit is calculated using the market interest rate of 10%.
This is case 4 where A is calculated for a given F.
A ⫽ 1000(A兾F,10%,5) ⫽ $163.80
Comment
The real interest rate is i ⫽ 1.85% as determined using Equation [14.9]. To put these calculations into perspective, if the inflation rate is zero when the real interest rate is 1.85%, the future
amount of money with the same purchasing power as $680.58 today is obviously $680.58.
Then the annual amount required to accumulate this future amount in 5 years is A ⫽
680.58(A兾F,1.85%,5) ⫽ $131.17. This is $32.63 lower than the $163.80 calculated above for
f ⫽ 8%. This difference is due to the fact that during inflationary periods, dollars deposited now
have more purchasing power than the dollars returned at the end of the period. To make up the
purchasing power difference, more higher-value dollars are required. That is, to maintain
equivalent purchasing power at f ⫽ 8% per year, an extra $32.63 per year is required.
The logic discussed here explains why, in times of increasing inflation, lenders of money
(credit card companies, mortgage companies, and banks) tend to further increase their market
interest rates. People tend to pay off less of their incurred debt at each payment because they
use any excess money to purchase additional items before the price is further inflated. Also, the
lending institutions must have more money in the future to cover the expected higher costs of
lending money. All this is due to the spiraling effect of increasing inflation. Breaking this cycle
is difficult to do at the individual level and much more difficult to alter at a national level.
CHAPTER SUMMARY
Inflation, treated computationally as an interest rate, makes the cost of the same product or service increase over time due to the decreased value of money. There are several ways to consider
inflation in engineering economy computations in terms of today’s (constant-value) dollars and
in terms of future dollars. Some important relations are the following:
Inflated interest rate: if ⫽ i ⫹ f ⫹ if
Real interest rate: i ⫽ (if − f )兾(1 ⫹ f)
PW of a future amount with inflation considered: P ⫽ F(P兾F, if, n)
Future worth in constant-value dollars of a present amount with the same purchasing power:
F ⫽ P(F兾P,i,n)
Future amount to cover a current amount with inflation only: F ⫽ P(F兾P, f, n)
Future amount to cover a current amount with inflation and interest: F ⫽ P(F兾P,if , n)
Annual equivalent of a future amount: A ⫽ F(A兾F,if , n)
Annual equivalent of a present amount in future dollars: A ⫽ P(A兾P,if , n)
Hyperinflation implies very high f values. Available funds are expended immediately because
costs increase so rapidly that larger cash inflows cannot offset the fact that the currency is losing
value. This can, and usually does, cause a national financial disaster when it continues over extended periods of time.
Problems
379
PROBLEMS
Interest Rate and Currency Considerations with
Inflation
14.1 What is the difference between today’s dollars and
constant-value dollars (a) when using today as the
reference point in time and (b) when using 2 years
ago as the reference point?
14.2 State the conditions under which the market interest rate is (a) higher than, (b) lower than, and
(c) the same as the real interest rate.
14.3 What annual inflation rate is implied from an
inflation-adjusted interest rate of 10% per year,
when the real interest rate is 4% per year?
14.4 Determine the inflation-adjusted interest rate for a
growth company that wants to earn a real rate of
return of 20% per year when the inflation rate is
5% per year.
14.5 For a high-growth company that wants to make a
real rate of return of 30% per year, compounded
monthly, determine the inflation-adjusted nominal
interest rate per year. Assume the inflation rate is
1.5% per month.
14.6 A high-tech company whose stock trades on the
NASDAQ stock exchange uses a MARR of 35% per
year. If the chief financial officer (CFO) said the
company expects to make a real rate of return of
25% per year on its investments over the next 3-year
period, what is the company expecting the inflation
rate per year to be over that time period?
14.7 Calculate the inflation-adjusted interest rate per
quarter when the real interest rate is 4% per quarter
and the inflation rate is 1% per quarter.
14.8 Calculate the real interest rate per month if the
nominal inflation-adjusted interest rate per year,
compounded monthly, is 18% and the inflation rate
per month is 0.5%.
14.9 A southwestern city has a contract with Firestone
Vehicle Fleet Maintenance to provide maintenance
services on its fleet of city-owned vehicles such as
street sweepers, garbage trucks, backhoes, and
carpool vehicles. The contract price is fixed at
$45,000 per year for 4 years. If you were asked to
convert the future amounts into constant-value
amounts per today’s dollars, what would the respective amounts be? Assume the current market
interest rate of 10% per year and inflation rate of
5% per year are expected to remain the same over
the next 4-year period.
14.10 When the inflation rate is 5% per year, how many
inflated dollars will be required 10 years from now
to buy the same things that $10,000 buys now?
14.11 Assume that you want to retire 30 years from now
with an amount of money that will have the same
value (same purchasing power) as $1.5 million
today. If you estimate the inflation rate will be 4%
per year, how many future (then-current) dollars
will you need?
Adjusting for Inflation
14.12 If the inflation rate is 7% per year, how many years
will it take for the cost of something to double when
prices increase at exactly the same rate as inflation?
14.13 The inflation rate in a Central American country is
6% per year. What real rate of return will an investor make on a $100,000 investment in a copper
mine stock that yields an overall internal rate of
return of 28% per year?
14.14 During periods of hyperinflation, prices increase
rapidly over short periods of time. In 1993, the
Consumer Price Index (CPI) in Brazil was
113.6 billion. In 1994, the CPI was 2472.4 billion.
(a) What was the inflation rate per year between
1993 and 1994?
(b) Assuming the inflation rate calculated in
part (a) occurred uniformly throughout the
year and represented a nominal rate, what
were the monthly and daily inflation rates
over that time period?
14.15 A trust was set up by your grandfather that states
you are to receive $250,000 exactly 5 years from
today. Determine the buying power of the $250,000
in terms of today’s dollars if the market interest
rate is 10% per year and the inflation rate is 4% per
year.
14.16 Assume the inflation rate is 4% per year and the
market interest rate is 5% above the inflation rate.
Determine (a) the number of constant-value dollars 5 years in the future that is equivalent to
$30,000 now and (b) the number of future dollars
that will be equivalent to $30,000 now.
14.17 The Pell Grant program of the federal government
provides financial aid to needy college students. If
the average grant is slated to increase from $4050
to $5400 over the next 5 years “to keep up with
inflation,” what is the average inflation rate per
year expected to be?
380
Effects of Inflation
Chapter 14
14.18 Ford Motor Company announced that the price of
its F-150 pickup trucks is going to increase by only
the inflation rate for the next 3 years. If the current
price of a well-equipped truck is $28,000 and the
inflation rate averages 2.1% per year, what is the
expected price of a comparably equipped truck
next year? 3 years from now?
14.19 In an effort to reduce pipe breakage, water hammer,
and product agitation, a chemical company plans to
install several chemically resistant pulsation dampeners. The cost of the dampeners today is $120,000,
but the company has to wait until a permit is approved for its bidirectional port-to-plant product
pipeline. The permit approval process will take at
least 2 years because of the time required for preparation of an environmental impact statement.
Because of intense foreign competition, the manufacturer plans to increase the price only by the inflation rate each year. If the inflation rate is 2.8% per
year and the company’s MARR is 20% per year,
estimate the cost of the dampeners in 2 years in
terms of (a) today’s dollars and (b) future dollars.
14.20 A machine currently under consideration by
Holzmann Industries has a cost of $45,000. When
the purchasing manager complained that a similar
machine the company purchased 5 years ago was
much cheaper, the salesman responded that the
cost of the machine has increased solely in accordance with the inflation rate, which has averaged
3% per year. When the purchasing manager
checked the invoice for the machine he purchased
5 years ago, he saw that the price was $29,000.
Was the salesman telling the truth about the increase in the cost of the machine? If not, what
should the machine cost now, provided the price
increased by only the inflation rate?
14.21 A report by the National Center for Public Policy
and Higher Education stated that tuition and fees
(T&F) at public colleges and universities increased
by 439% over the last 25 years. During this same
time period, the median family income (MFI) rose
147%. When the report was written, T&F at a
4-year public university constituted 28% of the
MFI of $52,000 (tuition and fees at a private university constituted 76% of MFI).
(a) What was the tuition and fee cost per year
when the report was written?
(b) What was the T&F cost 25 years ago?
(c) What percentage was T&F of the MFI
25 years ago?
14.22 The headline on a Chronicle of Higher Education
article reads “College Costs Rise Faster than Inflation.” The article states that tuition at public colleges and universities increased by 58% over the
past 5 years. (a) What was the average annual percentage increase over that period of time? (b) If the
real increase in tuition (i.e., without inflation) was
5% per year, what was the inflation rate per year?
14.23 Stadium Capital Financing Group helps cashstrapped sports programs through the marketing of
its “sports mortgage.” At the University of Kansas,
Jayhawks fans can sign up to pay $105,000 over
10 years for the right to buy top seats for football
during the next 30 years. In return, the season tickets will stay locked in at current-year prices. Season tickets in tier 1 are currently selling for $350.
A fan plans to purchase the sports mortgage along
with season tickets now and each year for the next
30 years (31 seasons). What is the dollar amount of
the savings on the tickets (with no interest considered), if ticket prices rise at a rate of 3% per year
for the next 30 years?
Present Worth Calculations with Inflation
14.24 There are two ways to account for inflation in present worth calculations. What are they?
14.25 An environmental testing company needs to purchase equipment 2 years from now and expects to
pay $50,000 at that time. At a real interest rate of
10% per year and inflation rate of 4% per year, what
is the present worth of the cost of the equipment?
14.26 Carlsbad Gas and Electric is planning to purchase
a degassing tower for removing CO2 from acidified saltwater. The supplier quoted a price of
$125,000 if the unit is purchased within the next
3 years. Your supervisor has asked you to calculate
the present worth of the tower when considering
inflation. Assuming the tower will not be purchased for 3 years, calculate the present worth at
an interest rate of 10% per year and an inflation
rate of 4% per year.
14.27 How much can the manufacturer of superconducting magnetic energy storage systems afford to
spend now on new equipment in lieu of spending
$75,000 four years from now? The company’s real
MARR is 12% per year, and the inflation rate is
3% per year.
14.28 Find the present worth of the cash flows shown.
Some are expressed as constant-value (CV) dollars, and others are inflated dollars. Assume a real
interest rate of 8% per year and an inflation rate of
6% per year.
Year
Cash Flow, $
Stated as
1
2
3
4
5
3000
CV
6000
Inflated
8000
Inflated
4000
CV
5000
CV
Problems
14.29 A doctor is on contract to a medium-sized oil company to provide medical services at remotely located, widely separated refineries. The doctor is
considering the purchase of a private plane to reduce
the total travel time between refineries. The doctor
can buy a used Lear jet now or wait for a new very
light jet (VLJ) that will be available 3 years from
now. The cost of the VLJ will be $1.9 million, payable when the plane is delivered in 3 years. The doctor has asked you, his friend, to determine the present
worth of the VLJ so that he can decide whether to
buy the used Lear now or wait for the VLJ. If the
MARR is 15% per year and the inflation rate is projected to be 3% per year, what is the present worth of
the VLJ with inflation considered?
14.30 A regional infrastructure building and maintenance contractor is trying to decide whether to buy
a new compact horizontal directional drilling
(HDD) machine now or wait to buy it 2 years from
now (when a large pipeline contract will require
the new equipment). The HDD machine will include an innovative pipe loader design and
maneuverable undercarriage system. The cost of
the system is $68,000 if purchased now or $81,000
if purchased 2 years from now. At a real MARR of
10% per year and an inflation rate of 5% per year,
determine if the company should buy now or later
(a) without any adjustment for inflation and
(b) with inflation considered.
14.31 An engineer must recommend one of two rapidprototyping machines for integration into an upgraded manufacturing line. She obtained estimates
from salespeople from two companies. Salesman
A gave her the estimates in constant-value
(today’s) dollars, while saleswoman B provided
the estimates in future (then-current) dollars. The
company’s MARR is equal to the real rate of return of 20% per year, and inflation is estimated at
4% per year. Use PW analysis to determine which
machine the engineer should recommend.
First cost, $
AOC, $ per year
Life, years
A
(in CV Dollars)
B
(in Future Dollars)
⫺140,000
⫺25,000
10
⫺155,000
⫺40,000
10
14.32 A salesman from Industrial Water Services (IWS),
who is trying to get his foot in the door for a large
account in Fremont, offered water chlorination
equipment for $2.1 million. This is $400,000 more
than the price offered by a competing saleswoman
from AG Enterprises. However, IWS said the company would not have to pay for the equipment until
the warranty runs out. If the equipment has
381
a 2-year warranty, determine which offer is better.
The company’s real MARR is 12% per year, and
the inflation rate is 4% per year.
14.33 A chemical engineer is considering two sizes of
pipes, small (S) and large (L), for moving distillate
from a refinery to the tank farm. A small pipeline
will cost less to purchase (including valves and
other appurtenances) but will have a high head loss
and, therefore, a higher pumping cost. In writing
the report, the engineer compared the alternatives
based on future worth values, but the company
president wants the costs expressed as present dollars. Determine present worth values if future
worth values are FWS ⫽ $2.3 million and FWL ⫽
$2.5 million. The company uses a real interest rate
of 1% per month and an inflation rate of 0.4% per
month. Assume the future worth values were for a
10-year project period.
14.34 As an innovative way to pay for various software
packages, a new high-tech service company has
offered to pay your company, Custom Computer
Services (CCS), in one of three ways: (1) pay
$480,000 now, (2) pay $1.1 million 5 years from
now, or (3) pay an amount of money 5 years from
now that will have the same purchasing power as
$850,000 now. If you, as president of CCS, want to
earn a real interest rate of 10% per year when the
inflation rate is 6% per year, which offer should
you accept?
Future Worth and Other Calculations with Inflation
14.35 If the inflation rate is 6% per year and a person
wants to earn a true (real) interest rate of 10% per
year, determine the number of future dollars she
has to receive 10 years from now if the present investment is $10,000.
14.36 How many future dollars would you need 5 years
from now just to have the same buying power as
$50,000 now, if the deflation rate is 3% per year?
14.37 If a company deposits $100,000 into an account
that earns a market interest rate of 10% per year at
a time when the deflation rate is 1% per year, what
will be the purchasing power of the accumulated
amount (with respect to today’s dollars) at the end
of 15 years?
14.38 Harmony Corporation plans to set aside $60,000
per year beginning 1 year from now for replacing
equipment 5 years from now. What will be the purchasing power (in terms of current-value dollars)
of the amount accumulated, if the investment
grows by 10% per year, but inflation averages
4% per year?
382
Effects of Inflation
Chapter 14
14.39 The strategic plan of a solar energy company that
manufactures high-efficiency solar cells includes
an expansion of its physical plant in 4 years. The
engineer in charge of planning estimates the expenditure required now to be $8 million, but in
4 years, the cost will be higher by an amount equal
to the inflation rate. If the company sets aside
$7,000,000 now into an account that earns interest
at 7% per year, what will the inflation rate have to
be in order for the company to have exactly the
right amount of money for the expansion?
14.40 Five years ago, an industrial engineer deposited
$10,000 into an account and left it undisturbed
until now. The account is now worth $25,000.
(a) What was the overall ROR during the
5 years?
(b) If the inflation over that period was 4% per
year, what was the real ROR?
(c) What is the buying power of the $25,000 now
compared to the buying power 5 years ago?
14.41 A Toyota Tundra can be purchased today for
$32,350. A civil engineering firm is going to need
three more trucks in 2 years because of a land development contract it just won. If the price of the
truck increases exactly in accordance with an estimated inflation rate of 3.5% per year, determine
how much the three trucks will cost in 2 years.
14.42 The Nobel Prize is administered by The Nobel
Foundation, a private institution that was founded
in 1900 based on the will of Alfred Nobel, the inventor of dynamite. In part, his will stated: “The
capital shall be invested by my executors in safe
securities and shall constitute a fund, the interest
on which shall be annually distributed in the form
of prizes to those who, during the preceding year,
shall have conferred the greatest benefit on mankind.” The will further stated that the prizes were
to be awarded in physics, chemistry, peace, physiology or medicine, and literature. In addition to a
gold medal and a diploma, each recipient receives
a substantial sum of money that depends on the
Foundation’s income that year. The first Nobel
Prize was awarded in 1901 in the amount of
$150,000. In 1996, the award was $653,000; it was
$1.4 million in 2009.
(a) If the increase between 1996 and 2009 was
strictly due to inflation, what was the average
inflation rate per year during that 13-year
period?
(b) If the Foundation expects to invest money
with a return of 5% above the inflation rate,
how much will a laureate receive in 2020,
provided the inflation rate averages 3% per
year between 2009 and 2020?
14.43 Timken Roller Bearing is a manufacturer of
seamless tubes for drill bit collars. The company
is planning to add larger-capacity robotic arms to
one of its assembly lines 3 years from now. If it
is done now, the cost of the equipment with installation is $2.4 million. If the company’s real
MARR is 15% per year, determine the equivalent amount the company can spend 3 years from
now in then-current dollars if the inflation rate is
2.8% per year.
14.44 The data below show two patterns of inflation that
are exactly the opposite of each other over a
20-year time period.
(a) If each machine costs $10,000 in year 0 and
they both increase in cost exactly in accordance with the inflation rate, how much will
each machine cost at the end of year 20?
(b) What is the average inflation rate over the
time period for machine A (that is, what single inflation rate would result in the same
final cost for machine A)?
(c) In which years will machine A cost more
than machine B?
Year
Machine A, %
Machine B, %
1
2
3
4
5
6
7
8
10
10
2
2
10
10
2
2
2
2
10
10
2
2
10
10
19
20
2
2
10
10
14.45 Factors that increase costs and prices—especially
for materials and manufacturing costs sensitive to
market, technology, and labor availability—can be
considered separately using the real interest rate i,
the inflation rate f, and additional increases that
grow at a geometric rate g. The future amount is
calculated based on a current estimate by using the
relation
F ⫽ P(1 ⫹ i)n(1 ⫹ f )n(1 ⫹ g)n
⫽ P[(1 ⫹ i) (1 ⫹ f ) (1 ⫹ g)]n
The product of the first two terms enclosed in parentheses results in the inflated interest rate if . The
geometric rate is the same one used in the geometric series (Chapter 2). It commonly applies to
maintenance and repair cost increases as machinery ages. This is over and above the inflation rate.
If the current cost to manufacture an electronic
subcomponent is $300,000 per year, what is the
383
Problems
equivalent cost in 3 years, provided the average
annual rates for i, f, and g are 10%, 3%, and 2%,
respectively?
Capital Recovery with Inflation
14.46 An electric utility is considering two alternatives
for satisfying state regulations regarding pollution control for one of its generating stations.
This particular station is located at the outskirts
of a major U.S. city and a short distance from a
large city in a neighboring country. The station is
currently producing excess VOCs and oxides of
nitrogen. Two plans have been proposed for satisfying the regulators. Plan A involves replacing
the burners and switching from fuel oil to natural
gas. The cost of the option will be $300,000 initially and an extra $900,000 per year in fuel
costs. Plan B involves going to the foreign city
and running gas lines to many of the “backyard”
brick-making sites that now use wood, tires, and
other combustible waste materials for firing the
bricks. The idea behind plan B is that by reducing the particulate pollution responsible for smog
in the neighboring city, there would be greater
benefit to U.S. citizens than would be achieved
through plan A. The initial cost of plan B will be
$1.2 million for installation of the lines. Additionally, the electric company would subsidize
the cost of gas for the brick makers to the extent
of $200,000 per year. Extra air monitoring associated with this plan will cost an additional
$150,000 per year. For a 10-year project period
and no salvage value for either plan, which one
should be selected on the basis of an annual
worth analysis at a real interest rate of 7% per
year and an inflation rate of 4% per year?
14.47 A chemical engineer who is smitten with wanderlust wants to build a reserve fund that will be sufficient to permit her to take time off from work to
travel around the world. Her goal is to save enough
money over the next 3 years so that when she begins her trip, the amount she has accumulated will
have the same buying power as $72,000 today. If
she expects to earn a market rate of 12% per year
on her investments and inflation averages 5% per
year over the 3 years, how much must she save
each year to reach her goal?
14.48 An entrepreneur engaged in wildcat oil well drilling is seeking investors who will put up $500,000
for an opportunity to reap high returns if the venture is successful. The prospectus states that a
return of at least 22% per year for 5 years is
likely. How much will the investors have to
receive each year to recover their money if an
inflation rate of 5% per year is to be included in
the calculation?
14.49 Veri-Trol, Inc. manufactures in situ calibration
verification systems that confirm flow measurement accuracies without removing the meters. The
company has decided to modify the main assembly line with one of the enhancement processes
shown below. If the company’s real MARR is 15%
per year, which process has the lower annual cost
when inflation of 5% per year is considered?
First cost, $
Operating cost, $ per year
Salvage value, $
Life, years
Process X
Process Y
⫺65,000
⫺40,000
0
5
⫺90,000
⫺34,000
10,000
5
14.50 Johnson Thermal Products used austenitic
nickel-chromium alloys to manufacture resistance
heating wire. The company is considering a new
annealing-drawing process to reduce costs. If the
new process will cost $3.7 million now, how much
must be saved each year to recover the investment
in 5 years if the company’s MARR is a real 12%
per year and the inflation rate is 3% per year?
14.51 The costs associated with a small X-ray inspection
system are $40,000 now and $24,000 per year,
with a $6000 salvage value after 3 years. Determine the equivalent cost of the system if the real
interest rate is 10% per year and the inflation rate
is 4% per year.
14.52 Maintenance costs for pollution control equipment
on a pulverized coal cyclone furnace are expected
to be $180,000 now and another $70,000 three
years from now. The CFO of Monongahela Power
wants to know the equivalent annual cost of the
equipment in years 1 through 5. If the company
uses a real interest rate of 9% per year and the inflation rate averages 3% per year, what is the
equivalent annual cost of the equipment?
14.53 MetroKlean LLC, a hazardous waste soil cleaning
company, borrowed $2.5 million for 5 years to finance start-up costs for a new project involving
site reclamation. The company expects to earn a
real rate of return of 20% per year. The average
inflation rate is 5% per year.
(a) Determine the capital recovery required each
year with inflation considered.
(b) Determine the capital recovery if the company is satisfied with accumulating $2.5 million at the end of the 5 years with inflation
considered.
(c) Determine the capital recovery in part (b)
without considering inflation.
384
Effects of Inflation
Chapter 14
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
14.54 Inflation occurs when:
(a) Productivity increases
(b) The value of the currency decreases
(c) The value of the currency increases
(d) The price of gold decreases
14.55 “Then-current” (future) dollars can be converted
into constant-value dollars by:
(a) Multiplying them by (1 ⫹ f )n
(b) Multiplying them by (1 ⫹ f )n兾(1 ⫹ i)n
(c) Dividing them by (1 ⫹ f )n
(d) Dividing them by (1 ⫹ if )n
14.56 To calculate how much something will cost if you
expect its cost to increase by exactly the inflation
rate, you should:
(a) Multiply by (1 ⫹ f )n
(b) Multiply by [(1 ⫹ f )n兾(1 ⫹ i)n]
(c) Divide by (1 ⫹ f )n
(d) Multiply by (1 ⫹ if )n
14.57 If the market interest rate is 16% per year when the
inflation rate is 9% per year, the real interest rate is
closest to:
(a) 6.4%
(b) 7%
(c) 9%
(d) 15.6%
14.58 The market interest rate if is 12% per year, compounded semiannually. For an inflation rate of 2%
per 6 months, the effective semiannual interest rate
is closest to:
(a) 2%
(b) 3%
(c) 4%
(d) 6%
14.59 Construction equipment has a cost today of
$40,000. If its cost has increased only by the inflation rate of 6% per year, when the market interest
rate was 10% per year, its cost 10 years ago was
closest to:
(a) $15,420
(b) $22,335
(c) $27,405
(d) $71,630
14.60 The amount of money that would be accumulated
now from an investment of $1000 25 years ago at
a market rate of 5% per year and an inflation rate
averaging 2% per year over that time period is
closest to:
(a) $1640
(b) $3385
(c) $5430
(d) Over $5500
14.61 If $1000 is invested now, the number of future dollars required 10 years from now to earn a real interest rate of 6% per year, when the inflation rate is
4% per year, is closest to:
(a) $1480
(b) $1790
(c) $2650
(d) Over $2700
14.62 If you expect to receive a gift of $50,000 six years
from now, the present worth of the gift at a real
interest rate of 4% per year and an inflation rate of
3% per year is closest to:
(a) $27,600
(b) $29,800
(c) $33,100
(d) $37,200
14.63 New state-mandated emission testing equipment
for annual inspection of automobiles at Charlie’s
Garage has a first cost of $30,000, an annual operating cost of $7000, and a $5000 salvage value
after its 20-year life. For a real interest rate of 5%
per year and an inflation rate of 4% per year, the
annual capital recovery requirement for the equipment (in future dollars) is determined by:
(a) AW ⫽ ⫺30,000(A兾P,4%,20) ⫺ 7000
⫹ 5000(A兾F,4%,20)
(b) AW ⫽ ⫺30,000(A兾P,5%,20) ⫺ 7000
⫹ 5000(A兾F,5%,20)
(c) AW ⫽ ⫺30,000(A兾P,9%,20) ⫺ 7000
⫹ 5000(A兾F,9%,20)
(d) AW ⫽ ⫺30,000(A兾P,9.2%,20) ⫺ 7000
⫹ 5000(A兾F,9.2%,20)
14.64 Temporary price deflation may occur in specific
sectors of the economy for all of the following reasons except:
(a) Improved technology
(b) Excessive demand
(c) Dumping
(d) Increased productivity
Case Study
385
CASE STUDY
INFLATION VERSUS STOCK AND BOND INVESTMENTS
Background
The savings and investments that an individual maintains
should have some balance between equity (corporate stocks
that rely on market growth and dividend income) and fixedincome investments (bonds that pay dividends to the
purchaser and a guaranteed amount upon maturity). When
inflation is moderately high, bonds offer a low return relative
to stocks, because the potential for market growth is not present with bonds. Additionally, the forces of inflation make the
dividends worth less in future years, because there is no inflation adjustment made in the amount the dividend pays as time
passes. However, bonds do offer a steady income that may be
important to an individual, and they serve to preserve the
principal invested in the bond, because the face value is returned at maturity.
Information
Earl is an engineer who wants a predictable flow of money
for travel and vacations. He has a collection of stocks in his
retirement portfolio, but no bonds. He has accumulated a
total of $50,000 of his own funds in low-yielding savings accounts and wants to improve his long-term return from this
nonretirement program “nest egg.” He can choose additional
stocks or bonds, but has decided to not split the $50,000 between the two forms of investments. There are two choices he
has outlined, with the best estimates he can make at this time.
He assumes the effects of federal and state income taxes will
be the same for both forms of investment.
Stock purchase: Stocks purchased through a mutual fund
would pay an estimated 2% per year dividend and appreciate in value at 5% per year.
Bond purchase: If he purchased a bond, he would have a
predictable income of 5% per year and the $50,000 face
value after the 12-year maturity period.
Case Study Questions
The analysis that Earl has laid out has the following questions. Can you answer them for him for both choices?
1. What is the overall rate of return after 12 years?
2. If he decided to sell the stock or bond immediately after
the fifth annual dividend, what is his minimum selling
price to realize a 7% real return? Include an adjustment
of 4% per year for inflation.
3. If Earl needed some money in the future, say, immediately after the fifth dividend payment, what would be
the minimum selling price in future dollars, if he were
only interested in recovering an amount that maintained
the purchasing power of the original price?
4. As a follow-on to question 3, what happens to the
selling price (in future dollars) 5 years after purchase,
if Earl is willing to remove (net out) the future purchasing power of each of the dividends in the computation to determine the required selling price 5 years
hence?
5. Earl plans to keep the stocks or bonds for 12 years, that
is, until the bond matures. However, he wants to make
the 7% per year real return and make up for the expected 4% per year inflation. For what amount must he
sell the stocks after 12 years, or buy the bonds now to
ensure he realizes this return? Do these amounts seem
reasonable to you, given your knowledge of the way
that stocks and bonds are bought and sold?
CHAPTER 15
Cost
Estimation and
Indirect Cost
Allocation
L E A R N I N G
O U T C O M E S
Purpose: Make cost estimates using different methods; demonstrate the allocation of indirect costs using traditional and
activity-based costing rates.
SECTION
TOPIC
LEARNING OUTCOME
15.1
Approach
• Explain the bottom-up and design-to-cost (topdown) approaches to cost estimation.
15.2
Unit method
• Use the unit method to make a preliminary cost
estimate.
15.3
Cost index
• Use a cost index to estimate a present cost based
on historical data.
15.4
Cost capacity
• Use a cost-capacity equation to estimate
component, system, or plant costs.
15.5
Factor method
• Estimate total plant cost using the factor
method.
15.6
Indirect cost rates
• Allocate indirect costs using traditional indirect
cost rates.
15.7
ABC allocation
• Use the Activity-Based Costing (ABC) method to
allocate indirect costs.
15.8
Ethics and profit
• Describe how biased estimation can become an
ethical dilemma.
U
p to this point, cost and revenue cash flow values have been stated or assumed
as known. In reality, they are not; they must be estimated. This chapter explains
what cost estimation involves and applies cost estimation techniques. Cost estimation is important in all aspects of a project, but especially in the stages of project conception, preliminary design, detailed design, and economic analysis. When a project is developed in the private or the public sector, questions about costs and revenues will be posed by
individuals representing many different functions: management, engineering, construction,
production, quality, finance, safety, environmental, legal, and marketing, to name some. In
engineering practice, the estimation of costs receives much more attention than revenue
estimation; costs are the topic of this chapter.
Unlike direct costs for labor and materials, indirect costs are not easily traced to a specific
department, machine, or processing line. Therefore, allocation of indirect costs for functions
such as utilities, safety, management and administration, purchasing, and quality is made
using some rational basis. Both the traditional method of allocation and the Activity-Based
Costing (ABC) method are covered in this chapter.
15.1 Understanding How Cost Estimation
Is Accomplished
Cost estimation is a major activity performed in the initial stages of virtually every effort in
industry, business, and government. In general, most cost estimates are developed for either a
project or a system; however, combinations of these are very common. A project usually involves physical items, such as a building, bridge, manufacturing plant, or offshore drilling
platform, to name just a few. A system is usually an operational design that involves processes, services, software, and other nonphysical items. Examples might be a purchase order
system, a software package, an Internet-based remote-control system, or a health care delivery system. Of course, many projects will have major elements that are not physical, so estimates of both types must be developed. For example, consider a computer network system.
There would be no operational system if only the costs of computer hardware plus wire and
wireless connectors were estimated; it is equally important to estimate the software, personnel, and maintenance costs.
Thus far virtually all cash flow estimates in the examples, problems, progressive examples,
and case studies were stated or assumed to be known. In real-world practice, the cash flows for
costs and revenues must be estimated prior to the evaluation of a project or comparison of alternatives. We concentrate on cost estimation because costs are the primary values estimated for the
economic analysis. Revenue estimates utilized by engineers are usually developed in marketing,
sales, and other departments.
Costs are comprised of direct costs and indirect costs. Normally direct costs are estimated with some detail, then the indirect costs are added using standard rates and factors.
However, direct costs in many industries, including manufacturing and assembly settings,
have become a small percentage of overall product cost, while indirect costs have become
much larger. Accordingly, many industrial settings require some estimating for indirect costs
as well. Indirect cost allocation is discussed in detail in later sections of this chapter. Primarily, direct costs are discussed here.
Because cost estimation is a complex activity, the following questions form a structure for the
discussion that follows.
•
•
•
•
What cost components must be estimated?
What approach to cost estimation will be applied?
How accurate should the estimates be?
What estimation techniques will be utilized?
Costs to Estimate If a project revolves around a single piece of equipment, for example, an
industrial robot, the cost components will be significantly simpler and fewer than the components
for a complete system such as the manufacturing and testing line for a new product. Therefore, it
is important to know up front how much the cost estimation task will involve. Examples of cost
components are the first cost P and the annual operating cost (AOC), also called the M&O costs
Direct兾Indirect costs
388
Chapter 15
Cost Estimation and Indirect Cost Allocation
(maintenance and operating) of equipment. Each component will have several cost elements.
Listed below are sample elements of the first cost and AOC components.
First cost component P:
Elements: Equipment cost
Delivery charges
Installation cost
Insurance coverage
Initial training of personnel for equipment use
Delivered-equipment cost is the sum of the first two elements in the list above; installedequipment cost adds the third element.
AOC component (part of the equivalent annual cost A):
Elements: Direct labor cost for operating personnel
Direct materials
Maintenance costs (daily, periodic, repairs, etc.)
Rework and rebuild
Some of these elements, such as equipment cost, can be determined with high accuracy; others,
such as maintenance costs, are harder to estimate. When costs for an entire system must be estimated, the number of cost components and elements is likely to be in the hundreds. It is then
necessary to prioritize the estimation tasks.
For familiar projects (houses, office buildings, highways, and some chemical plants) there are
standard cost estimation software packages available. For example, state highway departments
utilize software that prompts for the cost components (bridges, pavement, cut-and-fill profiles,
etc.) and estimates costs with time-proven, built-in relations. Once these components are estimated, exceptions for the specific project are added. However, there are no “canned” software
packages for a large percentage of industrial, business, service and public sector projects.
Cost Estimation Approach Historically in industry, business, and the public sector, a
bottom-up approach to cost estimation was applied. For a simple rendition of this approach, see
Figure 15–1 (left side). The progression is as follows: cost components and their elements are
identified, cost elements are estimated, and estimates are summed to obtain total direct cost. The
price is then determined by adding indirect costs and the profit margin, which is usually a percentage of the total cost.
The bottom-up approach treats the required price as an output variable and the cost estimates
as input variables. This approach works well when competition is not a dominant factor in pricing the product or service.
Figure 15–1 (right side) shows a simplistic progression for the design-to-cost, or top-down, approach. The competitive price establishes the target cost.
The design-to-cost, or top-down, approach treats the competitive price as an input variable
and the cost estimates as output variables. This approach is useful in encouraging innovation,
new design, manufacturing process improvement, and efficiency. These are some of the essentials of value engineering and value-added systems engineering.
This approach places greater emphasis on the accuracy of the price estimation activity. The target
cost must be realistic, or else it can be a disincentive to design and engineering staff. The designto-cost approach is best applied in the early stages of a new or enhanced product design. The
detailed design and specific equipment options are not yet known, but the price estimates assist
in establishing target costs for different components.
Usually, the resulting approach is some combination of these two philosophies. However, it is
helpful to understand up front what approach is to be emphasized. Historically, the bottom-up
approach was more predominant in Western engineering cultures, especially in North America.
The design-to-cost approach is considered routine in Eastern engineering cultures; however,
389
Understanding How Cost Estimation Is Accomplished
15.1
Figure 15–1
Top-down approach
Required
price
Competitive
price
Desired
profit
⫹
Allowed
profit
⫺
Total cost ⫽
⫽ Target cost
Indirect
costs
⫹
⫹
Indirect
costs
Direct
costs
Maintenance
and operations
⫹
⫹
Maintenance
and operations
Direct
labor
⫹
⫹
Direct
labor
Direct
materials
⫹
⫹
Direct
materials
Equipment and
⫹
capital recovery
⫹
Equipment
capital recovery
Cost
component
estimates
At design stage
Before design stage
Bottom-up approach
Design-to-cost approach
Cost
component
estimates
globalization of engineering design has speeded the adoption of the design-to-cost approach
worldwide.
Accuracy of the Estimates No cost estimates are expected to be exact; however, they are
expected to be reasonable and accurate enough to support economic scrutiny. The accuracy required increases as the project progresses from preliminary design to detailed design and on to
economic evaluation. Cost estimates made before and during the preliminary design stage are
expected to be good “first-cut” estimates that serve as input to the project budget.
When utilized at early and conceptual design stages, estimates are referred to as order-ofmagnitude estimates and generally range within ⫾20% of actual cost. At the detailed design
stage, cost estimates are expected to be accurate enough to support economic evaluation for a
go–no go decision. Every project setting has its own characteristics, but a range of ⫾5% of actual
costs is expected at the detailed design stage. Figure 15–2 shows the general range of estimate
accuracy for the construction cost of a building versus time spent in preparing the estimate. Obviously, the desire for better accuracy has to be balanced against the cost of obtaining it.
⫾25%
}
Figure 15–2
Scoping/feasibility
Order of magnitude estimate
⫾20%
Partially designed
⫾15%
Design 60 –100% complete
Detailed estimate
⫾10%
Time Spent on Estimate
3
w
ee
ks
ee
k
w
1
m
in
u
1 tes
ho
ur
1
da
y
⫾5%
10
Accuracy of Estimate
⫾30%
Characteristic curve of
estimate accuracy versus
time spent to estimate
construction cost of a
building.
Simplified cost estimation
processes for bottom-up
and top-down approaches.
390
Chapter 15
Cost Estimation and Indirect Cost Allocation
Cost Estimation Techniques Methods such as expert opinion and comparison with comparable installations serve as excellent estimators. The use of the unit method and cost indexes base
the present estimate on past cost experiences, with inflation considered. Models such as costcapacity equations and the factor method are simple mathematical techniques applied at the
preliminary design stage. They are called cost estimating relationships (CERs). There are many
additional methods discussed in the handbooks and publications of different industries.
Most cost estimates made in a professional setting are accomplished in part or wholly using
software packages linked to updated databases that contain cost indexes and rates for the locations, products, or processes being studied. There are a wide variety of estimators, cost trackers,
and cost compliance software systems, most of them developed for specific industries. Corporations usually standardize on one or two packages to ensure consistency over time and projects.
15.2 Unit Method
The unit method is a popular preliminary estimation technique applicable to virtually all professions. The total estimated cost CT is obtained by multiplying the number of units N by a per unit
cost factor u.
CT ⫽ u ⫻ N
[15.1]
Unit cost factors must be updated frequently to remain current with changing costs, areas,
and inflation. Some sample unit cost factors (and values) are
Total average cost of operating an automobile (52¢ per mile)
Cost to bury fiber cable in a suburban area ($30,000 per mile)
Cost to construct a parking space in a parking garage ($4500 per space)
Cost of constructing interstate highway ($6.2 million per mile)
Cost of house construction per livable area ($225 per square foot)
Applications of the unit method to estimate costs are easily found. If house construction costs
average $225 per square foot, a preliminary cost estimate for an 1800-square-foot house, using
Equation [15.1], is $405,000. Similarly, a 200-mile trip should cost about $104 for the car only
at 52¢ per mile.
When there are several components to a project or system, the unit cost factors for each component are multiplied by the amount of resources needed, and the results are summed to obtain
the total cost CT. This is illustrated in Example 15.1.
EXAMPLE 15.1
Justin, an ME with Dynamic Castings, has been asked to make a preliminary estimate of the
total cost to manufacture 1500 sections of high-pressure gas pipe using an advanced centrifugal casting method. Since a ⫾20% estimate is acceptable at this preliminary stage, a unit
method estimate is sufficient. Use the following resource and unit cost factor estimates to
help Justin.
Materials: 3000 tons at $45.90 per ton
Machinery and tooling: 1500 hours at $120 per hour
Direct labor in plant:
Casting and treating: 3000 hours at $55 per hour
Finishing and shipping: 1200 hours at $45 per hour
Indirect labor: 400 hours at $75 per hour
Solution
Apply Equation [15.1] to each of the five areas and sum the results to obtain the total cost
estimate of $566,700. Table 15–1 provides the details.
Cost Indexes
15.3
TABLE 15–1
Total Cost Estimate Using Unit Cost Factors for Several Resource Areas,
Example 15.1
Resource
Amount N
Unit Cost Factor u, $
Cost Estimate, u ⴛ N, $
Materials
Machinery, tooling
Labor, casting
Labor, finishing
Labor, indirect
3000 tons
1500 hours
3000 hours
1200 hours
400 hours
45.90 per ton
120 per hour
55 per hour
45 per hour
75 per hour
137,700
180,000
165,000
54,000
30,000
Total cost estimate
566,700
15.3 Cost Indexes
This section explains indexes and their use in cost estimation. An index is a ratio or other number
based on observation and used as an indicator or measure. A preliminary cost estimate is often
based on a cost index.
A cost index is a ratio of the cost of something today to its cost sometime in the past. The index is
dimensionless and measures relative cost change over time. Because these indexes are sensitive to
technological change, the predefined quantity and quality of elements used to define the index may be
hard to retain over time, thus causing “index creep.” Timely updating of the index is very important.
One such index that most people are familiar with is the Consumer Price Index (CPI), which
shows the relationship between present and past costs for many of the things that “typical” consumers must buy. This index includes such items as rent, food, transportation, and certain services. Other indexes track the costs of equipment, and goods and services that are more pertinent
to the engineering disciplines. Table 15–2 is a listing of some of the more common indexes.
TABLE 15–2
Types and Sources of Various Cost Indexes
Type of Index
Overall prices
Consumer (CPI)
Producer (wholesale)
Construction
Chemical plant overall
Equipment, machinery, and supports
Construction labor
Buildings
Engineering and supervision
Engineering News Record overall
Construction
Building
Common labor
Skilled labor
Materials
EPA treatment plant indexes
Large-city advanced treatment (LCAT)
Small-city conventional treatment (SCCT)
Federal highway
Contractor cost
Equipment
Marshall and Swift (M&S) overall
M&S specific industries
Labor
Output per worker-hour by industry
Source
Bureau of Labor Statistics
U.S. Department of Labor
Chemical Engineering
Engineering News Record (ENR)
Environmental Protection Agency
and ENR
Marshall & Swift
U.S. Department of Labor
391
392
Cost Estimation and Indirect Cost Allocation
Chapter 15
TABLE 15–3
Year
1995
1996
1997
1998
1999
2000
2001
2002
2003
2004
2005
2006
2007
2008
2009
2010 (midyear)
Values for Selected Indexes
CE Plant
Cost Index
ENR Construction
Cost Index
M&S Equipment
Cost Index
381.1
381.8
386.5
389.5
390.6
394.1
394.3
395.6
401.7
444.2
468.2
499.6
525.4
575.4
521.9
555.3
5471
5620
5826
5920
6059
6221
6343
6538
6694
7115
7446
7751
7967
8310
8570
8837
1027.5
1039.2
1056.8
1061.9
1068.3
1089.0
1093.9
1104.2
1123.6
1178.5
1244.5
1302.3
1373.3
1449.3
1468.6
1461.3
The general equation for updating costs through the use of a cost index over a period from
time t ⫽ 0 (base) to another time t is
()
I
Ct ⴝ C0 —t
I0
[15.2]
where Ct ⫽ estimated cost at present time t
C0 ⫽ cost at previous time t0
It ⫽ index value at time t
I0 ⫽ index value at time t0
Generally, the indexes for equipment and materials are made up of a mix of components that
are assigned certain weights, with the components sometimes further subdivided into more basic
items. For example, the equipment, machinery, and support component of the chemical plant cost
index is subdivided into process machinery, pipes, valves and fittings, pumps and compressors,
and so forth. These subcomponents, in turn, are built up from even more basic items such as pressure pipe, black pipe, and galvanized pipe. Table 15–3 shows the Chemical Engineering plant
cost index (CEPCI), the Engineering News Record (ENR) construction cost index, and the
Marshall and Swift (M&S) equipment cost index for several years. The base period of 1957 to
1959 is assigned a value of 100 for the CEPCI, 1913 ⫽ 100 for the ENR index, and 1926 ⫽ 100
for the M&S equipment cost index.
Current and past values of several of the indexes may be obtained from the Internet (usually
for a fee). For example, the CE plant cost index is available at www.che.com兾pci. The ENR construction cost index is found at www.construction.com. This latter site offers a comprehensive
series of construction-related resources, including several ENR cost indexes and cost estimation
systems. A website used by many engineering professionals in the form of a “technical chat
room” for all types of topics, including estimation, is www.eng-tips.com.
EXAMPLE 15.2
In evaluating the feasibility of a major construction project, an engineer is interested in estimating the cost of skilled labor for the job. The engineer finds that a project of similar complexity and magnitude was completed 5 years ago at a skilled labor cost of $360,000. The ENR
skilled labor index was 3496 then and is now 5127. What is the estimated skilled labor cost for
the new project?
Cost Indexes
15.3
Solution
The base time t0 is 5 years ago. Using Equation [15.2], the present cost estimate is
(
5127
Cr ⫽ 360,000 ———
3496
⫽ $527,952
)
In the manufacturing and service industries, tabulated cost indexes are not readily available.
The cost index will vary, perhaps with the region of the country, the type of product or service,
and many other factors. When estimating costs for a manufacturing system, for example, it is
often necessary to develop the cost index for high-priority items such as subcontracted components, selected materials, and labor costs. The development of the cost index requires the actual
cost at different times for a prescribed quantity and quality of the item. The base period is a selected time when the index is defined with a basis value of 100 (or 1). The index each year (period) is determined as the cost divided by the base-year cost and multiplied by 100 (or 1). Future
index values may be forecast using simple extrapolation or more refined mathematical techniques, such as time-series analysis.
EXAMPLE 15.3
Sean, owner of Alamo Pictures, makes fact-based documentaries about the Old West and
sells them in a variety of outlets, by mail, and online. He has decided to expand into new
areas and wants to make cost estimates for three of the more significant labor costs involved
in making these types of films. The Director of Finance of Alamo Pictures generated the
annual average hourly costs (Table 15–4).
(a) Make 2008 the base year, and determine the cost indexes using a basis of 1.00. Comment
on the trend of each index over the years.
(b) Sean expects to utilize a lot of graphics services in 2014 for a planned documentary; however, it is the most rapidly increasing cost component. The cost in 2010 was $78 per hour;
assume a worst-case scenario is that the graphics index continues the same arithmetic trend
it had from 2010 to 2011. Determine the hourly cost that he should budget for in 2014.
Solution
(a) For each type of service, calculate It兾I0 where t ⫽ 2005, 2006, . . . with 2008 as the base
year 0. Table 15–5 presents the indexes. Observations about trend are as follows:
Graphics labor cost: Constantly increasing over all years.
Stuntmen labor cost: Rising until 2009, then stable.
Actors labor cost: Higher in 2007 and 2008; comparatively lower and stable in other years.
TABLE 15–4
Average Hourly Costs for Three Services, Example 15.3
Cost of Service, Average $ per Hour
Type of Service
Graphics
Stuntmen
Actors
TABLE 15–5
2005
2006
2007
2008
2009
2010
2011
50
50
80
56
55
80
65
60
90
67
70
90
70
87
80
78
83
75
90
85
85
Index Values with 2008 as Base Year, Example 15.3
Index It 兾I0
Type of Service
2005
2006
2007
2008
2009
2010
2011
Graphics
Stuntmen
Actors
0.75
0.71
0.89
0.84
0.79
0.89
0.97
0.86
1.00
1.00
1.00
1.00
1.04
1.24
0.89
1.16
1.19
0.83
1.34
1.21
0.94
393
394
Cost Estimation and Indirect Cost Allocation
Chapter 15
(b) The index in 2010 is 1.16. The increase to 2011 is 0.18; the index value in 2014 will be
1.34 ⫹ 3(0.18) ⫽ 1.88. Equation [15.2] finds the expected, worst-case cost in 2014.
C2014 ⫽ C2010(I2014兾I2010) ⫽ 78(1.88兾1.16)
⫽ 78(1.62)
⫽ $126 per hour
15.4 Cost-Estimating Relationships:
Cost-Capacity Equations
Design variables (speed, weight, thrust, physical size, etc.) for plants, equipment, and construction are determined in the early design stages. Cost-estimating relationships (CERs) use these
design variables to predict costs. Thus, a CER is generically different from the cost index method,
because the index is based on the cost history of a defined quantity and quality of a variable.
One of the most widely used CER models is a cost-capacity equation. As the name implies,
an equation relates the cost of a component, system, or plant to its capacity. This is also known
as the power law and sizing model. Since many cost-capacity equations plot as a straight line on
log-log paper, a common form is
( )
Q
C2 ⴝ C1 ——2
Q1
x
[15.3]
where C1 ⫽ cost at capacity Q1
C2 ⫽ cost at capacity Q2
x ⫽ correlating exponent
The value of the exponent for various components, systems, or entire plants can be obtained or
derived from a number of sources, including Plant Design and Economics for Chemical Engineers, Preliminary Plant Design in Chemical Engineering, Chemical Engineers’ Handbook,
technical journals (especially Chemical Engineering), the U.S. Environmental Protection Agency,
professional or trade organizations, consulting firms, handbooks, and equipment companies.
Table 15–6 is a partial listing of typical values of the exponent for various units. When an exponent value for a particular unit is not known, it is common practice to use the value of x ⫽ 0.6. In
fact, in the chemical processing industry, Equation [15.3] is referred to as the six-tenths model.
The exponent x in the cost-capacity equation is commonly in the range 0 ⬍ x ⱕ 1.
If x ⬍ 1, economies of scale provide a cost advantage for larger sizes.
If x ⫽ 1, a linear relationship is present.
If x ⬎ 1, there are diseconomies of scale present in that a larger size is more costly than that
of a linear relation.
It is especially powerful to combine the time adjustment of the cost index (It兾I0) from Equation [15.2] with a cost-capacity equation to estimate costs that change over time. If the index is
embedded into the cost-capacity computation in Equation [15.3], the cost at time t and capacity
level 2 may be written as the product of two independent terms.
C2,t ⫽ (cost at time 0 of level 2) ⫻ (time adjustment cost index)
[ ( ) ] ( II )
Q
⫽ C1,0 ——2
Q1
x
—t
0
This is commonly expressed without the time subscripts. Thus,
( ) ( II )
Q
C2 ⴝ C1 ——2
Q1
x
—t
0
The following example illustrates the use of this relation.
[15.4]
Cost-Estimating Relationships: Factor Method
15.5
TABLE 15–6
Sample Exponent Values for Cost-Capacity Equations
Size Range
Exponent
1–100 MGD
0.2–40 MGD
1000–7000 ft兾min
40–60 in
3000–350,000 tons兾year
0.1–100 MGD
5–300 hp
200–2100 hp
20–8000 ft3兾min
15–400 ft2
0.5–200 MGD
500–3000 ft2
500–20,000 scfd
0.05–50 MGD
0.05–20 MGD
10–200 hp
50–4000 gal
0.04–5 MGD
0.01–0.2 MGD
100–2000 gal
0.84
0.14
0.46
0.71
0.44
0.98
0.90
0.32
0.64
0.71
0.82
0.55
0.56
1.02
1.13
0.69
0.74
1.35
0.14
0.67
Component兾System兾Plant
Activated sludge plant
Aerobic digester
Blower
Centrifuge
Chlorine plant
Clarifier
Compressor, reciprocating (air service)
Compressor
Cyclone separator
Dryer
Filter, sand
Heat exchanger
Hydrogen plant
Laboratory
Lagoon, aerated
Pump, centrifugal
Reactor
Sludge drying beds
Stabilization pond
Tank, stainless
Note: MGD ⫽ million gallons per day; hp ⫽ horsepower; scfd ⫽ standard cubic feet per day.
EXAMPLE 15.4
The total design and construction cost for a digester to handle a flow rate of 0.5 million gallons
per day (MGD) was $1.7 million in 2010. Estimate the cost today for a flow rate of 2.0 MGD.
The exponent from Table 15–6 for the MGD range of 0.2 to 40 is 0.14. The cost index in 2010
of 131 has been updated to 225 for this year.
Solution
Equation [15.3] can estimate the cost of the larger system in 2010, but it must be updated by
the cost index to today’s dollars. Equation [15.4] performs both operations at once. The estimated cost is
2.0 0.14 ——
225
C2 ⫽ 1,700,000 ——
131
0.5
⫽ 1,700,000(1.214)(1.718) ⫽ $3.546 million
( ) ( )
15.5 Cost-Estimating Relationships:
Factor Method
Another widely used model for preliminary cost estimates of process plants is called the factor
method. While the methods discussed above can be used to estimate the costs of major items of
equipment, processes, and the total plant costs, the factor method was developed specifically for
total plant costs. The method is based on the premise that fairly reliable total plant costs can be
obtained by multiplying the cost of the major equipment by certain factors. Since major equipment costs are readily available, rapid plant estimates are possible if the appropriate factors are
known. These factors are commonly referred to as Lang factors after Hans J. Lang, who first
proposed the method.
395
396
Cost Estimation and Indirect Cost Allocation
Chapter 15
In its simplest form, the factor method is expressed in the same form as the unit method
CT ⴝ hCE
[15.5]
where CT ⫽ total plant cost
h ⫽ overall cost factor or sum of individual cost factors
CE ⫽ total cost of major equipment
Direct兾Indirect costs
The h may be one overall cost factor or, more realistically, the sum of individual cost components
such as construction, maintenance, direct labor, materials, and indirect cost elements. This follows the cost estimation approaches presented in Figure 15–1.
In his original work, Lang showed that direct cost factors and indirect cost factors can be
combined into one overall factor for some types of plants as follows: solid process plants, 3.10;
solid-fluid process plants, 3.63; and fluid process plants, 4.74. These factors reveal that the total
installed-plant cost is many times the first cost of the major equipment.
EXAMPLE 15.5
An engineer with Valero Petroleum has learned that an expansion of the solid-fluid process
plant is expected to have a delivered equipment cost of $2.08 million. If the overall cost factor
for this type of plant is 3.63, estimate the plant’s total cost.
Solution
The total plant cost is estimated by Equation [15.5].
CT ⫽ 3.63(2,080,000)
⫽ $7,550,400
Subsequent refinements of the factor method have led to the development of separate factors
for direct and indirect cost components. Direct costs as discussed in Section 15.1 are specifically
identifiable with a product, function, or process. Indirect costs are not directly attributable to a
single function, but are shared by several because they are necessary to perform the overall objective. Examples of indirect costs are general administration, computer services, quality, safety,
taxes, security, and a variety of support functions. The factors for both direct and indirect costs
are sometimes developed for use with delivered-equipment costs and other times for installedequipment costs, as defined in Section 15.1. In this text, we assume that all factors apply to delivered-equipment costs, unless otherwise specified.
For indirect costs, some of the factors apply to equipment costs only, while others apply to the
total direct cost. In the former case, the simplest procedure is to add the direct and indirect cost
factors before multiplying by the delivered-equipment cost. The overall cost factor h can be
written as
n
h⫽1⫹
兺f
[15.6]
i
i⫽1
where fi ⫽ factor for each cost component
i ⫽ 1 to n components, including indirect cost
If the indirect cost factor is applied to the total direct cost, only the direct cost factors are added
to obtain h. Therefore, Equation [15.5] is rewritten as
[ (
CT ⴝ CE 1 ⴙ
n
兺f
i
iⴝ1
where fI ⫽ indirect cost factor
fi ⫽ factors for direct cost components
Examples 15.6 and 15.7 illustrate these equations.
)]
(1 ⴙ fI)
[15.7]
Traditional Indirect Cost Rates and Allocation
15.6
EXAMPLE 15.6
The delivered-equipment cost for a small chemical process plant is expected to be $2 million.
If the direct cost factor is 1.61 and the indirect cost factor is 0.25, determine the total plant cost.
Solution
Since all factors apply to the delivered-equipment cost, they are added to obtain h, the total cost
factor in Equation [15.6].
h ⫽ 1 ⫹ 1.61 ⫹ 0.25 ⫽ 2.86
The total plant cost is
CT ⫽ 2.86(2,000,000) ⫽ $5,720,000
EXAMPLE 15.7
A new container-handling crane at the Port of Singapore is expected to have a deliveredequipment cost of $875,000. The cost factor for the installation of tracks, concrete, steel, noise
abatement, supports, etc., is 0.49. The construction factor is 0.53, and the indirect cost factor is
0.21. Determine the total cost if (a) all cost factors are applied to the cost of the delivered
equipment and (b) the indirect cost factor is applied to the total direct cost.
Solution
(a) Total equipment cost is $875,000. Since both the direct and indirect cost factors are applied
to only the equipment cost, the overall cost factor from Equation [15.6] is
h ⫽ 1 ⫹ 0.49 ⫹ 0.53 ⫹ 0.21 ⫽ 2.23
The total cost is
CT ⫽ 2.23(875,000) ⫽ $1,951,250
(b) Now the total direct cost is calculated first, and Equation [15.7] is used to estimate total
cost.
n
h⫽1⫹
兺 f ⫽ 1 ⫹ 0.49 ⫹ 0.53 ⫽ 2.02
i
i⫽1
CT ⫽ [875,000(2.02)](1.21) ⫽ $2,138,675
Comment
Note the decrease in estimated cost when the indirect cost is applied to the equipment cost only
in part (a). This illustrates the importance of determining exactly what the factors apply to
before they are used.
15.6 Traditional Indirect Cost Rates and Allocation
Costs incurred in the production of an item or delivery of a service are tracked and assigned by a
cost accounting system. For the manufacturing environment, it can be stated generally that the
statement of cost of goods sold (discussed in Appendix B) is one end product of this system. The
cost accounting system accumulates material costs, labor costs, and indirect costs (also called
overhead costs or factory expenses) by using cost centers. All costs incurred in one department
or process line are collected under a cost center title, for example, Department 3X. Since direct
materials and direct labor are usually directly assignable to a cost center, the system need only
identify and track these costs. Of course, this in itself is no easy chore, and the cost of the tracking
system may prohibit collection of all direct cost data to the level of detail desired.
One of the primary and more difficult tasks of cost accounting is the allocation of indirect costs
when it is necessary to allocate them separately to departments, processes, and product lines.
397
398
Cost Estimation and Indirect Cost Allocation
Chapter 15
TABLE 15–7
Indirect costs
Sample Indirect Cost Allocation Bases
Cost Category
Possible Allocation Basis
Taxes
Heat, light
Power
Receiving, purchasing
Personnel, machine shop
Building maintenance
Software
Quality control
Space occupied
Space, usage, number of outlets
Space, direct labor hours, horsepower-hours, machine hours
Cost of materials, number of orders, number of items
Direct labor hours, direct labor cost
Space occupied, direct labor cost
Number of accesses
Number of inspections
Indirect costs are costs associated with property taxes, service and maintenance departments, personnel, legal, quality, supervision, purchasing, utilities, software development,
etc. They must all be allocated to the using cost center. Detailed collection of these data is
cost-prohibitive and often impossible; thus, allocation schemes are utilized to distribute the
expenses on a reasonable basis.
A listing of possible allocation bases is included in Table 15–7. Historically, common bases have
been direct labor cost, direct labor hours, machine-hours, number of employees, space, and direct
materials.
Most allocation is accomplished utilizing a predetermined indirect cost rate, computed by
using the general relation.
estimated total indirect costs
Indirect cost rate ⴝ —————————————
estimated basis level
[15.8]
The estimated indirect cost is the amount allocated to a cost center. For example, if a division has
two producing departments, the total indirect cost allocated to a department is used as the numerator in Equation [15.8] to determine the department rate. Example 15.8 illustrates allocation
when the cost center is a machine.
EXAMPLE 15.8
The manager of beauty products at BestWay wants to determine allocation rates for $150,000
of indirect costs for the three machines used to process beauty lotions. The following information was obtained from last year’s budget for the three machines. Determine rates for each
machine if the amount is equally distributed.
Cost Source
Allocation Basis
Machine 1
Machine 2
Machine 3
Direct labor cost
Direct labor hours
Direct material cost
Estimated Activity Level
$100,000
2000 hours
$250,000
Solution
Applying Equation [15.8] for each machine, the annual rates are
indirect budget
50,000
Machine 1 rate ⫽ ——————— ⫽ ————
direct labor cost 100,000
⫽ $0.50 per direct labor dollar
indirect budget
50,000
Machine 2 rate ⫽ ———————— ⫽ ———
2000
direct labor hours
⫽ $25 per direct labor hour
indirect budget
50,000
Machine 3 rate ⫽ ——————— ⫽ ————
material cost
250,000
⫽ $0.20 per direct material dollar
Traditional Indirect Cost Rates and Allocation
15.6
Now the actual direct labor costs and hours and material costs are determined for this
year, and each dollar of direct labor cost spent on machine 1 implies that $0.50 in indirect
cost will be added to the cost of the product. Similarly, indirect costs are added for machines 2 and 3.
When the same allocation basis is used to distribute indirect costs to several cost centers, a
blanket rate may be determined. For example, if direct materials are the basis for allocation to
four separate processing lines, the blanket rate is
total indirect costs
Indirect cost rate ⫽ ———————————
total direct materials cost
If $500,000 in indirect costs and $3 million in materials are estimated for the four lines, the blanket indirect rate is 500,000兾3,000,000 ⫽ $0.167 per materials cost dollar. Blanket rates are easier
to calculate and apply, but they do not account for differences in the type of activities accomplished in each cost center.
In most cases, machinery or processes add value to the end product at different rates per unit
or hour of use. For example, light machinery may contribute less per hour than heavy, more expensive machinery. This is especially true when advanced technology processing, for example,
an automated manufacturing cell, is used along with traditional methods, for example, nonautomated finishing equipment. The use of blanket rates in these cases is not recommended, as the
indirect cost will be incorrectly allocated. The lower-value-contribution machinery will accumulate too much of the indirect cost. The approach should be the application of different bases for
different machines, activities, etc., as discussed earlier and illustrated in Example 15.8. The use
of different, appropriate bases is often called the productive hour rate method since the cost
rate is determined based on the value added, not a uniform or blanket rate. Realization that more
than one basis should be normally used in allocating indirect costs has led to the use of activitybased costing methods, as discussed in the next section.
Once a period of time (month, quarter, or year) has passed, the indirect cost rates are applied
to determine the indirect cost charge, which is then added to direct costs. This results in the total
cost of production, which is called the cost of goods sold, or factory cost. These costs are all accumulated by cost center.
If the total indirect cost budget is correct, the indirect costs charged to all cost centers for the
period of time should equal this budget amount. However, since some error in budgeting always
exists, there will be overallocation or underallocation relative to actual charges, which is termed
allocation variance. Experience in indirect cost estimation assists in reducing the variance at the
end of the accounting period.
EXAMPLE 15.9
Since the manager determined indirect cost rates for BestWay (Example 15.8), she can now
compute the total factory cost for a month. Perform the computations using the data in
Table 15–8. Also calculate the variance for indirect cost allocation for the month.
Solution
Start with the cost of goods sold (factory cost) relation given by Equation [B.1] in Appendix B,
which is
Cost of goods sold ⫽ direct materials ⫹ direct labor ⫹ indirect costs
TABLE 15–8
Cost Source
Material
Labor
Actual Monthly Data Used for Indirect Cost Allocation
Machine Number
Actual Cost, $
Actual Hours
1
3
1
2
3
3,800
19,550
2,500
3,200
2,800
650
750
720
399
400
Cost Estimation and Indirect Cost Allocation
Chapter 15
To determine indirect cost, the rates from Example 15.8 are applied:
Machine 1 indirect ⫽ (labor cost)(rate) ⫽ 2500(0.50) ⫽ $1250
Machine 2 indirect ⫽ (labor hours)(rate) ⫽ 750(25.00) ⫽ $18,750
Machine 3 indirect ⫽ (material cost)(rate) ⫽ 19,550(0.20) ⫽ $3910
Total charged indirect cost ⫽ $23,910
Factory cost is the sum of actual material and labor costs from Table 15–8 of $31,850 and the
indirect cost charge for a total of $55,760.
Based on the annual indirect cost budget of $150,000, one month represents 1兾12 of the
total or
150,000
Monthly budget ⫽ ————
12
⫽ $12,500
The allocation variance for total indirect cost is
Variance ⫽ 12,500 ⫺ 23,910 ⫽ $⫺11,410
This is a large budget underallocation, since much more was actually charged than allocated.
The $12,500 budgeted for the three machines represents a 91.3% underallocation of indirect
costs. This analysis for only one month of a year will most likely prompt a rapid review of the
rates and the indirect cost budget.
Once estimates of indirect costs are determined, it is possible to perform an economic analysis of
the present operation versus a proposed operation. Such a study is described in Example 15.10.
EXAMPLE 15.10
For several years, Cuisinart Corporation has purchased the carafe assembly of its major coffeemaker line at an annual cost of $2.2 million. The suggestion to make the component in-house
has been made. For the three departments involved, the annual indirect cost rates, estimated
material, labor, and hours are found in Table 15–9. The allocated hours column is the time
necessary to produce the carafes for a year.
Equipment must be purchased with the following estimates: first cost of $2 million, salvage
value of $50,000, and life of 10 years. Perform an economic analysis for the make alternative,
assuming that a market rate of 15% per year is the MARR.
Solution
For making the components in-house, the AOC is comprised of direct labor, direct material,
and indirect costs. Use the data of Table 15–9 to calculate the indirect cost allocation.
25,000(10) ⫽ $250,000
25,000(5) ⫽ 125,000
10,000(15) ⫽ 150,000
$525,000
AOC ⫽ 500,000 ⫹ 300,000 ⫹ 525,000 ⫽ $1,325,000
Department A:
Department B:
Department C:
TABLE 15–9
Production Cost Estimates for Example 15.10
Indirect Costs
Department
Basis,
Hours
Rate
per Hour, $
Allocated
Hours
Material
Cost, $
Direct
Labor
Cost, $
A
B
C
Labor
Machine
Labor
10
5
15
25,000
25,000
10,000
200,000
50,000
50,000
200,000
200,000
100,000
300,000
500,000
Activity-Based Costing (ABC) for Indirect Costs
15.7
The make alternative annual worth is
AWmake ⫽ ⫺P(A兾P,i,n) ⫹ S(A兾F,i,n) ⫺ AOC
⫽ ⫺2,000,000(A兾P,15%,10) ⫹ 50,000(A兾F,15%,10) ⫺ 1,325,000
⫽ $⫺1,721,037
Currently, the carafes are purchased with an AW of
AWbuy ⫽ $⫺2,200,000
It is cheaper to make, because the AW of costs is less.
15.7 Activity-Based Costing (ABC) for Indirect Costs
As automation, software, and manufacturing technologies have advanced, the number of
direct labor hours necessary to manufacture a product has decreased substantially. Where
once as much as 35% to 45% of the final product cost was represented in labor, now the
labor component is commonly 5% to 15% of total manufacturing cost. However, the indirect cost may represent as much as 35% to 45% of the total manufacturing cost. The use of
bases, such as direct labor hours, to allocate indirect cost is not accurate enough for automated and technologically advanced environments. This has led to the development of
methods that replace or supplement traditional cost allocations that rely upon one form or
another of Equation [15.8]. Also, allocation bases different from traditional ones are commonly utilized.
A product that by traditional methods may have contributed a large portion to profit may actually be a loser when indirect costs are allocated more correctly. Companies that have a wide variety of products and produce some in small lots may find that traditional allocation methods have
a tendency to underallocate the indirect cost to small-lot products. This may indicate that they are
profitable, when in actuality they are losing money.
The best allocation method for high-overhead industries is Activity-Based Costing (ABC). It
is designed to identify cost centers, activities, and cost drivers. Descriptions of each follow.
Cost centers: The final products or services of the corporation are called cost centers or cost
pools. They receive the allocated indirect costs.
Activities: These are usually support departments (purchasing, quality, IT, maintenance, engineering, supervision) that generate the indirect costs which are then distributed to the cost
centers.
Cost drivers: Commonly expressed in volumes, these drive the consumption of a shared resource. Examples are the number of purchase orders, cost of engineering change orders,
number of machine setups, number of safety violations, and the like.
Implementing ABC involves several steps.
1. Identify each activity and its total cost.
2. Identify the cost drivers and their usage volumes.
3. Calculate the indirect cost rate for each activity.
total cost of activity
ABC indirect cost rate ⴝ ————————————
total volume of cost driver
[15.9]
4. Use the rate to allocate indirect cost to cost centers for each activity.
As an illustration, assume a company that produces two types of industrial lasers (cost centers) has three primary support departments (activities; step 1 in the procedure). The allocation
for costs generated by the purchasing department, for example, is based on the number of purchase orders (step 2) to support laser production. The ABC rate (step 3) in dollars per purchase
order is used to allocate indirect costs to the two laser products (step 4).
401
402
Cost Estimation and Indirect Cost Allocation
Chapter 15
EXAMPLE 15.11
A multinational aerospace firm uses traditional methods to allocate manufacturing and management support costs for its European division. However, accounts such as business travel
have historically been allocated on the basis of the number of employees at the plants in France,
Italy, Germany, and Spain.
The president recently stated that some product lines are likely generating much more travel
than others. The ABC system is chosen to augment the traditional method to more precisely
allocate travel costs to major product lines at each plant.
(a) First, assume that allocation of total observed travel expenses of $500,000 to the plants
using a traditional basis of workforce size is sufficient. If total employment of 29,100 is
distributed as follows, allocate the $500,000.
Paris, France plant
Florence, Italy plant
Hamburg, Germany plant
Barcelona, Spain plant
12,500 employees
8,600 employees
4,200 employees
3,800 employees
(b) Now, assume that corporate management wants to know more about travel expenses based
on product line, not merely plant location and workforce size. The ABC method will be
applied to allocate travel costs to major product lines. Annual plant support budgets indicate that the following percentages are expended for travel:
Paris
Florence
Hamburg
Barcelona
5% of $2 million
15% of $500,000
17.5% of $1 million
30% of $500,000
Further, the study indicates that in 1 year a total of 500 travel vouchers were processed by
the management of the major five product lines produced at the four plants. The distribution is as follows:
Paris
Florence
Hamburg
Barcelona
Product lines—1 and 2; number of vouchers—50 for line 1, 25 for 2.
Product lines—1, 3, and 5; vouchers—80 for line 1, 30 for 3, 30 for 5.
Product lines—1, 2 and 4; vouchers—100 for line 1, 25 for 2, 20 for 4.
Product line—5; vouchers—140 for line 5.
Use the ABC method to determine how the product lines drive travel costs at the plants.
Solution
(a) In this case, Equation [15.8] takes the form of a blanket rate per employee.
travel budget
Indirect cost rate ⫽ ———————
total workforce
$500,000
⫽ ———— ⫽ $17.1821 per employee
29,100
Using this traditional basis of rate times workforce size results in a plant-by-plant allocation.
Paris
Florence
Hamburg
Barcelona
$17.1821(12,500) ⫽ $214,777
$147,766
$72,165
$65,292
(b) The ABC method is more involved to apply, and the by-plant amounts will be different
from those in part (a) since completely different bases are applied. Use the 4-step procedure to allocate travel costs to the five products.
Step 1. The total amount to be allocated is determined from the percentages of each
plant’s support budget devoted to travel. The number is determined from the
percent-of-budget data as follows:
0.05(2,000,000) ⫹ . . . ⫹ 0.30(500,000) ⫽ $500,000
Making Estimates and Maintaining Ethical Practices
15.8
TABLE 15–10
ABC Allocation of Travel Cost ($ in Thousands), Example 15.11
Product Line
Paris
Florence
Hamburg
Barcelona
Total
1
2
50
80
100
25
$230
$50
3
4
30
25
5
30
20
$30
$20
140
$170
Total
75
140
145
140
$500
Step 2. The cost driver is the number of travel vouchers submitted by the management
unit responsible for each product line at each plant. The allocation will be to
the products directly, not to the plants. However, the travel allocation to the
plants can be determined afterward since we know what product lines are
produced at each plant.
Step 3. Equation [15.9] determines an ABC allocation rate.
total cost of travel
ABC allocation rate ⫽ ———————————
total number of vouchers
$500,000
⫽ ————
500
⫽ $1000 per voucher
Step 4. Table 15–10 summarizes the vouchers and allocation by product line and by city.
Product 1 ($230,000) and product 5 ($170,000) drive the travel costs based on
the ABC analysis. Comparison of the by-plant totals in Table 15–10 (far right
column) with the respective totals in part (a) indicates a substantial difference in
the amounts allocated, especially to Paris, Hamburg, and Barcelona. This comparison verifies the president’s statement that product lines, not plants, drive
travel requirements.
Comment
Let’s assume that product 1 has been produced in small lots at the Hamburg plant for a number
of years. This analysis, when compared to the traditional cost allocation method in part (a), reveals a very interesting fact. In the ABC analysis, Hamburg has a total of $145,000 travel dollars
allocated, $100,000 from product 1. In the traditional analysis based on workforce size, Hamburg
was allocated only $72,165—about 50% of the more precise ABC analysis amount. This indicates to management the need to examine the manufacturing lot size practices at Hamburg and
possibly other plants, especially when a product is currently manufactured at more than one plant.
Some proponents of the ABC method recommend discarding the traditional cost accounting
methods of a company and utilizing ABC exclusively. This is not a good approach, since ABC is
not a complete cost system. The ABC method provides information that assists in cost control,
while the traditional method emphasizes cost allocation and cost estimation. The two systems
work well together with the traditional methods allocating costs that have identifiable direct
bases, for example, direct labor. ABC analysis is usually more expensive and time-consuming
than a traditional cost allocation system, but in many cases it can assist in understanding the
economic impact of management decisions and in controlling certain types of indirect costs.
15.8 Making Estimates and Maintaining
Ethical Practices
Making estimates about the future of costs, revenues, cash flows, rates of return, and many other
parameters is routine when one is engaged in any type of economic analysis. Public agencies,
private corporations, and not-for-profit businesses all make economic decisions based on these
403
404
Chapter 15
Cost Estimation and Indirect Cost Allocation
estimates, most of which are made by employees of the organizations or by outside consultants
hired to perform specific activities under contract. The opportunities for bias, poor accuracy,
deception, and profit-driven or other motives are always present. The personal morals and adherence to codes of professional ethics discussed in Section 1.3 guide individuals in their work to
make fair and believable estimates for the analyses and decisions that follow.
The NSPE Code of Ethics for Engineers (Appendix C) referenced previously starts with a list
of six Fundamental Canons. One very relevant to estimation integrity is “Avoid deceptive acts.”
Acts that bias the results from experimental samples, previous cost data, or survey results for the
purposes of personal gain, increased profits, or favoritism are examples of unethical behavior.
Estimates of all types should be founded on practices such as the following:
Base estimates on sound information gathered over a range of situations representative of the
current one.
Use accepted theory and techniques in taking statistical samples, building budget elements,
and drawing conclusions that are included in proposals, applications, and recommendations.
As a consultant or contractor, keep personal and working relationships separate when making
estimates and delivering the final documentation to a client or sponsor.
The second case study at the end of this chapter presents an example of some ethical challenges
present when preparing estimates and proposals for contract work.
CHAPTER SUMMARY
Cost estimates are not expected to be exact, but they should be accurate enough to support a
thorough economic analysis using an engineering economy approach. There are bottom-up and
top-down approaches; each treats price and cost estimates differently.
Costs can be updated via a cost index, which is a ratio of costs for the same item at two separate times. The Consumer Price Index (CPI) is an often-quoted example of cost indexing. Cost
estimating may also be accomplished with a variety of models called cost-estimating relationships. Two of them are
Cost-capacity equation—good for estimating costs from design variables for equipment, materials, and construction
Factor method—good for estimating total plant cost
Traditional cost allocation uses an indirect cost rate determined for a machine, department, product line, etc. Bases such as direct labor cost, direct material cost, and direct labor hours are used.
With increased automation and information technology, different techniques of indirect cost allocation have been developed. The Activity-Based Costing method is an excellent technique to
augment the traditional allocation method.
The ABC method allocates indirect costs on the rationale that purchase orders, inspections,
machine setups, reworks, etc. drive the costs accumulated in departments or functions, such as
quality, purchasing, accounting, and maintenance. Improved understanding of how the company
or plant accumulates indirect costs is a major by-product of implementing the ABC method.
PROBLEMS
Understanding Cost Estimation
15.1 Rank the following estimate types in terms of time
spent to carry out the estimate (most time to least
time): partially designed, design 60% to 100%
complete, order of magnitude, scoping兾feasibility,
detailed estimate.
15.2 Classify the following cost elements as first cost
(FC) components or annual operating cost (AOC)
components for a piece of equipment on the
shop floor: supplies, insurance, equipment cost,
utility cost, installation, delivery charges, labor
cost.
15.3 State whether actual (A) or estimated (E) costs are
more likely to be used to carry out the following
activities: calculate taxes, make bids, pay bonuses,
determine profit or loss, predict sales, set prices,
405
Problems
evaluate proposals, distribute resources, plan production, and set goals.
15.4 Identify the output and input variables in both the
bottom-up and top-down approaches to cost estimating.
15.5 Classify the following costs as typically direct (D)
or indirect (I):
Project staff
Utilities
Raw materials
Project supplies
Administrative staff
Audit and legal
Rent
Equipment training
Labor
Miscellaneous office supplies
15.6 Identify each of the following costs associated
with owning an automobile as direct or indirect
(assume a direct cost is one that is directly attributable to the number of miles driven): license plate,
driver’s license, gasoline, highway toll fee, oil
change, repairs after collision, gasoline tax,
monthly loan payment, annual inspection fee, and
garage rental.
15.7 In the early and conceptual design stages of a project, what are the cost estimates called? Approximately how close should they be to the actual cost?
Unit Costs
15.8 Use the unit cost method to determine the preliminary cost of a guardrail for a small bridge if a total
of 120 linear feet will be required at a cost of
$58.19 per linear foot for material, equipment, and
labor.
15.9 Sun-Metro Bus is planning to construct a 600-car
parking garage on the outskirts of the city to encourage people to use the “Park ’n Ride” to work
or for shopping downtown. Prepare a preliminary
cost estimate for the garage if the cost per parking
space is $4700.
15.10 The Office of the Undersecretary of Defense periodically releases unit cost data for use in military
construction programs. If the unit cost for a satellite communications center (with shielding and
generator) is $496 per square foot, what would be
the estimated cost of a 6000-square-foot building?
15.11 The Department of Defense uses area cost factors
(ACFs) to adjust for differences in construction costs
in different parts of the country (and world). The
area cost factor for Andros Island in the Bahamas is
1.70 while the ACF for Rapid City, South Dakota, is
0.93. If a cold storage processing warehouse costs
$1,350,000 in Rapid City, estimate the cost for
Andros Island.
15.12 Preliminary cost estimates for jails can be made
using costs based on either unit area (square feet)
or unit volume (cubic feet). If the unit area cost is
$185 per square foot and the average height of the
ceilings is 10 feet, what is the unit volume cost?
15.13 The unit area and unit volume total project costs
for a library are $114 per square foot and $7.55 per
cubic foot, respectively. Based on these numbers,
what is the average height of library rooms?
15.14 The equipment required to place 160 cubic yards of
concrete by experienced workers is two gasoline engine vibrators and one concrete pump. If the vibrators cost $76 per day and the concrete pump costs
$580 per day, estimate (a) the cost of the equipment
per cubic yard of concrete and (b) the equipment
cost for placing 56 cubic yards of concrete.
15.15 A labor crew for placing concrete consists of
1 labor foreman at $25.85 per hour, 1 cement finisher at $28.60 per hour, 5 laborers at $23.25 per
hour each, and 1 equipment operator at $31.45 per
hour. Such a crew, called a C20 crew, can place 160
cubic yards of concrete per 8-hour day. Determine
(a) the cost per day of labor for the C20 crew, (b) the
cost of the C20 crew per cubic yard of concrete, and
(c) the cost to place 250 cubic yards of concrete.
15.16 Site work activities associated with constructing a
small bridge are shown in the table below. The
table includes the quantity of each activity, the unit
of measurement associated with each activity, and
the unit cost of each activity. Use the data to determine (a) the total cost for structural excavation, (b)
the total cost for the pile-driving rig, and (c) the
total labor cost for the site work.
Activity
Excavation,
unclassified
Unit of
Labor
Equipment
Material
Quantity Measurement Unit Cost, $ Unit Cost, $ Unit Cost, $
1667
cy
1.35
1.43
0
Excavation,
structural
120
cy
21.31
5.00
0
Backfill,
compacted
340
cy
7.78
1.72
0
job
ls
5688
6420
300
2240
lf
3.13
2.93
16.57
Pile-driving rig
Piling, steel,
driving
Legend: cy ⫽ cubic yard; ls ⫽ lump sum; lf ⫽ linear foot
Cost Indexes
15.17 From historical data, you discover that the national
average construction cost of middle schools is
$10,500 per student. If the state index for Texas is
76.9 and for California it is 108.5, estimate the
total construction cost of a middle school for 800
students in each state.
406
Chapter 15
15.18 A consulting engineering firm is preparing a preliminary cost estimate for a design-construct project of a coal processing plant. The firm, which
completed a similar project in 2001 with a construction cost of $30 million, wants to use the ENR
construction cost index to update the cost. Use the
values in Table 15–3 to estimate the cost of construction for a similar-size plant in mid-2010.
15.19 If the editors at ENR decide to redo the construction cost index so that the year 2000 has a base
value of 100, determine the value for the year (a)
1995 and (b) 2009.
15.20 (a) Estimate the value of the ENR construction
cost index by using the average (compounded)
percentage change in its value between 1995
and 2005 to predict the value in 2009.
(b) How much difference (numerical) is there between the estimated and actual 2009 values?
Is this an underestimate or overestimate?
15.21 An engineer who owns a construction company
that specializes in large commercial projects noticed that material costs increased at a rate of 1%
per month over the past 12 months. If a material
cost index were created for each month of this year
with the value of the index set at 100 for the beginning of the year, what would be the value of the
index at the end of the year? Express your answer
to two decimal places.
15.22 Using 1913 as the base year with a value of 100,
the ENR construction cost index (CCI) for August
2009 was 8563.35. For August 2010, the CCI
value was 8837.38. (a) What was the inflation rate
for construction for that 1-year period? (b) What
will be the CCI value in August 2013, provided the
inflation rate remains the same as it was in the
2009–2010 period?
15.23 Electropneumatic general-purpose pressure transducers convert supply pressure to regulated output
pressure in direct proportion to an electrical input
signal. The cost of a certain transducer was $194 in
1985 when the M&S equipment index was at
789.6. If the price increased exactly in proportion
to the M&S equipment index, what would its cost
be in 2010 (midyear) when the index value was
1461.3?
15.24 The ENR construction cost index for New York
City had a value of 12,381.40 in February 2007.
For Pittsburgh and Atlanta, the values were
7341.32 and 4874.06, respectively. If a general
contractor in Atlanta won construction jobs totaling $54.3 million, determine their equivalent total
value in New York City.
Cost Estimation and Indirect Cost Allocation
15.25 The ENR construction cost index (CCI) for August
2010 had a value of 8837.37 when the base year
was 1913 with a value of 100. If the base year is
1967 with a value of 1.0, the CCI for August 2010
would be 8.2272. In this case, what is the CCI
value for 1967 when the base year is 1913?
15.26 The ENR materials cost index (MCI) had a value
of 2708.51 in August 2010. In the same month, the
cost for cement was $96.55 per ton. If cement increased in price exactly in accordance with the
MCI, what was the cost of cement per ton in 1913
when the MCI index value was 100?
15.27 A contractor purchased equipment costing $40,000
in 2010 when the M&S equipment cost index was
at 1461.3. He remembers purchasing the same
equipment for $21,771 many years ago, but he does
not remember the year that he did so. If the M&S
equipment cost index increased by 2.68% per year
over that time period and the equipment increased
in price exactly in proportion to the index, (a) in
what year did he purchase the equipment and (b)
what was the value of the index in that year?
Cost-Capacity and Factor Methods
15.28 If the materials cost index decreased by 2% over
the same time period that the construction cost
index increased by 2%, what does that suggest
about the labor cost index, if labor represents 35%
of the cost of construction?
15.29 Use the exponent values in Table 15–6 to estimate
the cost for the following equipment to be placed
on an offshore drilling platform.
(a) The cost of a 125-hp centrifugal pump if a
200-hp pump costs $28,000.
(b) The cost of a 1700-gallon stainless steel tank
if a 900-gallon tank costs $4100.
15.30 A high-pressure stainless steel pump (1000 psi)
with a variable-frequency drive (VFD) is installed
in a seawater reverse-osmosis pilot plant that is recovering water from membrane concentrate at a
rate of 4 gallons per minute (gpm). The cost of the
pump was $13,000. Because of favorable results
from the pilot study, the city utility wants to go with
a full-scale system that will produce 500 gpm.
Determine the estimated cost of the larger pump, if
the exponent in the cost-capacity equation has a
value of 0.37.
15.31 A 0.75 million gallon per day (MGD) induceddraft packed tower for air-stripping trihalomethanes from drinking water costs $58,890. Estimate
the cost of a 2-MGD tower if the exponent in the
cost-capacity equation is 0.58.
Problems
15.32 The on-site manager told you that the variablefrequency drive (VFD) for a 300-hp motor costs
$20,000. Make the best cost estimate possible for
the VFD for a 100-hp motor. The exponent in the
cost-capacity equation is not available currently
where you are located.
15.33 The cost of a 68 m2 falling-film evaporator was
1.52 times the cost of the 30 m2 unit. What exponent value in the cost-capacity equation yielded
these results?
15.34 Reinforced concrete pipe (RCP) that is 12 inches
in diameter had a cost of $12.54 per linear foot in
Dallas, Texas in 2007. The cost for 24-inch RCP
was $27.23 per foot. If the cross-sectional area of
the pipe is considered the “capacity” in the costcapacity equation, determine the value of the exponent in the cost-capacity equation that exactly
relates the two pipe sizes.
15.35 A 100,000 barrel per day (bpd) fractionation tower
cost $1.2 million in 2001 when the Chemical Engineering plant cost index value was 394.3. How
much would a 450,000 bpd plant cost when the
index value is 575.8, provided the exponent in the
cost-capacity equation is 0.67?
15.36 A mini wind tunnel for calibrating vane or hotwire anemometers cost $3750 in 2002 when the
M&S equipment index value was 1104.2. If the
index value is now 1620.6, estimate the cost of a
tunnel twice as large. The cost-capacity equation
exponent is 0.89.
15.37 In 2008, a military engineer estimated the cost for
a classified laser-guided device to be $376,900.
The engineer used the M&S equipment index for
the years 1998 and 2008 and the cost-capacity
equation with an exponent value of 0.61. If the
original equipment had only one-fourth the capacity of the new equipment, what was the cost of the
original equipment in 1998?
15.38 Instead of using a cost-capacity equation for relating project size and construction cost, the Department of Defense uses Size Adjustment Factors
(SAFs) that are based on Size Relationship Ratios
(SRRs). For example, a SRR of 2.00 has a SAF of
0.942. Determine the required exponent value in
the cost-capacity equation in order for C2 to be
0.942C1 when Q2兾Q1 ⫽ 2.00.
15.39 The equipment cost for removing arsenic from a
well that delivers 800 gallons per minute (gpm) is
$1.8 million. If the overall cost factor for this type
of treatment system is 2.25, what is the total plant
cost expected to be?
407
15.40 A closed-loop filtration system for waterjet cutting
industries eliminates the cost of makeup water
treatments (water softeners, reverse osmosis, etc.)
while maximizing orifice life and machine performance. If the equipment cost is $225,000 and the
total plant cost is $1.32 million, what is the overall
cost factor for the system?
15.41 The equipment cost for a laboratory that plans to
specialize in analyzing for endocrine disrupters,
pharmaceuticals, and personal care products is
$870,000. If the direct cost factor is 1.32 and the
indirect cost factor is 0.45 (applies to equipment
only), determine the expected cost of the laboratory.
15.42 The equipment cost for a 10 gallon per minute
farm-scale ethanol fuel production plant is
$243,000. The direct cost factor for construction is
1.28 and for installation is 0.23. The indirect cost
factor for licenses, insurance, etc. is 0.84 (applied
to total direct cost). Determine the estimated total
plant cost.
15.43 A chemical engineer at Western Refining has estimated that the total cost for a diesel fuel desulfurization system will be $2.3 million. If the direct
cost factor is 1.35 and the indirect cost factor is
0.41, what is the total equipment cost? Both factors apply to delivered-equipment cost.
15.44 A mechanical engineer estimated that the equipment cost for a multitube cyclone system with a
capacity of 60,000 cfm would be $400,000. If the
direct cost factor is 3.1 and the indirect cost factor
is 0.38, what is the estimated total plant cost? The
indirect cost factor applies to the total direct cost.
15.45 Nicole is an engineer on temporary assignment at
a refinery operation in Seaside. She has reviewed a
cost estimate for $430,000, which covers some
new processing equipment for the ethylene line.
The equipment itself is estimated at $250,000 with
a construction cost factor of 0.30 and an installation cost factor of 0.30. No indirect cost factor is
listed, but she knows from other sites that indirect
cost is a sizable amount that increases the cost of
the line’s equipment. (a) If the indirect cost factor
should be 0.40, determine whether the current estimate includes a factor comparable to this value
(b) Determine the cost estimate if the 0.40 indirect
cost factor is used.
Indirect Cost Allocation
15.46 The company you work for currently allocates insurance costs on the basis of cost per direct labor
hour. This indirect cost component for the year is
budgeted at $36,000. If the direct labor hours for
408
Cost Estimation and Indirect Cost Allocation
Chapter 15
departments A, B, and C are expected to be 2000,
8000, and 5000, respectively, this year, determine
the allocation to each department.
of allocation is not indicated, and the company
accountant has no record of the basis used. However, the accountant advises the manager to not
be concerned because the allocation rates have
decreased each month.
15.47 The director of public works needs to distribute the
indirect cost allocation of $1.2 million to the three
branches around the city. The recorded amounts
for this year are as follows:
Indirect Cost, $
Records for This Year
Branch
Miles Driven
Direct Labor
Hours
275,000
247,000
395,000
38,000
31,000
55,500
North
South
Midtown
Branch
Direct
Labor
Hours
Basis
Indirect Cost
Allocation
Last Year, $
North
South
Midtown
350,000
200,000
500,000
40,000
20,000
64,000
Miles
Labor
Labor
300,000
200,000
450,000
(a)
(b)
Determine the rates for this year for each
branch.
Use the rate to distribute this year’s total indirect cost. What percentage of this year’s
budget is now distributed?
15.48 A company has a processing department with 10
stations. Because of the nature and use of three of
these stations, each is considered a separate cost
center for indirect cost allocation. The remaining
seven are grouped as one center, CC190. Machine
operating hours are used as the allocation basis for
all machines. A total of $250,000 is allocated to the
department for next year. Use the data collected
this year to determine the indirect cost rate for
each center.
Cost Center
Indirect
Cost Allocated, $
Estimated
Machine Hours
CC100
CC110
CC120
CC190
25,000
50,000
75,000
100,000
800
200
1200
1600
15.49 All the indirect costs are allocated by accounting
for a department. The manager has obtained records of allocation rates and actual charges for
the prior 3 months and estimates for this month
(May) and next month (see the table). The basis
Rate
Allocated
Charged
February
March
April
May
June
1.40
1.33
1.37
1.03
0.92
2800
3400
3500
3600
6000
2600
3800
3500
During the evaluation, the following additional
information from departmental and accounting
records is obtained.
The director plans to use the allocation and information from last year to determine the rates for
this year. This information follows:
Miles
Driven
Month
Direct Labor
Month
February
March
April
May
June
(a)
(b)
Hours
Cost, $
Material
Cost, $
Departmental
Space, ft2
640
640
640
640
800
2560
2560
2560
2720
3320
5400
4600
5700
6300
6500
2000
2000
3500
3500
3500
With this information determine the allocation basis used each month.
Comment on the accountant’s statement
about decreasing allocation rates.
15.50 The mechanical components division manager
asks you to recommend a make/buy decision on a
major automotive subassembly that is currently
purchased externally for a total of $3.9 million
this year. This cost is expected to continue rising
at a rate of $300,000 per year. Your manager asks
that both direct and indirect costs be included
when in-house manufacturing (make alternative)
is evaluated. New equipment will cost $3 million
and will have a salvage of $0.5 million and a life
of 6 years. Estimates of materials, labor costs, and
other direct costs are $1.5 million per year. Typical indirect rates, bases, and expected usage are
shown below. Perform the AW evaluation at
MARR ⫽ 12% per year over a 6-year study period. Show both hand and spreadsheet solutions, as directed.
Department
Basis
Rate
X
Y
Z
Direct labor cost
Materials cost
Number of
inspections
$2.40 per $
$0.50 per $
$20 per
inspection
Expected
Usage
$450,000
$850,000
4,500
409
Problems
ABC Method
15.51 The municipal water and desalinization utility in a
California city currently allocates some costs for
maintenance shop workers to pumping stations
based on the number of pumps at each station. At
the last director’s semiannual meeting, a suggestion was made to change the allocation basis to the
number of trips that pump service personnel make
to each station, because some stations have old
pumps that require more maintenance. Information about the stations is below. The indirect cost
budget is $20,000 per pump.
(a) Allocate the budget to each station based on
the number of service trips.
(b) Determine the old allocation on the basis of
the number of pumps, and comment on any
significant differences in the amounts allocated to the stations.
Station ID
No. of Pumps
Service Trips per Year
Sylvester
Laurel
7th St
Spicewood
5
7
3
4
190
55
38
104
15.52 Factory Direct manufactures and sells manufactured
homes. Traditionally, it has distributed indirect costs
to its three construction plants based on materials
cost. Each plant builds different models and floor
plans. Advances in weight and shape of plastic and
wood composite components have decreased cost
and time to produce a unit. Because of these advances, the CFO plans to use build-time per unit as
the new basis. However, he initially wants to determine what the allocation would have been this year
had build-time been the basis prior to incorporation
of the new materials. The data shown represents average costs and times. Use this data and the three
bases indicated to determine the allocation rates and
indirect cost distribution of $900,000 for this year.
Plant
Texas
Oklahoma
Kansas
Direct material cost,
$ per unit
Previous build-time
per unit, work-hours
New build-time per
unit, work-hours
20,000
12,700
18,600
400
415
580
425
355
480
Problems 15.53 through 15.55 use the following
information.
Jet Green Airways historically distributes the indirect
costs of lost and damaged baggage to its three major hubs
using a basis of annual number of flights in and out of
each hub. Last year $667,500 was distributed as follows:
Hub Airport
Flights
Rate, $ per Flight
Allocation, $
DFW
YYZ
MEX
55,000
20,833
15,000
6
9
10
330,000
187,500
150,000
The airline’s baggage management director suggests that
an allocation on the basis of baggage traffic, not flights,
will better represent the distribution, primarily based on
the fact that the high fees now charged to passengers to
check luggage have significantly changed the number of
bags handled at the major hubs. Total number of bags
handled during the year are 2,490,000 at DFW, 1,582,400
at YYZ, and 763,500 at MEX.
15.53 What are the activity and the cost driver for the
suggested baggage-traffic basis?
15.54 Using the baggage-traffic basis, determine the allocation rate using last year’s total of $667,500,
and distribute this amount to the hubs this year.
15.55 What are the percentage changes in allocation at
each hub using the two different bases?
15.56 On-line Vacation distributes advertising costs to its
four resort sites in the Caribbean on the basis of
the size of the resort budget. For this year, in round
numbers, the budgets and allocation of $1 million
advertising indirect costs are as follows:
Site
Budget, $
Allocation, $
(a)
A
B
C
D
2 million
200,000
3 million
300,000
4 million
400,000
1 million
100,000
Determine the allocation if the ABC method is
used with a new basis. Define the activity as the
advertising department at each resort. The cost
driver is the number of guests during the year.
Site
Guests
(b)
A
B
C
D
3500
4000
8000
1000
Again use the ABC method, but now make the
cost driver the total number of guest nights at
each resort. The average number of lodging
nights for guests at each site is as follows:
Site
Length of stay, nights
(c)
A
B
C
D
3.0
2.5
1.25
4.75
Comment on the distribution of advertising
costs using the two methods. Identify any
other cost drivers that might be considered
for the ABC approach that may reflect a
realistic allocation of the costs.
410
Chapter 15
Cost Estimation and Indirect Cost Allocation
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
15.57 In the bottom-up approach to cost estimating:
(a) Required price is an input variable.
(b) Cost estimates are an output variable.
(c) Required price is an output variable.
(d) Both (a) and (b) are correct.
15.58 A ratio of the cost of something today to its cost at
some time in the past is called a:
(a) Cost-capacity index
(b) Cost index
(c) Buyer’s guide
(d) Bluebook index
15.59 Index values could probably be obtained from all
of the following places except:
(a) Trade organizations
(b) Engineering News Record magazine
(c) Government organizations
(d) Home Depot or Lowe’s home improvement
stores
15.60 A 50-hp turbine pump was purchased for $2100. If
the exponent in the cost-capacity equation has a
value of 0.76, a 200-hp turbine pump could be expected to cost about:
(a) $6020
(b) $5320
(c) $4890
(d) $4260
100,000 units per day was $3 million, the value of
the exponent in the cost-capacity equation is:
(a) 0.26
(b) 0.39
(c) 0.45
(d) 0.60
15.64 The equipment for applying specialty coatings
that provide a high angle of skid for the paperboard and corrugated box industries has a delivered cost of $390,000. If the overall cost factor
for the complete system is 2.96, the total plant
cost is approximately:
(a) $954,400
(b) $1,054,400
(c) $1,154,400
(d) $1,544,400
15.65 The delivered-equipment cost for setting up a
production and assembly line for high-sensitivity,
gas-damped accelerometers is $650,000. If the direct cost and indirect cost factors are 1.82 and
0.31, respectively, and both factors apply to
delivered-equipment cost, the total plant cost estimate is approximately:
(a) $2,034,500
(b) $1,734,500
(c) $1,384,500
(d) $1,183,000
15.61 The city built a recreation park in 1980 for
$500,000. The ENR construction cost index had
a value of 3378.17 at that time. If the city is
planning to construct a similar recreation park
when the index value is 5542.16, the estimated
cost is closest to:
(a) $695,800
(b) $750,700
(c) $820,300
(d) $910,500
15.66 A police department wants to allocate the indirect
cost of speed monitoring to the three toll roads
around the city. An allocation basis that may not be
reasonable is:
(a) Miles of toll road monitored
(b) Average number of cars patrolling per hour
(c) Amount of car traffic per section of toll
road
(d) Cost to operate a patrol car
15.62 A small company bought a 250-hp compressor in
1998 for $3000 when the M&S equipment cost
index had a value of 1061.9. If the exponent in the
cost-capacity equation is 0.32 and the M&S index
value was 1449.3 in 2008, the cost of a 500-hp
compressor in 2008 was closest to:
(a) $3744
(b) $4094
(c) $4627
(d) $5110
15.67 The IT department allocates indirect costs to user
departments on the basis of CPU time at the rate of
$2000 per second. For the first quarter, the two
heaviest-use departments logged 900 and 1300 seconds, respectively. If the IT indirect budget for the
year is $8.0 million, the percentage of this year’s
allocation consumed by these departments is
closest to:
(a) 32%
(b) 22.5%
(c) 55%
(d) Not enough information to determine
15.63 The cost for implementing a manufacturing process that has a capacity of 6000 units per day was
$550,000. If the cost for a plant with a capacity of
15.68 If the engineering department is the activity to
receive indirect cost allocation for the year, cost
411
Case Study
drivers for the ABC method that seem reasonable
may be:
1.
2.
3.
(a)
(b)
(c)
(d)
15.69 Advantages of the ABC method are:
1.
Cost of engineering changes processed
Size of the workforce
Administrative cost to process a change
order
1
2
3
1 and 3
2.
3.
(a)
(b)
(c)
(d)
It is an excellent replacement for a traditional
cost accounting system.
It is always cheaper to operate than a traditional allocation system.
It can help explain the economic impact of
management decisions.
1
2
3
1 and 3
CASE STUDY
INDIRECT COST ANALYSIS OF MEDICAL EQUIPMENT MANUFACTURING COSTS
Background
Three years ago Medical Dynamics, a medical equipment
unit of Johnson and Sons, Inc., initiated the manufacture and
sales of a portable sterilization unit (Quik-Sterz) that can be
placed in the hospital room of a patient. This unit sterilizes
and makes available at the bedside some of the reusable instruments that nurses and doctors usually obtain by walking
to or receiving delivery from a centralized area. This new unit
makes the instruments available at the point and time of use
for burn and severe wound patients who are in a regular patient room.
There are two models of Quik-Sterz sold. The standard
version sells for $10.75, and a premium version with customized trays and a battery backup system sells for $29.75. The
product has sold well to hospitals, convalescent units, and
nursing homes at the level of about 1 million units per year.
Information
Medical Dynamics has historically used an indirect cost allocation system based upon direct hours to manufacture for
TABLE 15–11
all its other product lines. The same was applied when QuikSterz was priced. However, Arnie, the person who performed
the indirect cost analysis and set the sales price, is no longer
at the company, and the detailed analysis is no longer available. Through e-mail and telephone conversations, Arnie said
the current price was set at about 10% above the total manufacturing cost determined 2 years ago, and that some records
were available in the design department files. A search of
these files revealed the manufacturing and cost information in
Table 15–11. It is clear from these and other records that
Arnie used traditional indirect cost analysis based on direct
labor hours to estimate the total manufacturing costs of
$9.73 per unit for the standard model and $27.07 per unit for
the premium model.
Last year management decided to place the entire plant on
the ABC system of indirect cost allocation. The costs and
sales figures collected for Quik-Sterz the year before were
still accurate. Five activities and their cost drivers were identified for the Medical Dynamics manufacturing operations
(Table 15–12). Also, the volumes for each model are summarized in this table.
Historical Records of Direct and Indirect Cost Analyses for Quik-Sterz
Quik-Sterz Direct Cost (DC) Evaluation
Model
Standard
Premium
Direct Labor,
$兾Unit1
Direct Material,
$兾Unit
Direct Labor,
Hours兾Unit
Total Direct
Labor Hours
5.00
10.00
2.50
3.75
0.25
0.50
187,500
125,000
Allocated
IDC, $
Sales,
Units/Year
1.67 million
750,000
3.33 million
250,000
Quik-Sterz Indirect Cost (IDC) Evaluation
Model
Direct Labor,
Hours兾Unit
Standard
0.25
Premium
0.50
1
Average direct labor rate is $20 per hour.
Fraction IDC
Allocated
1
—
3
2
—
3
412
Cost Estimation and Indirect Cost Allocation
Chapter 15
TABLE 15–12
Quik-Sterz Activities, Cost Drivers, and Volume Levels for ABC-Based
Indirect Cost Allocation
Activity
Quality
Purchasing
Scheduling
Production setup
Machine operations
Cost Driver
Volume兾Year
Actual
Cost, $/Year
Inspections
Purchase orders
Change orders
Setups
Machine hours
20,000 inspections
40,000 orders
1,000 orders
5,000 setups
10,000 hours
800,000
1,200,000
800,000
1,000,000
1,200,000
Volume Level for the Year
Cost Driver
Quality inspections
Purchase orders
Number of change orders
Production setup
Machine hours
Standard
Premium
8,000
30,000
400
1,500
7,000
12,000
10,000
600
3,500
3,000
The ABC method will be used henceforth, with the intention
of determining the total cost and price based on its results. The
first impression of the production manager is that the new system will show that indirect costs for Quik-Sterz are about the
same as they have been for other products over the last several
years when a standard model and an upgrade (premium) model
were sold. Predictably, they state, the standard model will receive about 1兾3 of the indirect cost, and the premium will receive the remaining 2兾3. Fundamentally, there are two reasons
why the production manager does not like to produce premium
versions: They are less profitable for the company, and they require significantly more time and operations to manufacture.
Case Study Exercises
1. Use traditional indirect cost allocation to verify Arnie’s
cost and price estimates.
2. Use the ABC method to estimate the indirect cost allocation and total cost for each model.
3. If the prices and number of units sold are the same
next year (750,000 standard and 250,000 premium),
and all other costs remain constant, compare the
profit from Quik-Sterz under the ABC method with
the profit using the traditional indirect cost allocation method.
4. What prices should Medical Dynamics charge next year
based on the ABC method and a 10% markup over cost?
What is the total profit from Quik-Sterz predicted to be
if sales hold steady?
5. Using the results above, comment on the production
manager’s prediction of indirect costs using ABC (1兾3
standard; 2兾3 premium) and the two reasons given to
not produce the premium version of Quik-Sterz.
CASE STUDY
DECEPTIVE ACTS CAN GET YOU IN TROUBLE
Contributed by Dr. Paul Askenasy, Agronomist, Texas Commission on Environmental Quality
Background
Surface mining of coal is the removal of soil and sediments
from underlying strata that lie above the material to be mined.
The law requires that land disturbed by these types of mining
activities be returned to a productive capacity that is as good
as or better than its productive capacity before mining.
The productive capacity of soils is directly correlated to
the textural (sand, silt, and clay content) and chemical
characteristics of the soil (for example, pH). To this end,
mining companies must sample the different soils found in
the areas to be disturbed by mining activities. The purpose
of the sampling is to establish a baseline characterizing the
textural and chemical makeup of the soils prior to mining.
Soils in a low pH range (pH values ⬍ 5) are indicative of
low fertility. Once the natural resource, such as coal, is removed, the pit is backfilled with sediments and the terrain
surface is contoured to reestablish the premine drainages.
To meet the baseline for pH, the acreage of the mine soils
with low pH should not exceed the acreage of the unmined
soils with low pH.
Case Study
Information
Yucatan Mining Company (not the actual name) planned to
disturb 600 acres due to mining activities. The different soils
within the 600 acres were depicted in the County Soil Survey
where the mining activities were to take place. Prior to mining, the company obtained soil samples from 10 different locations within each soil type and had them analyzed for a
number of parameters including pH. Assessment of the data
indicated that 30% of the area (180 acres) occupied by the
soils in the area to be disturbed had pH values between 4.0
and 4.9. The application for mining was approved by the
State Department of Mining and Reclamation.
Six years later, 450 acres had been mined and the terrain
surface had been leveled to reestablish premine slopes. Of the
450 acres leveled, 175 acres had pH values between 4.0 and
4.9. The president of Yucatan indicated that the company
would submit a revised soil baseline based on new sampling
in the remaining 150 acres of unmined soils because, in his
opinion, the first soil baseline was biased.
The request to do more soil analyses to augment their existing soil baseline was approved. The company quickly hired
a consultant to develop the new baseline, and subsequently
Yucatan submitted the final report from its consultant to the
State Department of Mining and Reclamation. This revised
premine soil baseline indicated that 45% of the premine soils
had pH values between 4.0 and 4.9. Comparative results between the old and new samples can be expressed as follows:
Percent and Acreage of Area
Soil Baselines
Old Soil Baseline
Revised Soil Baseline
pH: 4.0 ⫺ 4.9
30%
180 acres
45%
270 acres
A rough statistical check between the old and revised soil
baselines indicated that the results were mixed. Based on this
preliminary result and the fact that there was a significant increase in the percent of area with low-productivity soil, an
in-depth analysis of the revised baseline sample study was
performed. Contained in the submitted new-sample package
was a letter from the Yucatan consultant. It indicated to
Yucatan’s management that 100 separate soil samples had
been obtained and analyzed and that the revised premine soil
baseline had been developed using the data from the 30 samples with the lowest pH values.
413
The State Department of Mining and Reclamation staff
concluded that the revised soil pH sample data had been carefully “screened” to reduce the amount of remediation work
that Yucatan Mining would have to complete. Within a week,
Yucatan was notified that further review of the revised soil
baseline could not be pursued, because it appeared the revised soil baseline was developed using a technique that
skewed results in favor of lower pH values. It was also noted
that should Yucatan Mining disagree with this response, the
case would be filed with the legal staff as a contested case.
Within several days, the Yucatan president responded indicating that the company was withdrawing the new application from consideration by the department.
Case Study Questions
1. Assume you are the director of the State Department of
Mining and Reclamation and were informed of the findings on the new samples versus the old samples. What
actions would you direct your staff to take concerning
this situation?
2. Suppose there had been several cases of deceptive acts
similar to this one over the last few years. What type of
“audit” procedures might you want implemented to
identify these possibly unethical activities?
3. Yucatan clearly would state that both the old and new
samples were randomly located about the entire mining
area. When the 30 lowest pH samples were used to establish the new baseline, were the samples still random,
according to experimental design standards? If so, why?
If not, why not?
4. You and the president of Yucatan Mining have been acquaintances for some years. You have golfed together
several times, your and his children are on the same soccer team at school, and your families are members of
the same community swimming pool club. What effect
would this event have upon your and your family’s relationships with the family of the Yucatan president? How
would you handle this situation?
5. As a matter of principle and practice, do you believe
there is some amount of data-altering or bias-making
that is allowed before an application (such as the one
described here) should be considered the result of professionally unethical acts? How would you define such
a threshold limit?
CHAPTER 16
Depreciation
Methods
L E A R N I N G
O U T C O M E S
Purpose: Use depreciation or depletion methods to reduce the book value of a capital investment in an asset and natural
resource.
SECTION
TOPIC
LEARNING OUTCOME
16.1
Terminology
• Define and use the basic terms of asset
depreciation.
16.2
Straight line
• Apply the straight line (SL) method of
depreciation.
16.3
Declining balance
• Apply the declining balance (DB) and double
declining balance (DDB) methods of
depreciation.
16.4
MACRS
• Apply the modified accelerated cost recovery
system for tax depreciation purposes for U.S.based corporations.
16.5
Recovery period
• Select the asset recovery period for MACRS
depreciation.
16.6
Depletion
• Explain depletion; apply cost depletion and
percentage depletion methods.
16A.1
Historical methods
• Apply the sum-of-years-digits (SYD) and unitof-production (UOP) methods of depreciation.
16A.2
Switching
• Switch between classical depreciation methods;
explain how MACRS provides for switching.
16A.3
MACRS and switching
• Calculate MACRS rates using switching between
classical methods and MACRS rules.
Chapter 16 Appendix
T
he capital investments of a corporation in tangible assets—equipment, computers, vehicles, buildings, and machinery—are commonly recovered on the books
of the corporation through depreciation. Although the depreciation amount is
not an actual cash flow, the process of depreciating an asset on the books of the corporation accounts for the decrease in an asset’s value because of age, wear, and obsolescence.
Even though an asset may be in excellent working condition, the fact that it is worth less
through time is taken into account in after-tax economic evaluation studies. An introduction to depreciation types, terminology, and classical methods is followed by a discussion
of the Modified Accelerated Cost Recovery System (MACRS), which is the standard in the
United States for tax purposes. Other countries commonly use the classical methods for tax
computations.
Why is depreciation important to engineering economy? Depreciation is a tax-allowed
deduction included in tax calculations in virtually all industrialized countries. Depreciation
lowers income taxes via the general relation
Taxes ⫽ (income ⫺ deductions)(tax rate)
Income taxes are discussed further in Chapter 17.
This chapter concludes with an introduction to two methods of depletion, which are used
to recover capital investments in deposits of natural resources such as oil, gas, minerals, ores,
and timber.
The chapter appendix describes two historically useful methods of depreciation—
sum-of-years-digits and unit-of-production. Additionally, the appendix includes an in-depth
derivation of the MACRS depreciation rates from the straight line and declining balance
rates. This is accomplished using a procedure called switching between classical depreciation methods.
16.1 Depreciation Terminology
The concept and types of depreciation are defined here. Most descriptions are applicable to corporations as well as individuals who own depreciable assets.
Depreciation is a book method (noncash) to represent the reduction in value of a tangible asset.
The method used to depreciate an asset is a way to account for the decreasing value of the asset
to the owner and to represent the diminishing value (amount) of the capital funds invested in it.
The annual depreciation amount is not an actual cash flow, nor does it necessarily reflect the
actual usage pattern of the asset during ownership.
Though the term amortization is sometimes used interchangeably with the term depreciation,
they are different. Depreciation is applied to tangible assets, while amortization is used to reflect
the decreasing value of intangibles, such as loans, mortgages, patents, trademarks, and goodwill.
In addition, the term capital recovery is sometimes used to identify depreciation. This is clearly
a different use of the term than what we learned in Chapter 5. The term depreciation is used
throughout this book.
There are two different purposes for using the depreciation methods we will cover in this
chapter:
Book depreciation Used by a corporation or business for internal financial accounting to
track the value of an asset or property over its life.
Tax depreciation Used by a corporation or business to determine taxes due based on current tax laws of the government entity (country, state, province, etc.). Even though depreciation itself is not a cash flow, it can result in actual cash flow changes because the amount of
tax depreciation is a deductible item when calculating annual income taxes for the corporation
or business.
The methods applied for these two purposes may or may not utilize the same formulas. Book
depreciation indicates the reduced investment in an asset based upon the usage pattern and
expected useful life of the asset. There are classical, internationally accepted depreciation methods used to determine book depreciation: straight line, declining balance, and the historical
416
Chapter 16
Depreciation Methods
sum-of-years-digits method. The amount of tax depreciation is important in an after-tax engineering economy study and will vary among nations.
In most industrialized countries, the annual tax depreciation is tax-deductible; that is, it is subtracted from income when calculating the amount of taxes due each year. However, the tax
depreciation amount must be calculated using a government-approved method.
Tax depreciation may be calculated and referred to differently in countries outside the United
States. For example, in Canada the equivalent is CCA (capital cost allowance), which is calculated based on the undepreciated value of all corporate properties that form a particular
class of assets, whereas in the United States depreciation may be determined for each asset
separately.
Where allowed, tax depreciation is usually based on an accelerated method, whereby the
depreciation for the first years of use is larger than that for later years. In the United States this
method is called MACRS, as covered in later sections. In effect, accelerated methods defer
some of the income tax burden to later in the asset’s life; they do not reduce the total tax
burden.
Common terms used in depreciation are explained here.
First cost P or unadjusted basis B is the delivered and installed cost of the asset including
purchase price, delivery and installation fees, and other depreciable direct costs incurred to
prepare the asset for use. The term unadjusted basis, or simply basis, is used when the asset is
new, with the term adjusted basis used after some depreciation has been charged. When the
first cost has no added, depreciable costs, the basis is the first cost, that is, P ⫽ B.
Book value BVt represents the remaining, undepreciated capital investment on the books after
the total amount of depreciation charges to date has been subtracted from the basis. The book
value is determined at the end of each year t (t ⫽ 1, 2, . . . , n), which is consistent with the
end-of-year convention.
Recovery period n is the depreciable life of the asset in years. Often there are different
n values for book and tax depreciation. Both of these values may be different from the asset’s
estimated productive life.
Market value MV, a term also used in replacement analysis, is the estimated amount realizable if the asset were sold on the open market. Because of the structure of depreciation laws,
the book value and market value may be substantially different. For example, a commercial
building tends to increase in market value, but the book value will decrease as depreciation
charges are taken. However, a computer workstation may have a market value much lower
than its book value due to rapidly changing technology.
Salvage value S is the estimated trade-in or market value at the end of the asset’s useful life.
The salvage value, expressed as an estimated dollar amount or as a percentage of the first cost,
may be positive, zero, or negative due to dismantling and carry-away costs.
Depreciation rate or recovery rate dt is the fraction of the first cost removed by depreciation
each year t. This rate may be the same each year, which is called the straight line rate d, or
different for each year of the recovery period.
Personal property, one of the two types of property for which depreciation is allowed is
the income-producing, tangible possessions of a corporation used to conduct business.
Included is most manufacturing and service industry property—vehicles, manufacturing
equipment, materials handling devices, computers and networking equipment, communications equipment, office furniture, refining process equipment, construction assets, and
much more.
Real property includes real estate and all improvements—office buildings, manufacturing
structures, test facilities, warehouses, apartments, and other structures. Land itself is considered real property, but it is not depreciable.
Half-year convention assumes that assets are placed in service or disposed of in midyear,
regardless of when these events actually occur during the year. This convention is utilized in
this text and in most U.S.-approved tax depreciation methods. There are also midquarter and
midmonth conventions.
417
Depreciation Terminology
16.1
Figure 16–1
General shape of book
value curves for different
depreciation methods.
Book value, $
SL
MACRS
DB
Estimated
salvage
value
Time, years
Recovery period
As mentioned before, there are several models for depreciating assets. The straight line (SL)
method is used historically and internationally. Accelerated models, such as the declining balance (DB) method, decrease the book value to zero (or to the salvage value) more rapidly than
the straight line method, as shown by the general book value curves in Figure 16–1.
For each of the methods—straight line, declining balance, MACRS, and sum-of-yearsdigits—there are spreadsheet functions available to determine annual depreciation. Each function is introduced and illustrated as the method is explained.
As expected, there are many rules and exceptions to the depreciation laws of a country. One
that may be of interest to a U.S.-based small or medium-sized business performing an economic
analysis is the Section 179 Deduction. This is an economic incentive that changes over the years
and encourages businesses to invest capital in equipment directly used in the company. Up to a
specified amount, the entire basis of an asset is treated as a business expense in the year of purchase. This tax treatment reduces federal income taxes, just as depreciation does, but it is allowed
in lieu of depreciating the first cost over several years. The limit changes with time; it was
$24,000 in 2002; $102,000 in 2004; $125,000 in 2007; and $250,000 in 2008–2010. The economic stimulus efforts in the United States and around the world during the latter part of the decade made many attempts to put investment capital to work in small and medium-sized businesses. Investments above these limits must be depreciated using MACRS.
In the 1980s the U.S. government standardized accelerated methods for federal tax depreciation
purposes. In 1981, all classical methods, including straight line, declining balance, and sum-of-yearsdigits depreciation, were disallowed as tax deductible and replaced by the Accelerated Cost Recovery System (ACRS). In a second round of standardization, MACRS (Modified ACRS) was made the
required tax depreciation method in 1986. To this date, the following is the law in the United States.
Tax depreciation must be calculated using MACRS; book depreciation may be calculated
using any classical method or MACRS.
MACRS has the DB and SL methods, in slightly different forms, embedded in it, but these two
methods cannot be used directly if the annual depreciation is to be tax deductible. Many U.S.
companies still apply the classical methods for keeping their own books, because these methods
are more representative of how the usage patterns of the asset reflect the remaining capital invested in it. Most other countries still recognize the classical methods of straight line and declining balance for tax or book purposes. Because of the continuing importance of the SL and DB
methods, they are explained in the next two sections prior to MACRS. Appendix Section 16A.1
discusses two historical methods of depreciation.
418
Depreciation Methods
Chapter 16
Tax law revisions occur often, and depreciation rules are changed from time to time in the
United States and other countries. For more depreciation and tax law information, consult the
U.S. Department of the Treasury, Internal Revenue Service (IRS), website at www.irs.gov. Pertinent publications can be downloaded. Publication 946, How to Depreciate Property, is especially
applicable to this chapter. MACRS and most corporate tax depreciation laws are discussed in it.
16.2 Straight Line (SL) Depreciation
Straight line depreciation derives its name from the fact that the book value decreases linearly
with time. The depreciation rate is the same (1兾n) each year of the recovery period n.
Straight line depreciation is considered the standard against which any depreciation model is
compared. For book depreciation purposes, it offers an excellent representation of book value for
any asset that is used regularly over an estimated number of years. For tax depreciation, as mentioned earlier, it is not used directly in the United States, but it is commonly used in most countries for tax purposes. However, the U.S. MACRS method includes a version of SL depreciation
with a larger n value than that prescribed by regular MACRS (see Section 16.5).
The annual SL depreciation is determined by multiplying the first cost minus the salvage value
by dt. In equation form,
Dt ⴝ (B ⴚ S)dt
BⴚS
ⴝ ———
n
[16.1]
where t ⫽ year (t ⫽ 1, 2, . . . , n)
Dt ⫽ annual depreciation charge
B ⫽ first cost or unadjusted basis
S ⫽ estimated salvage value
n ⫽ recovery period
dt ⫽ depreciation rate ⫽ 1兾n
Book Value
B
Since the asset is depreciated by the same amount each year, the book value after t years of service, denoted by BVt, will be equal to the first cost B minus the annual depreciation times t.
S
BVt ⴝ B ⴚ tDt
1
Time
[16.2]
Earlier we defined dt as a depreciation rate for a specific year t. However, the SL model has the
same rate for all years, that is,
1
[16.3]
d ⫽ dt ⫽ —
n
The format for the spreadsheet function to display the annual depreciation Dt in a single-cell
operation is
ⴝ SLN(B, S, n)
[16.4]
EXAMPLE 16.1
If an asset has a first cost of $50,000 with a $10,000 estimated salvage value after 5 years,
(a) calculate the annual depreciation and (b) calculate and plot the book value of the asset after
each year, using straight line depreciation.
Solution
(a) The depreciation each year for 5 years can be found by Equation [16.1].
50,000 ⫺ 10,000
B ⫺ S ⫽ ———————
⫽ $8000
Dt ⫽ ———
n
5
Enter the function ⫽ SLN(50000,10000,5) in any cell to display the Dt of $8000.
16.3
419
Declining Balance (DB) and Double Declining Balance (DDB) Depreciation
Figure 16–2
Book value BVt ⫻ $1000
50
Book value of an asset
using straight line depreciation, Example 16.1.
Dt = $8000
40
30
20
10
S = $10,000
0
1
2
3
4
5
t
Year t
(b) The book value after each year t is computed using Equation [16.2]. The BVt values are
plotted in Figure 16–2. For years 1 and 5, for example,
BV1 ⫽ 50,000 ⫺ 1(8000) ⫽ $42,000
BV5 ⫽ 50,000 ⫺ 5(8000) ⫽ $10,000 ⫽ S
16.3 Declining Balance (DB) and Double Declining
Balance (DDB) Depreciation
dmax ⴝ 2兾n
[16.5]
In this case the method is called double declining balance (DDB). If n ⫽ 10 years, the DDB rate
is 2兾10 ⫽ 0.2; so 20% of the book value is removed annually. Another commonly used percentage for the DB method is 150% of the SL rate, where d ⫽ 1.5兾n.
The depreciation for year t is the fixed rate d times the book value at the end of the previous year.
Dt ⴝ (d)BVtⴚ1
[16.6]
The actual depreciation rate for each year t, relative to the basis B, is
dt ⫽ d(1 ⫺ d)t⫺1
[16.7]
If BVt⫺1 is not known, the depreciation in year t can be calculated using B and d.
Dt ⴝ dB(1 ⴚ d)tⴚ1
[16.8]
Book value in year t is determined in one of two ways: by using the rate d and basis B or by
subtracting the current depreciation charge from the previous book value. The equations are
BVt ⴝ B(1 ⴚ d)t
BVt ⴝ BVtⴚ1 ⴚ Dt
[16.9]
[16.10]
B
Book Value
The declining balance method is commonly applied as the book depreciation method. Like the
SL method, DB is embedded in the MACRS method, but the DB method itself cannot be used to
determine the annual tax-deductible depreciation in the United States. This method is used routinely in most other countries for tax and book depreciation purposes.
Declining balance is also known as the fixed percentage or uniform percentage method.
DB depreciation accelerates the write-off of asset value because the annual depreciation is determined by multiplying the book value at the beginning of a year by a fixed (uniform) percentage d,
expressed in decimal form. If d ⫽ 0.1, then 10% of the book value is removed each year. Therefore, the depreciation amount decreases each year.
The maximum annual depreciation rate for the DB method is twice the straight line rate, that is,
Time
420
Depreciation Methods
Chapter 16
It is important to understand that the book value for the DB method never goes to zero, because
the book value is always decreased by a fixed percentage. The implied salvage value after n
years is the BVn amount, that is,
Implied S ⫽ BVn ⫽ B(1 ⫺ d)n
[16.11]
If a salvage value is estimated for the asset, this estimated S value is not used in the DB or
DDB method to calculate annual depreciation. However, if the implied S ⬍ estimated S, it is
necessary to stop charging further depreciation when the book value is at or below the estimated
salvage value. In most cases, the estimated S is in the range of zero to the implied S value. (This
guideline is important when the DB method can be used directly for tax depreciation purposes.)
If the fixed percentage d is not stated, it is possible to determine an implied fixed rate using the
estimated S value, if S ⬎ 0. The range for d is 0 ⬍ d ⬍ 2兾n.
( )
S
Implied d ⫽ 1 ⫺ —
B
1兾n
[16.12]
The spreadsheet functions DDB and DB are used to display depreciation amounts for specific
years. The function is repeated in consecutive spreadsheet cells because the depreciation amount
Dt changes with t. For the double declining balance method, the format is
ⴝ DDB(B, S, n, t, d)
[16.13]
The entry d is the fixed rate expressed as a number between 1 and 2. If omitted, this optional
entry is assumed to be 2 for DDB. An entry of d ⫽ 1.5 makes the DDB function display 150%
declining balance method amounts. The DDB function automatically checks to determine when
the book value equals the estimated S value. No further depreciation is charged when this
occurs. (To allow full depreciation charges to be made, ensure that the S entered is between zero
and the implied S from Equation [16.11].) Note that d ⫽ 1 is the same as the straight line rate
1兾n, but Dt will not be the SL amount because declining balance depreciation is determined as a
fixed percentage of the previous year’s book value, which is completely different from the SL
calculation in Equation [16.1].
The format for the DB function is ⫽ DB(B,S,n,t). Caution is needed when using this function. The fixed rate d is not entered in the DB function; d is an embedded calculation using
a spreadsheet equivalent of Equation [16.12]. Also, only three significant digits are maintained for d, so the book value may go below the estimated salvage value due to round-off
errors. Therefore, if the depreciation rate is known, always use the DDB function to ensure
correct results. Examples 16.2 and 16.3 illustrate DB and DDB depreciation and their spreadsheet functions.
EXAMPLE 16.2
Underwater electroacoustic transducers were purchased for use in SONAR applications. The
equipment will be DDB depreciated over an expected life of 12 years. There is a first cost of
$25,000 and an estimated salvage of $2500. (a) Calculate the depreciation and book value for
years 1 and 4. Write the spreadsheet functions to display depreciation for years 1 and 4.
(b) Calculate the implied salvage value after 12 years.
Solution
(a) The DDB fixed depreciation rate is d ⫽ 2兾n ⫽ 2兾12 ⫽ 0.1667 per year. Use Equations [16.8]
and [16.9].
Year 1:
Year 4:
D1 ⫽ (0.1667)(25,000)(1 ⫺ 0.1667)1⫺1 ⫽ $4167
BV1 ⫽ 25,000(1 ⫺ 0.1667)1 ⫽ $20,833
D4 ⫽ (0.1667)(25,000)(1 ⫺ 0.1667)4⫺1 ⫽ $2411
BV4 ⫽ 25,000(1 ⫺ 0.1667)4 ⫽ $12,054
The DDB functions for D1 and D4 are, respectively, ⫽ DDB(25000,2500,12,1) and
⫽ DDB(25000,2500,12,4).
Declining Balance (DB) and Double Declining Balance (DDB) Depreciation
16.3
(b) From Equation [16.11], the implied salvage value after 12 years is
Implied S ⫽ 25,000(1 − 0.1667)12 ⫽ $2803
Since the estimated S ⫽ $2500 is less than $2803, the asset is not fully depreciated when
its 12-year expected life is reached.
EXAMPLE 16.3
Freeport-McMoRan Copper and Gold has purchased a new ore grading unit for $80,000. The
unit has an anticipated life of 10 years and a salvage value of $10,000. Use the DB and DDB
methods to compare the schedule of depreciation and book values for each year. Solve by hand
and by spreadsheet.
Solution by Hand
An implied DB depreciation rate is determined by Equation [16.12].
(
10,000
d ⫽ 1 ⫺ ———
80,000
)
1兾10
⫽ 0.1877
Note that 0.1877 ⬍ 2兾n ⫽ 0.2, so this DB model does not exceed twice the straight line rate.
Table 16–1 presents the Dt values using Equation [16.6] and the BVt values from Equation [16.10] rounded to the nearest dollar. For example, in year t ⫽ 2, the DB results are
D2 ⫽ d(BV1) ⫽ 0.1877(64,984) ⫽ $12,197
BV2 ⫽ 64,984 ⫺ 12,197 ⫽ $52,787
Because we round off to even dollars, $2312 is calculated for depreciation in year 10, but
$2318 is deducted to make BV10 ⫽ S ⫽ $10,000 exactly. Similar calculations for DDB with
d ⫽ 0.2 result in the depreciation and book value series in Table 16–1.
TABLE 16–1
Dt and BVt Values for DB and DDB Depreciation, Example 16.3
Declining Balance, $
Double Declining Balance, $
Year t
Dt
BVt
Dt
BVt
0
1
2
3
4
5
6
7
8
9
10
—
15,016
12,197
9,908
8,048
6,538
5,311
4,314
3,504
2,846
2,318
80,000
64,984
52,787
42,879
34,831
28,293
22,982
18,668
15,164
12,318
10,000
—
16,000
12,800
10,240
8,192
6,554
5,243
4,194
3,355
2,684
737
80,000
64,000
51,200
40,960
32,768
26,214
20,972
16,777
13,422
10,737
10,000
Solution by Spreadsheet
The spreadsheet in Figure 16–3 displays the results for the DB and DDB methods. The chart
plots book values for each year. Since the fixed rates are close—0.1877 for DB and 0.2 for
DDB—the annual depreciation and book value series are approximately the same for the two
methods.
The depreciation rate (cell B5) is calculated by Equation [16.12], but note in the cell tags
that the DDB function is used in both columns B and D to determine annual depreciation.
As mentioned earlier, the DB function automatically calculates the implied rate by Equation [16.12] and maintains it to only three significant digits. Therefore, if the DB function
421
422
Depreciation Methods
Chapter 16
⫽ 1⫺(B3/B2)^0.1
⫽ DDB(B$2,B$3,B$4,$A18,10*$B$5)
⫽ DDB(B$2,B$3,B$4,$A18)
Figure 16–3
Annual depreciation and book value using DB and DDB methods, Example 16.3.
were used in column B (Figure 16–3), the fixed rate applied would be 0.188. The resulting
Dt and BVt values for years 8, 9, and 10 would be as follows:
t
D t, $
BVt, $
8
9
10
3,501
2,842
2,308
15,120
12,277
9,969
Also noteworthy is the fact that the DB function uses the implied rate without a check to halt
the book value at the estimated salvage value. Thus, BV10 will go slightly below S ⫽ $10,000,
as shown above. However, the DDB function uses a relation different from that of the DB function to determine annual depreciation—one that correctly stops depreciating at the estimated
salvage value, as shown in Figure 16–3, cells E17–E18.
16.4 Modified Accelerated Cost Recovery
System (MACRS)
In the 1980s, the United States introduced MACRS as the required tax depreciation method for
all depreciable assets. Through MACRS, the 1986 Tax Reform Act defined statutory depreciation
rates that take advantage of the accelerated DB and DDB methods. Corporations are free to apply
any of the classical methods for book depreciation. When developed, MACRS and its predecessor ACRS were intended to create economic growth through the investment of new capital and
the tax advantages that accelerated depreciation methods offer corporations and businesses.1
Many aspects of MACRS deal with the specific depreciation accounting aspects of tax law.
This section covers only the elements that materially affect after-tax economic analysis. Additional information on how the DDB, DB, and SL methods are embedded into MACRS and how
to derive the MACRS depreciation rates is presented and illustrated in the chapter appendix,
Sections 16A.2 and 16A.3.
MACRS determines annual depreciation amounts using the relation
Dt ⴝ dtB
1
[16.14]
R. Lundquist, “The Pedagogy of Taxes and Tax Purpose Depreciation,” Proceedings, ASEE Annual
Conference, Austin, TX, June 2009.
Modified Accelerated Cost Recovery System (MACRS)
16.4
where the depreciation rate dt is provided in tabulated form. As for other methods, the book value
in year t is determined by subtracting the depreciation amount from the previous year’s book
value
BVt ⫽ BVt⫺1 ⫺ Dt
[16.15]
or by subtracting the total depreciation from the first cost.
BVt ⴝ first cost ⴚ sum of accumulated depreciation
jⴝt
ⴝBⴚ
兺D
[16.16]
j
jⴝ1
MACRS has standardized and simplified many of the decisions and calculations of depreciation.
The basis B (or first cost P) is completely depreciated; salvage is always assumed to be zero,
or S ⫽ $0.
Recovery periods are standardized to specific values:
n ⫽ 3, 5, 7, 10, 15, or 20 years
for personal property (e.g., equipment or vehicles)
n ⫽ 27.5 or 39 years
for real property (e.g., rental property or structures)
Depreciation rates provide accelerated write-off by incorporating switching between classical
methods.
Section 16.5 explains how to determine an allowable MACRS recovery period. The MACRS
personal property depreciation rates (dt values) for n ⫽ 3, 5, 7, 10, 15, and 20 for use in Equation [16.14] are included in Table 16–2.
MACRS depreciation rates incorporate the DDB method (d ⫽ 2兾n) and switch to SL depreciation during the recovery period as an inherent component for personal property depreciation.
The MACRS rates start with the DDB rate or the 150% DB rate and switch when the SL method
offers faster write-off.
TABLE 16–2
Depreciation Rates dt Applied to the Basis B for the MACRS Method
Depreciation Rate (%) for Each MACRS Recovery Period in Years
Year
nⴝ3
nⴝ5
nⴝ7
n ⴝ 10
n ⴝ 15
n ⴝ 20
1
2
3
4
5
33.33
44.45
14.81
7.41
20.00
32.00
19.20
11.52
11.52
14.29
24.49
17.49
12.49
8.93
10.00
18.00
14.40
11.52
9.22
5.00
9.50
8.55
7.70
6.93
3.75
7.22
6.68
6.18
5.71
5.76
8.92
8.93
4.46
7.37
6.55
6.55
6.56
6.55
6.23
5.90
5.90
5.91
5.90
5.29
4.89
4.52
4.46
4.46
3.28
5.91
5.90
5.91
5.90
5.91
4.46
4.46
4.46
4.46
4.46
2.95
4.46
4.46
2.23
6
7
8
9
10
11
12
13
14
15
16
17–20
21
423
424
Depreciation Methods
Chapter 16
For real property, MACRS utilizes the SL method for n ⫽ 39 throughout the recovery period.
The annual percentage depreciation rate is d ⫽ 1兾39 ⫽ 0.02564. However, MACRS forces
partial-year recovery in years 1 and 40. The MACRS real property rates in percentage amounts are
Year 1
Years 2–39
Year 40
100d1 ⫽ 1.391%
100dt ⫽ 2.564%
100d40 ⫽ 1.177%
The real property recovery period of 27.5 years, which applies only to residential rental property,
uses the SL method in a similar fashion.
Note that all MACRS depreciation rates in Table 16–2 are presented for 1 year longer than the
stated recovery period. Also note that the extra-year rate is one-half of the previous year’s rate.
This is so because a built-in half-year convention is imposed by MACRS. This convention assumes that all property is placed in service at the midpoint of the tax year of installation. Therefore, only 50% of the first-year DB depreciation applies for tax purposes. This removes some of
the accelerated depreciation advantage and requires that one-half year of depreciation be taken in
year n ⫹ 1.
No specially designed spreadsheet function is present for MACRS depreciation. However, the
variable declining balance (VDB) function, which is used to determine when to switch between
classical methods, can be adapted to display MACRS deprecation for each year. (The VDB function is explained in detail in Section 16A.2 of this chapter and Appendix A of the text.) The
MACRS depreciation format of the VDB function requires embedded MAX and MIN functions,
as follows:
ⴝ VDB(B,0, n, MAX(0, tⴚ1.5), MIN(n, tⴚ0.5),d)
[16.17]
where B ⫽ first cost
0 ⫽ salvage value of S ⫽ 0
n ⫽ recovery period
if MACRS n ⫽ 3, 5, 7, or 10
d⫽ 2
1.5
if MACRS n ⫽ 15 or 20
{
The MAX and MIN functions ensure that the MACRS half-year conventions are followed; that
is, only one-half of the first year’s depreciation is charged in year 1, and one-half of the last year’s
charge is carried over to year n ⫹ 1.
EXAMPLE 16.4
Chevron Phillips Chemical Company in Baytown, Texas, acquired new equipment for its
polyethylene processing line. This chemical is a resin used in plastic pipe, retail bags, blow
molding, and injection molding. The equipment has an unadjusted basis of B ⫽ $400,000, a
life of only 3 years, and a salvage value of 5% of B. The chief engineer asked the finance director to provide an analysis of the difference between (1) the DDB method, which is the internal book depreciation and book value method used at the plant, and (2) the required
MACRS tax depreciation and its book value. He is especially curious about the differences
after 2 years of service for this short-lived, but expensive asset. Use hand and spreadsheet
solutions to do the following:
(a) Determine which method offers the larger total depreciation after 2 years.
(b) Determine the book value for each method after 2 years and at the end of the recovery
period.
Solution by Hand
The basis is B ⫽ $400,000 and the estimated S ⫽ 0.05(400,000) ⫽ $20,000. The MACRS rates
for n ⫽ 3 are taken from Table 16–2, and the depreciation rate for DDB is dmax ⫽ 2兾3 ⫽
0.6667. Table 16–3 presents the depreciation and book values. Year 3 depreciation for DDB
would be $44,444(0.6667) ⫽ $29,629, except this would make BV3 ⬍ $20,000. Only the
remaining amount of $24,444 is removed.
Modified Accelerated Cost Recovery System (MACRS)
16.4
TABLE 16–3
Comparing MACRS and DDB Depreciation, Example 16.4
MACRS
DDB
Year
Rate
Tax
Depreciation, $
Book
Value, $
Book
Depreciation, $
0
1
2
3
4
0.3333
0.4445
0.1481
0.0741
133,320
177,800
59,240
29,640
400,000
266,680
88,880
29,640
0
266,667
88,889
24,444
Book
Value, $
400,000
133,333
44,444
20,000
(a) The 2-year accumulated depreciation values from Table 16–3 are
MACRS:
DDB:
D1 ⫹ D2 ⫽ $133,320 ⫹ 177,800 ⫽ $311,120
D1 ⫹ D2 ⫽ $266,667 ⫹ 88,889 ⫽ $355,556
The DDB depreciation is larger. (Remember that for tax purposes, the company does not
have the choice in the United States of DDB as applied here.)
(b) After 2 years the book value for DDB at $44,444 is 50% of the MACRS book value of $88,880.
At the end of recovery (4 years for MACRS due to the built-in half-year convention, and
3 years for DDB), the MACRS book value is BV4 ⫽ 0 and for DDB, BV3 ⫽ $20,000. This
occurs because MACRS always removes the entire first cost, regardless of the estimated salvage value. This is a tax depreciation advantage of the MACRS method (unless the asset is
disposed of for more than the MACRS-depreciated book value, as discussed in Section 17.4).
Solution by Spreadsheet
Figure 16–4 presents the spreadsheet solution using the VDB function (column B) for MACRS
depreciation (in lieu of the MACRS rates) and applying the DDB function in column D.
(a) The 2-year accumulated depreciation values are
MACRS (add cells B6 ⫹ B7):
DDB (add cells D6 ⫹ D7):
$133,333 ⫹ 177,778 ⫽ $311,111
$266,667 ⫹ 88,889 ⫽ $355,556
(b) Book values after 2 years are
MACRS (cell C7):
DDB (cell E7):
$88,889
$44,444
The book values are plotted in Figure 16–4. Observe that MACRS goes to zero in year 4, while
DDB stops at $20,000 in year 3.
Comment
It is advisable to set up a spreadsheet template for use with depreciation problems in this and
future chapters. The format and functions of Figure 16–4 are a good template for MACRS and
DDB methods.
Figure 16–4
Spreadsheet screen shot of MACRS and DDB depreciation and book value, Example 16.4.
425
426
Depreciation Methods
Chapter 16
MACRS simplifies depreciation computations, but it removes much of the flexibility of
method selection for a business or corporation. In general, an economic comparison that includes
depreciation may be performed more rapidly and usually without altering the final decision by
applying the classical straight line method in lieu of MACRS.
16.5 Determining the MACRS Recovery Period
The expected useful life of property is estimated in years and used as the n value in alternative
evaluation and in depreciation computations. For book depreciation the n value should be the
expected useful life. However, when the depreciation will be claimed as tax deductible, the
n value should be lower. There are tables that assist in determining the life and recovery period
for tax purposes.
The advantage of a recovery period shorter than the anticipated useful life is leveraged by the
accelerated depreciation methods that write off more of the basis B in the initial years.
The U.S. government requires that all depreciable property be classified into a property class
which identifies its MACRS-allowed recovery period. Table 16–4, a summary of material from
IRS Publication 946, gives examples of assets and the MACRS n values. Virtually any property
considered in an economic analysis has a MACRS n value of 3, 5, 7, 10, 15, or 20 years.
Table 16–4 provides two MACRS n values for each property. The first is the general depreciation system (GDS) value, which we use in examples and problems. The depreciation rates in Table
16–2 correspond to the n values for the GDS column and provide the fastest write-off allowed. The
rates utilize the DDB method or the 150% DB method with a switch to SL depreciation. Note that
any asset not in a stated class is automatically assigned a 7-year recovery period under GDS.
The far right column of Table 16–4 lists the alternative depreciation system (ADS) recovery period range. This alternative method allows the use of SL depreciation over a longer
TABLE 16–4
Example MACRS Recovery Periods for Various Asset Descriptions
MACRS
n Value, Years
Asset Description (Personal and Real Property)
GDS
ADS Range
Special manufacturing and handling devices, tractors,
racehorses
3
3–5
Computers and peripherals, oil and gas drilling
equipment, construction assets, autos, trucks, buses,
cargo containers, some manufacturing equipment
5
6–9.5
Office furniture; some manufacturing equipment;
railroad cars, engines, tracks; agricultural machinery; petroleum and natural gas equipment; all
property not in another class
7
10–15
Equipment for water transportation, petroleum refining, agriculture product processing, durable-goods
manufacturing, shipbuilding
10
15–19
Land improvements, docks, roads, drainage, bridges,
landscaping, pipelines, nuclear power production
equipment, telephone distribution
15
20–24
Municipal sewers, farm buildings, telephone switching buildings, power production equipment (steam
and hydraulic), water utilities
20
25–50
Residential rental property (house, mobile home)
27.5
40
Nonresidential real property attached to the land, but
not the land itself
39
40
Depletion Methods
16.6
recovery period than the GDS. The half-year convention applies, and any salvage value is neglected,
as it is in regular MACRS. The use of ADS is generally a choice left to a company, but it is required
for some special asset situations. Since it takes longer to depreciate the asset, and since the SL model
is required (thus removing the advantage of accelerated depreciation), ADS is usually not considered an option for the economic analysis. This electable SL option is, however, sometimes chosen
by businesses that are young and do not need the tax benefit of accelerated depreciation during the
first years of operation and asset ownership. If ADS is selected, tables of dt rates are available.
16.6 Depletion Methods
Previously, for all assets, facilities, and equipment that can be replaced we have applied depreciation. We now turn to irreplaceable natural resources and the equivalent of depreciation, which
is called depletion.
Depletion is a book method (noncash) to represent the decreasing value of a natural resource
as it is recovered, removed, or felled. The two methods of depletion for book or tax purposes are
used to write off the first cost, or value of the estimated quantity, of resources in mines, wells,
quarries, geothermal deposits, forests, and the like.
The two methods of depletion are cost and percentage depletion, as described below. Details for
U.S. taxes on depletion are found in IRS Publication 535, Business Expenses.
Cost depletion Sometimes referred to as factor depletion, cost depletion is based on the level
of activity or usage, not time, as in depreciation. Cost depletion may be applied to most types of
natural resources and must be applied to timber production. The cost depletion factor for year t, denoted by CDt, is the ratio of the first cost of the resource to the estimated number of units recoverable.
first cost
CDt ⴝ ————————
resource capacity
[16.18]
The annual depletion charge is CDt times the year’s usage or volume. The total cost depletion
cannot exceed the first cost of the resource. If the capacity of the property is reestimated some
year in the future, a new cost depletion factor is determined based upon the undepleted amount
and the new capacity estimate.
Percentage depletion This is a special consideration given for natural resources. A constant,
stated percentage of the resource’s gross income may be depleted each year provided it does not
exceed 50% of the company’s taxable income. The depletion amount for year t is calculated as
Percentage depletion t ⴝ percentage depletion rate
ⴛ gross income from property
ⴝ PD ⴛ GIt
[16.19]
Using percentage depletion, total depletion charges may exceed first cost with no limitation. The
U.S. government does not generally allow percentage depletion to be applied to oil and gas wells
(except small independent producers).
The annual percentage depletion rates for some common natural deposits are listed below per
U.S. tax law.
Deposit
Sulfur, uranium, lead, nickel, zinc,
and some other ores and minerals
Gold, silver, copper, iron ore, and
some oil shale
Oil and natural gas wells (varies)
Coal, lignite, sodium chloride
Gravel, sand, peat, some stones
Most other minerals, metallic ores
Percentage of
Gross Income, PD
22
15
15–22
10
5
14
427
428
Depreciation Methods
Chapter 16
EXAMPLE 16.5
Temple-Inland Corporation has negotiated the rights to cut timber on privately held forest acreage for $700,000. An estimated 350 million board feet of lumber is harvestable.
(a) Determine the depletion amount for the first 2 years if 15 million and 22 million board feet
are removed.
(b) After 2 years the total recoverable board feet was reestimated upward to be 450 million from
the time the rights were purchased. Compute the new cost depletion factor for years 3 and later.
Solution
(a) Use Equation [16.18] for CDt in dollars per million board feet.
700,000
CDt ⫽ ———— ⫽ $2000 per million board feet
350
Multiply CDt by the annual harvest to obtain depletion of $30,000 in year 1 and $44,000
in year 2. Continue until a total of $700,000 is written off.
(b) After 2 years, a total of $74,000 has been depleted. A new CDt value must be calculated
based on the remaining 700,000 ⫺ 74,000 ⫽ $626,000 investment. Additionally, with the
new estimate of 450 million board feet, a total of 450 ⫺ 15 ⫺ 22 ⫽ 413 million board feet
remains. For years t ⫽ 3, 4, . . . , the cost depletion factor is
626,000
CDt ⫽ ———— ⫽ $1516 per million board feet
413
EXAMPLE 16.6
A gold mine was purchased for $10 million. It has an anticipated gross income of $5.0 million
per year for years 1 to 5 and $3.0 million per year after year 5. Assume that depletion charges
do not exceed 50% of taxable income. Compute annual depletion amounts for the mine. How
long will it take to recover the initial investment at i ⫽ 0%?
Solution
The rate for gold is PD ⫽ 0.15. Depletion amounts are
Years 1 to 5:
Years thereafter:
0.15(5.0 million) ⫽ $750,000
0.15(3.0 million) ⫽ $450,000
A total of $3.75 million is written off in 5 years, and the remaining $6.25 million is written off
at $450,000 per year. The total number of years is
$6.25 million
5 ⫹ —————— ⫽ 5 ⫹ 13.9 ⫽ 18.9
$450,000
In 19 years, the initial investment could be fully depleted.
In many of the natural resource depletion situations, the tax law allows the larger of the two
depletion amounts to be claimed each year. This is allowed provided the percentage depletion
amount does not exceed 50% of taxable income. Therefore, it is wise to calculate both depletion
amounts and select the larger. Use the following terminology for year t (t ⫽1, 2, . . .).
CDAt ⫽ cost depletion amount
PDAt ⫽ percentage depletion amount
TIt ⫽ taxable income
The guideline for the tax-allowed depletion amount for year t is
{
Depletion ⫽ max[CDAt, PDAt]
max[CDAt, 50% of TIt]
if PDAt ⱕ 50% of TIt
if PDAt ⬎ 50% of TIt
Chapter Summary
For example, assume a medium-sized quarry owner calculates the following for 1 year.
TI ⫽ $500,000
CDA ⫽ $275,000
PDA ⫽ $280,000
Since 50% of TI is $250,000, the PDA is too large and, therefore, is not allowed. For tax purposes, apply the guideline above and use the cost depletion of $275,000, since it is larger than
50% of TI.
CHAPTER SUMMARY
Depreciation may be determined for internal company records (book depreciation) or for income tax purposes (tax depreciation). In the United States, the MACRS method is the only
one allowed for tax depreciation. In many other countries, straight line and declining balance
methods are applied for both tax and book depreciation. Depreciation does not result in cash
flow directly. It is a book method by which the capital investment in tangible property is recovered. The annual depreciation amount is tax deductible, which can result in actual cash
flow changes.
Some important points about the straight line, declining balance, and MACRS methods are
presented below. Common relations for each method are summarized in Table 16–5.
Straight Line (SL)
• It writes off capital investment linearly over n years.
• The estimated salvage value is always considered.
• This is the classical, nonaccelerated depreciation model.
Declining Balance (DB)
• The method accelerates depreciation compared to the straight line method.
• The book value is reduced each year by a fixed percentage.
• The most used rate is twice the SL rate, which is called double declining balance (DDB).
• It has an implied salvage that may be lower than the estimated salvage.
• It is not an approved tax depreciation method in the United States. It is frequently used for
book depreciation purposes.
Modified Accelerated Cost Recovery System (MACRS)
• It is the only approved tax depreciation system in the United States.
• It automatically switches from DDB or DB to SL depreciation.
• It always depreciates to zero; that is, it assumes S ⫽ 0.
• Recovery periods are specified by property classes.
• Depreciation rates are tabulated.
• The actual recovery period is 1 year longer due to the imposed half-year convention.
• MACRS straight line depreciation is an option, but recovery periods are longer than those for
regular MACRS.
Cost and percentage depletion methods recover investment in natural resources. The annual
cost depletion factor is applied to the amount of resource removed. No more than the initial investment can be recovered with cost depletion. Percentage depletion, which can recover more
than the initial investment, reduces the investment value by a constant percentage of gross income each year.
TABLE 16–5
Summary of Common Depreciation Method Relations
Method
MACRS
SL
DDB
Fixed depreciation rate d
Not defined
1
—
n
2
—
n
Annual rate dt
Table 16–2
1
—
n
d(1 ⫺ d)t⫺1
dtB
B⫺S
———
n
d(BVt⫺1)
BVt⫺1 ⫺ Dt
B ⫺ tDt
B(1 ⫺ d)t
Annual depreciation Dt
Book value BVt
429
430
Depreciation Methods
Chapter 16
CHAPTER 16
APPENDIX
16A.1 Sum-of-Years-Digits (SYD) and Unit-ofProduction (UOP) Depreciation
The SYD method is a historical accelerated depreciation technique that removes much of the
basis in the first one-third of the recovery period; however, write-off is not as rapid as for DDB
or MACRS. This technique may be used in an engineering economy analysis in the book depreciation of multiple-asset accounts (group and composite depreciation).
The mechanics of the method involve the sum of the year’s digits from 1 through the recovery
period n. The depreciation charge for any given year is obtained by multiplying the basis of the
asset, less any salvage value, by the ratio of the number of years remaining in the recovery period
to the sum of the year’s digits SUM.
depreciable years remaining
Dt ⴝ ————————————— (basis ⴚ salvage value)
sum of years digits
n ⴚ t ⴙ 1 (B ⴚ S)
Dt ⴝ —————
SUM
[16A.1]
where SUM is the sum of the digits 1 through n.
j⫽n
SUM ⫽
n(n ⫹ 1)
兺 j ⫽ ————
2
j⫽1
The book value for any year t is calculated as
t(n ⴚ t兾2 ⴙ 0.5)
BVt ⴝ B ⴚ ——————— (B ⴚ S)
SUM
[16A.2]
The rate of depreciation decreases each year and equals the multiplier in Equation [16A.1].
n⫺t⫹1
dt ⫽ ————
SUM
[16A.3]
The SYD spreadsheet function displays the depreciation for the year t. The function format is
⫽ SYD(B,S,n,t)
EXAMPLE 16A.1
Calculate the SYD depreciation charges for year 2 for electro-optics equipment with
B ⫽ $25,000, S ⫽ $4000, and an 8-year recovery period.
Solution
The sum of the year’s digits is 36, and the depreciation amount for the second year by Equation [16A.1] is
7 (21,000) ⫽ $4083
D2 ⫽ ——
36
The SYD function is ⫽ SYD(25000,4000,8,2).
Figure 16A–1 is a plot of the book values for an $80,000 asset with S ⫽ $10,000 and n ⫽
10 years using the four depreciation methods that we have learned. The MACRS, DDB, and SYD
curves track closely except for year 1 and years 9 through 11.
16A.1
Sum-of-Years-Digits (SYD) and Unit-of-Production (UOP) Depreciation
80
MACRS
Book value BVt ⫻ $1000
70
60
50
SL
DDB
40
SYD
30
20
10
S = $10,000
0
1
2
3
4
5
6
7
8
9
10
11
Year
Figure 16A–1
Comparison of book values using SL, SYD, DDB, and MACRS depreciation.
A second depreciation method that is not allowed for tax purposes, but useful in some situations is the unit-of-production (UOP) method. When the decreasing value of equipment is
based on usage, not time, the UOP method is quite applicable. Suppose a highway contractor
has a series of state highway department contracts that will last several years and that earth moving equipment is purchased for use on all contracts. If the equipment usage goes up and down
significantly over the years, the UOP method is ideal for book depreciation. For year t, UOP
deprecation is calculated as
actual usage for year t
Dt ⫽ —————————— (basis ⫺ salvage)
total lifetime usage
[16A.4]
EXAMPLE 16A.2
Zachry Contractors purchased an $80,000 mixer for use during the next 10 years for contract
work on IH-10 in San Antonio. The mixer will have a negligible salvage value after 10 years,
and the total amount of material to process is estimated at 2 million m3. Use the actual usage per
year shown in Table 16A–1 and the unit-of-production method to determine annual depreciation.
Solution
The actual usage each year is placed in the numerator of Equation [16A.4] to determine the annual depreciation based on the estimated total lifetime amount of material, 2 million m3 in this
case. Table 16A–1 shows the annual and cumulative depreciation over the 10 years. If the mixer
is continued in service after the 2 million m3 is processed, no further depreciation is allowed.
TABLE 16A–1
Unit-of-Production Method of Depreciation, Example 16A.2
Year t
Actual
Usage, 1000 m3
Annual
Depreciation Dt , $
Cumulative
Depreciation, $
1
2–8
9–10
400
200
100
16,000
8,000
4,000
16,000
72,000
80,000
Total
2000
80,000
431
432
Depreciation Methods
Chapter 16
16A.2 Switching between Depreciation Methods
Switching between depreciation methods may assist in accelerated reduction of the book value.
It also maximizes the present value of accumulated and total depreciation over the recovery period. Therefore, switching usually increases the tax advantage in years where the depreciation is
larger. The approach below is an inherent part of MACRS.
Switching from a DB method to the SL method is the most common switch because it usually
offers a real advantage, especially if the DB method is DDB. General rules of switching are summarized here.
1. Switching is recommended when the depreciation for year t by the currently used method is
less than that for a new method. The selected depreciation Dt is the larger amount.
2. Only one switch can take place during the recovery period.
3. Regardless of the (classical) depreciation methods, the book value cannot go below the
estimated salvage value. When switching from a DB method, the estimated salvage value,
not the DB-implied salvage value, is used to compute the depreciation for the new method;
we assume S ⫽ 0 in all cases. (This does not apply to MACRS, since it already includes
switching.)
4. The undepreciated amount, that is, BVt, is used as the new adjusted basis to select the larger
Dt for the next switching decision.
In all situations, the criterion is to maximize the present worth of the total depreciation PWD.
The combination of depreciation methods that results in the largest present worth is the best
switching strategy.
tⴝn
PWD ⴝ
兺 D (P兾F, i, t)
t
[16A.5]
tⴝ1
This logic minimizes tax liability in the early part of an asset’s recovery period.
Switching is most advantageous from a rapid write-off method such as DDB to the SL model.
This switch is predictably advantageous if the implied salvage value computed by Equation
[16.11] exceeds the salvage value estimated at purchase time; that is, switch if
BVn ⫽ B(1 ⫺ d)n ⬎ estimated S
[16A.6]
Since we assume that S will be zero per rule 3 above, and since BVn will be greater than zero, for
a DB method a switch to SL is always advantageous. Depending upon the values of d and n, the
switch may be best in the later years or last year of the recovery period, which removes the implied S inherent to the DDB model.
The procedure to switch from DDB to SL depreciation is as follows:
1. For each year t, compute the two depreciation charges.
For DDB:
For SL:
DDDB ⫽ d(BVt⫺1)
BVt⫺1
DSL ⫽ ————
n⫺t⫹1
[16A.7]
[16A.8]
2. Select the larger depreciation value. The depreciation for each year is
Dt ⫽ max[DDDB, DSL]
[16A.9]
3. If needed, determine the present worth of total depreciation, using Equation [16A.5].
It is acceptable, though not usually financially advantageous, to state that a switch will take
place in a particular year, for example, a mandated switch from DDB to SL in year 7 of a 10-year
recovery period. This approach is usually not taken, but the switching technique will work
correctly for all depreciation methods.
To use a spreadsheet for switching, first understand the depreciation model switching rules
and practice the switching procedure from declining balance to straight line. Once these are understood, the mechanics of the switching can be speeded up by applying the spreadsheet function
Switching between Depreciation Methods
16A.2
VDB (variable declining balance). This is a quite powerful function that determines the depreciation for 1 year or the total over several years for the DB-to-SL switch. The function format is
⫽ VDB(B, S, n,start_t,end_t, d,no_switch)
[16A.10]
Appendix A explains all the fields in detail, but for simple applications, where the DDB and SL
annual Dt values are needed, the following are correct entries:
start_t is the year (t⫺1)
end_t is year t
d is optional; 2 for DDB is assumed, the same as in the DDB function
no_switch is an optional logical value:
FALSE or omitted—switch to SL occurs, if advantageous
TRUE—DDB or DB method is applied with no switching to SL depreciation considered.
Entering TRUE for the no_switch option obviously causes the VDB function to display the same
depreciation amounts as the DDB function. This is discussed in Example 16A.3d. You may notice that the VDB function is the same one used to calculate annual MACRS depreciation.
EXAMPLE 16A.3
The Outback Steakhouse main office has purchased a $100,000 online document imaging system with an estimated useful life of 8 years and a tax depreciation recovery period of 5 years.
Compare the present worth of total depreciation for (a) the SL method, (b) the DDB method,
and (c) DDB-to-SL switching. (d) Perform the DDB-to-SL switch using a spreadsheet and plot
the book values. Use a rate of i ⫽ 15% per year.
Solution by Hand
The MACRS method is not involved in this solution.
(a) Equation [16.1] determines the annual SL depreciation.
100,000 ⫺ 0
Dt ⫽ —————— ⫽ $20,000
5
Since Dt is the same for all years, the P兾A factor replaces P兾F to compute PWD.
PWD ⫽ 20,000(P兾A,15%,5) ⫽ 20,000(3.3522) ⫽ $67,044
(b) For DDB, d ⫽ 2兾5 ⫽ 0.40. The results are shown in Table 16A–2. The value PWD ⫽
$69,915 exceeds $67,044 for SL depreciation. As is predictable, the accelerated depreciation of DDB increases PWD.
(c) Use the DDB-to-SL switching procedure.
1. The DDB values for Dt in Table 16A–2 are repeated in Table 16A–3 for comparison
with the DSL values from Equation [16A.8]. The DSL values change each year because
BVt⫺1 is different. Only in year 1 is DSL ⫽ $20,000, the same as computed in part (a).
For illustration, compute DSL values for years 2 and 4. For t ⫽ 2, BV1 ⫽ $60,000 by the
DDB method and
60,000 ⫺ 0
DSL ⫽ ————— ⫽ $15,000
5⫺2⫹1
For t ⫽ 4, BV3 ⫽ $21,600 by the DDB method and
21,600 ⫺ 0
DSL ⫽ ————— ⫽ $10,800
5⫺4⫹1
2. The column “Larger Dt” indicates a switch in year 4 with D4 ⫽ $10,800. The DSL ⫽
$12,960 in year 5 would apply only if the switch occurred in year 5. Total depreciation
with switching is $100,000 compared to the DDB amount of $92,224.
3. With switching, PWD ⫽ $73,943, which is an increase over both the SL and DDB
methods.
433
434
Depreciation Methods
Chapter 16
TABLE 16A–2
DDB Model Depreciation and Present Worth Computations,
Example 16A.3b
Year
t
Dt , $
BVt , $
(P兾F,15%,t)
Present Worth
of Dt , $
0
1
2
3
4
5
40,000
24,000
14,400
8,640
5,184
100,000
60,000
36,000
21,600
12,960
7,776
0.8696
0.7561
0.6575
0.5718
0.4972
34,784
18,146
9,468
4,940
2,577
Totals
92,224
TABLE 16A–3
Year
t
69,915
Depreciation and Present Worth for DDB-to-SL Switching,
Example 16A.3c
DDB Method, $
DDDB
BVt
0
1
2
3
4*
5
—
40,000
24,000
14,400
8,640
5,184
Totals
92,224
100,000
60,000
36,000
21,600
12,960
7,776
SL Method
DSL, $
Larger
Dt , $
P兾F
Factor
Present
Worth of
Dt, $
20,000
15,000
12,000
10,800
12,960
40,000
24,000
14,400
10,800
10,800
0.8696
0.7561
0.6575
0.5718
0.4972
34,784
18,146
9,468
6,175
5,370
100,000
73,943
*Indicates year of switch from DDB to SL depreciation.
Solution by Spreadsheet
(d) In Figure 16A–2, column D entries are the VDB functions to determine that the DDB-toSL switch should take place in year 4. The entries “2,FALSE” at the end of the VDB function are optional (see the VDB function description). If TRUE were entered, the declining
balance model would be maintained throughout the recovery period, and the annual depreciation amounts would be equal to those in column B. The plot in Figure 16A–2 indicates
another difference in depreciation methods. The terminal book value in year 5 for the DDB
method is BV5 ⫽ $7776, while the DDB-to-SL switch reduces the book value to zero.
The NPV function determines the PW of depreciation (row 9). The results here are the
same as in parts (b) and (c) above. The DDB-to-SL switch has the larger PWD value.
⫽ DDB($C$3,0,5,$A7)
⫽ VDB($E$3,0,5,$A6,$A7,2,FALSE)
Figure 16A–2
Depreciation for DDB-to-SL switch using the VDB function, Example 16A.3.
Determination of MACRS Rates
16A.3
In MACRS, recovery periods of 3, 5, 7, and 10 years apply DDB depreciation with half-year
convention switching to SL. When the switch to SL takes place, which is usually in the last 1 to
3 years of the recovery period, any remaining basis is charged off in year n ⫹ 1 so that the book
value reaches zero. Usually 50% of the applicable SL amount remains after the switch has occurred. For recovery periods of 15 and 20 years, 150% DB with the half-year convention and the
switch to SL apply.
The present worth of depreciation PWD will always indicate which method is the most advantageous. Only the MACRS rates for the GDS recovery periods (Table 16–4) utilize the DDBto-SL switch. The MACRS rates for the alternative depreciation system (ADS) have longer
recovery periods and impose the SL model for the entire recovery period.
EXAMPLE 16A.4
In Example 16A.3, parts (c) and (d), the DDB-to-SL switching method was applied to a
$100,000, n ⫽ 5 years asset resulting in PWD ⫽ $73,943 at i ⫽ 15%. Use MACRS to depreciate
the same asset for a 5-year recovery period, and compare PWD values.
Solution
Table 16A–4 summarizes the computations for depreciation (using Table 16–2 rates), book
value, and present worth of depreciation. The PWD values for all four methods are as follows:
DDB-to-SL switching
Double declining balance
MACRS
Straight line
$73,943
$69,916
$69,016
$67,044
MACRS provides a slightly less accelerated write-off. This is so, in part, because the half-year
convention disallows 50% of the first-year DDB depreciation (which amounts to 20% of the
basis). Also the MACRS recovery period extends to year 6, further reducing PWD.
TABLE 16A–4
Depreciation and Book Value Using MACRS,
Example 16A.4
t
dt
Dt , $
BVt, $
0
1
2
3
4
5
6
—
0.20
0.32
0.192
0.1152
0.1152
0.0576
—
20,000
32,000
19,200
11,520
11,520
5,760
100,000
80,000
48,000
28,800
17,280
5,760
0
100,000
1.000
t⫽6
PWD ⫽
兺 D (P兾F,15%,t) ⫽ $69,016
t
t⫽1
16A.3 Determination of MACRS Rates
The depreciation rates for MACRS incorporate the DB-to-SL switching for all GDS recovery
periods from 3 to 20 years. In the first year, some adjustments have been made to compute the
MACRS rate. The adjustments vary and are not usually considered in detail in economic analyses. The half-year convention is always imposed, and any remaining book value in year n is removed in year n ⫹ 1. The value S ⫽ 0 is assumed for all MACRS schedules.
Since different DB depreciation rates apply for different n values, the following summary may
be used to determine Dt and BVt values. The symbols DDB and DSL are used to identify DB and
SL depreciation, respectively.
435
436
Depreciation Methods
Chapter 16
For n ⫽ 3, 5, 7, and 10 Use DDB depreciation with the half-year convention, switching to SL
depreciation in year t when DSL ⱖ DDB. Use the switching rules of Section 16A.2, and add onehalf year when computing DSL to account for the half-year convention. The yearly depreciation
rates are
dt ⫽
{
1
—
n
t⫽1
2
—
n
t ⫽ 2, 3, . . .
[16A.11]
Annual depreciation values for each year t applied to the adjusted basis, allowing for the halfyear convention, are
DDB ⫽ dt(BVt⫺1)
DSL ⫽
{
()
1 —
1B
—
[16A.12]
t⫽1
2 n
BVt⫺1
—————
n ⫺ t ⫹ 1.5
[16A.13]
t ⫽ 2, 3, . . . , n
After the switch to SL depreciation takes place—usually in the last 1 to 3 years of the recovery
period—any remaining book value in year n is removed in year n ⫹ 1.
For n ⫽ 15 and 20 Use 150% DB with the half-year convention and the switch to SL when
DSL ⱖ DDB. Until SL depreciation is more advantageous, the annual DB depreciation is computed using a form of Equation [16A.7]
DDB ⫽ dt(BVt⫺1)
where
dt ⫽
{
0.75
——
n
t⫽1
1.50
——
n
t ⫽ 2, 3, . . .
[16A.14]
EXAMPLE 16A.5
A wireless tracking system for shop floor control with a MACRS 5-year recovery period has
been purchased for $10,000. (a) Use Equations [16A.11] through [16A.13] to obtain the annual
depreciation and book value. (b) Determine the resulting annual depreciation rates and compare them with the MACRS rates in Table 16–2 for n ⫽ 5.
Solution
(a) With n ⫽ 5 and the half-year convention, use the DDB-to-SL switching procedure to obtain the results in Table 16A–5. The switch to SL depreciation, which occurs in year 4
when both depreciation values are equal, is indicated by
DDB ⫽ 0.4(2880) ⫽ $1152
2880
⫽ $1152
DSL ⫽ —————
5 ⫺ 4 ⫹ 1.5
The SL depreciation of $1000 in year 1 results from applying the half-year convention included in the first relation of Equation [16A.13]. Also, the SL depreciation of $576 in
year 6 is the result of the half-year convention.
(b) The actual rates are computed by dividing the “Larger Dt” column values by the first cost
of $10,000. The rates below are the same as the Table 16–2 rates.
t
1
2
3
4
5
6
dt
0.20
0.32
0.192
0.1152
0.1152
0.0576
Determination of MACRS Rates
16A.3
Depreciation Amounts Used to Determine MACRS Rates
for n ⫽ 5, Example 16A.5
TABLE 16A–5
Years
DDB
t
dt
DDB, $
SL Depreciation
DSL, $
Larger
Dt, $
BVt, $
0
1
2
3
4
5
6
—
0.2
0.4
0.4
0.4
0.4
—
—
2,000
3,200
1,920
1,152
691
—
—
1,000
1,777
1,371
1,152
1,152
576
—
2,000
3,200
1,920
1,152
1,152
576
10,000
8,000
4,800
2,880
1,728
576
0
10,000
It is clearly easier to use the rates in Table 16–2 or the VDB spreadsheet function than to determine each MACRS rate using the switching logic above. But the logic behind the MACRS
rates is described here for those interested. The annual MACRS rates may be derived by using
the applicable rate for the DB method. The subscripts DB and SL have been inserted along with
the year t. For the first year t ⫽ 1,
1
dDB,1 ⫽ —
n
()
1
1 —
dSL,1 ⫽ —
2 n
or
For summation purposes only, we introduce the subscript i (i ⫽ 1, 2, . . . , t) on d. Then the depreciation rates for years t ⫽ 2, 3, . . . , n are
(
i⫽t⫺1
dDB,t ⫽ d 1 ⫺
(
兺d
i
i⫽1
i⫽t⫺1
1⫺
兺d
i
)
[16A.15]
)
i⫽1
dSL,t ⫽ —————
[16A.16]
n ⫺ t ⫹ 1.5
Also, for year n ⫹ 1, the MACRS rate is one-half the SL rate of the previous year n.
dSL,n⫹1 ⫽ 0.5(dSL,n)
[16A.17]
The DB and SL rates are compared each year to determine which is larger and when the switch
to SL depreciation should occur.
EXAMPLE 16A.6
Verify the MACRS rates in Table 16–2 for a 3-year recovery period. The rates in percent are
33.33, 44.45, 14.81, and 7.41.
Solution
The fixed rate for DDB with n ⫽ 3 is d ⫽ 2兾3 ⫽ 0.6667. Using the half-year convention in year 1
and Equations [16A.15] through [16A.17], the results are as follows:
d1:
dDB,1 ⫽ 0.5d ⫽ 0.5(0.6667) ⫽ 0.3333
d2: Cumulative depreciation rate is 0.3333.
dDB,2 ⫽ 0.6667(1 ⫺ 0.3333) ⫽ 0.4445
1 ⫺ 0.3333 ⫽ 0.2267
dSL,2 ⫽ —————
3 ⫺ 2 ⫹ 1.5
(larger value)
437
438
Depreciation Methods
Chapter 16
d3: Cumulative depreciation rate is 0.3333 ⫹ 0.4445 ⫽ 0.7778.
dDB,3 ⫽ 0.6667(1 ⫺ 0.7778) ⫽ 0.1481
1 ⫺ 0.7778 ⫽ 0.1481
dSL,2 ⫽ —————
3 ⫺ 3 ⫹ 1.5
Both values are the same; switch to straight line depreciation.
d4: This rate is 50% of the last SL rate.
d4 ⫽ 0.5(dSL,3) ⫽ 0.5(0.1481) ⫽ 0.0741
PROBLEMS
Fundamentals of Depreciation
16.1 How does depreciation affect a company’s cash
flow?
16.2 What is the difference between book value and
market value?
16.3 State the difference between book depreciation
and tax depreciation.
16.4 State the difference between unadjusted and adjusted basis.
16.5 Explain why the recovery period used for tax depreciation purposes may be different from the estimated n value in an engineering economy study.
16.6 Visit the U.S. Internal Revenue Service website at
www.irs.gov and answer the following questions
about depreciation and MACRS by consulting
Publication 946, How to Depreciate Property.
(a) What is the definition of depreciation according to the IRS?
(b) What is the description of the term salvage
value?
(c) What are the two depreciation systems within
MACRS, and what are the major differences
between them?
(d) What are the properties listed that cannot be
depreciated under MACRS?
(e) When does depreciation begin and end?
(f ) What is a Section 179 deduction?
Salvage value ⫽ 10% of purchase price
Operating cost (with technician) ⫽ $185,000
per year
The manager of the department asked the newest
hire to enter the appropriate data in the tax accounting program. For the MACRS method, what
are the values of B, n, and S in depreciating the
asset for tax purposes?
16.8 Stahmann Products paid $350,000 for a numerical
controller during the last month of 2007 and had it
installed at a cost of $50,000. The recovery period
was 7 years with an estimated salvage value of
10% of the original purchase price. Stahmann sold
the system at the end of 2011 for $45,000.
(a) What numerical values are needed to develop
a depreciation schedule at purchase time?
(b) State the numerical values for the following:
remaining life at sale time, market value in
2011, book value at sale time if 65% of the
basis had been depreciated.
16.9 An asset with an unadjusted basis of $50,000 was
depreciated over ntax ⫽ 10 years for tax depreciation purposes and nbook ⫽ 5 years for book depreciation purposes. The annual depreciation was
1兾n using the relevant life value. Use a spreadsheet to plot on one graph the annual book value
for both methods of depreciation.
Straight Line Depreciation
16.7 A major energy production company has the following information regarding the acquisition of
new-generation equipment.
16.10 What is the depreciation rate dt per year for an
asset that has an 8-year useful life and is straight
line depreciated?
Purchase price ⫽ $580,000
Transoceanic shipping and delivery cost ⫽ $4300
Installation cost (1 technician at $1600 per day
for 4 days) ⫽ $6400
Tax recovery period ⫽ 15 years
Book depreciation recovery period ⫽ 10 years
16.11 Pneumatics Engineering purchased a machine that
had a first cost of $40,000, an expected useful life
of 8 years, a recovery period of 10 years, and a
salvage value of $10,000. The operating cost of the
machine is expected to be $15,000 per year. The
inflation rate is 6% per year and the company’s
Problems
MARR is 11% per year. Determine (a) the depreciation charge for year 3, (b) the present worth of
the third-year depreciation charge in year 0, the
time of asset purchase, and (c) the book value for
year 3 according to the straight line method.
16.12 An asset that is book-depreciated over a 5-year period by the straight line method has BV3 ⫽ $62,000
with a depreciation charge of $26,000 per year.
Determine (a) the first cost of the asset and (b) the
assumed salvage value.
16.13 Lee Company of Westbrook, Connecticut, manufactures pressure relief inserts for thermal relief
and low-flow hydraulic pressure relief applications
where zero leakage is required. A machine purchased 3 years ago has been book-depreciated by
the straight line method using a 5-year useful life.
If the book value at the end of year 3 is $30,000 and
the company assumed that the machine would be
worthless at the end of its 5-year useful life,
(a) what is the book depreciation charge each year
and (b) what was the first cost of the machine?
16.14 An asset has an unadjusted basis of $200,000, a
salvage value of $10,000, and a recovery period of
7 years. Write a single-cell spreadsheet function to
display the book value after 5 years of straight line
depreciation. Use your function to determine the
book value.
16.15 Bristol Myers Squibb purchased a tablet-forming
machine in 2008 for $750,000. The company
planned to use the machine for 10 years; however,
due to rapid obsolescence it will be retired after
only 4 years in 2012. Develop a spreadsheet for
depreciation and book value amounts necessary to
answer the following.
(a) What is the amount of capital investment remaining when the asset is prematurely retired?
(b) If the asset is sold at the end of 4 years for
$175,000, what is the amount of capital investment lost based on straight line depreciation?
(c) If the new-technology machine has an estimated cost of $300,000, how many more
years should the company retain and depreciate the currently owned machine to make its
book value and the first cost of the new machine equal to each other?
16.16 A special-purpose graphics workstation acquired
by Busbee Consultants has B ⫽ $50,000 with a
4-year recovery period. Tabulate the values for SL
depreciation, accumulated depreciation, and book
value for each year if (a) S ⫽ 0 and (b) S ⫽ $16,000.
(c) Use a spreadsheet to plot the book value over
the 4 years on one chart for both salvage value
estimates.
439
16.17 A company owns the same asset in a U.S. plant
(New York) and in a EU plant (Paris). It has B ⫽
$2,000,000 and a salvage value of 20% of B. For
tax depreciation purposes, the United States allows a straight line write-off over 5 years, while
the EU allows SL write-off over 8 years. The general managers of the two plants want to know the
difference in (a) the depreciation amount for year
5 and (b) the book value after 5 years. Using a
spreadsheet, write cell functions in only two cells
to answer both questions.
Declining Balance Depreciation
16.18 When declining balance (DB) depreciation is applied, there can be three different depreciation
rates involved—d, dmax, and dt. Explain the differences between these rates.
16.19 Equipment for immersion cooling of electronic
components has an installed value of $182,000
with an estimated trade-in value of $40,000 after
15 years. For years 2 and 10, use DDB book depreciation to determine (a) the depreciation charge
and (b) the book value.
16.20 A cooling-water pumping station at the LCRA
plant costs $600,000 to construct, and it is projected to have a 25-year life with an estimated
salvage value of 15% of the construction cost.
However, the station will be book-depreciated to
zero over a recovery period of 30 years. Calculate
the annual depreciation charge for years 4, 10,
and 25, using (a) straight line depreciation and
(b) DDB depreciation. (c) What is the implied
salvage value for DDB? (d) Use a spreadsheet to
build the depreciation and book value schedules
for both methods to verify your answers.
16.21 A video recording system was purchased 3 years
ago at a cost of $30,000. A 5-year recovery period
and DDB depreciation have been used to write off
the basis. The system is to be replaced this year with
a trade-in value of $5000. What is the difference between the book value and the trade-in value?
16.22 An engineer with Accenture Middle East BV in
Dubai was asked by her client to help him understand
the difference between 150% DB and DDB depreciation. Answer these questions if B ⫽ $180,000,
n ⫽ 12 years, and S ⫽ $30,000.
(a) What are the book values after 12 years for
both methods?
(b) How do the estimated salvage and these book
values compare in value after 12 years?
(c) Which of the two methods, when calculated
correctly considering S ⫽ $30,000, writes off
more of the first cost over 12 years?
440
Chapter 16
Depreciation Methods
16.23 Exactly 10 years ago, Boyditch Professional Associates purchased $100,000 in depreciable assets with an estimated salvage of $10,000. For tax
depreciation, the SL method with n ⫽10 years
was used, but for book depreciation, Boyditch applied the DDB method with n ⫽ 7 years and neglected the salvage estimate. The company sold
the assets today for $12,500.
(a) Compare the sales price today with the book
values using the SL and DDB methods.
(b) If the salvage of $12,500 had been estimated
exactly 10 years ago, determine the depreciation for each method in year 10.
16.29 A plant manager for a large cable company knows
that the remaining invested value of certain types
of manufacturing equipment is more closely approximated when the equipment is depreciated
linearly by the SL method compared to a rapid
write-off method such as MACRS. Therefore, he
keeps two sets of books, one for tax purposes
(MACRS) and one for equipment management
purposes (SL). For an asset that has a first cost of
$80,000, a depreciable life of 5 years, and a salvage value equal to 25% of the first cost, determine the difference in the book values shown in
the two sets of books at the end of year 4.
16.24 Shirley is studying depreciation in her engineering
management course. The instructor asked her to
graphically compare the total percent of first cost
depreciated for an asset costing B dollars over a
life of n ⫽ 5 years for DDB and 125% DB depreciation. Help her by developing the plots of percent
of B depreciated versus years. Use a spreadsheet
unless otherwise instructed.
16.30 The manager of a Glidden Paint manufacturing
plant is aware that MACRS and DDB are both accelerated depreciation methods; however, out of curiosity, she wants to determine which one provides
the faster write-off in the first 3 years for a recently
purchased mixer that has a first cost of $300,000, a
5-year recovery period, and a $60,000 salvage
value. Determine which method yields the lower
book value and by how much after 3 years. The annual MACRS depreciation rates are 20%, 32%, and
19.2% for years 1, 2, and 3, respectively.
MACRS Depreciation
16.25 Explain the difference between an accelerated depreciation method and one that is not accelerated.
Give an example of each.
16.26 What was one of the prime reasons that MACRS
depreciation was initiated in the mid-1980s?
16.27 A company just purchased an intelligent robot,
which has a first cost of $80,000. Since the robot is
unique in its capabilities, the company expects to
be able to sell it in 4 years for $95,000.
(a) If the company spends $10,000 per year in
maintenance and operation of the robot, what
will the company’s MACRS depreciation
charge be in year 2? Assume the recovery period for robots is 5 years and the company’s
MARR is 16% per year when the inflation
rate is 9% per year.
(b) Determine the book value of the robot at the
end of year 2.
16.28 Animatics Corp. of Santa Clara, California, makes
small servo systems with built-in controllers, amplifiers, and encoders so that they can control entire
machines. The company purchased an asset 2 years
ago that has a 5-year recovery period. The depreciation
charge by the MACRS method for year 2 is $24,320.
(a) What was the first cost of the asset?
(b) How much was the depreciation charge in
year 1?
(c) Develop the complete MACRS depreciation
and book value schedule using the VDB
function.
16.31 Railroad cars used to transport coal from Wyoming
mines to Texas power plants cost $1.2 million and
have an estimated salvage value of $300,000. Develop the depreciation and book value schedules for
the GDS MACRS method by using two methods on
a spreadsheet—the VDB function and the MACRS
rates. Are the book value series the same?
16.32 A 120-metric-ton telescoping crane that cost
$320,000 is owned by Upper State Power. Salvage
is estimated at $75,000. (a) Compare book values
for MACRS and standard SL depreciation over a
7-year recovery period. (b) Explain how the estimated salvage is treated using MACRS.
16.33 Youngblood Shipbuilding Yard just purchased
$800,000 in capital equipment for ship repairing
functions on dry-docked ships. Estimated salvage
is $150,000 for any year after 5 years of use. Compare the depreciation and book value for year 3 for
each of the following depreciation methods.
(a) GDS MACRS where a recovery period of
10 years is allowed
(b) Double declining balance with a recovery
period of 15 years
(c) ADS straight line as an alternative to MACRS,
with a recovery period of 15 years
16.34 Basketball.com has installed $100,000 worth of
depreciable software and equipment that represents the latest in Internet teaming and basket
competition, intended to allow anyone to enjoy
441
Problems
the sport on the Web or in the alley. No salvage
value is estimated. The company can depreciate
using MACRS for a 5-year recovery period or opt
for the ADS alternate system over 10 years using
the straight line method. The SL rates require the
half-year convention; that is, only 50% of the regular annual rate applies for years 1 and 11.
(a) Construct the book value curves for both
methods on one graph. Show hand or spreadsheet computations as instructed.
(b) After 3 years of use, what percentage of the
$100,000 basis is removed for each method?
Compare the two percentages.
16.35 A company has purchased special-purpose equipment for the manufacture of rubber products
(asset class 30.11 in IRS Publication 946) and expects to use it predominately outside the United
States. In this case, the ADS alternative to
MACRS is required for tax depreciation purposes.
The manager wants to understand the difference
in yearly recovery rates for classical SL, MACRS,
and the ADS alternative to MACRS. Using a recovery period of 3 years, except for the ADS alternative, which requires a 4-year recovery with
half-year convention included, prepare a single
graph showing the annual recovery rates (in
percent) for the three methods.
16.38 A sand and gravel pit purchased for $900,000 is
expected to yield 50,000 tons of gravel and 80,000
tons of sand per year. The gravel will sell for
$6 per ton and the sand for $9 per ton.
(a) Determine the depletion charge according to
the percentage depletion method. The percentage depletion rate for sand and gravel
is 5%.
(b) If taxable income is $100,000 for the year, is
this depletion charge allowed? If not, how
much is allowed?
16.39 Vesco Mineral Resources purchased mineral rights
to land in the foothills of the Santa Cristo mountains. The cost of the purchase was $9 million.
Vesco originally thought that it would be able to
extract 200,000 tons of lignite from the land, but
further exploration revealed that 280,000 tons
could be economically removed. If the company
sold 20,000 tons in year 1 and 30,000 tons in year
2, what would the depletion charges be each year
according to the cost depletion method?
16.40 A company owns gold mining operations in the
United States, Australia, and South Africa. The
Colorado mine has the taxable income and sales
results summarized below. Determine the annual
percentage depletion amount for the gold mine.
Year
Taxable
Income, $
Sales,
Ounces
Average
Sales Price,
$ per Ounce
1
2
3
1,500,000
2,000,000
800,000
1800
5500
2300
1115
1221
1246
Depletion
16.36 A coal mine purchased 3 years ago for $7 million
was estimated to contain 4,000,000 tons of coal.
During the past 3 years the amount of coal removed was 21,000, 18,000, and 20,000 tons,
respectively. The gross income obtained in these 3
years was $257,000 for the first year, $320,000 for
the second year, and $340,000 for the third year.
Determine (a) the cost depletion allowance for
each year and (b) the percentage of the purchase
price depleted thus far.
16.37 NA Forest Resources purchased forest acreage
for $500,000 from which an estimated 200 million board feet of lumber is recoverable. The
company will sell the lumber for $0.10 per board
foot. No lumber will be sold for the next 2 years
because an environmental impact statement
must be completed before harvesting can begin.
In years 3 to 10, however, the company expects
to remove 20 million board feet per year. The
inflation rate is 8%, and the company’s MARR
is 10%.
(a) Determine the depletion amount in year 2 by
the cost depletion method.
(b) Determine the depletion amount in year 5 by
the percentage depletion method.
16.41 A highway construction company operates a
quarry. During the last 5 years, the amount extracted each year was 60,000, 50,000, 58,000
60,000, and 65,000 tons. The mine is estimated to
contain a total of 2.5 million tons of usable stones
and gravel. The quarry land had an initial cost of
$3.2 million. The company had a per-ton gross
income of $30 for the first year, $25 for the second year, $35 for the next 2 years, and $40 for the
last year.
(a) Determine the depletion charge each year,
using the larger of the values for the two depletion methods. Assume all depletion amounts
are less than 50% of taxable income.
(b) Compute the percent of the initial cost that
has been written off in these 5 years, using
the depletion charges in part (a).
(c) If the quarry operation is reevaluated after
the first 3 years of operation and estimated to
contain a total of 1.5 million tons remaining,
rework parts (a) and (b).
442
Depreciation Methods
Chapter 16
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
16.42 All of the following types of real property are depreciable except:
(a) Warehouses
(b) Land
(c) Office buildings
(d) Test facilities
16.43 According to information on the IRS website, a
taxpayer can take a depreciation deduction as long
as the property meets all of the following requirements except:
(a) The taxpayer must own the property.
(b) The taxpayer must use the property in an
income-producing activity.
(c) The taxpayer must use the property for personal purposes.
(d) The property must have a determinable useful life of more than 1 year.
16.44 A machine with a 5-year life has a first cost of
$20,000 and a $2000 salvage value. Its annual operating cost is $8000 per year. According to the
classical straight line method, the depreciation
charge in year 2 is nearest to:
(a) $2800
(b) $3600
(c) $4500
(d) $5300
16.45 A machine with a 10-year life is MACRSdepreciated. The machine has a first cost of $40,000
with a $5000 salvage value. Its annual operating
cost is $7000 per year, and dt for years 1, 2, and 3 is
10.00%, 18.00%, and 14.40%, respectively. The
depreciation charge in year 3 is nearest to:
(a) $5800
(b) $7200
(c) $8500
(d) $9300
16.46 A coal mine purchased for $5 million has enough
coal to operate for 10 years. The annual cost is expected to be $200,000 per year. The coal is
expected to sell for $150 per ton, with annual production expected to be 10,000 tons. Coal has a
depletion percentage rate of 10%. The depletion
charge for year 6 according to the percentage depletion method would be closest to:
(a) $75,000
(b) $100,000
(c) $125,000
(d) $150,000
16.47 The depreciation charge for a 5-year, straight line
depreciated vehicle is $3000 in year 4. If the first
cost was $20,000, the salvage value used in the depreciation calculation was closest to:
(a) $0
(b) $2500
(c) $5000
(d) $7500
16.48 The book value of an asset that was DDBdepreciated over a 10-year period was $5832 at the
end of year 4. If the first cost of the asset was
$80,000, the salvage value that was used in the depreciation calculation was closest to:
(a) $0
(b) $2000
(c) $5000
(d) $8000
16.49 For an asset that has B ⫽ $100,000, S ⫽ $40,000,
and a 10-year depreciable life, the book value at
the end of year 4 according to the MACRS method
would be closest to (dt values for years 1, 2, 3, 4,
and 5 are 10.00%, 18.00%, 14.40%, 11.52%, and
9.22%, respectively):
(a) $58,700
(b) $62,400
(c) $53,900
(d) $46,100
16.50 An asset that was depreciated over a 5-year period
by the MACRS method had a BV of $33,025 at the
end of year 3. If the MACRS depreciation rates for
years 1, 2, and 3, were 0.20, 0.32, and 0.192, respectively, the basis of the asset was closest to:
(a) $158,000
(b) $172,000
(c) $185,000
(d) $193,000
16.51 A lumber company purchased a tract of land for
$70,000 that contained an estimated 25,000 usable
trees. The value of the land was estimated at
$20,000. In the first year of operation, the lumber
company cut down 5000 trees. According to the
cost depletion method, the depletion deduction for
year 1 is closest to:
(a) $2000
(b) $7000
(c) $10,000
(d) $14,000
16.52 An asset with a first cost of $50,000 and an estimated salvage value of $10,000 is depreciated by
the MACRS method. If its book value at the end of
year 3 is $21,850 and its market value is $25,850,
the total amount of depreciation charged against
the asset up to this time is closest to:
(a) $18,850
(b) $21,850
(c) $25,850
(d) $28,150
16.53 Under the General Depreciation System (GDS) of
asset classification, any asset that is not in a stated
class is automatically assigned a recovery period of:
(a) 5 years
(b) 7 years
(c) 10 years
(d) 15 years
16.54 All of the following statements about the Alternative Depreciation System (ADS) are true except:
(a) The half-year convention applies.
(b) Salvage value is neglected.
(c) The recovery periods are shorter than in GDS.
(d) The straight line method is required.
Appendix Problems
443
APPENDIX PROBLEMS
depreciation, and calculate the present worth of
depreciation at i ⫽ 18% per year.
Sum-of-Years-Digits Depreciation
16A.1
16A.2
16A.3
A European manufacturing company has new
equipment with a first cost of 12,000 euros, an
estimated salvage value of 2000 euros, and a
recovery period of 8 years. Use the SYD method
to tabulate annual depreciation and book value.
Earthmoving equipment with a first cost of
$150,000 is expected to have a life of 10 years.
The salvage value is expected to be 10% of the
first cost. Calculate (a) by hand and (b) by
spreadsheet the depreciation charge and book
value for years 2 and 7 using the SYD method.
If B ⫽ $12,000, n ⫽ 6 years, and S is estimated
at 15% of B, use the SYD method to determine
(a) the book value after 3 years and (b) the rate
of depreciation and the depreciation amount in
year 4.
Unit-of-Production Depreciation
16A.4
16A.5
A robot used in simulated car crashes cost
$70,000, has no salvage value, and has an
expected capacity of tests not to exceed 10,000
according to the manufacturer. Volvo Motors decided to use the unit-of-production depreciation
method because the number of test crashes per
year in which the robot would be involved was
not estimable. Determine the annual depreciation
and book value for the first 3 years if the number
of tests were 3810, 2720, and 5390 per year.
A new hybrid car was purchased by Pedernales
Electric Cooperative as a courier vehicle to
transport items between its 12 city offices. The
car cost $30,000 and was retained for 5 years.
Alternatively, it could have been retained for
100,000 miles. Salvage value is nil. Five-year
DDB depreciation was applied. The car pool
manager stated that he prefers UOP depreciation on vehicles because it writes off the first
cost faster. Use the actual annual miles driven,
listed below, to plot the book values for both
methods. Determine which method would have
removed the $30,000 faster. Show hand or
spreadsheet solution, as instructed.
Year
1
2
3
4
5
Miles Driven, 1000
15
22
16
18
24
Switching Methods
16A.6
An asset has a first cost of $45,000, a recovery
period of 5 years, and a $3000 salvage value.
Use the switching procedure from DDB to SL
16A.7
If B ⫽ $45,000, S ⫽ $3000, and n ⫽ 5-year
recovery period, use a spreadsheet and i ⫽
18% per year to maximize the present worth of
depreciation, using the following methods:
DDB-to-SL switching (this was determined in
Problem 16A.6) and MACRS. Given that
MACRS is the required depreciation system in
the United States, comment on the results.
16A.8
Hempstead Industries has a new milling machine
with B ⫽ $110,000, n ⫽ 10 years, and S ⫽ $10,000.
Determine the depreciation schedule and present
worth of depreciation at i ⫽ 12% per year, using
the 175% DB method for the first 5 years and
switching to the classical SL method for the last 5
years. Use a spreadsheet to solve this problem.
16A.9
Reliant Electric Company has erected a large
portable building with a first cost of $155,000
and an anticipated salvage of $50,000 after
25 years. (a) Should the switch from DDB to
SL depreciation be made? (b) For what values
of the uniform depreciation rate in the DB
method would it be advantageous to switch
from DB to SL depreciation at some point in
the life of the building?
MACRS Rates
16A.10 Verify the 5-year recovery period rates for
MACRS given in Table 16–2. Start with the
DDB method in year 1, and switch to SL depreciation when it offers a larger recovery rate.
16A.11 A video recording system was purchased
3 years ago at a cost of $30,000. A 5-year recovery
period and MACRS depreciation have been used
to write off the basis. The system is to be prematurely replaced with a trade-in value of $5000.
Determine the MACRS depreciation, using the
switching rules to find the difference between the
book value and the trade-in value after 3 years.
16A.12 Use the computations in Equations [16A.11]
through [16A.13] to determine the MACRS annual depreciation for the following asset data:
B ⫽ $50,000 and a recovery period of 7 years.
16A.13 The 3-year MACRS recovery rates are 33.33%,
44.45%, 14.81%, and 7.41%, respectively.
(a) What are the corresponding rates for the alternative MACRS straight line ADS method
with the half-year convention imposed?
(b) Compare the PWD values for these two methods if B ⫽ $80,000 and i ⫽ 15% per year.
CHAPTER 17
After-Tax
Economic
Analysis
L E A R N I N G
O U T C O M E S
Purpose: Perform an after-tax economic evaluation considering the impact of pertinent tax regulations, income taxes,
and depreciation.
SECTION
TOPIC
LEARNING OUTCOME
17.1
Terminology and rates
• Know the fundamental terms and relations of after-tax
analysis; use a marginal tax rate table.
17.2
CFBT and CFAT
• Determine cash flow series before taxes and after taxes.
17.3
Taxes and depreciation
• Demonstrate the tax advantage of accelerated
depreciation and shortened recovery periods.
17.4
Depreciation recapture
• Calculate the tax impact of DR; explain the use of
capital gains and capital losses.
17.5
After-tax analysis
• Evaluate one project or multiple alternatives using
after-tax PW, AW, and ROR analysis.
17.6
After-tax replacement
• Evaluate a defender and challenger in an after-tax
replacement study.
17.7
EVA analysis
• Evaluate an alternative using after-tax economic valueadded analysis; compare to CFAT analysis.
17.8
Taxes outside the United States
• Understand the fundamental practices for depreciation
and tax rates in international settings.
17.9
VAT
• Demonstrate the use and computation of a valueadded tax on manufactured products.
T
his chapter provides an overview of tax terminology, income tax rates, and tax
equations pertinent to an after-tax economic analysis. The change from estimating cash flow before taxes (CFBT) to cash flow after taxes (CFAT) involves a consideration of significant tax effects that may alter the final decision, and estimates the magnitude of the effect on cash flow over the life of the alternative that taxes may have.
Mutually exclusive alternative comparisons using after-tax PW, AW, and ROR methods
are explained with major tax implications considered. Replacement studies are discussed
with tax effects that occur at the time that a defender is replaced. Also, the after-tax economic value added by an alternative is discussed in the context of annual worth analysis. All
these methods use the procedures learned in earlier chapters, except now with tax effects
considered.
An after-tax evaluation using any method requires more computations than in previous
chapters. Templates for tabulation of cash flow after taxes by hand and by spreadsheet
are developed. Additional information on U.S. federal taxes—tax law and annually updated tax rates—is available through Internal Revenue Service publications and, more readily,
on the IRS website www.irs.gov. Publications 542, Corporations, and 544, Sales and Other
Dispositions of Assets, are especially applicable to this chapter. Some differences in tax considerations outside the United States are summarized.
17.1 Income Tax Terminology and Basic Relations
The perspective taken in engineering economy when performing an after-tax evaluation is that of
the project and how relevant tax rules and allowances influence the economic decision. The perspective of a financial study is that of the corporation and how the tax structure and laws affect
profitability. We take the engineering economy viewpoint in the sections that follow.
There are many types of taxes levied upon corporations and individuals in all countries, including the United States. Some are sales tax, valued-added tax, import tax, income tax, highway
tax, gasoline tax, and property (real estate) tax. Federal governments rely on income taxes for a
significant portion of their annual revenue. States, provinces, and municipal-level governments
rely on sales, value-added, and property taxes to maintain services, schools, etc. for the citizenry.
For a starting point on corporate income taxes and how they are used when performing an aftertax economic evaluation of a project or multiple alternatives, this section covers basic definitions,
terms, and relations.
Income tax is the amount of the payment (taxes) on income or profit that must be delivered to a
federal (or lower-level) government unit. Taxes are real cash flows; however, for corporations
tax computation requires some noncash elements, such as depreciation. Corporate income taxes
are usually submitted quarterly, and the last payment of the year is submitted with the annual tax
return.
The Internal Revenue Service (IRS), a part of the U.S. Department of the Treasury, collects the
taxes and enforces tax laws. The website www.irs.gov provides information on tax laws, rates,
publications, etc. that are referenced in this chapter.
Though the formulas are much more complex when applied to a specific situation, two fundamental relations form the basis for income tax computations. The first involves only actual cash
flows:
Net operating income ⫽ revenue ⫺ operating expenses
The second involves actual cash flows and noncash deductibles, such as depreciation.
Taxable income ⫽ revenue ⫺ operating expenses ⫺ depreciation
These terms and relations for corporations are now described. Since each term is calculated for
1 year, there can be a subscript t (t ⫽ 1, 2, . . .) added; t is omitted here for simplicity.
Operating revenue R, also commonly called gross income GI, is the total income realized
from all revenue-producing sources. These incomes are listed in the income statement. (See
Appendix B on accounting reports.) Other, nonoperating revenues such as sale of assets,
license fee income, and royalties are considered separately for tax purposes.
446
After-Tax Economic Analysis
Chapter 17
Operating expenses OE include all costs incurred in the transaction of business. These expenses are tax-deductible for corporations. For after-tax economic evaluations, the AOC (annual operating costs) and M&O (maintenance and operating) costs are applicable here. Depreciation is not included here since it is not an operating expense.
Net operating income NOI, often called EBIT (earnings before interest and income taxes),
is the difference between gross income and operating expenses.
NOI ⴝ EBIT ⴝ GI ⴚ OE
[17.1]
Taxable income TI is the amount of income upon which taxes are based. A corporation is
allowed to remove depreciation, depletion and amortization, and some other deductibles
from net operating income in determining the taxable income for a year. For our evaluations,
we define taxable income as
TI ⴝ gross income ⴚ operating expenses ⴚ depreciation
ⴝ GI ⴚ OE ⴚ D
[17.2]
Though there may be subtleties and varying interpretations over time, in essence, the differences between NOI and TI are tax-law-allowed deductibles, such as depreciation. (In keeping
with the project view of engineering economics, we will primarily use the TI relation when
conducting an after-tax evaluation.)
Tax rate T is a percentage, or decimal equivalent, of TI that is owed in taxes. The tax rates
in many countries (including the United States) are graduated (or progressive) by level of
TI; that is, higher rates apply as the TI increases. The marginal tax rate is the percentage
paid on the last dollar of income. The average tax rate paid is calculated separately from the
highest marginal rate used, as shown later. The general tax computation relation is
Income taxes ⴝ applicable tax rate ⴛ taxable income
ⴝ (T)(TI)
[17.3]
Net operating profit after taxes NOPAT is the amount remaining each year after taxes are
subtracted from taxable income.
NOPAT ⴝ TI ⴚ taxes ⴝ TI ⴚ (T)(TI)
ⴝ TI(1 ⴚ T)
[17.4]
Basically, NOPAT represents the money remaining in the corporation as a result of the capital
invested during the year. It is also called net profit after taxes (NPAT).
The graduated tax rate schedule for corporations is presented in Table 17–1 as taken from IRS
Publication 542, Corporations. These are rates for the entire corporation, not for an individual
project, though they are often applied in the after-tax analysis of a single project. The rates can
change annually based upon government legislation; however, the corporate tax rate schedule has
remained the same for some years. To illustrate the use of the graduated tax rate, assume a company is expected to generate a taxable income of $500,000 in 1 year. From Table 17–1, the marginal tax rate for the last dollar of TI is 34%, but the graduated rates become progressively larger
as TI increases. In this case, for TI ⫽ $500,000,
Taxes ⫽ 113,900 ⫹ 0.34(500,000 ⫺ 335,000)
⫽ 113,900 ⫹ 56,100
⫽ $170,000
Alternatively, the rates for each TI level can be used to calculate taxes the longer way.
Taxes ⫽ 0.15(50,000) ⫹ 0.25(75,000 ⫺ 50,000) ⫹ 0.34(100,000 ⫺ 75,000)
⫹ 0.39(335,000 ⫺ 100,000) ⫹ 0.34(500,000 ⫺ 335,000)
⫽ 7500 ⫹ 6250 ⫹ 8500 ⫹ 91,650 ⫹ 56,100
⫽ $170,000
Income Tax Terminology and Basic Relations
17.1
TABLE 17–1
U.S. Corporate Income Tax Rate Schedule (2010)
If Taxable Income ($) Is:
Over
But Not
over
Tax Is
Of the
Amount over
0
50,000
75,000
100,000
335,000
10,000,000
15,000,000
18,333,333
50,000
75,000
100,000
335,000
10,000,000
15,000,000
18,333,333
—
15%
7,500 ⫹ 25%
13,750 ⫹ 34%
22,250 ⫹ 39%
113,900 ⫹ 34%
3,400,000 ⫹ 35%
5,150,000 ⫹ 38%
35%
0
50,000
75,000
100,000
335,000
10,000,000
15,000,000
0
Smaller businesses (with TI ⬍ $335,000) receive a slight tax advantage compared to large corporations. Once the TI exceeds $335,000, an effective federal tax rate of 34% applies, and when
TI ⬎ $18.33 million, there is a flat tax rate of 35%.
As we move forward with after-tax analysis, it is important to keep the following in mind.
The corporate tax rates apply to a corporation as a whole, not to a specific project, unless the
project is the company. Tax rates are usually graduated by level of taxable income. Therefore, the
last dollar of TI is taxed at a marginal rate. The average federal tax rate actually paid is lower than
the highest marginal rate paid because the rates, in general, graduate to higher percentages as
TI increases.
Because the marginal tax rates change with TI, it is not possible to quote directly the percent of TI paid in income taxes. Alternatively, a single-value number, the average tax rate, is
calculated as
total taxes paid taxes
[17.5]
Average tax rate ⫽ ——————— ⫽ ———
TI
taxable income
Referring to Table 17–1, for a small business with TI ⫽ $100,000, the federal income tax
burden averages $22,250兾100,000 ⫽ 22.25%. If TI ⫽ $15 million, the average tax rate is
$5.15 million兾15 million ⫽ 34.33%.
As mentioned earlier, there are federal, state, and local taxes imposed. For the sake of simplicity, the tax rate used in an economy study is often a single-figure effective tax rate Te, which
accounts for all taxes. Effective tax rates are in the range of 35% to 50%. One reason to use the
effective tax rate is that state taxes are deductible for federal tax computation. The effective tax
rate and taxes are calculated as
Te ⫽ state rate ⫹ (1 ⫺ state rate)(federal rate)
Taxes ⫽ (Te)(TI)
[17.6]
[17.7]
EXAMPLE 17.1
REI (Recreational Equipment Incorporated) sells outdoor equipment and sporting goods through
retail outlets, the Internet, and catalogs. Assume that for 1 year REI has the following financial
results in the state of Kentucky, which has a flat tax rate of 6% on corporate taxable income.
Total revenue
Operating expenses
Depreciation and other allowed deductions
$19.9 million
$8.6 million
$1.8 million
(a) Determine the state taxes and federal taxes due using Table 17–1 rates.
(b) Find the average federal tax rate paid for the year.
(c) Determine a single-value tax rate useful in economic evaluations using the average federal
tax rate determined in part (b).
(d) Estimate federal and state taxes using the single-value rate, and compare their total with the
total in part (a).
447
448
After-Tax Economic Analysis
Chapter 17
Solution
(a) Calculate TI by Equation [17.2] and use Table 17–1 rates for federal taxes due.
Kentucky state TI ⫽ GI ⫺ OE ⫺ D ⫽ 19.9 million ⫺ 8.6 million ⫺ 1.8 million
⫽ $9.5 million
Kentucky state taxes ⫽ 0.06(TI) ⫽ 0.06(9,500,000) ⫽ $570,000
Federal TI ⫽ GI ⫺ OE ⫺ D ⫺ state taxes ⫽ 9,500,000 ⫺ 570,000
⫽ $8,930,000
Federal taxes ⫽ 113,900 ⫹ 0.34(8,930,000 ⫺ 335,000) ⫽ $3,036,200
Total federal and state taxes ⫽ 3,036,200 ⫹ 570,000 ⫽ $3,606,200
[17.8]
(b) From Equation [17.5], the average tax rate paid is approximately 32% of TI.
Average federal tax rate ⫽ 3,036,200兾9,500,000 ⫽ 0.3196
(c) By Equation [17.6], Te is slightly over 36% per year for combined state and federal taxes.
Te ⫽ 0.06 ⫹ (1 ⫺ 0.06)(0.3196) ⫽ 0.3604
(36.04%)
(d) Use the effective tax rate and TI ⫽ $9.5 million from part (a) in Equation [17.7] to
approximate total taxes.
Taxes ⫽ 0.3604(9,500,000) ⫽ $3,423,800
Compared to Equation [17.8], this approximation is $182,400 low, a 5.06% underestimate.
It is interesting to understand how corporate tax and individual tax computations differ. Gross
income for an individual taxpayer is comparable if revenue is replaced by salaries and wages.
However, for an individual’s taxable income, most of the expenses for living and working are not
tax deductible to the same degree as operating expenses are for corporations. For individual taxpayers,
GI ⫽ salaries ⫹ wages ⫹ interest and dividends ⫹ other income
TI ⫽ GI ⫺ personal exemption ⫺ standard or itemized deductions
Taxes ⫽ (T )(TI)
For TI, operating expenses are replaced by individual exemptions and specific deductions. Exemptions are yourself, your spouse, your children, and your other dependents. Each exemption
reduces TI by $3500 to $4000 per year, depending upon current exemption allowances.
In the United States, the tax rates for individuals, like those for corporations, are graduated by
level of TI. In 2010, the marginal rates ranged from 10% to 35%; however, the top marginal rates
are increasing for individuals with larger TI amounts. These rates and TI levels are the subject of
ongoing political debates at the U.S. national level depending upon the balance of power in the
congressional bodies and the economic condition of the country. Once the marginal rates are set,
the TI levels are adjusted each year to account for inflation and other factors. This process is
called indexing. Clearly, tax rates for individuals change much more frequently than the rates for
corporations change. Current information is available on the IRS website www.irs.gov through
Publication 17, Your Federal Income Tax. The current rate schedule is published in the back of
this document for four filing status categories:
Unmarried individuals (single)
Married filing jointly
Married filing separately
Head of household
17.2 Calculation of Cash Flow after Taxes
Early in the text, the term net cash flow (NCF) was identified as the best estimate of actual cash
flow each year. The NCF is calculated as cash inflows minus cash outflows. Since then, the annual NCF amounts have been used many times to perform alternative evaluations via the PW,
AW, ROR, and B/C methods. Now that the impact on cash flow of depreciation and related taxes
Calculation of Cash Flow after Taxes
17.2
will be considered, it is time to expand our terminology. NCF is replaced by the term cash flow
before taxes (CFBT), and we introduce the new term cash flow after taxes (CFAT).
CFBT and CFAT are actual cash flows; that is, they represent the estimated actual flow of
money into and out of the corporation that will result from the alternative. The remainder of
this section explains how to transition from before-tax to after-tax cash flows for solutions by
hand and by spreadsheet, using tax regulations described in the next few sections. Once the
CFAT estimates are developed, the economic evaluation is performed using the same methods
and selection guidelines applied previously. However, the analysis is performed on the CFAT
estimates.
We learned that net operating income (NOI) does not include the purchase or sale of capital
assets. However, the annual CFBT estimate must include the initial capital investment and salvage value for the years in which they occur. Incorporating the definitions of gross income and
operating expenses from NOI, CFBT for any year is defined as
CFBT ⴝ gross income ⴚ operating expenses ⴚ initial investment ⴙ salvage value
ⴝ GI ⴚ OE ⴚ P ⴙ S
[17.9]
As in previous chapters, P is the initial investment (year 0) and S is the estimated salvage value
in year n. Therefore, only in year 0 will the CFBT include P, and only in year n will an S value
be present. Once all taxes are estimated, the annual after-tax cash flow is simply
CFAT ⴝ CFBT ⴚ taxes
[17.10]
where taxes are estimated using the relation (T)(TI) or (Te)(TI).
We know from Equation [17.2] that depreciation D is considered when calculating TI. It is
very important to understand the different roles of depreciation for income tax computations and
in CFAT estimation.
Depreciation is not an operating expense and is a noncash flow. Depreciation is tax deductible for
determining the amount of income taxes only, but it does not represent a direct, after-tax cash flow.
Therefore, the after-tax engineering economy study must be based on actual cash flow estimates,
that is, annual CFAT estimates that do not include depreciation as an expense (negative cash flow).
Accordingly, if the CFAT expression is determined using the TI relation, depreciation must not
be included outside of the TI component. Equations [17.9] and [17.10] are now combined.
CFAT ⴝ GI ⴚ OE ⴚ P ⴙ S ⴚ (GI ⴚ OE ⴚ D)(Te)
[17.11]
Suggested table column headings for CFBT and CFAT calculations by hand or by spreadsheet are
shown in Table 17–2. The equations are shown in column numbers, with the effective tax rate Te
used for income tax estimation. Expenses OE and initial investment P carry negative signs in all
tables and spreadsheets.
A negative TI value may occur in some years due to a depreciation amount that is larger than
(GI − OE). It is possible to account for this in a detailed after-tax analysis using carry-forward
and carry-back rules for operating losses. It is the exception that the engineering economy study
will consider this level of detail. Rather, the associated negative income tax is considered as a tax
savings for the year. The assumption is that the negative tax will offset taxes for the same year
in other income-producing areas of the corporation.
TABLE 17–2
Year
Suggested Column Headings for Calculation of CFAT
Gross
Income
GI
Operating
Expenses
OE
Investment
and
Salvage
P and S
(1)
(2)
(3)
CFBT
(4) ⫽
(1) ⫹ (2) ⫹ (3)
Depreciation
D
(5)
Taxable
Income
TI
Taxes
CFAT
(6) ⫽
(7) ⫽ (8) ⫽
(1) ⫹ (2) ⫺ (5) Te(6) (4) ⫺ (7)
449
450
After-Tax Economic Analysis
Chapter 17
EXAMPLE 17.2
Wilson Security has received a contract to provide additional security for corporate and government personnel along the Arizona-Mexico border. Wilson plans to purchase listening and detection
equipment for use in the 6-year contract. The equipment is expected to cost $550,000 and have a
resale value of $150,000 after 6 years. Based on the incentive clause in the contract, Wilson estimates that the equipment will increase contract revenue by $200,000 per year and require an additional M&O expense of $90,000 per year. MACRS depreciation allows recovery in 5 years, and
the effective corporate tax rate is 35% per year. Tabulate and plot the CFBT and CFAT series.
Solution
The spreadsheet in Figure 17–1 presents before-tax and after-tax cash flows using the format
of Table 17–2. The functions for year 6 are detailed in row 11. Discussion and sample calculations follow.
CFBT: The operating expenses OE and initial investment P are shown as negative cash flows.
The $150,000 salvage (resale) is a positive cash flow in year 6. CFBT is calculated by Equation
[17.9]. In year 6, for example, when the equipment is sold, the function in row 11 indicates that
CFBT6 ⫽ 200,000 ⫺ 90,000 ⫹ 150,000 ⫽ $260,000
CFAT: Column F for MACRS depreciation, which is determined using the VDB function
over the 6-year period, writes off the entire $550,000 investment. Taxable income, taxes,
and CFAT are calculated, using year 4 as an example.
TI4 ⫽ GI ⫺ OE ⫺ D ⫽ 200,000 ⫺ 90,000 ⫺ 63,360 ⫽ $46,640
Taxes4 ⫽ (0.35) (TI) ⫽ (0.35) (46,640) ⫽ $16,324
CFAT4 ⫽ GI ⫺ OE ⫺ taxes ⫽ 200,000 − 90,000 ⫺ 16,324 ⫽ $93,676
In year 2, MACRS depreciation is large enough to cause TI to be negative ($−66,000). As
mentioned above, the negative tax ($−23,100) is considered a tax savings in year 2, thus
increasing CFAT.
Comment
MACRS depreciates to a salvage value of S ⫽ 0. Later we will learn about a tax implication
due to “recapturing of depreciation” when an asset is sold for an amount larger than zero and
MACRS was applied to fully depreciate the asset to zero.
Figure 17–1
Computation of CFBT and CFAT using MACRS depreciation and Te ⫽ 35%, Example 17.2.
17.3 Effect on Taxes of Different Depreciation
Methods and Recovery Periods
It is important to understand why accelerated depreciation rates give the corporation a tax advantage relative to the straight line method with the same recovery period. Larger rates in earlier
years of the recovery period require less taxes due to the larger reductions in taxable income. The
criterion of minimizing the present worth of taxes is used to demonstrate the tax effect. For the
451
Effect on Taxes of Different Depreciation Methods and Recovery Periods
17.3
recovery period n, choose the depreciation rates that result in the minimum present worth value
for taxes.
tⴝn
PWtax ⴝ
兺 (taxes in year t)(P兾F, i, t)
[17.12]
tⴝ1
This is equivalent to maximizing the present worth of total depreciation PWD.
Compare any two depreciation methods. Assume the following: (1) There is a constant singlevalue tax rate, (2) CFBT exceeds the annual depreciation amount, (3) both methods reduce book
value to the same salvage value, and (4) the same recovery period is used. On the basis of these
assumptions, the following statements are correct:
The total taxes paid are equal for all depreciation methods.
The present worth of taxes is less for accelerated depreciation methods.
As we learned in Chapter 16, MACRS is the prescribed tax depreciation method in the United
States, and the only alternative is MACRS straight line depreciation with an extended recovery
period. The accelerated write-off of MACRS always provides a smaller PWtax compared to less
accelerated methods. If the DDB method were still allowed directly, rather than embedded in
MACRS, DDB would not fare as well as MACRS. This is so because DDB does not reduce the
book value to zero. This is illustrated in Example 17.3.
EXAMPLE 17.3
An after-tax analysis for a new $50,000 machine proposed for a fiber optics manufacturing line
is in process. The CFBT for the machine is estimated at $20,000. If a recovery period of 5 years
applies, use the present worth of taxes criterion, an effective tax rate of 35%, and a return of
8% per year to compare the following: classical straight line, classical DDB, and required
MACRS depreciation. Use a 6-year period for the comparison to accommodate the half-year
convention imposed by MACRS.
Solution
Table 17–3 presents a summary of annual depreciation, taxable income, and taxes for each
method. For classical straight line depreciation with n ⫽ 5, Dt ⫽ $10,000 for 5 years and D6 ⫽
0 (column 3). The CFBT of $20,000 is fully taxed at 35% in year 6.
The classical DDB percentage of d ⫽ 2兾n ⫽ 0.40 is applied for 5 years. The implied salvage
value is $50,000 − 46,112 ⫽ $3888, so not all $50,000 is tax-deductible. The taxes using classical DDB will be $3888(0.35) ⫽ $1361 larger than for the classical SL method.
TABLE 17–3
Comparison of Taxes and Present Worth of Taxes for Different Depreciation Methods
Classical Straight Line
Classical Double Declining
Balance
MACRS
(1)
Year t
(2)
CFBT, $
(3)
D t, $
(4)
TI, $
(5) ⴝ 0.35(4)
Taxes, $
(6)
D t, $
(7)
TI, $
(8) ⴝ 0.35(7)
Taxes, $
(9)
D t, $
(10)
TI, $
1
2
3
4
5
6
⫹20,000
⫹20,000
⫹20,000
⫹20,000
⫹20,000
⫹20,000
10,000
10,000
10,000
10,000
10,000
0
10,000
10,000
10,000
10,000
10,000
20,000
3,500
3,500
3,500
3,500
3,500
7,000
20,000
12,000
7,200
4,320
2,592
0
0
8,000
12,800
15,680
17,408
20,000
0
2,800
4,480
5,488
6,093
7,000
10,000
16,000
9,600
5,760
5,760
2,880
10,000
4,000
10,400
14,240
14,240
17,120
24,500
18,386
46,112
25,861*
18,549
50,000
Totals
PWtax
50,000
*Larger than other values since there is an implied salvage value of $3888 not recovered.
(11) ⴝ 0.35(10)
Taxes, $
3,500
1,400
3,640
4,984
4,984
5,992
24,500
18,162
452
After-Tax Economic Analysis
Chapter 17
25,000
Accumulated taxes, $
20,000
15,000
SL
method
10,000
MACRS
method
DDB method
5,000
0
1
2
3
Year, t
4
5
6
Figure 17–2
Cumulative taxes incurred by different depreciation rates for a
6-year comparison period, Example 17.3.
MACRS writes off $50,000 in 6 years using the rates of Table 16–2. Total taxes are $24,500,
the same as for classical SL depreciation.
The annual taxes (columns 5, 8, and 11) are accumulated year by year in Figure 17–2. Note
the pattern of the curves, especially the lower total taxes relative to the SL model after year 1
for MACRS and in years 1 through 4 for DDB. These higher tax values for SL cause PWtax for
SL depreciation to be larger. The PWtax values at the bottom of Table 17–3 are calculated using
Equation [17.12]. The MACRS PWtax value is the smallest at $18,162.
To compare taxes for different recovery periods, change only assumption 4 at the beginning
of this section to read: The same depreciation method is applied. It can be shown that a shorter
recovery period will offer a tax advantage over a longer period using the criterion to minimize
PWtax. Comparison will indicate that
The total taxes paid are equal for all n values.
The present worth of taxes is less for smaller n values.
This is why corporations want to use the shortest MACRS recovery period allowed for income
tax purposes. Example 17.4 demonstrates these conclusions for classical straight line depreciation, but the conclusions are correct for MACRS or any other tax depreciation method.
EXAMPLE 17.4
Grupo Grande Maquinaría, a diversified manufacturing corporation based in Mexico, maintains parallel records for depreciable assets in its European operations in Germany. This is
common for multinational corporations. One set is for corporate use that reflects the estimated useful life of assets. The second set is for foreign government purposes, such as
depreciation and taxes.
The company just purchased an asset for $90,000 with an estimated useful life of 9 years;
however, a shorter recovery period of 5 years is allowed by German tax law. Demonstrate the
tax advantage for the smaller n if net operating income (NOI) is $30,000 per year, an effective
tax rate of 35% applies, invested money is returning 5% per year after taxes, and classical SL
depreciation is allowed. Neglect the effect of any salvage value.
Solution
Determine the annual TI and taxes by Equations [17.2] through [17.3] and the present worth of
taxes using Equation [17.12] for both n values.
Depreciation Recapture and Capital Gains (Losses)
17.4
Useful life n ⫽ 9 years:
90,000
D ⫽ ——— ⫽ 10,000
9
TI ⫽ 30,000 ⫺ 10,000 ⫽ $20,000 per year
Taxes ⫽ (0.35)(20,000) ⫽ $7000 per year
PWtax ⫽ 7000(P兾A,5%,9) ⫽ $49,755
Total taxes ⫽ (7000)(9) ⫽ $63,000
Recovery period n ⫽ 5 years:
Use the same comparison period of 9 years, but depreciation occurs only during the first 5 years.
90,000
——— ⫽ $18,000
t ⫽ 1 to 5
5
Dt ⫽
0
t ⫽ 6 to 9
{
Taxes ⫽
⫺ 18,000) ⫽ $4200
{ (0.35)(30,000
(0.35)(30,000) ⫽ $10,500
t ⫽ 1 to 5
t ⫽ 6 to 9
PWtax ⫽ 4200(P兾A,5%,5) ⫹ 10,500(P兾A,5%,4)(P兾F,5%,5)
⫽ $47,356
Total taxes ⫽ 4200(5) ⫹ 10,500(4) ⫽ $63,000
A total of $63,000 in taxes is paid in both cases. However, the more rapid write-off for
n ⫽ 5 results in a present worth tax savings of nearly $2400 (49,755 ⫺ 47,356).
17.4 Depreciation Recapture and Capital
Gains (Losses)
All the tax implications discussed here are the result of disposing of a depreciable asset before, at,
or after its recovery period. In an after-tax economic analysis of large investment assets, these tax
effects should be considered. The key is the size of the selling price (or salvage or market value)
relative to the current book value at disposal time and relative to the first cost, which is called the
unadjusted basis B in depreciation terminology. There are three relevant terms.
Depreciation recapture DR, also called ordinary gain, occurs when a depreciable asset is
sold for more than the current book value BVt. As shown in Figure 17–3,
Depreciation recapture ⴝ selling price ⴚ book value
DR ⴝ SP2 ⴚ BVt
If selling price
(SP) at time of
sale is:
The capital gain,
depreciation recapture,
or capital loss is:
SP1
CG
First cost P
or basis B
[17.13]
For an after-tax
study, the tax
effect is:
CG: Taxed at Te
(after CL offset)
plus
DR
DR: Taxed at Te
SP2
DR
Book
value BVt
CL
SP3
CL: Can only offset
CG
Zero, $0
Figure 17–3
Summary of calculations and tax treatment for depreciation recapture and capital gains (losses).
453
454
After-Tax Economic Analysis
Chapter 17
Depreciation recapture is often present in the after-tax study. In the United States, an amount
equal to the estimated salvage value can always be anticipated as DR when the asset is disposed of after the MACRS recovery period. This is correct simply because MACRS depreciates every asset to zero in n ⫹ 1 years. The amount of DR is treated as ordinary taxable income
in the year of asset disposal.
Capital gain CG is an amount incurred when the selling price exceeds its (unadjusted) basis
B. See Figure 17–3.
Capital gain ⫽ selling price ⫺ basis
CG ⫽ SP1 ⫺ B
[17.14]
Since future capital gains are difficult to predict, they are usually not detailed in an after-tax economy study. An exception is for assets that historically increase in value, such as buildings and land.
When the selling price exceeds B, the TI due to the sale is the capital gain plus the depreciation recapture, as shown in Figure 17–3. The DR is now the total amount of depreciation
taken thus far, that is, B ⫺ BV.
Capital loss CL occurs when a depreciable asset is disposed of for less than its current book
value. In Figure 17–3,
Capital loss ⫽ book value ⫺ selling price
CL ⫽ BVt ⫺ SP3
[17.15]
An economic analysis does not commonly account for capital loss, simply because it is not
estimable for a specific alternative. However, an after-tax replacement study should account
for any capital loss if the defender must be traded at a “sacrifice” price. For the purposes of the
economic study, this provides a tax savings in the year of replacement. Use the effective tax
rate to estimate the tax savings. These savings are assumed to be offset elsewhere in the corporation by other income-producing assets that generate taxes.
There are several additional points worth mentioning about capital gains and capital losses for
a corporation, apart from their presence in an economic evaluation.
• U.S. tax law defines capital gains as long-term (items retained for more than 1 year) or short-term.
• Capital gains actually take place for property that is not depreciated or amortized. The term
capital gain correctly applies at sale time to property such as investments (stocks and bonds),
art, jewelry, land, and the like. When a depreciable asset’s selling price is higher than the
original cost (its basis), it is correctly termed an ordinary gain. Corporate tax treatment at this
time is the same for both: taxed as ordinary income. All said, it is common to classify an
expected ordinary gain on a depreciable asset as a capital gain, without altering the economic
decision.
• Capital gains are taxed as ordinary taxable income at the corporation’s regular tax rates.
• Capital losses do not directly reduce annual income taxes because they can only be netted
against capital gains to the maximum extent of the capital gains for the year. The terms used
then are net capital gains (losses).
• When capital losses exceed capital gains, the corporation can take advantage of carry-back
and carry-forward tax laws for the excess, something of value to a finance or tax officer, not
an engineer doing an economic analysis.
• When an asset is disposed of, the tax treatment is referred to as a Section 1231 transaction,
which is the IRS rule section of the same number.
• IRS Publication 544, Sales and Other Dispositions of Assets, may be helpful if gains and
losses are present in a study.
• All these rules apply to corporations. Individual tax rules and rates are different concerning
asset disposition.
If the three additional income and tax elements covered here are incorporated into Equation [17.2], taxable income is defined as
TI ⴝ gross income ⴚ operating expenses ⴚ depreciation
ⴙ depreciation recapture ⴙ net capital gain ⴚ net capital loss
ⴝ GI ⴚ OE ⴚ D ⴙ DR ⴙ CG ⴚ CL
[17.16]
Depreciation Recapture and Capital Gains (Losses)
17.4
In keeping with our perspective of an engineering economy study rather than that of a financial
study, deprecation recapture (i.e., ordinary gain) will be the primary element considered in aftertax evaluations. Only when a capital gain or loss must be included due to the nature of the problem will the calculations involve it.
EXAMPLE 17.5
Biotech, a medical imaging and modeling company, must purchase a bone cell analysis system for
use by a team of bioengineers and mechanical engineers studying bone density in athletes. This
particular part of a 3-year contract with the NBA will provide additional gross income of $100,000
per year. The effective tax rate is 35%. Estimates for two alternatives are summarized below.
Analyzer 1
Analyzer 2
150,000
30,000
5
225,000
10,000
5
Basis B, $
Operating expenses, $ per year
MACRS recovery, years
Answer the following questions, solving by hand and spreadsheet.
(a) The Biotech president, who is very tax conscious, wishes to use a criterion of minimizing
total taxes incurred over the 3 years of the contract. Which analyzer should be purchased?
(b) Assume that 3 years have now passed, and the company is about to sell the analyzer. Using
the same total tax criterion, did either analyzer have an advantage? Assume the selling
price is $130,000 for analyzer 1, or $225,000 for analyzer 2.
Solution by Hand
(a) Table 17–4 details the tax computations. First, the yearly MACRS depreciation is determined. Equation [17.2], TI ⫽ GI − OE − D, is used to calculate TI, after which the 35% tax
rate is applied each year. Taxes for the 3-year period are summed, with no consideration of
the time value of money.
Analyzer 1 tax total: $36,120
Analyzer 2 tax total: $38,430
The two analyzers are very close, but analyzer 1 wins with $2310 less in total taxes.
(b) When the analyzer is sold after 3 years of service, there is a depreciation recapture (DR)
that is taxed at the 35% rate. This tax is in addition to the third-year tax. For each analyzer, account for the DR by Equation [17.13], SP ⫺ BV3; then determine the TI, using
Equation [17.16], TI ⫽ GI − OE − D ⫹ DR. Again, find the total taxes over 3 years, and
select the analyzer with the smaller total.
TABLE 17–4
Year
Gross
Income
GI, $
Comparison of Total Taxes for Two Alternatives, Example 17.5a
Operating
Expenses
OE, $
MACRS
Depreciation
D, $
Basis
B, $
Book
Value
BV, $
Taxable
Income
TI, $
Taxes
at
0.35TI, $
150,000
120,000
72,000
43,200
40,000
22,000
41,200
14,000
7,700
14,420
Analyzer 1
0
1
2
3
150,000
100,000
100,000
100,000
30,000
30,000
30,000
30,000
48,000
28,800
36,120
Analyzer 2
0
1
2
3
225,000
100,000
100,000
100,000
10,000
10,000
10,000
45,000
72,000
43,200
225,000
180,000
108,000
64,800
45,000
18,000
46,800
15,750
6,300
16,380
38,430
455
456
After-Tax Economic Analysis
Chapter 17
Analyzer 1:
DR ⫽ 130,000 ⫺ 43,200 ⫽ $86,800
Year 3 TI ⫽ 100,000 ⫺ 30,000 ⫺ 28,800 ⫹ 86,800 ⫽ $128,000
Year 3 taxes ⫽ (0.35)(128,000) ⫽ $44,800
Total taxes ⫽ 14,000 ⫹ 7700 ⫹ 44,800 ⫽ $66,500
Analyzer 2:
DR ⫽ 225,000 ⫺ 64,800 ⫽ $160,200
Year 3 TI ⫽ 100,000 ⫺ 10,000 ⫺ 43,200 ⫹ 160,200 ⫽ $207,000
Year 3 taxes ⫽ (0.35)(207,000) ⫽ $72,450
Total taxes ⫽ 15,750 ⫹ 6300 ⫹ 72,450 ⫽ $94,500
Now, analyzer 1 has a considerable advantage in total taxes ($94,500 versus $66,500).
Solution by Spreadsheet
(a) Rows 5 through 9 of Figure 17–4 perform the same computations as the hand solution for
analyzer 1 with total taxes of $36,120. Similar analysis in rows 14 to 18 results in total
taxes of $38,430 for analyzer 2, indicating that the company should select analyzer 1, based
on taxes only.
(b) Revised year 3 entries for analyzer 1 in row 10 show the sales price of $130,000, the updated TI of $128,000, and a 3-year tax total of $66,500. The new TI in year 3 has the depreciation recapture incorporated as DR ⫽ selling price ⫺ book value ⫽ SP ⫺ BV3, which
is shown in the cell tag as the last term (D10 ⫺ F10). With a similar updating for analyzer
2 (row 19), total taxes of $94,500 now show a significantly larger tax advantage for analyzer 1 over 3 years.
Comment
Note that no time value of money is considered in these analyses, as we have used in previous
alternative evaluations. In Section 17.5 below we will rely upon PW, AW, and ROR analyses
at an established MARR to make an after-tax decision based upon CFAT values.
⫽ B6⫹C6⫺E6
⫽ B10⫹C10⫺E10⫹(D10⫺F10)
Last term calculates the
depreciation recapture
Total tax advantage
of analyzer 1 over 2
without sale and
with sale in year 3
Figure 17–4
Impact of depreciation recapture on total taxes, Example 17.5.
17.5 After-Tax Evaluation
The required after-tax MARR is established using the market interest rate, the corporation’s effective tax rate, and its weighted average cost of capital. The CFAT estimates are used to compute
the PW or AW at the after-tax MARR. When positive and negative CFAT values are present, a
PW or AW ⬍ 0 indicates the MARR is not met. For a single project or mutually exclusive alternatives, apply the same logic as in Chapters 5 and 6. The guidelines are:
457
After-Tax Evaluation
17.5
One project. PW or AW ⱖ 0, the project is financially viable because the after-tax MARR is
met or exceeded.
Two or more alternatives. Select the alternative with the best (numerically largest) PW or AW
value.
If only cost CFAT amounts are estimated, calculate the after-tax savings generated by the operating expenses and depreciation. Assign a plus sign to each saving and apply the guidelines above.
Remember, the equal-service assumption requires that the PW analysis be performed over the
least common multiple (LCM) of alternative lives. This requirement must be met for every
analysis—before or after taxes.
Since the CFAT estimates usually vary from year to year in an after-tax evaluation, the spreadsheet offers a much speedier analysis than solution by hand.
For AW analysis: Use the PMT function with an embedded NPV function over one life cycle.
The general format is as follows, with the NPV function in italics for the CFAT series.
⫽ ⫺PMT(MARR,n,NPV(MARR,year_1: year_n) ⫹ year_0)
[17.17]
For PW analysis: Obtain the PMT function results first, followed by the PV function taken
over the LCM. (There is an LCM function in Excel.) The cell containing the PMT function result
is entered as the A value. The general format is
⫽ ⫺PV(MARR,LCM_years,PMT_result_cell )
[17.18]
EXAMPLE 17.6
Paul is designing the interior walls of an industrial building. In some places, it is important
to reduce noise transmission across the wall. Two construction options—stucco on metal
lath (S) and bricks (B)—each have about the same transmission loss, approximately
33 decibels. This will reduce noise attenuation costs in adjacent office areas. Paul has estimated the first costs and after-tax savings each year for both designs. (a) Use the CFAT
values and an after-tax MARR of 7% per year to determine which is economically better.
(b) Use a spreadsheet to select the alternative and determine the required first cost for the
plans to break even.
Plan S
Plan B
Year
CFAT, $
Year
CFAT, $
0
1–6
7–10
10
⫺28,800
5,400
2,040
2,792
0
1
2
3
4
5
⫺50,000
14,200
13,300
12,400
11,500
10,600
Solution by Hand
(a) In this example, both AW and PW analyses are shown. Develop the AW relations using
the CFAT values over each plan’s life. Select the larger value.
AWS ⫽ [−28,800 ⫹ 5400(P兾A,7%,6) ⫹ 2040(P兾A,7%,4)(P兾F,7%,6)
⫹ 2792(P兾F,7%,10)](A兾P,7%,10)
⫽ $422
AWB ⫽ [−50,000 ⫹ 14,200(P兾F,7%,1) ⫹ · · · ⫹ 10,600(P兾F,7%,5)](A兾P,7%,5)
⫽ $327
Both plans are financially viable; select plan S because AWS is larger.
For the PW analysis, the LCM is 10 years. Use the AW values and the P兾A factor for the
LCM of 10 years to select stucco on metal lath, plan S.
PWS ⫽ AWS(P兾A,7%,10) ⫽ 422(7.0236) ⫽ $2964
PWB ⫽ AWB(P兾A,7%,10) ⫽ 327(7.0236) ⫽ $2297
ME alternative selection
Equal service
458
After-Tax Economic Analysis
Chapter 17
⫽ ⫺PMT(7%,5,NPV(7%,C4:C8)+C3)
⫽ ⫺PV(7%,10,$C$14)
(a)
(b)
Figure 17–5
(a) After-tax AW and PW analysis, and (b) breakeven first cost using Goal Seek, Example 17.6.
Solution by Spreadsheet
(b) Figure 17–5a, row 14, displays the AW value calculated using the PMT function defined by
Equation [17.17], and row 15 shows the 10-year PW that results from the PV function in
Equation [17.18]. Plan S is chosen by a relatively small margin.
Figure 17–5b shows the Goal Seek template used to equate the AW values and determine the plan B first cost of $⫺49,611 that causes the plans to break even. This is a small
reduction from the $−50,000 first cost initially estimated.
Comment
It is important to remember the minus signs in PMT and PV functions when utilizing them to
obtain the corresponding PW and AW values. If the minus is omitted, the AW and PW values
have the wrong sign and it appears that the plans are not financially viable in that they do not
return at least the after-tax MARR. That would happen in this example.
To utilize the ROR method, apply exactly the same procedures as in Chapter 7 (single project) and Chapter 8 (two or more alternatives) to the CFAT series. A PW or AW relation is developed to estimate the rate of return i* for a project, or ⌬i* for the incremental CFAT between two
alternatives. Multiple roots may exist in the CFAT series, as they can for any cash flow series. For
a single project, set the PW or AW equal to zero and solve for i*.
t⫽n
Present worth:
0⫽
兺 CFAT (P兾F,i*,t)
t
[17.19]
t⫽1
t⫽n
Project evaluation
Annual worth:
0⫽
兺 CFAT (P兾F,i*,t)(A兾P,i*,n)
t
t⫽1
If i* ⱖ after-tax MARR, the project is economically justified.
[17.20]
After-Tax Evaluation
17.5
Spreadsheet solution for i* is faster for most CFAT series. It is performed using the IRR function
with the general format
⫽ IRR(year_0_CFAT:year_n_CFAT)
[17.21]
If the after-tax ROR is important to the analysis, but the details of an after-tax study are not of
interest, the before-tax ROR (or MARR) can be adjusted with the effective tax rate Te by using
the approximating relation
after-tax ROR
Before-tax ROR ⴝ ———————
1 ⴚ Te
[17.22]
For example, assume a company has an effective tax rate of 40% and normally uses an after-tax
MARR of 12% per year for economic analyses that consider taxes explicitly. To approximate the
effect of taxes without performing the details of an after-tax study, the before-tax MARR can be
estimated as
0.12 ⫽ 20% per year
Before-tax MARR ⫽ ————
1 ⫺ 0.40
If the decision concerns the economic viability of a project and the resulting PW or AW value is
close to zero, the details of an after-tax analysis should be developed.
EXAMPLE 17.7
A fiber optics manufacturing company operating in Hong Kong has spent $50,000 for a
5-year-life machine that has a projected $20,000 annual NOI and annual depreciation of
$10,000 for years 1 through 5. The company has a Te of 40%. (a) Determine the after-tax rate
of return. (b) Approximate the before-tax return.
Solution
(a) The CFAT in year 0 is $⫺50,000. For years 1 through 5 there is no capital purchase or sale,
so NOI ⫽ CFBT. (See Equations [17.1] and [17.9].) Determine CFAT.
TI ⫽ NOI ⫺ D ⫽ 20,000 ⫺ 10,000 ⫽ $10,000
Taxes ⫽ Te (TI) ⫽ 0.4(10,000) ⫽ $4000
CFAT ⫽ CFBT ⫺ taxes ⫽ 20,000 ⫺ 4000 ⫽ $16,000
Since the CFAT for years 1 through 5 has the same value, use the P兾A factor in Equation [17.19].
0 ⫽ ⫺50,000 ⫹ 16,000(P兾A,i*,5)
(P兾A,i*,5) ⫽ 3.125
Solution gives i* ⫽ 18.03% as the after-tax rate of return.
(b) Use Equation [17.22] for the before-tax return estimate.
0.1803 ⫽ 0.3005
Before-tax ROR ⫽ ————
1 ⫺ 0.40
(30.05%)
The actual before-tax i* using CFBT ⫽ $20,000 for 5 years is 28.65% from the relation
0 ⫽ ⫺50,000 ⫹ 20,000(P兾A,i*,5)
The tax effect will be slightly overestimated if a MARR of 30.05% is used in a before-tax
analysis.
A rate of return evaluation performed by hand on two or more alternatives utilizes a PW or
AW relation to determine the incremental return ⌬i* of the incremental CFAT series between
two alternatives. Solution by spreadsheet is accomplished using the incremental CFAT values
and the IRR function. The equations and procedures applied are the same as in Chapter 8
(Sections 8.4 through 8.6) for selection from mutually exclusive alternatives using the ROR
method. You should review and understand these sections before proceeding with this section.
459
460
Chapter 17
After-Tax Economic Analysis
From this review, several important facts can be recalled:
Selection guideline: The fundamental rule of incremental ROR evaluation at a stated MARR
is as follows:
Select the one alternative that requires the largest initial investment, provided the extra investment is justified relative to another justified alternative.
Incremental ROR: Incremental analysis must be performed. Overall i* values cannot be
depended upon to select the correct alternative, unlike the PW or AW method at the MARR,
which will always indicate the correct alternative.
Equal-service requirement: Incremental ROR analysis requires that the alternatives be evaluated over equal time periods. The LCM of the two alternative lives must be used to find the
PW or AW of incremental cash flows. (The only exception, mentioned in Section 8.5, occurs
when the AW analysis is performed on actual cash flows, not the increments; then onelife-cycle analysis is acceptable over the respective alternative lives.)
Revenue and cost alternatives: Revenue alternatives (positive and negative cash flows) may
be treated differently from cost alternatives (cost-only cash flow estimates). For revenue alternatives, the overall i* may be used to perform an initial screening. Alternatives with
i* ⬍ MARR can be removed from further evaluation. An i* for cost-only alternatives cannot
be determined, so incremental analysis is required with all alternatives included.
Breakeven ROR
Once the CFAT series are developed, the breakeven ROR can be obtained using a plot of PW
versus i* by solving the PW relation for each alternative over the LCM at several interest rates.
For any after-tax MARR greater than the breakeven ROR, the extra investment is not justified.
The next examples solve CFAT problems using incremental ROR analysis and the breakeven
ROR plot of PW versus i.
EXAMPLE 17.8
In Example 17.6, Paul estimated the CFAT for interior wall materials to reduce sound transmission; plan S is to construct with stucco on metal lath, and plan B is to construct using brick.
Figure 17–5a presented both a PW analysis over 10 years and an AW analysis over the respective lives. Plan S was selected. After reviewing this earlier solution, (a) perform an ROR
evaluation at the after-tax MARR of 7% per year and (b) plot the PW versus ⌬i graph to determine the breakeven ROR.
Solution by Spreadsheet
(a) The LCM is 10 years for the incremental ROR analysis, and plan B requires the extra investment that must be justified. Apply the procedure in Section 8.6 for incremental ROR
analysis. Figure 17–6 shows the estimated CFAT for each alternative and the incremental
CFAT series. Since these are revenue alternatives, the overall ⌬i* is calculated first to ensure that they both make at least the MARR of 7%. Row 14 indicates they do. The IRR
function (cell E14) is applied to the incremental CFAT, indicating that ⌬i* ⫽ 6.35%. Since
this is lower than the MARR, the extra investment in brick walls is not justified. Plan S is
selected, the same as with the PW and AW methods.
(b) The NPV function is used to find the PW of the incremental CFAT series at various i values.
The graph indicates that the breakeven ⌬i* occurs at 6.35%—the same that the IRR function
found. Whenever the after-tax MARR is above 6.35%, as is the case here with MARR ⫽
7%, the extra investment in plan B is not justified.
Comment
Note that the incremental CFAT series has three sign changes. The cumulative series also has
three sign changes (Norstrom’s criterion). Accordingly, there may be multiple ⌬i* values. The
application of the IRR function using the “guess” option finds no other real number roots in the
normal rate of return range.
After-Tax Evaluation
17.5
Breakeven ROR
⬇ 6.35%
Figure 17–6
Incremental evaluation of CFAT and determination of breakeven ROR, Example 17.8.
EXAMPLE 17.9
In Example 17.5 an after-tax analysis of two bone cell analyzers was initiated due to a new
3-year NBA contract. The criterion used to select analyzer 1 was the total taxes for the 3 years.
The complete solution is in Table 17–4 (hand) and Figure 17–4 (spreadsheet).
Continue the spreadsheet analysis by performing an after-tax ROR evaluation, assuming the
analyzers are sold after 3 years for the amounts estimated in Example 17.5: $130,000 for analyzer 1 and $225,000 for analyzer 2. The after-tax MARR is 10% per year.
Solution
A spreadsheet solution is presented here, but a hand solution is equivalent, just slower.
Figure 17–7 is an updated version of the spreadsheet in Figure 17–4 to include the sale of the
analyzers in year 3. The CFAT series (column I) are determined by the relation CFAT ⫽ CFBT ⫺
taxes, with the taxable income determined using Equation [17.16], where DR is included. For
example, in year 3 when analyzer 2 is sold for S ⫽ $225,000, the CFAT calculation is
CFAT3 ⫽ CFBT ⫺ (TI)(Te) ⫽ GI ⫺ OE ⫺ P ⫹ S ⫺ (GI ⫺ OE ⫺ D ⫹ DR)(Te)
The depreciation recapture DR is the amount above the year 3 book value received at sale time.
Using the book value after 3 years (F14),
DR ⫽ selling price ⫺ BV3 ⫽ 225,000 ⫺ 64,800 ⫽ $160,200
Now the CFAT in year 3 for analyzer 2 can be determined.
CFAT3 ⫽ 100,000 ⫺ 10,000 ⫹ 0 ⫹ 225,000
⫺(100,000 ⫺ 10,000 ⫺ 43,200 ⫹ 160,200)(0.35)
⫽ 315,000 ⫺ 207,000(0.35) ⫽ $242,550
The cell tags in row 14 of Figure 17–7 follow this same progression. The incremental CFAT is
calculated in column J, ready for the after-tax incremental ROR analysis.
These are revenue alternatives, so the overall i* values indicate that both CFAT series are
acceptable. The value ⌬i* ⫽ 23.6% (cell J17) also exceeds MARR ⫽ 10%, so analyzer 2 is
selected. This decision applies the ROR method guideline: Select the alternative that requires
the largest, incrementally justified investment.
461
462
After-Tax Economic Analysis
Chapter 17
Overall i*
⫽ B14⫹C14⫺E14⫹(D14⫺F14)
This term
calculates
DR ⫽ $160,200
CFAT calculation
⫽ B14⫹C14⫹D14⫺H14
Incremental i*
⫽ IRR(J11:J14)
Figure 17–7
Incremental ROR analysis of CFAT with depreciation recapture, Example 17.9.
Comment
In Section 8.4, we demonstrated the fallacy of selecting an alternative based solely on the
overall i*, because of the ranking inconsistency problem of the ROR method. The incremental
ROR must be used. The same fact is demonstrated in this example. If the larger i* alternative
is chosen, analyzer 1 is incorrectly selected. When ⌬i* exceeds the MARR, the larger investment is correctly chosen—analyzer 2 in this case. For verification, the PW at 10% is calculated
for each analyzer (column I). Again, analyzer 2 is the winner, based on its larger PW of $93,905.
17.6 After-Tax Replacement Study
When a currently installed asset (the defender) is challenged with possible replacement, the effect of taxes can have an impact upon the decision of the replacement study. The final decision
may not be reversed by taxes, but the difference between before-tax AW values may be significantly different from the after-tax difference. Tax considerations in the year of the replacement
are as follows:
Depreciation recapture or tax savings due to a sizable capital loss are possible, if it is necessary
to trade the defender at a sacrifice price. Additionally, the after-tax replacement study considers
tax-deductible depreciation and operating expenses not accounted for in a before-tax analysis.
The effective tax rate Te is used to estimate the amount of annual taxes (or tax savings) from TI.
The same procedure as the before-tax replacement study in Chapter 11 is applied here, but for
CFAT estimates. The procedure should be thoroughly understood before proceeding. Special attention to Sections 11.3 and 11.5 is recommended.
Example 17.10 presents a solution by hand of an after-tax replacement study using a simplifying assumption of classical SL (straight line) depreciation. Example 17.11 solves the same problem by spreadsheet, but includes the detail of MACRS depreciation. This provides an opportunity
to observe the difference in the AW values between the two depreciation methods.
EXAMPLE 17.10
Midcontinent Power Authority purchased emission control equipment 3 years ago for $600,000.
Management has discovered that it is technologically and legally outdated now. New equipment has been identified. If a market value of $400,000 is offered as the trade-in for the current
equipment, perform a replacement study using (a) a before-tax MARR of 10% per year and
After-Tax Replacement Study
17.6
463
(b) a 7% per year after-tax MARR. Assume an effective tax rate of 34%. As a simplifying assumption, use classical straight line depreciation with S ⫽ 0 for both alternatives.
Defender
Market value, $
First cost, $
Annual cost, $兾year
Recovery period, years
Challenger
400,000
⫺100,000
8 (originally)
⫺1,000,000
⫺15,000
5
Solution
Assume that an ESL (economic service life) analysis has determined the best life values to be
5 more years for the defender and 5 years total for the challenger.
(a) For the before-tax replacement study, find the AW values. The defender AW uses the market value as the first cost, PD ⫽ $⫺400,000.
AWD ⫽ ⫺400,000(A兾P,10%,5) ⫺ 100,000 ⫽ $⫺205,520
AWC ⫽ ⫺1,000,000(A兾P,10%,5) ⫺ 15,000 ⫽ $⫺278,800
Applying step 1 of the replacement study procedure (Section 11.3), we select the better AW
value. The defender is retained now with a plan to keep it for the 5 remaining years. The
defender has a $73,280 lower equivalent annual cost compared to the challenger. This complete solution is included in Table 17–5 (left half) for comparison with the after-tax study.
(b) For the after-tax replacement study, there are no tax effects other than income tax for the
defender. The annual SL depreciation is $75,000, determined when the equipment was
purchased 3 years ago.
Dt ⫽ 600,000兾8 ⫽ $75,000
t ⫽ 1 to 8 years
Table 17–5 shows the TI and taxes at 34%. The taxes are actually tax savings of $59,500
per year, as indicated by the minus sign. (Remember that for tax savings in an economic
TABLE 17–5
Before-Tax and After-Tax Replacement Analyses, Example 17.10
Before Taxes
Defender
Age
Year
Expenses
OE, $
P and S, $
After Taxes
CFBT, $
Depreciation
D, $
Taxable
Income
TI, $
Taxes*
at
0.34TI, $
75,000
75,000
75,000
75,000
75,000
AW at 7%
⫺175,000
⫺175,000
⫺175,000
⫺175,000
⫺175,000
⫺59,500
⫺59,500
⫺59,500
⫺59,500
⫺59,500
⫹25,000†
⫺215,000
⫺215,000
⫺215,000
−215,000
⫺215,000‡
8,500
⫺73,100
⫺73,100
⫺73,100
⫺73,100
⫺73,100
CFAT, $
Defender
3
4
5
6
7
8
AW at 10%
0
1
2
3
4
5
⫺400,000
⫺100,000
⫺100,000
⫺100,000
⫺100,000
⫺100,000
0
⫺400,000
⫺100,000
⫺100,000
⫺100,000
⫺100,000
⫺100,000
⫺205,520
⫺400,000
⫺40,500
⫺40,500
⫺40,500
⫺40,500
⫺40,500
⫺138,056
Challenger
0
1
2
3
4
5
AW at 10%
*
⫺1,000,000
⫺15,000
⫺15,000
⫺15,000
⫺15,000
⫺15,000
0
⫺1,000,000
⫺15,000
⫺15,000
⫺15,000
⫺15,000
⫺15,000
⫺278,800
Minus sign indicates a tax savings for the year.
Depreciation recapture on defender trade-in.
‡
Assumes challenger’s salvage actually realized is S ⫽ 0; no tax.
†
200,000
200,000
200,000
200,000
200,000
AW at 7%
⫺1,008,500
⫹58,100
⫹58,100
⫹58,100
⫹58,100
⫹58,100
⫺187,863
464
After-Tax Economic Analysis
Chapter 17
analysis it is assumed that there is positive taxable income elsewhere in the corporation to
offset the saving.) Since only costs are estimated, the annual CFAT is negative, but the
$59,500 tax savings has reduced it. The CFAT and AW at 7% per year are
CFAT ⫽ CFBT ⫺ taxes ⫽ ⫺100,000 ⫺ (⫺59,500) ⫽ $⫺40,500
AWD ⫽ ⫺400,000(A兾P,7%,5) ⫺ 40,500 ⫽ $⫺138,056
For the challenger, depreciation recapture on the defender occurs when it is replaced
because the trade-in amount of $400,000 is larger than the current book value. In year
0 for the challenger, Table 17–5 includes the following computations to arrive at a tax
of $8500.
Defender book value, year 3:
Depreciation recapture:
Taxes on trade-in, year 0:
BV3 ⫽ 600,000 ⫺ 3(75,000) ⫽ $375,000
DR3 ⫽ TI ⫽ 400,000 ⫺ 375,000 ⫽ $25,000
Taxes ⫽ 0.34(25,000) ⫽ $8500
The SL depreciation is $1,000,000兾5 ⫽ $200,000 per year. This results in tax saving and
CFAT as follows:
Taxes ⫽ (⫺15,000 ⫺ 200,000)(0.34) ⫽ $⫺73,100
CFAT ⫽ CFBT ⫺ taxes ⫽ ⫺15,000 ⫺ (⫺73,100) ⫽ $⫹58,100
In year 5, it is assumed the challenger is sold for $0; there is no depreciation recapture. The
AW for the challenger at the 7% after-tax MARR is
AWC ⫽ ⫺1,008,500(A兾P,7%,5) ⫹ 58,100 ⫽ $⫺187,863
The defender is again selected; however, the equivalent annual advantage has decreased
from $73,280 before taxes to $49,807 after taxes.
Conclusion: By either analysis, retain the defender now and plan to keep it for 5 more
years. Additionally, plan to evaluate the estimates for both alternatives 1 year hence. If and
when cash flow estimates change significantly, perform another replacement analysis.
Comment
If the market value (trade-in) had been less than the current defender book value of $375,000, a
capital loss, rather than depreciation recapture, would occur in year 0. The resulting tax savings
would decrease the CFAT (which is to reduce costs if CFAT is negative) of the challenger. For
example, a trade-in amount of $350,000 would result in a TI of $350,000 ⫺ 375,000 ⫽ $⫺25,000
and a tax savings of $⫺8500 in year 0. The CFAT is then $⫺1,000,000 ⫺ (⫺8500) ⫽
$⫺991,500.
EXAMPLE 17.11
Repeat the after-tax replacement study of Example 17.10b using 7-year MACRS depreciation for the defender and 5-year MACRS depreciation for the challenger. Assume either
asset is sold after 5 years for exactly its book value. Determine if the answers are significantly different from those obtained when the simplifying assumption of classical SL depreciation was made.
Solution
Figure 17–8 shows the complete analysis. MACRS requires more computation than SL depreciation, but this effort is easily reduced by the use of a spreadsheet. Again the defender is selected for retention, but now by an advantage of $44,142 annually. This compares to the
$49,807 advantage using classical SL depreciation and the $73,280 before-tax advantage of the
defender. Therefore, taxes and MACRS have reduced the defender’s economic advantage, but
not enough to reverse the decision to retain it.
Several other differences in the results between SL and MACRS depreciation are worth
noting. There is depreciation recapture in year 0 of the challenger due to trade-in of the defender at $400,000, a value larger than the book value of the 3-year-old defender. This amount,
465
After-Tax Value-Added Analysis
17.7
Depreciation recapture
⫽⫺C5 ⫺ F11
Challenger sales price
⫽ B15 ⫺ F24
Figure 17–8
After-tax replacement study with MACRS depreciation and depreciation recapture, Example 17.11.
$137,620 (cell G18), is treated as ordinary taxable income. The calculations for the DR and
associated tax, by hand, are as follows:
BV3 ⫽ first cost ⫺ MACRS depreciation for 3 years
⫽ total MACRS depreciation for years 4 through 8
⫽ $262,380
(cell F11)
DR ⫽ TI0 ⫽ trade-in ⫺ BV3
⫽ 400,000 − 262,380 ⫽ $137,620 (cell G18)
Taxes ⫽ (0.34)(137,620) ⫽ $46,791
(cell H18)
See the cell tags and table notes that duplicate this logic.
The assumption that the challenger is sold after 5 years at its book value implies a positive
cash flow in year 5. The entry $57,600 (C23) reflects this assumption, since the forgone
MACRS depreciation in year 6 would be 1,000,000(0.0576) ⫽ $57,600. The spreadsheet
relation ⫽ B15 ⫺ F24 determines this value using the accumulated depreciation in F24.
[Note: If the salvage S ⫽ 0 is anticipated after 5 years, then a capital loss of $57,600 will be
incurred. This implies an additional tax saving of 57,600(0.34) ⫽ $19,584 in year 5. Conversely, if the salvage value exceeds the book value, a depreciation recapture and associated
tax should be estimated.]
17.7 After-Tax Value-Added Analysis
When a person or company is willing to pay more for an item, it is likely that some processing
has been performed on an earlier version of the item to make it more valuable now to the purchaser. This is value added.
Value added is a term used to indicate that a product or service has added worth from the
perspective of a consumer, owner, investor, or purchaser. It is common to leverage valueadding activities on a product or service.
Value added
466
Chapter 17
After-Tax Economic Analysis
For an example of highly leveraged value-added activities, consider onions that are grown and
sold at the farm level for cents per pound. They may be purchased by the shopper in a store at
50 cents to $1.25 per pound. But when onions are cut and coated with a special batter, they may
be fried in hot oil and sold as onion rings for several dollars per pound. Thus, from the perspective of the consumer, there has been a large amount of value added by the processing from raw
onions in the ground into onion rings sold at a restaurant or fast-food shop.
The value-added measure was briefly introduced in conjunction with AW analysis before
taxes. When value-added analysis is performed after taxes, the approach is somewhat different
from that of CFAT analysis developed previously in this chapter. However, as shown below,
The decision about an alternative will be the same for both the value-added and CFAT methods,
because the AW of economic value-added estimates is the same as the AW of CFAT estimates.
Value-added analysis starts with Equation [17.4], net operating profit after taxes (NOPAT), which
includes the depreciation for year 1 through year n. Depreciation D is included in that TI ⫽ GI ⫺
OE − D. This is different from CFAT, where the depreciation has been specifically removed so that
only actual cash flow estimates are used for year 0 through year n.
The term economic value added (EVA) indicates the monetary worth added by an alternative
to the corporation’s bottom line. (The term EVA is a trademark of Stern Stewart & Co.) The technique discussed below was first publicized in several articles1 in the mid-1990s, and it has since
become very popular as a means to evaluate the ability of a corporation to increase its economic
worth, especially from the shareholders’ viewpoint.
The annual EVA is the amount of NOPAT remaining on corporate books after removing the cost
of invested capital during the year. That is, EVA indicates the project’s contribution to the net
profit of the corporation after taxes.
The cost of invested capital is the after-tax rate of return (usually the MARR value) multiplied by the book value of the asset during the year. This is the interest incurred by the current
level of capital invested in the asset. (If different tax and book depreciation methods are used, the
book depreciation value is used here, because it more closely represents the remaining capital
invested in the asset from the corporation’s perspective.) Computationally,
EVA ⴝ NOPAT ⴚ cost of invested capital
ⴝ NOPAT ⴚ (after-tax interest rate)(book value in year t ⴚ 1)
ⴝ TI(1 ⴚ Te) ⴚ (i)(BVtⴚ1)
[17.23]
Since both TI and the book value consider depreciation, EVA is a measure of worth that mingles
actual cash flow with noncash flows to determine the estimated financial worth contribution to
the corporation. This financial worth is the amount used in public documents of the corporation
(balance sheet, income statement, stock reports, etc.). Because corporations want to present the
largest value possible to the stockholders and other owners, the EVA method is often more appealing than the AW method from the financial perspective.
The result of an EVA analysis is a series of annual EVA estimates. Two or more alternatives
are compared by calculating the AW of EVA estimates and selecting the alternative with the
larger AW value. If only one project is evaluated, AW ⬎ 0 means the after-tax MARR is exceeded, thus making the project value-adding.
Sullivan and Needy2 have demonstrated that the AW of EVA and the AW of CFAT are identical
in amount. Thus, either method can be used to make a decision. The annual EVA estimates indicate added worth to the corporation generated by the alternative, while the annual CFAT estimates describe how cash will flow. This comparison is made in Example 17.12.
1
A. Blair, “EVA Fever,” Management Today, Jan. 1997, pp. 42–45; W. Freedman, “How Do You Add Up?”
Chemical Week, Oct. 9, 1996, pp. 31–34.
2
W. G. Sullivan and K. L. Needy, “Determination of Economic Value Added for a Proposed Investment in
New Manufacturing.” The Engineering Economist, vol. 45, no. 2 (2000), pp. 166–181.
After-Tax Value-Added Analysis
17.7
EXAMPLE 17.12
Biotechnics Engineering has developed two mutually exclusive plans for investing in new
capital equipment with the expectation of increased revenue from its medical diagnostic services to cancer patients. The estimates are summarized below. (a) Use classical straight line
depreciation, an after-tax MARR of 12%, and an effective tax rate of 40% to perform two annual worth after-tax analyses: EVA and CFAT. (b) Explain the fundamental difference between
the results of the two analyses.
Plan A
Plan B
Initial investment, $
Gross income ⫺ expenses, $
⫺500,000
170,000 per year
Estimated life, years
Salvage value
None
⫺1,200,000
600,000 in year 1, decreasing by 100,000
per year thereafter
4
None
4
Solution by Spreadsheet
(a) Refer to the spreadsheet and function cells (row 22) in Figure 17–9.
EVA evaluation: All the necessary information for EVA estimation is determined in
columns B through G. The net operating profit after taxes (NOPAT) in column H is calculated by Equation [17.4], TI ⫺ taxes. The book values (column E) are used to determine the
cost of invested capital in column I, using the second term in Equation [17.23], that is,
i(BVt⫺1), where i is the 12% after-tax MARR. This represents the amount of interest at
12% per year, after taxes, for the currently invested capital as reflected by the book value
at the beginning of the year. The EVA estimate is the sum of columns H and I for years 1
through 4. Notice there is no EVA estimate for year 0, since NOPAT and the cost of invested
capital are estimated for years 1 through n. Finally, the larger AW of the EVA value is selected, which indicates that plan B is better and that plan A does not make the 12% return.
CFAT evaluation: As shown in function row 22 (plan B for year 3), CFAT estimates
(column K) are calculated as (GI ⫺ OE) ⫺ P ⫺ taxes. The AW of CFAT again concludes
that plan B is better and that plan A does not return the after-tax MARR of 12% (K10).
(b) What is the fundamental difference between the EVA and CFAT series in columns J and K?
They are clearly equivalent from the time value of money perspective since the AW values
are numerically the same. To answer the question, consider plan A, which has a constant
CFAT estimate of $152,000 per year. To obtain the AW of EVA estimate of $⫺12,617 for
Figure 17–9
Comparison of two plans using EVA and CFAT analyses, Example 17.12.
467
468
Chapter 17
After-Tax Economic Analysis
years 1 through 4, the initial investment of $500,000 is distributed over the 4-year life using
the A兾P factor at 12%. That is, an equivalent amount of $500,000(A兾P,12%,4) ⫽ $164,617
is “charged” against the cash inflows in each of years 1 through 4. In effect, the yearly
CFAT is reduced by this charge.
CFAT ⫺ (initial investment)(A兾P,12%,4) ⫽ 152,000 − 500,000(A兾P,12%,4)
152,000 ⫺ 164,617 ⫽ $⫺12,617
⫽ AW of EVA
This is the AW value for both series, demonstrating that the two methods are economically
equivalent. However, the EVA method indicates an alternative’s yearly estimated contribution to the value of the corporation, whereas the CFAT method estimates the actual cash
flows to the corporation. This is why the EVA method is often more popular than the cash
flow method with corporate executives.
Comment
The calculation P(A兾P,i,n) ⫽ $500,000(A兾P,12%,4) is exactly the same as the capital recovery
in Equation [6.3], assuming an estimated salvage value of zero. Thus, the cost of invested
capital for EVA is the same as the capital recovery discussed in Chapter 6. This further demonstrates why the AW method is economically equivalent to the EVA evaluation.
17.8 After-Tax Analysis for International Projects
Primary questions to be answered prior to performing a corporate-based after-tax analysis for
international settings revolve around tax-deductible allowances—depreciation, business expenses, capital asset evaluation—and the effective tax rate needed for Equation [17.7], taxes ⫽
(Te)(TI). As discussed in Chapter 16, most governments of the world recognize and use the
straight line (SL) and declining balance (DB) methods of depreciation with some variations to
determine the annual tax-deductible allowance. Expense deductions vary widely from country to
country. By way of example, some of these are summarized here.
Canada
Depreciation: This is deductible and is normally based on DB calculations, although SL may be
used. An equivalent of the half-year convention is applied in the first year of ownership. The
annual tax-deductible allowance is termed capital cost allowance (CCA). As in the U.S. system, recovery rates are standardized, so the depreciation amount does not necessarily reflect
the useful life of an asset.
Class and CCA rate: Asset classes are defined and annual depreciation rates are specified by
class. No specific recovery period (life) is identified, in part because assets of a particular class
are grouped together and the annual CCA is determined for the entire class, not individual assets. There are some 44 classes, and CCA rates vary from 4% per year (the equivalent of a
25-year-life asset) for buildings (class 1) to 100% (1-year life) for applications software, chinaware, dies, etc. (class 12). Most rates are in the range of 10% to 30% per year.
Expenses: Business expenses are deductible in calculating TI. Expenses related to capital investments are not deductible, since they are accommodated through the CCA.
Internet: Further details are available on the Revenue Canada website at www.cra.gc.ca in the
Forms and Publications section.
China (PRC)
Depreciation: Officially, SL is the primary method of tax-deductible depreciation; however, assets employed in selected industries or types of assets can utilize accelerated DB or SYD
(sum-of-years-digits) depreciation, when approved by the government. The selected industries and assets can change over time; currently favored industries serve areas such as technology and oil exploration, and equipment subjected to large vibrations during normal usage is
allowed accelerated depreciation.
17.8
After-Tax Analysis for International Projects
Recovery period: Standardized recovery periods are published that vary from 3 years (electronic
equipment) to 10 years (aircraft, machinery, and other production equipment) to 20 years (buildings). Shortened periods can be approved, but the minimum recovery period cannot be less
than 60% for the normal period defined by current tax law.
Expenses: Business expenses are deductible with some limitations and some special incentives.
Limitations are placed, for example, on advertising expense deductions (15% of sales for the
year). Incentives are generous in some cases; for example, 150% of actual expenses is deductible for new technology and new product R&D activities.
Internet: Summary information for China and several other countries is available at
www.worldwide-tax.com.
Mexico
Depreciation: This is a fully deductible allowance for calculating TI. The SL method is applied
with an index for inflation considered each year. For some asset types, an immediate deduction of a percentage of the first cost is allowed. (This is a close equivalent to the Section 179
Deduction in the United States.)
Class and rates: Asset types are identified, though not as specifically defined as in some countries. Major classes are identified, and annual recovery rates vary from 5% for buildings (the
equivalent of a 20-year life) to 100% for environmental machinery. Most rates range from
10% to 30% per year.
Profit tax: The income tax is levied on profits on income earned from carrying on business in
Mexico. Most business expenses are deductible. Corporate income is taxed only once, at the
federal level; no state-level taxes are imposed.
Tax on Net Assets (TNA): Under some conditions, a tax of 1.8% of the average value of assets
located in Mexico is paid annually in addition to income taxes.
Internet: The best information is via websites for companies that assist international corporations located in Mexico. One example is PriceWaterhouseCoopers at www.pwcglobal
.com/mx/eng.
Japan
Depreciation: This is fully deductible and based on classical SL and DB methods. To encourage
capital investment for long-term economic growth, in 2007 Japan allowed assets to be depreciated using a 250% DB method; that is, a significantly increased accelerated rate is allowed
compared to historical Japanese allowances. Switching to classical SL depreciation must take
place in the year that the accelerated rate amount falls below the corresponding SL amount.
Class and life: A statutory useful life ranging from 4 to 24 years, with a 50-year life for reinforced
concrete buildings, is specified.
Expenses: Business expenses are deductible in calculating TI.
Internet: Further details are available on the Japanese Ministry of Finance website at
www.mof.go.jp.
The effective tax rate varies considerably among countries. Some countries levy taxes only
at the federal level, while others impose taxes at several levels of government (federal, state or
provincial, prefecture, county, and city). A summary of international corporate average tax rates
is presented in Table 17–6 for a wide range of industrialized countries. These include income
taxes at all reported levels of government within each country; however, other types of taxes
may be imposed by a particular government. Although these average rates of taxation will vary
from year to year, especially as tax reform is enacted, it can be surmised that most corporations
face effective rates of about 20% to 40% of taxable income. A close examination of international rates shows that they have decreased significantly over the last decade. In fact, the KPMG
report noted in Table 17–6 indicates that the global average corporate tax rate on TI has decreased from 32.7% (1999) to 25.5% (2009). This has encouraged corporate investment and
business expansion within country borders and helped soften the massive economic downturn
experienced in recent years.
469
470
After-Tax Economic Analysis
Chapter 17
TABLE 17–6
Summary of International Corporate Average Tax Rates
Tax Rate Levied on
Taxable Income, %
ⱖ 40
35 to ⬍ 40
32 to ⬍ 35
28 to ⬍ 32
24 to ⬍ 28
20 to ⬍ 24
⬍ 20
For These Countries
United States, Japan
South Africa, Pakistan
Canada, France, India
Australia, United Kingdom, New Zealand, Spain, Germany,
Mexico, Indonesia
China, Taiwan, Republic of Korea
Russia, Turkey, Saudi Arabia
Singapore, Hong Kong, Chile, Ireland, Iceland, Hungary
Sources: Extracted from KPMG’s Corporate and Indirect Tax Survey 2009 (www.kpmg.
com兾Global兾en兾IssuesAndInsights兾Pages兾default.aspx) and from country websites on corporate taxation.
One of the prime ways that governments have been able to reduce corporate tax rates is by
shifting to indirect taxes on goods and services for additional tax revenue. These taxes are usually in the form of a value-added tax (VAT), goods and services tax (GST), and taxes on products imported from outside its borders. As corporate tax rates have declined, in general the indirect tax rates have increased. This has been especially true during the first decade of the 21st
century. However, the worldwide economic slump experienced in recent years has made it necessary for governments around the world to be more cautious of how they tax corporations and
maintain a reasonable balance between regular tax rates (as listed in Table 17–6) and indirect tax
rates. The VAT system is explained now.
17.9 Value-Added Tax
A value-added tax (VAT) has facetiously been called a sales tax on steroids, because the VAT
rates on some items in some countries that impose a VAT can be as high as 90%. It is also called
a GST (goods and service tax).
A value-added tax is an indirect tax; that is, it is a tax on goods and services rather than on
people or corporations. It differs from a sales tax in two ways: (1) when it is charged and (2) who
pays (as explained below). A specific percentage, say 10%, is a charge added to the price of the
item and paid by the buyer. The seller then sends this 10% VAT to the taxing entity, usually a
government unit. This process of 10% VAT continues every time the item is resold—as purchased or in a modified form—thus the term value-added.
A value-added tax is commonly used throughout the world. In some countries, VAT is used in
lieu of business or individual income taxes. There is no VAT system in the United States yet. In
fact, the United States is the only major industrialized country in the world that does not have a
VAT system, though other forms of indirect taxation are used liberally. There is mounting evidence, however, that a VAT兾GST system will be necessary in the United States in the near future,
but not without a great amount of political discord.
A sales tax is used by the U.S. government, by nearly all of the states, and by many local entities. A sales tax is charged on goods and services at the time the goods and services reach the end
user or consumer. That is, businesses do not pay a sales tax on raw material, unfinished goods,
or items they purchase that will ultimately be sold to an end user; only the end user pays the sales
tax. Businesses do pay a sales tax on items for which they are the end user. Total sales tax percentages imposed by multiple government levels can range from 5% to 11%, sometimes larger
on specific items. For example, when Home Depot (HD) purchases microwave ovens from JennAir, Inc., Home Depot does not pay a sales tax on the microwave ovens because they will be sold
to HD customers who will pay the sales tax. On the other hand, if Home Depot purchases a forklift from Caterpillar for loading and unloading merchandise in one of its stores, HD will pay the
sales tax on the forklift, since HD is the end user. Thus, a sales tax is paid only one time, and that
17.9
Value-Added Tax
time is when the goods or services are purchased by the end user. The sales tax is the responsibility of the merchant to collect and remand to the taxing entity.
A value-added tax (VAT), on the other hand, is charged to the buyer at purchase time, whether
the buyer is a business or end user. The seller sends the collected VAT to the taxing entity. If the
buyer subsequently resells the goods to another buyer (as is or a modification of it), another VAT
is collected by the seller. Now, this second seller will send to the taxing entity an amount that is
equal to the total tax collected minus the amount of VAT already paid.
As an illustration, assume the U.S. government charged a 10% VAT. Here is how the VAT
might work.
Northshore Mining Corporation of Babbitt, Minnesota, sells $100,000 worth of iron ore to
Westfall Steel. As part of the price, Northshore collects $110,000, that is, $100,000 ⫹ 0.1 ⫻
100,000, from Westfall Steel. Northshore remits the $10,000 VAT to the U.S. Treasury.
Westfall Steel sells all of the steel it made from the iron ore at a price of $300,000 to General
Electric (GE). Westfall collects $330,000 from GE and then sends $20,000 to the U.S. Treasury, that is, $30,000 it collected in VAT from GE minus $10,000 it paid in taxes to Northshore
Mining.
GE uses the steel to make refrigerators that it sells for $700,000 to retailers, such as Home
Depot, Lowe’s, and others. GE collects $770,000 and then remits $40,000 to the U.S. Treasury, that is, $70,000 it collected in taxes from retailers minus $30,000 it paid in VAT to
Westfall Steel.
If GE purchased machines, tools, or other items to make the refrigerators during this accounting period, and it paid taxes on those items, the taxes paid would also be deducted from the
VAT that GE collected before sending the money to the U.S. Treasury. For example, if GE
paid $5000 in taxes on motors it purchased for the refrigerators, the amount GE would remit
to the U.S. Treasury would be $35,000 (that is, $70,000 it collected from retailers minus
$30,000 it paid in taxes to Westfall Steel minus $5000 it paid in taxes on the motors).
The retailers sell the refrigerators for $950,000 and collect $95,000 in taxes from end users—
consumers. The retailers remit $25,000 to the U.S. Treasury (that is, $95,000 they collected
minus $70,000 they paid previously).
Through this process, the U.S Treasury has received $10,000 from Northshore, $20,000
from Westfall Steel, $35,000 from GE, $5000 from the supplier of the motors, and $25,000
from the retailers, for a total of $95,000. This is 10% of the final sales price of $950,000. The
VAT money was deposited into the Treasury at several different times from several different
companies.
The taxes that a company pays for materials or items it purchases in order to produce goods
or services that will subsequently be sold to another business or end user are called input taxes,
and the company is able to recover them when it collects the VAT from the sale of its products.
The taxes that a company collects are called output taxes, and these are forwarded to the taxing
entity, less the amount of input taxes the company paid. Hence, the businesses incur no taxes
themselves, the same as with a sales tax.
Several dimensions of a VAT distinguish it from a sales tax or corporate income taxes. Some
are as follow:
• Value-added taxes are taxes on consumption, not production or taxable income.
• The end user pays all of the value-added taxes, but VATs are not as obvious as a sales tax that
is added to the price of the item at the time of purchase (and displayed on the receipt). Therefore, VAT taxing entities encounter less resistance from consumers.
• Value-added taxes are generally considerably higher than sales taxes, with the average
European VAT rate at 20% and the worldwide average at 15.25%.
• The VAT is essentially a “sales tax,” but it is charged at each stage of the product development
process instead of when the product is sold.
• With VAT, there is less evasion of taxes because it is harder for multiple entities to evade collecting and paying the taxes than it is for one entity to do so.
• VAT rates vary from country to country and from category to category. For example, in some
countries, food has a 0% VAT rate, while aviation fuel is taxed at 32%.
471
472
After-Tax Economic Analysis
Chapter 17
EXAMPLE 17.13
Tata Motors is a major player in automobile manufacturing in India. It has three different
manufacturing units that specialize in manufacturing different transportation-related products, such as trucks, engines and axles, commercial vehicles, utility vehicles, and passenger cars. The company buys products that fall under different sections of the Indian government VAT tax code, and therefore the products have different VAT rates. In one
particular accounting period, Tata had invoices from four different suppliers (vendors A,
B, C, and D) in the respective amounts of $1.5 million, $3.8 million, $1.1 million, and
$900,000. The products Tata purchased were subject to VAT rates of 4%, 4%, 12.5%, and
22%, respectively.
(a) How much total VAT did Tata pay to its vendors?
(b) Assume that Tata’s products have a VAT rate of 12.5%. If Tata’s sales during the period
were $9.2 million, how much VAT did the Indian Treasury receive from Tata?
Solution
(a) Let X equal the product price before the VAT is added. Solve for X and then subtract it
from the purchase amount to determine the VAT charged by each vendor. Table 17–7
shows the VAT that Tata paid its four vendors. An example computation for vendor A is
as follows:
X ⫹ 0.04X ⫽ 1,500,000
1.04X ⫽ 1,500,000
X ⫽ $1,442,308
VATA ⫽ 1,500,000 ⫺ 1,442,308
⫽ $57,692
Total VAT paid ⫽ 57,692 ⫹ 146,154 ⫹ 122,222 ⫹ 162,295
⫽ $488,363
(b) Total from Tata ⫽ total VAT ⫺ VAT paid by vendors
⫽ 9,200,000(0.125) ⫺ 488,363
⫽ $661,637
TABLE 17–7
VAT Computation, Example 17.13
Vendor
Purchases, $
VAT Rate, %
Price before VAT, X, $
VAT, $
A
B
C
D
1,500,000
3,800,000
1,100,000
900,000
4.0
4.0
12.5
22.0
1,442,308
3,653,846
977,778
737,705
57,692
146,154
122,222
162,295
Total
488,363
CHAPTER SUMMARY
After-tax analysis does not usually change the decision to select one alternative over another;
however, it does offer a much clearer estimate of the monetary impact of taxes. After-tax PW,
AW, and ROR evaluations of one or more alternatives are performed on the CFAT series using
exactly the same procedures as in previous chapters.
Income tax rates for U.S. corporations and individual taxpayers are graduated or progressive—
higher taxable incomes pay higher income taxes. A single-value, effective tax rate Te is usually
applied in an after-tax economic analysis. Taxes are reduced because of tax deductible items,
such as depreciation and operating expenses. Because depreciation is a noncash flow, it is important to consider depreciation only in TI computations, and not directly in the CFBT and
Problems
473
CFAT calculations. Accordingly, key general cash flow after-tax relations for each year are as
follows:
NOI ⫽ gross income ⫺ expenses
TI ⫽ gross income ⫺ operating expenses ⫺ depreciation ⫹ depreciation recapture
CFBT ⫽ gross income ⫺ operating expenses ⫺ initial investment ⫹ salvage value
CFAT ⫽ CFBT ⫺ taxes ⫽ CFBT ⫺ (Te)(TI)
If an alternative’s estimated contribution to corporate financial worth is the economic measure,
the economic value added (EVA) should be determined. Unlike CFAT, the EVA includes the effect of depreciation. The equivalent annual worths of CFAT and EVA estimates are the same
numerically, because they interpret the annual cost of the capital investment in different, but
equivalent manners when the time value of money is taken into account.
In a replacement study, the tax impact of depreciation recapture, which may occur when the
defender is traded for the challenger, is accounted for in an after-tax analysis. The replacement
study procedure of Chapter 11 is applied. The tax analysis may not reverse the decision to replace
or retain the defender, but the effect of taxes will likely reduce (possibly by a significant amount)
the economic advantage of one alternative over the other.
International corporate tax rates have steadily decreased, but indirect taxes, such as valueadded tax (VAT), have increased. The mechanism of a VAT is explained and compared to a sales
tax. The United States currently has no VAT.
PROBLEMS
Basic Tax Computations
17.1 (a)
(b)
Define the following tax terms: graduated tax
rates, marginal tax rate, and indexing.
Describe a fundamental difference between
each of the following terms: net operating income (NOI), taxable income (TI), and net
operating profit after taxes (NOPAT).
17.2 Determine the average tax rate for a corporation
that has taxable income of (a) $150,000 and
(b) $12,000,000.
17.3 Determine the single-value effective tax rate for a
corporation that has a federal tax rate of 35% and a
state tax rate of 5%.
17.4 Identify the primary term described by each event
below: gross income, depreciation, operating expense, taxable income, income tax, or net operating profit after taxes.
(a) A new machine had a first-year write-off of
$10,500.
(b) A public corporation estimates that it will report a $−750,000 net profit on its annual income statement.
(c) An asset with a book value of $8000 was retired and sold for $8450.
(d) An over-the-counter software system will
generate $420,000 in revenue this quarter.
(e) An asset with a MACRS recovery period of 7
years has been owned for 10 years. It was
just sold for $2750.
(f )
(g)
(h)
(i)
The cost of goods sold in the past year was
$3,680,200.
A convenience store collected $33,550 in lottery ticket sales last month. Based on winners holding these tickets, a rebate of $350
was sent to the manager.
An asset with a first cost of $65,000 was utilized on a new product line to increase sales
by $150,000.
The cost to maintain equipment during the
past year was $641,000.
17.5 Two companies have the following values on their
annual tax returns.
Sales revenue, $
Interest revenue, $
Expenses, $
Depreciation, $
(a)
(b)
(c)
Company 1
Company 2
1,500,000
31,000
⫺754,000
48,000
820,000
25,000
⫺591,000
18,000
Calculate the federal income tax for the year
for each company.
Determine the percent of sales revenue each
company will pay in federal income tax.
Estimate the taxes using an effective rate of
34% of the entire TI. Determine the percentage error made relative to the exact taxes in
part (a).
17.6 Last year, one division of Hagauer.com, a dot-com
sports industry service firm that provides real-time
474
After-Tax Economic Analysis
Chapter 17
analysis of mechanical stress due to athlete injury,
had $250,000 in taxable income. This year, TI is
estimated to be $600,000. Calculate the federal income taxes and answer the following.
(a) What was the average federal tax rate paid
last year?
(b) What is the marginal federal tax rate on the
additional TI this year?
(c) What will be the average federal tax rate this
year?
(d) What will be the NOPAT on just the additional $350,000 in taxable income?
17.7 Yamachi and Nadler of Hawaii has a gross income
of $7.5 million for the year. Depreciation and operating expenses total $4.3 million. The combined
state and local tax rate is 7.2%. If an effective federal rate of 35% applies, estimate the income taxes
using the effective tax rate equation.
17.8 Workman Tools reported a TI of $90,000 last year.
If the state income tax rate is 7%, determine the
(a) average federal tax rate, (b) overall effective
tax rate, (c) total taxes to be paid based on the effective tax rate, and (d) total taxes paid to the state
and paid to the federal government.
17.9 The taxable income for a motorcycle sales and repair business is estimated to be $150,000 this year.
A single-value tax rate of 39% is used. A new engine diagnostics system will cost $40,000 and
have an average annual depreciation of $8000. The
equipment will increase gross income by an estimated $9000 and expenses by $2000 for the year.
Compute the expected change in income taxes for
the year if the new system is purchased.
17.10 C.F. Jordon Construction Services has operated for
the last 26 years in a northern U.S. state where the
state income tax on corporate revenue is 6% per
year. C.F. Jordon pays an average federal tax of
23% and reports taxable income of $7 million. Because of pressing labor cost increases, the president wants to move to another state to reduce the
total tax burden. The new state may have to be
willing to offer tax allowances or an interest-free
grant for the first couple of years in order to attract
the company. You are an engineer with the company and are asked to do the following.
(a) Determine the effective tax rate for C.F.
Jordon.
(b) Estimate the state tax rate that would be necessary to reduce the overall effective tax rate
by 10% per year.
(c) Determine what the new state would have to
do financially for C.F. Jordon to move there
and to reduce its effective tax rate to 22% per
year.
CFBT and CFAT
17.11 Identify which of the following items are included
in the calculation of cash flow before taxes
(CFBT): operating expenses, salvage value, depreciation, initial investment, gross income, tax rate.
17.12 What is the basic difference between cash flow
after taxes (CFAT) and net operating profit after
taxes (NOPAT)?
17.13 For a year in which there is no initial investment P
or salvage value S, derive an equation for CFBT
that contains only the following terms: CFAT,
CFBT, D, and Te.
17.14 Determine the cash flow before taxes for Anderson
Consultants when the cash flow after taxes was
$600,000, asset depreciation was $350,000 and the
company’s effective tax rate was 36%.
17.15 Four years ago Sierra Instruments of Monterey,
California spent $200,000 for equipment to manufacture standard gas flow calibrators. The equipment was depreciated by MACRS using a 3-year
recovery period. The gross income for year 4 was
$100,000, with operating expenses of $50,000.
Use an effective tax rate of 40% to determine the
CFAT in year 4 if the asset was (a) discarded with
no salvage value in year 4 and (b) sold for $20,000
at the end of year 4 (neglect any taxes that may be
incurred on the sale of the equipment). The
MACRS depreciation rate for year 4 is 7.41%.
17.16 Four years ago a division of Harcourt-Banks purchased an asset that was depreciated by the MACRS
method using a 3-year recovery period. The total
revenue for year 2 was $48 million, depreciation
was $8.2 million, and operating expenses were
$28 million. Use a federal tax rate of 35% and a
state tax rate of 6.5% to determine (a) CFAT, (b)
percentage of total revenue expended on taxes, and
(c) net profit after taxes for the year.
17.17 Advanced Anatomists, Inc., researchers in medical science, is contemplating a commercial venture concentrating on proteins based on the new
X-ray technology of free-electron lasers. To recover the huge investment needed, an annual
$2.5 million CFAT is needed. A favored average
federal tax rate of 20% is expected; however, state
taxing authorities will levy an 8% tax on TI. Over
a 3-year period, the deductible expenses and depreciation are estimated to total $1.3 million the
first year, increasing by $500,000 per year thereafter. Of this, 50% is expenses and 50% is depreciation. What is the required gross income each
year?
475
Problems
The following information is used in Problems 17.18
through 17.21. (Show hand and spreadsheet solutions, as
instructed)
After 4 years of use, Procter and Gamble has decided to
replace capital equipment used on its Zest bath soap line.
The equipment was MACRS-depreciated over a 3-year
recovery period. After-tax MARR is 10% per year, and Te
is 35% in the United States. The cash flow data is tabulated in $1000 units.
Year
0
1
2
3
4
Purchase, $
⫺1900
Gross income, $
800
950
600
300
Operating expenses, $
⫺100 ⫺150 ⫺200 ⫺250
Salvage, $
700
17.18 Utilize the CFBT series and AW value to determine whether the equipment investment exceeded
the MARR.
year 2 if the depreciation method had been straight
line instead of MACRS.
17.24 Cheryl, an electrical engineering student who is
working on a business minor, is studying depreciation and finance in her engineering management
course. The assignment is to demonstrate that
shorter recovery periods require the same total
taxes, but they offer a time value of taxes advantage for depreciable assets. Help her, using asset
estimates made for a 6-year study period: P ⫽
$65,000, S ⫽ $5000, GI ⫽ $32,000 per year, AOC
is $10,000 per year, SL depreciation, i ⫽ 12% per
year, Te ⫽ 31%. Make the comparison using recovery periods of 3 and 6 years.
17.25 Complete the last four columns of the table below
using an effective tax rate of 40% for an asset that
has a first cost of $20,000 and a 3-year recovery
period with no salvage value, using (a) straight
line depreciation and (b) MACRS depreciation.
All cash flows are in $1000 units.
17.19 Calculate MACRS depreciation and estimate the
CFAT series over 4 years. Neglect any tax impact
caused by the $700,000 salvage received in year 4.
Year
GI
P
OE
D
TI
Taxes
CFAT
0
1
2
3
4
—
8
15
12
10
⫺20
—
⫺2
⫺4
⫺3
⫺5
—
—
—
⫺20
17.20 Utilize the CFAT series and AW value to determine
whether the investment exceeded the MARR.
17.21 Compare the after-tax ROR values using both
methods—CFAT series and approximation from
the CFBT values using the before-tax ROR and Te.
17.22 A Wal-Mart Distribution Center has put into service
forklifts and conveyors purchased for $250,000.
Use a spreadsheet to tabulate CFBT, CFAT, NOPAT,
and i* before and after taxes for 6 years of ownership. The effective tax rate is 40%, and the estimated cash flow and depreciation amounts are
shown. Salvage is expected to be zero.
Year
Gross
Income, $
Operating
Expenses, $
MACRS
Depreciation, $
1
2
3
4
5
6
210,000
210,000
160,000
160,000
160,000
140,000
⫺120,000
⫺120,000
⫺122,000
⫺124,000
⫺126,000
⫺128,000
50,000
80,000
48,000
28,800
28,800
14,400
Taxes and Depreciation
17.23 An asset purchased by Stratasys, Inc. had a first
cost of $70,000 with an expected salvage value of
$10,000 at the end of its 5-year life. In year 2, the
revenue was $490,000 with operating expenses of
$140,000. If the company’s effective tax rate was
36%, determine the difference in taxes paid in
Estimates, $
17.26 Use an effective tax rate of 32% to determine the
CFAT and NOPAT associated with the asset shown
below under two different scenarios: (a) with depreciation at $6000 per year and (b) with depreciation at $6000, $9600, $5760, and $3456 in years 1
through 4, respectively. All monetary amounts are
in $1000.
Estimates, $
Year GI OE
0
1
2
3
4
—
8
15
12
10
P
D
TI
— ⫺30 — —
⫺2
⫺4
⫺3
⫺5
Taxes CFAT NOPAT
—
⫺30
—
17.27 J. B. Hunt, Inc., an overland freight company, has
purchased new trailers for $150,000 and expects to
realize a net $80,000 in gross income over operating expenses for each of the next 3 years. The trailers have a recovery period of 3 years. Assume an
effective tax rate of 35% and an interest rate of
15% per year.
(a) Show the advantage of accelerated depreciation by calculating the present worth of taxes
for the MACRS method versus the classical
476
After-Tax Economic Analysis
Chapter 17
(b)
SL method. Since MACRS takes an additional year to fully depreciate the basis, assume no CFBT beyond year 3, but include
any negative tax as a tax savings.
Show that the total taxes are the same for
both methods.
17.28 A bioengineer is evaluating methods used to
apply adhesive onto microporous paper tape that
is commonly used after surgery. The machinery
costs $200,000, has no salvage value, and the
CFBT estimate is $75,000 per year for up to 10
years. The Te ⫽ 38% and i ⫽ 8% per year. The
two depreciation methods to consider are:
MACRS with n ⫽ 5 years and SL with n ⫽
8 years (neglect the half-year convention effect).
For a study period of 8 years, (a) determine
which depreciation method offers the better tax
advantage, and (b) demonstrate that the same
total taxes are paid for MACRS and SL.
17.29 An asset with a first cost of $9000 is depreciated
by MACRS over a 5-year recovery period. The
CFBT is estimated at $10,000 for the first 4 years
and $5000 thereafter as long as the asset is retained. The effective tax rate is 40%, and money is
worth 10% per year. In present worth dollars, how
much of the cash flow generated by the asset over
its recovery period is lost to taxes?
Depreciation Recapture and Capital Gains (Losses)
17.30 Last month, a company specializing in wind power
plant design and construction made a capital investment of $400,000 in physical simulation
equipment that will be used for at least 5 years,
after which it is expected to be sold for approximately 25% of its first cost. According to tax law,
the simulation is MACRS-depreciated using a
3-year recovery period.
(a) Explain why there is a predictable tax implication when the simulator is sold.
(b) Determine by how much the sale will cause
TI and taxes to change in year 5 if Te ⫽ 35%.
The following information is used in Problems 17.31
through 17.34.
Open Access, Inc. is an international provider of computer network communications gear. Different depreciation, recovery period, and tax law practices in the three
countries where depreciable assets are located are summarized in the table. Also, information is provided about
assets purchased 5 years ago at each location and sold this
year. After-tax MARR ⫽ 9% per year and Te ⫽ 30% can
be used for all countries.
Practice or Estimate
Country 1
Country 2
Country 3
Depreciation
SL with
MACRS
DDB with
method
n⫽5
with n ⫽ 3 n ⫽ 5
Depreciation recapture Not taxed Taxed as TI Taxed as TI
First cost, $
⫺100,000
⫺100,000
⫺100,000
GI ⫺ OE, $ per year
25,000
25,000
25,000
0 in year 5 0 in year 5 20,000 in
Estimated
salvage, $
year 5
Life, years
5
5
5
20,000 in
20,000 in
20,000 in
Actual selling
price, $
year 5
year 5
year 5
17.31 For country 1, SL depreciation is $20,000 per year.
Determine the (a) CFAT series and (b) PW of depreciation, taxes, and CFAT series.
17.32 For country 2, MACRS depreciation for 4 years is
$33,333, $44,444, $14,815, and $7407, respectively. Determine the (a) CFAT series and (b) PW
of depreciation, taxes, and CFAT series.
17.33 For country 3, DDB depreciation for 5 years is
$40,000, $24,000, $14,400, $1600, and 0, respectively. Determine the (a) CFAT series and (b) PW
of depreciation, taxes, and CFAT series.
17.34 If you worked Problems 17.31 through 17.33, develop a table that summarizes, for each country,
the total taxes paid and the PW values of the depreciation, taxes, and CFAT series. For each criterion, select the country that provides the best PW
value. Explain why the same country is not selected for all three criteria. (Hint: The PW should
be minimized for some criteria and maximized for
other criteria. Review previous material first to be
sure you choose correctly.)
17.35 An asset with a first cost of $350,000 three years
ago is sold for $385,000. The asset was depreciated by the MACRS method and has a book value
of $100,800 at the time of sale. Determine the capital gain and depreciation recapture, if any.
17.36 Sun-Tex Truck Stop is located in the desert southwest and is 5 miles from the nearest municipal water
system. In order to have fresh water at the site, the
company purchased a turnkey reverse osmosis (RO)
system for $355,000. The company depreciated the
RO system using the MACRS method with a
10-year recovery period. Four years after the system
was purchased, water lines from a local water system were extended to the truck stop, so Sun-Tex
sold the RO system for $190,000. Determine which
of the following apply and the amount, if any, to
include in an after-tax analysis of the project: depreciation recapture, capital gain, capital loss. The
MACRS depreciation rates are 10%, 18%, 14.4%,
and 11.52% for years 1 through 4, respectively.
Problems
17.37 Freeman Engineering paid $28,500 for specialized
equipment for use with its new GPS兾GIS system.
The equipment was depreciated over a 3-year recovery period using MACRS depreciation. The
company sold the equipment after 2 years for
$5000 when it purchased an upgraded system.
(a) Determine the amount of the depreciation recapture or capital loss involved in selling the asset.
(b) What tax effect will this amount have?
17.38 Determine any depreciation recapture, capital gain,
or capital loss generated by each event described
below. Use them to determine the amount of income tax effect, if the effective tax rate is 35%.
(a) A MACRS-depreciated asset with a 7-year
recovery period has been sold prematurely
after 4 years at an amount equal to 40% of its
first cost, which was $150,000.
(b) A hi-tech machine was sold internationally
for $10,000 more than its purchase price just
after it was in service 1 year. The asset had
P ⫽ $100,000, S ⫽ $1000, and n ⫽ 3 years
and was depreciated by the MACRS method
for the 1 year.
(c) Land purchased 4 years ago for $1.8 million
was sold at a 10% profit.
(d) A 21-year-old asset was removed from
service and sold for $500. When purchased,
the asset was entered on the books with a
basis of P ⫽ $180,000, S ⫽ $5000, and n ⫽
18 years. Classical straight line depreciation
was used for the entire recovery period.
(e) A corporate car was depreciated using MACRS
over a 3-year recovery period. It was sold in
the fourth year of use for $2000.
17.39 Sunnen Products Co. of St. Louis, Missouri, makes
actuator hones for gas meter tubes where light-duty
metal removal is needed. The company purchased
land, a building, and two depreciable assets from
MPG Automation Systems Corporation, all of which
have recently been disposed of. Use the information
shown to determine the presence and amount of any
depreciation recapture, capital gain, or capital loss.
Asset
Purchase
Price, $
Recovery
Period,
Years
Current
Book
Value, $
Sales
Price, $
Land
Building
Asset 1
Asset 2
⫺220,000
⫺900,000
⫺50,500
⫺20,000
—
27.5
3
3
300,000
15,500
5,000
295,000
275,000
19,500
12,500
After-Tax Economic Analysis
17.40 Compute the required before-tax return if an aftertax return of 7% per year is expected and the state
477
and local tax rates total 4.2%. The effective federal
tax rate is 34%.
17.41 Estimate the approximate after-tax rate of return
(ROR) for a project that has a before-tax ROR of
24%. Assume that the company has an effective
tax rate of 35% and it uses MACRS depreciation
for an asset that has a $27,000 salvage value.
17.42 Estimate the approximate after-tax rate of return
for a project that has a first cost of $500,000, a
salvage value of 20% of the first cost after 3 years,
and annual gross income less operating expenses
of (GI⫺OE) ⫽ $230,000. Assume the company
has a 35% effective tax rate.
17.43 An engineer is making an annual return of 12%
before taxes on the retirement investments placed
through her employer. No taxes are paid on retirement earnings until they are withdrawn; however,
she was told by her brother, an accountant, that this
is the equivalent of an 8% per year after-tax return.
What percent of taxable income is her brother assuming to be taken by income taxes?
17.44 Bart is an economic consultant to the textile industry. In both a small business and a large corporation he performed economic evaluations that have
an average 18% per year before-tax return. If the
stated MARR in both companies is 12% per year
after taxes, determine if management at both companies should accept the projects. The before-tax
return is used to approximate the after-tax return.
Effective tax rates are 34% for the larger corporation and 28% for the small company.
17.45 Elias wants to perform an after-tax evaluation of
equivalent methods A and B to electrostatically remove airborne particulate matter from clean rooms
used to package liquid pharmaceutical products.
Using the information shown, MACRS depreciation
with n ⫽ 3 years, a 5-year study period, after-tax
MARR ⫽ 7% per year, and Te ⫽ 34% and a spreadsheet, he obtained the results AWA ⫽ $−2176 and
AWB ⫽ $3545. Any tax effects when the equipment
is salvaged were neglected. Thus, with MACRS depreciation, method B is the better method. Now, use
classical SL depreciation with n ⫽ 5 years to evaluate the alternatives. Is the decision different from
that reached using MACRS?
First cost, $
Salvage value, $
Savings, $ per year
AOC, $ per year
Expected life, years
Method A
Method B
⫺100,000
10,000
35,000
⫺15,000
5
⫺150,000
20,000
45,000
⫺6,000
5
478
After-Tax Economic Analysis
Chapter 17
17.46 A corporation uses the following: before-tax
MARR of 14% per year, after-tax MARR of
7% per year, and Te of 50%. Two new machines
have the following estimates.
First cost, $
Salvage value, $
AOC, $ per year
Life, years
Machine A
Machine B
⫺15,000
3,000
⫺3,000
10
⫺22,000
5,000
⫺1,500
10
The machine is retained in use for a total of 10 years,
then sold for the estimated salvage value. Select one
machine under the following conditions:
(a) Before-tax PW analysis.
(b) After-tax PW analysis, using classical SL depreciation over the 10-year life.
(c) After-tax PW analysis, using MACRS depreciation with a 5-year recovery period.
17.47 Choose between alternatives A and B below if the
after-tax MARR is 8% per year, MACRS depreciation is used, and Te ⫽ 40%. The GI⫺OE estimate
is made for only 3 years; it is zero when each asset
is sold in year 4.
First cost, $
Actual salvage value, $
Gross income ⫺ expenses, $
Recovery period, years
Alternative
A
Alternative
B
⫺8,000
0
3,500
3
⫺13,000
2,000
5,000
3
17.48 Offshore platform safety equipment, designed for
special jobs, will cost $2,500,000, will have no
salvage value, and will be kept in service for exactly 5 years, according to company policy. Operating revenue minus expenses is estimated to be
$1,500,000 in year 1 and only $300,000 each additional year. The effective tax rate for the multinational oil company is 30%. Show hand and
spreadsheet solutions, as instructed, to find the
after-tax ROR using (a) classical SL depreciation
and (b) MACRS 5-year depreciation, neglecting
any tax effects at the end of 5 years of service.
17.49 Automatic inspection equipment purchased for
$78,000 by Stimson Engineering generated an average of $26,080 annually in before-tax cash flow
during its 5-year estimated life. This represents a
return of 20%. However, the corporate tax expert
said the CFAT was only $15,000 per year. If the
corporation wants to realize an after-tax return of
10% per year, for how many more years must the
equipment remain in service?
After-Tax Replacement
17.50 In a replacement study between a defender and a
challenger, there may be a capital gain or loss
when the defender asset is sold. (a) How is the
gain or loss calculated, and (b) how does it affect
the AW values in the study?
17.51 In an after-tax replacement study of cost alternatives involving one challenger and one defender,
how will a capital loss when selling the defender
affect the AW of each alternative?
17.52 An asset that was purchased 2 years ago was expected
to be kept in service for its projected life of 5 years, but
a new version (the challenger) of this asset promises
to be more efficient and have lower operating costs.
You have been asked to figure out if it would be more
economically attractive to replace the defender now
or keep it for 3 more years as originally planned. The
defender had a first cost of $300,000, but its market
value now is only $150,000. It has operating expenses
of $120,000 per year and no estimated salvage value
after 3 more years. To simplify calculations for this
problem only, assume that SL depreciation was
charged at $60,000 per year and that it will continue
at that rate for the next 3 years.
The challenger will cost $420,000, will have no
salvage value after its 3-year life, will have chargeable expenses of $30,000 per year, and will be depreciated at $140,000 per year (again, using SL
depreciation for simplicity in this case). Assume
the company’s effective tax rate is 35%, and its
after-tax MARR is 15% per year.
(a) Determine the CFAT in year 0 for the challenger and defender.
(b) Determine the CFAT in years 1 through 3 for
the challenger and defender.
(c) Conduct an AW evaluation to determine if
the defender should be kept for 3 more years
or replaced now.
17.53 The defender in a catalytic oxidizer manufacturing
plant has a market value of $130,000 and expected
annual operating costs of $70,000 with no salvage
value after its remaining life of 3 years. The depreciation charges for the next 3 years will be $69,960,
$49,960, and $35,720. Using an effective tax rate
of 35%, determine the CFAT for next year only
that should be used in a present worth equation to
compare this defender against a challenger that
also has a 3-year life. Assume the company’s aftertax MARR is 12%.
17.54 Perform a PW replacement study (hand and
spreadsheet solutions, if instructed) from the information shown using an after-tax MARR ⫽ 12%
per year, Te ⫽ 35%, and a study period of 4 years.
Problems
479
(Assume that the assets will be salvaged at their
original salvage estimates. Since no revenues are
estimated, all taxes are negative and considered
“savings” to the alternative.) All monetary values
are in $1000 units.
The effective tax rate is 35%, and the after-tax
MARR is 10% per year. Perform an after-tax
AW analysis, and determine which vice president has the better economic strategy over the
next 3 years.
Defender Challenger
17.57 Nuclear safety devices installed several years
ago have been depreciated from a first cost of
$200,000 to zero using MACRS. The devices
can be sold on the used equipment market for an
estimated $15,000. Or they can be retained in
service for 5 more years with a $9000 upgrade
now and an AOC of $6000 per year. The upgrade
investment will be depreciated over 3 years with
no salvage value. The challenger is a replacement with newer technology at a first cost of
$40,000, n ⫽ 5 years, and S ⫽ 0. The new units
will have operating expenses of $7000 per year.
(a) Use a 5-year study period, an effective tax
rate of 40%, an after-tax MARR of 12% per
year, and an assumption of classical straight line
depreciation (no half-year convention) to perform an after-tax replacement study. (b) If the
challenger is known to be salable after 5 years
for an amount between $2000 and $4000, will
the challenger AW value become more or less
costly? Why?
First cost, $1000
Estimated S at purchase, $1000
Market value now, $1000
AOC, $1000 per year
Depreciation method
Recovery period, years
Useful life, years
Years owned
⫺45
5
35
⫺7
SL
8
8
3
⫺24
0
—
⫺8
MACRS
3
5
—
17.55 After 8 years of use, the heavy truck engine overhaul equipment at Pete’s Truck Repair was evaluated for replacement. Pete’s accountant used an
after-tax MARR of 8% per year, Te ⫽ 30%, and a
current market value of $25,000 to determine
AW ⫽ $2100. The new equipment costs $75,000,
uses SL depreciation over a 10-year recovery period, and has a $15,000 salvage estimate. Estimated CFBT is $15,000 per year. Pete asked his
engineer son Ramon to determine if the new equipment should replace what is owned currently.
From the accountant, Ramon learned the current
equipment cost $20,000 when purchased and
reached a zero book value several years ago. Help
Ramon answer his father’s question.
17.56 Apple Crisp Foods signed a contract some years
ago for maintenance services on its fleet of
trucks and cars. The contract is up for renewal
now for a period of 1 year or 2 years only. The
contract quote is $300,000 per year if taken for
1 year and $240,000 per year if taken for 2 years.
The finance vice president wants to renew the
contract for 2 years without further analysis, but
the vice president for engineering believes it is
more economical to perform the maintenance
in-house. Since much of the fleet is aging and
must be replaced in the near future, a fixed
3-year study period has been agreed upon. The
estimates for the in-house (challenger) alternative are as follows:
First cost, $
AOC, $ per year
Life, years
Estimated salvage
MACRS depreciation
−800,000
−120,000
4
Loses 25% of P annually:
End year 1, S ⫽ $600,000
End year 2, S ⫽ $400,000
End year 3, S ⫽ $200,000
End year 4, S ⫽ $0
3-year recovery period
17.58 Three years ago, Silver House Steel purchased a
new quenching system for $550,000. The
salvage value after 10 years at that time was estimated to be $50,000. Currently the expected
remaining life is 7 years with an AOC of $27,000
per year. The new president has recommended
early replacement of the system with one that
costs $400,000 and has a 12-year life, a $35,000
salvage value, and an estimated AOC of $50,000
per year. The MARR for the corporation is 12%
per year. The president wishes to know the
replacement value that will make the recommendation to replace now economically advantageous. Use a spreadsheet and Solver to find the
minimum trade-in value (a) before taxes and
(b) after taxes, using an effective tax rate of
30%. For solution purposes, use classical SL depreciation for both systems. Comment on the
difference in replacement value made by the
consideration of taxes.
Economic Value Added
17.59 While an engineering manager may prefer to use
CFAT estimates to evaluate the AW of a project, a
financial manager may select AW of EVA estimates. Why are these preferences predictable?
17.60 Cardenas and Moreno Engineering is evaluating
a very large flood control program for several
480
After-Tax Economic Analysis
Chapter 17
southern U.S. cities. One component is a 4-year
project for a special-purpose transport ship-crane
for use in building permanent storm surge protection against hurricanes on the New Orleans coastline. The estimates are P ⫽ $300,000, S ⫽ 0, and
n ⫽ 3 years. MACRS depreciation with a 3-year
recovery is indicated. Gross income and operating expenses are estimated at $200,000 and
$80,000, respectively, for each of 4 years. The
CFAT is shown below. Calculate the AW values
of the CFAT and EVA series. They should have
the same value. The after-tax MARR is 9.75%
and Te ⫽ 35%.
Year
0
1
2
3
4
GI, $
OE, $
P and S, $
D, $
TI, $
Taxes, $
CFAT, $
7,003
⫺4,673
26,450
34,220
⫺300,000
112,997
124,673
93,551
85,781
⫺300,000
200,000
200,000
200,000
200,000
⫺80,000
⫺80,000
⫺80,000
⫺80,000
99,990
20,010
133,350 ⫺13,350
0 44,430
75,570
22,230
97,770
17.61 Triple Play Innovators Corporation (TPIC) plans
to offer IPTV (Internet Protocol TV) service to
North American customers starting soon. Perform
an AW analysis of the EVA series for the two alternative suppliers available for the hardware and
software. Let Te ⫽ 30% and after-tax MARR ⫽
8%; use SL depreciation (neglect half-year convention and MACRS, for simplicity) and a study
period of 8 years.
Vendor
Hong Kong
Japan
First cost, $
4.2 million
3.6 million
Recovery period, years
8
5
Salvage value, $
0
0
GI − OE, $ per year
1,500,000 in year 1; increasing
by 300,000 per year up to 8 years
17.62 Sun Microsystems has developed partnerships
with several large manufacturing corporations to
use Java software in their consumer and industrial
products. A new corporation will be formed to
manage these applications. One major project involves using Java in commercial and industrial
appliances that store and cook food. The gross
income and operating expenses are expected to
follow the relations shown for the estimated life of
6 years. For t ⫽ 1 to 6 years,
Annual gross income, GI ⫽ 2,800,000 ⫺ 100,000t
Annual operating expenses, OE ⫽ 950,000 ⫹ 50,000t
The effective tax rate is 30%, the interest rate is
12% per year, and the depreciation method chosen
for the $3,000,000 capital investment is a 5-year
MACRS ADS alternative that allows straight line
write-off with the half-year convention in years 1
and 6. Using a spreadsheet, estimate (a) the annual
economic contribution of the project to the new
corporation and (b) the equivalent annual worth of
these contributions.
Value-Added Tax
17.63 What is the primary difference between a sales tax
and a value-added tax?
17.64 In Denmark, VAT is applied at a rate of 25%, with
few exceptions. Vendor A sells raw materials to
vendor B for $60,000 plus VAT, vendor B sells a
product to vendor C for $130,000 plus VAT, and
vendor C sells an improved, value-added product
to an end user for $250,000 plus VAT. Determine
the following.
(a) The amount of VAT collected by vendor B.
(b) The amount of tax vendor B sends to
Denmark’s Treasury.
(c) The total amount of tax collected by the
Treasury department.
The following information is used in Problems 17.65
through 17.70
Ajinkya Electronic Systems, a company in India that
manufactures many different electronic products, has to
purchase goods and services from a variety of suppliers
(wire, diodes, LED displays, plastic components, etc.).
The table below shows several suppliers and the VAT
rates associated with each. It also shows the purchases in
$1000 units that Ajinkya made (before taxes) from each
supplier in the previous accounting period. Ajinkya’s
sales to end users was $9.2 million, and Ajinkya’s products carry a VAT of 12.5%.
Supplier
VAT Rate, %
Purchases by
Ajinkya, $1000
A
B
C
D
E
4.0
12.5
12.5
21.3
32.6
350
870
620
90
50
17.65 How much VAT did supplier C collect?
17.66 How much tax did Ajinkya keep from the tax it
collected based on the purchases it made from
supplier A?
481
Additional Problems and FE Exam Review Questions
17.67 What was the total amount of taxes paid by Ajinkya
to the suppliers?
17.69 What was the amount of taxes Ajinkya sent to the
Treasury of India?
17.68 What was the average VAT rate paid by Ajinkya in
purchasing goods and services?
17.70 What was the total amount of taxes collected by
the Treasury of India from Ajinkya and Ajinkya’s
suppliers?
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
17.71 A graduated income tax system means:
(a) Only taxable incomes above a certain level
pay any taxes.
(b) A higher flat rate goes with all of the taxable
income.
(c) Higher tax rates go with higher taxable
incomes.
(d) Rates are indexed each year to keep up with
inflation.
17.72 All of the following are characteristics of a valueadded tax system except:
(a) Value-added taxes are taxes on consumption.
(b) The end user pays value-added taxes.
(c) Value-added taxes are charged at each stage
of product development.
(d) Value-added taxes are charged only on the
raw materials for product development.
17.73 A small company has a taxable income that places
it in the 35% tax bracket. The amount of taxes that
a depreciation charge of $16,000 would save is
closest to:
(a) $0
(b) $3200
(c) $5600
(d) $10,400
17.74 A company that has a 50% effective tax rate had
income of $200 million in each of the last 2 years.
In one of those years, the company had deductions
of $100 million. In the other year, the company
had deductions of only $80 million. The difference
in income taxes paid by the company in those
2 years was closest to:
(a) $10 million
(b) $20 million
(c) $50 million
(d) $60 million
17.75 Taxable income (TI) is defined as:
(a) TI ⫽ revenue ⫹ operating expenses
⫺ depreciation
(b) TI ⫽ revenue ⫺ operating expenses
⫹ depreciation
(c)
(d)
TI ⫽ revenue ⫺ operating expenses
⫺ depreciation ⫹ amortization
TI ⫽ revenue ⫺ operating expenses
⫺ depreciation
17.76 The marginal tax rate is defined as:
(a) The percentage paid on the last dollar of
income
(b) The tax rate that applies to a questionable
investment
(c) The tax rate that includes federal, state, and
local taxes
(d) The percentage paid on the first dollar of
income
17.77 A subcontractor with an effective tax rate of 25%
has gross income of $55,000, other income of
$4000, operating expenses of $13,000, and other
deductions and exemptions of $11,000. The income tax due is closest to:
(a) $11,750
(b) $8,750
(c) $10,750
(d) $13,750
17.78 When a depreciable asset is disposed of for less than
its current book value, the transaction is known as:
(a) An after-tax expense
(b) Capital loss
(c) Capital gain
(d) Depreciation recapture
17.79 The after-tax analysis for a $60,000 investment
with associated gross income minus expenses
(GI⫺OE) is shown below for the first 2 years only.
If the effective tax rate is 40%, the values for depreciation (D), taxable income (TI), and taxes for
year 1 are closest to:
Year
0
1
2
Investment, GI − OE,
$
$
D, $
TI, $
Taxes,
$
⫺60,000
30,000
35.000
15,000 6000
CFAT,
$
⫺60,000
26,000
29,000
482
After-Tax Economic Analysis
Chapter 17
(a)
(b)
(c)
(d)
D ⫽ $5,000, TI ⫽ $25,000, taxes ⫽ $10,000
D ⫽ $30,000, TI ⫽ $30,000, taxes ⫽ $4,000
D ⫽ $20,000, TI ⫽ $50,000, taxes ⫽ $20,000
D ⫽ $20,000, TI ⫽ $10,000, taxes ⫽ $4,000
17.80 If the after-tax rate of return for a cash flow series
is 11.9% and the corporate effective tax rate is
34%, the approximated before-tax rate of return is
closest to:
(a) 6.8%
(b) 5.4%
(c) 18.0%
(d) 28.7%
17.81 An asset purchased for $100,000 with S ⫽ $20,000
after 5 years was depreciated using the 5-year
MACRS rates. Expenses averaged $18,000 per
year, and the effective tax rate is 30%. The asset
was actually sold after 5 years of service for
$22,000. MACRS rates in years 5 and 6 are 11.53%
and 5.76%, respectively. The after-tax cash flow
from the sale is closest to:
(a)
(b)
(c)
(d)
$27,760
$17,130
$26,870
$20,585
17.82 When accelerated depreciation methods or
shortened recovery periods are applied, there
are impacts on the income taxes due. Of the following, the statements that are commonly incorrect are:
1. Total taxes paid are the same for all depreciation methods.
2. Present worth of taxes is lower for shorter
recovery periods.
3. Accelerated depreciation imposes more taxes
in the later years of the recovery period.
4. Present worth of taxes is higher for shorter
recovery periods.
(a) 1, 2, and 3
(b) 1 and 4
(c) 2
(d) 4
CASE STUDY
AFTER-TAX ANALYSIS FOR BUSINESS EXPANSION
Background
Charles was always a hands-on type of person. Within a couple
of years of graduating from college, he started his own business.
After some 20 years, it has grown significantly. He owns and
operates Pro-Fence, Inc. in the Metroplex, specializing in custom-made metal and stone fencing for commercial and residential sites. For some time, Charles has thought he should expand
into a new geographic region, with the target area being another
large metropolitan area about 500 miles north, called Victoria.
Pro-Fence is privately owned by Charles; therefore, the question of how to finance such an expansion has been, and still is,
the major challenge. Debt financing would not be a problem in
that the Victoria Bank has already offered a loan of up to $2 million. Taking capital from the retained earnings of Pro-Fence is a
second possibility, but taking too much will jeopardize the current business, especially if the expansion were not an economic
success and Pro-Fence were stuck with a large loan to repay.
This is where you come in as a long-time friend of
Charles. He knows you are quite economically oriented and
that you understand the rudiments of debt and equity financing and economic analysis. He wants you to advise him on
the balance between using Pro-Fence funds and borrowed
funds. You have agreed to help him, as much as you can.
Information
Charles has collected some information that he shares with
you. Between his accountant and a small market survey of the
business opportunities in Victoria, the following generalized
estimates seem reasonable.
Initial capital investment ⫽ $1.5 million
Annual gross income ⫽ $700,000
Annual operating expenses ⫽ $100,000
Effective income tax rate for Pro-Fence ⫽ 35%
Five-year MACRS depreciation for all $1.5 million investment
The terms of the Victoria Bank loan would be 6% per year
simple interest based on the initial loan principal. Repayment would be in 5 equal payments of interest and principal. Charles comments that this is not the best loan arrangement he hopes to get, but it is a good worst-case
scenario upon which to base the debt portion of the analysis. A range of D-E mixes should be analyzed. Between
Charles and yourself, you have developed the following
viable options.
Debt
Loan
Percentage Amount, $
0
50
70
90
750,000
1,050,000
1,350,000
Equity
Percentage
100
50
30
10
Investment
Amount, $
1,500,000
750,000
450,000
150,000
Case Study
Case Study Exercises
1. For each funding option, perform a spreadsheet analysis
that shows the total CFAT and its present worth over a
6-year period, the time it will take to realize the full
advantage of MACRS depreciation. An after-tax return
of 10% is expected. Which funding option is best for
Pro-Fence? (Hint: For the spreadsheet, sample column
headings are: Year, GI − OE, Loan interest, Loan principal, Equity investment, Depreciation rate, Depreciation,
Book value, TI, Taxes, and CFAT.)
2. Observe the changes in the total 6-year CFAT as the
D-E percentages change. If the time value of money
483
is neglected, what is the constant amount by which
this sum changes for every 10% increase in equity
funding?
3. Charles noticed that the CFAT total and PW values go in
opposite directions as the equity percentage increases.
He wants to know why this phenomenon occurs. How
should you explain this to Charles?
4. After deciding on the 50-50 split of debt and equity financing, Charles wants to know what additional bottomline contributions to the economic worth of the company
may be added by the new Victoria site. What are the best
estimates at this time?
CHAPTER 18
Sensitivity
Analysis and
Staged
Decisions
L E A R N I N G
O U T C O M E S
Purpose: Perform a sensitivity analysis of parameters; use expected values to evaluate staged funding options.
SECTION
TOPIC
LEARNING OUTCOME
18.1
Sensitivity to variation
• Use a measure of worth to explain sensitivity to
variation in one or more parameters.
18.2
Three-estimate variation
• Choose the better alternative using three
estimates of variation in selected parameters.
18.3
E(X)
• Calculate the expected value of a variable.
18.4
E(X) of cash flows
• Evaluate a project or alternatives using expected
values of cash flows.
18.5
Decision trees
• Use a decision tree to evaluate alternatives stage
by stage.
18.6
Real options
• Explain a real option in engineering economics
and evaluate a staged decision using real
options analysis.
T
his chapter includes several related topics about alternative evaluation. Initially, we expand our capability to perform a sensitivity analysis of one or
more parameters and of an entire alternative. Then the determination and use
of the expected value of a cash flow series are treated. The techniques of decision trees
help make a series of economic decisions for alternatives that have different, but closely
connected stages.
Economic decisions that involve staged funding are very common in professional and
everyday life. The last topic of real options analysis introduces a method useful in these
circumstances.
18.1 Determining Sensitivity to Parameter Variation
The term parameter is used in this chapter to represent any variable or factor for which an estimate or stated value is necessary. Example parameters are first cost, salvage value, AOC, estimated life, production rate, and materials costs. Estimates such as the loan interest rate and the
inflation rate can also be parameters of the analysis.
Economic analysis uses estimates of a parameter’s future value to assist decision makers.
Since future estimates are always incorrect to some degree, inaccuracy is present in the economic
projections. The effect of variation may be determined by using sensitivity analysis.
Sensitivity analysis determines how a measure of worth—PW, AW, FW, ROR, B兾C, or
CER—is altered when one or more parameters vary over a selected range of values. Usually
one parameter at a time is varied, and independence with other parameters is assumed. Though
this approach is an oversimplification in real-world situations, since the dependencies are difficult to accurately model, the end results are usually correct.
Sensitivity analysis
In reality, we have applied this approach (informally) throughout previous chapters to determine
the response to variation in a variety of parameters. Variation in a parameter such as MARR will
not alter the decision to select an alternative when all compared alternatives return considerably
more than the MARR; thus, the decision is relatively insensitive to the MARR. However, variation in the n or AOC value may indicate that the alternative’s measure of worth is very sensitive
to the estimated life or annual operating costs.
Usually the variations in life, annual costs, and revenues result from variations in selling price,
operation at different levels of capacity, inflation, etc. For example, if an operating level of 90%
of airline seating capacity for a domestic route is compared with 70% for a proposed international
route, the operating cost and revenue per passenger-mile will increase, but anticipated aircraft
life will probably decrease only slightly. Usually several important parameters are studied to
learn how the uncertainty of estimates affects the economic analysis.
Sensitivity analysis routinely concentrates on the variation expected in estimates of P, AOC,
S, n, unit costs, unit revenues, and similar parameters. These parameters are often the result of
design questions and their answers, as discussed in Chapter 15. Parameters that are interest rate–
based are not treated in the same manner.
Parameters such as MARR and other interest rates (loan rates, inflation rate) are more stable from
project to project. If performed, sensitivity analysis on them is for specific values or over a
narrow range of values. This point is important to remember if simulation is used for decision
making under risk (Chapter 19).
Plotting the sensitivity of PW, AW, or ROR versus the parameter(s) studied is very helpful.
Two alternatives can be compared with respect to a given parameter and the breakeven point.
This is the value at which the two alternatives are economically equivalent. However, the breakeven chart commonly represents only one parameter per chart. Thus, several charts are constructed, and independence of each parameter is assumed. In previous uses of breakeven analysis,
we often computed the measure of worth at only two values of a parameter and connected the
points with a straight line. However, if the results are sensitive to the parameter value, several
intermediate points should be used to better evaluate the sensitivity, especially if the relationships are not linear.
When several parameters are studied, sensitivity analysis can become quite complex. It may
be performed one parameter at a time using a spreadsheet or computations by hand or calculator.
PW
Low
High
Parameter
486
Sensitivity Analysis and Staged Decisions
Chapter 18
The computer facilitates comparison of multiple parameters and multiple measures of worth, and
the software can rapidly plot the results.
Here is a general procedure to follow when conducting a thorough sensitivity analysis.
1.
2.
3.
4.
5.
Determine which parameter(s) of interest might vary from the most likely estimated value.
Select the probable range and an increment of variation for each parameter.
Select the measure of worth.
Compute the results for each parameter, using the measure of worth as a basis.
To better interpret the sensitivity, graphically display the parameter versus the measure of
worth.
This sensitivity analysis procedure should indicate the parameters that warrant closer study
or require additional information. When there are two or more alternatives, it is better to use
the PW or AW measure of worth in step 3. If ROR is used, it requires the extra efforts of
incremental analysis between alternatives. Example 18.1 illustrates sensitivity analysis for
one project.
EXAMPLE 18.1
Wild Rice, Inc. expects to purchase a new asset for automated rice handling. Most likely estimates are a first cost of $80,000, zero salvage value, and a cash flow before taxes (CFBT) per
year t that follows the relation $27,000 − 2000t. The MARR for the company varies over a
wide range from 10% to 25% per year for different types of investments. The economic life of
similar machinery varies from 8 to 12 years. Evaluate the sensitivity of PW by varying
(a) MARR, while assuming a constant n value of 10 years, and (b) n, while MARR is constant
at 15% per year. Perform the analysis by hand and by spreadsheet.
Solution by Hand
(a) Follow the procedure above to understand the sensitivity of PW to MARR variation.
1. MARR is the parameter of interest.
2. Select 5% increments to evaluate sensitivity to MARR; the range is 10% to 25%.
3. The measure of worth is PW.
4. Set up the PW relation for 10 years. When MARR ⫽ 10%,
PW ⫽ ⫺80,000 ⫹ 25,000(P兾A,10%,10) ⫺ 2000(P兾G,10%,10)
⫽ $27,830
The PW for all MARR values at 5% intervals is as follows:
MARR, %
PW, $
10
15
20
25
27,830
11,512
⫺962
⫺10,711
5. A plot of MARR versus PW is shown in Figure 18–1. The steep negative slope indicates that the decision to accept the proposal based on PW is quite sensitive to
variations in the MARR. If the MARR is established at the upper end of the range, the
investment is not attractive.
(b) 1. Asset life n is the parameter.
2. Select 2-year increments to evaluate PW sensitivity over the range 8 to 12 years.
3. The measure of worth is PW.
4. Set up the same PW relation as in part (a) at i ⫽ 15%. The PW results are
n
PW, $
8
10
12
7,221
11,511
13,145
487
Determining Sensitivity to Parameter Variation
18.1
Figure 18–1
30,000
RR
MA
Plot of PW versus MARR
and n for sensitivity
analysis, Example 18.1.
20,000
Present worth, $
n
10,000
10
15
6
8
20
25
MARR %
12
Life n
0
10
– 10,000
– 20,000
5. Figure 18–1 presents the plot of PW versus n. Since the PW measure is positive for all
values of n, the decision to invest is not materially affected by the estimated life. The PW
curve levels out above n ⫽ 10. This insensitivity to changes in cash flow in the distant
future is a predictable observation, because the P兾F factor gets smaller as n increases.
Solution by Spreadsheet
Figure 18–2 presents two spreadsheets and accompanying plots of PW versus MARR (fixed n)
and PW versus n (fixed MARR). The NPV function calculates PW for i values from 10% to
25% and n values from 8 to 12 years. As the solution by hand indicated, so do the charts; PW
is sensitive to changes in MARR values, but not very sensitive to variations in n.
Figure 18–2
Sensitivity analysis of
PW to variation in
(a) MARR values and
(b) life estimates,
Example 18.1.
PW computation
⫽ NPV(C6,B$4:B$13)
⫹B$3
(a)
PW computation
⫽ NPV(15%,B$4:B$15)
⫹B$3
(b)
488
Sensitivity Analysis and Staged Decisions
Chapter 18
Figure 18–3
50%
Sensitivity analysis graph
of percent variation from
the most likely estimate.
40%
+
+
+
Rate of return on capital
30%
+
+
20%
+
+
+
10%
+
0%
– 10%
– 20%
– 50%
– 40%
– 30%
– 20%
0%
10%
20%
30%
40%
50%
% Change in individual parameter
Parameter Legend
Sales price
Direct material
+ Sales volume
– 10%
Indirect cost
Direct labor
Capital
When the sensitivity of several parameters is considered for one alternative using a single
measure of worth, it is helpful to graph percentage change for each parameter versus the measure of worth. This is sometimes called a spider graph. Figure 18–3 illustrates ROR versus six
different parameters for one alternative. The variation in each parameter is indicated as a percentage deviation from the most likely estimate on the horizontal axis. If the ROR response curve is
flat and approaches horizontal over the range of total variation graphed for a parameter, there is
little sensitivity of ROR to changes in the parameter’s value. This is the conclusion for indirect
cost in Figure 18–3. On the other hand, ROR is very sensitive to sales price. A reduction of 30%
from the expected sales price reduces the ROR from approximately 20% to −10%, whereas a
10% increase in price raises the ROR to about 30%.
If two alternatives are compared and the sensitivity to one parameter is sought, the graph
may show quite nonlinear results. Observe the general shape of the sample sensitivity graphs in
Figure 18–4. The plots are shown as linear segments between specific computation points. The
graph indicates that the PW of each plan is a nonlinear function of hours of operation. Plan A is
Figure 18–4
150
Sample PW sensitivity to
hours of operation for two
alternatives.
PW, $ ⫻ 1000
Plan B
100
Pl
an
A
50
0
1000
2000
Hours of operation per year
3000
489
Determining Sensitivity to Parameter Variation
18.1
very sensitive in the range of 0 to 2000 hours, but it is comparatively insensitive above
2000 hours. Plan B is more attractive due to its relative insensitivity. The breakeven point is at
about 1750 hours per year. It may be necessary to plot the measure of worth at intermediate
points to better understand the nature of the sensitivity.
EXAMPLE 18.2
Columbus, Ohio needs to resurface a 3-kilometer stretch of highway. Knobel Construction has
proposed two methods of resurfacing. The first method is a concrete surface for a cost of
$1.5 million and an annual maintenance cost of $10,000. The second method is an asphalt
covering with a first cost of $1 million and a yearly maintenance of $50,000. However, Knobel
requests that every third year the asphalt highway be touched up at a cost of $75,000.
The city uses the interest rate on bonds, 6% on its last bond issue, as the discount rate.
(a) Determine the breakeven number of years of the two methods. If the city expects an interstate to replace this stretch of highway in 10 years, which method should be selected?
(b) If the touch-up cost increases by $5000 per kilometer every 3 years, is the decision sensitive to this increase?
Solution
(a) Use PW analysis to determine the breakeven n value.
PW of concrete ⫽ PW of asphalt
⫺1,500,000 ⫺ 10,000(P兾A,6%,n) ⫽ ⫺1,000,000 ⫺ 50,000(P兾A,6%,n)
⫺75,000
[兺
(P兾F,6%, j )
j
]
where j ⫽ 3, 6, 9, . . . , n. The relation can be rewritten to reflect the incremental cash flows.
⫺500,000 ⫹ 40,000(P兾A,6%,n) ⫹ 75,000
[兺
]
(P兾F,6%, j ) ⫽ 0
j
[18.1]
The breakeven n value can be determined by hand solution by increasing n until Equation
[18.1] switches from negative to positive PW values. Alternatively, a spreadsheet solution
using the NPV function can find the breakeven n value (Figure 18–5). The NPV functions
in column C are the same each year, except that the cash flows are extended 1 year for each
Figure 18–5
Sensitivity of the breakeven life between two alternatives, Example 18.2.
PW for 11 years
⫽ NPV(6%,$B$5:$B15)⫹$B$4
PW for 12 years
⫽ NPV(6%,$B$5:$B16)⫹$B$4
Year counter advances by 1
490
Sensitivity Analysis and Staged Decisions
Chapter 18
present worth calculation. At approximately n ⫽ 11.4 years, concrete and asphalt resurfacing break even economically. Since the road is needed for 10 more years, the extra cost of
concrete is not justified; select the asphalt alternative.
(b) The total touch-up cost will increase by $15,000 every 3 years. Equation [18.1] is now
[
(
j⫺3
⫺500,000 ⫹ 40,000(P兾A,6%,n) ⫹ 75,000 ⫹15,000 ———
3
) ] [ 兺(P兾F,6%, j ) ] ⫽ 0
j
Now the breakeven n value is between 10 and 11 years—10.8 years using linear interpolation (Figure 18–5, column E). The decision has become marginal for asphalt, since the
interstate is planned for 10 years hence.
Noneconomic considerations may be used to determine if asphalt is still the better alternative. One conclusion is that the asphalt decision becomes more questionable as the
asphalt alternative maintenance costs increase; that is, the PW value is sensitive to increasing touch-up costs.
18.2 Sensitivity Analysis Using Three Estimates
We can thoroughly examine the economic advantages and disadvantages among two or more
alternatives by borrowing from the field of project scheduling the concept of making three estimates for each parameter: a pessimistic, a most likely, and an optimistic estimate. Depending
upon the nature of a parameter, the pessimistic estimate may be the lowest value (alternative life
is an example) or the largest value (such as asset first cost).
This approach allows us to study measure of worth and alternative selection sensitivity within a
predicted range of variation for each parameter. Usually the most likely estimate is used for all other
parameters when the measure of worth is calculated for one particular parameter or one alternative.
EXAMPLE 18.3
An engineer is evaluating three alternatives for new equipment at Emerson Electronics. She
has made three estimates for the salvage value, annual operating cost, and life. The estimates
are presented on an alternative-by-alternative basis in Table 18–1. For example, alternative B
has pessimistic estimates of S ⫽ $500, AOC ⫽ $⫺4000, and n ⫽ 2 years. The first costs are
known, so they have the same value. Perform a sensitivity analysis and determine the most
economical alternative, using AW analysis at a MARR of 12% per year.
TABLE 18–1
Strategy
Alternative A
P
Estimates ML
O
Alternative B
P
Estimates ML
O
Alternative C
P
Estimates ML
O
Competing Alternatives with Three Estimates Made for Salvage Value,
AOC, and Life Parameters
First Cost,
$
Salvage Value S,
$
AOC,
$ per Year
Life n,
Years
⫺20,000
⫺20,000
⫺20,000
0
0
0
⫺11,000
⫺9,000
⫺5,000
3
5
8
⫺15,000
⫺15,000
⫺15,000
500
1,000
2,000
⫺4,000
⫺3,500
⫺2,000
2
4
7
⫺30,000
⫺30,000
⫺30,000
3,000
3,000
3,000
⫺8,000
⫺7,000
⫺3,500
3
7
9
P ⫽ pessimistic; ML ⫽ most likely; O ⫽ optimistic.
Estimate Variability and the Expected Value
18.3
TABLE 18–2 Annual Worth Values, Example 18.3
Alternative AW Values, $
Estimates
A
B
C
P
ML
O
⫺19,327
⫺14,548
⫺9,026
⫺12,640
⫺8,229
⫺5,089
⫺19,601
⫺13,276
⫺8,927
20
18
AW of costs, ⫺$1000
16
Alternative C
14
Alternative A
12
10
8
Alternative B
6
4
2
0
1
2
3
4
5
6
7
8
9
Life n, years
Figure 18–6
Plot of AW of costs for different-life estimates, Example 18.3.
Solution
For each alternative in Table 18–1, calculate the AW value of costs. For example, the AW relation for alternative A, pessimistic estimates, is
AW ⫽ ⫺20,000(A兾P,12%,3) ⫺ 11,000 ⫽ $⫺19,327
Table 18–2 presents all AW values. Figure 18–6 is a plot of AW versus the three estimates of
life for each alternative. Since the AW calculated using the ML estimates for alternative B
($−8229) is economically better than even the optimistic AW value for alternatives A and C,
alternative B is clearly favored.
Comment
While the alternative that should be selected here is quite obvious, this is not normally the case.
For example, in Table 18–2, if the pessimistic alternative B equivalent AW were much higher,
say, $⫺21,000 per year (rather than $⫺12,640), and the optimistic AW values for alternatives A
and C were less than that for B ($−5089), the choice of B would not be apparent or correct. In
this case, it would be necessary to select one set of estimates (P, ML, or O) upon which to base
the decision. Alternatively, the different estimates can be used in an expected value analysis,
which is introduced next.
18.3 Estimate Variability and the Expected Value
Engineers and economic analysts usually deal with estimate variation and risk about an uncertain
future by placing appropriate reliance on past data, if any exist. This means that probability and
samples are used. Actually the use of probabilistic analysis is not as common as might be expected. The reason is not that the computations are difficult to perform or understand, but that
realistic probabilities associated with cash flow estimates are difficult to assign. Experience and
491
492
Sensitivity Analysis and Staged Decisions
Chapter 18
judgment can often be used in conjunction with probabilities and expected values to evaluate the
desirability of an alternative.
The expected value can be interpreted as a long-run average observable if the project is repeated
many times. Since a particular alternative is evaluated or implemented only once, the expected
value results in a point estimate. However, even for a single occurrence, the expected value is a
meaningful number.
The expected value E(X) is computed using the relation
iⴝm
E(X ) ⴝ
兺 X P(X )
i
i
[18.2]
iⴝ1
where
Xi ⫽ value of the variable X for i from 1 to m different values
P(Xi) ⫽ probability that a specific value of X will occur
Probabilities are always correctly stated in decimal form, but they are routinely spoken of in
percentages and often referred to as chance, such as the chances are about 10%. When placing
the probability value in Equation [18.2] or any other relation, use the decimal equivalent of 10%,
that is, 0.1. In all probability statements the P(Xi) values for a variable X must total to 1.0.
i⫽m
兺 P(X ) ⫽ 1.0
i
i⫽1
We may frequently omit the subscript i on X for simplicity.
If X represents the estimated cash flows, some will be positive and others will be negative. If
a cash flow sequence includes revenues and costs, and the measure of worth is present worth
calculated at the MARR, the result is the expected value of the discounted cash flows E(PW). If
the expected value is negative, the overall outcome is expected to be a cash outflow. For example,
if E(PW) ⫽ $⫺1500, this indicates that the proposal is not expected to return the MARR.
EXAMPLE 18.4
ANA airlines plans to offer several new electronic services on flights between Tokyo and selected European destinations. The marketing director estimates that for a typical 24-hour period there is a 50% chance of having a net cash flow of $5000 and a 35% chance of $10,000.
He also estimates there is a small 5% chance of no cash flow and a 10% chance of a loss of
$1000, which is the estimated extra personnel and utility costs to offer the services. Determine
the expected net cash flow.
Solution
Let NCF be the net cash flow in dollars, and let P(NCF) represent the associated probabilities.
Using Equation [18.2],
E(NCF) ⫽ 5000(0.5) ⫹ 10,000(0.35) ⫹ 0(0.05) ⫺ 1000(0.1) ⫽ $5900
Although the “no cash flow” possibility does not increase or decrease E(NCF), it is included
because it makes the probability values sum to 1.0 and it makes the computation complete.
18.4 Expected Value Computations for Alternatives
The expected value computation E(X) is utilized in a variety of ways. Two prime ways are to:
• Prepare information for use in an economic analysis.
• Evaluate the expected viability of a fully formulated alternative.
Example 18.5 illustrates the first situation, and Example 18.6 determines the expected PW when
the entire cash flow series and its probabilities are estimated.
Expected Value Computations for Alternatives
18.4
EXAMPLE 18.5
There are many government incentives to become more energy-efficient. Installing solar
panels on homes, business buildings, and multiple-family dwellings is one of them. The
owner pays a portion of the total installation costs, and the government agency pays the
rest. Nichole works for the Department of Energy and is responsible for approving solar
panel incentive payouts. She has exceeded the annual budgeted amount of $50 million per
year in each of the previous 2 years. Disappointed with this situation, Nichole and her boss
decided to collect data to determine what size increase in annual budget the incentive program needs in the future. Over the last 36 months, the amount of average monthly payout
and number of months are shown in Table 18–3. She categorized by level the monthly
averages according to her experience with the program. Provided the same pattern continues, what is the expected value of the dollar increase in annual budget that is needed to
meet the requests?
TABLE 18–3
Level
Very high
High
Moderate
Low
Solar Panel Incentive Payouts, Example 18.5
Average Payout,
$ Million per Month
Months over
Past 3 Years
6.5
4.7
3.2
2.9
15
10
7
4
Solution
Use the 36 months of payouts POj ( j ⫽ low, . . . ,very high) to estimate the probability P(POj)
for each level, and make sure the total is 1.0.
Level, j
Very high
High
Moderate
Low
Probability of Payout Level, P(POj)
P(PO1) ⫽ 15兾36 ⫽ 0.417
P(PO2) ⫽ 10兾36 ⫽ 0.278
P(PO3) ⫽ 7兾36 ⫽ 0.194
P(PO4) ⫽ 4兾36 ⫽ 0.111
1.000
The expected monthly payout is calculated using Equation [18.2]. In $ million units,
E[PO] ⫽ 6.5(0.417) ⫹ 4.7(0.278) ⫹ 3.2(0.194) ⫹ 2.9(0.111)
⫽ 2.711 ⫹ 1.307 ⫹ 0.621 ⫹ 0.322
⫽ $4.961 ($4,961,000)
The annual expected budget need is 12 ⫻ 4.961 million ⫽ $59.532 million. The current budget
of $50 million should be increased by an average of $9.532 million per year.
EXAMPLE 18.6
Lite-Weight Wheelchair Company has a substantial investment in tubular steel bending equipment. A new piece of equipment costs $5000 and has a life of 3 years. Estimated cash flows
(Table 18–4) depend on economic conditions classified as receding, stable, or expanding. A
probability is estimated that each of the economic conditions will prevail during the 3-year
period. Apply expected value and PW analysis to determine if the equipment should be purchased. Use a MARR of 15% per year.
493
494
Sensitivity Analysis and Staged Decisions
Chapter 18
TABLE 18–4
Equipment Cash Flow and Probabilities, Example 18.6
Economic Condition
Receding
(Prob. ⫽ 0.4)
Year
Stable
(Prob. ⫽ 0.4)
Expanding
(Prob. ⫽ 0.2)
Annual Cash Flow Estimates, $
0
1
2
3
⫺5000
⫹2500
⫹2000
⫹1000
⫺5000
⫹2500
⫹2500
⫹2500
⫺5000
⫹2000
⫹3000
⫹3500
Solution
First determine the PW of the cash flows in Table 18–4 for each economic condition, and then
calculate E(PW) using Equation [18.2]. Define subscripts R for receding economy, S for stable,
and E for expanding. The PW values for the three scenarios are
PWR ⫽ ⫺5000 ⫹ 2500(P兾F,15%,1) ⫹ 2000(P兾F,15%,2) ⫹ 1000(P兾F,15%,3)
⫽ ⫺5000 ⫹ 4344 ⫽ $⫺656
PWS ⫽ ⫺5000 ⫹ 5708 ⫽ $⫹708
PWE ⫽ ⫺5000 ⫹ 6309 ⫽ $⫹1309
Only in a receding economy will the cash flows not return the 15% to justify the investment.
The expected present worth is
E(PW) ⫽
兺 PW [P(j)]
j
j⫽R,S,E
⫽ ⫺656(0.4) ⫹ 708(0.4) ⫹ 1309(0.2)
⫽ $283
At 15%, E(PW) ⬎ 0; the equipment is justified, using an expected value analysis.
Comment
It is also correct to calculate the E(cash flow) for each year and then determine PW of the
E(cash flow) series, because the PW computation is a linear function of cash flows. Computing
E(cash flow) first may be easier in that it reduces the number of PW computations. In this
example, calculate E(CFt) for each year, then determine E(PW).
E(CF0) ⫽ $⫺5000
E(CF1) ⫽ 2500(0.4) ⫹ 2500(0.4) ⫹ 2000(0.2) ⫽ $2400
E(CF2) ⫽ $2400
E(CF3) ⫽ $2100
E(PW) ⫽ ⫺5000 ⫹ 2400(P兾F,15%,1) ⫹ 2400(P兾F,15%,2) ⫹ 2100(P兾F,15%,3)
⫽ $283
18.5 Staged Evaluation of Alternatives
Using a Decision Tree
Alternative evaluation may require a series of decisions in which the outcome from one stage is important to the next stage of decision making. When each alternative is clearly defined and probability estimates can be made to account for risk, it is helpful to perform the evaluation using a decision tree.
A decision tree includes:
•
•
•
•
•
•
More than one stage of alternative selection.
Selection of an alternative at one stage that leads to another stage.
Expected results from a decision at each stage.
Probability estimates for each outcome.
Estimates of economic value (cost or revenue) for each outcome.
Measure of worth as the selection criterion, such as E(PW).
Staged Evaluation of Alternatives Using a Decision Tree
18.5
Decision
node
Alternatives
D
(a) Decision node
Outcomes
0.5
Probability
node
0.2
Probabilities
0.3
(b) Probability node with outcomes
D
D
D
Final outcomes
(c) Tree structure
Figure 18–7
Decision and probability nodes used to construct a decision tree.
The decision tree is constructed left to right and includes each possible decision and outcome.
• A square represents a decision node with the possible alternatives indicated on the branches
from the decision node (Figure 18–7a).
• A circle represents a probability node with the possible outcomes and estimated probabilities
on the branches (Figure 18–7b).
• The treelike structure in Figure 18–7c results, with outcomes following a decision.
Usually each branch of a decision tree has some estimated economic value (often referred to
as payoff) in cost, revenue, saving, or benefit. These cash flows are expressed in terms of PW,
AW, or FW values and are shown to the right of each final outcome branch. The cash flow and
probability estimates on each outcome branch are used in calculating the expected economic
value of each decision branch. This process, called solving the tree or rollback, is explained
after Example 18.7, which illustrates the construction of a decision tree.
EXAMPLE 18.7
Jerry Hill is president and CEO of a U.S.-based food processing company, Hill Products and
Services. He was recently approached by an international supermarket chain that wants to
market in-country its own brand of frozen microwaveable dinners. The offer made to Jerry by
the supermarket corporation requires that a series of two decisions be made, now and 2 years
hence. The current decision involves two alternatives: (1) Lease a facility in the United Arab
Emirates (UAE) from the supermarket chain, which has agreed to convert a current processing
facility for immediate use by Jerry’s company; or (2) build and own a processing and packaging facility in the UAE. Possible outcomes of this first decision stage are good market or poor
market depending upon the public’s response.
The decision choices 2 years hence are dependent upon the lease-or-own decision made
now. If Hill decides to lease, good market response means that the future decision alternatives
are to produce at twice, equal to, or one-half of the original volume. This will be a mutual decision between the supermarket chain and Jerry’s company. A poor market response will indicate
495
496
Sensitivity Analysis and Staged Decisions
Chapter 18
a one-half level of production, or complete removal from the UAE market. Outcomes for the
future decisions are, again, good and poor market responses.
As agreed by the supermarket company, the current decision for Jerry to own the facility
will allow him to set the production level 2 years hence. If market response is good, the decision alternatives are four or two times original levels. The reaction to poor market response
will be production at the same level or no production at all.
Construct the tree of decisions and outcomes for Hill Products and Services.
Solution
This is a two-stage decision tree that has alternatives now and 2 years hence. Identify the decision nodes and branches, and then develop the tree using the branches and the outcomes of
good and poor market for each decision. Figure 18–8 details the decision stages and outcome
branches.
Stage 1 (decision now):
Label it D1.
Alternatives: lease (L) and own (O).
Outcomes: good and poor markets.
Good
Poor
Good
2⫻
D2
1⫻
Poor
Good
0.5⫻
Poor
Good
Good
0.5⫻
Poor
D3
Poor
0⫻
Out of
business
Lease
(L)
D1
Own
(O)
Good
4⫻
D4
Poor
2⫻
Good
Good
Poor
Good
Poor
1⫻
D5
Poor
0⫻
Out of
business
Figure 18–8
A two-stage decision tree identifying alternatives and possible outcomes.
Staged Evaluation of Alternatives Using a Decision Tree
18.5
Stage 2 (Decisions 2 years hence):
Label them D2 through D5.
Outcomes: good market, poor market, and out of business.
Choice of production levels for D2 through D5:
Quadruple production (4⫻); double production (2⫻); level production
(1⫻); one-half production (0.5⫻); stop production (0⫻)
The alternatives for future production levels (D2 through D5) are added to the tree and followed by the market responses of good and poor. If the stop-production (0⫻) decision is made
at D3 or D5, the only outcome is out of business.
To utilize the decision tree for alternative evaluation and selection, the following additional
information is necessary for each branch:
• The estimated probability that each outcome may occur. These probabilities must sum to 1.0
for each set of outcomes (branches) that result from a decision.
• Economic information for each decision alternative and possible outcome, such as initial
investment and estimated cash flows.
Decisions are made using the probability estimate and economic value estimate for each outcome branch. Commonly the present worth at the MARR is used in an expected value computation of the type in Equation [18.2]. This is the general procedure to solve the tree using PW
analysis:
1. Start at the top right of the tree. Determine the PW value for each outcome branch considering the time value of money.
2. Calculate the expected value for each decision alternative.
E(decision) ⴝ 兺(outcome estimate)P(outcome)
[18.3]
where the summation is taken over all possible outcomes for each decision alternative.
3. At each decision node, select the best E(decision) value—minimum cost or maximum value
(if both costs and revenues are estimated).
4. Continue moving to the left of the tree to the root decision in order to select the best alternative.
5. Trace the best decision path through the tree.
EXAMPLE 18.8
A decision is needed to either market or sell a new invention. If the product is marketed, the
next decision is to take it international or national. Assume the details of the outcome branches
result in the decision tree of Figure 18–9. The probabilities for each outcome and PW of CFBT
(cash flow before taxes) are indicated. These payoffs are in millions of dollars. Determine the
best decision at the decision node D1.
Solution
Use the procedure above to determine that the D1 decision alternative to sell the invention
should maximize E(PW of CFBT).
1. Present worth of CFBT is supplied.
2. Calculate the expected PW for alternatives from nodes D2 and D3, using Equation [18.3].
In Figure 18–9, to the right of decision node D2, the expected values of 14 and 0.2 in ovals
are determined as
E(international decision) ⫽ 12(0.5) ⫹ 16(0.5) ⫽ 14
E(national decision) ⫽ 4(0.4) ⫺ 3(0.4) ⫺ 1(0.2) ⫽ 0.2
The expected PW values of 4.2 and 2 for D3 are calculated in a similar fashion.
497
498
Sensitivity Analysis and Staged Decisions
Chapter 18
14
14
International
D2
High
6.16
0.8
Low
Market
4.2
0.4
0.4
0.2
4.2
0.8
0.2
2
9
National
D1
0.2
International
D3
0.5
0.5
National
0.2
PW of
CFBT
($ million)
0.4
0.4
0.2
12
16
4
–3
–1
6
–3
6
–2
2
9
Sell
1.0
9
Expected values for each
decision alternative
Figure 18–9
Solution of a decision tree with present worth of estimated CFBT values,
Example 18.8.
3. Select the larger expected value at each decision node. These are 14 (international) at D2
and 4.2 (international) at D3.
4. Calculate the expected PW for the two D1 branches.
E(market decision) ⫽ 14(0.2) ⫹ 4.2(0.8) ⫽ 6.16
E(sell decision) ⫽ 9(1.0) ⫽ 9
The expected value for the sell decision is simple since the one outcome has a payoff of 9.
The sell decision yields the larger expected PW of 9.
5. The largest expected PW of CFBT path is to select the sell branch at D1 for a guaranteed
$9,000,000.
18.6 Real Options in Engineering Economics
As we learned with decision trees, many of the problems in engineering economy can be viewed
as staged decisions. When the decision to invest more or less can be delayed into the future, the
problem is called staged funding. As an illustration, assume a large company expects to sell an
energy-saving, window-mounted residential air conditioning unit at the rate of 100,000 per
month by the end of 2 years on the market. Decision makers may opt to (1) build the capacity to
supply100,000 per month to market immediately or (2) build capacity to supply 25,000 per
month now and test the market’s receptivity. If positive, they can stage the increase by 25,000
additional units each 6 months to meet current demand. Of course, if an aggressive competitor
enters the scene, or the economy falters, the staged funding decision will change as warranted.
These alternatives provide time-based options to the company. Before we go further, some definitions are needed.
An option is a purchase or investment that contractually provides the privilege to take a
stated action by some stated time in the future, or the right to not accept the offer and forfeit
the option.
18.6
Real Options in Engineering Economics
A real option, in engineering economy terms, is the investment (cost) in a project, process, or
system. The options usually involve physical (real) assets, buildings, equipment, materials, and
the like, thus, the word real. Options may also be leases, subcontracts, or franchises. The investment alternatives present varying amounts of risk, which is estimated by probabilities of occurrence for predictable future events.
Real options analysis is the application of techniques to determine the economic consequences
of delaying the funding decisions as allowed by the option. The estimated cash flows and other
consequences of these delays are analyzed with risk taken into account to the degree possible. A
measure of worth, e.g., PW or AW, is the criterion used to make the staged funding decisions. A
decision may be to expand, continue as is, contract, abandon, or replicate the alternative at the
time the option must be exercised.
An inherent part of real options analysis is the uncertainty of future estimates, as it is for most
economic analyses. After some illustrations of real options, we will discuss the probabilistic
dimensions. Samples from industry and everyday personal life that can be formulated as real
options follow.
Industrial Setting
New markets—Purchase equipment and staff to enter an expanding international market over
the next 5 years.
New planes—Purchase commercial airplanes now with an option to buy an additional 5 planes
over the next 3 years at the same price as that paid for the current order.
Removing car models—Ford Motor Company can decide to maintain production on an established car model with dwindling sales for the next 3 years or can opt to discontinue the model
in stages over a 1- or 2-year period.
Drilling lease—Buy a drilling option contract from landowners to drill for oil and gas at some
time in the next 10 years. The drilling may not be justified at this time, but the contract offers
the option to drill were it to become economically advantageous based on events such as increased oil prices or improved recovery technology.
Personal Decision Making
Extended car warranty—When purchasing a new car, the option to buy an extended-coverage
warranty beyond the manufacturer’s warranty is always an option. The price of the option is
the cost of the extended warranty. The uncertainties and risks are the future unknown costs for
repairs and failed components.
House insurance—When a homeowner has no mortgage to pay, maintaining house insurance
is an option. Deductibles are high enough, e.g., 1% to 5% of the fully appraised value, that
insurance primarily covers only catastrophic damage to the structure. Self-insurance, where
money is set aside for potential damages while accepting the risk that a major event will take
place, is an option for the homeowner.
Some of the primary characteristics (with an example) of a real options analysis performed within
the context of engineering economics are as follows:
• Cost to obtain the option to delay a decision (PW of initial investment, lease cost, or future
investment amount).
• Anticipated future options and cash flow estimates (double production with annual net cash
flows estimated).
• Time period for follow-on decisions (staged decision time, such as 1 year or a 3-year test period).
• Market and risk-free interest rates (expected market MARR of 12% per year and inflation
rate estimate of 4% per year).
• Estimates of risk and future uncertainty for each option (probability that an estimated cash
flow series will actually occur, if a specific option is selected).
• Economic criterion used to make a decision (PW, ROR, or other measure of worth).
499
500
Sensitivity Analysis and Staged Decisions
Chapter 18
It is common to use a decision tree to record and understand the options prior to performing a real
options analysis with risk included. Example 18.9 demonstrates the use of a decision tree and PW
analysis.
EXAMPLE 18.9
A start-up company in the solar energy production business, SolarScale Energy, Inc., has
developed and field-tested a modularized, scalar solar thermal electric (STE) generation
system that is relatively inexpensive to purchase and has an efficiency considerably better
than traditional photovoltaic (PV) panels. The technology is promising enough that Capital
Investor Funds (CIF) has provided $10 million for manufacturing. Additionally, a contract
with a consortium of sunbelt states has been offered, but not accepted thus far, for a total of
$1.5 million per year for a 2-year test period. By contract, the units will be marketed through
the state energy departments with all revenue going to the state treasuries. The lead engineer at SolarScale, the manager of CIF, and a conservation representative for the state
consortium have developed the following staged-funding options, based on the delayed
decision to increase manufacturing production level until preliminary results of the 2-year
contract are in hand.
Condition
Option
CIF Funding
Consortium Contract
2 ⫻ production
level
Additional $10 million in
year 2
Additional 8 years;
$4 million in years 3–10
Sales are excellent
1 ⫻ production
(3000–5000 units兾year)
level
Nothing; no salvage after
10 years
Additional 8 years;
$1.5 million in
years 3–10
Sales are poor
Reduce to ½ ⫻
(2000–3000 units兾year)
production
Nothing; sell for
$2.5 million after 5 years
Additional 3 years;
$1.5 million in
years 3–5
Sales are poor
(⬍ 2000 units兾year)
Nothing; sell for
$5 million after 2 years
Nothing
Sales are excellent
(⬎ 5000 units兾year)
Stop after
2 years
(a) Develop the two-stage decision tree for the options described.
(b) The base case is the 1 ⫻ production level with the 8-year follow-on contract from the consortium. If the estimates for this option are considered the most likely (expected value)
estimates, determine the present worth at a MARR of 10% per year.
(c) Determine the PW values for each possible final outcome at 10% per year, and identify the
best economic option when the stage 2 funding decision must be made.
Solution
(a) Figure 18–10 details the options with the year shown at the bottom. There are 2 outcome branches initially (accept option; decline option) and four final branches for the
accept decision at D1, based upon sales level. The decline option has a $0 outcome.
(SolarScale has other ways to pursue revenue that are not represented in this abbreviated example.)
(b) Perform a PW evaluation as we have in all previous chapters, assuming the estimates are
point estimates over the 10-year life of the project. The resulting PW1⫻ ⬍ 0 as shown
below indicates the contract is not justified economically. In $ millions,
PW1⫻ ⫽ ⫺10 ⫹ 1.5(P兾A,10%,10)
⫽ $⫺0.78 ($⫺780,000)
(c) Figure 18–11 is a spreadsheet screen shot that calculates the PW for each option using the
NPV function. The i* values are also shown using the IRR function. Note that the sale of
production assets after 5 years for $2.5 million (½ ⫻ level) or after 2 years for $5 million
(stop) is included. Also, the extra $10 million investment in year 2 for the 2 ⫻ level option
is included.
501
Real Options in Engineering Economics
18.6
2⫻
> 5000
D2
Sales
excellent
Accept
option
Sales
poor
3000–5000
1⫻
1/2⫻
2000–3000
D3
D1
Decline
option
Now
< 2000
Stop
now
No
sales
Year 2
Year 5
Year 10
Time
Figure 18–10
Decision tree showing real options over 10-year period, Example 18.9.
Sale of assets in this year
Figure 18–11
PW analysis of real options without risk considered, Example 18.9.
Only the 2 ⫻ production-level option is justified at MARR ⫽ 10% per year. If SolarScale and CIF, the financial backers, are not convinced that the sales level will exceed
5000 units per year, the contract option should be declined. Some marketing survey information and risk analysis may be very helpful before this important decision is made.
Of the characteristics listed above for real option situations, the primary one absent in
Example 18.9 is that of estimate variation and some measure of risk. Decision making under
risk is covered more extensively in the next chapter; however, we can use the following definition for this discussion of real options analysis.
When a parameter can take on more than one value and there is any estimate of chance or
probability about the opportunity that each value may be observed, risk is present.
Risk
502
Sensitivity Analysis and Staged Decisions
Chapter 18
A coin has two sides. If it is a perfectly balanced coin, a flip of the coin should result in heads 50%
of the time and tails 50% of the time. If the coin is intentionally biased in weight, such that 58% of
the time it lands heads up, then the long-run probability of heads is P(heads) ⫽ 0.58. Since the sum
of probabilities across all possible values must add to 1, the biased coin has P(tails) ⫽ 0.42.
There are a couple of points worth mentioning about risk and the calculated PW values for real
options analysis.
• When the risk is higher and the stakes are larger, the real options analysis is often more valuable. (We shall see this in Example 18.10.)
• Since one of the objectives of real options analysis is to evaluate the economic consequences
of delaying a decision, a PW value of the base case that is moderately positive means that the
project is justified and should be accepted immediately, without the decision delay. On the
other hand, if the PW is largely negative, the delay is likely not worthwhile, as it would take
a very large positive PW to result in E(PW) ⬎ 0. Thus, the project should be rejected now, not
delayed for a future decision.
We return to the previous example with some risk assessment added to determine if the contract
option should be declined, as indicated by the base case.
EXAMPLE 18.10
Before the stage 1 decision is made in Example 18.9 about SolarScale’s state-consortium contract offer, some expected sales information was collected. Once the results were reviewed by
the three individuals in attendance at the final decision-making meeting—one representative
each for SolarScale, CIF, and the consortium—each person recorded her or his estimated probability as a measure of risk that the sales would be excellent or poor, which represents the two
outcome possibilities were the option accepted. The results are as follows:
Probability of Outcome
SolarScale
CIF
Consortium
Excellent
Poor
0.5
0.8
0.6
0.5
0.2
0.4
Use these probability estimates to determine the expected PW value, provided equal weighting
is given to each representative’s input.
Solution
For each outcome (excellent and poor), select the best PW value from Figure 18–11, and then
find E(PW) for each representative.
Excellent: From 2 ⫻ level and 1 ⫻ level, select 2 ⫻ with PW ⫽ $1.97 million.
Poor: From ½ ⫻ level and stop now, select ½ ⫻ with PW ⫽ $⫺2.76 million.
In $ million, E(PW) for each organization is
E(PW for SolarScale) ⫽ 1.97(0.5) ⫺ 2.76(0.5) ⫽ $⫺0.40
E(PW for CIF) ⫽ 1.97(0.8) ⫺ 2.76(0.2) ⫽ $1.02
E(PW for consortium) ⫽ 1.97(0.6) ⫺ 2.76(0.4) ⫽ $0.08
With a 1/3 chance assigned to each representative, the overall E(PW of stage 2 decision) is
E(PW of stage 2 decision) ⫽ 0.33(⫺0.40 ⫹ 1.02 ⫹ 0.08)
⫽ $0.23 ($230,000)
The base case of 1 ⫻ level production in part (b) of Example 18.9 resulted in E(PW) ⫽
$−780,000. When compared with the positive E(PW) result here, we see that with consideration of the different options of production level and probabilities for sales level, the expected
PW has increased to a positive value. All other things being equal, the state consortium offer
should be accepted; that is, accept the real option of the contract.
Problems
503
There are many other examples and dimensions of real options analysis in engineering economics and in the area of financial analysis, where options analysis got its start some years ago.
If you are interested in this new and interesting area of analysis, consult more advanced texts and
journal articles on the topic of real options.
CHAPTER SUMMARY
In this chapter the emphasis is on sensitivity to variation in one or more parameters using a
specific measure of worth. When two alternatives are compared, compute and graph the
measure of worth for different values of the parameter to determine when each alternative is
better.
When several parameters are expected to vary over a predictable range, the measure of worth
is plotted and calculated using three estimates for a parameter—most likely, pessimistic, and
optimistic. This approach can help determine which alternative is best among several. Independence between parameters is assumed in all these analyses.
The combination of parameter and probability estimates results in the expected value relation
E(X) ⫽ ⌺XP(X)
This expression is also used to calculate E(revenue), E(cost), E(cash flow), and E(PW) for the
entire cash flow sequence of an alternative.
Decision trees are used to make a series of alternative selections. This is a way to explicitly
take risk into account. It is necessary to make several types of estimates for a decision tree: outcomes for each possible decision, cash flows, and probabilities. Expected value computations are
coupled with those for the measure of worth to solve the tree and find the best alternatives stage
by stage.
Staged funding over time can be approached using the evolving area of real options. Delaying
an investment decision and considering the risks of the future can improve the overall E(PW) of
a project, process, or system.
PROBLEMS
Sensitivity to Parameter Variation
18.1 Kahn Instruments is considering an investment of
$500,000 in a new product line. The company will
make the investment only if it will result in a rate
of return of 15% per year or higher. If the revenue
is expected to be between $135,000 and $165,000
per year for 5 years, determine if the decision to
invest is sensitive to the projected range of income
using a present worth analysis.
18.2 A young couple planning ahead for their retirement has decided that $2,600,000 is the amount
they will need to retire comfortably 20 years
from now. For the past 5 years they have been
able to invest one of their salaries ($50,000 per
year, which includes employer contributions)
while living off the other one. They plan to start
a family sometime in the next 10 years, and
when they have their first child, one of the parents will quit working, causing the savings to
decrease to $15,000 per year thereafter. If they
have gotten a rate of return of 10% per year on
their investments and expect to continue at this
ROR, is reaching their goal of $2.6 million
20 years from now sensitive to when they have
their first child (i.e., between now and 10 years
from now)? Use an FW analysis.
18.3 A company that manufactures high-speed submersible rotary indexing spindles is considering
upgrading the production equipment to reduce
costs over a 6-year planning horizon. The company can invest $80,000 now, 1 year from now,
or 2 years from now. Depending on when the investment is made, the savings will vary. That is,
the savings will be $25,000, $26,000, or $29,000
per year if the investment is made now (year 0),
in 1 year, or in 2 years, respectively. Will the
timing of the investment affect the request to
make at least a 20% per year return? Use future
worth analysis.
18.4 Different membrane systems are under consideration for treating 3 million gallons per day (MGD)
of cooling tower blowdown water to reduce its
504
Sensitivity Analysis and Staged Decisions
Chapter 18
volume. Option 1 is a low-pressure seawater reverse osmosis (SWRO) system that will operate at
500 psi with a fixed cost of $465 per day and an
operating cost of $0.67 per 1000 gallons. A second
option is a higher-pressure SWRO system offered
by vendor X that operates at 800 psi and will have
a lower fixed cost of $328 per day (because of
fewer membranes); however, its operating cost
will be $1.35 per 1000 gallons. A third option is
also a high-pressure SWRO system from vendor Y,
who claims that its system will have a lower operating cost of $1.28 per 1000 gallons and the same
fixed cost as that of vendor X. Determine if the
selection of a low- or high-pressure system is dependent on the lower operating cost offered by
vendor Y.
18.5 A machine that is currently used in manufacturing
circuit board card locks has AW ⫽ $–63,000 per
year. A possible replacement is under consideration with a first cost of $64,000 and an operating
cost of $38,000 per year for the next 3 years. Three
different engineers have given their opinion, about
what the salvage value of the new machine will be
3 years from now: $10,000, $13,000, and $18,000.
Is the decision to replace the machine sensitive to
the salvage value estimates at the company’s
MARR of 15% per year?
18.6 An equipment alternative is being economically
evaluated separately by three engineers at
Raytheon. The first cost will be $77,000, and the
life is estimated at 6 years with a salvage value of
$10,000. The engineers disagree, however, on the
estimated revenue the equipment will generate.
Joe has made an estimate of $10,000 per year. Jane
states that this is too low and estimates $14,000,
while Carlos estimates $18,000 per year. If the
before-tax MARR is 8% per year, use PW to determine if these different estimates will change the
decision to purchase the equipment.
18.7 The owner of a small construction company is
planning to purchase specialized equipment to
complete a contract he just received. The first
cost of the equipment is $250,000, and it will
likely have a salvage value of $90,000 in 3 years,
at which time he will not need the equipment
anymore. The operating cost is expected to be
$75,000 per year. Alternatively, the owner can
subcontract the work for $175,000 per year. Because the equipment is specialized, the owner is
not sure about the salvage value. He thinks it
might be worth as little as $10,000 in 3 years
(a scrap value). If his minimum attractive rate of
return is 15% per year, determine if the decision
to buy the equipment is sensitive to the salvage
value.
18.8 A company planning to borrow $10.5 million for a
plant expansion is not sure what the interest rate
will be when it applies for the loan. The rate could
be as low as 10% per year or as high as 12% per
year for a 5-year loan. The company will only
move forward with the project if the annual worth
of the expansion is below $5.7 million. The M&O
cost is fixed at $3.1 million per year. The salvage
could be $2 million if the interest rate is 10% or
$2.5 million if it is 12% per year. Is the decision to
move forward with the project sensitive to the interest rate and salvage value estimates?
18.9 A company that manufactures clear PVC pipe is
investigating the production options of batch and
continuous processes. Estimated cash flows are as
follows:
First cost, $
Annual cost, $ per year
Salvage value for any year, $
Life, years
Batch
Continuous
⫺80,000
⫺55,000
10,000
3–10
⫺130,000
⫺30,000
40,000
5
The chief operating officer (COO) has asked you
to determine if the batch option would ever have a
lower annual worth than the continuous flow system, using interest rates over a range of 5% to 15%
for the batch option but only 15% for the continuous flow system. (Note: The continuous flow process was previously determined to have its lowest
cost over a 5-year life cycle; the batch process can
be used from 3 to 10 years.)
18.10 An engineer collected average cost and revenue data
for Arenson’s FC1 handheld financial calculator.
Fixed cost ⫽ $300,000 per year
Cost per unit ⫽ $40
Revenue per unit ⫽ $70
(a)
(b)
What is the range in breakeven quantity if
there is possible variation in the fixed cost
from $200,000 to $400,000 per year? (Use
$50,000 increments.)
What is the incremental change in the breakeven quantity for each $50,000 change in
fixed cost?
The following information is used for Problems 18.11
through 18.14.
A new online patient diagnostics system for surgeons will
cost $200,000 to install, cost $5000 annually to maintain
and will have an expected life of 5 years. The added revenue is estimated to be $60,000 per year, and the MARR
is 10% per year. Examine the sensitivity of present worth
to variation in selected parameter estimates, while others
remain constant.
505
Problems
18.11 Sensitivity to first cost variation: $150,000 to
$250,000 (−25% to ⫹25%).
18.12 Sensitivity to revenue variation: $45,000 to $75,000
(−25% to ⫹25%).
18.13 Sensitivity to life variation: 4 years to 7 years (−20%
to ⫹40%).
Company A Company B
First cost, $
AOC, $ per year
Savings best estimate, $ per year
Salvage, $
Life, years
18.18 (a)
18.14 Plot the results on a graph similar to Figure 18–3
and comment on the relative sensitivity of each
parameter.
18.15 Charlene plans to place an annual savings
amount of A ⫽ $27,185 into a retirement program at the end of each year for 20 years starting
next year. She expects to retire and start to draw
a total of R ⫽ $60,000 per year 1 year after the
20th deposit. Assume an effective earning rate
of i ⫽ 6% per year on the retirement investments
and an infinite life. Determine and comment on
the sensitivity of the size of the annual withdrawal R for variations in A and i. Show hand
and spreadsheet solutions.
(a) Variation of ⫾5% in the annual deposit A.
(b) Variation of ⫾1% in the effective earning rate
i, that is, ranging from 5% to 7% per year.
18.16 Ned Thompson Labs performs tests on super
alloys, titanium, aluminum, and most metals. Tests
on metal composites that rely upon scanning electron microscope results can be subcontracted, or
the labs can purchase new equipment. Evaluate the
sensitivity of the economic decision to purchase
the equipment over a range of ⫾20% (in 10% increments) of the estimates for P, AOC, R, n, and
MARR (range on MARR is 12% to 16%). Use the
AW method and plot the results on a sensitivity
graph (like Figure 18–3). For which parameter(s)
is the AW most sensitive? Least sensitive?
First cost P ⫽ $−220,000
Salvage S ⫽ $20,000
Life n ⫽ 10 years
Annual operating cost AOC ⫽ $−30,000/year
Annual revenue R ⫽ $70,000 per year
MARR i ⫽ 15% per year
18.17 Titan manufactures and sells gas-powered electricity generators. It can purchase a new line of fuel
injectors from either of two companies. Cost and
savings estimates are made, but the savings estimate is unreliable at this time. Use an AW analysis
at 10% per year to determine if the selection between company A and company B changes when
the savings per year may vary as much as 40%
from the best estimates made thus far.
(b)
⫺50,000
⫺7,500
15,000
5,000
5
⫺37,500
⫺8,000
13,000
3,700
5
Graph the sensitivity of what a person
should be willing to pay now for a 9%,
$10,000 bond due in 10 years if there is a
30% change in (1) face value, (2) dividend
rate, or (3) required nominal rate of return,
which is expected to be 8% per year, compounded semiannually. The bond pays dividends semiannually.
If the investor did purchase the $10,000 face
value bond at a premium of 5% (i.e., 5%
above face value) and all your other estimates were correct, that is, 0% change, did
he pay too much or too little? How much?
Three Estimates
18.19 DVH Technologies purchases several parts for
the instruments it makes via a fixed-price contract of $190,000 per year from a local supplier.
The company is considering making the parts
in-house through the purchase of equipment that
will have a first cost of $240,000 with an estimated salvage value of $30,000 after 5 years. The
operating cost is difficult to estimate, but company engineers have made optimistic, most likely,
and pessimistic estimates of $60,000, $85,000,
and $120,000 per year, respectively. Determine if
the company should purchase the equipment
under any of the operating cost scenarios. The
MARR is 20% per year.
18.20 Astor Engineering recently merged with another
firm and could lease additional office space or
purchase its own building. The $30,000 per year
lease agreement will be a net, net, net lease,
which means that the lessee (Astor) will pay the
real estate taxes on the leased space, the building
insurance on the leased space, and the common
area maintenance. Since these costs are about the
same if Astor owned the building, they do not
need to be considered in the analysis. A new
building will cost $880,000 to purchase, but there
is considerable uncertainty about what it will be
worth in 20 years, which is the planning period
selected. The individuals involved in the discussion made optimistic, most likely, and pessimistic
estimates of $2,400,000, $1,400,000, and
$900,000, respectively. Determine if Astor should
purchase the building under any of the estimated
resale values at i ⫽ 10% per year.
506
Sensitivity Analysis and Staged Decisions
Chapter 18
18.21 Holly Farms is considering two environmental
chambers to accomplish detailed laboratory confirmations of online bacteria tests in chicken
meat for the presence of E. coli 0157:H7 and Listeria monocytogenes. There is some uncertainty
about how long the D103 chamber will be useful.
A realistic estimate is 3 years, but pessimistic
and optimistic estimates of 2 years and 6 years,
respectively, are also possible. The estimated
salvage value will remain the same. Using an interest rate of 10% per year, determine if any of
the D103 estimates would result in a lower cost
than that of the 490G chamber for a 6-year planning period.
Chamber D103 Chamber 490G
Installed cost, $
AOC, $ per year
Salvage value at 10% of P, $
Life, years
⫺400,000
⫺4,000
40,000
2, 3, or 6
⫺250,000
⫺3,000
25,000
2
18.22 When the country’s economy is expanding, AB Investment Company is optimistic and expects a
MARR of 15% for new investments. However, in
a receding economy the expected return is 8%.
Normally a 10% return is required. An expanding
economy causes the estimates of asset life to go
down about 20%, and a receding economy makes
the n values increase about 10%. Calculate and observe or plot the sensitivity of PW values versus
(a) the MARR and (b) the life values for the two
plans detailed below, using the most likely estimates for the other factors. (c) Considering all the
analyses, under which scenario, if any, should plan
M or Q be rejected?
Initial investment, $
Cash flow, $ per year
Life, years
Plan M
Plan Q
⫺100,000
⫹15,000
20
⫺110,000
⫹19,000
20
Expected Value
18.23 Determine the expected net operating income
(NOI) from sales of micro turbine components.
The probabilities are 20%, 50%, and 30% for revenues of $800,000, $1,000,000, and $1,100,000
per year, respectively, and operating expenses are
constant at $200,000 per year.
18.24 A company that manufactures amplified pressure
transducers is trying to decide between a dualspeed and a variable-speed machine. The engineers are not sure about the salvage value of the
variable-speed machine, so they have asked several different used-equipment dealers for
estimates. The results can be summarized as follows: there is a 32% chance of getting $20,000, a
45% chance of getting $28,000, and a 13%
chance of getting $34,000. Also, there is a 10%
chance that the company may have to pay $5000
to dispose of the equipment. Calculate the
expected salvage value.
18.25 The average success probability for a wildcat oil
well drilled in the Wind River basin 7 miles from
the nearest existing production well is estimated
to be 13%. If the value of the oil has equal
chances of being $1.5 million, $1.9 million, and
$2.4 million, what is the expected income from
the well?
18.26 Nationwide income from monthly sales data
(rounded to the nearest $100,000) of Stay Flat
vacuum hold-down tables for last year is shown
below. Determine the expected value of the
monthly income, if economic conditions remain
the same.
Income, $ per Month
Number
of Months
500,000
600,000
700,000
800,000
900,000
4
2
1
2
3
18.27 Determine the expected maximum rainfall intensity in El Paso, Texas for the month of July using
the estimated probabilities shown.
Rainfall Rate, inches per hour
Probability
3
4
5
6
0.4
0.3
0.2
0.1
18.28 There are four estimates made for the anticipated
cycle time to produce a subcomponent. The estimates, in seconds, are 10, 20, 30, and 50. (a) If
equal weight is placed on each estimate, what is
the expected cycle time? (b) If the largest time is
disregarded, what is the percent reduction in the
expected time?
18.29 The PW value for an alternative is expected to be
one of two values based on bids from two vendors.
Your office partner told you that the low bid is
$3200 per year. If she indicates a chance of 70% of
accepting the high bid and that her expected PW is
$5875, what is the PW of the high bid?
18.30 A total of 40 different proposals were evaluated by
the IRAD (Industrial Research and Development)
committee during the past year. Twenty were
funded. Their rate of return estimates are summarized with the i* values rounded to the nearest
507
Problems
integer. For the accepted proposals, calculate the
expected rate of return E(i).
Proposal
ROR, i*%
Number of
Proposals
⫺8
⫺5
0
5
8
10
15
1
1
5
5
2
3
3
20
18.34 A flagship hotel in Cedar Falls must construct a
retaining wall next to its parking lot due to the
widening of the city’s main thoroughfare located
in front of the hotel. The amount of rainfall experienced in a short time may cause damage in varying amounts, and the wall increases in cost in
order to protect against larger and faster rainfalls.
The probabilities of a specific amount of rainfall
in a 30-minute period and wall cost estimates are
as follows:
18.31 Beckman Electronics has performed an economic
analysis of proposed service in a new region of the
country. The three-estimate approach to sensitivity
analysis has been applied. The optimistic and pessimistic values each have an estimated 20% chance
of occurring. Use the FW values shown to determine the expected FW.
FW value, $
$35,000 per year. Can the company expect to make
a return of 8% per year on its investment? Use
present worth analysis.
Optimistic
Most Likely
Pessimistic
300,000
50,000
⫺25,000
18.32 A very successful health and recreation club
wants to construct a mock mountain for climbing and exercise outside for its customers’ use.
Because of its location, there is a 30% chance of
a 120-day season of good outdoor weather, a
50% chance of a 150-day season, and a 20%
chance of a 165-day season. The mountain will
be used by an estimated 350 persons each day of
the 4-month (120-day) season, but by only 100
per day for each extra day the season lasts. The
feature will cost $375,000 to construct and require a $25,000 rework each 4 years; and the annual maintenance and insurance costs will be
$56,000. The climbing fee will be $5 per person.
If a life of 10 years is anticipated and a 12% per
year return is expected, determine if the addition
is economically justified.
18.33 The owner of Ace Roofing may invest $200,000 in
new equipment. A life of 6 years and a salvage
value of 12% of first cost are anticipated. The annual extra revenue will depend upon the state of
the housing and construction industry. The extra
revenue is expected to be only $20,000 per year if
the current slump in the industry continues. Real
estate economists estimate a 50% chance of the
slump lasting 3 years and they give it a 20% chance
of continuing for 3 additional years. However, if
the depressed market does improve, during either
the first or second 3-year period, the revenue of the
investment is expected to increase by a total of
Rainfall, Inches
per 30 Minutes
Probability of
Greater Rainfall
First Cost
of Wall, $
2.0
2.25
2.5
3.0
3.25
0.3
0.1
0.05
0.01
0.005
200,000
225,000
300,000
400,000
450,000
The wall will be financed through a 6% per year
loan. The principal and interest will be repaid over
a 10-year period. Records indicate an average
damage of $50,000 has occurred with heavy rains,
due to the relatively poor cohesive properties of
the soil along the thoroughfare. A discount rate of
6% per year is applicable. Find the amount of rainfall to protect against by choosing the retaining
wall with the smallest AW value over the 10-year
period.
Decision Trees
18.35 For the decision tree branch shown, determine the
expected values of the two outcomes if decision
D3 is already selected and the maximum outcome
value is sought. (This decision branch is part of a
larger tree.)
Probability
0.4
0.3
0.3
Value, $
55
– 30
10
D3
0.6
0.4
– 17
0
18.36 A large decision tree has an outcome branch detailed (next page). If decisions D1, D2, and D3 are
all options in a 1-year period, find the decision
508
Sensitivity Analysis and Staged Decisions
Chapter 18
path that maximizes the outcome value. There are
specific investments necessary for decision nodes
D1, D2, and D3, as indicated on each branch.
annual equivalent cost for each alternative is dependent upon specific circumstances of the plant, producer, or contractor. The information shown details
the circumstance, a probability of occurrence, and
the estimated annual cost. Construct and solve a decision tree to determine the least-cost alternative to
provide the subassemblies.
Value, $
30
0.9
D3
0.4
Investment
$50
0.6
0.1
D1
100
– 50
Decision
Alternative
1. Make
500
$80
90
2. Buy off
the shelf
$25
D2
Value $
150
– 30
75
0.3
0.3
0.4
0.5
0.5
$30
3. Contract
50
18.37 Decision D4, which has three possible alternatives—x, y, or z—must be made in year 3 of a 6-year
study period in order to maximize the expected
value of present worth. Using a rate of return of 15%
per year, the investment required in year 3 and the
estimated cash flows for years 4 through 6, determine which decision should be made in year 3.
Investment Cash flow, $1000
Required, $
Outcome
Years
6 probability
3
4
5
50
50
50
0.7
40
30
20
0.3
30
40
50
0.45
30
30
30
0.55
– 350,000 190
170
150
0.7
– 30
– 30
– 30
0.3
– 200,000
Low
x
High
D4
y
Low
z
High
Low
– 75,000
Plant:
A
B
C
Quantity:
⬍5000, pay premium
5000 available
⬎5000, forced to buy
Delivery:
Timely delivery
Late delivery, then
buy some off shelf
Probability
0.3
0.5
0.2
⫺250,000
⫺400,000
⫺350,000
0.2
0.7
0.1
⫺550,000
⫺250,000
⫺290,000
0.5
0.5
⫺175,000
⫺450,000
200
– 100
$20
High
Outcomes
Annual Cost for
5000 Units, $
per Year
18.38 A total of 5000 mechanical subassemblies are
needed annually on a final assembly line. The subassemblies can be obtained in one of three ways:
(1) Make them in one of three plants owned by the
company; (2) buy them off the shelf from the one
and only manufacturer; or (3) contract to have them
made to specifications by a vendor. The estimated
18.39 The president of ChemTech is trying to decide
whether to start a new product line or purchase a
small company. It is not financially possible to do
both. To make the product for a 3-year period will
require an initial investment of $250,000. The expected annual cash flows with probabilities in
parentheses are: $75,000 (0.5), $90,000 (0.4), and
$150,000 (0.1). To purchase the small company
will cost $450,000 now. Market surveys indicate a
55% chance of increased sales for the company and
a 45% chance of severe decreases with an annual
cash flow of $25,000. If decreases are experienced
in the first year, the company will be sold immediately (during year 1) at a price of $200,000. Increased sales could be $100,000 the first 2 years. If
this occurs, a decision to expand after 2 years at an
additional investment of $100,000 will be considered. This expansion could generate cash flows
with indicated probabilities as follows: $120,000
(0.3), $140,000 (0.3), and $175,000 (0.4). If expansion is not chosen, the current size will be maintained with anticipated sales to continue. Assume
there are no salvage values on any investments. Use
the description given and a 15% per year return to
do the following.
(a) Construct a decision tree with all values and
probabilities shown.
(b) Determine the expected PW values at the
“expansion/no expansion” decision node after
2 years, provided sales are up.
(c) Determine what decision should be made
now to offer the greatest return possible for
ChemTech.
509
Additional Problems and FE Exam Review Questions
(d)
Explain in words what would happen to the
expected values at each decision node if the
planning horizon were extended beyond
3 years and all cash flow values continued as
forecasted in the description.
Real Options
18.40 A privately held company that makes chips that are
essential for high-volume data storage is valued at
$3 billion. A computer company that wants to get
into cloud computing is considering purchasing
the company, but because of the uncertain economy, it would prefer to purchase an option that will
allow it to buy the company for up to 1 year from
now at a cost of $3.1 billion. What is the maximum
amount the company should be willing to pay for
the option, if its MARR is 12% per year?
18.41 A company that is considering adding a new product line has determined that the first cost would be
$80 million. The company is not sure about how
the product will be received, so it has projected revenues using optimistic, most likely, and pessimistic
estimates of $35 million, $25 million, and $10 million, respectively, with equal probability for each.
Instead of expanding now, the company could implement a test program for 1 year in a limited area
that will cost $4 million. (The full-scale project will
still cost $80 million if implemented after the test
program is over.) This will provide the company
with the option to move forward or cancel the project. The criterion identified to move ahead with
full-scale implementation is that revenues must exceed $900,000. In this case, the pessimistic estimate will be eliminated, and equal probability will
be placed on the remaining revenue projections. If
the company uses a 5-year planning horizon and a
MARR of 12% per year, should the company go
ahead with the full-scale project now or take the
option to implement the test program for 1 year?
18.42 Dow Chemical is considering licensing a low liquid discharge (LLD) water treatment system from a
small company that developed the process. Dow
can purchase a 1-year option for $150,000 that will
give it time to pilot-test the LLD process, or Dow
can acquire the license now at a cost of $1.8 million
plus 25% of sales. If Dow waits 1 year, the cost
will increase to $1.9 million plus 30% of sales. If
Dow projects the sales to be $1,000,000 per year
over the 5-year license period, should the company
license the process now or purchase the option to
license it after the 1-year test period? Assume the
MARR is 15% per year.
18.43 Abby has just negotiated a $15,000 price on a
2-year-old car and is with the salesman closing the
deal. There is a 1-year sales warranty with the purchase; however, an extended warranty is available
for $2500 that will cover the same repairs and component failures as the 1-year warranty for 3 additional years. Abby understands this to be a real options situation with the price of the option ($2500)
paid to avoid future, unknown costs. To help with
her decision, the salesman provided three typical
sets of historical data on estimated repair costs for
used cars. The first-year costs are shown as zero
because they will be covered by the sales warranty.
Year
Repair cost, $ per year:
A
B
C
1
2
3
4
0
0
0
⫺500
⫺1000
0
⫺1200
⫺1400
⫺500
⫺850
⫺400
⫺2000
The salesman said case C is the base case, since it
shows that the extended warranty is not needed because the cost of repairs equals the warranty cost.
Abby immediately recognized this to be the case
only when i ⫽ 0%.
(a) If Abby assumes that each repair cost scenario
has equal probability of occurring with her
car, and money is worth 5% per year to her,
how much should she be willing to pay for the
extended warranty that is offered at $2500?
(b) If the base case actually occurs for her car
and she does not purchase the warranty, what
is the PW value of the expected future costs
at i ⫽ 5% per year?
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
18.44 In conducting a sensitivity analysis, all of the
following could be used as a measure of worth
except:
(a) Present worth
(b) Cost-capacity equations
(c) Annual worth
(d) Benefit-cost ratio
18.45 When the measure of worth is plotted versus percent
change for several parameters, the parameter that is
the most sensitive in the economic analysis is the one:
(a) That has the steepest curve
(b) That has the flattest curve
(c) With the largest present worth
(d) With the shortest life
510
Sensitivity Analysis and Staged Decisions
Chapter 18
18.46 In conducting a formalized sensitivity analysis
using three estimates, the three estimates should be:
(a) Strategic, pessimistic, real likely:
(b) Deterministic, realistic, optimistic
(c) Optimistic, pessimistic, most likely
(d) Authentic, realistic, probabilistic
18.47 For annual worth values of $30,000, $40,000, and
$50,000 with chances of 20%, 20%, and 60%, respectively, the expected AW is closest to:
(a) $34,000
(b) $40,000
(c) $44,000
(d) $48,000
18.48 The AW of a 3-year-old machine that is used in the
manufacture of modular magnetic encoders, which
provide data on the speed and positioning of rotating
motor shafts, is $−48,000. A challenger will have a
first cost of $90,000, an operating cost of $29,000,
and a salvage value after 5 years that may vary considerably. For an interest rate of 10% per year and
optimistic, most likely, and pessimistic salvage values of $15,000, $10,000, and $2000, respectively,
the salvage value(s) for which the challenger AW
will be lower than that of the defender are:
(a) All of them
(b) Only the optimistic one
(c)
(d)
Only the optimistic and pessimistic ones
None of them
18.49 A decision tree includes all of the following except:
(a) Probability estimates for each outcome
(b) Measure of worth as the selection criterion
(c) Expected results from a decision at each
stage
(d) The MARR
18.50 A real options analysis is most valuable when:
(a) The risk is low and stakes are high
(b) The stakes are low and risk is high
(c) The stakes are high and risk is high
(d) The stakes are low and risk is low
18.51 A small manufacturing company needs to purchase a machine that will have a first cost of
$70,000. The company wants to buy an option that
will allow it to purchase the machine for the same
price of $70,000 for up to 1 year from now. If the
company’s MARR is 10% per year, the maximum
amount the company should pay for the option is
closest to:
(a) $5850
(b) $6365
(c) $6845
(d) $7295
CASE STUDY
SENSITIVITY TO THE ECONOMIC ENVIRONMENT
Background and Information
Case Study Questions
Berkshire Controllers usually finances its engineering projects with a combination of debt and equity capital. The resulting MARR ranges from a low of 4% per year, if business is
slow, to a high of 10% per year. Normally, a 7% per year return is expected. Also the life estimates for assets tend to go
down about 20% from normal in a vigorous business environment and up about 10% in a receding economy. The following estimates are the most likely values for two expansion
plans currently being evaluated. Plan A will be executed at
one location; Plan B will require two locations. All monetary
estimates are in $1000 units.
At the weekly meeting, you were asked to examine the following questions from Berkshire’s president.
1. Are the PW values for plans A and B sensitive to
changes in the MARR?
2. Are the PW values sensitive to varying life estimates?
3. Is the breakeven point for the first cost of plan A sensitive to the changes in MARR as business goes from vigorous to receding?
Plan B
First cost, $
AOC, $ per year
Salvage value, $
Estimated life, years
Plan A
Location 1
Location 2
⫺10,000
⫺500
1,000
40
⫺30,000
⫺100
5,000
40
⫺5,000
⫺200
⫺200
20
511
Case Study
CASE STUDY
SENSITIVITY ANALYSIS OF PUBLIC SECTOR PROJECTS—WATER SUPPLY PLANS
Background
Information
One of the most basic services provided by municipal governments is the delivery of a safe, reliable water supply. As
cities grow and extend their boundaries to outlying areas,
they often inherit water systems that were not constructed
according to city codes. The upgrading of these systems is
sometimes more expensive than installing one correctly in
the first place. To avoid these problems, city officials sometimes install water systems beyond the existing city limits
in anticipation of future growth. This case study was extracted from such a countywide water and wastewater management plan and is limited to only some of the water supply alternatives.
From about a dozen suggested plans, five methods were
developed by an executive committee as alternative ways
of providing water to the study area. These methods were
then subjected to a preliminary evaluation to identify the
most promising alternatives. Six attributes or factors were
used in the initial rating: ability to serve the area, relative
cost, engineering feasibility, institutional issues, environmental considerations, and lead time requirement. Each
factor carried the same weighting and had values ranging
from 1 to 5, with 5 being best. After the top three alternatives were identified, each was subjected to a detailed economic evaluation for selection of the best alternative.
These detailed evaluations included an estimate of the capital investment of each alternative amortized over 20 years
at 8% per year interest and the annual maintenance and operation (M&O) costs. The annual cost (an AW value) was
then divided by the population served to arrive at a monthly
cost per household.
Table 18–5 presents the results of the screening using the six
factors rated on a scale of 1 to 5. Alternatives 1A, 3, and 4
were determined to be the three best and were chosen for further evaluation.
TABLE 18–5
Detailed Cost Estimates
All amounts are cost estimates.
Alternative 1A
Capital cost
Land with water rights: 1720 hectares
@ $5000 per hectare
Primary treatment plant
Booster station at plant
Reservoir at booster station
Site cost
Transmission line from river
Transmission line right-of-way
Percolation beds
Percolation bed piping
Production wells
Well field gathering system
Distribution system
Additional distribution system
Reservoirs
Reservoir site, land, and development
Subtotal
Engineering and contingencies
$8,600,000
2,560,000
221,425
50,325
40,260
3,020,000
23,350
2,093,500
60,400
510,000
77,000
1,450,000
3,784,800
250,000
17,000
22,758,060
5,641,940
Total capital investment
$28,400,000
Results of Rating Six Factors for Each Alternative, Case Study
Factors
Alternative
Description
1A
Receive city
water and
recharge wells
Joint city and
county plant
County treatment
plant
Desalt
groundwater
Develop military
water
3
4
8
12
Ability
to
Supply Relative Engineering Institutional Environmental
Lead Time
Area
Cost
Feasibility
Issues
Considerations Requirement Total
5
4
3
4
5
3
24
5
4
4
3
4
3
23
4
4
3
3
4
3
21
1
2
1
1
3
4
12
5
5
4
1
3
1
19
512
Sensitivity Analysis and Staged Decisions
Chapter 18
Alternative 4
Maintenance and operation costs (annual)
Pumping 9,812,610 kWh per year
@ $0.08 per kWh
Fixed operating cost
Variable operating cost
Taxes for water rights
Total annual M&O cost
Total capital investment ⫽ $29,000,000
$ 785,009
180,520
46,730
48,160
Total annual M&O cost ⫽ $1,063,449
Total annual cost
⫽ 2,953,650 ⫹ 1,063,449
$1,060,419
⫽ $4,017,099
Household cost
Total annual cost ⫽ equivalent capital
investment ⫹ M&O cost
⫽ 29,000,000(A兾P,8%,20)
⫹ 1,063,449
⫽ $70.76 per month
⫽ 28,400,000(A兾P,8%,20)
⫹ 1,060,419
On the basis of the lowest monthly household cost, alternative 3 (joint city and county plant) is the most economically
attractive.
⫽ 2,892,540 ⫹ 1,060,419
Case Study Exercises
⫽ $3,952,959
Average monthly household cost to serve 95% of 4980
households is
( )(
)(
1 ——
1
1
——
Household cost ⫽ (3,952,959) —
12 4980 0.95
⫽ $69.63 per month
)
Alternative 3
Total capital investment ⫽ $29,600,000
Total annual M&O cost ⫽ $867,119
Total annual cost
⫽ 29,600,000(A兾P,8%,20)
⫹ 867,119
⫽ 3,014,760 ⫹ 867,119
⫽ $3,881,879
Household cost
⫽ $68.38 per month
TABLE 18–6
Alternative 1A
Pessimistic
Most likely
Optimistic
Alternative 3
Pessimistic
Most likely
Optimistic
Alternative 4
Pessimistic
Most likely
Optimistic
1. If the environmental considerations factor is to have a
weighting of twice as much as any of the other five factors, what is its percentage weighting?
2. If the ability to supply area and relative cost factors were
each weighted 20% and the other four factors 15% each,
which alternatives would be ranked in the top three?
3. By how much would the capital investment of alternative 4 have to decrease to make it more attractive than
alternative 3?
4. If alternative 1A served 100% of the households instead
of 95%, by how much would the monthly household
cost decrease?
5. (a) Perform a sensitivity analysis on the two parameters of M&O costs and number of households to
determine if alternative 3 remains the best economic choice. Three estimates are made for each
parameter in Table 18–6. M&O costs may vary up
Pessimistic, Most Likely, and Optimistic
Estimates for Two Parameters
Annual
M&O
Costs
Number
of
Households
⫹1%
$1,060,419
−1%
4980
⫹2%
⫹5%
⫹5%
$867,119
0%
4980
⫹2%
⫹5%
⫹2%
$1,063,449
−10%
4980
⫹2%
⫹5%
Case Study
(pessimistic) or down (optimistic) from the most
likely estimates presented in the case statement.
The estimated number of households (4980) is determined to be the pessimistic estimate. Growth of
2% up to 5% (optimistic) will tend to lower the
monthly cost per household.
(b) Consider the monthly cost per household for alternative 4, the optimistic estimate. The number of
513
households is 5% above 4980, or 5230. What is
the number of households that would have to be
available in order for this option to have exactly
the same monthly household cost as that for alternative 3 at the optimistic estimate of 5230 households?
CHAPTER 19
More on
Variation and
Decision
Making under
Risk
L E A R N I N G
O U T C O M E S
Purpose: Incorporate decision making under risk into an engineering economy evaluation using probability, sampling,
and simulation.
SECTION
TOPIC
LEARNING OUTCOME
19.1
Risk versus certainty
• Understand the approaches to decision making
under risk and certainty.
19.2
Probability and distributions
• Construct a probability distribution and
cumulative distribution for one variable.
19.3
Random sample
• Obtain a random sample from a cumulative
distribution using a random number table.
19.4
, , and 2
• Estimate the population expected value, standard
deviation, and variance from a random sample.
19.5
Simulation
• Use Monte Carlo sampling and spreadsheetbased simulation for alternative evaluation.
his chapter further expands our ability to analyze variation in estimates, to consider probability, and to make decisions under risk. Fundamentals discussed include variables; probability distributions, especially their graphs and properties
of expected value and dispersion; random sampling; and the use of simulation to
account for estimate variation in engineering economy studies.
Through coverage of variation and probability, this chapter complements topics in the first
sections of Chapter 1: the role of engineering economy in decision making and economic
analysis in the problem-solving process. These techniques are more time-consuming than using estimates made with certainty, so they should be used primarily for critical parameters.
T
19.1 Interpretation of Certainty,
Risk, and Uncertainty
All things in the world vary—one from another, over time, and with different environments. We
are guaranteed that variation will occur in engineering economy due to its emphasis on decision
making for the future. Except for the use of breakeven analysis, sensitivity analysis, and a very
brief introduction to expected values, virtually all our estimates have been certain; that is, no
variation in the amount has entered into the computations of PW, AW, ROR, or any relations
used. For example, the estimate that cash flow next year will be $⫹4500 is one of certainty. Decision making under certainty is, of course, not present in the real world now and surely not in the
future. We can observe outcomes with a high degree of certainty, but even this depends upon the
accuracy and precision of the scale or measuring instrument.
To allow a parameter of an engineering economy study to vary implies that risk, and possibly
uncertainty, is introduced.
When there may be two or more observable values for a parameter and it is possible to estimate the chance that each value may occur, risk is present. Virtually all decision making is
performed under risk.
As an illustration, decision making under risk is introduced when an annual cash flow estimate
has a 50-50 chance of being either $−1000 or $⫹500.
Decision making under uncertainty means there are two or more values observable, but the
chances of their occurring cannot be estimated or no one is willing to assign the chances. The
observable values in uncertainty analysis are often referred to as states of nature.
For example, consider the states of nature to be the rate of national inflation in a particular
country during the next 2 to 4 years: remain low, increase 2% to 6% annually, or increase 6%
to 8% annually. If there is absolutely no indication that the three values are equally likely, or
that one is more likely than the others, this is a statement that indicates decision making under
uncertainty.
Example 19.1 explains how a parameter can be described and graphed to prepare for decision
making under risk.
EXAMPLE 19.1
CMS in Fairfield, Virginia received three bids each from vendors for two different pieces of
large equipment, A and B. One of each piece of equipment must be purchased. Tom, an engineer
at CMS, performed an evaluation of each bid and assigned it a rating between 0 and 100, with
100 points being the best of the three. The total for each piece of equipment is 100%. The bid
amounts and ratings are shown at the top of Figure 19–1.
(a) Consider the ratings as the chance out of 100 that the bid will be chosen, and plot cost
versus chance for each vendor.
(b) Since one each of A and B must be purchased, the total cost will vary somewhere between the
sum of the lowest bids ($11 million) and the sum of the highest bids ($25 million). Plot this
range with an equal chance of 1 in 14 that any amount in between these limits is possible.
(c) Discuss the significant difference between the values of the cost (x axis values) in the
graphs in (a) and (b) above and how the chances are stated (y axis values).
Risk
516
Figure 19–1
Equipment A
Equipment B
Bid, $1000
Rating, %
Bid, $1000
Rating, %
3,000
5,000
10,000
65
25
10
8,000
10,000
15,000
33.3
33.3
33.3
80
Equipment A
65
Chance, %
60
40
25
20
10
0
1
2
3
4
5
6
7
8
Cost, $ million
9
10
11
12
Equipment B
Chance, %
40
33.3
33.3
33.3
7
8
9
Cost, $ million
10
11
12
25
27
29
20
0
1
2
3
4
5
6
13
14
15
16
(a) Specific values
Total cost
1
—
14
Chance
Plot of cost estimates
versus chance for (a)
each piece of equipment and (b) total cost
range, Example 19.1.
More on Variation and Decision Making under Risk
Chapter 19
0
7
9
11
13
15
17
19
21
Cost, $ million
23
(b) Continuous range
Solution
(a) Figure 19–1a plots the specific bids for equipment A and B. The chances (ratings) for A and
for B add to 100%. No values between the specific bids have any chance of occurring, according to the single-estimate bids from the three vendors.
(b) The range of total cost is between $11 million and $25 million, as shown in Figure 19–1b.
Tom decided to make his estimate of total cost continuous between these two extremes.
This means that the discrete sums of bids ($11 million, $15 million, and $25 million) are
no longer used. Rather the entire range from $11 million to $25 million with a chance for
every total cost in between is included. Every value has a chance of 1 in 14 of being observed. Now, the sum is a continuous value.
(c) In the graph for bid values (Figure 19–1a), only specific or discrete estimates are included
on the x axis. In the graph for the sum of the cost for equipment A and B (Figure 19–1b),
the y axis values are continuous over a specific range.
In the next section, the term variable is defined and two types of variables are explained—
discrete and continuous—as illustrated here in an elementary form.
19.1
Interpretation of Certainty, Risk, and Uncertainty
Before initiating an engineering economy study, it is important to decide if the analysis will
be conducted with certainty for all parameters or if risk will be introduced. A summary of the
meaning and use for each type of analysis follows.
Decision Making under Certainty This is what we have done in most analyses thus far.
Deterministic estimates are made and entered into measure of worth relations—PW, AW, FW,
ROR, B兾C—and decision making is based on the results. The values estimated can be considered
the most likely to occur with all chance placed on the single-value estimate. A typical example is
an asset’s first cost estimate made with certainty, say, P ⫽ $50,000. A plot of P versus chance has
the general form of Figure 19–1a with one vertical bar at $50,000 and 100% chance placed on it.
The term deterministic, in lieu of certainty, is often used when single-value or single-point
estimates are used exclusively.
In fact, sensitivity analysis using different values of an estimate is simply another form of
analysis with certainty, except that the analysis is repeated with different values, each estimated with certainty. The resulting measure of worth values are calculated and graphically
portrayed to determine the decision’s sensitivity to different estimates for one or more
parameters.
Decision Making under Risk Now the element of chance is formally taken into account.
However, it is more difficult to make a clear decision because the analysis attempts to accommodate variation. One or more parameters in an alternative will be allowed to vary. The estimates will be expressed as in Example 19.1 or in slightly more complex forms. Fundamentally,
there are two ways to consider risk in an analysis:
Expected value analysis. Use the chance and parameter estimates to calculate expected
values E(parameter) via formulas such as Equation [18.2]. Analysis results in E(cash flow),
E(AOC), and the like; and the final result is the expected value for a measure of worth, such
as E(PW), E(AW), E(ROR), E(B兾C). To select the alternative, choose the most favorable
expected value of the measure of worth. In an elementary form, this is what we learned about
expected values in Chapter 18. The computations may become more elaborate, but the principle is fundamentally the same.
Simulation analysis. Use the chance and parameter estimates to generate repeated computations of the measure of worth relation by randomly sampling from a plot for each varying
parameter similar to those in Figure 19–1. When a representative and random sample is complete, an alternative is selected utilizing a table or plot of the results. Usually, graphics are an
important part of decision making via simulation analysis. Basically, this is the approach discussed in the rest of this chapter.
Decision Making under Uncertainty When chances are not known for the identified
states of nature (or values) of the uncertain parameters, the use of expected value–based decision making under risk as outlined above is not an option. In fact, it is difficult to determine
what criterion to use to even make the decision. If it is possible to agree that each state is
equally likely, then all states have the same chance, and the situation reduces to one of decision making under risk, because expected values can be determined. Because of the relatively
inconclusive approaches necessary to incorporate decision making under uncertainty into an
engineering economy study, the techniques can be quite useful but are beyond the intended
scope of this text.
In an engineering economy study, observed parameter values will vary from the value estimated at the time of the study. However, when performing the analysis, not all parameters should
be considered as probabilistic (or at risk). Those that are estimable with a relatively high degree
of certainty should be fixed for the study. Accordingly, the methods of sampling, simulation, and
statistical data analysis are selectively used on parameters deemed important to the decisionmaking process. Parameters such as P, AOC, material and unit costs, sales price, revenues, etc.,
are the targets of decision making under risk. Anticipated variation in interest rates is more commonly addressed by sensitivity analysis.
The remainder of this chapter concentrates on decision making under risk as applied in an
engineering economy study. Sections 19.2 to 19.4 provide foundation material necessary to design and correctly conduct a simulation analysis (Section 19.5).
517
(a) Discrete and continuous variable scales and
(b) scales for a variable
versus its probability.
Discrete
variable
3
5
10
15
Years
Estimated life, n
Continuous
variable
…
–100%
∞
Rate of return, i
(a)
1.0
0.8
P(n)
Figure 19–2
More on Variation and Decision Making under Risk
Chapter 19
Discrete variable
0.6
Estimated life,
probability vs. years
0.4
0.2
0
3
5
10
15
n, years
1.0
0.8
P(i )
518
Continuous variable
0.6
Rate of return,
probability vs. percent
0.4
0.2
10 5
0
5
10
15
20
25
30
35 …
i, %
(b)
19.2 Elements Important to Decision
Making under Risk
Some basics of probability and statistics are essential to correctly perform decision making under
risk via expected value or simulation analysis. They are the random variable, probability, probability distribution, and cumulative distribution, as defined here. (If you are already familiar with
them, this section will provide a review.)
A random variable or variable is a characteristic or parameter that can take on any one of several values. Variables are classified as discrete or continuous. Discrete variables have several
specific, isolated values, while continuous variables can assume any value between two stated
limits, called the range of the variable.
The estimated life of an asset is a discrete variable. For example, n may be expected to
have values of n ⫽ 3, 5, 10, or 15 years, and no others. The rate of return is an example of a
continuous variable; i can vary from −100% to , that is, −100% i . The ranges of possible values for n (discrete) and i (continuous) are shown as the x axes in Figure 19–2a. (In
probability texts, capital letters symbolize a variable, say X, and small letters x identify a
specific value of the variable. Though correct, this level of rigor in terminology is not applied
in this chapter.)
Probability is a number between 0 and 1.0 that expresses the chance in decimal form that a random variable (discrete or continuous) will take on any value from those identified for it. Probability is simply the amount of chance, divided by 100.
Probabilities are commonly identified by P(Xi) or P(X Xi), which is read as the probability that
the variable X takes on the value Xi. (Actually, for a continuous variable, the probability at a
single value is zero, as shown in a later example.) The sum of all P(Xi) for a variable must be 1.0,
519
Elements Important to Decision Making under Risk
19.2
a requirement already discussed. The probability scale, like the percentage scale for chance in
Figure 19–1, is indicated on the ordinate (y axis) of a graph. Figure 19–2b shows the 0 to 1.0
range of probability for the variables n and i.
A probability distribution describes how probability is distributed over the different values of
a variable. Discrete variable distributions look significantly different from continuous variable
distributions, as indicated by the inset at the right.
P(Xi )
Discrete
Xi
P(Xi )
The individual probability values are stated as
Continuous
P(Xi) ⴝ probability that X equals Xi
[19.1]
The distribution may be developed in one of two ways: by listing each probability value for each
possible variable value (see Example 19.2) or by a mathematical description or expression that
states probability in terms of the possible variable values (Example 19.3).
Xi
F(Xi )
Discrete
Cumulative distribution, also called the cumulative probability distribution, is the accumulation of probability over all values of a variable up to and including a specified value.
Identified by F(Xi), each cumulative value is calculated as
F(Xi) ⴝ sum of all probabilities through the value Xi
ⴝ P ( X ⱕ X i)
[19.2]
As with a probability distribution, cumulative distributions appear differently for discrete (stairstepped) and continuous variables (smooth curve). Examples 19.2 and 19.3 illustrate cumulative
distributions that correspond to specific probability distributions. These fundamentals about
F(Xi) are applied in the next section to develop a random sample.
EXAMPLE 19.2
Alvin is a medical doctor and biomedical engineering graduate who practices at Medical Center Hospital. He is planning to start prescribing an antibiotic that may reduce infection in patients with flesh wounds. Tests indicate the drug has been applied up to 6 times per day without
harmful side effects. If no drug is used, there is always a positive probability that the infection
will be reduced by a person’s own immune system.
Published drug test results provide good probability estimates of positive reaction (i.e., reduction in the infection count) within 48 hours for increased treatments per day. Use the probabilities listed below to construct a probability distribution and a cumulative distribution for
the total number of treatments per day.
Number of Added
Treatments per Day
Probability of Infection Reduction
for Each Added Treatment
0
1
2
3
4
5
6
0.07
0.08
0.10
0.12
0.13
0.25
0.25
Solution
Define the random variable T as the number of added treatments per day. Since T can take on
only seven different values, it is a discrete variable. The probability of infection reduction is
listed for each value in column 2 of Table 19–1. The cumulative probability F(Ti) is determined
using Equation [19.2] by adding all P(Ti) values through Ti, as indicated in column 3.
Figure 19–3a and b shows plots of the probability distribution and cumulative distribution,
respectively. The summing of probabilities to obtain F(Ti) gives the cumulative distribution the
stair-stepped appearance, and in all cases the final F(Ti) ⫽ 1.0, since the total of all P(Ti) values
must equal 1.0.
F(Xi)
Continuous
Xi
Xi
520
More on Variation and Decision Making under Risk
Chapter 19
TABLE 19–1
Figure 19–3
(a) Probability distribution P(Ti) and
(b) cumulative distribution F(Ti) for
Example 19.2.
Probability Distribution and Cumulative
Distribution for Example 19.2
(1)
(2)
Number per Day
Ti
Probability
P(Ti)
(3)
Cumulative
Probability
F(Ti)
0
1
2
3
4
5
6
0.07
0.08
0.10
0.12
0.13
0.25
0.25
0.07
0.15
0.25
0.37
0.50
0.75
1.00
F(Ti)
P(Ti)
0.3
1.00
1.0
0.25 0.25
0.8
0.75
0.2
0.6
0.50
0.12 0.13
0.1
0.37
0.10
0.07 0.08
0.3
0.25
0.15
0.07
0
0
0
1
2
3
4
5
6
Ti
(a)
0
1
2
3
4
5
6
Ti
(b)
Comment
Rather than use a tabular form as in Table 19–1 to state P(Ti) and F(Ti) values, it is possible to
express them for each value of the variable.
{
0.07
0.08
0.10
P(Ti) ⫽ 0.12
0.13
0.25
0.25
T1 ⫽ 0
T2 ⫽ 1
T3 ⫽ 2
T4 ⫽ 3
T5 ⫽ 4
T6 ⫽ 5
T7 ⫽ 6
{
0.07
0.15
0.25
F(Ti) ⫽ 0.37
0.50
0.75
1.00
T1 ⫽ 0
T2 ⫽ 1
T3 ⫽ 2
T4 ⫽ 3
T5 ⫽ 4
T6 ⫽ 5
T7 ⫽ 6
In basic engineering economy situations, the probability distribution for a continuous variable
is commonly expressed as a mathematical function, such as a uniform distribution, a triangular
distribution (both discussed in Example 19.3 in terms of cash flow), or the more complex, but
commonly used, normal distribution. For continuous variable distributions, the symbol f(X) is
routinely used instead of P(Xi), and F(X) is used instead of F(Xi), simply because the point probability for a continuous variable is zero. Thus, f(X) and F(X) are continuous lines and curves.
EXAMPLE 19.3
As president of a manufacturing systems consultancy, Sallie has observed the monthly cash
flows that have occurred over the last 3 years into company accounts from two longstanding
clients. Sallie has concluded the following about the distribution of these monthly cash flows:
521
Elements Important to Decision Making under Risk
19.2
Client 1
Estimated low cash flow: $10,000
Estimated high cash flow: $15,000
Most likely cash flow: same for all values
Distribution of probability: uniform
Client 2
Estimated low cash flow: $20,000
Estimated high cash flow: $30,000
Most likely cash flow: $28,000
Distribution of probability: mode at $28,000
The mode is the most frequently observed value for a variable. Sallie assumes cash flow to be
a continuous variable referred to as C. (a) Write and graph the two probability distributions and
cumulative distributions for monthly cash flow, and (b) determine the probability that monthly
cash flow is no more than $12,000 for client 1 and at least $25,000 for client 2.
Solution
All cash flow values are expressed in $1000 units.
Client 1: monthly cash flow distribution
(a) The distribution of cash flows for client 1, identified by the variable C1, follows the uniform
distribution. Probability and cumulative probability take the following general forms.
1
f(C1) ⫽ —————
high low
low value C1 high value
1
f(C1) ———
HL
L C1 H
value low
F(C1) ——————
high low
C1 L
F(C1) ———
HL
[19.3]
low value C1 high value
L C1 H
[19.4]
For client 1, monthly cash flow is uniformly distributed with L $10, H $15, and $10
C1 $15. Figure 19–4 is a plot of f(C1) and F(C1) from Equations [19.3] and [19.4].
1 0.2
f(C1) —
5
C1 10
F(C1) ————
5
$10 C1 $15
$10 C1 $15
(b) The probability that client 1 has a monthly cash flow of no more than $12 is easily determined from the F(C1) plot as 0.4, or a 40% chance. If the F(C1) relation is used directly, the
computation is
12 10 0.4
F($12) P(C1 $12) ————
5
Figure 19–4
F(C1 )
Uniform distribution for
client 1 monthly cash
flow, Example 19.3.
1.0
0.8
0.6
0.4
f (C1)
0.2
0.2
0
0
10
15
C1
10
12
15
C1
522
More on Variation and Decision Making under Risk
Chapter 19
Client 2: monthly cash flow distribution
(a) The distribution of cash flows for client 2, identified by the variable C2, follows the triangular distribution. This probability distribution has the shape of an upward-pointing triangle with the peak at the mode M, and downward-sloping lines joining the x axis on either
side at the low (L) and high (H) values. The mode of the triangular distribution has the
maximum probability value.
2
f(mode) ⫽ f(M) ⫽ ———
HL
[19.5]
The cumulative distribution is comprised of two curved line segments from 0 to 1 with a
break point at the mode, where
ML
F(mode) F(M) ———
HL
[19.6]
For C2, the low value is L $20, the high is H $30, and the most likely cash flow is the
mode M $28. The probability at M from Equation [19.5] is
2 0.2
2
——
f(28) ————
30 20 10
The break point in the cumulative distribution occurs at C2 28. Using Equation [19.6],
28 20 0.8
F(28) ————
30 20
Figure 19–5 presents the plots for f(C2) and F(C2). Note that f(C2) is skewed, since the
mode is not at the midpoint of the range H L, and F(C2) is a smooth S-shaped curve with
an inflection point at the mode.
(b) From the cumulative distribution in Figure 19–5, there is an estimated 31.25% chance that
cash flow is $25 or less. Therefore,
F($30) F($25) P(C2 $25) 1 0.3125 0.6875
Comment
The general relations f(C2) and F(C2) are not developed here. The variable C2 is not a uniform
distribution; it is triangular. Therefore, it requires the use of an integral to find cumulative probability values from the probability distribution f(C2).
F(C2)
Mode
1.0
0.8
0.6
0.4
0.3125
Mode
f (C2)
0.2
0.2
0
0
28
20
30
20
C2
Figure 19–5
Triangular distribution for client 2 monthly cash flow, Example 19.3.
25
C2
28
30
Random Samples
19.3
19.3 Random Samples
Estimating a parameter with a single value in previous chapters is the equivalent of taking a random sample of size 1 from an entire population of possible values. As an illustration, assume that
estimates of first cost, annual operating cost, interest rate, and other parameters are used to compute one PW value in order to accept or reject an alternative. Each estimate is a sample of size 1
from an entire population of possible values for each parameter. Now, if a second estimate is
made for each parameter and a second PW value is determined, a sample of size 2 has been taken.
If all values in the population were known, the probability distribution and cumulative distribution would be known. Then a sample would not be necessary.
When we perform an engineering economy study and utilize decision making under certainty,
we use one estimate for each parameter to calculate a measure of worth (i.e., a sample of size 1
for each parameter). The estimate is the most likely value, that is, one estimate of the expected
value. We know that all parameters will vary somewhat; yet some are important enough, or will
vary enough, that a probability distribution should be determined or assumed for it and the parameter treated as a random variable. This is using risk, and a sample from the parameter’s probability distribution—P(X ) for discrete or f(X ) for continuous—helps formulate probability statements about the estimates. This approach complicates the analysis somewhat; however, it also
provides a sense of confidence (or possibly a lack of confidence in some cases) about the decision
made concerning the economic viability of the alternative based on the varying parameter. (We
will further discuss this aspect later, after we learn how to correctly take a random sample from
any probability distribution.)
A random sample of size n is the selection in a random fashion of n values from a population
with an assumed or known probability distribution, such that the values of the variable have the
same chance of occurring in the sample as they are expected to occur in the population.
Suppose Yvon is an engineer with 20 years of experience working for the Aircraft Safety
Commission. For a two-crew aircraft, there are three parachutes on board. The safety standard
states that 99% of the time, all three chutes must be “fully ready for emergency deployment.”
Yvon is relatively sure that nationwide the probability distribution of N, the specific number of
chutes fully ready, may be described by the probability distribution
{
0.005
0.015
P(N ⫽ Ni) ⫽
0.060
0.920
N ⫽ 0 chutes ready
N ⫽ 1 chute ready
N ⫽ 2 chutes ready
N ⫽ 3 chutes ready
This means that the safety standard is clearly not met nationwide. Yvon is in the process of
sampling 200 (randomly selected) corporate and private aircraft across the nation to determine how many chutes are classified as fully ready. If the sample is truly random and Yvon’s
probability distribution is a correct representation of actual parachute readiness, the observed N values in the 200 aircraft will approximate the same proportions as the population
probabilities, that is, 1 aircraft with 0 chutes ready, etc. Since this is a sample, it is likely that
the results won’t track the population exactly. However, if the results are relatively close, the
study indicates that the sample results may be useful in predicting parachute safety across
the nation.
To develop a random sample, use random numbers (RN) generated from a uniform probability distribution for the discrete numbers 0 through 9, that is,
P(Xi) ⫽ 0.1
for Xi ⫽ 0, 1, 2, … , 9
In tabular form, the random digits so generated are commonly clustered in groups of two digits,
three digits, or more. Table 19–2 is a sample of 264 random digits clustered into two-digit numbers. This format is very useful because the numbers 00 to 99 conveniently relate to the cumulative distribution values 0.01 to 1.00. This makes it easy to select a two-digit RN and enter F(X )
to determine a value of the variable with the same proportions as it occurs in the probability distribution. To apply this logic manually and develop a random sample of size n from a known
523
524
More on Variation and Decision Making under Risk
Chapter 19
TABLE 19–2
Random Digits Clustered into Two-Digit Numbers
51 82 88 18 19 81 03 88 91 46 39 19 28 94 70 76 33 15 64 20 14 52
73 48 28 59 78 38 54 54 93 32 70 60 78 64 92 40 72 71 77 56 39 27
10 42 18 31 23 80 80 26 74 71 03 90 55 61 61 28 41 49 00 79 96 78
45 44 79 29 81 58 66 70 24 82 91 94 42 10 61 60 79 30 01 26 31 42
68 65 26 71 44 37 93 94 93 72 84 39 77 01 97 74 17 19 46 61 49 67
75 52 14 99 67 74 06 50 97 46 27 88 10 10 70 66 22 56 18 32 06 24
discrete probability distribution P(X ) or a continuous variable distribution f(X ), the following
procedure may be used.
1. Develop the cumulative distribution F(X) from the probability distribution. Plot F(X).
2. Assign the RN values from 00 to 99 to the F(X) scale (the y axis) in the same proportion as
the probabilities. For the parachute safety example, the probabilities from 0.0 to 0.15 are
represented by the random numbers 00 to 14. Indicate the RNs on the graph.
3. To use a table of random numbers, determine the scheme or sequence of selecting RN
values—down, up, across, diagonally. Any direction and pattern is acceptable, but the
scheme should be used consistently for one entire sample.
4. Select the first number from the RN table, enter the F(X ) scale, and observe and record the
corresponding variable value. Repeat this step until there are n values of the variable that
constitute the random sample.
5. Use the n sample values for analysis and decision making under risk. These may include
•
•
•
•
•
Plotting the sample probability distribution.
Developing probability statements about the parameter.
Comparing sample results with the assumed population distribution.
Determining sample statistics (Section 19.4).
Performing a simulation analysis (Section 19.5).
EXAMPLE 19.4
Develop a random sample of size 10 for the variable N, number of months, as described by the
probability distribution
{
0.20
P(N ⫽ Ni) ⫽ 0.50
0.30
N ⫽ 24
N ⫽ 30
N ⫽ 36
[19.7]
Solution
Apply the procedure above, using the P(N ⫽ Ni) values in Equation [19.7].
1. The cumulative distribution, Figure 19–6, is for the discrete variable N, which can assume
three different values.
2. Assign 20 numbers (00 through 19) to N1 ⫽ 24 months, where P(N ⫽ 24) ⫽ 0.2; 50 numbers to N2 ⫽ 30; and 30 numbers to N3 ⫽ 36.
3. Initially select any position in Table 19–2, and go across the row to the right and onto the
row below toward the left. (Any routine can be developed, and a different sequence for
each random sample may be used.)
4. Select the initial number 45 (4th row, 1st column), and enter Figure 19–6 in the RN range
of 20 to 69 to obtain N ⫽ 30 months.
5. Select and record the remaining nine values from Table 19–2 as shown below.
RN
45
44
79
29
81
58
66
70
24
82
N
30
30
36
30
36
30
30
36
30
36
525
Random Samples
19.3
Figure 19–6
Assigned
random
numbers
Cumulative
probability
1.0
Cumulative distribution
with random number values assigned in proportion to probabilities,
Example 19.4.
70–99
F(Ni )
0.7
45
20–69
0.2
00–19
24
30
36
Ni, months
Now, using the 10 values, develop the sample probabilities.
Months
N
Times in
Sample
Sample
Probability
Equation [19.7]
Probability
24
30
36
0
6
4
0.00
0.60
0.40
0.2
0.5
0.3
With only 10 values, we can expect the sample probability estimates to be different from the
values in Equation [19.7]. Only the value N ⫽ 24 months is significantly different, since no RN
of 19 or less occurred. A larger sample will definitely make the probabilities closer to the
original data.
To take a random sample of size n for a continuous variable, the procedure above is applied,
except the random number values are assigned to the cumulative distribution on a continuous
scale of 00 to 99 corresponding to the F(X ) values. As an illustration, consider Figure 19–4,
where C1 is the uniformly distributed cash flow variable for client 1 in Example 19.3. Here L ⫽
$10, H ⫽ $15, and f(C1) ⫽ 0.2 for all values between L and H (all values are divided by $1000).
The F(C1) is repeated as Figure 19–7 with the assigned random number values shown on the right
scale. If the two-digit RN of 45 is chosen, the corresponding C1 is graphically estimated to be
$12.25. It can also be linearly interpolated as $12.25 ⫽ 10 (45兾100)(15 – 10).
F(C1)
Figure 19–7
Assigned
random
numbers
Cumulative
probability
1.0
99
0.8
79
0.6
59
0.4
39
0.2
19
Random numbers
assigned to the
continuous variable of
client 1 cash flows in
Example 19.3.
45
00
10
11
12
12.25
13
14
15
C1, $1000
526
More on Variation and Decision Making under Risk
Chapter 19
For greater accuracy when developing a random sample, especially for a continuous variable,
it is possible to use 3-, 4-, or 5-digit RNs. These can be developed from Table 19–2 simply by
combining digits in the columns and rows or obtained from tables with RNs printed in larger
clusters of digits. In computer-based sampling, most simulation software packages have an RN
generator built in that will generate values in the range of 0 to 1 from a continuous variable uniform distribution, usually identified by the symbol U(0, 1). The RN values, usually between
0.00000 and 0.99999, are used to sample directly from the cumulative distribution employing
essentially the same procedure explained here. The Excel functions RAND and RANDBETWEEN are described in Appendix A, Section A.3.
An initial question in random sampling usually concerns the minimum size of n required to
ensure confidence in the results. Without detailing the mathematical logic, sampling theory,
which is based upon the law of large numbers and the central limit theorem (check a basic statistics book to learn about these), indicates that an n of 30 is sufficient. However, since reality does
not follow theory exactly, and since engineering economy often deals with sketchy estimates,
samples in the range of 100 to 200 are the common practice. But samples as small as 10 to 25
provide a much better foundation for decision making under risk than the single-point estimate
for a parameter that is known to vary widely.
19.4 Expected Value and Standard Deviation
Two very important measures or properties of a random variable are the expected value and standard deviation. If the entire population for a variable were known, these properties would be
calculated directly. Since they are usually not known, random samples are commonly used to
estimate them via the sample mean and the sample standard deviation, respectively. The following is a brief introduction to the interpretation and calculation of these properties using a random
sample of size n from the population.
The usual symbols are Greek letters for the true population measures and English letters for
the sample estimates.
True Population Measure
Symbol
Name
Sample Estimate
Symbol
Name
__
Expected value
or E(X)
______
___
√Var(X)
Standard deviation or
or √2
Mu or true mean
Sample mean
X
__
Sigma or true
standard deviation
s or √ s2
Sample standard
deviation
The expected value E(X) is the long-run expected average if the variable is sampled many times.
The population expected value is not known exactly, since the population
itself is not known
__
completely, so µ is estimated either by E(X) from a distribution or by X , the sample mean. Equation [18.2], repeated here as Equation [19.8], is used to compute the E(X) of a probability distribution, and Equation [19.9] is the sample mean, also called the sample average.
Population:
Probability distribution:
Sample:
E(X) ⴝ ⌺Xi P(Xi)
__
sum of sample values
X ⴝ ——————————
sample size
⌺Xi ⌺fi Xi
ⴝ ——
n ⴝ ———
n
[19.8]
[19.9]
The fi in the second form of Equation [19.9] is__the frequency of Xi, that is, the number of times
each value occurs in the sample. The resulting X is not necessarily an observed value of the variable; it is the long-run average value and can take on any value within the range of the variable.
(We omit the subscript i on X and f when there is no confusion introduced.)
Expected Value and Standard Deviation
19.4
EXAMPLE 19.5
Kayeu, an engineer with Pacific NW Utilities, is planning to test several hypotheses about
residential electricity bills in North American and Asian countries. The variable of interest is
X, the monthly residential bill in U.S. dollars (rounded to the nearest dollar). Two small samples have been collected from different countries of North America and Asia. Estimate the
population expected value. Do the samples (from a nonstatistical viewpoint) appear to be
drawn from one population of electricity bills or from two different populations?
North American, Sample 1, $
40
66
75
92
107
Asian, Sample 2, $
84
90
104
187
190
159
275
Solution
Use Equation [19.9] for the sample mean.
n7
n5
Sample 1:
Sample 2:
Xi 814
Xi 655
__
X
__ $116.29
X $131.00
Based solely on the small sample averages, the approximate $15 difference, which is
only11% of the larger average bill, does not seem sufficiently large to conclude that the two
populations are different. There are several statistical tests available to determine if samples come from the same or different populations. (Check a basic statistics text to learn
about them.)
Comment
There are three commonly used measures of central tendency for data. The sample average
is the most popular, but the mode and the median are also good measures. The mode, which
is the most frequently observed value, was utilized in Example 19.3 for a triangular distribution. There is no specific mode in Kayeu’s two samples, since all values are different. The
median is the middle value of the sample. It is not biased by extreme sample values, as is the
mean. The two medians in the samples are $92 and $104. Based solely on the medians, the
conclusion is still that the samples do not necessarily come from two different populations
of electricity bills.
The standard deviation __
s or s(X ) is the dispersion or spread of values about the expected value
E(X ) or sample average X .
The sample standard deviation s estimates the property , which is the population measure of
dispersion about the expected value of the variable. A probability distribution for data with strong
central tendency is more closely clustered about the center of the data, and has a smaller s, than
a wider, more dispersed distribution. In Figure 19–8, the samples with larger s values—s1 and
s4—have a flatter, wider probability distribution.
f (X)
P(X)
s2
s3
s4
s1
X1
s1
X2
s2
s4
s3
Figure 19–8
Sketches of distributions with different separate lines standard deviation values.
527
528
More on Variation and Decision Making under Risk
Chapter 19
Actually, the variance s2 is often quoted as the measure of dispersion. The standard deviation
is simply the square root of the variance, so either measure can be used. The s value is what we
use routinely in making computations about risk and probability. Mathematically, the formulas
and symbols for variance and standard deviation of a discrete variable and a random sample of
size n are as follows:
___
Population:
2 ⴝ Var(X)
and
_______
ⴝ √ 2 ⴝ √ Var(X)
Probability distribution:
Var(X) ⴝ ⌺[Xi ⫺ E(X)]2P(Xi)
sum of (sample value ⫺ sample average)2
Sample:
s2 ⴝ ——————————————————
sample size ⫺ 1
[19.10]
__
⌺(Xi ⫺ X)2
ⴝ —————
n⫺1
[19.11]
__
s ⴝ √ s2
Equation [19.11] for sample variance is usually applied in a more computationally convenient
form.
__
__
⌺Xi2
⌺fiXi2
n X
n X
2
2
s2 ⴝ ———
ⴝ ———
ⴚ ———
ⴚ ———
nⴚ1 nⴚ1
nⴚ1 nⴚ1
[19.12]
The standard deviation uses the sample average
as a basis about which to measure the spread or
__
dispersion of data via the calculation (X ⫺ X ), which can have a minus or plus sign.
__ To accurately
measure the dispersion in both directions from the average, the quantity (X ⫺ X ) is squared. To
return to the__dimension of the variable itself, the square root of Equation [19.11] is extracted. The
term (X ⫺ X )2 is called the mean-squared deviation, and s has historically also been referred to
as the root-mean-square deviation. The fi in the second form of Equation [19.12] uses the frequency of each Xi value to calculate s2.
One simple way to combine the average and standard deviation is to determine the percentage
or fraction of the sample that is within ⫾1, ⫾2, or ⫾3 standard deviations of the average, that is,
__
X ⫾ ts
for t ⫽ 1, 2, or 3
[19.13]
In probability terms, this is stated as
__
__
P(X ⴚ ts ⱕ X ⱕ X ⴙ ts)
[19.14]
__
Virtually all the sample values will always be within the ⫾3s range of
__ X, but the percent within
⫾1s will vary depending on how the data points are distributed about X. Example 19.6
__ illustrates
the calculation of s to estimate and incorporates s with the sample average using X ⫾ ts.
EXAMPLE 19.6
(a) Use the two samples of Example 19.5 to estimate population variance and standard deviation for electricity bills. (b) Determine the percentages of each sample that are inside the ranges
of 1 and 2 standard deviations from the mean.
Solution
(a) For illustration purposes only, apply the two different relations to calculate s for the two
samples. For sample 1 (North American)
with n ⫽ 7, use X to identify
__
__ the values. Table 19–3
presents the computation of (X X)2 for Equation [19.11], with X $116.29. The resulting s2 and s values are
37,743.40
s2 ————— 6290.57
6
s $79.31
Expected Value and Standard Deviation
19.4
TABLE 19–3 Computation of Standard Deviation
__
Using Equation [19.11] with X
$116.29, Example 19.6
__
X, $
X−X
40
66
75
92
107
159
275
76.29
50.29
41.29
24.29
9.29
42.71
158.71
__
(X − X)2
5,820.16
2,529.08
1,704.86
590.00
86.30
1,824.14
25,188.86
814
37,743.40
TABLE 19–4 Computation of Standard Deviation
__
Using Equation [19.12] with Y $131,
Example 19.6
Y, $
Y2
84
90
104
187
190
7,056
8,100
10,816
34,969
36,100
655
97,041
__
For sample 2 (Asian), use Y to identify the values. With n ⫽ 5 and Y ⫽ 131, Table 19–4
shows Y2 for Equation [19.12]. Then
97,041 5
(131)2 42,260.25 1.25(17,161) 2809
s2 ——— —
4
4
s $53
The dispersion is smaller for the Asian sample ($53) than for the North American sample
($79.31).
__
__
(b) Equation [19.13] determines the ranges of X 1s and X 2s. Count the number of sample
data points between the limits, and calculate the corresponding percentage. See Figure 19–9
for a plot of the data and the standard deviation ranges.
–
X = 116.29
– 80
– 40
0
40
80
120
–
X ± 1s
–
X ± 2s
160
200
240
280
X
240
Y
(a)
–
Y = 131
0
40
80
120
–
Y ± 1s
–
Y ± 2s
160
200
(b)
Figure 19–9
Values, averages, and standard deviation ranges for (a) North American and
(b) Asian samples, Example 19.6.
529
530
More on Variation and Decision Making under Risk
Chapter 19
North American sample
__
1s 116.29
X
79.31
for a range of $36.98 to $195.60
Six out of seven values are within this range, so the percentage is 85.7%.
__
X
2s 116.29
158.62
for a range of $42.33 to $274.91
__
There are still six of the seven values within the X 2s range. The limit $42.33 is meaningful only from the probabilistic perspective; from the practical viewpoint, use zero, that
is, no amount billed.
Asian sample
__
Y
1s 131
53
for a range of $78 to $184
There are three of five values, or 60%, within the range.
__
Y
2s 131
106
for a range of $25 to $237
__
All five of the values are within the Y
2s range.
Comment
A second common measure of dispersion is the range, which is simply the largest minus the
smallest sample values. In the two samples here, the range estimates are $235 and $106.
Only the hand computations for E(X), s, and s2 have been demonstrated here. Calculators and
spreadsheets all have functions to determine these values by simply entering the data.
Before we perform simulation analysis in engineering economy, it may be of use to summarize the expected value and standard deviation relations for a continuous variable, since Equations [19.8] through [19.12] address only discrete variables. The primary differences are that the
summation symbol is replaced by the integral over the defined range of the variable, which we
identify as R, and that P(X) is replaced by the differential element f (X) dX. For a stated continuous probability distribution f(X), the formulas are
E(X) ⫽ 兰R Xf(X) dX
Expected value:
[19.15]
Var(X) ⫽ 兰R X2f(X) dX [E(X)]2
Variance:
[19.16]
For a numerical example, again use the uniform distribution in Example 19.3 (Figure 19–4) over
the range R from $10 to $15. If we identify the variable as X, rather than C1, the following are
correct.
1 0.2
$10 X $15
f(X ) —
5
E(X) 兰 X(0.2) dX 0.1X2 10 0.1(225 100) $12.5
|
15
R
0.2X3 15 (12.5)2
Var(X) 兰 X2(0.2) dX (12.5)2 ——
10
R
3
0.06667(3375 1000) 156.25 2.08
|
____
√ 2.08 $1.44
Therefore, the uniform distribution between L $10 and H $15 has an expected value of
$12.5 (the midpoint of the range, as expected) and a standard deviation of $1.44.
EXAMPLE 19.7
Christy is the regional safety engineer for a chain of franchise-based gasoline and food stores.
The home office has had many complaints and several legal actions from employees and customers about slips and falls due to liquids (water, oil, gas, soda, etc.) on concrete surfaces.
Expected Value and Standard Deviation
19.4
Corporate management has authorized each regional engineer to contract locally to apply to all
exterior concrete surfaces a newly marketed product that absorbs up to 100 times its own
weight in liquid and to charge a home office account for the installation. The authorizing letter
to Christy states that, based upon their simulation and random samples that assume a normal
population, the cost of the locally arranged installation should be about $10,000 and almost
always is within the range of $8000 to $12,000.
You have been asked to write a brief but thorough summary about the normal distribution,
explain the $8000 to $12,000 range statement, and explain the phrase “random samples that
assume a normal population.”
Solution
The following summaries about the normal distribution and sampling will help explain the
authorization letter.
Normal distribution, probabilities, and random samples
The normal distribution is also referred to as the bell-shaped curve, the Gaussian distribution, or the error distribution. It is, by far, the most commonly used probability distribution
in all applications. It places exactly one-half of the probability on either side of the mean or
expected value. It is used for continuous variables over the entire range of numbers. The
normal distribution is found to accurately predict many types of outcomes, such as IQ values; manufacturing errors about a specified size, volume, weight, etc.; and the distribution of
sales revenues, costs, and many other business parameters around a specified mean, which is
why it may apply in this situation.
The normal distribution, identified by the symbol N(, 2), where is the expected value or
mean and 2 is the variance, or measure of spread, can be described as follows:
• The mean locates the probability distribution (Figure 19–10a), and the spread of the distribution varies with variance (Figure 19–10b), growing wider and flatter for larger variance
values.
__
• When a sample is taken, the estimates are identified as sample mean X for and sample
standard deviation s for .
• The normal probability distribution f(X) for a variable X is quite complicated, because its
formula is
{[
(X )2
1___ exp ————
f(X) ⫽ ———
22
√ 2
]}
where exp represents the number e 2.71828.
Since f(X) is so unwieldy, random samples and probability statements are developed using a
transformation,
called the standard normal distribution (SND), which uses and (popula__
tion) or X and s (sample) to compute values of the variable Z.
Population :
X
deviation from mean ———
Z —————————
standard deviation
[19.17]
__
Sample:
XX
Z ———
s
[19.18]
The SND for Z (Figure 19–10c) is the same as for X, except that it always has a mean of 0 and
a standard deviation of 1, and it is identified by the symbol N(0, 1). Therefore, the probability
values under the SND curve can be stated exactly. It is always possible to transfer back to the
original values from sample data by solving Equation [19.17] for X:
X Z
[19.19]
531
532
Figure 19–10
More on Variation and Decision Making under Risk
Chapter 19
f(X)
Normal distribution
showing (a) different
mean values µ, (b)
different standard deviation values , and
(c) relation of
normal X to standard
normal Z.
1
2
2
1
• Equal dispersion
(1 = 2 = 3)
• Increasing means
3
X
3
(a)
f (X)
• Increasing
dispersion
(1 < 2 < 3)
• Two different
means
1
2
3
1 = 3
X
2
(b)
f (X)
Normal distribution
– 3
+ 3
X
0.9974
0.9546
f(Z)
0.6826
Standard normal
distribution
0.3413 0.3413
0.0013 0.0214
–3
0.1360
–2
0.1360
–1
0
1
0.0214 0.0013
2
3 Z
(c)
Several probability statements for Z and X are summarized in the following table and are
shown on the distribution curve for Z in Figure 19–10c.
Variable X Range
Probability
Variable Z Range
1
1
2
2
3
3
0.3413
0.6826
0.4773
0.9546
0.4987
0.9974
0 to 1
1 to 1
0 to 2
2 to 2
0 to 3
3 to 3
As an illustration, probability statements from this tabulation and Figure 19–10c for X and Z
are as follows:
The probability that X is within 2 of its mean is 0.9546.
The probability that Z is within 2 of its mean, which is the same as between the values
2 and 2, is also 0.9546.
Monte Carlo Sampling and Simulation Analysis
19.5
In order to take a random sample from a normal N(µ, 2) population, a specially prepared table of
SND random numbers is used. (Tables of SND values are available in many statistics books.) The
numbers are actually values from the Z or N(0, 1) distribution and have values such as 2.10,
1.24, etc. Translation from the Z value back to the sample values for X is via Equation [19.19].
Interpretation of the home office memo
The statement that virtually all the local contract amounts should be between $8000 and
$12,000 may be interpreted as follows: A normal distribution is assumed with a mean of µ
$10,000 and a standard deviation for $667, or a variance of 2 ($667)2; that is, an
N[$10,000, ($667)2] distribution is assumed. The value $667 is calculated using the fact
that virtually all the probability (99.74%) is within 3 of the mean, as stated above. Therefore,
3 $2000
and
$667
(rounded off)
As an illustration, if six SND random numbers are selected and used to take a sample of size 6
from the normal distribution N[$10,000, ($667)2], the results are as follows:
SND Random Number
Z
X Using Equation [19.19]
X ⴝ Z ⴙ
2.10
3.12
0.23
1.24
2.61
0.99
X (2.10)(667) 10,000 $8599
X (3.12)(667) 10,000 $12,081
X (0.23)(667) 10,000 $9847
X (1.24)(667) 10,000 $10,827
X (2.61)(667) 10,000 $8259
X (0.99)(667) 10,000 $9340
In this sample of six typical concrete surfacing contract amounts for sites in our region, the
average is $9825 and five of six values are within the range of $8000 to $12,000, with the sixth
being only $81 above the upper limit.
19.5 Monte Carlo Sampling and Simulation Analysis
Up to this point, all alternative selections have been made using estimates with certainty, possibly followed by some testing of the decision via sensitivity analysis or expected values. In this
section, we will use a simulation approach that incorporates the material of the previous sections to facilitate the engineering economy decision about one alternative or between two or
more alternatives.
The random sampling technique discussed in Section 19.3 is called Monte Carlo sampling.
The general procedure outlined below uses Monte Carlo sampling to obtain samples of size n for
selected parameters of formulated alternatives. These parameters, expected to vary according to a
stated probability distribution, warrant decision making under risk. All other parameters in an alternative are considered certain; that is, they are known, or they can be estimated with enough precision to consider them certain. An important assumption is made, usually without realizing it.
All parameters are independent; that is, one variable’s distribution does not affect the value of
any other variable of the alternative. This is referred to as the property of independent random
variables.
The simulation approach to engineering economy analysis is summarized in the following
basic steps.
Step 1. Formulate alternative(s). Set up each alternative in the form to be considered
using engineering economic analysis, and select the measure of worth upon which
to base the decision. Determine the form of the relation(s) to calculate the measure
of worth.
Step 2. Parameters with variation. Select the parameters in each alternative to be treated
as random variables. Estimate values for all other (certain) parameters for the analysis.
533
534
More on Variation and Decision Making under Risk
Chapter 19
Step 3.
Step 4.
Step 5.
Step 6.
Step 7.
Determine probability distributions. Determine whether each variable is discrete or continuous, and describe a probability distribution for each variable in each
alternative. Use standard distributions, where possible, to simplify the sampling process and to prepare for computer-based simulation.
Random sampling. Incorporate the random sampling procedure of Section 19.3
(the first four steps) into this procedure. This results in the cumulative distribution,
assignment of RNs, selection of the RNs, and a sample of size n for each variable.
Measure of worth calculation. Compute n values of the selected measure of
worth from the relation(s) determined in step 1. Use the estimates made with certainty and the n sample values for the varying parameters. (This is when the property
of independent random variables is actually applied.)
Measure of worth description. Construct the probability distribution of the measure
of
__
__ worth, using between 10 and 20 cells of data, and calculate measures such as
X, s, X ⫾ ts, and relevant probabilities.
Conclusions. Draw conclusions about each alternative, and decide which is to be
selected. If the alternative(s) has (have) been previously evaluated under the assumption
of certainty for all parameters, comparison of results may help with the final decision.
Example 19.8 illustrates this procedure using an abbreviated manual simulation analysis, and
Example 19.9 utilizes spreadsheet simulation for the same estimates.
EXAMPLE 19.8
Yvonne Ramos is the CEO of a chain of 50 fitness centers in the United States and Canada. An
equipment salesperson has offered Yvonne two long-term opportunities on new aerobic exercise systems, for which the usage is charged to customers on a per-use basis on top of the
monthly fees paid by customers. As an enticement, the offer includes a guarantee of annual
revenue for one of the systems for the first 5 years.
Since this is an entirely new and risky concept of revenue generation, Yvonne wants to do a
careful analysis of each alternative. Details for the two systems follow:
System 1. First cost is P ⫽ $12,000 for a set period of n ⫽ 7 years with no salvage value.
No guarantee for annual net revenue is offered.
System 2. First cost is P ⫽ $8000, there is no salvage value, and there is a guaranteed
annual net revenue of $1000 for each of the first 5 years, but after this period, there is
no guarantee. The equipment with updates may be useful up to 15 years, but the exact
number is not known. Cancellation anytime after the initial 5 years is allowed, with no
penalty.
For either system, new versions of the equipment will be installed with no added costs. If the
MARR is 15% per year, use PW analysis to determine if neither, one, or both of the systems
should be installed.
Solution by Hand
Estimates that Yvonne makes to use the simulation analysis procedure are included in the following steps.
Step 1.
Formulate alternatives. Using PW analysis, the relations for system 1 and system 2 are developed. The symbol NCF identifies the net cash flows (revenues), and
NCFG is the guaranteed NCF of $1000 for system 2.
PW1 ⫽ P1 NCF1(P兾A,15%,n1)
[19.20]
PW2 P2 NCFG(P兾A,15%,5)
NCF2(P兾A,15%,n25)(P兾F,15%,5)
[19.21]
Step 2. Parameters with variation. Yvonne summarizes the parameters estimated with
certainty and makes distribution assumptions about three parameters treated as random variables.
Monte Carlo Sampling and Simulation Analysis
19.5
System 1
Certainty. P1 ⫽ $12,000; n1 ⫽ 7 years.
Variable. NCF1 is a continuous variable, uniformly distributed between
L ⫽ $−4000 and H ⫽ $6000 per year, because this is considered a high-risk
venture.
System 2
Certainty. P2 ⫽ $8000; NCFG ⫽ $1000 for first 5 years.
Variable. NCF2 is a discrete variable, uniformly distributed over the
values L ⫽ $1000 to H ⫽ $6000 only in $1000 increments, that is, $1000,
$2000, etc.
Variable. n2 is a continuous variable that is uniformly distributed between
L ⫽ 6 and H ⫽ 15 years.
Now, rewrite Equations [19.20] and [19.21] to reflect the estimates made with certainty.
PW1 ⫽ 12,000 NCF1(P兾A,15%,7)
12,000 NCF1(4.1604)
[19.22]
PW2 8000 1000(P兾A,15%,5)
NCF2(P兾A,15%,n25)(P兾F,15%,5)
4648 NCF2(P兾A,15%,n25)(0.4972)
[19.23]
Step 3.
Determine probability distributions. Figure 19–11 (left side) shows the assumed
probability distributions for NCF1, NCF2, and n2.
Step 4. Random sampling. Yvonne decides on a sample of size 30 and applies the first
four of the random sample steps in Section 19.3. Figure 19–11 (right side) shows the
cumulative distributions (step 1) and assigns RNs to each variable (step 2). The RNs
for NCF2 identify the x axis values so that all net cash flows will be in even $1000
amounts. For the continuous variable n2, three-digit RN values are used to make the
numbers come out evenly, and they are shown in cells only as “indexers” for easy
reference when a RN is used to find a variable value. However, we round the number
to the next higher value of n2 because it is likely the contract may be canceled on an
anniversary date. Also, now the tabulated compound interest factors for (n2 5)
years can be used directly (see Table 19–5).
Once the first RN is selected randomly from Table 19–2, the sequence (step 3)
used will be to proceed down the RN table column and then up the column to the left.
Table 19–5 shows only the first five RN values selected for each sample and the corresponding variable values taken from the cumulative distributions in Figure 19–11
(step 4).
Step 5. Measure of worth calculation. With the five sample values in Table 19–5, calculate the PW values using Equations [19.22] and [19.23].
1.
2.
3.
4.
5.
1.
2.
3.
4.
5.
PW1 12,000 (2200)(4.1604)
PW1 12,000 2000(4.1604)
PW1 12,000 (1100)(4.1604)
PW1 12,000 (900)(4.1604)
PW1 12,000 3100(4.1604)
PW2 4648 1000(P兾A,15%,7)(0.4972)
PW2 4648 1000(P兾A,15%,5)(0.4972)
PW2 4648 5000(P兾A,15%,8)(0.4972)
PW2 4648 3000(P兾A,15%,10)(0.4972)
PW2 4648 4000(P兾A,15%,3)(0.4972)
$21,153
$3679
$16,576
$15,744
$897
$2579
$2981
$6507
$2838
$107
Now, 25 more RNs are selected for each variable from Table 19–2, and the PW values are calculated.
Step 6. Measure of worth description. Figure 19–12a and b presents the PW1 and PW2
probability distributions for the 30 samples with
__ 14 and 15 cells, respectively, as well
as the range of individual PW values and the X and s values.
535
536
More on Variation and Decision Making under Risk
Chapter 19
Figure 19-11
F(NCF1 )
Distributions used for
random samples,
Example 19.8.
RN
1.0
80–99
0.8
60–79
0.6
40–59
0.4
20–39
f (NCF1)
Continuous variable
1
10
0.2
00–19
0
–4
–2
0
2
4
6
–4
–2
0
2
4
6
NCF1, $1000
NCF1, $1000
F(NCF2)
RN
1.00
83–99
0.83
67–82
0.67
50–66
0.50
33–49
0.33
P(NCF2)
17–32
Discrete variable
1
6
0.17
00–16
0
0
1
2
3
4
5
6
1
2
NCF2, $1000
3
4
5
6
NCF2 , $1000
F(n2)
RN
1.0
888–999
0.8
666–887
0.6
444–665
0.4
222–443
f (n2)
Continuous variable
1
9
0.2
000–221
0
6
10
n2, years
15
6
8
10
12
n2, years
14 15
TABLE 19–5 Random Numbers and Variable Values for NCF1, NCF2, and
n2, Example 19.8
NCF1
NCF2
†
n2
RN*
Value, $
RN
Value, $
RN
Value, Year
Rounded§
18
59
31
29
71
2200
2000
1100
900
3100
10
10
77
42
55
1000
1000
5000
3000
4000
586
379
740
967
144
11.3
9.4
12.7
14.4
7.3
12
10
13
15
8
*Randomly start with row 1, column 4 in Table 19–2.
†
Start with row 6, column 14.
‡
Start with row 4, column 6.
§
The n2 value is rounded up.
‡
Monte Carlo Sampling and Simulation Analysis
19.5
X1 ± 1s1
Frequency
n1 = 30
s1 = $10,190
Range = $37,443
X1 = $–7729
5
4
PW
0
3
2
1
0
– 27 – 24 – 21 – 18 – 15 – 12 – 9
–6
–3
0
3
6
9
12
15
PW1, $1000
(a) System 1
X2 ± 1s2
Frequency
6
5
X2 = $2724
4
PW
n2 = 30
s2 = $4336
Range = $13,355
0
3
2
1
0
–4
–3
–2
–1
0
1
2
3
4
5
6
7
8
9
10
11
PW2, $1000
(b) System 2
Figure 19–12
Probability distributions of simulated PW values for a sample of size 30, Example 19.8.
PW1. Sample values range from $−24,481 to $12,962. The calculated measures
of the 30 values are
__
X1 $7729
s1 $10,190
PW2. Sample values range from $−3031 to $10,324. The sample measures
are
__
X2 $2724
s2 $4336
Step 7.
Conclusions. Additional sample values will surely make the central tendency of
the PW distributions more evident and may reduce the s values, which are quite
large. Of course, many conclusions are possible once the PW distributions are
known, but the following seem clear at this point.
System 1. Based on this small sample of 30 observations, do not accept this
alternative. The likelihood of making the MARR 15% is relatively small, since
the sample indicates
a probability of 0.27 (8 out of 30 values) that the PW will be
__
positive, and X1 is a large negative. Though appearing large, the standard deviation may be used to determine
__ that about 20 of the 30 sample PW values (twothirds) are within the limits X
1s, which are $17,919 and $2461. A larger
sample may alter this analysis somewhat.
System 2. If Yvonne is willing to accept the longer-term commitment that may
increase the NCF some years out, the sample of 30 observations indicates to accept this alternative. At a MARR of 15%, the simulation approximates the chance
for a positive PW as 67% (20 of the 30 PW values in Figure
19–12b are positive).
__
However, the probability of observing PW within the X 1s limits ($1612 and
$7060) is 0.53 (16 of 30 sample values).
Conclusion at this point. Reject system 1; accept system 2; and carefully
watch net cash flow, especially after the initial 5-year period.
537
538
Chapter 19
More on Variation and Decision Making under Risk
Comment
The estimates in Example 13.5 are very similar to those here, except all estimates were made
with certainty (NCF1 ⫽ $3000, NCF2 ⫽ $3000, and n2 ⫽ 14 years). The alternatives were
evaluated by the payback period method at MARR ⫽ 15%, and the first alternative was selected. However, the subsequent PW analysis in Example 13.5 selected alternative 2 based, in
part, upon the anticipated larger cash flow in the later years.
EXAMPLE 19.9
Help Yvonne Ramos set up a spreadsheet simulation for the three random variables and PW
analysis in Example 19.8. Does the PW distribution vary appreciably from that developed
using manual simulation? Do the decisions to reject the system 1 proposal and accept the system 2 proposal still seem reasonable?
Solution by Spreadsheet
Figures 19–13 and 19–14 are spreadsheet screen shots that accomplish the simulation portion
of the analysis described above in steps 3 (determine probability distribution) through 6 (measure of worth description). Most spreadsheet systems are limited in the variety of distributions
they can accept for sampling, but common ones such as uniform and normal are available.
Figure 19–13 shows the results of a small sample of 30 values from the three distributions
using the RAND and IF functions. (See Section A.3 in Appendix A.)
NCF1: Continuous uniform from $−4000 to $6000. The spreadsheet relation in column B
translates RN1 values (column A) into NCF1 amounts.
NCF2: Discrete uniform in $1000 increments from $1000 to $6000. Column D cells display
NCF2 in the $1000 increments using the logical IF function to translate from the RN2 values.
n2: Continuous uniform from 6 to 15 years. The results in column F are integer values obtained using the INT function operating on the RN3 values.
Figure 19–13
Random sample of 30 values generated for spreadsheet simulation, Example 19.9.
Monte Carlo Sampling and Simulation Analysis
19.5
⫽SUM(G13:G42)
⫽AVERAGE(F13:F42)
⫽STDEV(F13:F42)
Figure 19–14
Simulation results for 30 PW values, Example 19.9.
Figure 19–14 presents the two alternatives’ estimates in the top section. The PW1 and PW2
computations for the 30 repetitions of NCF1, NCF2, and N2 are the spreadsheet equivalent of
Equations [19.22] and [19.23]. The tabular approach used here tallies the number of PW values
below zero ($0) and equal to or exceeding zero using the IF operator. For example, cell C17
contains a 1, indicating PW1 ⬎ 0 when NCF1 ⫽ $3100 (in cell B7 of Figure 19–13), which was
used to calculate PW1 ⫽ $897 by Equation [19.22]. Cells in rows 7 and 8 show the number of
times in the 30 samples that system 1 and system 2 may return at least the MARR ⫽ 15% because the corresponding PW 0. Sample averages and standard deviations are also indicated.
Comparison between the hand and spreadsheet simulations is presented below.
System 1 PW
X, $
s, $
No. of
PW 0
X, $
s, $
No. of
PW 0
7,729
7,105
10,190
13,199
8
10
2,724
1,649
4,336
3,871
20
19
__
Hand
Spreadsheet
System 2 PW
__
For the spreadsheet simulation, 10 (33%) of the PW1 values exceed zero, while the manual simulation included 8 (27%) positive values. These comparative results will change every time this
spreadsheet is activated since the RAND function is set up (in this case) to produce a new RN each
time. (It is possible to define RAND to keep the same RN values. See the Excel User’s Guide.)
The conclusion to reject the system 1 proposal and accept system 2 is still appropriate for
the spreadsheet simulation as it was for the hand solution, since there are comparable chances
that PW 0.
539
540
Chapter 19
More on Variation and Decision Making under Risk
CHAPTER SUMMARY
To perform decision making under risk implies that some parameters of an engineering alternative are treated as random variables. Assumptions about the shape of the variable’s probability
distribution are used to explain how the estimates of parameter values may vary. Additionally,
measures such as the expected value and standard deviation describe the characteristic shape of
the distribution. In this chapter, we learned several of the simple, but useful, discrete and continuous population distributions used in engineering economy—uniform and triangular—as well
as specifying our own distribution or assuming the normal distribution.
Since the population’s probability distribution for a parameter is not fully known, a random
sample of size n is usually taken, and its sample average and standard deviation are determined.
The results are used to make probability statements about the parameter, which help make the
final decision with risk considered.
The Monte Carlo sampling method is combined with engineering economy relations for a
measure of worth such as PW to implement a simulation approach to risk analysis. The results of
such an analysis can then be compared with decisions when parameter estimates are made with
certainty.
PROBLEMS
Certainty, Risk, and Uncertainty
19.1 Identify the following variables as either discrete
or continuous.
(a) The interest rates available in the marketplace for jumbo certificates of deposit
(b) Optimistic, most likely, and pessimistic estimates of salvage value
(c) The number of cars that are red in the first
100 that pass through a certain intersection
(d ) The weight of the purse or wallet carried when
a person leaves her or his residence
(e) The gallons of water that evaporate from
Lake Erie in a given day
2000. He wants to use E(output) in the decisionmaking process. Identify at least two additional
pieces of information that must be obtained or
assumed to finalize the output information for
this use.
Probability and Distributions
19.4 Royalties received by an investor in an oil well
vary according to the price of oil. Data collected
from stripper wells in an established oil field were
used to develop the probability-royalty relationship shown below.
Royalties, $ per Year 6200 8500 9600 10,300 12,600 15,500
19.2 For each situation below, determine (1) if the variable is discrete or continuous and (2) if the information involves certainty, risk, and兾or uncertainty.
(a) The first cost of a new front-end loader is
$34,000 or $38,000 depending on the size
purchased.
(b) The raises for engineers and technical staff
employees will be 3%, or 5%, with one-half
getting 3% and one-half getting 5%.
(c) Revenue from a new product line is expected
to be between $350,000 and $475,000 per
year.
(d ) The salvage value for an old machine will be
$500 (i.e., its asking price) or $0 (it will be
thrown away).
(e) Profits are equally likely to be up anywhere
from 25% to 60% this year.
19.3 An engineer learned that production output is between 1000 and 2000 units per week 90% of the
time, and it may fall below 1000 or go above
Probability
(a)
(b)
0.10 0.21 0.32
0.24
0.09
0.04
Calculate the expected value of royalty income (RI) per year.
Determine the probability that the royalty
income will be at least $12,600 per year.
19.5 Daily revenue from vending machines placed in
various buildings of a major university is as follows:
20, 75, 43, 62, 51, 52, 78, 33, 28, 39, 61, 56, 43, 49, 48, 49,
71, 53, 57, 46, 42, 41, 63, 36, 51, 59, 40, 32, 37, 29, 26
(a)
(b)
(c)
(d)
Construct a frequency distribution table with
a cell size of 12 starting with 19.5 (i.e., first
cell is 19.5–31.5, next is 31.5–43.5, etc.).
Determine the probability distribution.
What is the probability of revenue from a
machine being less than $44?
What is the probability that revenue from a
machine will equal or exceed $44?
541
Problems
19.6 A survey of households included a question about
the number of operating automobiles N currently
owned by people living at the residence and the
interest rate i on the lowest-rate loan for the cars.
The results for 100 households are shown.
Number of Cars N
Households
0
1
2
3
4
12
56
26
3
3
Loan Rate i, %
(b)
(c)
(d)
19.9 An alternative to buy and an alternative to lease
hydraulic lifting equipment have been formulated.
Use the parameter estimates and assumed distribution data shown to plot the probability distributions
on one graph for each corresponding parameter.
Label the parameters carefully.
Households
Purchase Alternative
13
14
19
38
12
4
Estimated Value
0.02
2.014
4.016
6.018
8.0110
10.0112
(a)
State whether each variable is discrete or
continuous.
Plot the probability distributions and cumulative distributions for N and i.
From the data collected, what is the probability that a household has 1 or 2 cars? Three or
more cars?
Use the data for i to estimate the chances that
the interest rate is between 7% and 11% per
year.
19.7 An officer of the state lottery commission sampled
lottery ticket purchasers over a 1-week period at
one location. The amounts distributed back to the
purchasers and the associated probabilities for
5000 tickets are as follows:
Distribution, $
Probability
(a)
(b)
(c)
5 months, which is the design life. Some batteries
fail early, but 2 months is the smallest life experienced thus far. (a) Write out and plot the probability
distributions and cumulative distributions for Bob.
(b) Determine the probability of N being 1, 2, or 3
consecutive units above the weight limit.
0
2
5
10
100
0.91
0.045
0.025
0.013
0.007
Plot the cumulative distribution of winnings.
Calculate the expected value of the distribution of dollars per ticket.
If tickets cost $2, what is the expected longterm income to the state per ticket, based
upon this sample?
19.8 Bob is working on two separate probability-related
projects. The first involves a variable N, which is the
number of consecutively manufactured parts that
weigh in above the weight specification limit. The
variable N is described by the formula (0.5)N because each unit has a 50-50 chance of being below
or above the limit. The second involves a battery
life L that varies between 2 and 5 months. The probability distribution is triangular with the mode at
Parameter
High
Low
First cost, $
25,000
20,000
Salvage value, $
3,000
2,000
8
4
9,000
5,000
Life, years
AOC, $ per year
Assumed
Distribution
Uniform;
continuous
Triangular;
mode at $2500
Triangular;
mode at 6
Uniform;
continuous
Lease Alternative
Estimated Value
Parameter
High
Low
Lease first cost, $
2000
1800
AOC, $ per year
9000
5000
Lease term, years
3
3
Assumed
Distribution
Uniform;
continuous
Triangular;
mode at $7000
Certainty
19.10 Carla is a statistician with a bank. She has collected debt-to-equity mix data on mature (M) and
young (Y) companies. The debt percentages vary
from 20% to 80% in her sample. Carla has defined
DM as a variable for the mature companies from
0 to 1, with DM 0 interpreted as the low of 20%
debt and DM 1.0 as the high of 80% debt. The
variable for young corporation debt percentages
DY is similarly defined. The probability distributions used to describe DM and DY are
f (DM) 3(1DM)2
f (DY) 2DY
0 DM 1
0 DY 1
(a) Use different values of the debt percentage between 20% and 80% to calculate values for the probability distributions and then plot them. (b) What can
you comment about the probability that a mature
company or a young company will have a low debt
percentage? A high debt percentage?
542
More on Variation and Decision Making under Risk
Chapter 19
19.11 A discrete variable X can take on integer values of
1 to 10. A sample of size 50 results in the following
probability estimates:
Xi
1
2
3
6
9
10
P(Xi)
0.2
0.2
0.2
0.1
0.2
0.1
(a)
(b)
(c)
Write out and graph the cumulative
distribution.
Calculate the following probabilities using
the cumulative distribution: X is between 6
and 10, and X has the values 4, 5, or 6.
Use the cumulative distribution to show that
P(X ⫽ 7 or 8) ⫽ 0.0. Even though this probability is zero, the statement about X is that it
can take on integer values of 1 to 10. How do
you explain the apparent contradiction in
these two statements?
19.14 Develop a discrete probability distribution of your
own for the variable G, the expected grade in this
course, where G A, B, C, D, F, or I (incomplete).
Assign random numbers to F(G), and take a sample from it. Now plot the probability values from
the sample for each G value.
19.15 Use the RAND function in Excel to generate
100 values from a U(0,1) distribution.
(a) Calculate the average and compare it to 0.5,
the expected value for a random sample between 0 and 1.
(b) For the RAND function sample, cluster the
results into cells of 0.1 width, that is 0.0–0.1,
0.1–0.2, etc., where the upper-limit value is
excluded from each cell. Determine the probability for each grouping from the results.
Does your sample come close to having
approximately 10% in each cell?
Random Samples
19.12 A discrete variable X can take on integer values of
1 to 5. A sample of size 100 results in the following
probability estimates.
(a)
(b)
Xi
1
2
3
4
5
P(Xi)
0.2
0.3
0.1
03
0.1
Use the following random numbers to estimate the probabilities for each value of X.
Determine the sample probabilities for X ⫽ 1
and X ⫽ 5. Compare the sample results with
the probabilities in the problem statement.
RN:
10, 42, 18, 31, 23, 80, 80, 26, 74, 71, 03, 90,
55, 61, 61, 28, 41, 49, 00, 79, 96, 78, 42, 31, 26
19.13 The percent price increase p on a variety of retail
food prices over a 1-year period varied from 5% to
10% in all cases. Because of the distribution of
p values, the assumed probability distribution for
the next year is
f(X) ⫽ 2X
where
X
{
Sample Estimates
19.16 An engineer was asked to determine whether the
average air quality in a vehicle assembly plant was
within OSHA guidelines. The following air quality
readings were collected:
81, 86, 80, 91, 83, 83, 96, 85, 89
(a)
(b)
(c)
19.17 Carol sampled the monthly maintenance costs for automated soldering machines a total of 100 times during 1 year. She clustered the costs into $200 cells, for
example, $500 to $700, with cell midpoints of $600,
$800, $1000, etc. She indicated the number of times
(frequency) each cell value was observed. The costs
and frequency data are as follows.
0X1
0
when p 5%
1
when p 10%
For a continuous variable the cumulative distribution F(X) is the integral of f (X) over the same
range of the variable. In this case
F(X) X2
(a)
(b)
0X1
Graphically assign RNs to the cumulative
distribution, and take a sample of size 30 for
the variable. Transform the X values into interest rates.
Calculate the average p value for the sample.
Determine the sample mean.
Calculate the standard deviation.
Determine the number of values and percent
of values that fall within 1 standard deviation of the mean.
(a)
(b)
Cell
Midpoint, $
Frequency
600
800
1000
1200
1400
1600
1800
2000
6
10
7
15
28
15
9
10
Estimate the expected value and standard deviation of the maintenance costs the company
should anticipate based on Carol’s sample.
What is the best estimate of the percentage of
costs that will fall within 2 standard deviations of the mean?
543
Additional Problems and FE Exam Review Questions
(c)
(d)
Develop a probability distribution of the
monthly maintenance costs from Carol’s
sample, and indicate the answers to the previous two questions on it.
Use a spreadsheet to display the sample mean.
Simulation
19.23 Carl, an engineering colleague, estimated net cash
flow after taxes (CFAT) for the project he is working on. The additional CFAT of $2800 in year 10 is
the salvage value of capital assets.
19.18 (a) Determine the values of sample average and
standard deviation of the data in Problem 19.11.
(b) Determine the values 1 and 2 standard deviations from the mean. Of the 50 sample points, how
many fall within these two ranges?
19.19 (a) Use the relations in Section 19.4 for continuous
variables to determine the expected value and
standard deviation for the distribution of f (DY) in
Problem 19.10. (b) It is possible to calculate the
probability of a continuous variable X between
two points (a, b) using the following integral:
P(a X b) 冮a f (X) dx
b
Determine the probability that DY is within 2 standard deviations of the expected value.
19.20 (a) Use the relations in Section 19.4 for continuous variables to determine the expected value
and variance for the distribution of DM in
Problem 19.10.
f (DM) 3(1 DM)2
(b)
0 DM 1
Determine the probability that DM is within 2
standard deviations of the expected value.
Use the relation in Problem 19.19.
19.21 Calculate the expected value for the variable N in
Problem 19.8.
19.22 A newsstand manager is tracking Y, the number of
weekly magazines left on the shelf when the new
edition is delivered. Data collected over a 30-week
period are summarized by the following probability distribution. Plot the distribution and the estimates for expected value and 1 standard deviation
on either side of E(Y) on the plot.
Y copies
P(Y)
3
7
10
12
1兾3
1兾4
1兾3
1兾12
Year
CFAT, $
0
16
710
10
28,800
5,400
2,040
2,800
The PW value at the current MARR of 7% per
year is
PW 28,800 5400(P兾A,7%,6)
2040(P兾A,7%,4)(P兾F,7%,6)
2800(P兾F,7%,10)
$2966
Carl believes the MARR will vary over a relatively
narrow range, as will the CFAT, especially during
the out years of 7 through 10. He is willing to accept the other estimates as certain. Use the following probability distribution assumptions for MARR
and CFAT to perform a simulation—hand- or
spreadsheet-based.
MARR. Uniform distribution over the range
6% to 10%.
CFAT, years 7 through 10. Uniform distribution over the range $1600 to $2400 for each
year.
Plot the resulting PW distribution. Should the plan
be accepted using decision making under certainty? Under risk?
19.24 Repeat Problem 19.23, except use the normal distribution for the CFAT in years 7 through 10 with
an expected value of $2000 and a standard deviation of $500.
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
19.25 When there are at least two values for a parameter
and it is possible to estimate the chance that each
may occur, this situation is known as:
(a) Uncertainty
(b) Risk
(c) Standard deviation
(d) Cost estimating
19.26 A deterministic economic analysis is one wherein:
(a) Single-value estimates are used exclusively
(b) Risk is taken into account
(c) A range of values for each parameter is included in the analysis
(d) Taxes and inflation are not considered in the
cash flow estimates
544
More on Variation and Decision Making under Risk
Chapter 19
19.27 Decision making under risk includes all of the following except:
(a) Expected value analysis
(b) Simulation
(c) Using only single-value estimates
(d) Probability
19.28 All of the following are elements in decision making under risk except:
(a) Random variable
(b) Cost indexes
(c) Probability
(d) Cumulative distribution
19.31 The revenue from an oil dispersant product has
averaged $15,000 per month
for the past 12 months.
__
If the value of (Xi − X)2 is $1,600,000, the standard deviation is closest to:
(a) $381
(b) $652
(c) $958
(d) $1265
19.32 A survey of the types of cars parked at an NFL
football stadium revealed that there were equal
probabilities of finding cars identified as type A, B,
C, and D. Car types were assigned random numbers as follows.
19.29 For the income and probability values shown in
the table, the probability that the income in any
year will be less than $9600 is closest to:
(a) 0.15
(b) 0.23
(c) 0.32
(d) 0.38
Income, $ per Year 6200 8500 9600 10,300 12,600 15,500
Probability
0.15
0.23
0.32
0.24
0.09
0.04
19.30 The symbol that represents the true population
mean is:
(a)
(b) s
(c)
__
(d) X
Car Type
Assigned RN
A
B
C
D
0 through 24
25 through 49
50 through 74
75 through 99
The sample probability of a type B car from the
12 random numbers shown is closest to:
RN: 75, 52, 14, 99, 67, 74, 06, 50, 97, 46, 27, 88
(a)
(b)
(c)
(d )
0.17
0.25
0.33
0.42
CASE STUDY
USING SIMULATION AND THREE-ESTIMATE SENSITIVITY ANALYSIS
Background
The Knox Brewing company makes specialty-named sodas
and flavored drinks for retail grocery chains throughout
North and Central America. During the past year, it has become obvious that a new bottle-capping machine is needed to
replace the current 10-year-old system. Dr. Knox, the owner
and president, knows the business quite well. You just handed
him the first cost bids from three vendors for the machine. He
looked carefully at the numbers and asked you to sit down.
You were quite surprised, as this was the first time you had
been in his office, and most other engineers at Knox have a
great fear of “the Old Man.”
Information
As he examined the three bids on first cost of the machine, he
started to write some numbers, which, he explained, were his
estimates of the annual operating cost, useful life, and possible
salvage value for each of the machines sold by the three vendors.
After a few minutes, he told you to take these numbers
and use some of that “new engineering knowledge” you
acquired in college to determine which, if any, of these three
bids made the best economic sense. He also told you to be
innovative and use a computer and some probability to come
up with a robust recommendation by tomorrow at 2 P.M.
You have used the estimates from the president to develop
Table 19–6 of pessimistic (P), most likely (ML), and optimistic (O) estimates for each vendor’s machine. In addition, you
developed some possible distributions for the parameters that
Dr. Knox estimated, namely, AOC, life, and salvage value.
These are summarized in Table 19–7. You plan to use a simple Monte Carlo simulation to help formulate your recommendation for tomorrow.
Case Study Exercises
First, learn to use the RNG (random number generator) in
Excel, if you have not already done so. It is necessary to
545
Case Study
TABLE 19–6
Vendor 1
P
ML
O
Vendor 2
P
ML
O
Vendor 3
P
ML
O
TABLE 19–7
Parameter Estimates for Bottle-Capping Machine
First Cost, $
AOC, $ per Year
Salvage, $
Life, Years
200,000
200,000
−200,000
11,000
10,000
6,000
0
0
0
3
5
8
150,000
150,000
150,000
5,000
3,500
2,000
0
5,000
8,000
2
4
7
300,000
300,000
300,000
8,000
6,000
4,500
5,000
5,000
8,000
5
7
9
Distribution Assumptions about AOC, Life, and Salvage
Parameter
Vendor 1
Vendor 2
Vendor 3
AOC, $ per year
Normal
Mean: 10,000
Std. dev.: 500
Uniform
0 to 1000
Discrete uniform
3 to 8, equal probability
Normal
Mean: 3500
Std. dev.: 500
Uniform
0 to 8000
Discrete uniform
2 to 7, equal probability
Normal
Mean: 6000
Std. dev.: 500
Uniform
5000 to 8000
Discrete uniform
5 to 9, equal probability
Salvage, $
Life, years
sample from the normal distributions that you have specified in Table 19–7. RNG is part of the Analysis Tool-Pak
accessed through the Office Button, Excel Options, AddIns path.
1. Prepare the simulation using a spreadsheet; determine
which of the vendors offers the best machine from an
economic perspective, and take into account the estimates made by Dr. Knox. Use a sample size of at least
50, and base your conclusions on the AW measure of
worth.
2. Prepare a short presentation for Dr. Knox (and class)
using your analysis.
APPENDIX A
USING SPREADSHEETS
AND MICROSOFT EXCEL©
This appendix explains the layout of a spreadsheet and the use of Microsoft Excel (hereafter
called Excel) functions in engineering economy. Refer to the Excel help system for your particular computer and version of Excel. Some specific commands and entries refer to Excel 2007 and
may differ slightly from your version.
A.1 Introduction to Using Excel
Enter a Formula or Use an Excel Function
The ⴝ sign is required to perform any formula or function computation in a cell. The formulas
and functions on the worksheet can be displayed by simultaneously pressing Ctrl and `. The symbol ` is usually in the upper left of the keyboard with the ~ (tilde) symbol. Pressing Ctrl+` a second
time hides the formulas and functions.
1. Run Excel.
2. Move to cell C3. (Move the pointer to C3 and left-click.)
3. Type ⫽ PV(5%,12,10) and ⬍Enter⬎. This function will calculate the present value of
12 payments of $10 at a 5% per year interest rate.
Another example: To calculate the future value of 12 payments of $10 at 6% per year interest, do
the following:
1.
2.
3.
4.
5.
6.
7.
8.
Move to cell B3, and type INTEREST.
Move to cell C3, and type 6% or ⫽ 6兾100.
Move to cell B4, and type PAYMENT.
Move to cell C4, and type 10 (to represent the size of each payment).
Move to cell B5, and type NUMBER OF PAYMENTS.
Move to cell C5, and type 12 (to represent the number of payments).
Move to cell B7, and type FUTURE VALUE.
Move to cell C7, and type ⫽ FV(C3,C5,C4) and hit ⬍Enter⬎. The answer will appear
in cell C7.
To edit the values in cells
1. Move to cell C3 and type ⫽5兾100 (the previous value will be replaced).
2. The value in cell C7 will update.
Cell References in Formulas and Functions
If a cell reference is used in lieu of a specific number, it is possible to change the number once
and perform sensitivity analysis on any variable that is referenced by the cell number, such as C5.
This approach defines the referenced cell as a global variable for the worksheet. There are two
types of cell references—relative and absolute.
Relative References If a cell reference is entered, for example, A1, into a formula or function
that is copied or dragged into another cell, the reference is changed relative to the movement of
the original cell. If the formula in C5 is ⫽ A1 and it is copied into cell C6, the formula is changed
to ⫽ A2. This feature is used when dragging a function through several cells, and the source entries must change with the column or row.
Absolute References
If adjusting cell references is not desired, place a $ sign in front of
the part of the cell reference that is not to be adjusted—the column, row, or both. For example,
548
Appendix A
Using Spreadsheets and Microsoft Excel©
⫽ $A$1 will retain the formula when it is moved anywhere on the worksheet. Similarly, ⫽ $A1
will retain the column A, but the relative reference on 1 will adjust the row number upon movement around the worksheet.
Absolute references are used in engineering economy for sensitivity analysis of parameters
such as MARR, first cost, and annual cash flows. In these cases, a change in the absolutereference cell entry can help determine the sensitivity of a result, such as PW or AW.
Print the Spreadsheet
First define the portion (or all) of the spreadsheet to be printed.
1. Move the pointer to the top left cell of your spreadsheet.
2. Hold down the left-click button. (Do not release the left-click button.)
3. Drag the mouse to the lower right corner of your spreadsheet or to wherever you want to stop
printing.
4. Release the left-click button. (It is ready to print.)
5. Left-click the Office button (see Figure A–1).
6. Move the pointer down to select Print and left-click.
7. In the dialog box, left-click the Print option (or similar command).
Depending on your computer environment, you may have to select a network printer and queue
your printout through a server.
Save the Spreadsheet
You can save your spreadsheet at any time during or after completing your work. It is recommended that you save your work regularly.
1. Left-click the Office button.
2. To save the spreadsheet the first time, left-click the Save As . . . option.
3. Type the file name, e.g., Prob 7.9, and left-click the Save button.
To save the spreadsheet after it has been saved the first time, i.e., a file name has been assigned
to it, left-click the Office button, move the pointer down, and left-click on Save.
Create an xy (Scatter) Chart
This chart is one of the most commonly used in scientific analysis, including engineering economy. It plots pairs of data and can place multiple series of entries on the Y axis. The xy scatter
chart is especially useful for results such as the PW versus i graph, where i is the X axis and the
Y axis displays the results of the NPV function for several alternatives.
1. Run Excel.
2. Enter the following numbers in columns A, B, and C, respectively.
Column A, cell A1 through A6: Rate i%, 4, 6, 8, 9, 10
Column B, cell B1 through B6: $ for A, 40, 55, 60, 45, 10
Column C, cell C1 through C6: $ for B, 100, 70, 65, 50, 30.
3. Move the mouse to A1, left-click, and hold while dragging to cell C6. All cells will be highlighted, including the title cell for each column.
4. If not all the columns for the chart are adjacent to one another, first press and hold the Control key on the keyboard during the entirety of step 3. After dragging over one column of
data, momentarily release the left click, then move to the top of the next (nonadjacent) column of the chart. Do not release the Control key until all columns to be plotted have been
highlighted.
5. Left-click on the Insert button on the toolbar.
6. Select the Scatter option and choose a subtype of scatter chart. The graph appears with a
legend (Figure A–1).
Now a large number of styling effects can be introduced for axis titles, legend, data series, etc.
Note that only the bottom row of the title can be highlighted. If titles are not highlighted, the data
sets are generically identified as series 1, series 2, etc. on the legend.
Organization (Layout) of the Spreadsheet
A.2
Office button
Insert for
scatter chart
Figure A–1
Scatter chart for data entries and location of commonly used buttons.
Obtain Help While Using Excel
1. To get general help information, left-click on the “?” (upper right).
2. Enter the topic or phrase. For example, if you want to know more about how to save a file,
type the word Save.
3. Select the appropriate matching words. You can browse through the options by left-clicking
on any item.
A.2 Organization (Layout) of the Spreadsheet
A spreadsheet can be used in several ways to obtain answers to numerical questions. The first is
as a rapid solution tool, often with the entry of only a few numbers or one predefined function.
For example, to find the future worth in a single-cell operation, move the pointer to any cell and
enter ⫽ FV(8%,5,-2500). The display of $14,666.50 is the 8% future worth at the end of year 5
of five equal payments of $2500 each.
A second use is more formal; it presents data, solutions, graphs, and tables developed on the
spreadsheet and ready for presentation to others. Some fundamental guidelines in spreadsheet
organization are presented here. A sample layout is presented in Figure A–2. As solutions become
more complex, organization of the spreadsheet becomes increasingly important, especially for
presentation to an audience via PowerPoint or similar software.
Cluster the data and the answers. It is advisable to organize the given or estimated data in the
top left of the spreadsheet. A very brief label should be used to identify the data, for example,
MARR ⫽ in cell A1 and the value, 12%, in cell B1. Then B1 can be the referenced cell for all
entries requiring the MARR. Additionally, it may be worthwhile to cluster the answers into one
area and frame it. Often, the answers are best placed at the bottom or top of the column of entries
used in the formula or predefined function.
Enter titles for columns and rows. Each column or row should be labeled so its entries are
clear to the reader. It is very easy to select from the wrong column or row when no brief title is
present at the head of the data.
Enter income and cost cash flows separately. When there are both income and cost cash flows
involved, it is strongly recommended that the cash flow estimates for revenue (usually positive)
549
550
Using Spreadsheets and Microsoft Excel©
Appendix A
Function used to find FW
Figure A–2
Spreadsheet layout with cash flow estimates, results of functions, function formula detailed, and a scatter chart.
and first cost, salvage value, and annual costs (usually negative, with salvage a positive number)
be entered into two adjacent columns. Then a formula combining them in a third column displays
the net cash flow. There are two immediate advantages to this practice: fewer errors are made
when performing the summation and subtraction mentally, and changes for sensitivity analysis
are more easily made.
Use cell references. The use of absolute and relative cell references is a must when any changes
in entries are expected. For example, suppose the MARR is entered in cell B1 and three separate
references are made to the MARR in functions on the spreadsheet. The absolute cell reference
entry $B$1 in the three functions allows the MARR to be changed one time, not three.
Obtain a final answer through summing and embedding. When the formulas and functions
are kept relatively simple, the final answer can be obtained using the SUM function. For example, if the present worth values (PW) of two columns of cash flows are determined separately,
then the total PW is the SUM of the subtotals. This practice is especially useful when the cash
flow series are complex.
Prepare for a chart. If a chart (graph) will be developed, plan ahead by leaving sufficient
room on the right of the data and answers. Charts can be placed on the same worksheet or on a
separate worksheet. Placement on the same worksheet is recommended, especially when the
results of sensitivity analysis are plotted.
A.3 Excel Functions Important to Engineering
Economy (alphabetical order)
DB (Declining Balance)
Calculates the depreciation amount for an asset for a specified period n using the declining balance method. The depreciation rate d used in the computation is determined from asset values
S (salvage value) and B (basis or first cost) as d ⫽ 1 ⫺ (S兾B)1兾n. This is Equation [16.12]. Threedecimal-place accuracy is used for d.
ⴝ DB(cost, salvage, life, period, month)
cost
salvage
life
period
month
First cost or basis of the asset.
Salvage value.
Depreciation life (recovery period).
The period, year, for which the depreciation is to be calculated.
(optional entry) If this entry is omitted, a full year is assumed for the
first year.
Excel Functions Important to Engineering Economy
A.3
Example A new machine costs $100,000 and is expected to last 10 years. At the end of
10 years, the salvage value of the machine is $50,000. What is the depreciation of the machine in
the first year and the fifth year?
Depreciation for the first year: ⫽ DB(100000,50000,10,1)
Depreciation for the fifth year: ⫽ DB(100000,50000,10,5)
Because of the manner in which the DB function determines the fixed percentage d and the accuracy
of the computations, it is recommended that the DDB function (below) be used for all declining
balance depreciation rates. Simply use the optional factor entry for rates other than d ⫽ 2兾n.
DDB (Double Declining Balance)
Calculates the depreciation of an asset for a specified period n using the double declining balance
method. A factor can also be entered for some other declining balance depreciation method by
specifying a factor in the function.
ⴝ DDB(cost, salvage, life, period, factor)
cost
salvage
life
period
factor
First cost or basis of the asset.
Salvage value of the asset.
Depreciation life.
The period, a year, for which the depreciation is to be calculated.
(optional entry) If this entry is omitted, the function will use a double
declining method with 2 times the straight line rate. If, for example,
the entry is 1.5, the 150% declining balance method will be used.
Example A new machine costs $200,000 and is expected to last 10 years. The salvage value is
$10,000. Calculate the depreciation of the machine for the first and the eighth years. Finally,
calculate the depreciation for the fifth year using the 175% declining balance method.
Depreciation for the first year: ⫽ DDB(200000,10000,10,1)
Depreciation for the eighth year: ⫽ DDB(200000,10000,10,8)
Depreciation for the fifth year using 175% DB: ⫽ DDB(200000,10000,10,5,1.75)
EFFECT (Effective Interest Rate)
Calculates the effective annual interest rate for a stated nominal annual rate and a given number
of compounding periods per year. Excel uses Equation [4.7] to calculate the effective rate.
ⴝ EFFECT(nominal, npery)
nominal
npery
Nominal interest rate for the year.
Number of times interest is compounded per year.
Example Claude has applied for a $10,000 loan. The bank officer told him that the interest rate
is 8% per year and that interest is compounded monthly to conveniently match his monthly payments. What effective annual rate will Claude pay?
Effective annual rate: ⫽ EFFECT(8%,12)
EFFECT can also be used to find effective rates other than annually. Enter the nominal rate for
the time period of the required effective rate; npery is the number of times compounding occurs
during the time period of the effective rate.
Example Interest is stated as 3.5% per quarter with quarterly compounding. Find the effective
semiannual rate.
The 6-month nominal rate is 7%, and compounding is 2 times per 6 months.
Effective semiannual rate: ⫽ EFFECT(7%,2)
551
552
Using Spreadsheets and Microsoft Excel©
Appendix A
FV (Future Value)
Calculates the future value (worth) based on periodic payments at a specific interest rate.
ⴝ FV(rate, nper, pmt, pv, type)
rate
nper
pmt
pv
Interest rate per compounding period.
Number of compounding periods.
Constant payment amount.
The present value amount. If pv is not specified, the function will
assume it to be 0.
(optional entry) Either 0 or 1. A 0 represents payments made at the
end of the period, and 1 represents payments at the beginning of the
period. If omitted, 0 is assumed.
type
Example Jack wants to start a savings account that can be increased as desired. He will deposit
$12,000 to start the account and plans to add $500 to the account at the beginning of each month
for the next 24 months. The bank pays 0.25% per month. How much will be in Jack’s account at
the end of 24 months?
Future value in 24 months: ⫽ FV(0.25%,24,500,12000,1)
IF (IF Logical Function)
Determines which of two entries is entered into a cell based on the outcome of a logical check on
the outcome of another cell. The logical test can be a function or a simple value check, but it must
use an equality or inequality sense. If the response is a text string, place it between quote marks
(“ ”). The responses can themselves be IF functions. Up to seven IF functions can be nested for
very complex logical tests.
ⴝ IF(logical_test,value_if_true,value_if_false)
logical_test
value_if_true
value_if_false
Any worksheet function can be used here, including a
mathematical operation.
Result if the logical_test argument is true.
Result if the logical_test argument is false.
Example The entry in cell B4 should be “selected” if the PW value in cell B3 is greater than
or equal to zero and “rejected” if PW ⬍ 0.
Entry in cell B4:
⫽ IF(B3 ⬎⫽0,“selected”,“rejected”)
Example The entry in cell C5 should be “selected” if the PW value in cell C4 is greater than
or equal to zero, “rejected” if PW ⬍ 0, and “fantastic” if PW ⱖ 200.
Entry in cell C5:
⫽ IF(C4⬍0,“rejected”,IF(C4⬎⫽200,“fantastic”,“selected”))
IPMT (Interest Payment)
Calculates the interest accrued for a given period n based on constant periodic payments and
interest rate.
ⴝ IPMT(rate, per, nper, pv, fv, type)
rate
per
nper
pv
fv
type
Interest rate per compounding period.
Period for which interest is to be calculated.
Number of compounding periods.
Present value. If pv is not specified, the function will assume it to be 0.
Future value. If fv is omitted, the function will assume it to be 0. The fv
can also be considered a cash balance after the last payment is made.
(optional entry) Either 0 or 1. A 0 represents payments made at the
end of the period, and 1 represents payments made at the beginning of
the period. If omitted, 0 is assumed.
Excel Functions Important to Engineering Economy
A.3
Example Calculate the interest due in the 10th month for a 48-month, $20,000 loan. The interest rate is 0.25% per month.
Interest due: ⫽ IPMT(0.25%,10,48,20000)
IRR (Internal Rate of Return)
Calculates the internal rate of return between ⫺100% and infinity for a series of cash flows at
regular periods.
ⴝ IRR(values, guess)
values
guess
A set of numbers in a spreadsheet column (or row) for which the rate of
return will be calculated. The set of numbers must consist of at least one
positive and one negative number. Negative numbers denote a payment
made or cash outflow, and positive numbers denote income or cash
inflow.
(optional entry) To reduce the number of iterations, a guessed rate of
return can be entered. In most cases, a guess is not required, and a
10% rate of return is initially assumed. If the #NUM! error appears, try
using different values for guess. Inputting different guess values makes
it possible to determine the multiple roots for the rate of return equation
of a nonconventional cash flow series.
Example John wants to start a printing business. He will need $25,000 in capital and anticipates that the business will generate the following incomes during the first 5 years. Calculate his
rate of return after 3 years and after 5 years.
Year 1
Year 2
Year 3
Year 4
Year 5
$5,000
$7,500
$8,000
$10,000
$15,000
Set up an array in the spreadsheet.
In cell A1, type ⫺25000 (negative for payment).
In cell A2, type 5000 (positive for income).
In cell A3, type 7500.
In cell A4, type 8000.
In cell A5, type 10000.
In cell A6, type 15000.
Therefore, cells A1 through A6 contain the array of cash flows for the first 5 years, including the
capital outlay. Note that any years with a zero cash flow must have a zero entered to ensure that
the year value is correctly maintained for computation purposes.
To calculate the internal rate of return after 3 years, move to cell A7, and type ⫽ IRR(A1:A4).
To calculate the internal rate of return after 5 years and specify a guess value of 5%, move to
cell A8, and type ⫽ IRR(A1:A6,5%).
MIRR (Modified Internal Rate of Return)
Calculates the modified internal rate of return for a series of cash flows and reinvestment of income and interest at a stated rate.
ⴝ MIRR(values, finance_rate, reinvest_rate)
values
Refers to an array of cells in the spreadsheet. Negative
numbers represent payments, and positive numbers represent
553
554
Using Spreadsheets and Microsoft Excel©
Appendix A
finance_rate
reinvest_rate
income. The series of payments and income must occur at
regular periods and must contain at least one positive number
and one negative number.
Interest rate on funds borrowed from external sources
(ib in Equation [7.9]).
Interest rate for reinvestment on positive cash flows (ii in
Equation [7.9]). (This is not the same reinvestment rate on the
net investments when the cash flow series is nonconventional.
See Section 7.5 for comments.)
Example Jane opened a hobby store 4 years ago. When she started the business, Jane borrowed $50,000 from a bank at 12% per year. Since then, the business has yielded $10,000 the
first year, $15,000 the second year, $18,000 the third year, and $21,000 the fourth year. Jane
reinvests her profits, earning 8% per year. What is the modified rate of return after 3 years and
after 4 years?
In cell A1, type ⫺50000.
In cell A2, type 10000.
In cell A3, type 15000.
In cell A4, type 18000.
In cell A5, type 21000.
To calculate the modified rate of return after 3 years, move to cell A6, and type
⫽ MIRR(A1:A4,12%,8%).
To calculate the modified rate of return after 4 years, move to cell A7, and type
⫽ MIRR(A1:A5,12%,8%).
NOMINAL (Nominal Interest Rate)
Calculates the nominal annual interest rate for a stated effective annual rate and a given number
of compounding periods per year. This function is designed to display only nominal annual rates.
ⴝ NOMINAL(effective, npery)
effective
npery
Effective interest rate for the year.
Number of times that interest is compounded per year.
Example Last year, a corporate stock earned an effective return of 12.55% per year. Calculate
the nominal annual rate, if interest is compounded quarterly and compounded continuously.
Nominal annual rate, quarterly compounding: ⫽ NOMINAL(12.55%,4)
Nominal annual rate, continuous compounding: ⫽ NOMINAL(12.55%,100000)
NPER (Number of Periods)
Calculates the number of periods for the present worth of an investment to equal the future value
specified, based on uniform regular payments and a stated interest rate.
ⴝ NPER(rate, pmt, pv, fv, type)
rate
pmt
pv
fv
type
Interest rate per compounding period.
Amount paid during each compounding period.
Present value (lump-sum amount).
(optional entry) Future value or cash balance after the last payment.
If fv is omitted, the function will assume a value of 0.
(optional entry) Enter 0 if payments are due at the end of the
compounding period and 1 if payments are due at the beginning of the
period. If omitted, 0 is assumed.
Excel Functions Important to Engineering Economy
A.3
Example Sally plans to open a savings account that pays 0.25% per month. Her initial deposit
is $3000, and she plans to deposit $250 at the beginning of every month. How many payments
does she have to make to accumulate $25,000 to buy a new car?
Number of payments: ⫽ NPER(0.25%,⫺250,⫺3000,25000,1)
NPV (Net Present Value)
Calculates the net present value of a series of future cash flows at a stated interest rate.
ⴝ NPV(rate, series)
rate
series
Interest rate per compounding period.
Series of costs and incomes set up in a range of cells in the
spreadsheet.
Example Mark is considering buying a sports store for $100,000 and expects to receive the
following income during the next 6 years of business: $25,000, $40,000, $42,000, $44,000,
$48,000, $50,000. The interest rate is 8% per year.
In cells A1 through A7, enter ⫺100,000, followed by the six annual incomes.
Present value: ⫽ NPV(8%,A2:A7) ⫹ A1
The cell A1 value is already a present value. Any year with a zero cash flow must have a 0 entered
to ensure a correct result.
PMT (Payments)
Calculates equivalent periodic amounts based on present value and/or future value at a constant
interest rate.
ⴝ PMT(rate, nper, pv, fv, type)
rate
nper
pv
fv
type
Interest rate per compounding period.
Total number of periods.
Present value.
Future value.
(optional entry) Enter 0 for payments due at the end of the
compounding period and 1 if payment is due at the start of the
compounding period. If omitted, 0 is assumed.
Example Jim plans to take a $15,000 loan to buy a new car. The interest rate is 7% per year.
He wants to pay the loan off in 5 years (60 months). What are his monthly payments?
Monthly payments: ⫽ PMT(7%兾12,60,15000)
PPMT (Principal Payment)
Calculates the payment on the principal based on uniform payments at a specified interest
rate.
ⴝ PPMT(rate, per, nper, pv, fv, type)
rate
per
nper
pv
fv
type
Interest rate per compounding period.
Period for which the payment on the principal is required.
Total number of periods.
Present value.
Future value.
(optional entry) Enter 0 for payments that are due at the end of the
compounding period and 1 if payments are due at the start of the
compounding period. If omitted, 0 is assumed.
555
556
Using Spreadsheets and Microsoft Excel©
Appendix A
Example Jovita is planning to invest $10,000 in equipment which is expected to last 10 years
with no salvage value. The interest rate is 5%. What is the principal payment at the end of
year 4 and year 8?
At the end of year 4: ⫽ PPMT(5%,4,10,⫺10000)
At the end of year 8: ⫽ PPMT(5%,8,10,⫺10000)
PV (Present Value)
Calculates the present value of a future series of equal cash flows and a single lump sum in the
last period at a constant interest rate.
ⴝ PV(rate, nper, pmt, fv, type)
rate
nper
pmt
fv
type
Interest rate per compounding period.
Total number of periods.
Cash flow at regular intervals. Negative numbers represent payments
(cash outflows), and positive numbers represent income.
Future value or cash balance at the end of the last period.
(optional entry) Enter 0 if payments are due at the end of the
compounding period and 1 if payments are due at the start of each
compounding period. If omitted, 0 is assumed.
There are two primary differences between the PV function and the NPV function: PV allows for
end or beginning of period cash flows, and PV requires that all amounts have the same value,
whereas they may vary for the NPV function.
Example Jose is considering leasing a car for $300 a month for 3 years (36 months). After the
36-month lease, he can purchase the car for $12,000. Using an interest rate of 8% per year, find
the present value of this option.
Present value: ⫽ PV(8%兾12,36,⫺300,⫺12000)
Note the minus signs on the pmt and fv amounts.
RAND (Random Number)
Returns an evenly distributed number that is (1) ⱖ 0 and ⬍ 1; (2) ⱖ 0 and ⬍ 100; or (3) between
two specified numbers.
ⴝ RAND()
for range 0 to 1
ⴝ RAND()*100
for range 0 to 100
ⴝ RAND()*(bⴚa)ⴙa
for range a to b
a ⫽ minimum integer to be generated
b ⫽ maximum integer to be generated
The Excel function RANDBETWEEN(a,b) may also be used to obtain a random number between two values.
Example Grace needs random numbers between 5 and 10 with 3 digits after the decimal. What
is the Excel function? Here a ⫽ 5 and b ⫽ 10.
Random number:
⫽ RAND()*5 ⫹ 5
Example Randi wants to generate random numbers between the limits of ⫺10 and 25. What
is the Excel function? The minimum and maximum values are a ⫽ ⫺10 and b ⫽ 25, so b ⫺ a ⫽
25 ⫺ (⫺10) ⫽ 35.
Random number:
⫽ RAND()*35 ⫺ 10
Excel Functions Important to Engineering Economy
A.3
RATE (Interest Rate)
Calculates the interest rate per compounding period for a series of payments or incomes.
ⴝ RATE(nper, pmt, pv, fv, type, guess)
nper
pmt
pv
fv
type
guess
Total number of periods.
Payment amount made each compounding period.
Present value.
Future value (not including the pmt amount).
(optional entry) Enter 0 for payments due at the end of the
compounding period and 1 if payments are due at the start of each
compounding period. If omitted, 0 is assumed.
(optional entry) To minimize computing time, include a guessed
interest rate. If a value of guess is not specified, the function will
assume a rate of 10%. This function usually converges to a solution if
the rate is between 0% and 100%.
Example Alysha wants to start a savings account at a bank. She will make an initial deposit of
$1000 to open the account and plans to deposit $100 at the beginning of each month. She plans
to do this for the next 3 years (36 months). At the end of 3 years, she wants to have at least $5000.
What is the minimum interest required to achieve this result?
Interest rate: ⫽ RATE(36,⫺100,⫺1000,5000,1)
SLN (Straight Line Depreciation)
Calculates the straight line depreciation of an asset for a given year.
ⴝ SLN(cost, salvage, life)
cost
salvage
life
First cost or basis of the asset.
Salvage value.
Depreciation life.
Example Maria purchased a printing machine for $100,000. The machine has an allowed depreciation life of 8 years and an estimated salvage value of $15,000. What is the depreciation
each year?
Depreciation: ⫽ SLN(100000,15000,8)
SYD (Sum-of-Years-Digits Depreciation)
Calculates the sum-of-years-digits depreciation of an asset for a given year.
ⴝ SYD(cost, salvage, life, period)
cost
salvage
life
period
First cost or basis of the asset.
Salvage value.
Depreciation life.
The year for which the depreciation is sought.
Example Jack bought equipment for $100,000 that has a depreciation life of 10 years. The
salvage value is $10,000. What is the depreciation for year 1 and year 9?
Depreciation for year 1: ⫽ SYD(100000,10000,10,1)
Depreciation for year 9: ⫽ SYD(100000,10000,10,9)
VDB (Variable Declining Balance)
Calculates the depreciation using the declining balance method with a switch to straight line
depreciation in the year in which straight line has a larger depreciation amount. This function
557
558
Using Spreadsheets and Microsoft Excel©
Appendix A
automatically implements the switch from DB to SL depreciation, unless specifically instructed
to not switch.
ⴝ VDB (cost, salvage, life, start_period, end_period, factor, no_switch)
cost
salvage
life
start_period
end_period
factor
no_switch
First cost of the asset.
Salvage value.
Depreciation life.
First period for depreciation to be calculated.
Last period for depreciation to be calculated.
(optional entry) If omitted, the function will use the double
declining rate of 2兾n, or twice the straight line rate. Other entries
define the declining balance method, for example, 1.5 for 150%
declining balance.
(optional entry) If omitted or entered as FALSE, the function will
switch from declining balance to straight line depreciation when
the latter is greater than DB depreciation. If entered as TRUE, the
function will not switch to SL depreciation at any time during the
depreciation life.
Example Newly purchased equipment with a first cost of $300,000 has a depreciable life of
10 years with no salvage value. Calculate the 175% declining balance depreciation for the first year
and the ninth year if switching to SL depreciation is acceptable and if switching is not permitted.
Depreciation for first year, with switching: ⫽ VDB(300000,0,10,0,1,1.75)
Depreciation for ninth year, with switching: ⫽ VDB(300000,0,10,8,9,1.75)
Depreciation for first year, no switching: ⫽ VDB(300000,0,10,0,1,1.75,TRUE)
Depreciation for ninth year, no switching: ⫽ VDB(300000,0,10,8,9,1.75,TRUE)
VDB (for MACRS Depreciation)
The VDB function can be adapted to generate the MACRS annual depreciation amount, when the
start_period and end_period are replaced with the MAX and MIN functions, respectively. As
above, the factor option should be entered if other than DDB rates start the MACRS depreciation.
The VDB format is
ⴝ VDB(cost,0,life,MAX(0,tⴚ1.5),MIN(life,tⴚ0.5),factor)
Example Determine the MACRS depreciation for year 4 for a $350,000 asset that has a 20%
salvage value and a MACRS recovery period of 3 years. D4 ⫽ $25,926 is the display.
Depreciation for year 4: ⫽ VDB(350000,0,3,MAX(0,4⫺1.5),MIN(3,4⫺0.5),2)
Example Find the MACRS depreciation in year 16 for a $350,000-asset with a recovery
period of n ⫽ 15 years. The optional factor 1.5 is required here, since MACRS starts with
150% DB for n ⫽ 15-year and 20-year recovery periods. D16 ⫽ $10,334.
Depreciation for year 16: ⫽ VDB(350000,0,15,MAX(0,16⫺1.5),MIN(15,16⫺0.5),1.5)
Other Functions
There are numerous additional financial functions available on Excel, as well as engineering,
mathematics, trigonometry, statistics, data and time, logical, and information functions. These
can be viewed by clicking the Formulas tab on the Excel toolbar.
A.4 Goal Seek—A Tool for Breakeven and
Sensitivity Analysis
Goal Seek is found on the Excel toolbar labeled Data, followed by What-if Analysis. This tool
changes the value in a specific cell based on a numerical value in another (changing) cell as input
by the user. It is a good tool for sensitivity analysis, breakeven analysis, and “what if?” questions
A.5
Solver—An Optimizing Tool for Capital Budgeting, Breakeven, and Sensitivity Analysis
Figure A–3
Goal Seek template used
to specify a cell, a value,
and the changing cell.
⫽ B$5
⫽ B$5⫹500
Figure A–4
Use of Goal Seek to determine an annual cash flow to increase the rate of return.
when no constraint relations or inequalities are needed. The initial Goal Seek template is pictured
in Figure A–3. One of the cells (set or changing cell) must contain an equation or spreadsheet
function that uses the other cell to determine a numeric value. Only a single cell can be identified
as the changing cell; however, this limitation can be avoided by using equations rather than specific numerical inputs in any additional cells also to be changed. This is demonstrated below.
Example A new asset will cost $25,000, generate an annual cash flow of $6000 over its 5-year
life, and have an estimated $500 salvage value. The rate of return using the IRR function is
6.94%. Determine the annual cash flow necessary to raise the return to 10% per year.
Figure A–4 (top left) shows the cash flows and return displayed using the function
⫽ IRR(B4:B9) prior to the use of Goal Seek. Note that the initial $6000 is input in cell B5, but
other years’ cash flows are input as equations that refer to B5. The $500 salvage is added for the
last year. This format allows Goal Seek to change only cell B5 while making the other cash flows
have the same value. The tool finds the required cash flow of $6506 to approximate the 10% per
year return. The Goal Seek Status inset indicates that a solution is found. Clicking OK saves all
changed cells; clicking Cancel returns to the original values.
A.5 Solver—An Optimizing Tool for Capital Budgeting,
Breakeven, and Sensitivity Analysis
Solver is a powerful spreadsheet tool to change the value in multiple (one or more) cells based on
the value in a specific (target) cell. It is excellent when solving a capital budgeting problem to
select from independent projects where budget constraints are present. (Section 12.4 details this
application.) The initial Solver template is shown in Figure A–5.
559
560
Using Spreadsheets and Microsoft Excel©
Appendix A
Figure A–5
Solver template used to
specify optimization in a
target cell, multiple
changing cells, and
constraint relations.
Set Target Cell box. Enter a cell reference or name. The target cell itself must contain a formula or function. The value in the cell can be maximized (Max), minimized (Min), or restricted
to a specified value (Value of).
By Changing Cells box. Enter the cell reference for each cell to be adjusted, using commas
between nonadjacent cells. Each cell must be directly or indirectly related to the target cell.
Solver proposes a value for the changing cell based on input provided about the target cell. The
Guess button will list all possible changing cells related to the target cell.
Subject to the Constraints box. Enter any constraints that may apply, for example,
$C$1 ⬍ $50,000. Integer and binary variables are determined in this box.
Options box. Choices here allow the user to specify various parameters of the solution: maximum time and number of iterations allowed, the precision and tolerance of the values determined, and the convergence requirements as the final solution is determined. Also, linear and
nonlinear model assumptions can be set here. If integer or binary variables are involved, the
tolerance option must be set to a small number, say, 0.0001. This is especially important for the
binary variables when selecting from independent projects (Chapter 12). If tolerance remains at
the default value of 5%, a project may be incorrectly included in the solution set at a very low
level.
Solver Results box. This appears after Solve is clicked and a solution appears. It is possible,
of course, that no solution can be found for the scenario described. It is possible to update the
spreadsheet by clicking Keep Solver Solution, or return to the original entries using Restore
Original Values.
A.6 Error Messages
If Excel is unable to complete a formula or function computation, an error message is displayed.
Some of the common messages are:
#DIV兾0!
#N兾A
#NAME?
#NULL!
#NUM!
#REF!
#VALUE!
#####
Requires division by zero.
Refers to a value that is not available.
Uses a name that Excel doesn’t recognize.
Specifies an invalid intersection of two areas.
Uses a number incorrectly.
Refers to a cell that is not valid.
Uses an invalid argument or operand.
Produces a result, or includes a constant numeric value,
that is too long to fit in the cell. (Widen the column.)
APPENDIX B
BASICS OF ACCOUNTING
REPORTS AND BUSINESS RATIOS
This appendix provides a fundamental description of financial statements. The documents discussed here will assist in reviewing or understanding basic financial statements and in gathering
information useful in an engineering economy study.
B.1 The Balance Sheet
The fiscal year and the tax year are defined identically for a corporation or an individual—
12 months in length. The fiscal year (FY) is commonly not the calendar year (CY) for a corporation. The U.S. government uses October through September as its FY. For example, October
2011 through September 2012 is FY2012. The fiscal or tax year is always the calendar year for
an individual citizen.
At the end of each fiscal year, a company publishes a balance sheet. A sample balance sheet
for JAGBA Corporation is presented in Table B–1. This is a yearly presentation of the state of the
firm at a particular time, for example, May 31, 2012; however, a balance sheet is also usually
prepared quarterly and monthly. Three main categories are used.
Assets. This section is a summary of all resources owned by or owed to the company. There
are two main classes of assets. Current assets represent shorter-lived working capital (cash,
accounts receivable, etc.), which is more easily converted to cash, usually within 1 year. Longerlived assets are referred to as fixed assets (land, equipment, etc.). Conversion of these holdings
to cash in a short time would require a major corporate reorientation.
Liabilities. This section is a summary of all financial obligations (debts, mortgages, loans,
etc.) of a corporation. Bond indebtedness is included here.
Net worth. Also called owner’s equity, this section provides a summary of the financial
value of ownership, including stocks issued and earnings retained by the corporation.
TABLE B–1
Sample Balance Sheet
JAGBA CORPORATION
Balance Sheet
May 31, 2012
Assets
Current
Cash
Accounts receivable
Interest accrued receivable
Inventories
Total current assets
Fixed
Land
Building and equipment
Less: Depreciation
allowance $82,000
Total fixed assets
Total assets
Liabilities
$10,500
18,700
500
52,000
$81,700
Accounts payable
Dividends payable
Long-term notes payable
Bonds payable
Total liabilities
$19,700
7,000
16,000
20,000
$62,700
Net Worth
$25,000
438,000
356,000
381,000
$462,700
Common stock
Preferred stock
Retained earnings
$275,000
100,000
25,000
Total net worth
Total liabilities and net worth
400,000
$462,700
562
Basics of Accounting Reports and Business Ratios
Appendix B
The balance sheet is constructed using the relation
Assets ⴝ liabilities ⴙ net worth
In Table B–1 each major category is further divided into standard subcategories. For example,
current assets is comprised of cash, accounts receivable, etc. Each subdivision has a specific interpretation, such as accounts receivable, which represents all money owed to the company by its
customers.
B.2 Income Statement and Cost of Goods
Sold Statement
A second important financial statement is the income statement (Table B–2). The income statement summarizes the profits or losses of the corporation for a stated period of time. Income statements always accompany balance sheets. The major categories of an income statement are
Revenues. This includes all sales and interest revenue that the company has received in the
past accounting period.
Expenses. This is a summary of all expenses (operating and others, including taxes) for
the period. Some expense amounts are itemized in other statements, for example, cost of
goods sold.
The final result of an income statement is the net profit after taxes (NPAT), or NOPAT (O for
operating), the amount used in Chapter 17, Sections 17.1 and 17.7. The income statement, published at the same time as the balance sheet, uses the basic equation
Revenues ⴚ expenses ⴝ profit (or loss)
The cost of goods sold is an important accounting term. It represents the net cost of producing
the product marketed by the firm. Cost of goods sold may also be called factory cost. A statement
of the cost of goods sold, such as that shown in Table B–3, is useful in determining exactly how
much it costs to make a particular product over a stated time period, usually a year. The total of
the cost of goods sold statement is entered as an expense item on the income statement. This total
is determined using the relations
Cost of goods sold ⴝ prime cost ⴙ indirect cost
Prime cost ⴝ direct materials ⴙ direct labor
TABLE B–2
Sample Income Statement
JAGBA CORPORATION
Income Statement
Year Ended May 31, 2012
Revenues
Sales
Interest revenue
Total revenues
Expenses
Cost of goods sold (from Table B–3)
Selling
Administrative
Other
Total expenses
Income before taxes
Taxes for year
Net profit after taxes (NPAT)
$505,000
3,500
$508,500
$290,000
28,000
35,000
12,000
365,000
143,500
64,575
$ 78,925
[B.1]
Business Ratios
B.3
TABLE B–3
Sample Cost of Goods Sold Statement
JAGBA CORPORATION
Statement of Cost of Goods Sold
Year Ended May 31, 2012
Materials
Inventory June 1, 2011
Purchases during year
Total
Less: Inventory May 31, 2012
Cost of materials
Direct labor
Prime cost
Indirect costs
Factory cost
Less: Increase in finished goods inventory during year
Cost of goods sold (into Table B–2)
$ 54,000
174,500
$228,500
50,000
$178,500
110,000
288,500
7,000
295,500
5,500
$290,000
Indirect costs include all indirect and overhead charges made to a product, process, or cost center.
Indirect cost allocation methods are discussed in Chapter 15.
B.3 Business Ratios
Accountants, financial analysts, and engineering economists frequently utilize business ratio
analysis to evaluate the financial health (status) of a company over time and in relation to industry norms. Because the engineering economist must continually communicate with others, she or
he should have a basic understanding of several ratios. For comparison purposes, it is necessary
to compute the ratios for several companies in the same industry. Industrywide median ratio values are published annually by firms such as Dun and Bradstreet in Industry Norms and Key Business Ratios. The ratios are classified according to their role in measuring the corporation.
Solvency ratios. Assess ability to meet short-term and long-term financial obligations.
Efficiency ratios. Measure management’s ability to use and control assets.
Profitability ratios. Evaluate the ability to earn a return for the owners of the corporation.
Numerical data for several important ratios are discussed here and are extracted from the JAGBA
balance sheet and income statement, Tables B–1 and B–2.
Current Ratio This ratio is utilized to analyze the company’s working capital condition. It is
defined as
current assets
Current ratio ⫽ ———————
current liabilities
Current liabilities include all short-term debts, such as accounts and dividends payable. Note that
only balance sheet data are utilized in the current ratio; that is, no association with revenues or
expenses is made. For the balance sheet of Table B–1, current liabilities amount to $19,700 ⫹
$7000 ⫽ $26,700 and
81,700
Current ratio ⫽ ——— ⫽ 3.06
26,700
Since current liabilities are those debts payable in the next year, the current ratio value of 3.06
means that the current assets would cover short-term debts approximately 3 times. Current ratio
values of 2 to 3 are common.
The current ratio assumes that the working capital invested in inventory can be converted to
cash quite rapidly. Often, however, a better idea of a company’s immediate financial position can
be obtained by using the acid test ratio.
563
564
Appendix B
Basics of Accounting Reports and Business Ratios
Acid Test Ratio (Quick Ratio) This ratio is
quick assets
Acid-test ratio ⫽ ———————
current liabilities
current assets ⫺ inventories
⫽ ————————————
current liabilities
It is meaningful for the emergency situation when the firm must cover short-term debts using its
readily convertible assets. For JAGBA Corporation,
81,700 ⫺ 52,000
Acid test ratio ⫽ ——————— ⫽ 1.11
26,700
Comparison of this and the current ratio shows that approximately 2 times the current debt of the
company is invested in inventories. However, an acid test ratio of approximately 1.0 is generally
regarded as a strong current position, regardless of the amount of assets in inventories.
Debt Ratio This ratio is a measure of financial strength since it is defined as
total liabilities
Debt ratio ⫽ ——————
total assets
For JAGBA Corporation,
62,700
Debt ratio ⫽ ———— ⫽ 0.136
462,700
JAGBA is 13.6% creditor-owned and 86.4% stockholder-owned. A debt ratio in the range of 20%
or less usually indicates a sound financial condition, with little fear of forced reorganization because
of unpaid liabilities. However, a company with virtually no debts, that is, one with a very low debt
ratio, may not have a promising future, because of its inexperience in dealing with short-term and
long-term debt financing. The debt-equity (D-E) mix is another measure of financial strength.
Return on Sales Ratio This often quoted ratio indicates the profit margin for the company. It
is defined as
net profit
Return on sales ⫽ ———— (100%)
net sales
Net profit is the after-tax value from the income statement. This ratio measures profit earned per
sales dollar and indicates how well the corporation can sustain adverse conditions over time, such
as falling prices, rising costs, and declining sales. For JAGBA Corporation,
78,925
Return on sales ⫽ ———— (100%) ⫽ 15.6%
505,000
Corporations may point to small return on sales ratios, say, 2.5% to 4.0%, as indications of sagging economic conditions. In truth, for a relatively large-volume, high-turnover business, an income ratio of 3% is quite healthy. Of course, a steadily decreasing ratio indicates rising company
expenses, which absorb net profit after taxes.
Return on Assets Ratio This is the key indicator of profitability since it evaluates the ability of the corporation to transfer assets into operating profit. The definition and value for
JAGBA are
net profit
Return on assets ⫽ ————— (100%)
total assets
78,925
⫽ ———— (100%) ⫽ 17.1%
462,700
Efficient use of assets indicates that the company should earn a high return, while low returns
usually accompany lower values of this ratio compared to the industry group ratios.
Inventory Turnover Ratio Two different ratios are used here. They both indicate the number
of times the average inventory value passes through the operations of the company. If turnover of
inventory to net sales is desired, the formula is
net sales
Net sales to inventory ⫽ ————————
average inventory
Business Ratios
B.3
where average inventory is the figure recorded in the balance sheet. For JAGBA Corporation this
ratio is
505,000
Net sales to inventory ⫽ ———— ⫽ 9.71
52,000
This means that the average value of the inventory has been sold 9.71 times during the year.
Values of this ratio vary greatly from one industry to another.
If inventory turnover is related to cost of goods sold, the ratio to use is
cost of goods sold
Cost of goods sold to inventory ⫽ ————————
average inventory
Now, average inventory is computed as the average of the beginning and ending inventory values
in the statement of cost of goods sold. This ratio is commonly used as a measure of the inventory
turnover rate in manufacturing companies. It varies with industries, but management likes to see
it remain relatively constant as business increases. For JAGBA, using the values in Table B–3,
290,000
Cost of goods sold to inventory ⫽ —————————
⫽ 5.58
_1 (54,000 ⫹ 50,000)
2
There are, of course, many other ratios to use in various circumstances; however, the ones
presented here are commonly used by both accountants and economic analysts.
EXAMPLE B.1
Sample values for financial ratios or percentages of four industry sectors are presented below.
Compare the corresponding JAGBA Corporation values with these norms, and comment on
differences and similarities.
Ratio or
Percentage
Current ratio
Quick ratio
Debt ratio
Return on
assets
Motor Vehicles
and Parts
Manufacturing
336105*
2.4
1.6
59.3%
40.9%
Air
Transportation
(Medium-Sized)
481000*
0.4
0.3
96.8%
8.1%
Industrial
Machinery
Home
Manufacturing Furnishings
333200*
442000*
2.2
1.5
49.1%
8.0%
2.6
1.2
52.4%
5.1%
*North American Industry Classification System (NAICS) code for this industry sector.
SOURCE: L. Troy, Almanac of Business and Industrial Financial Ratios, CCH, Wolters Kluwer, USA.
Solution
It is not correct to compare ratios for one company with indexes in different industries, that is,
with indexes for different NAICS codes. So the comparison below is for illustration purposes
only. The corresponding values for JAGBA are
Current ratio ⫽ 3.06
Quick ratio ⫽ 1.11
Debt ratio ⫽ 13.5%
Return on assets ⫽ 17.1%
JAGBA has a current ratio larger than all four of these industries, since 3.06 indicates it can
cover current liabilities 3 times compared with 2.6 and much less in the case of the “average”
air transportation corporation. JAGBA has a significantly lower debt ratio than that of any of
the sample industries, so it is likely more financially sound. Return on assets, which is a measure of ability to turn assets into profitability, is not as high at JAGBA as motor vehicles, but
JAGBA competes well with the other industry sectors.
To make a fair comparison of JAGBA ratios with other values, it is necessary to have norm
values for its industry type as well as ratio values for other corporations in the same NAICS
category and about the same size in total assets. Corporate assets are classified in categories by
$100,000 units, such as 100 to 250, 1001 to 5000, over 250,000, etc.
565
APPENDIX C
CODE OF ETHICS FOR ENGINEERS
Source: National Society of Professional Engineers (www.nspe.org).
Appendix C
Code of Ethics for Engineers
567
Code of Ethics for Engineers
Preamble
Engineering is an important and learned profession. As members of this
profession, engineers are expected to exhibit the highest standards of honesty
and integrity. Engineering has a direct and vital impact on the quality of life for
all people. Accordingly, the services provided by engineers require honesty,
impartiality, fairness, and equity, and must be dedicated to the protection of the
public health, safety, and welfare. Engineers must perform under a standard of
professional behavior that requires adherence to the highest principles of ethical
conduct.
I. Fundamental Canons
Engineers, in the fulfillment of their professional duties, shall:
1. Hold paramount the safety, health, and welfare of the public.
2. Perform services only in areas of their competence.
3. Issue public statements only in an objective and truthful manner.
4. Act for each employer or client as faithful agents or trustees.
5. Avoid deceptive acts.
6. Conduct themselves honorably, responsibly, ethically, and
lawfully so as to enhance the honor, reputation, and usefulness
of the profession.
II. Rules of Practice
1. Engineers shall hold paramount the safety, health, and welfare
of the public.
a. If engineers’ judgment is overruled under circumstances that
endanger life or property, they shall notify their employer or client
and such other authority as may be appropriate.
b. Engineers shall approve only those engineering documents that are
in conformity with applicable standards.
c. Engineers shall not reveal facts, data, or information without the
prior consent of the client or employer except as authorized or
required by law or this Code.
d. Engineers shall not permit the use of their name or associate in
business ventures with any person or firm that they believe is
engaged in fraudulent or dishonest enterprise.
e. Engineers shall not aid or abet the unlawful practice of engineering
by a person or firm.
f. Engineers having knowledge of any alleged violation of this Code
shall report thereon to appropriate professional bodies and, when
relevant, also to public authorities, and cooperate with the proper
authorities in furnishing such information or assistance as may be
required.
2. Engineers shall perform services only in the areas of their
competence.
a. Engineers shall undertake assignments only when qualified by
education or experience in the specific technical fields involved.
b. Engineers shall not affix their signatures to any plans or documents
dealing with subject matter in which they lack competence, nor to
any plan or document not prepared under their direction and
control.
c. Engineers may accept assignments and assume responsibility for
coordination of an entire project and sign and seal the engineering
documents for the entire project, provided that each technical
segment is signed and sealed only by the qualified engineers who
prepared the segment.
3. Engineers shall issue public statements only in an objective and
truthful manner.
a. Engineers shall be objective and truthful in professional reports,
statements, or testimony. They shall include all relevant and
pertinent information in such reports, statements, or testimony,
which should bear the date indicating when it was current.
b. Engineers may express publicly technical opinions that are founded
upon knowledge of the facts and competence in the subject matter.
c. Engineers shall issue no statements, criticisms, or arguments on
technical matters that are inspired or paid for by interested parties,
unless they have prefaced their comments by explicitly identifying
the interested parties on whose behalf they are speaking, and by
revealing the existence of any interest the engineers may have in the
matters.
4. Engineers shall act for each employer or client as faithful agents or
trustees.
a. Engineers shall disclose all known or potential conflicts of interest
that could influence or appear to influence their judgment or the
quality of their services.
b. Engineers shall not accept compensation, financial or otherwise,
from more than one party for services on the same project, or for
services pertaining to the same project, unless the circumstances are
fully disclosed and agreed to by all interested parties.
c. Engineers shall not solicit or accept financial or other valuable
consideration, directly or indirectly, from outside agents in
connection with the work for which they are responsible.
d. Engineers in public service as members, advisors, or employees
of a governmental or quasi-governmental body or department shall
not participate in decisions with respect to services solicited or
provided by them or their organizations in private or public
engineering practice.
e. Engineers shall not solicit or accept a contract from a governmental
body on which a principal or officer of their organization serves as
a member.
5. Engineers shall avoid deceptive acts.
a. Engineers shall not falsify their qualifications or permit
misrepresentation of their or their associates’ qualifications. They
shall not misrepresent or exaggerate their responsibility in or for the
subject matter of prior assignments.
Brochures or other
presentations incident to the solicitation of employment shall not
misrepresent pertinent facts concerning employers, employees,
associates, joint venturers, or past accomplishments.
b. Engineers shall not offer, give, solicit, or receive, either directly or
indirectly, any contribution to influence the award of a contract by
public authority, or which may be reasonably construed by the
public as having the effect or intent of influencing the awarding of a
contract.
They shall not offer any gift or other valuable
consideration in order to secure work. They shall not pay a
commission, percentage, or brokerage fee in order to secure work,
except to a bona fide employee or bona fide established commercial
or marketing agencies retained by them.
III. Professional Obligations
1. Engineers shall be guided in all their relations by the highest standards
of honesty and integrity.
a. Engineers shall acknowledge their errors and shall not distort or
alter the facts.
b. Engineers shall advise their clients or employers when they believe
a project will not be successful.
c. Engineers shall not accept outside employment to the detriment of
their regular work or interest. Before accepting any outside
engineering employment, they will notify their employers.
d. Engineers shall not attempt to attract an engineer from another
employer by false or misleading pretenses.
e. Engineers shall not promote their own interest at the expense of the
dignity and integrity of the profession.
2. Engineers shall at all times strive to serve the public interest.
a. Engineers are encouraged to participate in civic affairs; career
guidance for youths; and work for the advancement of the safety,
health, and well-being of their community.
b. Engineers shall not complete, sign, or seal plans and/or
specifications that are not in conformity with applicable engineering
standards. If the client or employer insists on such unprofessional
conduct, they shall notify the proper authorities and withdraw from
further service on the project.
c. Engineers are encouraged to extend public knowledge and
appreciation of engineering and its achievements.
d. Engineers are encouraged to adhere to the principles of sustainable
development1 in order to protect the environment for future
generations.
568
Code of Ethics for Engineers
Appendix C
3. Engineers shall avoid all conduct or practice that deceives the public.
a. Engineers shall avoid the use of statements containing a material
misrepresentation of fact or omitting a material fact.
b. Consistent with the foregoing, engineers may advertise for
recruitment of personnel.
c. Consistent with the foregoing, engineers may prepare articles for
the lay or technical press, but such articles shall not imply credit to
the author for work performed by others.
4. Engineers shall not disclose, without consent, confidential information
concerning the business affairs or technical processes of any present or
former client or employer, or public body on which they serve.
a. Engineers shall not, without the consent of all interested parties,
promote or arrange for new employment or practice in connection
with a specific project for which the engineer has gained particular
and specialized knowledge.
b. Engineers shall not, without the consent of all interested parties,
participate in or represent an adversary interest in connection with a
specific project or proceeding in which the engineer has gained
particular specialized knowledge on behalf of a former client or
employer.
5. Engineers shall not be influenced in their professional duties by
conflicting interests.
a. Engineers shall not accept financial or other considerations,
including free engineering designs, from material or equipment
suppliers for specifying their product.
b. Engineers shall not accept commissions or allowances, directly or
indirectly, from contractors or other parties dealing with clients or
employers of the engineer in connection with work for which the
engineer is responsible.
6. Engineers shall not attempt to obtain employment or advancement or
professional engagements by untruthfully criticizing other engineers,
or by other improper or questionable methods.
a. Engineers shall not request, propose, or accept a commission on a
contingent basis under circumstances in which their judgment may
be compromised.
b. Engineers in salaried positions shall accept part-time engineering
work only to the extent consistent with policies of the employer and
in accordance with ethical considerations.
c. Engineers shall not, without consent, use equipment, supplies,
laboratory, or office facilities of an employer to carry on outside
private practice.
7.
Engineers shall not attempt to injure, maliciously or falsely, directly
or indirectly, the professional reputation, prospects, practice, or
employment of other engineers. Engineers who believe others are
guilty of unethical or illegal practice shall present such information
to the proper authority for action.
a. Engineers in private practice shall not review the work of another
engineer for the same client, except with the knowledge of such
engineer, or unless the connection of such engineer with the work
has been terminated.
b. Engineers in governmental, industrial, or educational employ are
entitled to review and evaluate the work of other engineers when so
required by their employment duties.
c. Engineers in sales or industrial employ are entitled to make
engineering comparisons of represented products with products of
other suppliers.
8. Engineers shall accept personal responsibility for their professional
activities, provided, however, that engineers may seek indemnification
for services arising out of their practice for other than gross
negligence, where the engineer’s interests cannot otherwise be
protected.
a. Engineers shall conform with state registration laws in the practice
of engineering.
b. Engineers shall not use association with a nonengineer, a
corporation, or partnership as a “cloak” for unethical acts.
9. Engineers shall give credit for engineering work to those to whom
credit is due, and will recognize the proprietary interests of others.
a. Engineers shall, whenever possible, name the person or persons
who may be individually responsible for designs, inventions,
writings, or other accomplishments.
b. Engineers using designs supplied by a client recognize that the
designs remain the property of the client and may not be duplicated
by the engineer for others without express permission.
c. Engineers, before undertaking work for others in connection with
which the engineer may make improvements, plans, designs,
inventions, or other records that may justify copyrights or patents,
should enter into a positive agreement regarding ownership.
d. Engineers’ designs, data, records, and notes referring exclusively to
an employer’s work are the employer’s property. The employer
should indemnify the engineer for use of the information for any
purpose other than the original purpose.
e. Engineers shall continue their professional development throughout
their careers and should keep current in their specialty fields by
engaging in professional practice, participating in continuing
education courses, reading in the technical literature, and attending
professional meetings and seminars.
Footnote 1 “Sustainable development” is the challenge of meeting human
needs for natural resources, industrial products, energy, food,
transportation, shelter, and effective waste management while
conserving and protecting environmental quality and the natural
resource base essential for future development.
As Revised July 2007
“By order of the United States District Court for the District of Columbia,
former Section 11(c) of the NSPE Code of Ethics prohibiting competitive
bidding, and all policy statements, opinions, rulings or other guidelines
interpreting its scope, have been rescinded as unlawfully interfering with the
legal right of engineers, protected under the antitrust laws, to provide price
information to prospective clients; accordingly, nothing contained in the NSPE
Code of Ethics, policy statements, opinions, rulings or other guidelines prohibits
the submission of price quotations or competitive bids for engineering services
at any time or in any amount.”
Statement by NSPE Executive Committee
In order to correct misunderstandings which have been indicated in some
instances since the issuance of the Supreme Court decision and the entry of the
Final Judgment, it is noted that in its decision of April 25, 1978, the Supreme
Court of the United States declared: “The Sherman Act does not require
competitive bidding.”
It is further noted that as made clear in the Supreme Court decision:
1. Engineers and firms may individually refuse to bid for engineering services.
2. Clients are not required to seek bids for engineering services.
3. Federal, state, and local laws governing procedures to procure engineering
services are not affected, and remain in full force and effect.
4. State societies and local chapters are free to actively and aggressively seek
legislation for professional selection and negotiation procedures by public
agencies.
5. State registration board rules of professional conduct, including rules
prohibiting competitive bidding for engineering services, are not affected and
remain in full force and effect. State registration boards with authority to
adopt rules of professional conduct may adopt rules governing procedures to
obtain engineering services.
6. As noted by the Supreme Court, “nothing in the judgment prevents NSPE and
its members from attempting to influence governmental action . . .”
Note: In regard to the question of application of the Code to corporations vis-a-vis real persons, business form or type should not negate nor
influence conformance of individuals to the Code. The Code deals with professional services, which services must be performed by real
persons. Real persons in turn establish and implement policies within business structures. The Code is clearly written to apply to the Engineer,
and it is incumbent on members of NSPE to endeavor to live up to its provisions. This applies to all pertinent sections of the Code.
1420 King Street
Alexandria, Virginia 22314-2794
703/684-2800 • Fax:703/836-4875
www.nspe.org
Publication date as revised: July 2007 • Publication #1102
APPENDIX D
ALTERNATE METHODS FOR
EQUIVALENCE CALCULATIONS
Throughout the text, engineering economy factor formulas, tabulated factor values, or built-in
spreadsheet functions have been used to obtain a value of P, F, A, i, or n. Because of advances in
programmable and scientific calculators, many of the equivalence computations can be performed without the use of tables or spreadsheets, but rather with a handheld calculator. An overview of the possibilities is presented here.
Alternatively, the recognition that all equivalence calculations involve geometric series can
likewise remove the need for tabulated values or spreadsheet functions. From the summation of
the series, it is possible to perform calculator-based computations to obtain P, F, or A values. A
brief introduction to this technique is presented in Section D.2.
D.1 Using Programmable Calculators
A basic way to calculate one parameter, given the other four, is to use a calculator that allows a
present worth relation to be encoded. The software can then solve for any one of the parameters,
when the remaining four are entered. For example, consider a PW relation in which all five parameters are included.
A(P/A,i,n) F(P/F,i,n) P 0
The A, P, and F values can be positive (cash inflow) or negative (cash outflow) or zero, as long
as there is at least one value with each sign. The interest rate i can be coded for entry as a percent
or decimal. When the unknown variable is identified, the calculator’s software can solve the
equation for zero, thus providing the answer.
This is the approach taken by relatively simple scientific calculators, such as the HewlettPackard (HP) scientific series, for example, HP 33s. By substituting the formulas for the factors,
the actual relation entered into the calculator is
[
]
1 ⴚ (1 ⴙ i/100)ⴚn
A ———————— ⴙ F[1 ⴙ (i/100)]ⴚn ⴙ P ⴝ 0
i/100
The HP calculator uses a slightly different symbol set than we have used thus far. The initial investment is called B rather than P, and the equal uniform amount is termed P rather than A. Once
entered, the relation can be solved for any one variable, given values for the other four.
Another example that offers freedom from the spreadsheet and tables is an engineering calculator that has the same functions built in as those on a spreadsheet to determine P, F, A, i, or n. An
example of this higher level is Texas Instrument’s TI-Nspire series. The functions are basically
the same as those on a spreadsheet. They are clustered under the heading of tvm (time value of
money) functions. For example, the tvmPV function format is
tvmPV(n,i,Pmt,FV,PpY,CpY,PmtAt)
where
n number of periods
i annual interest rate as a percent
Pmt equal uniform periodic amount A
FV future amount F
PpY payments per year (optional; default is 1)
CpY compounding periods per year (optional; default is 1)
PmtAt beginning- or end-of-period payments (optional; default is 0 end)
570
Alternate Methods for Equivalence Calculations
Appendix D
It is easy to understand why it is possible to do a lot on a calculator with relatively wellbehaved cash flow series. As the series become more complex, it is necessary to move to a spreadsheet for speed and versatility. However, the use of tables or factor formulas is not necessary.
D.2 Using the Summation of a Geometric Series
A geometric progression is a series of n terms with a common ratio or base r. If c is a constant for
each term, the series is written in the form
j⫽n
⌺r
cr a cr a1 ⋅ ⋅ ⋅ cr n c
j
j⫽a
The sum S of a geometric series adds the terms using the closed-end form
r n⫹1 ⫺ ra
S ⫽ ————
r⫺1
[D.1]
Ristroph1 and others have explained how the recognition that equivalence computations are
simple applications of geometric series can be used to determine F, P, and A using Equation
[D.1] and a simple handheld calculator with exponentiation capability.
Before explaining how to apply this approach, we define the base r as follows when a future
worth F or present worth P is sought.
r⫽1⫹i
r ⫽ (1 ⫹ i)⫺1
To find F:
To find P:
A very familiar application of geometric series is the determination of the equivalent future worth
F in year n for a single present worth amount P in year 0. This is the same as using a geometric
series of only one term. As shown in Figure D–1, if P ⫽ $100, n ⫽ 10 years, and i ⫽ 10% per
year, when r ⫽ 1 ⫹ i,
F ⫽ P(1 ⫹ i)n ⫽ P(r)n ⫽ 100(1.1)10
⫽ 100(2.5937)
⫽ $259.37
This is identical to using the tabulated value (or formula) for the F/P factor.
F ⫽ P(F/P,i,n) ⫽ 100(F/P,10%,10) ⫽ 100(2.5937)
⫽ $259.37
Figure D–2 shows a uniform annual series A⫽ $100 for 10 years. To calculate F, we can move
each A value forward to year 10. Place a subscript j on each A value to indicate the year of occurrence, and determine F for each Aj value.
F ⫽ A1(1 ⫹ i)9 ⫹ A2(1 ⫹ i)8 ⫹ ⋅ ⋅ ⋅ ⫹ A9(1 ⫹ i)1 ⫹ A10(1 ⫹ i)0
Figure D–1
F=?
Future worth of a single
amount in year 0.
i = 10%
0
1
2
3
4
5
6
7
8
9 10
1
2
3
4
5
6
7
8
9
Year
10
Times
compounded
P = $100
1
J. H. Ristroph, “Engineering Economics: Time for New Directions?” Proceedings, ASEE
Annual Conference, Austin, TX, June 2009.
Using the Summation of a Geometric Series
D.2
F= ?
i = 10%
0
1
2
3
4
5
6
7
8
9
10
Year
A ⫽ $100
9
8
7
6
5
4
3
2
1
0
Times
compounded
Figure D–2
Future worth of an A series from year 1 through year 10.
P= ?
i = 10%
0 1
2
3
4
5
6
7
8
9
10
Year
A ⫽ $100
5
6
7
8
9
Times
discounted
10
Figure D–3
Present worth of a shifted uniform series.
Note that the first value A1 is compounded 9 times, not 10. With r ⫽ 1 ⫹ i, the geometric series
and its sum S (from Equation [D.1]) can be developed. Removing the subscript on A,
j⫽9
9
8
1
0
⌺r
F = A [r ⫹ r ⫹ ⋅ ⋅ ⋅ ⫹ r ⫹ r ] ⫽ A
j
j⫽0
(1.1)10 ⫺ 1
r10 ⫺ r0 ⫽ —————
⫽ 15.9374
S ⫽ ————
r⫺1
0.1
The F value is the same whether using the geometric series sum or the tabulated F/A factor.
F ⫽ 100(S) ⫽ 100(15.9374) ⫽ $1593.74
F ⫽ 100(F/A,10%,10) ⫽ 100(15.9374) ⫽ $1593.74
As a final demonstration of the geometric series approach to equivalence computations, consider the shifted cash flow series in Figure D–3. The A series is present from years 5 through 10,
and the P value is sought. If the tables are used, the solution is
P ⫽ A(P/A,10%,6)(P/F,10%,4) = 100(4.3553)(0.6830)
⫽ $297.47
571
572
Alternate Methods for Equivalence Calculations
Appendix D
As shown in the figure, this is a geometric series with the first A value discounted 5 years, therefore, a ⫽ 5. The last A term is discounted 10 years, making n ⫽ 10. Since P is sought, the geometric series base is r ⫽ (1 ⫹ i)⫺1. Again using Equation [D.1] for the summation, we have
[ ]
j⫽10
P⫽A
⌺r
j
j⫽5
[
[
[
]
(1/1.1)11 ⫺ (1/1.1)5
r11 ⫺ r5 ⫽ 100 ————————
⫽ 100 ————
r⫺1
(1/1.1) ⫺ 1
]
]
(0.9091)11 ⫺ (0.9091)5
⫽ 100 —————————— ⫽ 100(2.9747)
0.9091 ⫺ 1
⫽ $297.47
It is possible to develop similar relations to handle arithmetic and geometric series, conversions from P to A, and vice versa. Proponents of this approach point out the use of standard
mathematical notation; the removal of a need to derive any factors; no need to remember the
placement of P, F, and A values based on factor formula development; and the easy use of a calculator to determine the equivalence relations. As with the use of programmable calculators, the
technique is excellent for well-behaved and reasonably complex series. When the series become
quite involved, or when sensitivity analysis is required to reach an economic decision, it may
be beneficial to use a spreadsheet. This often helps in performing sidebar and ancillary calculations that assist in the understanding of the problem, not just the math computations necessary to
obtain an answer. However, once again, the use of tables and factors is not necessary.
APPENDIX E
GLOSSARY OF CONCEPTS
AND TERMS
E.1 Important Concepts and Guidelines
The following elements of engineering economy are identified throughout the text in the margin
by this checkmark and a title below it. The numbers in parentheses indicate chapters where the
concept or guideline is introduced or essential to obtaining a correct solution.
Time Value of Money It is a fact that money makes money. This concept explains the change
in the amount of money over time for both owned and borrowed funds. (1)
Economic Equivalence A combination of time value of money and interest rate that makes
different sums of money at different times have equal economic value. (1)
Cash Flow The flow of money into and out of a company, project, or activity. Revenues are
cash inflows and carry a positive (+) sign; expenses are outflows and carry a negative (−) sign.
If only costs are involved, the − sign may be omitted, e.g., benefit/cost (B/C) analysis. (1, 9)
End-of-Period Convention To simplify calculations, cash flows (revenues and costs) are
assumed to occur at the end of a time period. An interest period or fiscal period is commonly
1 year. A half-year convention is often used in depreciation calculations. (1)
Cost of Capital The interest rate incurred to obtain capital investment funds. COC is usually
a weighted average that involves the cost of debt capital (loans, bonds, and mortgages) and equity capital (stocks and retained earnings). (1, 10)
Minimum Attractive Rate of Return (MARR) A reasonable rate of return established for
the evaluation of an economic alternative. Also called the hurdle rate, MARR is based on cost of
capital, market trend, risk, etc. The inequality ROR ≥ MARR > COC is correct for an economically viable project. (1, 10)
Opportunity Cost A forgone opportunity caused by the inability to pursue a project. Numerically, it is the largest rate of return of all the projects not funded due to the lack of capital
funds. Stated differently, it is the ROR of the first project rejected because of unavailability of
funds. (1, 10)
Nominal or Effective Interest Rate (r or i) A nominal interest rate does not include any
compounding; for example, 1% per month is the same as nominal 12% per year. Effective interest
rate is the actual rate over a period of time because compounding is imputed; for example, 1% per
month, compounded monthly, is an effective 12.683% per year. Inflation or deflation is not considered. (4)
Placement of Present Worth (P; PW) In applying the (P/A,i%,n) factor, P or PW is always
located one interest period (year) prior to the first A amount. The A or AW is a series of equal,
end-of-period cash flows for n consecutive periods, expressed as money per time (say,
$/year; /year). (2, 3)
Placement of Future Worth (F; FW) In applying the (F/A,i%,n) factor, F or FW is always
located at the end of the last interest period (year) of the A series. (2, 3)
Title
574
Appendix E
Glossary of Concepts and Terms
Placement of Gradient Present Worth (PG ; Pg) The (P/G,i%,n) factor for an arithmetic
gradient finds the PG of only the gradient series 2 years prior to the first appearance of the constant gradient G. The base amount A is treated separately from the gradient series.
The (P/A,g,i,n) factor for a geometric gradient determines Pg for the gradient and initial amount
A1 two years prior to the appearance of the first gradient amount. The initial amount A1 is included in the value of Pg. (2, 3)
Equal-Service Requirement Identical capacity of all alternatives operating over the
same amount of time is mandated by the equal-service requirement. Estimated costs and revenues for equal service must be evaluated. PW analysis requires evaluation over the same
number of years (periods) using the LCM (least common multiple) of lives; AW analysis is
performed over one life cycle. Further, equal service assumes that all costs and revenues rise
and fall in accordance with the overall rate of inflation or deflation over the total time period
of the evaluation. (5, 6, 8)
LCM or Study Period To select from mutually exclusive alternatives under the equal-service
requirement for PW computations, use the LCM of lives with repurchase(s) as necessary. For a
stated study period (planning horizon), evaluate cash flows only over this period , neglecting any
beyond this time; estimated market values at termination of the study period are the salvage
values. (5, 6, 11)
Salvage/Market Value Expected trade-in, market, or scrap value at the end of the estimated
life or the study period. In a replacement study, the defender’s estimated market value at the end
of a year is considered its “first cost” at the beginning of the next year. MACRS depreciation always reduces the book value to a salvage of zero. (6, 11)
Do Nothing The DN alternative is always an option, unless one of the defined alternatives
must be selected. DN is status quo; it generates no new costs, revenues, or savings. (5)
Revenue or Cost Alternative Revenue alternatives have costs and revenues estimated; savings are considered negative costs and carry a + sign. Incremental evaluation requires comparison with DN for revenue alternatives. Cost (or service) alternatives have only costs estimated;
revenues and savings are assumed equal between alternatives. (5)
Rate of Return An interest rate that equates a PW or AW relation to zero. Also defined as the
rate on the unpaid balance of borrowed money, or rate earned on the unrecovered balance of an
investment such that the last cash flow brings the balance exactly to zero. (7, 8)
Project Evaluation For a specified MARR, determine a measure of worth for net cash flow
series over the life or study period. Guidelines for a single project to be economically justified at
the MARR (or discount rate) follow. (5, 6, 7, 9, 17)
Present worth: If PW ≥ 0
Future worth: If FW ≥ 0
Benefit/cost: If B/C ≥ 1.0
Annual worth: If AW ≥ 0
Rate of return: If i* ≥ MARR
Profitability index: If PI ≥ 1.0
ME Alternative Selection For mutually exclusive (select only one) alternatives, compare
two alternatives at a time by determining a measure of worth for the incremental (∆) cash flow
series over the life or study period, adhering to the equal-service requirement. (5, 6, 8, 9, 10, 17)
Present worth or annual worth: Find PW or AW values at MARR; select numerically largest
(least negative or most positive).
Rate of return: Order by initial cost, perform pairwise ∆i* comparison; if ∆i* ≥ MARR,
select larger cost alternative; continue until one remains.
Benefit/cost: Order by total equivalent cost, perform pairwise ∆B/C comparison; if
∆B/C ≥ 1.0, select larger cost alternative; continue until one remains.
E.1
Important Concepts and Guidelines
Cost-effectiveness ratio: For service sector alternatives; order by effectiveness measure;
perform pairwise ∆C/E comparison using dominance; select from nondominated alternatives
without exceeding budget.
Independent Project Selection No comparison between projects; only against DN. Calculate a measure of worth and select using the guidelines below. (5, 6, 8, 9, 12)
Present worth or annual worth: Find PW or AW at MARR; select all projects with PW or
AW ≥ 0.
Rate of return: No incremental comparison; select all projects with overall i * ≥ MARR.
Benefit/cost: No incremental comparison; select all projects with overall B/C ≥ 1.0.
Cost-effectiveness ratio: For service sector projects; no incremental comparison; order by
CER and select projects to not exceed budget.
When a capital budget limit is defined, independent projects are selected using the capital budgeting process based on PW values. The Solver spreadsheet tool is useful here.
Capital Recovery CR is the equivalent annual amount an asset or system must earn to recover
the initial investment plus a stated rate of return. Numerically, it is the AW value of the initial
investment at a stated rate of return. The salvage value is considered in CR calculations. (6)
Economic Service Life The ESL is the number of years n at which the total AW of costs,
including salvage and AOC, is at its minimum, considering all the years the asset may provide
service. (11)
Sunk Cost Capital (money) that is lost and cannot be recovered. Sunk costs are not included
when making decisions about the future. They should be handled using tax laws and write-off
allowances, not the economic study. (11)
Inflation Expressed as a percentage per time (% per year), it is an increase in the amount
of money required to purchase the same amount of goods or services over time. Inflation
occurs when the value of a currency decreases. Economic evaluations are performed using
either a market (inflation-adjusted) interest rate or an inflation-free rate (constant-value
terms). (1, 14)
Breakeven For a single project, the value of a parameter that makes two elements equal,
e.g., sales necessary to equate revenues and costs. For two alternatives, breakeven is the value
of a common variable at which the two are equally acceptable. Breakeven analysis is fundamental to make-buy decisions, replacement studies, payback analysis, sensitivity analysis,
breakeven ROR analysis, and many others. The Goal Seek spreadsheet tool is useful in breakeven analysis. (8, 13)
Payback Period Amount of time n before recovery of the initial capital investment is expected.
Payback with i > 0 or simple payback at i = 0 is useful for preliminary or screening analysis to
determine if a full PW, AW, or ROR analysis is needed. (13)
Direct / Indirect Costs Direct costs are primarily human labor, machines, and materials associated with a product, process, system, or service. Indirect costs, which include support functions, utilities, management, legal, taxes, and the like, are more difficult to associate with a
specific product or process. (15)
Value Added Activities have added worth to a product or service from the perspective of a
consumer, owner, or investor who is willing to pay more for an enhanced value. (17)
Sensitivity Analysis Determination of how a measure of worth is affected by changes in
estimated values of a parameter over a stated range. Parameters may be any cost factor, revenue,
life, salvage value, inflation rate, etc. (18)
575
576
Glossary of Concepts and Terms
Appendix E
Risk Variation from an expected, desirable, or predicted value that may be detrimental to the
product, process, or system. Risk represents an absence of or deviation from certainty. Probability estimates of variation (values) help evaluate risk and uncertainty using statistics and simulation. (10, 18, 19, 20)
E.2 Symbols and Terms
This section identifies and defines the common terms and their symbols used throughout the text.
The numbers in parentheses indicate sections where the term is introduced and used in various
applications.
Term
Symbol
Annual amount or
worth
A or AW
Equivalent uniform annual worth of all cash inflows and
outflows over estimated life (1.5, 6.l).
Description
Annual operating cost
AOC
Estimated annual costs to maintain and support an
alternative (1.3).
Benefit/cost ratio
B/C
Ratio of a project’s benefits to costs expressed in PW, AW,
or FW terms (9.2).
Book value
BV
Remaining capital investment in an asset after depreciation
is accounted for (16.1).
Breakeven point
QBE
Quantity at which revenues and costs are equal, or two
alternatives are equivalent (13.1).
Capital budget
b
Amount of money available for capital investment projects
(12.1).
Capital recovery
CR or A
Equivalent annual cost of owning an asset plus the required
return on the initial investment (6.2).
Capitalized cost
CC or P
Present worth of an alternative that will last forever (or a
long time) (5.5).
Cash flow
CF
Actual cash amounts that are receipts (inflow) and disbursements (outflow) (1.6).
Cash flow before or
after taxes
CFBT or
CFAT
Cash flow amount before relevant taxes or after taxes are
applied (17.2).
Compounding
frequency
m
Number of times interest is compounded per period (year)
(4.1).
Cost-effectiveness ratio
CER
Ratio of equivalent cost to effectiveness measure to evaluate service sector projects (9.5).
Cost estimating relationships
C2 or CT
Relations that use design variables and changing costs over
time to estimate current and future costs (15.3–4).
Cost of capital
WACC
Interest rate paid for the use of capital funds; includes both
debt and equity funds. For debt and equity considered, it is
weighted average cost of capital (1.9, 10.2).
Debt-equity mix
D-E
Percentages of debt and equity investment capital used by a
corporation (10.2).
Depreciation
D
Reduction in the value of assets using specific models and
rules; there are book and tax depreciation methods (16.1).
Depreciation rate
dt
Annual rate for reducing the value of assets using different
depreciation methods (16.1).
Economic service life
ESL or n
Number of years at which the AW of costs is a minimum
(11.2).
Effectiveness measure
E
A nonmonetary measure used in the cost-effectiveness ratio
for service sector projects (9.5).
Symbols and Terms
E.2
Term
Expected value
(average)
Symbol
–
X, , or E(X)
Description
Long-run expected average if a random variable is sampled
many times (18.3, 19.4).
Expenses, operating
OE
All corporate costs incurred in transacting business (17.1).
First cost
P
Total initial cost—purchase, construction, setup, etc.
(1.3, 16.1).
Future amount or worth
F or FW
Amount at some future date considering time value of
money (1.5, 5.4).
Gradient, arithmetic
G
Uniform change (+ or –) in cash flow each time period
(2.5).
Gradient, geometric
g
Constant rate of change (+ or –) each time period (2.6).
Gross income
GI
Income from all sources for corporations or individuals
(17.1).
Inflation rate
f
Rate that reflects changes in the value of a currency over
time (14.1).
Interest
I
Amount earned or paid over time based on an initial
amount and interest rate (1.4).
Interest rate
i or r
Interest expressed as a percentage of the original amount per
time period; nominal (r) and effective (i) rates (1.4, 4.1).
Interest rate, inflationadjusted
if
Interest rate adjusted to take inflation into account (14.1).
Life (estimated)
n
Number of years or periods over which an alternative or
asset will be used; the evaluation time (1.5).
Life-cycle cost
LCC
Evaluation of costs for a system over all stages: feasibility
to design to phaseout (6.5).
Measure of worth
Varies
Value, such as PW, AW, i*, used to judge economic
viability (1.1).
Minimum attractive
rate of return
MARR
Minimum value of the rate of return for an alternative to be
financially viable (1.9, 10.1).
Modified ROR
i' or MIRR
Unique ROR when a reinvestment rate ii and external borrowing rate ib are applied to multiple-rate cash flows (7.5).
Net cash flow
NCF
Resulting, actual amount of cash that flows in or out during
a time period (1.6).
Net operating income
NOI
Difference between gross income and operating expenses
(17.1).
Net operating profit
after taxes
NOPAT or
NPAT
Amount remaining after taxes are removed from taxable
income (17.1).
Net present value
NPV
Another name for the present worth, PW.
Payback period
np
Number of years to recover the initial investment and a
stated rate of return (13.3).
Present amount
or worth
P or PW
Amount of money at the current time or a time denoted as
present (1.5, 5.2).
Probability distribution
P(X)
Distribution of probability over different values of a
variable (19.2).
Profitability index
PI
Ratio of PW of net cash flows to initial investment used for
revenue projects; rewritten modified B/C ratio (9.2, 12.5).
Random variable
X
Parameter or characteristic that can take on any one of
several values; discrete and continuous (19.2).
Rate of return
i* or ROR
Compound interest rate on unpaid or unrecovered balances
such that the final amount results in a zero balance (7.1).
577
578
Glossary of Concepts and Terms
Appendix E
Term
Symbol
Description
Recovery period
n
Number of years to completely depreciate an asset (16.1).
Return on invested
capital
i'' or ROIC
Unique ROR when a reinvestment rate ii is applied to
multiple-rate cash flows (7.5).
Salvage/market value
S or MV
Expected trade-in or market value when an asset is traded
or disposed of (6.2, 11.1, 16.1).
Standard deviation
s or σ
Measure of dispersion or spread about the expected value
or average (19.4).
Study period
n
Specified number of years over which an evaluation takes
place (5.3, 11.5).
Taxable income
TI
Amount upon which income taxes are based (17.1).
Tax rate
T
Decimal rate, usually graduated, used to calculate corporate
or individual taxes (17.1).
Tax rate, effective
Te
Single-figure tax rate incorporating several rates and bases
(17.1).
Time
t
Indicator for a time period (1.7).
Unadjusted basis
B
Depreciable amount of first cost, delivery, and installation
costs of an asset (18.1).
Value added
EVA
Economic value added reflects net profit after taxes (NPAT)
after removing cost of invested capital during the year
(17.7).
Value-added tax
VAT
An indirect consumption tax collected at each stage of
production/distribution process; different from a sales tax
paid by end user at purchase time (17.9).
REFERENCE MATERIALS
Textbooks on Related Topics
Blank, L. T., and A. Tarquin: Basics of Engineering Economy, McGraw-Hill, New York, 2008.
Bowman, M. S.: Applied Economic Analysis for Technologists, Engineers, and Managers, 2d ed., Pearson
Prentice Hall, Upper Saddle River, NJ, 2003.
Bussey, L. E., and T. G. Eschenbach: The Economic Analysis of Industrial Projects, 2d ed., Pearson
Prentice Hall, Upper Saddle River, NJ, 1992.
Canada, J. R., W. G. Sullivan, D. J. Kulonda, and J. A. White: Capital Investment Analysis for Engineering and Management, 3d ed., Pearson Prentice Hall, Upper Saddle River, NJ, 2005.
Collier, C. A., and C. R. Glagola: Engineering Economic and Cost Analysis, 3d ed., Pearson Prentice Hall,
Upper Saddle River, NJ, 1999.
Cushman, R. F., and M. Loulakis: Design-Build Contracting Handbook, 2d ed., Aspen Publishers, New
York, 2001.
Eschenbach, T. G.: Engineering Economy: Applying Theory to Practice, 3d ed., Oxford University Press,
New York, 2010.
Fabrycky, W. J., G. J. Thuesen, and D. Verma: Economic Decision Analysis, 3d ed., Pearson Prentice Hall,
Upper Saddle River, NJ, 1998.
Fraser, N. M., E. M. Jewkes, I. Bernhardt, and M. Tajima: Engineering Economics in Canada, Pearson
Prentice Hall, Upper Saddle River, NJ, 2006.
Hartman, J. C.: Engineering Economy and the Decision Making Process, Pearson Prentice Hall, Upper
Saddle River, NJ, 2007.
Levy, S. M.: Build, Operate, Transfer: Paving the Way for Tomorrow’s Infrastructure, John Wiley & Sons,
New York, 1996.
Newnan, D. G., J. P. Lavelle, and T. G. Eschenbach: Engineering Economic Analysis, 10th ed., Oxford
University Press, New York, 2009.
Ostwald, P. F.: Construction Cost Analysis and Estimating, Pearson Prentice Hall, Upper Saddle River,
NJ, 2001.
Ostwald, P. F., and T. S. McLaren: Cost Analysis and Estimating for Engineering and Management,
Pearson Prentice Hall, Upper Saddle River, NJ, 2004.
Park, C. S.: Contemporary Engineering Economics, 5th ed., Pearson Prentice Hall, Upper Saddle River,
NJ, 2011.
Park, C. S.: Fundamentals of Engineering Economics, 2d ed., Pearson Prentice Hall, Upper Saddle River,
NJ, 2008.
Peurifoy, R. L., and G. D. Oberlender: Estimating Construction Costs, 5th ed., McGraw-Hill, New York,
2002.
Riggs, J. L., D. D. Bedworth, and S. U. Randhawa: Engineering Economics, 4th ed., McGraw-Hill,
New York, 1996.
Stewart, R. D., R. M. Wyskida, and J. D. Johannes: Cost Estimator’s Reference Manual, 2d ed., John
Wiley & Sons, New York, 1995.
Sullivan, W. G., E. M. Wicks, and C. P. Koelling: Engineering Economy, 15th ed., Pearson Prentice Hall,
Upper Saddle River, NJ, 2011.
Thuesen, G. J., and W. J. Fabrycky: Engineering Economy, 9th ed., Pearson Prentice Hall, Upper Saddle
River, NJ, 2001.
White, J. A., K. E. Case, D. B. Pratt, and M. H. Agee: Principles of Engineering Economic Analysis,
5th ed., John Wiley & Sons, New York, 2010.
Using Excel 2007
Gottfried, B. S.: Spreadsheet Tools for Engineers Using Excel® 2007, McGraw-Hill, New York, 2010.
Materials on Engineering Ethics
Harris, C. E., M. S. Pritchard, and M. J. Rabins: Engineering Ethics: Concepts and Cases, 4th ed.,
Wadsworth Cengage Learning, Belmont, CA, 2009.
Martin, M. W., and R. Schinzinger: Introduction to Engineering Ethics, 2d ed., McGraw-Hill,
New York, 2010.
580
Reference Materials
Websites
Construction cost estimation index: www.construction.com
The Economist: www.economist.com
For this textbook: www.mhhe.com/blank
Plant cost estimation index: www.che.com/pci
Revenue Canada: www.cra.gc.ca
U.S. Internal Revenue Service: www.irs.gov
Wall Street Journal: www.online.wsj.com
U.S. Government Publications (available at
www.irs.gov)
Corporations, Publication 544, Internal Revenue Service, GPO, Washington, DC, annually.
Sales and Other Dispositions of Assets, Publication 542, Internal Revenue Service, GPO, Washington,
DC, annually.
Your Federal Income Tax, Publication 17, Internal Revenue Service, GPO, Washington, DC, annually.
Selected Journals and Other Publications
The Engineering Economist, joint publication of ASEE and IIE, published by Taylor and Francis,
Philadelphia, quarterly.
Harvard Business Review, Harvard University Press, Boston, bimonthly.
Journal of Finance, American Finance Association, published by John Wiley & Sons, New York,
bimonthly.
581
Compound Interest Factor Tables
0.25%
TABLE 1
0.25%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
84
90
96
100
108
120
132
144
240
360
480
1.0025
1.0050
1.0075
1.0100
1.0126
1.0151
1.0176
1.0202
1.0227
1.0253
1.0278
1.0304
1.0330
1.0356
1.0382
1.0408
1.0434
1.0460
1.0486
1.0512
1.0538
1.0565
1.0591
1.0618
1.0644
1.0671
1.0697
1.0724
1.0751
1.0778
1.0941
1.1050
1.1273
1.1330
1.1386
1.1472
1.1616
1.1969
1.2059
1.2334
1.2520
1.2709
1.2836
1.3095
1.3494
1.3904
1.4327
1.8208
2.4568
3.3151
0.9975
0.9950
0.9925
0.9901
0.9876
0.9851
0.9827
0.9802
0.9778
0.9753
0.9729
0.9705
0.9681
0.9656
0.9632
0.9608
0.9584
0.9561
0.9537
0.9513
0.9489
0.9466
0.9442
0.9418
0.9395
0.9371
0.9348
0.9325
0.9301
0.9278
0.9140
0.9050
0.8871
0.8826
0.8782
0.8717
0.8609
0.8355
0.8292
0.8108
0.7987
0.7869
0.7790
0.7636
0.7411
0.7192
0.6980
0.5492
0.4070
0.3016
1.00000
0.49938
0.33250
0.24906
0.19900
0.16563
0.14179
0.12391
0.11000
0.09888
0.08978
0.08219
0.07578
0.07028
0.06551
0.06134
0.05766
0.05438
0.05146
0.04882
0.04644
0.04427
0.04229
0.04048
0.03881
0.03727
0.03585
0.03452
0.03329
0.03214
0.02658
0.02380
0.01963
0.01880
0.01803
0.01698
0.01547
0.01269
0.01214
0.01071
0.00992
0.00923
0.00881
0.00808
0.00716
0.00640
0.00578
0.00305
0.00172
0.00108
1.0000
2.0025
3.0075
4.0150
5.0251
6.0376
7.0527
8.0704
9.0905
10.1133
11.1385
12.1664
13.1968
14.2298
15.2654
16.3035
17.3443
18.3876
19.4336
20.4822
21.5334
22.5872
23.6437
24.7028
25.7646
26.8290
27.8961
28.9658
30.0382
31.1133
37.6206
42.0132
50.9312
53.1887
55.4575
58.8819
64.6467
78.7794
82.3792
93.3419
100.7885
108.3474
113.4500
123.8093
139.7414
156.1582
173.0743
328.3020
582.7369
926.0595
1.00250
0.50188
0.33500
0.25156
0.20150
0.16813
0.14429
0.12641
0.11250
0.10138
0.09228
0.08469
0.07828
0.07278
0.06801
0.06384
0.06016
0.05688
0.05396
0.05132
0.04894
0.04677
0.04479
0.04298
0.04131
0.03977
0.03835
0.03702
0.03579
0.03464
0.02908
0.02630
0.02213
0.02130
0.02053
0.01948
0.01797
0.01519
0.01464
0.01321
0.01242
0.01173
0.01131
0.01058
0.00966
0.00890
0.00828
0.00555
0.00422
0.00358
0.9975
1.9925
2.9851
3.9751
4.9627
5.9478
6.9305
7.9107
8.8885
9.8639
10.8368
11.8073
12.7753
13.7410
14.7042
15.6650
16.6235
17.5795
18.5332
19.4845
20.4334
21.3800
22.3241
23.2660
24.2055
25.1426
26.0774
27.0099
27.9400
28.8679
34.3865
38.0199
45.1787
46.9462
48.7048
51.3264
55.6524
65.8169
68.3108
75.6813
80.5038
85.2546
88.3825
94.5453
103.5618
112.3121
120.8041
180.3109
237.1894
279.3418
0.9950
2.9801
5.9503
9.9007
14.8263
20.7223
27.5839
35.4061
44.1842
53.9133
64.5886
76.2053
88.7587
102.2441
116.6567
131.9917
148.2446
165.4106
183.4851
202.4634
222.3410
243.1131
264.7753
287.3230
310.7516
335.0566
360.2334
386.2776
413.1847
592.4988
728.7399
1040.06
1125.78
1214.59
1353.53
1600.08
2265.56
2447.61
3029.76
3446.87
3886.28
4191.24
4829.01
5852.11
6950.01
8117.41
19399
36264
53821
0.4994
0.9983
1.4969
1.9950
2.4927
2.9900
3.4869
3.9834
4.4794
4.9750
5.4702
5.9650
6.4594
6.9534
7.4469
7.9401
8.4328
8.9251
9.4170
9.9085
10.3995
10.8901
11.3804
11.8702
12.3596
12.8485
13.3371
13.8252
14.3130
17.2306
19.1673
23.0209
23.9802
24.9377
26.3710
28.7514
34.4221
35.8305
40.0331
42.8162
45.5844
47.4216
51.0762
56.5084
61.8813
67.1949
107.5863
152.8902
192.6699
582
Compound Interest Factor Tables
0.5%
TABLE 2
0.5%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
84
90
96
100
108
120
132
144
240
360
480
1.0050
1.0100
1.0151
1.0202
1.0253
1.0304
1.0355
1.0407
1.0459
1.0511
1.0564
1.0617
1.0670
1.0723
1.0777
1.0831
1.0885
1.0939
1.0994
1.1049
1.1104
1.1160
1.1216
1.1272
1.1328
1.1385
1.1442
1.1499
1.1556
1.1614
1.1967
1.2208
1.2705
1.2832
1.2961
1.3156
1.3489
1.4320
1.4536
1.5204
1.5666
1.6141
1.6467
1.7137
1.8194
1.9316
2.0508
3.3102
6.0226
10.9575
0.9950
0.9901
0.9851
0.9802
0.9754
0.9705
0.9657
0.9609
0.9561
0.9513
0.9466
0.9419
0.9372
0.9326
0.9279
0.9233
0.9187
0.9141
0.9096
0.9051
0.9006
0.8961
0.8916
0.8872
0.8828
0.8784
0.8740
0.8697
0.8653
0.8610
0.8356
0.8191
0.7871
0.7793
0.7716
0.7601
0.7414
0.6983
0.6879
0.6577
0.6383
0.6195
0.6073
0.5835
0.5496
0.5177
0.4876
0.3021
0.1660
0.0913
1.00000
0.49875
0.33167
0.24813
0.19801
0.16460
0.14073
0.12283
0.10891
0.09777
0.08866
0.08107
0.07464
0.06914
0.06436
0.06019
0.05651
0.05323
0.05030
0.04767
0.04528
0.04311
0.04113
0.03932
0.03765
0.03611
0.03469
0.03336
0.03213
0.03098
0.02542
0.02265
0.01849
0.01765
0.01689
0.01584
0.01433
0.01157
0.01102
0.00961
0.00883
0.00814
0.00773
0.00701
0.00610
0.00537
0.00476
0.00216
0.00100
0.00050
1.0000
2.0050
3.0150
4.0301
5.0503
6.0755
7.1059
8.1414
9.1821
10.2280
11.2792
12.3356
13.3972
14.4642
15.5365
16.6142
17.6973
18.7858
19.8797
20.9791
22.0840
23.1944
24.3104
25.4320
26.5591
27.6919
28.8304
29.9745
31.1244
32.2800
39.3361
44.1588
54.0978
56.6452
59.2180
63.1258
69.7700
86.4089
90.7265
104.0739
113.3109
122.8285
129.3337
142.7399
163.8793
186.3226
210.1502
462.0409
1004.52
1991.49
1.00500
0.50375
0.33667
0.25313
0.20301
0.16960
0.14573
0.12783
0.11391
0.10277
0.09366
0.08607
0.07964
0.07414
0.06936
0.06519
0.06151
0.05823
0.05530
0.05267
0.05028
0.04811
0.04613
0.04432
0.04265
0.04111
0.03969
0.03836
0.03713
0.03598
0.03042
0.02765
0.02349
0.02265
0.02189
0.02084
0.01933
0.01657
0.01602
0.01461
0.01383
0.01314
0.01273
0.01201
0.01110
0.01037
0.00976
0.00716
0.00600
0.00550
0.9950
1.9851
2.9702
3.9505
4.9259
5.8964
6.8621
7.8230
8.7791
9.7304
10.6770
11.6189
12.5562
13.4887
14.4166
15.3399
16.2586
17.1728
18.0824
18.9874
19.8880
20.7841
21.6757
22.5629
23.4456
24.3240
25.1980
26.0677
26.9330
27.7941
32.8710
36.1722
42.5803
44.1428
45.6897
47.9814
51.7256
60.3395
62.4136
68.4530
72.3313
76.0952
78.5426
83.2934
90.0735
96.4596
102.4747
139.5808
166.7916
181.7476
0.9901
2.9604
5.9011
9.8026
14.6552
20.4493
27.1755
34.8244
43.3865
52.8526
63.2136
74.4602
86.5835
99.5743
113.4238
128.1231
143.6634
160.0360
177.2322
195.2434
214.0611
233.6768
254.0820
275.2686
297.2281
319.9523
343.4332
367.6625
392.6324
557.5598
681.3347
959.9188
1035.70
1113.82
1235.27
1448.65
2012.35
2163.75
2640.66
2976.08
3324.18
3562.79
4054.37
4823.51
5624.59
6451.31
13416
21403
27588
0.4988
0.9967
1.4938
1.9900
2.4855
2.9801
3.4738
3.9668
4.4589
4.9501
5.4406
5.9302
6.4190
6.9069
7.3940
7.8803
8.3658
8.8504
9.3342
9.8172
10.2993
10.7806
11.2611
11.7407
12.2195
12.6975
13.1747
13.6510
14.1265
16.9621
18.8359
22.5437
23.4624
24.3778
25.7447
28.0064
33.3504
34.6679
38.5763
41.1451
43.6845
45.3613
48.6758
53.5508
58.3103
62.9551
96.1131
128.3236
151.7949
583
Compound Interest Factor Tables
0.75%
TABLE 3
0.75%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
84
90
96
100
108
120
132
144
240
360
480
1.0075
1.0151
1.0227
1.0303
1.0381
1.0459
1.0537
1.0616
1.0696
1.0776
1.0857
1.0938
1.1020
1.1103
1.1186
1.1270
1.1354
1.1440
1.1525
1.1612
1.1699
1.1787
1.1875
1.1964
1.2054
1.2144
1.2235
1.2327
1.2420
1.2513
1.3086
1.3483
1.4314
1.4530
1.4748
1.5083
1.5657
1.7126
1.7514
1.8732
1.9591
2.0489
2.1111
2.2411
2.4514
2.6813
2.9328
6.0092
14.7306
36.1099
0.9926
0.9852
0.9778
0.9706
0.9633
0.9562
0.9490
0.9420
0.9350
0.9280
0.9211
0.9142
0.9074
0.9007
0.8940
0.8873
0.8807
0.8742
0.8676
0.8612
0.8548
0.8484
0.8421
0.8358
0.8296
0.8234
0.8173
0.8112
0.8052
0.7992
0.7641
0.7416
0.6986
0.6883
0.6780
0.6630
0.6387
0.5839
0.5710
0.5338
0.5104
0.4881
0.4737
0.4462
0.4079
0.3730
0.3410
0.1664
0.0679
0.0277
1.00000
0.49813
0.33085
0.24721
0.19702
0.16357
0.13967
0.12176
0.10782
0.09667
0.08755
0.07995
0.07352
0.06801
0.06324
0.05906
0.05537
0.05210
0.04917
0.04653
0.04415
0.04198
0.04000
0.03818
0.03652
0.03498
0.03355
0.03223
0.03100
0.02985
0.02430
0.02153
0.01739
0.01656
0.01580
0.01476
0.01326
0.01053
0.00998
0.00859
0.00782
0.00715
0.00675
0.00604
0.00517
0.00446
0.00388
0.00150
0.00055
0.00021
1.0000
2.0075
3.0226
4.0452
5.0756
6.1136
7.1595
8.2132
9.2748
10.3443
11.4219
12.5076
13.6014
14.7034
15.8137
16.9323
18.0593
19.1947
20.3387
21.4912
22.6524
23.8223
25.0010
26.1885
27.3849
28.5903
29.8047
31.0282
32.2609
33.5029
41.1527
46.4465
57.5207
60.3943
63.3111
67.7688
75.4241
95.0070
100.1833
116.4269
127.8790
139.8562
148.1445
165.4832
193.5143
224.1748
257.7116
667.8869
1830.74
4681.32
1.00750
0.50563
0.33835
0.25471
0.20452
0.17107
0.14717
0.12926
0.11532
0.10417
0.09505
0.08745
0.08102
0.07551
0.07074
0.06656
0.06287
0.05960
0.05667
0.05403
0.05165
0.04948
0.04750
0.04568
0.04402
0.04248
0.04105
0.03973
0.03850
0.03735
0.03180
0.02903
0.02489
0.02406
0.02330
0.02226
0.02076
0.01803
0.01748
0.01609
0.01532
0.01465
0.01425
0.01354
0.01267
0.01196
0.01138
0.00900
0.00805
0.00771
0.9926
1.9777
2.9556
3.9261
4.8894
5.8456
6.7946
7.7366
8.6716
9.5996
10.5207
11.4349
12.3423
13.2430
14.1370
15.0243
15.9050
16.7792
17.6468
18.5080
19.3628
20.2112
21.0533
21.8891
22.7188
23.5422
24.3595
25.1707
25.9759
26.7751
31.4468
34.4469
40.1848
41.5664
42.9276
44.9316
48.1734
55.4768
57.2027
62.1540
65.2746
68.2584
70.1746
73.8394
78.9417
83.6064
87.8711
111.1450
124.2819
129.6409
0.9852
2.9408
5.8525
9.7058
14.4866
20.1808
26.7747
34.2544
42.6064
51.8174
61.8740
72.7632
84.4720
96.9876
110.2973
124.3887
139.2494
154.8671
171.2297
188.3253
206.1420
224.6682
243.8923
263.8029
284.3888
305.6387
327.5416
350.0867
373.2631
524.9924
637.4693
886.8404
953.8486
1022.59
1128.79
1313.52
1791.25
1917.22
2308.13
2578.00
2853.94
3040.75
3419.90
3998.56
4583.57
5169.58
9494.12
13312
15513
0.4981
0.9950
1.4907
1.9851
2.4782
2.9701
3.4608
3.9502
4.4384
4.9253
5.4110
5.8954
6.3786
6.8606
7.3413
7.8207
8.2989
8.7759
9.2516
9.7261
10.1994
10.6714
11.1422
11.6117
12.0800
12.5470
13.0128
13.4774
13.9407
16.6946
18.5058
22.0691
22.9476
23.8211
25.1223
27.2665
32.2882
33.5163
37.1357
39.4946
41.8107
43.3311
46.3154
50.6521
54.8232
58.8314
85.4210
107.1145
119.6620
584
Compound Interest Factor Tables
1%
TABLE 4
1%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
84
90
96
100
108
120
132
144
240
360
480
1.0100
1.0201
1.0303
1.0406
1.0510
1.0615
1.0721
1.0829
1.0937
1.1046
1.1157
1.1268
1.1381
1.1495
1.1610
1.1726
1.1843
1.1961
1.2081
1.2202
1.2324
1.2447
1.2572
1.2697
1.2824
1.2953
1.3082
1.3213
1.3345
1.3478
1.4308
1.4889
1.6122
1.6446
1.6777
1.7285
1.8167
2.0471
2.1091
2.3067
2.4486
2.5993
2.7048
2.9289
3.3004
3.7190
4.1906
10.8926
35.9496
118.6477
0.9901
0.9803
0.9706
0.9610
0.9515
0.9420
0.9327
0.9235
0.9143
0.9053
0.8963
0.8874
0.8787
0.8700
0.8613
0.8528
0.8444
0.8360
0.8277
0.8195
0.8114
0.8034
0.7954
0.7876
0.7798
0.7720
0.7644
0.7568
0.7493
0.7419
0.6989
0.6717
0.6203
0.6080
0.5961
0.5785
0.5504
0.4885
0.4741
0.4335
0.4084
0.3847
0.3697
0.3414
0.3030
0.2689
0.2386
0.0918
0.0278
0.0084
1.00000
0.49751
0.33002
0.24628
0.19604
0.16255
0.13863
0.12069
0.10674
0.09558
0.08645
0.07885
0.07241
0.06690
0.06212
0.05794
0.05426
0.05098
0.04805
0.04542
0.04303
0.04086
0.03889
0.03707
0.03541
0.03387
0.03245
0.03112
0.02990
0.02875
0.02321
0.02046
0.01633
0.01551
0.01476
0.01373
0.01224
0.00955
0.00902
0.00765
0.00690
0.00625
0.00587
0.00518
0.00435
0.00368
0.00313
0.00101
0.00029
0.00008
1.0000
2.0100
3.0301
4.0604
5.1010
6.1520
7.2135
8.2857
9.3685
10.4622
11.5668
12.6825
13.8093
14.9474
16.0969
17.2579
18.4304
19.6147
20.8109
22.0190
23.2392
24.4716
25.7163
26.9735
28.2432
29.5256
30.8209
32.1291
33.4504
34.7849
43.0769
48.8864
61.2226
64.4632
67.7689
72.8525
81.6697
104.7099
110.9128
130.6723
144.8633
159.9273
170.4814
192.8926
230.0387
271.8959
319.0616
989.2554
3494.96
11765
1.01000
0.50751
0.34002
0.25628
0.20604
0.17255
0.14863
0.13069
0.11674
0.10558
0.09645
0.08885
0.08241
0.07690
0.07212
0.06794
0.06426
0.06098
0.05805
0.05542
0.05303
0.05086
0.04889
0.04707
0.04541
0.04387
0.04245
0.04112
0.03990
0.03875
0.03321
0.03046
0.02633
0.02551
0.02476
0.02373
0.02224
0.01955
0.01902
0.01765
0.01690
0.01625
0.01587
0.01518
0.01435
0.01368
0.01313
0.01101
0.01029
0.01008
0.9901
1.9704
2.9410
3.9020
4.8534
5.7955
6.7282
7.6517
8.5660
9.4713
10.3676
11.2551
12.1337
13.0037
13.8651
14.7179
15.5623
16.3983
17.2260
18.0456
18.8570
19.6604
20.4558
21.2434
22.0232
22.7952
23.5596
24.3164
25.0658
25.8077
30.1075
32.8347
37.9740
39.1961
40.3942
42.1472
44.9550
51.1504
52.5871
56.6485
59.1609
61.5277
63.0289
65.8578
69.7005
73.1108
76.1372
90.8194
97.2183
99.1572
0.9803
2.9215
5.8044
9.6103
14.3205
19.9168
26.3812
33.6959
41.8435
50.8067
60.5687
71.1126
82.4221
94.4810
107.2734
120.7834
134.9957
149.8950
165.4664
181.6950
198.5663
216.0660
234.1800
252.8945
272.1957
292.0702
312.5047
333.4863
355.0021
494.6207
596.8561
820.1460
879.4176
939.9175
1032.81
1192.81
1597.87
1702.73
2023.32
2240.57
2459.43
2605.78
2898.42
3334.11
3761.69
4177.47
6878.60
8720.43
9511.16
0.4975
0.9934
1.4876
1.9801
2.4710
2.9602
3.4478
3.9337
4.4179
4.9005
5.3815
5.8607
6.3384
6.8143
7.2886
7.7613
8.2323
8.7017
9.1694
9.6354
10.0998
10.5626
11.0237
11.4831
11.9409
12.3971
12.8516
13.3044
13.7557
16.4285
18.1776
21.5976
22.4363
23.2686
24.5049
26.5333
31.2386
32.3793
35.7170
37.8724
39.9727
41.3426
44.0103
47.8349
51.4520
54.8676
75.7393
89.6995
95.9200
585
Compound Interest Factor Tables
1.25%
TABLE 5
1.25%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
84
90
96
100
108
120
132
144
240
360
480
1.0125
1.0252
1.0380
1.0509
1.0641
1.0774
1.0909
1.1045
1.1183
1.1323
1.1464
1.1608
1.1753
1.1900
1.2048
1.2199
1.2351
1.2506
1.2662
1.2820
1.2981
1.3143
1.3307
1.3474
1.3642
1.3812
1.3985
1.4160
1.4337
1.4516
1.5639
1.6436
1.8154
1.8610
1.9078
1.9803
2.1072
2.4459
2.5388
2.8391
3.0588
3.2955
3.4634
3.8253
4.4402
5.1540
5.9825
19.7155
87.5410
388.7007
0.9877
0.9755
0.9634
0.9515
0.9398
0.9282
0.9167
0.9054
0.8942
0.8832
0.8723
0.8615
0.8509
0.8404
0.8300
0.8197
0.8096
0.7996
0.7898
0.7800
0.7704
0.7609
0.7515
0.7422
0.7330
0.7240
0.7150
0.7062
0.6975
0.6889
0.6394
0.6084
0.5509
0.5373
0.5242
0.5050
0.4746
0.4088
0.3939
0.3522
0.3269
0.3034
0.2887
0.2614
0.2252
0.1940
0.1672
0.0507
0.0114
0.0026
1.00000
0.49680
0.32920
0.24536
0.19506
0.16153
0.13759
0.11963
0.10567
0.09450
0.08537
0.07776
0.07132
0.06581
0.06103
0.05685
0.05316
0.04988
0.04696
0.04432
0.04194
0.03977
0.03780
0.03599
0.03432
0.03279
0.03137
0.03005
0.02882
0.02768
0.02217
0.01942
0.01533
0.01452
0.01377
0.01275
0.01129
0.00865
0.00812
0.00680
0.00607
0.00545
0.00507
0.00442
0.00363
0.00301
0.00251
0.00067
0.00014
0.00003
1.0000
2.0125
3.0377
4.0756
5.1266
6.1907
7.2680
8.3589
9.4634
10.5817
11.7139
12.8604
14.0211
15.1964
16.3863
17.5912
18.8111
20.0462
21.2968
22.5630
23.8450
25.1431
26.4574
27.7881
29.1354
30.4996
31.8809
33.2794
34.6954
36.1291
45.1155
51.4896
65.2284
68.8818
72.6271
78.4225
88.5745
115.6736
123.1035
147.1290
164.7050
183.6411
197.0723
226.0226
275.2171
332.3198
398.6021
1497.24
6923.28
31016
1.01250
0.50939
0.34170
0.25786
0.20756
0.17403
0.15009
0.13213
0.11817
0.10700
0.09787
0.09026
0.08382
0.07831
0.07353
0.06935
0.06566
0.06238
0.05946
0.05682
0.05444
0.05227
0.05030
0.04849
0.04682
0.04529
0.04387
0.04255
0.04132
0.04018
0.03467
0.03192
0.02783
0.02702
0.02627
0.02525
0.02379
0.02115
0.02062
0.01930
0.01857
0.01795
0.01757
0.01692
0.01613
0.01551
0.01501
0.01317
0.01264
0.01253
0.9877
1.9631
2.9265
3.8781
4.8178
5.7460
6.6627
7.5681
8.4623
9.3455
10.2178
11.0793
11.9302
12.7706
13.6005
14.4203
15.2299
16.0295
16.8193
17.5993
18.3697
19.1306
19.8820
20.6242
21.3573
22.0813
22.7963
23.5025
24.2000
24.8889
28.8473
31.3269
35.9315
37.0129
38.0677
39.6017
42.0346
47.2925
48.4890
51.8222
53.8461
55.7246
56.9013
59.0865
61.9828
64.4781
66.6277
75.9423
79.0861
79.7942
0.9755
2.9023
5.7569
9.5160
14.1569
19.6571
25.9949
33.1487
41.0973
49.8201
59.2967
69.5072
80.4320
92.0519
104.3481
117.3021
130.8958
145.1115
159.9316
175.3392
191.3174
207.8499
224.9204
242.5132
260.6128
279.2040
298.2719
317.8019
337.7797
466.2830
559.2320
759.2296
811.6738
864.9409
946.2277
1084.84
1428.46
1515.79
1778.84
1953.83
2127.52
2242.24
2468.26
2796.57
3109.35
3404.61
5101.53
5997.90
6284.74
0.4969
0.9917
1.4845
1.9752
2.4638
2.9503
3.4348
3.9172
4.3975
4.8758
5.3520
5.8262
6.2982
6.7682
7.2362
7.7021
8.1659
8.6277
9.0874
9.5450
10.0006
10.4542
10.9056
11.3551
11.8024
12.2478
12.6911
13.1323
13.5715
16.1639
17.8515
21.1299
21.9295
22.7211
23.8936
25.8083
30.2047
31.2605
34.3258
36.2855
38.1793
39.4058
41.7737
45.1184
48.2234
51.0990
67.1764
75.8401
78.7619
586
Compound Interest Factor Tables
1.5%
TABLE 6
1.5%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
84
90
96
100
108
120
132
144
240
360
480
1.0150
1.0302
1.0457
1.0614
1.0773
1.0934
1.1098
1.1265
1.1434
1.1605
1.1779
1.1956
1.2136
1.2318
1.2502
1.2690
1.2880
1.3073
1.3270
1.3469
1.3671
1.3876
1.4084
1.4295
1.4509
1.4727
1.4948
1.5172
1.5400
1.5631
1.7091
1.8140
2.0435
2.1052
2.1689
2.2679
2.4432
2.9212
3.0546
3.4926
3.8189
4.1758
4.4320
4.9927
5.9693
7.1370
8.5332
35.6328
212.7038
1269.70
0.9852
0.9707
0.9563
0.9422
0.9283
0.9145
0.9010
0.8877
0.8746
0.8617
0.8489
0.8364
0.8240
0.8118
0.7999
0.7880
0.7764
0.7649
0.7536
0.7425
0.7315
0.7207
0.7100
0.6995
0.6892
0.6790
0.6690
0.6591
0.6494
0.6398
0.5851
0.5513
0.4894
0.4750
0.4611
0.4409
0.4093
0.3423
0.3274
0.2863
0.2619
0.2395
0.2256
0.2003
0.1675
0.1401
0.1172
0.0281
0.0047
0.0008
1.00000
0.49628
0.32838
0.24444
0.19409
0.16053
0.13656
0.11858
0.10461
0.09343
0.08429
0.07668
0.07024
0.06472
0.05994
0.05577
0.05208
0.04881
0.04588
0.04325
0.04087
0.03870
0.03673
0.03492
0.03326
0.03173
0.03032
0.02900
0.02778
0.02664
0.02115
0.01843
0.01437
0.01357
0.01283
0.01183
0.01039
0.00781
0.00730
0.00602
0.00532
0.00472
0.00437
0.00376
0.00302
0.00244
0.00199
0.00043
0.00007
0.00001
1.0000
2.0150
3.0452
4.0909
5.1523
6.2296
7.3230
8.4328
9.5593
10.7027
11.8633
13.0412
14.2368
15.4504
16.6821
17.9324
19.2014
20.4894
21.7967
23.1237
24.4705
25.8376
27.2251
28.6335
30.0630
31.5140
32.9867
34.4815
35.9987
37.5387
47.2760
54.2679
69.5652
73.6828
77.9249
84.5296
96.2147
128.0772
136.9728
166.1726
187.9299
211.7202
228.8030
266.1778
331.2882
409.1354
502.2109
2308.85
14114
84580
1.01500
0.51128
0.34338
0.25944
0.20909
0.17553
0.15156
0.13358
0.11961
0.10843
0.09929
0.09168
0.08524
0.07972
0.07494
0.07077
0.06708
0.06381
0.06088
0.05825
0.05587
0.05370
0.05173
0.04992
0.04826
0.04673
0.04532
0.04400
0.04278
0.04164
0.03615
0.03343
0.02937
0.02857
0.02783
0.02683
0.02539
0.02281
0.02230
0.02102
0.02032
0.01972
0.01937
0.01876
0.01802
0.01744
0.01699
0.01543
0.01507
0.01501
0.9852
1.9559
2.9122
3.8544
4.7826
5.6972
6.5982
7.4859
8.3605
9.2222
10.0711
10.9075
11.7315
12.5434
13.3432
14.1313
14.9076
15.6726
16.4262
17.1686
17.9001
18.6208
19.3309
20.0304
20.7196
21.3986
22.0676
22.7267
23.3761
24.0158
27.6607
29.9158
34.0426
34.9997
35.9287
37.2715
39.3803
43.8447
44.8416
47.5786
49.2099
50.7017
51.6247
53.3137
55.4985
57.3257
58.8540
64.7957
66.3532
66.6142
0.9707
2.8833
5.7098
9.4229
13.9956
19.4018
25.6157
32.6125
40.3675
48.8568
58.0571
67.9454
78.4994
89.6974
101.5178
113.9400
126.9435
140.5084
154.6154
169.2453
184.3798
200.0006
216.0901
232.6310
249.6065
267.0002
284.7958
302.9779
321.5310
439.8303
524.3568
703.5462
749.9636
796.8774
868.0285
988.1674
1279.79
1352.56
1568.51
1709.54
1847.47
1937.45
2112.13
2359.71
2588.71
2798.58
3870.69
4310.72
4415.74
0.4963
0.9901
1.4814
1.9702
2.4566
2.9405
3.4219
3.9008
4.3772
4.8512
5.3227
5.7917
6.2582
6.7223
7.1839
7.6431
8.0997
8.5539
9.0057
9.4550
9.9018
10.3462
10.7881
11.2276
11.6646
12.0992
12.5313
12.9610
13.3883
15.9009
17.5277
20.6667
21.4277
22.1794
23.2894
25.0930
29.1893
30.1631
32.9668
34.7399
36.4381
37.5295
39.6171
42.5185
45.1579
47.5512
59.7368
64.9662
66.2883
587
Compound Interest Factor Tables
2%
TABLE 7
2%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
84
90
96
100
108
120
132
144
240
360
480
1.0200
1.0404
1.0612
1.0824
1.1041
1.1262
1.1487
1.1717
1.1951
1.2190
1.2434
1.2682
1.2936
1.3195
1.3459
1.3728
1.4002
1.4282
1.4568
1.4859
1.5157
1.5460
1.5769
1.6084
1.6406
1.6734
1.7069
1.7410
1.7758
1.8114
2.0399
2.2080
2.5871
2.6916
2.8003
2.9717
3.2810
4.1611
4.4158
5.2773
5.9431
6.6929
7.2446
8.4883
10.7652
13.6528
17.3151
115.8887
1247.56
13430
0.9804
0.9612
0.9423
0.9238
0.9057
0.8880
0.8706
0.8535
0.8368
0.8203
0.8043
0.7885
0.7730
0.7579
0.7430
0.7284
0.7142
0.7002
0.6864
0.6730
0.6598
0.6468
0.6342
0.6217
0.6095
0.5976
0.5859
0.5744
0.5631
0.5521
0.4902
0.4529
0.3865
0.3715
0.3571
0.3365
0.3048
0.2403
0.2265
0.1895
0.1683
0.1494
0.1380
0.1178
0.0929
0.0732
0.0578
0.0086
0.0008
0.0001
1.00000
0.49505
0.32675
0.24262
0.19216
0.15853
0.13451
0.11651
0.10252
0.09133
0.08218
0.07456
0.06812
0.06260
0.05783
0.05365
0.04997
0.04670
0.04378
0.04116
0.03878
0.03663
0.03467
0.03287
0.03122
0.02970
0.02829
0.02699
0.02578
0.02465
0.01923
0.01656
0.01260
0.01182
0.01111
0.01014
0.00877
0.00633
0.00586
0.00468
0.00405
0.00351
0.00320
0.00267
0.00205
0.00158
0.00123
0.00017
0.00002
1.0000
2.0200
3.0604
4.1216
5.2040
6.3081
7.4343
8.5830
9.7546
10.9497
12.1687
13.4121
14.6803
15.9739
17.2934
18.6393
20.0121
21.4123
22.8406
24.2974
25.7833
27.2990
28.8450
30.4219
32.0303
33.6709
35.3443
37.0512
38.7922
40.5681
51.9944
60.4020
79.3535
84.5794
90.0164
98.5865
114.0515
158.0570
170.7918
213.8666
247.1567
284.6467
312.2323
374.4129
488.2582
632.6415
815.7545
5744.44
62328
1.02000
0.51505
0.34675
0.26262
0.21216
0.17853
0.15451
0.13651
0.12252
0.11133
0.10218
0.09456
0.08812
0.08260
0.07783
0.07365
0.06997
0.06670
0.06378
0.06116
0.05878
0.05663
0.05467
0.05287
0.05122
0.04970
0.04829
0.04699
0.04578
0.04465
0.03923
0.03656
0.03260
0.03182
0.03111
0.03014
0.02877
0.02633
0.02586
0.02468
0.02405
0.02351
0.02320
0.02267
0.02205
0.02158
0.02123
0.02017
0.02002
0.02000
0.9804
1.9416
2.8839
3.8077
4.7135
5.6014
6.4720
7.3255
8.1622
8.9826
9.7868
10.5753
11.3484
12.1062
12.8493
13.5777
14.2919
14.9920
15.6785
16.3514
17.0112
17.6580
18.2922
18.9139
19.5235
20.1210
20.7069
21.2813
21.8444
22.3965
25.4888
27.3555
30.6731
31.4236
32.1449
33.1748
34.7609
37.9841
38.6771
40.5255
41.5869
42.5294
43.0984
44.1095
45.3554
46.3378
47.1123
49.5686
49.9599
49.9963
0.9612
2.8458
5.6173
9.2403
13.6801
18.9035
24.8779
31.5720
38.9551
46.9977
55.6712
64.9475
74.7999
85.2021
96.1288
107.5554
119.4581
131.8139
144.6003
157.7959
171.3795
185.3309
199.6305
214.2592
229.1987
244.4311
259.9392
275.7064
291.7164
392.0405
461.9931
605.9657
642.3606
678.7849
733.3527
823.6975
1034.06
1084.64
1230.42
1322.17
1409.30
1464.75
1569.30
1710.42
1833.47
1939.79
2374.88
2482.57
2498.03
0.4950
0.9868
1.4752
1.9604
2.4423
2.9208
3.3961
3.8681
4.3367
4.8021
5.2642
5.7231
6.1786
6.6309
7.0799
7.5256
7.9681
8.4073
8.8433
9.2760
9.7055
10.1317
10.5547
10.9745
11.3910
11.8043
12.2145
12.6214
13.0251
15.3809
16.8885
19.7556
20.4420
21.1164
22.1057
23.6961
27.2234
28.0434
30.3616
31.7929
33.1370
33.9863
35.5774
37.7114
39.5676
41.1738
47.9110
49.7112
49.9643
588
Compound Interest Factor Tables
3%
TABLE 8
3%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
84
85
90
96
108
120
1.0300
1.0609
1.0927
1.1255
1.1593
1.1941
1.2299
1.2668
1.3048
1.3439
1.3842
1.4258
1.4685
1.5126
1.5580
1.6047
1.6528
1.7024
1.7535
1.8061
1.8603
1.9161
1.9736
2.0328
2.0938
2.1566
2.2213
2.2879
2.3566
2.4273
2.5001
2.5751
2.6523
2.7319
2.8139
3.2620
3.7816
4.3839
5.0821
5.8916
6.8300
7.9178
9.1789
10.6409
11.9764
12.3357
14.3005
17.0755
24.3456
34.7110
0.9709
0.9426
0.9151
0.8885
0.8626
0.8375
0.8131
0.7894
0.7664
0.7441
0.7224
0.7014
0.6810
0.6611
0.6419
0.6232
0.6050
0.5874
0.5703
0.5537
0.5375
0.5219
0.5067
0.4919
0.4776
0.4637
0.4502
0.4371
0.4243
0.4120
0.4000
0.3883
0.3770
0.3660
0.3554
0.3066
0.2644
0.2281
0.1968
0.1697
0.1464
0.1263
0.1089
0.0940
0.0835
0.0811
0.0699
0.0586
0.0411
0.0288
1.00000
0.49261
0.32353
0.23903
0.18835
0.15460
0.13051
0.11246
0.09843
0.08723
0.07808
0.07046
0.06403
0.05853
0.05377
0.04961
0.04595
0.04271
0.03981
0.03722
0.03487
0.03275
0.03081
0.02905
0.02743
0.02594
0.02456
0.02329
0.02211
0.02102
0.02000
0.01905
0.01816
0.01732
0.01654
0.01326
0.01079
0.00887
0.00735
0.00613
0.00515
0.00434
0.00367
0.00311
0.00273
0.00265
0.00226
0.00187
0.00129
0.00089
1.0000
2.0300
3.0909
4.1836
5.3091
6.4684
7.6625
8.8923
10.1591
11.4639
12.8078
14.1920
15.6178
17.0863
18.5989
20.1569
21.7616
23.4144
25.1169
26.8704
28.6765
30.5368
32.4529
34.4265
36.4593
38.5530
40.7096
42.9309
45.2189
47.5754
50.0027
52.5028
55.0778
57.7302
60.4621
75.4013
92.7199
112.7969
136.0716
163.0534
194.3328
230.5941
272.6309
321.3630
365.8805
377.8570
443.3489
535.8502
778.1863
1123.70
1.03000
0.52261
0.35353
0.26903
0.21835
0.18460
0.16051
0.14246
0.12843
0.11723
0.10808
0.10046
0.09403
0.08853
0.08377
0.07961
0.07595
0.07271
0.06981
0.06722
0.06487
0.06275
0.06081
0.05905
0.05743
0.05594
0.05456
0.05329
0.05211
0.05102
0.05000
0.04905
0.04816
0.04732
0.04654
0.04326
0.04079
0.03887
0.03735
0.03613
0.03515
0.03434
0.03367
0.03311
0.03273
0.03265
0.03226
0.03187
0.03129
0.03089
0.9709
1.9135
2.8286
3.7171
4.5797
5.4172
6.2303
7.0197
7.7861
8.5302
9.2526
9.9540
10.6350
11.2961
11.9379
12.5611
13.1661
13.7535
14.3238
14.8775
15.4150
15.9369
16.4436
16.9355
17.4131
17.8768
18.3270
18.7641
19.1885
19.6004
20.0004
20.3888
20.7658
21.1318
21.4872
23.1148
24.5187
25.7298
26.7744
27.6756
28.4529
29.1234
29.7018
30.2008
30.5501
30.6312
31.0024
31.3812
31.9642
32.3730
0.9426
2.7729
5.4383
8.8888
13.0762
17.9547
23.4806
29.6119
36.3088
43.5330
51.2482
59.4196
68.0141
77.0002
86.3477
96.0280
106.0137
116.2788
126.7987
137.5496
148.5094
159.6566
170.9711
182.4336
194.0260
205.7309
217.5320
229.4137
241.3613
253.3609
265.3993
277.4642
289.5437
301.6267
361.7499
420.6325
477.4803
531.7411
583.0526
631.2010
676.0869
717.6978
756.0865
784.5434
791.3529
823.6302
858.6377
917.6013
963.8635
0.4926
0.9803
1.4631
1.9409
2.4138
2.8819
3.3450
3.8032
4.2565
4.7049
5.1485
5.5872
6.0210
6.4500
6.8742
7.2936
7.7081
8.1179
8.5229
8.9231
9.3186
9.7093
10.0954
10.4768
10.8535
11.2255
11.5930
11.9558
12.3141
12.6678
13.0169
13.3616
13.7018
14.0375
15.6502
17.1556
18.5575
19.8600
21.0674
22.1841
23.2145
24.1634
25.0353
25.6806
25.8349
26.5667
27.3615
28.7072
29.7737
589
Compound Interest Factor Tables
4%
TABLE 9
4%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
85
90
96
108
120
144
1.0400
1.0816
1.1249
1.1699
1.2167
1.2653
1.3159
1.3686
1.4233
1.4802
1.5395
1.6010
1.6651
1.7317
1.8009
1.8730
1.9479
2.0258
2.1068
2.1911
2.2788
2.3699
2.4647
2.5633
2.6658
2.7725
2.8834
2.9987
3.1187
3.2434
3.3731
3.5081
3.6484
3.7943
3.9461
4.8010
5.8412
7.1067
8.6464
10.5196
12.7987
15.5716
18.9453
23.0498
28.0436
34.1193
43.1718
69.1195
110.6626
283.6618
0.9615
0.9246
0.8890
0.8548
0.8219
0.7903
0.7599
0.7307
0.7026
0.6756
0.6496
0.6246
0.6006
0.5775
0.5553
0.5339
0.5134
0.4936
0.4746
0.4564
0.4388
0.4220
0.4057
0.3901
0.3751
0.3607
0.3468
0.3335
0.3207
0.3083
0.2965
0.2851
0.2741
0.2636
0.2534
0.2083
0.1712
0.1407
0.1157
0.0951
0.0781
0.0642
0.0528
0.0434
0.0357
0.0293
0.0232
0.0145
0.0090
0.0035
1.00000
0.49020
0.32035
0.23549
0.18463
0.15076
0.12661
0.10853
0.09449
0.08329
0.07415
0.06655
0.06014
0.05467
0.04994
0.04582
0.04220
0.03899
0.03614
0.03358
0.03128
0.02920
0.02731
0.02559
0.02401
0.02257
0.02124
0.02001
0.01888
0.01783
0.01686
0.01595
0.01510
0.01431
0.01358
0.01052
0.00826
0.00655
0.00523
0.00420
0.00339
0.00275
0.00223
0.00181
0.00148
0.00121
0.00095
0.00059
0.00036
0.00014
1.0000
2.0400
3.1216
4.2465
5.4163
6.6330
7.8983
9.2142
10.5828
12.0061
13.4864
15.0258
16.6268
18.2919
20.0236
21.8245
23.6975
25.6454
27.6712
29.7781
31.9692
34.2480
36.6179
39.0826
41.6459
44.3117
47.0842
49.9676
52.9663
56.0849
59.3283
62.7015
66.2095
69.8579
73.6522
95.0255
121.0294
152.6671
191.1592
237.9907
294.9684
364.2905
448.6314
551.2450
676.0901
827.9833
1054.30
1702.99
2741.56
7066.55
1.04000
0.53020
0.36035
0.27549
0.22463
0.19076
0.16661
0.14853
0.13449
0.12329
0.11415
0.10655
0.10014
0.09467
0.08994
0.08582
0.08220
0.07899
0.07614
0.07358
0.07128
0.06920
0.06731
0.06559
0.06401
0.06257
0.06124
0.06001
0.05888
0.05783
0.05686
0.05595
0.05510
0.05431
0.05358
0.05052
0.04826
0.04655
0.04523
0.04420
0.04339
0.04275
0.04223
0.04181
0.04148
0.04121
0.04095
0.04059
0.04036
0.04014
0.9615
1.8861
2.7751
3.6299
4.4518
5.2421
6.0021
6.7327
7.4353
8.1109
8.7605
9.3851
9.9856
10.5631
11.1184
11.6523
12.1657
12.6593
13.1339
13.5903
14.0292
14.4511
14.8568
15.2470
15.6221
15.9828
16.3296
16.6631
16.9837
17.2920
17.5885
17.8736
18.1476
18.4112
18.6646
19.7928
20.7200
21.4822
22.1086
22.6235
23.0467
23.3945
23.6804
23.9154
24.1085
24.2673
24.4209
24.6383
24.7741
24.9119
0.9246
2.7025
5.2670
8.5547
12.5062
17.0657
22.1806
27.8013
33.8814
40.3772
47.2477
54.4546
61.9618
69.7355
77.7441
85.9581
94.3498
102.8933
111.5647
120.3414
129.2024
138.1284
147.1012
156.1040
165.1212
174.1385
183.1424
192.1206
201.0618
209.9556
218.7924
227.5634
236.2607
244.8768
286.5303
325.4028
361.1638
393.6890
422.9966
449.2014
472.4789
493.0408
511.1161
526.9384
540.7369
554.9312
576.8949
592.2428
610.1055
0.4902
0.9739
1.4510
1.9216
2.3857
2.8433
3.2944
3.7391
4.1773
4.6090
5.0343
5.4533
5.8659
6.2721
6.6720
7.0656
7.4530
7.8342
8.2091
8.5779
8.9407
9.2973
9.6479
9.9925
10.3312
10.6640
10.9909
11.3120
11.6274
11.9371
12.2411
12.5396
12.8324
13.1198
14.4765
15.7047
16.8122
17.8070
18.6972
19.4909
20.1961
20.8206
21.3718
21.8569
22.2826
22.7236
23.4146
23.9057
24.4906
590
Compound Interest Factor Tables
5%
TABLE 10
5%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
85
90
95
96
98
100
1.0500
1.1025
1.1576
1.2155
1.2763
1.3401
1.4071
1.4775
1.5513
1.6289
1.7103
1.7959
1.8856
1.9799
2.0789
2.1829
2.2920
2.4066
2.5270
2.6533
2.7860
2.9253
3.0715
3.2251
3.3864
3.5557
3.7335
3.9201
4.1161
4.3219
4.5380
4.7649
5.0032
5.2533
5.5160
7.0400
8.9850
11.4674
14.6356
18.6792
23.8399
30.4264
38.8327
49.5614
63.2544
80.7304
103.0347
108.1864
119.2755
131.5013
0.9524
0.9070
0.8638
0.8227
0.7835
0.7462
0.7107
0.6768
0.6446
0.6139
0.5847
0.5568
0.5303
0.5051
0.4810
0.4581
0.4363
0.4155
0.3957
0.3769
0.3589
0.3418
0.3256
0.3101
0.2953
0.2812
0.2678
0.2551
0.2429
0.2314
0.2204
0.2099
0.1999
0.1904
0.1813
0.1420
0.1113
0.0872
0.0683
0.0535
0.0419
0.0329
0.0258
0.0202
0.0158
0.0124
0.0097
0.0092
0.0084
0.0076
1.00000
0.48780
0.31721
0.23201
0.18097
0.14702
0.12282
0.10472
0.09069
0.07950
0.07039
0.06283
0.05646
0.05102
0.04634
0.04227
0.03870
0.03555
0.03275
0.03024
0.02800
0.02597
0.02414
0.02247
0.02095
0.01956
0.01829
0.01712
0.01605
0.01505
0.01413
0.01328
0.01249
0.01176
0.01107
0.00828
0.00626
0.00478
0.00367
0.00283
0.00219
0.00170
0.00132
0.00103
0.00080
0.00063
0.00049
0.00047
0.00042
0.00038
1.0000
2.0500
3.1525
4.3101
5.5256
6.8019
8.1420
9.5491
11.0266
12.5779
14.2068
15.9171
17.7130
19.5986
21.5786
23.6575
25.8404
28.1324
30.5390
33.0660
35.7193
38.5052
41.4305
44.5020
47.7271
51.1135
54.6691
58.4026
62.3227
66.4388
70.7608
75.2988
80.0638
85.0670
90.3203
120.7998
159.7002
209.3480
272.7126
353.5837
456.7980
588.5285
756.6537
971.2288
1245.09
1594.61
2040.69
2143.73
2365.51
2610.03
1.05000
0.53780
0.36721
0.28201
0.23097
0.19702
0.17282
0.15472
0.14069
0.12950
0.12039
0.11283
0.10646
0.10102
0.09634
0.09227
0.08870
0.08555
0.08275
0.08024
0.07800
0.07597
0.07414
0.07247
0.07095
0.06956
0.06829
0.06712
0.06605
0.06505
0.06413
0.06328
0.06249
0.06176
0.06107
0.05828
0.05626
0.05478
0.05367
0.05283
0.05219
0.05170
0.05132
0.05103
0.05080
0.05063
0.05049
0.05047
0.05042
0.05038
0.9524
1.8594
2.7232
3.5460
4.3295
5.0757
5.7864
6.4632
7.1078
7.7217
8.3064
8.8633
9.3936
9.8986
10.3797
10.8378
11.2741
11.6896
12.0853
12.4622
12.8212
13.1630
13.4886
13.7986
14.0939
14.3752
14.6430
14.8981
15.1411
15.3725
15.5928
15.8027
16.0025
16.1929
16.3742
17.1591
17.7741
18.2559
18.6335
18.9293
19.1611
19.3427
19.4850
19.5965
19.6838
19.7523
19.8059
19.8151
19.8323
19.8479
0.9070
2.6347
5.1028
8.2369
11.9680
16.2321
20.9700
26.1268
31.6520
37.4988
43.6241
49.9879
56.5538
63.2880
70.1597
77.1405
84.2043
91.3275
98.4884
105.6673
112.8461
120.0087
127.1402
134.2275
141.2585
148.2226
155.1101
161.9126
168.6226
175.2333
181.7392
188.1351
194.4168
200.5807
229.5452
255.3145
277.9148
297.5104
314.3432
328.6910
340.8409
351.0721
359.6460
366.8007
372.7488
377.6774
378.5555
380.2139
381.7492
0.4878
0.9675
1.4391
1.9025
2.3579
2.8052
3.2445
3.6758
4.0991
4.5144
4.9219
5.3215
5.7133
6.0973
6.4736
6.8423
7.2034
7.5569
7.9030
8.2416
8.5730
8.8971
9.2140
9.5238
9.8266
10.1224
10.4114
10.6936
10.9691
11.2381
11.5005
11.7566
12.0063
12.2498
13.3775
14.3644
15.2233
15.9664
16.6062
17.1541
17.6212
18.0176
18.3526
18.6346
18.8712
19.0689
19.1044
19.1714
19.2337
591
Compound Interest Factor Tables
6%
TABLE 11
6%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
85
90
95
96
98
100
1.0600
1.1236
1.1910
1.2625
1.3382
1.4185
1.5036
1.5938
1.6895
1.7908
1.8983
2.0122
2.1329
2.2609
2.3966
2.5404
2.6928
2.8543
3.0256
3.2071
3.3996
3.6035
3.8197
4.0489
4.2919
4.5494
4.8223
5.1117
5.4184
5.7435
6.0881
6.4534
6.8406
7.2510
7.6861
10.2857
13.7646
18.4202
24.6503
32.9877
44.1450
59.0759
79.0569
105.7960
141.5789
189.4645
253.5463
268.7590
301.9776
339.3021
0.9434
0.8900
0.8396
0.7921
0.7473
0.7050
0.6651
0.6274
0.5919
0.5584
0.5268
0.4970
0.4688
0.4423
0.4173
0.3936
0.3714
0.3503
0.3305
0.3118
0.2942
0.2775
0.2618
0.2470
0.2330
0.2198
0.2074
0.1956
0.1846
0.1741
0.1643
0.1550
0.1462
0.1379
0.1301
0.0972
0.0727
0.0543
0.0406
0.0303
0.0227
0.0169
0.0126
0.0095
0.0071
0.0053
0.0039
0.0037
0.0033
0.0029
1.00000
0.48544
0.31411
0.22859
0.17740
0.14336
0.11914
0.10104
0.08702
0.07587
0.06679
0.05928
0.05296
0.04758
0.04296
0.03895
0.03544
0.03236
0.02962
0.02718
0.02500
0.02305
0.02128
0.01968
0.01823
0.01690
0.01570
0.01459
0.01358
0.01265
0.01179
0.01100
0.01027
0.00960
0.00897
0.00646
0.00470
0.00344
0.00254
0.00188
0.00139
0.00103
0.00077
0.00057
0.00043
0.00032
0.00024
0.00022
0.00020
0.00018
1.0000
2.0600
3.1836
4.3746
5.6371
6.9753
8.3938
9.8975
11.4913
13.1808
14.9716
16.8699
18.8821
21.0151
23.2760
25.6725
28.2129
30.9057
33.7600
36.7856
39.9927
43.3923
46.9958
50.8156
54.8645
59.1564
63.7058
68.5281
73.6398
79.0582
84.8017
90.8898
97.3432
104.1838
111.4348
154.7620
212.7435
290.3359
394.1720
533.1282
719.0829
967.9322
1300.95
1746.60
2342.98
3141.08
4209.10
4462.65
5016.29
5638.37
1.06000
0.54544
0.37411
0.28859
0.23740
0.20336
0.17914
0.16104
0.14702
0.13587
0.12679
0.11928
0.11296
0.10758
0.10296
0.09895
0.09544
0.09236
0.08962
0.08718
0.08500
0.08305
0.08128
0.07968
0.07823
0.07690
0.07570
0.07459
0.07358
0.07265
0.07179
0.07100
0.07027
0.06960
0.06897
0.06646
0.06470
0.06344
0.06254
0.06188
0.06139
0.06103
0.06077
0.06057
0.06043
0.06032
0.06024
0.06022
0.06020
0.06018
0.9434
1.8334
2.6730
3.4651
4.2124
4.9173
5.5824
6.2098
6.8017
7.3601
7.8869
8.3838
8.8527
9.2950
9.7122
10.1059
10.4773
10.8276
11.1581
11.4699
11.7641
12.0416
12.3034
12.5504
12.7834
13.0032
13.2105
13.4062
13.5907
13.7648
13.9291
14.0840
14.2302
14.3681
14.4982
15.0463
15.4558
15.7619
15.9905
16.1614
16.2891
16.3845
16.4558
16.5091
16.5489
16.5787
16.6009
16.6047
16.6115
16.6175
0.8900
2.5692
4.9455
7.9345
11.4594
15.4497
19.8416
24.5768
29.6023
34.8702
40.3369
45.9629
51.7128
57.5546
63.4592
69.4011
75.3569
81.3062
87.2304
93.1136
98.9412
104.7007
110.3812
115.9732
121.4684
126.8600
132.1420
137.3096
142.3588
147.2864
152.0901
156.7681
161.3192
165.7427
185.9568
203.1096
217.4574
229.3222
239.0428
246.9450
253.3271
258.4527
262.5493
265.8096
268.3946
270.4375
270.7909
271.4491
272.0471
0.4854
0.9612
1.4272
1.8836
2.3304
2.7676
3.1952
3.6133
4.0220
4.4213
4.8113
5.1920
5.5635
5.9260
6.2794
6.6240
6.9597
7.2867
7.6051
7.9151
8.2166
8.5099
8.7951
9.0722
9.3414
9.6029
9.8568
10.1032
10.3422
10.5740
10.7988
11.0166
11.2276
11.4319
12.3590
13.1413
13.7964
14.3411
14.7909
15.1601
15.4613
15.7058
15.9033
16.0620
16.1891
16.2905
16.3081
16.3411
16.3711
592
Compound Interest Factor Tables
7%
TABLE 12
7%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
85
90
95
96
98
100
1.0700
1.1449
1.2250
1.3108
1.4026
1.5007
1.6058
1.7182
1.8385
1.9672
2.1049
2.2522
2.4098
2.5785
2.7590
2.9522
3.1588
3.3799
3.6165
3.8697
4.1406
4.4304
4.7405
5.0724
5.4274
5.8074
6.2139
6.6488
7.1143
7.6123
8.1451
8.7153
9.3253
9.9781
10.6766
14.9745
21.0025
29.4570
41.3150
57.9464
81.2729
113.9894
159.8760
224.2344
314.5003
441.1030
618.6697
661.9766
757.8970
867.7163
0.9346
0.8734
0.8163
0.7629
0.7130
0.6663
0.6227
0.5820
0.5439
0.5083
0.4751
0.4440
0.4150
0.3878
0.3624
0.3387
0.3166
0.2959
0.2765
0.2584
0.2415
0.2257
0.2109
0.1971
0.1842
0.1722
0.1609
0.1504
0.1406
0.1314
0.1228
0.1147
0.1072
0.1002
0.0937
0.0668
0.0476
0.0339
0.0242
0.0173
0.0123
0.0088
0.0063
0.0045
0.0032
0.0023
0.0016
0.0015
0.0013
0.0012
1.00000
0.48309
0.31105
0.22523
0.17389
0.13980
0.11555
0.09747
0.08349
0.07238
0.06336
0.05590
0.04965
0.04434
0.03979
0.03586
0.03243
0.02941
0.02675
0.02439
0.02229
0.02041
0.01871
0.01719
0.01581
0.01456
0.01343
0.01239
0.01145
0.01059
0.00980
0.00907
0.00841
0.00780
0.00723
0.00501
0.00350
0.00246
0.00174
0.00123
0.00087
0.00062
0.00044
0.00031
0.00022
0.00016
0.00011
0.00011
0.00009
0.00008
1.0000
2.0700
3.2149
4.4399
5.7507
7.1533
8.6540
10.2598
11.9780
13.8164
15.7836
17.8885
20.1406
22.5505
25.1290
27.8881
30.8402
33.9990
37.3790
40.9955
44.8652
49.0057
53.4361
58.1767
63.2490
68.6765
74.4838
80.6977
87.3465
94.4608
102.0730
110.2182
118.9334
128.2588
138.2369
199.6351
285.7493
406.5289
575.9286
813.5204
1146.76
1614.13
2269.66
3189.06
4478.58
6287.19
8823.85
9442.52
10813
12382
1.07000
0.55309
0.38105
0.29523
0.24389
0.20980
0.18555
0.16747
0.15349
0.14238
0.13336
0.12590
0.11965
0.11434
0.10979
0.10586
0.10243
0.09941
0.09675
0.09439
0.09229
0.09041
0.08871
0.08719
0.08581
0.08456
0.08343
0.08239
0.08145
0.08059
0.07980
0.07907
0.07841
0.07780
0.07723
0.07501
0.07350
0.07246
0.07174
0.07123
0.07087
0.07062
0.07044
0.07031
0.07022
0.07016
0.07011
0.07011
0.07009
0.07008
0.9346
1.8080
2.6243
3.3872
4.1002
4.7665
5.3893
5.9713
6.5152
7.0236
7.4987
7.9427
8.3577
8.7455
9.1079
9.4466
9.7632
10.0591
10.3356
10.5940
10.8355
11.0612
11.2722
11.4693
11.6536
11.8258
11.9867
12.1371
12.2777
12.4090
12.5318
12.6466
12.7538
12.8540
12.9477
13.3317
13.6055
13.8007
13.9399
14.0392
14.1099
14.1604
14.1964
14.2220
14.2403
14.2533
14.2626
14.2641
14.2669
14.2693
0.8734
2.5060
4.7947
7.6467
10.9784
14.7149
18.7889
23.1404
27.7156
32.4665
37.3506
42.3302
47.3718
52.4461
57.5271
62.5923
67.6219
72.5991
77.5091
82.3393
87.0793
91.7201
96.2545
100.6765
104.9814
109.1656
113.2264
117.1622
120.9718
124.6550
128.2120
131.6435
134.9507
138.1353
152.2928
163.7559
172.9051
180.1243
185.7677
190.1452
193.5185
196.1035
198.0748
199.5717
200.7042
201.5581
201.7016
201.9651
202.2001
0.4831
0.9549
1.4155
1.8650
2.3032
2.7304
3.1465
3.5517
3.9461
4.3296
4.7025
5.0648
5.4167
5.7583
6.0897
6.4110
6.7225
7.0242
7.3163
7.5990
7.8725
8.1369
8.3923
8.6391
8.8773
9.1072
9.3289
9.5427
9.7487
9.9471
10.1381
10.3219
10.4987
10.6687
11.4233
12.0360
12.5287
12.9215
13.2321
13.4760
13.6662
13.8136
13.9273
14.0146
14.0812
14.1319
14.1405
14.1562
14.1703
593
Compound Interest Factor Tables
8%
TABLE 13
8%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
85
90
95
96
98
100
1.0800
1.1664
1.2597
1.3605
1.4693
1.5869
1.7138
1.8509
1.9990
2.1589
2.3316
2.5182
2.7196
2.9372
3.1722
3.4259
3.7000
3.9960
4.3157
4.6610
5.0338
5.4365
5.8715
6.3412
6.8485
7.3964
7.9881
8.6271
9.3173
10.0627
10.8677
11.7371
12.6760
13.6901
14.7853
21.7245
31.9204
46.9016
68.9139
101.2571
148.7798
218.6064
321.2045
471.9548
693.4565
1018.92
1497.12
1616.89
1885.94
2199.76
0.9259
0.8573
0.7938
0.7350
0.6806
0.6302
0.5835
0.5403
0.5002
0.4632
0.4289
0.3971
0.3677
0.3405
0.3152
0.2919
0.2703
0.2502
0.2317
0.2145
0.1987
0.1839
0.1703
0.1577
0.1460
0.1352
0.1252
0.1159
0.1073
0.0994
0.0920
0.0852
0.0789
0.0730
0.0676
0.0460
0.0313
0.0213
0.0145
0.0099
0.0067
0.0046
0.0031
0.0021
0.0014
0.0010
0.0007
0.0006
0.0005
0.0005
1.00000
0.48077
0.30803
0.22192
0.17046
0.13632
0.11207
0.09401
0.08008
0.06903
0.06008
0.05270
0.04652
0.04130
0.03683
0.03298
0.02963
0.02670
0.02413
0.02185
0.01983
0.01803
0.01642
0.01498
0.01368
0.01251
0.01145
0.01049
0.00962
0.00883
0.00811
0.00745
0.00685
0.00630
0.00580
0.00386
0.00259
0.00174
0.00118
0.00080
0.00054
0.00037
0.00025
0.00017
0.00012
0.00008
0.00005
0.00005
0.00004
0.00004
1.0000
2.0800
3.2464
4.5061
5.8666
7.3359
8.9228
10.6366
12.4876
14.4866
16.6455
18.9771
21.4953
24.2149
27.1521
30.3243
33.7502
37.4502
41.4463
45.7620
50.4229
55.4568
60.8933
66.7648
73.1059
79.9544
87.3508
95.3388
103.9659
113.2832
123.3459
134.2135
145.9506
158.6267
172.3168
259.0565
386.5056
573.7702
848.9232
1253.21
1847.25
2720.08
4002.56
5886.94
8655.71
12724
18702
20199
23562
27485
1.08000
0.56077
0.38803
0.30192
0.25046
0.21632
0.19207
0.17401
0.16008
0.14903
0.14008
0.13270
0.12652
0.12130
0.11683
0.11298
0.10963
0.10670
0.10413
0.10185
0.09983
0.09803
0.09642
0.09498
0.09368
0.09251
0.09145
0.09049
0.08962
0.08883
0.08811
0.08745
0.08685
0.08630
0.08580
0.08386
0.08259
0.08174
0.08118
0.08080
0.08054
0.08037
0.08025
0.08017
0.08012
0.08008
0.08005
0.08005
0.08004
0.08004
0.9259
1.7833
2.5771
3.3121
3.9927
4.6229
5.2064
5.7466
6.2469
6.7101
7.1390
7.5361
7.9038
8.2442
8.5595
8.8514
9.1216
9.3719
9.6036
9.8181
10.0168
10.2007
10.3711
10.5288
10.6748
10.8100
10.9352
11.0511
11.1584
11.2578
11.3498
11.4350
11.5139
11.5869
11.6546
11.9246
12.1084
12.2335
12.3186
12.3766
12.4160
12.4428
12.4611
12.4735
12.4820
12.4877
12.4917
12.4923
12.4934
12.4943
0.8573
2.4450
4.6501
7.3724
10.5233
14.0242
17.8061
21.8081
25.9768
30.2657
34.6339
39.0463
43.4723
47.8857
52.2640
56.5883
60.8426
65.0134
69.0898
73.0629
76.9257
80.6726
84.2997
87.8041
91.1842
94.4390
97.5687
100.5738
103.4558
106.2163
108.8575
111.3819
113.7924
116.0920
126.0422
133.7331
139.5928
144.0065
147.3000
149.7387
151.5326
152.8448
153.8001
154.4925
154.9925
155.3524
155.4112
155.5176
155.6107
0.4808
0.9487
1.4040
1.8465
2.2763
2.6937
3.0985
3.4910
3.8713
4.2395
4.5957
4.9402
5.2731
5.5945
5.9046
6.2037
6.4920
6.7697
7.0369
7.2940
7.5412
7.7786
8.0066
8.2254
8.4352
8.6363
8.8289
9.0133
9.1897
9.3584
9.5197
9.6737
9.8208
9.9611
10.5699
11.0447
11.4107
11.6902
11.9015
12.0602
12.1783
12.2658
12.3301
12.3772
12.4116
12.4365
12.4406
12.4480
12.4545
594
Compound Interest Factor Tables
9%
TABLE 14
9%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
85
90
95
96
98
100
1.0900
1.1881
1.2950
1.4116
1.5386
1.6771
1.8280
1.9926
2.1719
2.3674
2.5804
2.8127
3.0658
3.3417
3.6425
3.9703
4.3276
4.7171
5.1417
5.6044
6.1088
6.6586
7.2579
7.9111
8.6231
9.3992
10.2451
11.1671
12.1722
13.2677
14.4618
15.7633
17.1820
18.7284
20.4140
31.4094
48.3273
74.3575
114.4083
176.0313
270.8460
416.7301
641.1909
986.5517
1517.93
2335.53
3593.50
3916.91
4653.68
5529.04
0.9174
0.8417
0.7722
0.7084
0.6499
0.5963
0.5470
0.5019
0.4604
0.4224
0.3875
0.3555
0.3262
0.2992
0.2745
0.2519
0.2311
0.2120
0.1945
0.1784
0.1637
0.1502
0.1378
0.1264
0.1160
0.1064
0.0976
0.0895
0.0822
0.0754
0.0691
0.0634
0.0582
0.0534
0.0490
0.0318
0.0207
0.0134
0.0087
0.0057
0.0037
0.0024
0.0016
0.0010
0.0007
0.0004
0.0003
0.0003
0.0002
0.0002
1.00000
0.47847
0.30505
0.21867
0.16709
0.13292
0.10869
0.09067
0.07680
0.06582
0.05695
0.04965
0.04357
0.03843
0.03406
0.03030
0.02705
0.02421
0.02173
0.01955
0.01762
0.01590
0.01438
0.01302
0.01181
0.01072
0.00973
0.00885
0.00806
0.00734
0.00669
0.00610
0.00556
0.00508
0.00464
0.00296
0.00190
0.00123
0.00079
0.00051
0.00033
0.00022
0.00014
0.00009
0.00006
0.00004
0.00003
0.00002
0.00002
0.00002
1.0000
2.0900
3.2781
4.5731
5.9847
7.5233
9.2004
11.0285
13.0210
15.1929
17.5603
20.1407
22.9534
26.0192
29.3609
33.0034
36.9737
41.3013
46.0185
51.1601
56.7645
62.8733
69.5319
76.7898
84.7009
93.3240
102.7231
112.9682
124.1354
136.3075
149.5752
164.0370
179.8003
196.9823
215.7108
337.8824
525.8587
815.0836
1260.09
1944.79
2998.29
4619.22
7113.23
10951
16855
25939
39917
43510
51696
61423
1.09000
0.56847
0.39505
0.30867
0.25709
0.22292
0.19869
0.18067
0.16680
0.15582
0.14695
0.13965
0.13357
0.12843
0.12406
0.12030
0.11705
0.11421
0.11173
0.10955
0.10762
0.10590
0.10438
0.10302
0.10181
0.10072
0.09973
0.09885
0.09806
0.09734
0.09669
0.09610
0.09556
0.09508
0.09464
0.09296
0.09190
0.09123
0.09079
0.09051
0.09033
0.09022
0.09014
0.09009
0.09006
0.09004
0.09003
0.09002
0.09002
0.09002
0.9174
1.7591
2.5313
3.2397
3.8897
4.4859
5.0330
5.5348
5.9952
6.4177
6.8052
7.1607
7.4869
7.7862
8.0607
8.3126
8.5436
8.7556
8.9501
9.1285
9.2922
9.4424
9.5802
9.7066
9.8226
9.9290
10.0266
10.1161
10.1983
10.2737
10.3428
10.4062
10.4644
10.5178
10.5668
10.7574
10.8812
10.9617
11.0140
11.0480
11.0701
11.0844
11.0938
11.0998
11.1038
11.1064
11.1080
11.1083
11.1087
11.1091
0.8417
2.3860
4.5113
7.1110
10.0924
13.3746
16.8877
20.5711
24.3728
28.2481
32.1590
36.0731
39.9633
43.8069
47.5849
51.2821
54.8860
58.3868
61.7770
65.0509
68.2048
71.2359
74.1433
76.9265
79.5863
82.1241
84.5419
86.8422
89.0280
91.1024
93.0690
94.9314
96.6935
98.3590
105.3762
110.5561
114.3251
117.0362
118.9683
120.3344
121.2942
121.9646
122.4306
122.7533
122.9758
123.1287
123.1529
123.1963
123.2335
0.4785
0.9426
1.3925
1.8282
2.2498
2.6574
3.0512
3.4312
3.7978
4.1510
4.4910
4.8182
5.1326
5.4346
5.7245
6.0024
6.2687
6.5236
6.7674
7.0006
7.2232
7.4357
7.6384
7.8316
8.0156
8.1906
8.3571
8.5154
8.6657
8.8083
8.9436
9.0718
9.1933
9.3083
9.7957
10.1603
10.4295
10.6261
10.7683
10.8702
10.9427
10.9940
11.0299
11.0551
11.0726
11.0847
11.0866
11.0900
11.0930
595
Compound Interest Factor Tables
10%
TABLE 15
10%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
85
90
95
96
98
100
1.1000
1.2100
1.3310
1.4641
1.6105
1.7716
1.9487
2.1436
2.3579
2.5937
2.8531
3.1384
3.4523
3.7975
4.1772
4.5950
5.0545
5.5599
6.1159
6.7275
7.4002
8.1403
8.9543
9.8497
10.8347
11.9182
13.1100
14.4210
15.8631
17.4494
19.1943
21.1138
23.2252
25.5477
28.1024
45.2593
72.8905
117.3909
189.0591
304.4816
490.3707
789.7470
1271.90
2048.40
3298.97
5313.02
8556.68
9412.34
11389
13781
0.9091
0.8264
0.7513
0.6830
0.6209
0.5645
0.5132
0.4665
0.4241
0.3855
0.3505
0.3186
0.2897
0.2633
0.2394
0.2176
0.1978
0.1799
0.1635
0.1486
0.1351
0.1228
0.1117
0.1015
0.0923
0.0839
0.0763
0.0693
0.0630
0.0573
0.0521
0.0474
0.0431
0.0391
0.0356
0.0221
0.0137
0.0085
0.0053
0.0033
0.0020
0.0013
0.0008
0.0005
0.0003
0.0002
0.0001
0.0001
0.0001
0.0001
1.00000
0.47619
0.30211
0.21547
0.16380
0.12961
0.10541
0.08744
0.07364
0.06275
0.05396
0.04676
0.04078
0.03575
0.03147
0.02782
0.02466
0.02193
0.01955
0.01746
0.01562
0.01401
0.01257
0.01130
0.01017
0.00916
0.00826
0.00745
0.00673
0.00608
0.00550
0.00497
0.00450
0.00407
0.00369
0.00226
0.00139
0.00086
0.00053
0.00033
0.00020
0.00013
0.00008
0.00005
0.00003
0.00002
0.00001
0.00001
0.00001
0.00001
1.0000
2.1000
3.3100
4.6410
6.1051
7.7156
9.4872
11.4359
13.5795
15.9374
18.5312
21.3843
24.5227
27.9750
31.7725
35.9497
40.5447
45.5992
51.1591
57.2750
64.0025
71.4027
79.5430
88.4973
98.3471
109.1818
121.0999
134.2099
148.6309
164.4940
181.9434
201.1378
222.2515
245.4767
271.0244
442.5926
718.9048
1163.91
1880.59
3034.82
4893.71
7887.47
12709
20474
32980
53120
85557
94113
1.10000
0.57619
0.40211
0.31547
0.26380
0.22961
0.20541
0.18744
0.17364
0.16275
0.15396
0.14676
0.14078
0.13575
0.13147
0.12782
0.12466
0.12193
0.11955
0.11746
0.11562
0.11401
0.11257
0.11130
0.11017
0.10916
0.10826
0.10745
0.10673
0.10608
0.10550
0.10497
0.10450
0.10407
0.10369
0.10226
0.10139
0.10086
0.10053
0.10033
0.10020
0.10013
0.10008
0.10005
0.10003
0.10002
0.10001
0.10001
0.10001
0.10001
0.9091
1.7355
2.4869
3.1699
3.7908
4.3553
4.8684
5.3349
5.7590
6.1446
6.4951
6.8137
7.1034
7.3667
7.6061
7.8237
8.0216
8.2014
8.3649
8.5136
8.6487
8.7715
8.8832
8.9847
9.0770
9.1609
9.2372
9.3066
9.3696
9.4269
9.4790
9.5264
9.5694
9.6086
9.6442
9.7791
9.8628
9.9148
9.9471
9.9672
9.9796
9.9873
9.9921
9.9951
9.9970
9.9981
9.9988
9.9989
9.9991
9.9993
0.8264
2.3291
4.3781
6.8618
9.6842
12.7631
16.0287
19.4215
22.8913
26.3963
29.9012
33.3772
36.8005
40.1520
43.4164
46.5819
49.6395
52.5827
55.4069
58.1095
60.6893
63.1462
65.4813
67.6964
69.7940
71.7773
73.6495
75.4146
77.0766
78.6395
80.1078
81.4856
82.7773
83.9872
88.9525
92.4544
94.8889
96.5619
97.7010
98.4705
98.9870
99.3317
99.5606
99.7120
99.8118
99.8773
99.8874
99.9052
99.9202
0.4762
0.9366
1.3812
1.8101
2.2236
2.6216
3.0045
3.3724
3.7255
4.0641
4.3884
4.6988
4.9955
5.2789
5.5493
5.8071
6.0526
6.2861
6.5081
6.7189
6.9189
7.1085
7.2881
7.4580
7.6186
7.7704
7.9137
8.0489
8.1762
8.2962
8.4091
8.5152
8.6149
8.7086
9.0962
9.3740
9.5704
9.7075
9.8023
9.8672
9.9113
9.9410
9.9609
9.9742
9.9831
9.9889
9.9898
9.9914
9.9927
596
Compound Interest Factor Tables
11%
TABLE 16
11%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
85
1.1100
1.2321
1.3676
1.5181
1.6851
1.8704
2.0762
2.3045
2.5580
2.8394
3.1518
3.4985
3.8833
4.3104
4.7846
5.3109
5.8951
6.5436
7.2633
8.0623
8.9492
9.9336
11.0263
12.2392
13.5855
15.0799
16.7386
18.5799
20.6237
22.8923
25.4104
28.2056
31.3082
34.7521
38.5749
65.0009
109.5302
184.5648
311.0025
524.0572
883.0669
1488.02
2507.40
4225.11
7119.56
0.9009
0.8116
0.7312
0.6587
0.5935
0.5346
0.4817
0.4339
0.3909
0.3522
0.3173
0.2858
0.2575
0.2320
0.2090
0.1883
0.1696
0.1528
0.1377
0.1240
0.1117
0.1007
0.0907
0.0817
0.0736
0.0663
0.0597
0.0538
0.0485
0.0437
0.0394
0.0355
0.0319
0.0288
0.0259
0.0154
0.0091
0.0054
0.0032
0.0019
0.0011
0.0007
0.0004
0.0002
0.0001
1.00000
0.47393
0.29921
0.21233
0.16057
0.12638
0.10222
0.08432
0.07060
0.05980
0.05112
0.04403
0.03815
0.03323
0.02907
0.02552
0.02247
0.01984
0.01756
0.01558
0.01384
0.01231
0.01097
0.00979
0.00874
0.00781
0.00699
0.00626
0.00561
0.00502
0.00451
0.00404
0.00363
0.00326
0.00293
0.00172
0.00101
0.00060
0.00035
0.00021
0.00012
0.00007
0.00004
0.00003
0.00002
1.0000
2.1100
3.3421
4.7097
6.2278
7.9129
9.7833
11.8594
14.1640
16.7220
19.5614
22.7132
26.2116
30.0949
34.4054
39.1899
44.5008
50.3959
56.9395
64.2028
72.2651
81.2143
91.1479
102.1742
114.4133
127.9988
143.0786
159.8173
178.3972
199.0209
221.9132
247.3236
275.5292
306.8374
341.5896
581.8261
986.6386
1668.77
2818.20
4755.07
8018.79
13518
22785
38401
64714
1.11000
0.58393
0.40921
0.32233
0.27057
0.23638
0.21222
0.19432
0.18060
0.16980
0.16112
0.15403
0.14815
0.14323
0.13907
0.13552
0.13247
0.12984
0.12756
0.12558
0.12384
0.12231
0.12097
0.11979
0.11874
0.11781
0.11699
0.11626
0.11561
0.11502
0.11451
0.11404
0.11363
0.11326
0.11293
0.11172
0.11101
0.11060
0.11035
0.11021
0.11012
0.11007
0.11004
0.11003
0.11002
0.9009
1.7125
2.4437
3.1024
3.6959
4.2305
4.7122
5.1461
5.5370
5.8892
6.2065
6.4924
6.7499
6.9819
7.1909
7.3792
7.5488
7.7016
7.8393
7.9633
8.0751
8.1757
8.2664
8.3481
8.4217
8.4881
8.5478
8.6016
8.6501
8.6938
8.7331
8.7686
8.8005
8.8293
8.8552
8.9511
9.0079
9.0417
9.0617
9.0736
9.0806
9.0848
9.0873
9.0888
9.0896
0.8116
2.2740
4.2502
6.6240
9.2972
12.1872
15.2246
18.3520
21.5217
24.6945
27.8388
30.9290
33.9449
36.8709
39.6953
42.4095
45.0074
47.4856
49.8423
52.0771
54.1912
56.1864
58.0656
59.8322
61.4900
63.0433
64.4965
65.8542
67.1210
68.3016
69.4007
70.4228
71.3724
72.2538
75.7789
78.1551
79.7341
80.7712
81.4461
81.8819
82.1614
82.3397
82.4529
82.5245
0.4739
0.9306
1.3700
1.7923
2.1976
2.5863
2.9585
3.3144
3.6544
3.9788
4.2879
4.5822
4.8619
5.1275
5.3794
5.6180
5.8439
6.0574
6.2590
6.4491
6.6283
6.7969
6.9555
7.1045
7.2443
7.3754
7.4982
7.6131
7.7206
7.8210
7.9147
8.0021
8.0836
8.1594
8.4659
8.6763
8.8185
8.9135
8.9762
9.0172
9.0438
9.0610
9.0720
9.0790
597
Compound Interest Factor Tables
12%
TABLE 17
12%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
85
1.1200
1.2544
1.4049
1.5735
1.7623
1.9738
2.2107
2.4760
2.7731
3.1058
3.4785
3.8960
4.3635
4.8871
5.4736
6.1304
6.8660
7.6900
8.6128
9.6463
10.8038
12.1003
13.5523
15.1786
17.0001
19.0401
21.3249
23.8839
26.7499
29.9599
33.5551
37.5817
42.0915
47.1425
52.7996
93.0510
163.9876
289.0022
509.3206
897.5969
1581.87
2787.80
4913.06
8658.48
15259
0.8929
0.7972
0.7118
0.6355
0.5674
0.5066
0.4523
0.4039
0.3606
0.3220
0.2875
0.2567
0.2292
0.2046
0.1827
0.1631
0.1456
0.1300
0.1161
0.1037
0.0926
0.0826
0.0738
0.0659
0.0588
0.0525
0.0469
0.0419
0.0374
0.0334
0.0298
0.0266
0.0238
0.0212
0.0189
0.0107
0.0061
0.0035
0.0020
0.0011
0.0006
0.0004
0.0002
0.0001
0.0001
1.00000
0.47170
0.29635
0.20923
0.15741
0.12323
0.09912
0.08130
0.06768
0.05698
0.04842
0.04144
0.03568
0.03087
0.02682
0.02339
0.02046
0.01794
0.01576
0.01388
0.01224
0.01081
0.00956
0.00846
0.00750
0.00665
0.00590
0.00524
0.00466
0.00414
0.00369
0.00328
0.00292
0.00260
0.00232
0.00130
0.0074
0.00042
0.00024
0.00013
0.00008
0.00004
0.00002
0.00001
0.00001
1.0000
2.1200
3.3744
4.7793
6.3528
8.1152
10.0890
12.2997
14.7757
17.5487
20.6546
24.1331
28.0291
32.3926
37.2797
42.7533
48.8837
55.7497
63.4397
72.0524
81.6987
92.5026
104.6029
118.1552
133.3339
150.3339
169.3740
190.6989
214.5828
241.3327
271.2926
304.8477
342.4294
384.5210
431.6635
767.0914
1358.23
2400.02
4236.01
7471.64
13174
23223
40934
72146
1.12000
0.59170
0.41635
0.32923
0.27741
0.24323
0.21912
0.20130
0.18768
0.17698
0.16842
0.16144
0.15568
0.15087
0.14682
0.14339
0.14046
0.13794
0.13576
0.13388
0.13224
0.13081
0.12956
0.12846
0.12750
0.12665
0.12590
0.12524
0.12466
0.12414
0.12369
0.12328
0.12292
0.12260
0.12232
0.12130
0.12074
0.12042
0.12024
0.12013
0.12008
0.12004
0.12002
0.12001
0.12001
0.8929
1.6901
2.4018
3.0373
3.6048
4.1114
4.5638
4.9676
5.3282
5.6502
5.9377
6.1944
6.4235
6.6282
6.8109
6.9740
7.1196
7.2497
7.3658
7.4694
7.5620
7.6446
7.7184
7.7843
7.8431
7.8957
7.9426
7.9844
8.0218
8.0552
8.0850
8.1116
8.1354
8.1566
8.1755
8.2438
8.2825
8.3045
8.3170
8.3240
8.3281
8.3303
8.3316
8.3324
8.3328
0.7972
2.2208
4.1273
6.3970
8.9302
11.6443
14.4714
17.3563
20.2541
23.1288
25.9523
28.7024
31.3624
33.9202
36.3670
38.6973
40.9080
42.9979
44.9676
46.8188
48.5543
50.1776
51.6929
53.1046
54.4177
55.6369
56.7674
57.8141
58.7821
59.6761
60.5010
61.2612
61.9612
62.6052
65.1159
66.7342
67.7624
68.4082
68.8100
69.0581
69.2103
69.3031
69.3594
69.3935
0.4717
0.9246
1.3589
1.7746
2.1720
2.5512
2.9131
3.2574
3.5847
3.8953
4.1897
4.4683
4.7317
4.9803
5.2147
5.4353
5.6427
5.8375
6.0202
6.1913
6.3514
6.5010
6.6406
6.7708
6.8921
7.0049
7.1098
7.2071
7.2974
7.3811
7.4586
7.5302
7.5965
7.6577
7.8988
8.0572
8.1597
8.2251
8.2664
8.2922
8.3082
8.3181
8.3241
8.3278
598
Compound Interest Factor Tables
14%
TABLE 18
14%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
85
1.1400
1.2996
1.4815
1.6890
1.9254
2.1950
2.5023
2.8526
3.2519
3.7072
4.2262
4.8179
5.4924
6.2613
7.1379
8.1372
9.2765
10.5752
12.0557
13.7435
15.6676
17.8610
20.3616
23.2122
26.4619
30.1666
34.3899
39.2045
44.6931
50.9502
58.0832
66.2148
75.4849
86.0528
98.1002
188.8835
363.6791
700.2330
1348.24
2595.92
4998.22
9623.64
18530
35677
68693
0.8772
0.7695
0.6750
0.5921
0.5194
0.4556
0.3996
0.3506
0.3075
0.2697
0.2366
0.2076
0.1821
0.1597
0.1401
0.1229
0.1078
0.0946
0.0829
0.0728
0.0638
0.0560
0.0491
0.0431
0.0378
0.0331
0.0291
0.0255
0.0224
0.0196
0.0172
0.0151
0.0132
0.0116
0.0102
0.0053
0.0027
0.0014
0.0007
0.0004
0.0002
0.0001
0.0001
1.00000
0.46729
0.29073
0.20320
0.15128
0.11716
0.09319
0.07557
0.06217
0.05171
0.04339
0.03667
0.03116
0.02661
0.02281
0.01962
0.01692
0.01462
0.01266
0.01099
0.00954
0.00830
0.00723
0.00630
0.00550
0.00480
0.00419
0.00366
0.00320
0.00280
0.00245
0.00215
0.00188
0.00165
0.00144
0.00075
0.00039
0.00020
0.00010
0.00005
0.00003
0.00001
0.00001
1.0000
2.1400
3.4396
4.9211
6.6101
8.5355
10.7305
13.2328
16.0853
19.3373
23.0445
27.2707
32.0887
37.5811
43.8424
50.9804
59.1176
68.3941
78.9692
91.0249
104.7684
120.4360
138.2970
158.6586
181.8708
208.3327
238.4993
272.8892
312.0937
356.7868
407.7370
465.8202
532.0350
607.5199
693.5727
1342.03
2590.56
4994.52
9623.13
18535
35694
68733
1.14000
0.60729
0.43073
0.34320
0.29128
0.25716
0.23319
0.21557
0.20217
0.19171
0.18339
0.17667
0.17116
0.16661
0.16281
0.15962
0.15692
0.15462
0.15266
0.15099
0.14954
0.14830
0.14723
0.14630
0.14550
0.14480
0.14419
0.14366
0.14320
0.14280
0.14245
0.14215
0.14188
0.14165
0.14144
0.14075
0.14039
0.14020
0.14010
0.14005
0.14003
0.14001
0.14001
0.14000
0.14000
0.8772
1.6467
2.3216
2.9137
3.4331
3.8887
4.2883
4.6389
4.9464
5.2161
5.4527
5.6603
5.8424
6.0021
6.1422
6.2651
6.3729
6.4674
6.5504
6.6231
6.6870
6.7429
6.7921
6.8351
6.8729
6.9061
6.9352
6.9607
6.9830
7.0027
7.0199
7.0350
7.0482
7.0599
7.0700
7.1050
7.1232
7.1327
7.1376
7.1401
7.1414
7.1421
7.1425
7.1427
7.1428
0.7695
2.1194
3.8957
5.9731
8.2511
10.6489
13.1028
15.5629
17.9906
20.3567
22.6399
24.8247
26.9009
28.8623
30.7057
32.4305
34.0380
35.5311
36.9135
38.1901
39.3658
40.4463
41.4371
42.3441
43.1728
43.9289
44.6176
45.2441
45.8132
46.3297
46.7979
47.2218
47.6053
47.9519
49.2376
49.9963
50.4375
50.6912
50.8357
50.9173
50.9632
50.9887
51.0030
51.0108
0.4673
0.9129
1.3370
1.7399
2.1218
2.4832
2.8246
3.1463
3.4490
3.7333
3.9998
4.2491
4.4819
4.6990
4.9011
5.0888
5.2630
5.4243
5.5734
5.7111
5.8381
5.9549
6.0624
6.1610
6.2514
6.3342
6.4100
6.4791
6.5423
6.5998
6.6522
6.6998
6.7431
6.7824
6.9300
7.0188
7.0714
7.1020
7.1197
7.1298
7.1356
7.1388
7.1406
7.1416
599
Compound Interest Factor Tables
15%
TABLE 19
Single Payments
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
70
75
80
85
15%
Discrete Cash Flow: Compound Interest Factors
Uniform Series Payments
Arithmetic Gradients
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1.1500
1.3225
1.5209
1.7490
2.0114
2.3131
2.6600
3.0590
3.5179
4.0456
4.6524
5.3503
6.1528
7.0757
8.1371
9.3576
10.7613
12.3755
14.2318
16.3665
18.8215
21.6447
24.8915
28.6252
32.9190
37.8568
43.5353
50.0656
57.5755
66.2118
76.1435
87.5651
100.6998
115.8048
133.1755
267.8635
538.7693
1083.66
2179.62
4384.00
8817.79
17736
35673
71751
0.8696
0.7561
0.6575
0.5718
0.4972
0.4323
0.3759
0.3269
0.2843
0.2472
0.2149
0.1869
0.1625
0.1413
0.1229
0.1069
0.0929
0.0808
0.0703
0.0611
0.0531
0.0462
0.0402
0.0349
0.0304
0.0264
0.0230
0.0200
0.0174
0.0151
0.0131
0.0114
0.0099
0.0086
0.0075
0.0037
0.0019
0.0009
0.0005
0.0002
0.0001
0.0001
1.00000
0.46512
0.28798
0.20027
0.14832
0.11424
0.09036
0.07285
0.05957
0.04925
0.04107
0.03448
0.02911
0.02469
0.02102
0.01795
0.01537
0.01319
0.01134
0.00976
0.00842
0.00727
0.00628
0.00543
0.00470
0.00407
0.00353
0.00306
0.00265
0.00230
0.00200
0.00173
0.00150
0.00131
0.00113
0.00056
0.00028
0.00014
0.00007
0.00003
0.00002
0.00001
1.0000
2.1500
3.4725
4.9934
6.7424
8.7537
11.0668
13.7268
16.7858
20.3037
24.3493
29.0017
34.3519
40.5047
47.5804
55.7175
65.0751
75.8364
88.2118
102.4436
118.8101
137.6316
159.2764
184.1678
212.7930
245.7120
283.5688
327.1041
377.1697
434.7451
500.9569
577.1005
664.6655
765.3654
881.1702
1779.09
3585.13
7217.72
14524
29220
58779
1.15000
0.61512
0.43798
0.35027
0.29832
0.26424
0.24036
0.22285
0.20957
0.19925
0.19107
0.18448
0.17911
0.17469
0.17102
0.16795
0.16537
0.16319
0.16134
0.15976
0.15842
0.15727
0.15628
0.15543
0.15470
0.15407
0.15353
0.15306
0.15265
0.15230
0.15200
0.15173
0.15150
0.15131
0.15113
0.15056
0.15028
0.15014
0.15007
0.15003
0.15002
0.15001
0.15000
0.15000
0.15000
0.8696
1.6257
2.2832
2.8550
3.3522
3.7845
4.1604
4.4873
4.7716
5.0188
5.2337
5.4206
5.5831
5.7245
5.8474
5.9542
6.0472
6.1280
6.1982
6.2593
6.3125
6.3587
6.3988
6.4338
6.4641
6.4906
6.5135
6.5335
6.5509
6.5660
6.5791
6.5905
6.6005
6.6091
6.6166
6.6418
6.6543
6.6605
6.6636
6.6651
6.6659
6.6663
6.6665
6.6666
6.6666
0.7561
2.0712
3.7864
5.7751
7.9368
10.1924
12.4807
14.7548
16.9795
19.1289
21.1849
23.1352
24.9725
26.6930
28.2960
29.7828
31.1565
32.4213
33.5822
34.6448
35.6150
36.4988
37.3023
38.0314
38.6918
39.2890
39.8283
40.3146
40.7526
41.1466
41.5006
41.8184
42.1033
42.3586
43.2830
43.8051
44.0958
44.2558
44.3431
44.3903
44.4156
44.4292
44.4364
44.4402
0.4651
0.9071
1.3263
1.7228
2.0972
2.4498
2.7813
3.0922
3.3832
3.6549
3.9082
4.1438
4.3624
4.5650
4.7522
4.9251
5.0843
5.2307
5.3651
5.4883
5.6010
5.7040
5.7979
5.8834
5.9612
6.0319
6.0960
6.1541
6.2066
6.2541
6.2970
6.3357
6.3705
6.4019
6.5168
6.5830
6.6205
6.6414
6.6530
6.6593
6.6627
6.6646
6.6656
6.6661
600
Compound Interest Factor Tables
16%
TABLE 20
16%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
22
24
26
28
30
32
34
35
36
38
40
45
50
55
60
1.1600
1.3456
1.5609
1.8106
2.1003
2.4364
2.8262
3.2784
3.8030
4.4114
5.1173
5.9360
6.8858
7.9875
9.2655
10.7480
12.4677
14.4625
16.7765
19.4608
26.1864
35.2364
47.4141
63.8004
85.8499
115.5196
155.4432
180.3141
209.1643
281.4515
378.7212
795.4438
1670.70
3509.05
7370.20
0.8621
0.7432
0.6407
0.5523
0.4761
0.4104
0.3538
0.3050
0.2630
0.2267
0.1954
0.1685
0.1452
0.1252
0.1079
0.0930
0.0802
0.0691
0.0596
0.0514
0.0382
0.0284
0.0211
0.0157
0.0116
0.0087
0.0064
0.0055
0.0048
0.0036
0.0026
0.0013
0.0006
0.0003
0.0001
1.00000
0.46296
0.28526
0.19738
0.14541
0.11139
0.08761
0.07022
0.05708
0.04690
0.03886
0.03241
0.02718
0.02290
0.01936
0.01641
0.01395
0.01188
0.01014
0.00867
0.00635
0.00467
0.00345
0.00255
0.00189
0.00140
0.00104
0.00089
0.00077
0.00057
0.00042
0.00020
0.00010
0.00005
0.00002
1.0000
2.1600
3.5056
5.0665
6.8771
8.9775
11.4139
14.2401
17.5185
21.3215
25.7329
30.8502
36.7862
43.6720
51.6595
60.9250
71.6730
84.1407
98.6032
115.3797
157.4150
213.9776
290.0883
392.5028
530.3117
715.7475
965.2698
1120.71
1301.03
1752.82
2360.76
4965.27
10436
21925
46058
1.16000
0.62296
0.44526
0.35738
0.30541
0.27139
0.24761
0.23022
0.21708
0.20690
0.19886
0.19241
0.18718
0.18290
0.17936
0.17641
0.17395
0.17188
0.17014
0.16867
0.16635
0.16467
0.16345
0.16255
0.16189
0.16140
0.16104
0.16089
0.16077
0.16057
0.16042
0.16020
0.16010
0.16005
0.16002
0.8621
1.6052
2.2459
2.7982
3.2743
3.6847
4.0386
4.3436
4.6065
4.8332
5.0286
5.1971
5.3423
5.4675
5.5755
5.6685
5.7487
5.8178
5.8775
5.9288
6.0113
6.0726
6.1182
6.1520
6.1772
6.1959
6.2098
6.2153
6.2201
6.2278
6.2335
6.2421
6.2463
6.2482
6.2492
0.7432
2.0245
3.6814
5.5858
7.6380
9.7610
11.8962
13.9998
16.0399
17.9941
19.8472
21.5899
23.2175
24.7284
26.1241
27.4074
28.5828
29.6557
30.6321
32.3200
33.6970
34.8114
35.7073
36.4234
36.9930
37.4441
37.6327
37.8000
38.0799
38.2992
38.6598
38.8521
38.9534
39.0063
0.4630
0.9014
1.3156
1.7060
2.0729
2.4169
2.7388
3.0391
3.3187
3.5783
3.8189
4.0413
4.2464
4.4352
4.6086
4.7676
4.9130
5.0457
5.1666
5.3765
5.5490
5.6898
5.8041
5.8964
5.9706
6.0299
6.0548
6.0771
6.1145
6.1441
6.1934
6.2201
6.2343
6.2419
601
Compound Interest Factor Tables
18%
TABLE 21
18%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
22
24
26
28
30
32
34
35
36
38
40
45
50
55
60
1.1800
1.3924
1.6430
1.9388
2.2878
2.6996
3.1855
3.7589
4.4355
5.2338
6.1759
7.2876
8.5994
10.1472
11.9737
14.1290
16.6722
19.6733
23.2144
27.3930
38.1421
53.1090
73.9490
102.9666
143.3706
199.6293
277.9638
327.9973
387.0368
538.9100
750.3783
1716.68
3927.36
8984.84
20555
0.8475
0.7182
0.6086
0.5158
0.4371
0.3704
0.3139
0.2660
0.2255
0.1911
0.1619
0.1372
0.1163
0.0985
0.0835
0.0708
0.0600
0.0508
0.0431
0.0365
0.0262
0.0188
0.0135
0.0097
0.0070
0.0050
0.0036
0.0030
0.0026
0.0019
0.0013
0.0006
0.0003
0.0001
1.00000
0.45872
0.27992
0.19174
0.13978
0.10591
0.08236
0.06524
0.05239
0.04251
0.03478
0.02863
0.02369
0.01968
0.01640
0.01371
0.01149
0.00964
0.00810
0.00682
0.00485
0.00345
0.00247
0.00177
0.00126
0.00091
0.00065
0.00055
0.00047
0.00033
0.00024
0.00010
0.00005
0.00002
1.0000
2.1800
3.5724
5.2154
7.1542
9.4420
12.1415
15.3270
19.0859
23.5213
28.7551
34.9311
42.2187
50.8180
60.9653
72.9390
87.0680
103.7403
123.4135
146.6280
206.3448
289.4945
405.2721
566.4809
790.9480
1103.50
1538.69
1816.65
2144.65
2988.39
4163.21
9531.58
21813
49910
114190
1.18000
0.63872
0.45992
0.37174
0.31978
0.28591
0.26236
0.24524
0.23239
0.22251
0.21478
0.20863
0.20369
0.19968
0.19640
0.19371
0.19149
0.18964
0.18810
0.18682
0.18485
0.18345
0.18247
0.18177
0.18126
0.18091
0.18065
0.18055
0.18047
0.18033
0.18024
0.18010
0.18005
0.18002
0.18001
0.8475
1.5656
2.1743
2.6901
3.1272
3.4976
3.8115
4.0776
4.3030
4.4941
4.6560
4.7932
4.9095
5.0081
5.0916
5.1624
5.2223
5.2732
5.3162
5.3527
5.4099
5.4509
5.4804
5.5016
5.5168
5.5277
5.5356
5.5386
5.5412
5.5452
5.5482
5.5523
5.5541
5.5549
5.5553
0.7182
1.9354
3.4828
5.2312
7.0834
8.9670
10.8292
12.6329
14.3525
15.9716
17.4811
18.8765
20.1576
21.3269
22.3885
23.3482
24.2123
24.9877
25.6813
26.8506
27.7725
28.4935
29.0537
29.4864
29.8191
30.0736
30.1773
30.2677
30.4152
30.5269
30.7006
30.7856
30.8268
30.8465
0.4587
0.8902
1.2947
1.6728
2.0252
2.3526
2.6558
2.9358
3.1936
3.4303
3.6470
3.8449
4.0250
4.1887
4.3369
4.4708
4.5916
4.7003
4.7978
4.9632
5.0950
5.1991
5.2810
5.3448
5.3945
5.4328
5.4485
5.4623
5.4849
5.5022
5.5293
5.5428
5.5494
5.5526
602
Compound Interest Factor Tables
20%
TABLE 22
20%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
22
24
26
28
30
32
34
35
36
38
40
45
50
55
1.2000
1.4400
1.7280
2.0736
2.4883
2.9860
3.5832
4.2998
5.1598
6.1917
7.4301
8.9161
10.6993
12.8392
15.4070
18.4884
22.1861
26.6233
31.9480
38.3376
55.2061
79.4968
114.4755
164.8447
237.3763
341.8219
492.2235
590.6682
708.8019
1020.67
1469.77
3657.26
9100.44
22645
0.8333
0.6944
0.5787
0.4823
0.4019
0.3349
0.2791
0.2326
0.1938
0.1615
0.1346
0.1122
0.0935
0.0779
0.0649
0.0541
0.0451
0.0376
0.0313
0.0261
0.0181
0.0126
0.0087
0.0061
0.0042
0.0029
0.0020
0.0017
0.0014
0.0010
0.0007
0.0003
0.0001
1.00000
0.45455
0.27473
0.18629
0.13438
0.10071
0.07742
0.06061
0.04808
0.03852
0.03110
0.02526
0.02062
0.01689
0.01388
0.01144
0.00944
0.00781
0.00646
0.00536
0.00369
0.00255
0.00176
0.00122
0.00085
0.00059
0.00041
0.00034
0.00028
0.00020
0.00014
0.00005
0.00002
0.00001
1.0000
2.2000
3.6400
5.3680
7.4416
9.9299
12.9159
16.4991
20.7989
25.9587
32.1504
39.5805
48.4966
59.1959
72.0351
87.4421
105.9306
128.1167
154.7400
186.6880
271.0307
392.4842
567.3773
819.2233
1181.88
1704.11
2456.12
2948.34
3539.01
5098.37
7343.86
18281
45497
1.20000
0.65455
0.47473
0.38629
0.33438
0.30071
0.27742
0.26061
0.24808
0.23852
0.23110
0.22526
0.22062
0.21689
0.21388
0.21144
0.20944
0.20781
0.20646
0.20536
0.20369
0.20255
0.20176
0.20122
0.20085
0.20059
0.20041
0.20034
0.20028
0.20020
0.20014
0.20005
0.20002
0.20001
0.8333
1.5278
2.1065
2.5887
2.9906
3.3255
3.6046
3.8372
4.0310
4.1925
4.3271
4.4392
4.5327
4.6106
4.6755
4.7296
4.7746
4.8122
4.8435
4.8696
4.9094
4.9371
4.9563
4.9697
4.9789
4.9854
4.9898
4.9915
4.9929
4.9951
4.9966
4.9986
4.9995
4.9998
0.6944
1.8519
3.2986
4.9061
6.5806
8.2551
9.8831
11.4335
12.8871
14.2330
15.4667
16.5883
17.6008
18.5095
19.3208
20.0419
20.6805
21.2439
21.7395
22.5546
23.1760
23.6460
23.9991
24.2628
24.4588
24.6038
24.6614
24.7108
24.7894
24.8469
24.9316
24.9698
24.9868
0.4545
0.8791
1.2742
1.6405
1.9788
2.2902
2.5756
2.8364
3.0739
3.2893
3.4841
3.6597
3.8175
3.9588
4.0851
4.1976
4.2975
4.3861
4.4643
4.5941
4.6943
4.7709
4.8291
4.8731
4.9061
4.9308
4.9406
4.9491
4.9627
4.9728
4.9877
4.9945
4.9976
603
Compound Interest Factor Tables
22%
TABLE 23
22%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
22
24
26
28
30
32
34
35
36
38
40
45
50
55
1.2200
1.4884
1.8158
2.2153
2.7027
3.2973
4.0227
4.9077
5.9874
7.3046
8.9117
10.8722
13.2641
16.1822
19.7423
24.0856
29.3844
35.8490
43.7358
53.3576
79.4175
118.2050
175.9364
261.8637
389.7579
580.1156
863.4441
1053.40
1285.15
1912.82
2847.04
7694.71
20797
56207
0.8197
0.6719
0.5507
0.4514
0.3700
0.3033
0.2486
0.2038
0.1670
0.1369
0.1122
0.0920
0.0754
0.0618
0.0507
0.0415
0.0340
0.0279
0.0229
0.0187
0.0126
0.0085
0.0057
0.0038
0.0026
0.0017
0.0012
0.0009
0.0008
0.0005
0.0004
0.0001
1.00000
0.45045
0.26966
0.18102
0.12921
0.09576
0.07278
0.05630
0.04411
0.03489
0.02781
0.02228
0.01794
0.01449
0.01174
0.00953
0.00775
0.00631
0.00515
0.00420
0.00281
0.00188
0.00126
0.00084
0.00057
0.00038
0.00026
0.00021
0.00017
0.00012
0.00008
0.00003
0.00001
1.0000
2.2200
3.7084
5.5242
7.7396
10.4423
13.7396
17.7623
22.6700
28.6574
35.9620
44.8737
55.7459
69.0100
85.1922
104.9345
129.0201
158.4045
194.2535
237.9893
356.4432
532.7501
795.1653
1185.74
1767.08
2632.34
3920.20
4783.64
5837.05
8690.08
12937
34971
94525
1.22000
0.67045
0.48966
0.40102
0.34921
0.31576
0.29278
0.27630
0.26411
0.25489
0.24781
0.24228
0.23794
0.23449
0.23174
0.22953
0.22775
0.22631
0.22515
0.22420
0.22281
0.22188
0.22126
0.22084
0.22057
0.22038
0.22026
0.22021
0.22017
0.22012
0.22008
0.22003
0.22001
0.22000
0.8197
1.4915
2.0422
2.4936
2.8636
3.1669
3.4155
3.6193
3.7863
3.9232
4.0354
4.1274
4.2028
4.2646
4.3152
4.3567
4.3908
4.4187
4.4415
4.4603
4.4882
4.5070
4.5196
4.5281
4.5338
4.5376
4.5402
4.5411
4.5419
4.5431
4.5439
4.5449
4.5452
4.5454
0.6719
1.7733
3.1275
4.6075
6.1239
7.6154
9.0417
10.3779
11.6100
12.7321
13.7438
14.6485
15.4519
16.1610
16.7838
17.3283
17.8025
18.2141
18.5702
19.1418
19.5635
19.8720
20.0962
20.2583
20.3748
20.4582
20.4905
20.5178
20.5601
20.5900
20.6319
20.6492
20.6563
0.4505
0.8683
1.2542
1.6090
1.9337
2.2297
2.4982
2.7409
2.9593
3.1551
3.3299
3.4855
3.6233
3.7451
3.8524
3.9465
4.0289
4.1009
4.1635
4.2649
4.3407
4.3968
4.4381
4.4683
4.4902
4.5060
4.5122
4.5174
4.5256
4.5314
4.5396
4.5431
4.5445
604
Compound Interest Factor Tables
24%
TABLE 24
Single Payments
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
22
24
26
28
30
32
34
35
36
38
40
45
50
55
24%
Discrete Cash Flow: Compound Interest Factors
Uniform Series Payments
Arithmetic Gradients
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1.2400
1.5376
1.9066
2.3642
2.9316
3.6352
4.5077
5.5895
6.9310
8.5944
10.6571
13.2148
16.3863
20.3191
25.1956
31.2426
38.7408
48.0386
59.5679
73.8641
113.5735
174.6306
268.5121
412.8642
634.8199
976.0991
1500.85
1861.05
2307.71
3548.33
5455.91
15995
46890
0.8065
0.6504
0.5245
0.4230
0.3411
0.2751
0.2218
0.1789
0.1443
0.1164
0.0938
0.0757
0.0610
0.0492
0.0397
0.0320
0.0258
0.0208
0.0168
0.0135
0.0088
0.0057
0.0037
0.0024
0.0016
0.0010
0.0007
0.0005
0.0004
0.0003
0.0002
0.0001
1.00000
0.44643
0.26472
0.17593
0.12425
0.09107
0.06842
0.05229
0.04047
0.03160
0.02485
0.01965
0.01560
0.01242
0.00992
0.00794
0.00636
0.00510
0.00410
0.00329
0.00213
0.00138
0.00090
0.00058
0.00038
0.00025
0.00016
0.00013
0.00010
0.00007
0.00004
0.00002
0.00001
1.0000
2.2400
3.7776
5.6842
8.0484
10.9801
14.6153
19.1229
24.7125
31.6434
40.2379
50.8950
64.1097
80.4961
100.8151
126.0108
157.2534
195.9942
244.0328
303.6006
469.0563
723.4610
1114.63
1716.10
2640.92
4062.91
6249.38
7750.23
9611.28
14781
22729
66640
1.24000
0.68643
0.50472
0.41593
0.36425
0.33107
0.30842
0.29229
0.28047
0.27160
0.26485
0.25965
0.25560
0.25242
0.24992
0.24794
0.24636
0.24510
0.24410
0.24329
0.24213
0.24138
0.24090
0.24058
0.24038
0.24025
0.24016
0.24013
0.24010
0.24007
0.24004
0.24002
0.24001
0.24000
0.8065
1.4568
1.9813
2.4043
2.7454
3.0205
3.2423
3.4212
3.5655
3.6819
3.7757
3.8514
3.9124
3.9616
4.0013
4.0333
4.0591
4.0799
4.0967
4.1103
4.1300
4.1428
4.1511
4.1566
4.1601
4.1624
4.1639
4.1664
4.1649
4.1655
4.1659
4.1664
4.1666
4.1666
0.6504
1.6993
2.9683
4.3327
5.7081
7.0392
8.2915
9.4458
10.4930
11.4313
12.2637
12.9960
13.6358
14.1915
14.6716
15.0846
15.4385
15.7406
15.9979
16.4011
16.6891
16.8930
17.0365
17.1369
17.2067
17.2552
17.2734
17.2886
17.3116
17.3274
17.3483
17.3563
17.3593
0.4464
0.8577
1.2346
1.5782
1.8898
2.1710
2.4236
2.6492
2.8499
3.0276
3.1843
3.3218
3.4420
3.5467
3.6376
3.7162
3.7840
3.8423
3.8922
3.9712
4.0284
4.0695
4.0987
4.1193
4.1338
4.1440
4.1479
4.1511
4.1560
4.1593
4.1639
4.1653
4.1663
605
Compound Interest Factor Tables
25%
TABLE 25
Single Payments
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
22
24
26
28
30
32
34
35
36
38
40
45
50
55
25%
Discrete Cash Flow: Compound Interest Factors
Uniform Series Payments
Arithmetic Gradients
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1.2500
1.5625
1.9531
2.4414
3.0518
3.8147
4.7684
5.9605
7.4506
9.3132
11.6415
14.5519
18.1899
22.7374
28.4217
35.5271
44.4089
55.5112
69.3889
86.7362
135.5253
211.7582
330.8722
516.9879
807.7936
1262.18
1972.15
2465.19
3081.49
4814.82
7523.16
22959
70065
0.8000
0.6400
0.5120
0.4096
0.3277
0.2621
0.2097
0.1678
0.1342
0.1074
0.0859
0.0687
0.0550
0.0440
0.0352
0.0281
0.0225
0.0180
0.0144
0.0115
0.0074
0.0047
0.0030
0.0019
0.0012
0.0008
0.0005
0.0004
0.0003
0.0002
0.0001
1.00000
0.44444
0.26230
0.17344
0.12185
0.08882
0.06634
0.05040
0.03876
0.03007
0.02349
0.01845
0.01454
0.01150
0.00912
0.00724
0.00576
0.00459
0.00366
0.00292
0.00186
0.00119
0.00076
0.00048
0.00031
0.00020
0.00013
0.00010
0.00008
0.00005
0.00003
0.00001
1.0000
2.2500
3.8125
5.7656
8.2070
11.2588
15.0735
19.8419
25.8023
33.2529
42.5661
54.2077
68.7596
86.9495
109.6868
138.1085
173.6357
218.0446
273.5558
342.9447
538.1011
843.0329
1319.49
2063.95
3227.17
5044.71
7884.61
9856.76
12322
19255
30089
91831
1.25000
0.69444
0.51230
0.42344
0.37185
0.33882
0.31634
0.30040
0.28876
0.28007
0.27349
0.26845
0.26454
0.26150
0.25912
0.25724
0.25576
0.25459
0.25366
0.25292
0.25186
0.25119
0.25076
0.25048
0.25031
0.25020
0.25013
.025010
0.25008
0.25005
0.25003
0.25001
0.25000
0.25000
0.8000
1.4400
1.9520
2.3616
2.6893
2.9514
3.1611
3.3289
3.4631
3.5705
3.6564
3.7251
3.7801
3.8241
3.8593
3.8874
3.9099
3.9279
3.9424
3.9539
3.9705
3.9811
3.9879
3.9923
3.9950
3.9968
3.9980
3.9984
3.9987
3.9992
3.9995
3.9998
3.9999
4.0000
0.6400
1.6640
2.8928
4.2035
5.5142
6.7725
7.9469
9.0207
9.9870
10.8460
11.6020
12.2617
12.8334
13.3260
13.7482
14.1085
14.4147
14.6741
14.8932
15.2326
15.4711
15.6373
15.7524
15.8316
15.8859
15.9229
15.9367
15.9481
15.9651
15.9766
15.9915
15.9969
15.9989
0.4444
0.8525
1.2249
1.5631
1.8683
2.1424
2.3872
2.6048
2.7971
2.9663
3.1145
3.2437
3.3559
3.4530
3.5366
3.6084
3.6698
3.7222
3.7667
3.8365
3.8861
3.9212
3.9457
3.9628
3.9746
3.9828
3.9858
3.9883
3.9921
3.9947
3.9980
3.9993
3.9997
606
Compound Interest Factor Tables
30%
TABLE 26
30%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
22
24
25
26
28
30
32
34
35
1.3000
1.6900
2.1970
2.8561
3.7129
4.8268
6.2749
8.1573
10.6045
13.7858
17.9216
23.2981
30.2875
39.3738
51.1859
66.5417
86.5042
112.4554
146.1920
190.0496
321.1839
542.8008
705.6410
917.3333
1550.29
2620.00
4427.79
7482.97
9727.86
0.7692
0.5917
0.4552
0.3501
0.2693
0.2072
0.1594
0.1226
0.0943
0.0725
0.0558
0.0429
0.0330
0.0254
0.0195
0.0150
0.0116
0.0089
0.0068
0.0053
0.0031
0.0018
0.0014
0.0011
0.0006
0.0004
0.0002
0.0001
0.0001
1.00000
0.43478
0.25063
0.16163
0.11058
0.07839
0.05687
0.04192
0.03124
0.02346
0.01773
0.01345
0.01024
0.00782
0.00598
0.00458
0.00351
0.00269
0.00207
0.00159
0.00094
0.00055
0.00043
0.00033
0.00019
0.00011
0.00007
0.00004
0.00003
1.0000
2.3000
3.9900
6.1870
9.0431
12.7560
17.5828
23.8577
32.0150
42.6195
56.4053
74.3270
97.6250
127.9125
167.2863
218.4722
285.0139
371.5180
483.9734
630.1655
1067.28
1806.00
2348.80
3054.44
5164.31
8729.99
14756
24940
32423
1.30000
0.73478
0.55063
0.46163
0.41058
0.37839
0.35687
0.34192
0.33124
0.32346
0.31773
0.31345
0.31024
0.30782
0.30598
0.30458
0.30351
0.30269
0.30207
0.30159
0.30094
0.30055
0.30043
0.30033
0.30019
0.30011
0.30007
0.30004
0.30003
0.7692
1.3609
1.8161
2.1662
2.4356
2.6427
2.8021
2.9247
3.0190
3.0915
3.1473
3.1903
3.2233
3.2487
3.2682
3.2832
3.2948
3.3037
3.3105
3.3158
3.3230
3.3272
3.3286
3.3297
3.3312
3.3321
3.3326
3.3329
3.3330
0.5917
1.5020
2.5524
3.6297
4.6656
5.6218
6.4800
7.2343
7.8872
8.4452
8.9173
9.3135
9.6437
9.9172
10.1426
10.3276
10.4788
10.6019
10.7019
10.8482
10.9433
10.9773
11.0045
11.0437
11.0687
11.0845
11.0945
11.0980
0.4348
0.8271
1.1783
1.4903
1.7654
2.0063
2.2156
2.3963
2.5512
2.6833
2.7952
2.8895
2.9685
3.0344
3.0892
3.1345
3.1718
3.2025
3.2275
3.2646
3.2890
3.2979
3.3050
3.3153
3.3219
3.3261
3.3288
3.3297
607
Compound Interest Factor Tables
35%
TABLE 27
35%
Discrete Cash Flow: Compound Interest Factors
Single Payments
Uniform Series Payments
Arithmetic Gradients
n
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
22
24
25
26
28
30
32
34
35
1.3500
1.8225
2.4604
3.3215
4.4840
6.0534
8.1722
11.0324
14.8937
20.1066
27.1439
36.6442
49.4697
66.7841
90.1585
121.7139
164.3138
221.8236
299.4619
404.2736
736.7886
1342.80
1812.78
2447.25
4460.11
8128.55
14814
26999
36449
0.7407
0.5487
0.4064
0.3011
0.2230
0.1652
0.1224
0.0906
0.0671
0.0497
0.0368
0.0273
0.0202
0.0150
0.0111
0.0082
0.0061
0.0045
0.0033
0.0025
0.0014
0.0007
0.0006
0.0004
0.0002
0.0001
0.0001
1.00000
0.42553
0.23966
0.15076
0.10046
0.06926
0.04880
0.03489
0.02519
0.01832
0.01339
0.00982
0.00722
0.00532
0.00393
0.00290
0.00214
0.00158
0.00117
0.00087
0.00048
0.00026
0.00019
0.00014
0.00008
0.00004
0.00002
0.00001
0.00001
1.0000
2.3500
4.1725
6.6329
9.9544
14.4384
20.4919
28.6640
39.6964
54.5902
74.6967
101.8406
138.4848
187.9544
254.7385
344.8970
466.6109
630.9247
852.7483
1152.21
2102.25
3833.71
5176.50
6989.28
12740
23222
42324
77137
1.35000
0.77553
0.58966
0.50076
0.45046
0.41926
0.39880
0.38489
0.37519
0.36832
0.36339
0.35982
0.35722
0.35532
0.35393
0.35290
0.35214
0.35158
0.35117
0.35087
0.35048
0.35026
0.35019
0.35014
0.35008
0.35004
0.35002
0.35001
0.35001
0.7407
1.2894
1.6959
1.9969
2.2200
2.3852
2.5075
2.5982
2.6653
2.7150
2.7519
2.7792
2.7994
2.8144
2.8255
2.8337
2.8398
2.8443
2.8476
2.8501
2.8533
2.8550
2.8556
2.8560
2.8565
2.8568
2.8569
2.8570
2.8571
0.5487
1.3616
2.2648
3.1568
3.9828
4.7170
5.3515
5.8886
6.3363
6.7047
7.0049
7.2474
7.4421
7.5974
7.7206
7.8180
7.8946
7.9547
8.0017
8.0669
8.1061
8.1194
8.1296
8.1435
8.1517
8.1565
8.1594
8.1603
0.4255
0.8029
1.1341
1.4220
1.6698
1.8811
2.0597
2.2094
2.3338
2.4364
2.5205
2.5889
2.6443
2.6889
2.7246
2.7530
2.7756
2.7935
2.8075
2.8272
2.8393
2.8433
2.8465
2.8509
2.8535
2.8550
2.8559
2.8562
608
Compound Interest Factor Tables
40%
TABLE 28
Single Payments
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
22
24
25
26
28
30
32
34
35
40%
Discrete Cash Flow: Compound Interest Factors
Uniform Series Payments
Arithmetic Gradients
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1.4000
1.9600
2.7440
3.8416
5.3782
7.5295
10.5414
14.7579
20.6610
28.9255
40.4957
56.6939
79.3715
111.1201
155.5681
217.7953
304.9135
426.8789
597.6304
836.6826
1639.90
3214.20
4499.88
6299.83
12348
24201
47435
92972
0.7143
0.5102
0.3644
0.2603
0.1859
0.1328
0.0949
0.0678
0.0484
0.0346
0.0247
0.0176
0.0126
0.0090
0.0064
0.0046
0.0033
0.0023
0.0017
0.0012
0.0006
0.0003
0.0002
0.0002
0.0001
1.00000
0.41667
0.22936
0.14077
0.09136
0.06126
0.04192
0.02907
0.02034
0.01432
0.01013
0.00718
0.00510
0.00363
0.00259
0.00185
0.00132
0.00094
0.00067
0.00048
0.00024
0.00012
0.00009
0.00006
0.00003
0.00002
0.00001
1.0000
2.4000
4.3600
7.1040
10.9456
16.3238
23.8534
34.3947
49.1526
69.8137
98.7391
139.2348
195.9287
275.3002
386.4202
541.9883
759.7837
1064.70
1491.58
2089.21
4097.24
8033.00
11247
15747
30867
60501
1.40000
0.81667
0.62936
0.54077
0.49136
0.46126
0.44192
0.42907
0.42034
0.41432
0.41013
0.40718
0.40510
0.40363
0.40259
0.40185
0.40132
0.40094
0.40067
0.40048
0.40024
0.40012
0.40009
0.40006
0.40003
0.40002
0.40001
0.40000
0.40000
0.7143
1.2245
1.5889
1.8492
2.0352
2.1680
2.2628
2.3306
2.3790
2.4136
2.4383
2.4559
2.4685
2.4775
2.4839
2.4885
2.4918
2.4941
2.4958
2.4970
2.4985
2.4992
2.4994
2.4996
2.4998
2.4999
2.4999
2.5000
2.5000
0.5102
1.2391
2.0200
2.7637
3.4278
3.9970
4.4713
4.8585
5.1696
5.4166
5.6106
5.7618
5.8788
5.9688
6.0376
6.0901
6.1299
6.1601
6.1828
6.2127
6.2294
6.2347
6.2387
6.2438
6.2466
6.2482
6.2490
6.2493
0.4167
0.7798
1.0923
1.3580
1.5811
1.7664
1.9185
2.0422
2.1419
2.2215
2.2845
2.3341
2.3729
2.4030
2.4262
2.4441
2.4577
2.4682
2.4761
2.4866
2.4925
2.4944
2.4959
2.4977
2.4988
2.4993
2.4996
2.4997
609
Compound Interest Factor Tables
50%
TABLE 29
Single Payments
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
22
24
25
26
28
30
32
34
35
50%
Discrete Cash Flow: Compound Interest Factors
Uniform Series Payments
Arithmetic Gradients
Compound
Amount
F兾P
Present
Worth
P兾F
Sinking
Fund
A兾F
Compound
Amount
F兾A
Capital
Recovery
A兾P
Present
Worth
P兾A
Gradient
Present Worth
P兾G
Gradient
Uniform Series
A兾G
1.5000
2.2500
3.3750
5.0625
7.5938
11.3906
17.0859
25.6289
38.4434
57.6650
86.4976
129.7463
194.6195
291.9293
437.8939
656.8408
985.2613
1477.89
2216.84
3325.26
7481.83
16834
25251
37877
85223
0.6667
0.4444
0.2963
0.1975
0.1317
0.0878
0.0585
0.0390
0.0260
0.0173
0.0116
0.0077
0.0051
0.0034
0.0023
0.0015
0.0010
0.0007
0.0005
0.0003
0.0001
0.0001
1.00000
0.40000
0.21053
0.12308
0.07583
0.04812
0.03108
0.02030
0.01335
0.00882
0.00585
0.00388
0.00258
0.00172
0.00114
0.00076
0.00051
0.00034
0.00023
0.00015
0.00007
0.00003
0.00002
0.00001
0.00001
1.0000
2.5000
4.7500
8.1250
13.1875
20.7813
32.1719
49.2578
74.8867
113.3301
170.9951
257.4927
387.2390
581.8585
873.7878
1311.68
1968.52
2953.78
4431.68
6648.51
14962
33666
50500
75752
1.50000
0.90000
0.71053
0.62308
0.57583
0.54812
0.53108
0.52030
0.51335
0.50882
0.50585
0.50388
0.50258
0.50172
0.50114
0.50076
0.50051
0.50034
0.50023
0.50015
0.50007
0.50003
0.50002
0.50001
0.50001
0.50000
0.50000
0.50000
0.50000
0.6667
1.1111
1.4074
1.6049
1.7366
1.8244
1.8829
1.9220
1.9480
1.9653
1.9769
1.9846
1.9897
1.9931
1.9954
1.9970
1.9980
1.9986
1.9991
1.9994
1.9997
1.9999
1.9999
1.9999
2.0000
2.0000
2.0000
2.0000
2.0000
0.4444
1.0370
1.6296
2.1564
2.5953
2.9465
3.2196
3.4277
3.5838
3.6994
3.7842
3.8459
3.8904
3.9224
3.9452
3.9614
3.9729
3.9811
3.9868
3.9936
3.9969
3.9979
3.9985
3.9993
3.9997
3.9998
3.9999
3.9999
0.4000
0.7368
1.0154
1.2417
1.4226
1.5648
1.6752
1.7596
1.8235
1.8713
1.9068
1.9329
1.9519
1.9657
1.9756
1.9827
1.9878
1.9914
1.9940
1.9971
1.9986
1.9990
1.9993
1.9997
1.9998
1.9999
2.0000
2.0000
PHOTO CREDITS
Chapter 1
Opener: Royalty-Free/CORBIS.
Chapter 2
Opener: Royalty-Free/CORBIS.
Chapter 3
Opener: Getty Images.
Chapter 4
Opener: Chad Baker/Getty Images.
Chapter 5
Opener: © Digital Vision/PunchStock.
Chapter 6
Opener: Stockdisc.
Chapter 7
Opener: Getty Images.
Chapter 8
Opener: PhotoLink/Getty Images.
Chapter 9
Opener: Royalty-Free/CORBIS.
Chapter 10
Opener: © Digital Vision/PunchStock.
Chapter 11
Opener: (a) Royalty-Free/CORBIS; (b) Royalty-Free/CORBIS.
Chapter 12
Opener: Ariel Skelley/Blend Images/Getty Images.
Chapter 13
Opener: Randy Allbritton/Getty Images.
Chapter 14
Opener: Photodisc/Getty Images.
Chapter 15
Opener: Ryan McVay/Getty Images.
Chapter 16
Opener: Arthur S. Aubry/Getty Images.
Chapter 17
Opener: Photodisc/Getty Images.
Chapter 18
Opener: Rubberball/SuperStock.
Chapter 19
Opener: © Brand X/JupiterImages.
INDEX
A
A, 13, 43
Absolute cell referencing, 9, 547, 550
Accelerated Cost Recovery System (ACRS), 422
Accelerated write-off, 416, 450–52
Accounting
ratios, 563–65
statements, 561–63
Acid-test ratio, 564
Acquisition phase, 161
Activity based costing (ABC), 401–03
Additive weight technique, 282–83
A/F factor, 46
After-tax
and alternative selection, 456–62
cash flow, 448–50
debt versus equity financing, 267
and depreciation, 415, 453
international, 468–70
and MARR, 269, 456
and present worth, 456
rate of return, 458–62
spreadsheet analysis, 457–58
and WACC, 271
After-tax replacement analysis, 462–65
A/G factor, 53. See also Gradient, arithmetic
Alternative depreciation system (ADS), 426–27
Allocation variance, 399
Alternatives
cost, 131
defined, 6
do-nothing, 130–31
independent, 130–31, 132, 157, 203, 207, 244, 247 (see also
Capital budgeting)
infinite life, 138, 157–60
mutually exclusive, 130–31, 132, 155, 208, 238, 248
revenue, 131
selection, 6
service, 131
in simulation, 533–40
Amortization, 415
Annual interest rate
effective, 99–105
nominal, 99–105
Annual operating costs (AOC), 6, 153, 297–98
and estimation, 388
Annual Percentage Rate (APR), 97
Annual Percentage Yield (APY), 97
Annual worth
advantages, 151
after-tax analysis, 456–58
of annual operating costs, 297–98
and B/C analysis, 235, 238–42
and breakeven analysis, 345–48
and capital-recovery-plus-interest, 153–54
equivalent uniform, 151
evaluation by, 155–60
and EVA, 153, 465–68
and future worth, 151
and incremental rate of return, 213–14
of infinite-life projects, 157–60
and inflation, 151, 377–78
and present worth, 151
and rate of return, 175, 213–14
and replacement analysis, 302–06, 307–10, 462–65
spreadsheet solutions, 156, 159, 458
when to use, 262
AOC. See Annual operating costs
A/P factor, 43–45, 153
Arithmetic gradient. See Gradient, arithmetic
Assets. See also Book value; Depreciation; Life; Salvage value
in balance sheet, 561
capital recovery, 153–54
return on, 564
sunk cost, 295
Attributes
evaluating multiple, 282–83
identifying, 278–79
weighting, 279–81
Average. See Expected value
Average cost per unit, 345
Average tax rate, 447
B
Balance sheet, 267, 561–63
Base amount
defined, 50
and shifted gradients, 80–82
Basis, unadjusted, 416
B/C. See Benefit/cost ratio
Before-tax rate of return
and after-tax, 459
calculation, 173–90
Bell-shaped curve. See Normal distribution
Benefit and cost difference, 236
Benefit/cost ratio
calculation, 235–36
conventional, 235
incremental analysis, 238–39
modified, 235–36
for three or more alternatives, 242–46
for two alternatives, 238–42
when to use, 262
Benefits
direct versus implied, 242
in public projects, 231, 235, 242
 (beta), 274
Bonds
and debt financing, 271–72
and inflation, 372, 385
interest computation, 190
payment periods, 190
present worth, 191–92
for public sector projects, 232
rate of return, 190–92
types, 191
Book depreciation, 415, 417, 466
Book value
declining balance method, 419–22
defined, 416
double declining balance method, 419–22
and EVA, 466
612
Index
Book value (continued)
MACRS method, 423
versus market value, 416
straight line method, 418
sum-of-years-digits method, 430, 557
unit-of-production method, 431
Borrowed money, 267
Borrowing rate, 185–87
Bottom-up approach, 388–89
Breakeven analysis. See also PW vs. i graph
average cost per unit, 345
fixed costs, 341
and Goal Seek, 353–54
and make-buy decisions, 341, 347
and payback, 351–54
and rate of return, 210–12, 460
versus sensitivity analysis, 341, 485
single project, 341–45
spreadsheet application, 212, 352–54
three or more alternatives, 348
two alternatives, 345–47
variable costs, 341
Breakeven point, 341, 489
Budgeting. See Capital budgeting
Bundles, 325, 327
Business ratios, 563–65
C
Canada, depreciation and taxes, 468
Capital
cost of (see Cost of capital)
cost of invested, 466
debt versus equity, 27, 267
description, 3
limited, 268, 323
unrecovered, 295
Capital asset pricing model (CAPM), 274
Capital budgeting
description, 323–25
equal life projects, 325–26
linear programming, 329–31
mutually exclusive bundles, 325, 327
present-worth use, 325–29
reinvestment assumption, 324, 328–29
spreadsheet solution, 330–31
unequal life projects, 327–29
Capital financing. See also Cost of capital
debt, 27, 267, 271–73
equity, 27, 267, 273–75
mixed (debt and equity), 275–77
Capital gains
defined, 454
short-term and long-term, 454
taxes for, 454
Capital losses
defined, 454
taxes for, 454
Capital rationing. See Capital budgeting
Capital recovery, 153–54. See also A/P factor; Depreciation
defined, 153
and economic service life, 297–99
and EVA, 468
and inflation, 377–78
and replacement analysis, 306
Capital recovery factor, 43–45
and equivalent annual worth, 153
Capitalized cost
in alternative evaluation, 138–42
and annual worth, 138
and public projects, 244–46
CAPM. See Capital asset pricing model
Carry-back and carry-forward, 454
Case studies
after-tax analysis, 482–83
alternative description, 38
annual worth, 93, 171
breakeven analysis, 363–64
compound interest, 70–71
debt versus equity financing, 290–91, 482–83
economic service life, 321
energy, 36, 363–64
ethics, 412–13
house financing, 124–25
indirect costs, 411–12
inflation, 385
multiple attributes, 511–13
multiple interest rates, 200–01
public project, 259–60, 511–13
rate of return, 200–01, 385
replacement analysis, 321
sale of business, 227
sensitivity analysis, 226, 510, 544–45
simulation, 544–45
social security, 149
Cash flow
after tax, 448–50
before tax, 449–50
beyond study period, 133–34
continuous, 116
conventional series, 180
defined, 15
diagramming, 16–18
discounted, 129
estimating, 15, 231, 387–90
incremental, 203–06, 213
inflow and outflow, 15
net, 15
and payback period, 349–51
nonconventional, 180–90
and public sector projects, 231
recurring and nonrecurring, 139
and replacement analysis, 306
revenue versus cost, 131, 204
and rule of signs, 180–84, 219
and simulation, 533–38
using actual versus incremental, 213, 240, 460
zero, 78, 553, 555
Cash flow after taxes (CFAT), 448–50, 456–65
and EVA, 465–68
Cash flow before taxes (CFBT), 448–50, 473
Cash flow diagrams, 16–18
partitioned, 55
Cell references, spreadsheet, 29, 547, 550
CFAT. See Cash flow after taxes
CFBT. See Cash flow before taxes
Challenger
in B/C analysis, 242–46
in replacement analysis, 294, 302–12, 462–64
in ROR analysis, 214–19
in service sector projects, 248–49
China, depreciation and taxes, 468–69
Code of ethics, 7–8, 251, 404, 566–68
Common stock, 267, 273–75
Index
Compound amount factors
single payment (F/P), 40
uniform series (F/A), 46
Compound interest, 22, 24–25, 28. See also Compounding
Compounding
annual, 99–105
continuous, 114–16
and effective interest rate, 96
frequency, 97–98
interperiod, 113–14
period, 97–98
and simple interest, 21–23
Compounding period
continuous, 114–16
defined, 97
and effective annual rate, 102
and payment period, 106–14
Concepts, fundamental, summary, 573–76
Contingent projects, 323
Continuous compounding, 114–16
Contracts, types, 234
Conventional benefit/cost ratio, 235
Conventional cash flow series, 180
Conventional gradient, 51–57
Corporations
and capital, 4
financial worth, 466–68
leveraged, 275–77
Cost alternative, 131, 204, 216, 457, 460
Cost-effectiveness
analysis, 246–50
ratio, 246
Cost, life-cycle, 160–63
Cost-capacity equations, 394–95
Cost centers, 397, 401
Cost components, 387–88
Cost depletion, 427–29
Cost drivers, 401
Cost-estimating relationships, 394–97
Cost estimation
accuracy, 389
approaches, 388–89
cost-capacity method, 394–95
and cost indexes, 391–94
factor method, 395–97
and inflation, 12, 377
unit method, 390
Cost of capital
and debt-equity mix, 269–71
for debt financing, 271–73
defined, 26, 267
for equity financing, 273–75
versus MARR, 26, 267
weighted average, 27, 270, 275
Cost of goods sold, 397, 562–63, 565
Cost of invested capital, 466–67
Costs. See also Capital; Incremental cash flow; Opportunity cost
annual operating, 6, 153, 297–98
and annual worth, 153
of asset ownership, 153
direct, 387, 390–97
estimating, 388–97
EUAW(see Annual worth)
fixed, 341
indirect, 387, 396, 397–404
life-cycle, 160–63
marginal, 300–01
in public projects, 231, 235
sign convention, 15, 235
sunk, 295
variable, 341
Coupon rate, 190
Cumulative cash flow sign test, 181–84, 219
Cumulative distribution, 519–22
Current assets, 561
Current liabilities, 561–62
Current ratio, 563
D
DB function, 420, 550–51
DDB function, 420, 551
Debt capital, 267
Debt-equity mix, 269–71, 275–77
Debt financing, 269–71
on balance sheet, 267, 561
costs of, 271–73
and inflation, 374, 378
leveraging, 275–77
Debt ratio, 564
Decision making
attributes, 3–4, 278–83
under certainty, 517
and engineering economy role, 3–4
under risk, 517, 518–22
under uncertainty, 517
Decision trees, 494–98
Declining balance depreciation, 419–22
in Excel, 420, 550–51
Decreasing gradients, 51, 56, 83–86
Defender
in B/C analysis, 242–46
in replacement analysis, 294, 302–12, 462–64
in ROR analysis, 214–19
in service sector projects, 248–49
Deflation, 368–69
Delphi method, 278–79
Dependent projects, 323
Depletion
cost, 427–29
percentage, 427–29
Depreciation. See also Depreciation recapture; Rate of
depreciation; Replacement analysis
accelerated, 416, 419
ACRS, 422
alternative system, 426–27
and amortization, 415
basis, 416
book, 415, 417, 466
declining balance, 419–22, 550–51
defined, 415
double declining balance, 419–22, 551
and EVA, 466
general depreciation system (GDS), 426
half-year convention, 416, 424, 427
and income taxes, 415, 417, 445–68
MACRS, 417, 422–27
present worth of, 432–35
property class, 426
recovery period for, 416, 418, 426
rate of, 416
recovery rate, 416
613
614
Index
Depreciation (continued)
straight line, 418–19, 557
straight line alternative, 426–27
sum-of-years digits, 430, 557
switching methods, 432–38, 557–58
tax, 415, 417
unit-of-production, 431
Depreciation recapture
definition, 453
in replacement studies, 462–65
and taxes, 453, 461
Descartes’ rule, 181–84, 219
Design-build contracts, 234
Design stages, preliminary and detailed, 161, 163, 389
Design-to-cost approach, 388–89
Direct benefits, 242, 244
Direct costs, 387, 390–97
Disbenefits, 231, 235
Disbursements, 15, 177
Discount rate, 129, 232
Discounted cash flow, 129
Discounted payback analysis, 349
Discrete cash flows
compound interest factors (tables), 581–609
discrete versus continuous compounding, 114–16
and end-of-period convention, 15–16
Disposal phase, 161
Distribution. See also Probability distribution
defined, 519
normal, 520, 531–33
standard normal, 531–33
triangular, 520, 522
uniform, 520–21, 535, 538
Dividends
bonds, 190, 271–72
stocks, 273–75
Dominance, 248–49
Do-nothing alternative
and B/C analysis, 239, 242
defined, 6, 131
and independent projects, 132, 207, 244, 325–27
and present worth, 130–31
and rate of return, 207, 214, 216
Double declining balance, 419–22
in Excel, 420–22, 551
in switching, 432–38
and taxes, 451–52
Dumping, 368
E
Economic equivalence. See Equivalence
Economic service life (ESL), 294, 296–302
Economic value added, 153, 465–68
EFFECT function, 103, 551
Effective interest rate
annual, 99–100
for any time period, 105–06
of bonds, 191–92
and compounding periods, 100, 105
for continuous compounding, 114–16
defined, 96
and nominal rate, 96–97
Effective tax rate, 447, 462
Efficiency ratios, 563
End-of-period convention, 15–16
Engineering economy
Concepts, summary, 573–76
defined, 3
study approach, 4–7
terminology and symbols, 13
Equal service requirement, 131, 151, 213, 217, 240, 457
Equity financing, 26, 267
cost of, 273–75
Equivalence
calculations without tables, 569–72
compounding period greater than payment period, 112–14
compounding period less than payment period, 107–12
defined, 19–21
Equivalent uniform annual cost. See Annual worth
Equivalent uniform annual worth. See Annual worth
Error distribution. See Normal distribution
Estimation
and alternatives, 6
of cash flow, 6, 15–19
of costs, 388–97
and sensitivity analysis, 485, 490–91
before tax ROR, 459
Ethics
and cost estimation, 403–04
in public sector, 250–51
overview, 7–10
EUAC. See Annual worth
EUAW. See Annual worth
Evaluation criteria, 4, 262
Evaluation method, 261–64
Excel. See also Spreadsheet, usage in examples
absolute cell reference, 9, 547, 550
basics, 547–50
charts, 548–49
displaying functions, 30
embedding functions, 75, 156–57
error messages, 560
functions, in engineering economy, 28, 550–58
Goal Seek tool, 558–59
introduction, 27–30, 547–50
and linear programming, 330–31
random number generation, 556
Solver tool, 559–60
spreadsheet layout, 549–50
Expected value
computation, 492–94, 526–27, 530
and decisions under risk, 517
and decision trees, 497–98
defined, 492, 526
and real options, 501–02
in simulation, 538–39
Expenses, operating, 6, 153, 297–98
External rate of return, 185–90. See also Rate of return
F
F, 13
F/A factor, 46
F/G factor, 54
Face value, of bonds, 190
Factor method estimation, 395–97
Factors, compound interest
annual worth, 44, 46, 53
arithmetic gradient, 52–54
Index
capital recovery, 44
continuous compound interest, 114–16
derivations, 39–46, 52–54
equivalence without, 569–72
future worth, 40, 46, 54
geometric gradient, 58–59
interpolation, 48–50
multiple, 73–86
notation, 40, 44, 47, 53, 59
present worth, 40, 43, 52, 58
single payment, 39–42
sinking fund, 46
tables, 581–609
uniform series, 43–48, 53
Factory cost, 397, 562
Financial worth of corporations, 466, 561
First cost
and depreciation, 416
description, 6, 15
and estimation, 388–97
in replacement studies, 294–95, 462
Fiscal year, 561
Fixed assets, 561
Fixed costs, 341
Fixed income investment, 372, 385. See also Bonds
Fixed percentage method. See Declining balance depreciation
F/P factor, 40. See also Single payment factors
Future worth
and annual worth, 151
and effective interest rate, 100
evaluation by, 137
and inflation, 374–77
and multiple rates of return, 186–87
from present worth, 137
of shifted series, 73, 77–78
and spreadsheet solutions, 42, 48
when to use, 262
FV function, 28, 552
and shifted uniform series, 78–79
and single payments, 41
and uniform series, 46
G
Gains and losses, 454
Gaussian distribution. See Normal distribution
General depreciation system (GDS), 426
Geometric gradient
defined, 58
factors, 58–59
shifted, 82–86
Geometric series, and equivalence, 570–72
Goal Seek, 56–57, 86, 156–57, 189, 312, 354, 458, 558–59
Government projects. See Public sector projects
Gradient, arithmetic
base amount, 50
conventional, 51
decreasing, 51, 56
defined, 50
factors, 52–54
increasing, 51–55
shifted, 80–82, 83–86
spreadsheet use, 57
Graduated tax rates, 446–47, 448
Gross income, 445, 448
615
H
Half-year convention, 416, 424, 427
Highly leveraged corporations, 275–77
Hurdle rate. See Minimum attractive rate of return
Hyperinflation, 368, 377
I
IF function, 189, 245, 552
Implied benefits, 240, 242
Income
estimated annual, 6, 15
gross, 445, 448
net operating, 446, 459
taxable, 446, 454
Income statement, 562
basic equation, 562
ratios, 564
Income tax, 445–70
average tax rate, 447
and capital gains and losses, 454
and cash flow, 448–50
corporate, 447
and debt financing, 271–73
defined, 445
and depreciation, 450–56
effective rates, 447
international, 468–70
negative, 449
present worth of, 451–53
and rate of return, 458–62
rates, 446–48
and replacement studies, 462–65
state, 445, 447
tax savings, 449
and taxable income, 446
terminology, 445–46
Incremental benefit/cost analysis
for three or more alternatives, 242–46
for two alternatives, 238–42
Incremental cash flow
and benefit/cost analysis, 238–39
calculation, 203–06
and rate of return, 207–10, 213–14
Incremental rate of return
for multiple alternatives, 214–18
for two alternatives, 213–14
unequal lives, 207, 213
Independent projects
AW evaluation, 157
B/C evaluation, 244
and capital budgeting, 323–31
definition, 130
and do-nothing alternative, 130, 132, 157, 207, 244, 325, 327
and incremental cash flow, 207, 244
PW evaluation, 132
ROR evaluation, 207
service project evaluation, 247–48
Indexing, income taxes, 448
Indirect costs
and activity-based costing, 401–03
allocation variance, 399
in cost of goods sold statement, 562–63
defined, 398
616
Index
Indirect costs (continued)
and factor method, 396–97
rates, 398–99
Infinite life, 138–40, 157–60, 231, 244–45
Inflation
assumption, PW and AW analysis, 134, 151
and capital recovery, 377–78
constant-value, 367
definition, 12, 367
versus deflation, 368–69
and dumping, 368
and future worth, 374–77
high, 377
impact, 12, 367–68
and interest rates, 368
market adjusted, 368
and MARR, 368, 375–77
and present worth, 369–74
and sensitivity analysis, 485
Initial investment. See First cost
Installment financing, 175
Intangible factors, 6. See also Multiple attribute evaluation
Integer linear programming, 329–31
Interest. See also Interest rate(s)
compound, 22, 24, 29
continuous compounding, 114–16
defined, 10
interperiod, 112–14
rate, 10, 12
simple, 21, 24
Interest period, 10, 12
Interest rate(s). See also Effective interest rate; Rate of return
and breakeven analysis, 345
definition, 10–12
Excel functions, 61, 177, 553, 557
expressions, 98
inflation-adjusted, 368, 370
inflation free (real), 368, 374
interpolation, 61–63
market, 368, 370
multiple, 180–84
nominal versus effective, 96–99
for public sector, 232
and risk, 267, 274
and sensitivity analysis, 485
on unrecovered balance (ROR), 173–75
varying over time, 116–17
Interest tables
discrete compounding, 581–609
interpolation, 48–50
Internal rate of return. See also Rate of return
and annual worth, 175
definition, 173
versus external ROR, 185
and present worth, 175
spreadsheet solution, 177
International aspects
contracts, 234
corporate taxes, 468–70
cost estimation, 388–89
deflation, 368–69
depreciation, 416–17, 470–72
dumping, 368
inflation aspects, 368, 377
value-added tax, 470–72
Interperiod interest, 112–14
Interpolation, in interest rate tables, 48–50
Inventory turnover ratio, 564–65
Invested capital, cost of, 466–67
Investment(s). See also First cost
extra, 206–07, 214–18
fixed income, 372, 385
net, 188
permanent, 138–41, 157–60
Investment rate, 185–90
IPMT function, 552
IRR function, 61, 63, 177–79, 217, 460, 553
L
Land, 416
Lang factors, 395
Least common multiple
and annual worth, 151–52
assumptions, 134
in evaluation methods, 262–63
and future worth, 137
and incremental cash flow, 204
and incremental rate of return, 207–09, 213–14, 218–19
and independent projects, 324, 329
and present worth, 133–37
in spreadsheet analysis, 205
versus study period, 134, 136
Leveraging, 275–77
Liabilities, 561
Life
finite, 141
and income taxes, 452–53
infinite or very long, 138–41, 157–58
minimum cost, 297
recovery period, 416
in simulation, 534–35, 538
unknown, 61, 63–64
useful, 6, 416
Life cycle, and annual worth, 160–62
Life-cycle costs, 160–63
Likert scale, 281
Linear programming, 329–31
Lives
equal, 131, 155, 204, 240
perpetual, 138, 157
unequal, 133, 155, 204, 207
Loan repayment, 24–25
M
MACRS (Modified Accelerated Cost Recovery System)
in CFAT example, 450
depreciation rates, 423–24
PW of, depreciation, 432
recovery period, 423, 426–27
spreadsheet function, 424, 558
straight line alternative (ADS), 426–27
switching, 432, 435–38
U.S., required, 417, 422
Maintenance and operating (M&O) costs, 6, 153. See also Annual
operating cost
Make-or-buy decisions, 341, 347. See also Breakeven analysis
Marginal costs, 300–01
Marginal tax rates, 446–48
Index
Market interest rate, 368
Market value
and depreciation, 416
in ESL analysis, 297, 300–01
estimating, 301–02
in replacement analysis, 294, 302–06
as salvage value, 153, 294
and study period, 134
MARR. See Minimum attractive rate of return
Mean. See Expected value
Mean squared deviation, 528
Measure of worth, 4, 6, 262
Median, 527
Mexico, depreciation and taxes, 469
Minimum attractive rate of return
after-tax, 269, 456–58
and bonds, 192
and capital budgeting, 323–27, 328–29
definition, 26, 267
establishing, 26–27, 267–69, 275
as hurdle rate, 26
inflation-adjusted, 368, 375–76
and rate of return, 203, 206, 208, 213, 214–18
and reinvestment, 324, 328–29
in sensitivity analysis, 485–87
before tax, 269
and WACC, 27, 267, 275
Minimum cost life. See Economic service life
MIRR function, 186, 553–54
M&O costs. See Annual operating costs; Maintenance and
operating costs
Mode, 522, 527
Modified benefit/cost ratio, 235–36
Modified ROR approach, 185–87
Monte Carlo simulation, 533–39
Most likely estimate, 490–91
Multiple alternatives
breakeven analysis, 348
incremental benefit/cost analysis, 242–46
incremental rate of return, 214–18
Multiple attribute evaluation, 278–83
Multiple rate of return
definition, 179–80
determining, 181–84
presence of, 180–81
removing, 184–90
Municipal bonds, 190–92
Mutually exclusive alternatives, 130–31
and annual worth, 155–57
and B/C, 238–44
evaluation method selection, 261–64
and present worth, 132–37
and rate of return, 208–10
and service projects, 247–49
N
Natural resources. See Depletion
Net cash flow, 15
Net operating income (NOI), 446, 459
Net investment procedure, 187–90
Net operating profit after taxes (NOPAT), 446, 466
Net present value. See NPV function; Present worth
Net profit after taxes (NPAT), 446
Net worth, 561
NOMINAL function, 103, 554
Nominal interest rate
annual, 96, 103
of bonds, 190–92
definition, 96
and effective rates, 96–98
Nonconventional cash flow series, 180–84
Noneconomic factors, 6
Nonowner’s viewpoint, 295
Nonrecurring cash flows, 139
Nonsimple cash flow series, 180–84
No-return (simple) payback, 349–50
Normal distribution, 520, 531–33
Norstrom’s criterion, 181–84, 219, 460
Notation for factors, 40, 44, 47, 53, 59
NPER function, 554–55
and payback, 350, 353
and unknown n, 61, 63–64
NPV function, 555
for arithmetic gradient, 57, 86
embedding in PMT, 75, 156
geometric gradients 86
independent projects, 328
and present worth, 135–36
in PW vs. i graphs, 178–79
sensitivity analysis, 486–87, 489
and shifted series, 75, 86
NSPE (National Society of Professional Engineers), 7,
404, 566–68
O
Obsolescence, 294
One-additional-year replacement study, 302–05
Operating costs. See Annual operating costs
Operations phase, 161, 163
Opportunity cost, 27, 267, 324
and replacement analysis, 306
Optimistic estimate, 490–91
Order of magnitude, 389
Overhead rates. See Indirect costs
Owner’s equity, 267, 561
P
P, 13, 40
P/A factor,43, 53. See also Geometric gradient;
Uniform series
Payback analysis
and breakeven analysis, 351–52
calculation, 349–51
definition, 348–49
limitations, 349
spreadsheet analysis, 350–54
Payment period
of bonds, 190
defined, 106
equals compounding period, 106, 107, 109–11
longer than compounding period, 109
shorter than compounding period, 112–14
single amount, 107–09
Payout period. See Payback analysis
Percentage depletion. 427–29
Permanent investments, 138–41, 157–58
Personal property, 416, 423, 426
617
618
Index
Perspective
for public sector analysis, 232–34
for replacement analysis, 295
Pessimistic estimate, 490–91
P/F factor, 40
P/G factor, 53. See also Gradient, arithmetic
Phaseout phase, 161
Planning horizon. See Study period
PMT function, 28, 555
and after-tax analysis, 457
and annual worth, 44, 155–57
and arithmetic gradient, 54
and capital recovery, 154
and economic service life, 298–99
and embedded NPV, 75, 80, 156
and geometric gradient, 59
and random single amounts, 79–80
and shifted series, 75–76
and sinking fund factor, 44
and uniform series, 75, 79
Point estimates, 15, 517
Power law and sizing model, 394–95
PPMT function, 555–56
Preferred stock, 267, 273
Present value. See Present worth
Present worth
after-tax analysis, 456–58
and annual worth, 151
assumptions, 134
and B/C analysis, 235
of bonds, 191–92
and breakeven analysis, 345–46
and capital budgeting, 325–29
of depreciation, 432
for equal lives, 132–33
evaluation method, 261–62
geometric gradient series, 82–86
income taxes, 451–53
and independent projects, 325–29
index, 332
and inflation, 369–74
and multiple interest rates, 181–84
and profitability index, 237
and rate of return, 175–76, 182–84, 458–62
and sensitivity analysis, 486–90
in shifted series, 73, 76, 80–86
in simulation, 534–39
single-payment factor, 40–41
for unequal lives, 133–37
Present worth factors
gradient, 50–53, 58–60
single payment factor, 40–41
uniform series, 43–45
Probability
in decision trees, 495–98
defined, 492, 518
and expected value, 492–94, 526–27
and standard deviation, 528
Probability distribution
of continuous variables, 519–22
defined, 519
of discrete variables, 519–20
properties, 526–28, 530
and samples, 523–26
in simulation, 533–39
Probability node, 495
Productive hour rate, 399
Profitability index, 237, 332
Profitability ratios, 563
Profit-and-loss statement, 562
Project net-investment, 187–90
Property class, 426
Property of independent random variables, 533
Public-private partnerships, 234–35
Public sector projects, 230–35
and annual worth, 157–58
B/C analysis, 235–38
capitalized cost, 138–42
characteristics, 231–32
design-build contracts, 234–35
profitability index, 237
public-private partnerships, 234–35
Purchasing power, 367, 374, 376
PV function, 28, 44, 556
versus NPV function, 556
and present worth, 41, 44
and single payment, 41
and uniform series present worth, 44
PW vs. i graph, 177–79, 182–84, 210–13, 460–62
R
RAND function, 523–25, 556
Random numbers, 523–25, 556
generation, 538, 556
Random samples, 523–26, 535–39
Random variable
continuous, 519–20, 525
cumulative distribution, 519–22, 525–26
defined, 518
discrete, 519–20, 523
expected value, 526, 528, 530
probability distribution of, 519
standard deviation, 527–28, 530
Range, 15, 530
Rank and rate technique, 282
Ranking inconsistency, 213, 216, 462
RATE function, 61–62, 177, 557
Rate of depreciation
declining balance, 419
defined, 416
MACRS, 422–23, 435–38
straight line, 418, 420
sum-of-years digits, 430
Rate of return. See also Incremental rate of return
after-tax, 458–62
and annual worth, 175, 213–14, 218
of bonds, 190–92, 272
breakeven, 210–12, 460
in capital budgeting, 332–34
cautions, 179–80
on debt capital, 272–73, 276
defined, 12, 173, 175
on equity capital, 273–74, 276
evaluation method, 261–62
external, 185–90
on extra investment, 206
incremental, 206, 207–10
and independent projects, 207
and inflation, 12–13, 368, 374–75
Index
installment financing, 175
internal, 173, 175, 185
minimum attractive (see Minimum attractive rate of return)
modified ROR approach, 185–87
multiple, 180–90
and mutually exclusive alternatives, 206, 207–18
and present worth, 175, 177–79, 207–09, 218
ranking inconsistency, 213, 216, 462
return on invested capital (ROIC) approach, 185, 187–90
spreadsheet solution, 177, 179, 211–13, 216–19
Ratios, accounting, 563–65
Real interest rate, 368, 370, 374–75
Real options, 498–503
and decision trees, 500
definition, 499
Real property, 416, 423–24, 426
Recovery period
defined, 416
effect on taxes, 452–53
MACRS, 423–24, 426–27
straight line option, 426
Recovery rate. See Rate of depreciation
Recurring cash flows, 139
Reinvestment, assumption in capital budgeting,
324, 328–29
Reinvestment rate. See Investment rate
Repayment of loans, 24–25
Replacement analysis, 292–313, 462–65
after-tax, 462–65
annual worth, 295, 302–06
and capital losses, 454, 462
cash flow approach, 306
depreciation recapture, 462–64
and economic service life, 296–99, 305
first costs, 294–96
and marginal costs, 300–01
market value, 294, 301
need for, 294
one-additional year, 302–05
opportunity cost approach, 306
overview, 302
and study periods, 307–12
sunk costs, 295
terminology, 294
viewpoint, 295
Replacement life. See Economic service life
Replacement value, 312
Retained earnings, 27, 267, 273
Retirement life. See Economic service life
Return on assets ratio, 564
Return on invested capital, 187–90
Return on investment (ROI), 12, 173. See also Rate of return
Return on sales ratio, 564
Revenue alternatives, 131, 204, 214, 237, 460
Risk
and debt-equity mix, 275–77
and decision making, 517, 526–30, 533–39
and decision trees, 494–98
description, 515
and MARR, 267–69
and payback analysis, 349
and random sampling, 523
and real options, 498–502
Risk-free investment, 26, 274
ROI. See Return on investment
Root mean square deviation, 528
ROR. See Rate of return
Rule of signs, 181
S
Safe investment, 26, 190, 274
Sales, return on, 564
Salvage value. See also Market value
and capital recovery, 153, 297
defined, 6, 153
and depreciation, 416, 418, 420, 423, 430
and market value, 294, 297–98
and public projects, 235
in PW analysis, 132, 134–37
in replacement analysis, 294, 297, 303–06, 464
and trade-in value, 294, 464
Sampling, 523–26
Savings, tax, 449, 463–65
Scatter charts. See xy Excel charts
Screening projects, 349, 351
Section 179 deduction, 417
Section 1231 transactions, 454
Security, defined, 274
Sensitivity analysis. See also Breakeven analysis
description, 485
and Excel cell referencing, 29, 547
of one parameter, 485–87
spider graph, 488
with three estimates, 490–91
two alternatives, 488–90
Service alternative. See Cost alternative
Service sector projects
analysis, 246–50
definition, 246
dominance, 248–49
Shifted gradients, 80–86
Shifted series, 73–80
Sign changes, number of, 181–84
Simple cash flow series, 180
Simple interest, 21–23
Simulation, Monte Carlo, 517, 533–40
Single payment compound amount (F/P) factor, 40
Single payment factors, 39–42
Single payment present worth (P/F) factor, 40
Sinking fund (A/F) factor, 46
SLN function, 418, 557
Social discount rate, 232
Solvency ratios, 563
Solver, 330–31, 559–60
Spreadsheet, usage in examples. See also Excel
annual worth, 155–56, 159, 218, 299, 301, 304,
305, 353, 458, 467
B/C analysis, 245
breakeven analysis, 352, 353–54
cash flow after tax (CFAT), 450, 456, 458, 461, 462
compound interest, 29–30
depreciation, 422, 425, 434
EVA, 467
and factor values, 49
independent projects, 328, 331
inflation, 372
layout, 549–50
multiple attributes, 283
nominal and effective interest, 103–04
619
620
Index
Spreadsheet (continued)
present worth, 136, 209, 212, 218, 311, 354, 458,
461, 489, 501, 539
rate of return, 63, 182, 184, 189, 192, 209, 212,
217, 218, 461, 462, 501
replacement analysis, 299, 301, 304, 305, 311, 465
replacement value, 312
sensitivity analysis, 487, 489
simulation, 538–39
Staged funding, 494–503
Standard deviation
for continuous variable, 530
definition, 527–28
for discrete variable, 528–29
Standard normal distribution, 531–33
Stocks
CAPM model, 274
common, 267, 274
in equity financing, 273–75, 276
preferred, 267, 273
Straight line alternative, in MACRS, 426–27
Straight line depreciation, 418–19
Straight line rate, 418
Study period
and AW evaluation, 155
and equal service, 133
and FW analysis, 137
and PW evaluation, 133–34
and replacement analysis, 302, 307–11
and salvage value, 153
spreadsheet example, 136, 310–11
Sum-of-years digits depreciation, 430
Sunk costs, 295
SYD function, 430, 557
System, phases of, 160–61
T
Tax depreciation, 415–16, 417
Taxable income, 446–70
and CFAT, 448–50, 456
and depreciation, 446, 450–53
negative, 449, 463
and taxes, 446–48, 463–65
Taxes. See After-tax; Income tax; Taxable income
Time, 13
Time value of money
defined, 4
and equivalence, 19
factors to account for, 39–61
and no-return payback, 349–50
Total cost relation, 342–45. See also Breakeven analysis
Trade-in value, 6, 294, 416. See also Market value; Salvage value
Treasury securities, 26, 190
Triangular distribution, 520, 522
U
Unadjusted basis, 416
Uncertainty, 515, 517
Uniform distribution, 520–21, 535, 538
Uniform gradient. See Gradient, arithmetic
Uniform percentage method. See Declining balance depreciation
Uniform series
compound amount (F/A) factor, 46
compounding period greater than payment period, 112–14
compounding period less than payment period, 109–12
description, 13
present worth (P/A) factor, 43
shifted, 73–80
Unit method, 390–91
Unit-of-production depreciation, 431
Unknown interest rate, 61–63
Unknown years (life), 61, 63–64
Unrecovered balance, 173–74
V
Value, resale, 6. See also Salvage value; Trade-in value
Value added analysis, after tax. See Economic value added
Value-added tax, 470–72
Variable. See Random variable
Variable costs, 341
Variance
in cost allocation, 399
description, 528
formula for, 528, 530
and normal distribution, 531
VDB function, 424, 433–34, 557–58
W
WACC. See Weighted average cost of capital
Websites, 580
Weighted attribute method, 282–83
Weighted average cost of capital, 27, 270–71, 275–77
Working capital, 563
Worth, measures of, 4, 6, 129, 534
X
xy Excel charts, 182, 184, 212, 352, 461, 487, 548–49
Y
Year(s)
and end-of-period convention, 15–16
fiscal versus calendar, 561
half-year convention, 416, 424, 427
symbols, 13
unknown, 61, 63–64
Format for Spreadsheet Functions on Excel©
Present worth:
Contents of ( )
ⴝ PV(i%, n, A, F)
for constant A series; single F value
ⴝ NPV(i%,second_cell:last_cell) ⴙ first_cell
for varying cash flow series
Type
Single
Amount
Future worth:
ⴝ FV(i%, n, A, P)
Relations for Discrete Cash Flows with End-of-Period Compounding
for constant A series; single P value
Find/Given
F兾P
Compound
amount
P兾F
Present
worth
Annual worth:
ⴝ PMT(i%, n, P, F)
for single amounts with no A series
ⴝ PMT(i%, n, NPV)
to find AW from NPV; embed NPV function
P兾A
Present
worth
Number of periods (years):
ⴝ NPER(i%, A, P, F)
for constant A series; single P and F
Uniform
Series
A兾P
Capital
recovery
F兾A
Compound
amount
(Note: The PV, FV, and PMT functions change the sense of the sign. Place a minus in front of the
function to retain the same sign.)
Rate of return:
ⴝ RATE(n, A, P, F)
ⴝ IRR(first_cell:last_cell)
A兾F
Sinking
fund
for constant A series; single P and F
for varying cash flow series
Interest rate:
ⴝ EFFECT(r%, m)
for nominal r, compounded m times per period
ⴝ NOMINAL(i%, m)
for effective annual i, compounded m times
per year
Arithmetic
Gradient
PG兾G
Present
worth
AG兾G
Uniform
series
Factor Notation
and Formula
Relation
(F兾P,i,n) ⫽ (1 ⫹ i)n
F ⫽ P(F兾P,i,n)
ⴝ DDB(P, S, n, t, d)
straight line depreciation for each period
double declining balance depreciation for
period t at rate d (optional)
ⴝ DB(P,S,n,t)
declining balance, rate determined by the
function
ⴝ VBD(P,0, n,MAX(0, tⴚ1.5),
MIN(n, tⴚ0.5), d)
MACRS depreciation for year t at rate d for
DDB or DB method
Logical IF function:
ⴝ IF(logical_test,value_if_true,value_if_false)
for logical two-branch operations
Geometric
Gradient
Pg兾A1 and g
Present
worth
F
1
0
1
(P兾F,i,n) ⫽ ᎏn
(1 ⫹ i)
2
n–1
…
(1 ⫹ i)n ⫺ 1
(P兾A,i,n) ⫽ ᎏᎏ
i(1 ⫹ i)n
P
P ⫽ A(P兾A,i,n)
A
A … A
1
2
0
i(1 ⫹ i)n
(A兾P,i,n) ⫽ ᎏᎏ
(1 ⫹ i)n ⫺ 1
(1 ⫹ i)n ⫺ 1
(F兾A,i,n) ⫽ ᎏᎏ
i
F
F ⫽ A(F兾A,i,n)
(1 ⫹ i)n ⫺ in ⫺ 1
(P兾G,i,n) ⫽ ᎏᎏ
i2(1 ⫹ i)n
PG ⫽ G(P兾G,i,n)
PG
n
1
(A兾G,i,n) ⫽ ᎏ ⫺ ᎏᎏ
(1 ⫹ i)n ⫺ 1
i
AG ⫽ G(A兾G,i,n)
0
(Gradient only)
(Sec. 2.5)
Pg ⫽
冦
)]
n
A1 ———
1⫹i
(Gradient and base A1)
1
2
A
A…A
n
n–1
…
A ⫽ F(A兾F,i,n)
(Sec. 2.3)
1⫹g n
A1 1 ⫺ ———
1⫹i
———————
i⫺g
n–1 n
P
0
i
(A兾F,i,n) ⫽ ᎏᎏ
(1 ⫹ i)n ⫺ 1
A
…
A ⫽ P(A兾P,i,n)
(Sec. 2.2)
[ (
n
P ⫽ F(P兾F,i,n)
(Sec. 2.1)
Depreciation:
ⴝ SLN(P, S, n)
Sample Cash Flow
Diagram
A
AG AG AG … AG AG
1
…
2 3
…
G 2G
n
(n–1) G
n–1
A1
g⫽i
A1(1+g)
…
A1(1+g)
…
0
g⫽i
(Sec. 2.6)
Pg
1
2
n–1 n