Ninth Edition o. Compound Amount: Single Payment F To Find F Given P (F I P,i,n) F = P(1 + iy Present Worth: p To Find P Given F Uniform Series Series Compound Amount: (P I F,i,n) To Find F A A A A A P = F(1 + i)-n (F I A,i,n) Given A F = A [(1 + ir - 1J Sinking Fund: To Find A A . A A . I A . . Capital Recovery: A To Find A Given P A-F (AI F,i,n) F Given F [ (1 + i)n l - 1J (AI P,i,n) - 1J A = P [ (1i + (1 i)n + i)n Series Present Worth: p To Find P Given A (P I A,i,n) + i)n P = A [(1i(1 + i)n - 1J Arithmetic Gradient Uniform Series: Arithmetic Gradient (n - l)G 3G To Find A Given G (AIG,i,n) A = G (1 .+ i)n - in - 1 l (1 + i)n - i [ J 2G o A G A or A A 1 A n ' = G [ i - (1 + i)n - 1J A Arithmetic Gradient Present Worth: . (n - l)G 3G 2G G 0 To Find P Given G (P IG,i,n) t p ....... -- L;; Geometric Gradient Geometric Series Present Worth: To Find P Given AI, g (P j A,g,i,n) When i = g P = Al [n(1+ i)~IJ" To Find P Given AI, g (P j A,g,i,n) When i # g P = Al [ 1 - (1 +.g)n(1 + i)-n l-g ] p Continuous Compounding at Nominal Rate r SinglePayment: F Uniform Series: = p[ern] = F[e-rn] er-l A=F [ ern F P [ - 1 er - 1 ] ern =A - 1] Continuous, Uniform Cash flow (One Period) With Continuous Compo.unding at Nominal Rate r Present Worth: ToFind P GivenF [PjF,r,n] P=F Compound Amount: F ~ [ rern ] P ~LJ. /' p ", " To Find F Given P U 1 [Fj p,r,n] ",' " n r'mpound Interest i = Interestrate per interestperiod*. F P = Number of interest periods. = A present sum of money. F = A futuresum of money.The future sum F is an amount,n interestperiodsfrom the present, n that is equivalent to P with interest rate i. A = An end-of-period ' cash receipt or disbursement in a uniform series continuing for n periods, the entire series equivalent to P or F at interest rate i. G = Uniform p~riod-by-period increase or decrease in cash receipts or disbursements; the arithmetic gradient. g r m = Uniform rate of cash flow increase or decrease from period to period; the geometric gradient. = Nominal interest rate per interest period*. = Number of compounding subperiods per period*. P,F = Amount of money flowing continuously and uniformly during one given period. *Nonnally the interest period is one year, but it could be something . else. -- [ ' - I '," 1 ~:~... L:7~' .~_c:.::,.:;.~..jg{~j~..;~c.-::,.~':.: .~-~;~~::-':-:1":~-~'~.-~~:.-;';f~~:,. ._:-:..:'-:;~~,,-.:.-~ - - - ".:. _. .;.._'-. - -- ::.; - _i._,"_ __-: .'. -:~ ~::__-~ .~:;,~~:..::. - ENGINEERING ECONOMIC ANALYSIS I I I I /, -- -- - -- --- ENGINEERING ECONOMIC ANALYSIS NINTH EDITION Donald G. Newnan Professor Emeritus of Industrial and Systems Engineering Ted G. Eschenbach University of Alaska Anchorage Jerome P. Lavelle North Carolina State University New York Oxford OXFORD UNIVERSITY PRESS 2004 Oxford University Press Oxford New York Auckland Bangkok Buenos Aires Cape Town Chennai Dar es Salaam Delhi Hong Kong Istanbul Karachi Kolkata Kuala Lumpur Madrid Melbourne Mexico City Mumbai Nairobi SiloPaulo Shanghai Taipei Tokyo Toronto Copyright @ 2004 by Oxford University Pr(:ss, Inc. Published by Oxford University Press, Inc. 198 Madison Avenue, New York, New York 10016 WWW.oup.com Oxford is a registered trademark of Oxford University Press All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior permission of Oxford University Press. Library of Congress Cataloging-in-Publication Data Newnan, Donald G. Engineering economic analysis / Donald G. Newnan, Ted G. Eschenbach, Jerome P. Lavelle. - 9th ed. p.cm. Includes bibliographical references and index. ISBN 0-19-516807-0 (acid-free paper) 1. Engineering economy. 1. Eschenbach, Ted. n. Lavelle, Jerome P. m. TItle. TA177.4N482004 658.15-dc22 2003064973 Photos: Chapter 1 @ Getty Images; Chapter 2 @ SAN FRANCISCO CHRONICLE/CORBIS SABA; Chapter 3 @ Olivia Baumgartner/CORBIS SYGMA; Chapter 4 @ Pete Pacifica/Getty Images; Chapter 5 @ Boeing Management Company; Ch.apter 7 @ Getty Images; Chapter 8 @ Michael Nelson/Getty Images; Chapter 9 @ Guido Alberto Rossi/Getty Images; Chapter 10 @ Terry Donnelly/Getty Images; Chapter 11 @ Michael Kim/CORBIS; Chapter 12 @ Getty Images; Chapter 13 @ Richard T Nowitz/CORBIS; Chapter 14 @ CORBIS SYGMA; Chapter 15 @ Shephard Sherbell/CORBIS SABA; Chapter 16 @ Macduff Everton/CORBIS; Chapter 17 @ United Defense, L.P.; Chapter 18 @. Steve Cole/Getty Images. Printingnumber: 9 8 7 6 5 4 3 Printed in the United States of America on acid-free paper Eugene Grant and Dick Bernhardfor leading the field of engineering economic analysis from Don Richard Corey Eschenbach for his lifelong example of engineering leadership and working well with others from Ted My lovely wife and sweet daughters, who always support all that I do from Jerome I I - -- ",,~.,,,,..:.;~.:~, ~,,;~ .#;;., ,j~ -------- PREFACE XVII 1 'AAKING ECONOMICDECISIONS A Sea of Problems 4 Simple Problems 4 Intennediate Problems 4 Complex Problems 4 The Role of Engineering Economic Analysis 5 Examples of Engineering Economic Analysis 5 The Decision-Making Process 6 Rational Decision Making 6 Engineering Decision Making for Current Costs Summary 18 Problems 19 2 15 ENGINEERINGCOSTSAND COST ESTIMATING Engineering Costs 28 Fixed, Variable, Marginal, and Average Costs Sunk Costs 32 Opportunity Costs 32 Recurring and Nonrecurring Costs 34 Incremental Costs 34 Cash Costs Versus Book Costs 35 Life-Cycle Costs 36 28 vii viii CONTENTS Cost Estimating 38 Types of Estimate 38 Difficulties in Estimation Estimating Models Per-Unit Model 39 41 41 Segmenting Model 43 Cost Indexes 44 Power-Sizing Model 45 Triangulation 47 Improvement and the Learning Curve 47 Estimating Benefits 50 Cash Flow Diagrams 50 Categories of Cash Flows 51 Drawing a Cash Flow Diagram 51 Drawing Cash Flow Diagrams with a Spreadsheet 52 Summary 52 Problems 54 3 INTEREST AND EQUIVALENCE Computing Eash Flows 62 Time Value of Money 64 Simple Interest 64 Compound Interest 65 Repaying a Debt 66 Equivalence 68 Differencein RepaymentPlans 69 . Equivalence Is Dependent on Interest Rate 71 Application of Equivalence Calculations 72 Single Payment Compound Interest Formulas 73 Summary 81 Problems 82 4 MORE INTEREST FORMULAS Uniform Series Compound Interest Formulas 86 Relationships Between Compound Interest Factors 97 -~-~ ~---- CONTENTS ix Single Payment 97 Uniform Series 97 ArithmeticGradient 98 . Derivation of Arithmetic Gradient Factors Geometric Gradient 105 Nominal and Effective Interest 99 109 Continuous Compounding 115 Single Payment Interest Factors: Continuous Compounding 116 Uniform Payment Series: Continuous Compounding at Nominal Rate r per Period 118 Continuous, Uniform Cash Flow (One Period) with Continuous Compounding at Nominal Interest Rate r 120 Spreadsheets for Economic Analysis 122 Spreadsheet Annuity Functions 122 Spreadsheet Block Functions 123 Using Spreadsheets for Basic Graphing 124 Summary 126 Problems 129 5 PRESENTWORTH ANALYSIS Assumptions in Solving Economic Analysis Problems End-of-Year Convention 144 Viewpoint of Economic Analysis Studies 145 Sunk Costs 145 Borrowed Money Viewpoint 145 Effect of Inflation and Deflation 145 Income Taxes 146 144 Economic Criteria 146 Applying Present Worth Techniques 147 Useful Lives Equal the Analysis Period 147 Useful Lives Different from the Analysis Period Infinite Analysis Period: Capitalized Cost 154 Multiple Alternatives 158 151 Spreadsheetsand PresentWorth 162 Summary 164 Problems 165 ',/k ~. (~ ._~~~ x 6 CONTENTS ANNUAL CASH FLOWANALYSIS Annual Cash Flow Calculations 178 Resolving a Present Cost to an Annual Cost Treatment of Salvage Value .. 178 . 178 Annual Cash Flow Analysis 182 Analysis Period 184 Analysis Period Equal to Alternative Lives 186 Analysis Period a Common Multiple of Alternative Lives 186 Analysis Period for a Continuing Requirement 186 Infinite Analysis Period 187 Some Other Analysis Period 188 Using Spreadsheets to Analyze Loans 190 Building an Amortization Schedule 190 How Much to Interest? How Much to Principal? 191 Finding the Balance Due on a Loan 191 Pay Off Debt Sooner by Increasing Payments 192 Summary 193 Problems 194 7 RATE OF RETURN ANALYSIS Internal Rate of Return 204 Calculating Rate of Return 205 Plot ofNPW versus Interest Rate i 209 Rate of Return Analysis 212 Present Worth Analysis 216 Analysis Period 219 Spreadsheets and Rate of Return Analysis Summary 221 Problems 222 220 Appendix 7A Difficulties in Solving for an Interest Rate 8 INCREMENTAL ANALYSIS Graphical Solutions 246 IncrementalRateof ReturnAnalysis 252 229 CONTENTS Elements in Incremental Rate of Return Analysis 257 Incremental Analysis with Unlimited Alternatives 258 Present Worth Analysis with Benefit cost .Graphs Choosing an Analysis Method 261 Spreadsheets and Incremental Analysis 262 Summary 263 Problems 264 9 260 OTHER ANALYSISTECHNIOUES Future Worth Analysis 272 Benefit-Cost Ratio Analysis 274 Continuous Alternatives 279 Payback Period 280 Sensitivity and Breakeven Analysis 285 Graphing with Spreadsheets for Sensitivity and Breakeven Analysis Summary 293 Problems 293 10 UNCERTAINTYIN FUTURE EVENTS Estimates and Their Use in Economic Analysis 304 A Range of Estimates 306 Probability 308 Joint Probability Distributions 311 Expected Value 313 Economic Decision Trees 316 Risk 322 Risk Versus Return 324 Simulation 326 Summary 330 Problems 330 11 DEPRECIATION BasicAspectsof Depreciation 338 Deterioration and Obsolescence 338 Depreciation and Expenses 339 289 xi xii CONTENTS Types of Property 340 Depreciation Calculation Fundamentals 341 Historical Depreciation Methods 342 Straight-Line Depreciation 342 Sum-of-Years'-Digits Depreciation 344 Declining Balance Depreciation 346 Modified Accelerated Cost Recovery System (MACRS) 347 Cost Basis and Placed-in-Service Date 348 Property Class and Recovery,Period 348 Percentage Tables 349 Where MACRS Percentage Rates Crt)Come From 351 MACRS Method Examples 353 Comparing MACRS and Historical Methods 355 Depreciation and Asset Disposal Unit-of-Production Depreciation Depletion 360 Cost Depletion 360 Percentage Depletion 361 356 359 Spreadsheets and Depreciation 362 Using VDB for MACRS 363 Summary 364 Problems 365 12 INCOME TAXES A Partner in the Business 372 Calculation of Taxable Income 372 Taxable Income of Individuals 372 Classificatio~ of Business Expenditures 373 Taxable Income of Business Firms 374 Income Tax Rates 375 Individual Tax Rates 375 Corporate Tax Rates 377 Combined Federal and State Income Taxes 379 Selecting an Income Tax Rate for Economy Studies Economic Analysis TakingIncome Taxes into Account 380 380 CONTENTS Capital Gains and Losses for Nondepreciated Assets Investment Tax Credit 384 384 Estimating the After-Tax Rate of Return- -385 After-Tax Cash Flows and Spreadsheets 385 Summary 386 Problems 387 13 REPLACEMENTANALYSIS The Replacement Problem 400 Replacement Analysis Decision Maps 401 What Is the Basic Comparison? 401 Minimum Cost Life of the Challenger 402 Use of Marginal Cost Data 404 Lowest EUAC of the Defender 411 No Defender Marginal Cost Data Available 415 Repeatability Assumptions Not Acceptable 418 A Closer Look at Future Challengers 419 After-Tax Replacement Analysis 420 Marginal Costs on an After-Tax Basis 420 . After-Tax Cash Flows for the Challenger 422 Mter- Tax Cash Flows for the Defender 422 Minimum Cost Life Problems 427 Spreadsheets and Replacement Analysis Summary 429 Problems 431 429 14 INFLATIONAND PRICE CHANGE Meaning and Effect of Inflation 440 HowDoes InflationHappen? 440 . Definitions for Considering Inflation in Engineering Economy Analysis in Constant Dollars Versus Then-Current Dollars Price Change with Indexes 450 441 448 . \-- What Is a Price Index? 450 Composite Versus Commodity Indexes 453 How to Use Price Indexes in Engineering Economic Analysis 456 xiii xiv CONTENTS Cash Flows That Inflate at Different Rates 456 Different Inflation Rates per Period 458 Inflation Effect on After-TaxCf!'.c~l(lti9.ns 460 Using Spreadsheets for Inflation Calculations Summary 464 Problems 465 462 15 SELECTIONOF A MINIMUM ATTRACTIVERATE OF RETURN Sources of Capital 474 Money Generated from the Operation of the Firm External Sources of Money 474 Choice of Source of Funds 474 474 Cost of Funds 475 Cost of Borrowed Money Cost of Capital 475 Investment Opportunities Opportunity Cost 476 475 476 Selecting a Minimum Attractive Rate of Return 479 Adjusting MARRto Account for Riskand Uncertainty Inflation and the Cost of Borrowed Money 481 479 Representative Values of MARRUsed in Industry 482 Spreadsheets, Cumulative Investments, and the Opportunity Cost of Capital 483 Summary 485 Problems 485 16 ECONOMIC ANALYSIS IN THE PUBLIC SECTOR Investment Objective 490 Viewpoint for Analysis 492 Selecting an Interest Rate 493 No Time-Value-of-Money Concept Cost of Capital Concept 494 Opportunity Cost Concept 494 Recommended Concept 495 The Benefit-Cost Ratio 496 494 CONTENTS Incremental Benefit-Cost Analysis 498 Elements of the Incremental Benefit-Cost Ratio Method Other Effects of Public Projects Project Financing 505 Project Duration 506 Project Politics 507 Summary 509 Problems 510 499 . -505 17 RATIONINGCAPITALAMONG COMPETINGPROJECTS Capital Expenditure Project Proposals 518 Mutually Exclusive Alternatives and Single Project Proposals 519 Identifying and Rejecting Unattractive Alternatives 520 Selecting the Best Alternative from Each Project Proposal 521 Rationing Capital by Rate of Return 521 Significance of the Cutoff Rate of Return 523 Rationing Capital by Present Worth Methods Ranking Project Proposals 530 Summary 532 Problems 533 524 18 ACCOUNTINGAND ENGINEERINGECONOMY The Role of Accounting 540 Accounting for Business Transactions 540 The Balance Sheet 541 Assets 541 Liabilities 542 Equity 543 Financial Ratios Derived from Balance Sheet Data 543 The Income Statement 544 Financial Ratios Derived from Income Statement Data 546 Linking the Balance Sheet, Income Statement, and Capital Transactions Traditional Cost Accounting 547 Direct and Indirect Costs 548 Indirect Cost Allocation 548 546 xv xvi CONTENTS Problems with Traditional Cost Accounting Other Problems to Watch For 550 Problems 551 ApPENDIXA 549 INTRODUCTION TO SPREADSHEETS 554 The Elements of a Spreadsheet 554 Defining Variables in a Data Block 555 Copy Command 555 ApPENDIXB COMPOUND INTERESTTABLES 559 REFERENCE 591 INDEX 593 . In the first edition of this book we said: This book is designed to teach the fundamental concepts of engineering economy to engineers. By limiting the intended audience to engineers it is possible to provide an expanded presentation of engineering economic analysis and do it more concisely than if the book were written for a wider audience. Our goal was, and still is, to provide an easy to understand and up-to-date presentation of engineering economic analysis. That means the book's writing style must promote the reader's understanding. We most humbly find that our approach has been well received by engineering professors-and more importantly-by engineering students through eight previous editions. This edition has significant improvements in coverage: · Appendix7A (Difficultiesin Solving for anInterest Rate) has been thoroughlyrevised to use the power of spreadsheets to identify and resolve multiple root problems. · Chapter 10 (Probability and Uncertainty) has been completely rewritten to emphasize how to make good choices by considering the uncertainty that is part of every engineering economy application. · Chapter 12 (Income Taxes)has been updated to reflect 2003 tax legislation and rates. · Chapter 13 (Replacement Analysis) has been rewritten to clarify the comparison of existing assets with newer alternatives. Chapter 18 (Accounting and Engineering Economy) has been added in response to adopter requests. · In this edition, we have also made substantial changes to increase student interest and understanding. Thes~ include: · · ·· Chapter-opening vignettes have been added to illustrate real-world applications of the questions being studied. Chapter learning objectives are included to help students check their comprehension of the chapter material. The end-of-chapter problems have been reorganized and updated thro~ghout. The interior design is completely reworked, including the use of color, to improve readability and facilitate comprehension of the material. xvii In the first edition of this book we said: This book is designed to teach the fundamental concepts of engineering economy to engineers. By limiting the intended audience to engineers it is possible to provide an expanded presentation of engineering economic analysis and do it more concisely than if the book were written for a wider audience. Our goal was, and still is, to provide an easy to understand and up-to-date presentation of engineering economic analysis. That means the book's writing style must promote the reader's understanding. We most humbly find that our approach has been well received by engineering professors-and more importantly-by engineering students through eight previous editions. This edition has significant improvements in coverage: . . .. . Appendix7A (Difficultiesin Solvingfor anInterest Rate) has been thoroughlyrevised to use the power of spreadsheets to identify and resolve multiple root problems. Chapter 10 (Probability and Uncertainty) has been completely rewritten to emphasize how to make good choices by considering the uncertainty that is part of every engineering economy application. Chapter 12 (Income Taxes) has been updated to reflect 2003 tax legislation and rates. Chapter 13 (Replacement Analysis) has been rewritten to clarify the comparison of existing assets with newer alternatives. Chapter 18 (Accounting and Engineering Economy) has been added in response to adopter requests. In this edition, we have also made substantial changes to increase student interest and understanding. These include: . Chapter-opening vignettes have been added to illustrate real-world applications of the questions being studied. Chapter learning objectives are included to help students check their comprehension of the chapter material. The end-of-chapter problems have been reorganized and updated thr0tlghout. . The interior design is completely reworked, including the use of color, to improve readability and facilitate comprehension of the material. . . xvii xviii PREFACE The supplement package for this text has been updated and expanded for this edition. For students: · A completely rewritten Study Guide by Ed Wheeler of the University of Tennessee, Martin. . ,.' . · Spreadsheet problem modules on CD by Thomas Lacksonen of the University of Wisconsin-Stout. · Interactive multiple-choice problems on CD by William Smyer of Mississippi State University. For instructors: · · · · A substantially enlarged exam file edited by Meenakshi Sundaram of Tennessee Technological University. PowerPoiIltlecture notes for key chapters by David Mandeville of Oklahoma State University. Instructor's Manual by the authors with complete solutions to all end-of-chapter problems. The compoundinterest tables from the textbook are available in print or Excel format for adopting professors who prefer to give closed book exams. For students and instructors: · A companionwebsite is availablewith updates to these supplements at www.oup.comJ us/engineeringeconomy This ~ditionmaintains the approach to spreadsheets that was established in theprevious edition. Rather than relying on spreadsheet templates, the emphasis is on helping students learn to use the eOormouscapabilities of software that is available on every computer.This approach reinforces the traditional engineering economy factor approach, as the equivalent spreadsheet functions (PMT, PV, RATE, etc.) are used frequently. For those studentswho would benefit from a refresher or introduction on how to write good spreadsheets,there is an appendixto introduce spreadsheets.In Chapter 2, spreadsheets are used to draw cash flow diagrams. Then, from Chapter 4 to Chapter 15, every chapter has a concluding section on spreadsheet use. Each section is designed to support the other material in the chapterand to add to the student's knowledge of spreadsheets.If spreadsheets are used, the student will be very well prepared to apply this tool to real-world problems after graduation. This approachis designed to support a range of approaches to spreadsheets.Professors and students can rely on the traditional tools of engineering economy and, without loss of continuity, completelyignore the material on spreadsheets. Or at the other extreme,professors can introducethe concepts and require all computations to be done with spreadsheets. Or a mix of approaches depending on the professor, students, and particular chapter may be taken. Acknowledgments Many people have directly or indirectly contributed to the content of the book in its ninth edition. We have been influenced by our Stanford and North Carolina State University educations, our universitycolleagues, and students who have provided invaluablefeedback on content and form.We are particularly grateful to the following professors for their work ." ' . I . I '-. I : ~ I~;;;)~.:' .~ ,~'"'. i..'- ,... "~.~..,"'._' ".~_ """"~'-.... . - .-. - - - PREFACE xix on previous editions: Dick Bernhard, North Carolina State University Charles Burford, Texas Tech University Jeff Douthwaite, UniversitYofWilsbington Utpal Dutta, University of Detroit, Mercy Lou Freund, San Jose State University Vernon Hillsman, Purdue University Oscar Lopez, Florida International University Nic Nigro, Cogswell College North Ben Nwokolo, Grambling State University Cecil Peterson, GMI Engineering & Management Institute Malgorzata Rys, Kansas State University Robert Seaman, New England College R. Meenakshi Sundaram, Tennessee Technological University Roscoe Ward, Miami University Jan Wolski, New Mexico Institute of Mining and Technology and particularly Bruce Johnson, U.S. Naval Academy We would also like to thank the following professors for their contributions to this edition: Mohamed Aboul-Seoud, Rensselaer Polytechnic Institute V. Dean Adams, University of Nevada Reno Ronald Terry Cutwright, Florida State University Sandra Duni Eksioglu, University of Florida John Erjavec, University of North Dakota Ashok Kumar Ghosh, University of New Mexico Scott E. Grasman, University of Missouri-Rolla Ted Huddleston, University of South Alabama RJ. Kim, Louisiana Tech University C. Patrick Koelling, Virginia Polytechnic Institute and State University Hampton Liggett, University of Tennessee Heather Nachtmann, University of Arkansas T. Papagiannakis, Washington State University John A. Roth, Vanderbilt University William N. Smyer, Mississippi State University R. Meenakshi Sundaram, Tennessee Technological University Arnold L. Sweet, Purdue University Kevin Taaffe, University of Florida Robert E. Taylor, VIrginiaPolytechnic Institute and State University John Whittaker, University of Alberta Our largest thanks must go to the professors (and their students) who have developed the supplements for this text. These include: Thomas Lacksonen, University of Wisconsin-Stout David Mandeville, Oklahoma State University William Peterson, Old Dominion University xx PREFACE William Smyer, Mississippi State University R. Meenakshi Sundaram, Tennessee Technological University Ed Wheeler, University of Tennessee, Martin Textbooks are produced through the'efforts of many people. We would like to thank Brian Newnan for bringing us together and for his support. We would like to thank our previous . editors,PeterGordonand AndrewGyory,for theirguidance.We wouldalsolike to thank Peter for suggesting the addition of chapter-opening vignettes and Ginger Griffinfor drafting them. Our editor Danielle Christensen has pulled everything together so that this could be produced on schedule. Karen Shapiro effectivelymanaged the text's design and production. We would appreciate being informed of errors or receiving other comments about the book. Please write us c/o Oxford University Press, 198 Madison Avenue, New York, NY 10016 or through the book's website at www.oup.comlus/engineeringeconomy. ," ~ I I I ,', / I i~;;Jl.#;~i:.z;;,~~~,:::t;#ii~~~.i~L..' ~f~~-",";'~.;;1;,.;,~~;,;.~'.i>;i,::;;';~£'-c t:i£i;;.'.:':',.e '~iO"~"'i>h'''';<,~,~,,,,=~,:~~~!~. . ENGINEERING ECONOMIC ANALYSIS .. After Completing This Chapter... The student should be able to: · Distinguish between simple and complex problems. Discuss the role and purpose of engineering economic analysis. · Describe and give examples of the nine steps in the economic decision making process. · Select appropriate economic criteria for use with different type of problems. · · Solvesimpleproblemsassociatedwithengineeringdecisionmaking. QUESTIONS TO CONSIDER 1. How did the cost and weight of fireproofing material affect the engineers' decision making when the Twin Towers were being constructed? 2. How much should a builder be expected to spend on improved fireproofing, given the unlikelihood of an attack on the scale of 9/11? 3. How have perceptions of risk changed since the 9/11 attacks, and how might this affect future building design decisions? --.--------- , ~~-' .,..~.~'-1.\:~~c-.-~---r--;\. ~~lA\1J?1f~~-u Making '" Economic Decisions Could the World Trade Center Have Withstood the 9/11 Attacks? - - In the immediate aftermath of the terrorist attacks of September II, 200I, most commentators assumed that no structure, however well built, could have withstood the damage inflicted by fully fueled passenger jets traveling at top speed. But questions soon began to be raised. Investigators scrutinizing the towers' collapse noted that they had withstood the initial impact with amazing resiliency. What brought them down was the fires that followed. Knowledgeable investigators noted that the rapid progress of the Twin Towers' fires showed similarities with earlier high-rise blazes that had resulted from more mundane causes, suggesting that better fire prevention measures could have saved the ,buildingsfrom crumbling. In spring 2002, a report drafted by the Federal Emergency Management Agency and the American Society of Civil Engineers suggested that the light, fluffy spray-applied fireproofingused throughout the towers might have been particularly vulnerable to damage from an impact or bomb blast. Sturdier material had been available, but it would have added significant weight to the building. 4 MAKING ECONOMIC DECISIONS This book is about making decisions. Decision making is a broad topic, for it is a major aspect of everyday human existence. This book develops the tools to properly analyze and solve the economic problems that are commorny faced by engineers. Even very complex situations can be broken do~. into .c0l!lponents from which sensible solutions are produced. If one understands the decision-making process and has tools for obtaining realistic comparisons between alternatives, one can expect to make better decisions. Although we will focus on solving problems that confront firms in the marketplace, we will also use examples of how these techniques may be applied to the problems faced in daily life. Since decision making or problem solving is our objective, let us start by looking at some problems. A SEA OF PROBLEMS A careful look at the world around us clearly demonstrates that we are surrounded by a sea of problems. There does not seem to be any exact way of classifying them, simply because they are so diverse in complexity and "personality." One approach would be to arrange problems by their difficulty. Simple Problems On the lowerend of our classificationof problemsare simplesituations. · ·· Should I pay cash or use my credit card? · Do I buy a semester parking pass or use the parking meters? Shall we replace a burned-out motor? If we use three crates of an item a week, how many crates should we buy at a time? These are pretty simple problems, and good solutions do not require much time or effort. Intermediate Problems At the middle level of complexity we find problems that are primarily economic. ·· · ·· Shall I buy or lease my next car? Which equipment should be selected for a new assembly line? Which materials should be used as roofing, siding, and structural support for a new building? Shall I buy a 1- or 2-semester parking pass? Which printing press should be purchased? A low-cost press requiring three operators, or a m9re expensive one needing only two operators? Complex Problems At theupperendofourclassificationsystemwediscoverproblemsthatareindeedcomplex. Theyrepresenta mixtureof economic,political,anq humanisticelements. · The decision of Mercedes-Benz to build an automobile assembly plant in Thscaloosa, Alabama, illustrates a complex problem. Beside the economic aspects, MercedesBenz had to consider possible reactions in the American auto industry. Would the _.+--- -------- The Role of Engineering Economic Analysis 5 German government pass legislationto prevent the overseasplant? What about · · German labor unions? The selection of a girlfriend or a boyfriend (who may later become a spouse) is obviously complex. Economic ~alysis can be of little or no help. The annual budget ofa corPoration-isan allocation of resources, but the budget process is heavily influencedby noneconomic forces such as power struggles, geographical balancing, and impact on individuals, programs, and profits. For multinational corporations there are even national interests to be considered. THE ROLE OF ENGINEERING ECONOMIC ANALYSIS Engineering economic analysis is most suitablefor intermediate problems and the economic aspects of complex problems. They have these qualities: 1. The problem is important enough to justify our giving it serious thought and effort. 2. The problem can't be worked in one's head-that is, a careful analysis requires that we organize the problem and all the various consequences, and this is just too much to be done all at once. 3. The problem has economic aspects important in reaching a decision. When problems meet these three criteria, engineering economic analysisis an appropriate technique for seeking a solution. Since vast numbers of problems that one will encounter in the business world (and in one's personal life) meet these criteria, engineering economic analysis is often required. Examples of Engineering Economic Analysis Engineering economic analysis focuses on costs, revenues, and benefits that occur at different times. For example, when a civil engineer"designs a road, a dam, or a building, the construction costs occur in the near future; thebenefits to users begin only when construction is finished, but then the benefits continue for a long time. In fact nearly everything that engineers design calls for spending money in the design and building stages, and after completion revenues or benefits occur-usually for years. Thus the economic analysis of costs, benefits, and revenues occurring over time is called engineering economic analysis. Engineering economic analysis is used to answer many different questions. · · Which engineering projects are worthwhile? Has the mining or petroleum engineer shown that the mineral or oil deposit is worth developing? Which engineering projects should have a higher priority? Has the industrial engi- neershownwhichfactoryimprovementprojectsshouldbe fundedwiththeavailabledollars? · How should the engineering project be designed? Has the mechanical or electrical engineer chosen the most economical motor size? Has the civil or mechanical engineer chosen the best thickness for insulation? Has the aeronautical engineer made the best trade-offs between 1) lighter materials that are expensive to buy but cheaper to fly and 2) heavier materials that are cheap to buy and more expensive to fly? 6 MAKING ECONOMIC DECISIONS Engineering economic analysis can also be used to answer questions that are personally important. · How to achieve long-termfinancial goals: How much should you save each month to buy a house, retir~,offund a trip around the world? Is going to graduate school a good investment-Will your additional earnings in later years balance your lost income while in graduate school? · How to compare different ways to finance purchases: Is it better to finance your car purchase by using the dealer's low interest rate loan or by taking the rebate and borrowing money from your bank or credit union? · How to make short and long-term investment decisions: Is a higher salary better than stock options? Should you buy a 1- or 2-semester parking pass? TH,E DECISION-MAKING PROCESS Decision making may take place by default; that is, a person may not consciously recognize that an opportunity for decision making exists. This fact leads us to a first element in a definition of decision making. To have a decision-making situation, there must be at least two alternatives available.If only one course of action is available, there can be no decision making, for there is nothing to decide. There is no alternative but to proceed with the single available course of action. (It is rather unusual to find that there are no alternative courses of action. More frequently, alternatives simply are not recognized.) At this point we might conclude that the decision-making process consists of choosing from among alternativecourses of action. But this is an inadequate definition. Consider the following: At a race track, a bettor was uncertain about which of the five horses to bet on in the next race. He closed his eyes and pointed his finger at the list of horses printed in the racing program. Upon opening his eyes, he saw that he was pointing to horse number 4. He hurried off to place his bet on that horse. . Does the racehorse selection represent the process of decision making? Yes, it clearly was a process of choosing among alternatives (assuming the bettor had already ruled out the "do-nothing" alternative of placing no bet). But the particular method of deciding seems inadequate and irrational. We want to deal with rational decision making. Rational Decision Making . Rational decision making is a complex process that contains nine essential elements, which are shown sequentially in Figure 1-1. Although these nine steps are shown sequentially, it is common for decision making to repeat steps, take them out of order, and do steps simultaneously. For example, when a new alternative is identified, then more data will be required. Or when the outcomes are summarized, it may become clear that the problem needs to be redefined or new goals established. The value of this sequential diagram is to show all the steps that are usually required, and to show them in a logical order. Occasionally we will skip a step entirely. For example, a new alternative may be so clearly superior that it is immediately adopted at Step 4 without further analysis. The following sections describe the elements listed in Figure 1-1. The Decision-Making Process .. FIGURE I-lOne possible flowchartof the decision process. ~.'i ~ 7 'W 1. Recognize problel!l 2. Define the goal or objective 3. Assemble relevant data ...' ~-.. - ~ 4. Identify feasible alternatives ." - L ~ 5. Select the criterion to determine the best alternative 6. Constructa model ".., ~ , ~ ~-'''''''''''.- , "--- ~-- ~,.." ,- ~",-"->.-~ 7. Predict each alternative's outcomes or consequences 0;;< . _ __ 8. Choose the best alternative -, ..-- 9. Audit the result 1. Recognize the Problem The starting point in rational decision making is recognizing that a problem exists. Some years ago, for example, it was discovered that several species of ocean fish contained substantial concentrations of mercury. The decision-making process began with this recognition of a problem, and the rush was on to determine what should be done. Research revealedthat fishtaken from the ocean decadesbefore and preserved in laboratories also contained similar concentrations of mercury. Thus, the problem had existed for a long time but had not been recognized. In typical situations, recognition is obvious and immediate. An auto accident, an overdrawn check, a burned-out motor, an exhausted supply of parts all produce the recognition of a problem. Once we are aware of the problem, we can solve it as best we can. Many firms establish programs for total quality management (TQM) or continuous improvement (CI) that are designed to identify problems, so that they can be solved. 2. Define the Goal or Objective The goal or objective can be a grand, overall goal of a person or a firm. For example, a personal goal could be to lead a pleasant and meaningful life, and a firm's goal is usually to operate profitably. The presence of multiple, conflicting goals is often the foundation of complex problems. But an objective need not be a grand, overall goal of a business or an individual. It may be quite narrow and specific: "I want to payoff the loan on my car by May," or "The r' ~ !<. - -- i I .~.'~ .~"".; -_,.. .c-_ 8 MAKING ECONOMIC DECISIONS plant must produce 300 golf carts in the next 2 weeks," are more limited objectives. Thus, defining the objective is the act of exactly describing the task or goal. 3. Assemble Relevant Data' To make a good decision, one must first assemble good information. In addition to all the . publishedinformation,thereis a vastquantityof informationthatis not writtendownanywhere but is stored as individuals' knowledge and experience. There is also information that remains ungathered. A question like "How many people in your town would be interested in buying a pair of left-handed scissors?" cannot be answered by examining published data or by asking anyone person. Market research or other data gathering would be required to obtain the desired information. From all this information, what is relevant in a specific decision-making process? Deciding which data are important and which are not may be a complex task. The availability of data further complicates this task. Some data are available immediately at little or no cost in published form; other data are available by consulting with specific knowledgeable people; still other data require surveys or research to assemble the information. Some data will be of high quality-that is, precise and accurate, while other data may rely on individual judgment for an estimate. If there is a published price or a contract, the data may be known exactly. In most cases, the data is uncertain. What will it cost to build the dam? How many vehicles will use the bridge next year and in year 20? How fast will a competing firm introduce a competing product? How will demand depend on growth in the economy? Future costs and revenues are uncertain, and the range of likely values should be part of assembling relevant data. The problem's time horizon is part of the data that must be assembled. How long will the building or equipment last? How long will it be needed? Will it be scrapped, sold, or shifted to another use? In some cases, such as for a road or a tunnel, the life may be centuries with regular maintenance and occasional re-building. A shorter time period, such as 50 years, may be chosen as the problem's time horizon, so that decisions can be based on more reliable data. In engineering decision making, an important source of data is a firm's own accounting system. These data must be examined quite carefully. Accounting data focuses on past information, and engineering judgment must often be applied to estimate current and future values. For example, accounting records can show the past cost of buying computers, but engineering judgment is required to estimate the future cost of buying computers. Financial and cost accounting are designed to show accounting values and the flow of money-specifically costs and benefits-in a company's operations. Where costs.are directlyrelatedto specificoperations,thereis no difficulty;butthereare othercoststhatare . not related to specific operations. These indirect costs, or overhead, are usually allocated to a company's operations and products by some arbitrary method. The results are generally satisfactory for cost-accounting purposes but may be unreliable for use in economic analysis. To create a meaningful economic analysis, we must determine the true differences between alternatives, which might require some adjustment of cost-accounting data. The following example illustrates this situation. The Decision-Making Process 9 The cost-accounting records of a large company show the average monthly costs for the threeperson printing department. The wages of the three--departmentmembers and benefits, such as vacation and sick leave, make up the first category of direct labor. The company's indirect or overhead costs-such as heat, electricity, and employee insurance-must be distributed to its various departments in some manner and, like many other firms, this one usesfloor space as the basis for its allocations. $ 6,000 7,000 Direct labor (including employee benefits) Materials and supplies consumed Allocated overhead costs: 200 m2 of floor area at $25/m2 5,000 $18,000 The printing department charges the other departments for its services to recover its $18,000 monthly cost. For example, the charge to run 1000 copies of an announcement is: Direct labor Materials and supplies Overhead costs $ 7.60 9.80 9.05 Cost to other departments $26.45 The shipping department checks with a commercial printer which would print the same 1000 copies for $22.95. Although the shipping department needs only about 30,000 copies printed a month, its foreman decides to stop using the printing department and have the work done by the outside printer. The in-house printing department objects to this. As a result, the general manager has asked you to study the situation and recommend what should be done. Much of the printing department's output reveals the company's costs, prices, and other financial information. The company president considers the printing department necessary to prevent disclosing such information to people outside the company. A review of the cost-accounting charges reveals nothing unusual. The charges made by the printing department cover direct labor, materials and supplies, and overhead. The allocation of indirect costs is a customary procedure in cost-accounting systems, but it is potentially misleading for pe£isio!:lIllaIQng,~asthe.:followingdiscussion indicates~ =';;; ~ ~ = = =::;::Ii:c;r; ~ ~ c: ~ Printing Department 1000 Direct labor Materials and supplies Overhead costs ... I Outside Printer Copies 30,000 Copies 1000 Copies 30,000 Copies $ 7.60 9.80 9.05 $228.00 294.00 271.50 $22.95 $688.50 $26.45 $793.50 $22.95 $688.50 ,.. _ _ --- - r:;; . . - --- .----- I i 1 ---- \-- " ~.' --,~~'.'''' ..-,'';' ,...",., . ,.-. . -' "'.:,.-./£'~ .1 The Decision-Making Process 13 · One might wish to invest in the stock market, but the total cost of the investment is not fixed, and neither are the benefits. An automobile battery is needed. Batteries are available at different prices, and although each will provide the energy to start the vehicle, the useful lives of the · variousproductsare different. ' What should be the criterion in this category? Obviously, to be as economically efficient as possible, we must maximize the difference between the return from the investment (benefits) and the cost of the investment. Since the difference between the benefits and the costs is simply profit, a businessperson would define this criterion as maximizing profit. For the three categories, the proper economic criteria are: Category Fixed input Fixed output Neither input nor output fixed Economic Criterion Maximize the benefits or other outputs. Minimize the costs or other inputs. Maximize (benefits or other outputs minus costs or other inp:uts)or, stated another way,maximize profit. 6. Constructing the Model At some point in the decision-making process, the various elements must be brought together. The objective, relevant data, feasible alternatives, and selection criterion must be merged. For example, if one were considering borrowing money to pay for an automobile, there is a mathematical relationship between the following variables for the loan: amount, interest rate, duration, and monthly payment. Constructing the interrelationships between the decision-making elements is frequently called model building or constructing the model. Toan engineer, modeling may be a scaled physical representation of the real thing or system or a mathematical equation, or set of equations, describing the desired interrelationships. In a laboratory there may be a physical model, but in economic decision making, the model is usually mathematical. In modeling, it is helpful to represent only that part of the real system that is important to the problem at hand. Thus, the mathematical model of the student capacity of a classroom might be, lw Capacity= k where 1 = length of classroom, in meters w = widthof classroom,in meters k = classroom arrangement factor The equation for student capacity of a classroom is a very simple model; yet it may be adequate for the problem being solved. 7. Predicting the Outcomes for Each Alternative A model and the data are used to predict the outcomes for each feasible alternative. As was suggested earlier, each alternative might produce a variety of outcomes. Selecting a motorcycle, rather than a bicycle, for example, may make the fuel supplier happy, the ro.' i ~' " I I' ; "~ I ~ ~ E'"" ~ ." ">-,.I~-::--S;.~,~~~~~,;~",""".;,~;,,,,~:;,';;,,,i;;;,~'~-"';'~~';'~;"'::"':""_~'h£'- c.,-~.,;;~ 10 MAKING ECONOMIC DECISIONS The shipping department would reduce its cost from $793.50 to $688.50 by using the outside printer. In that case, how much would the printing department's costs decline? We will examine each of the cost components: \ __ '.' _ " _.. 1. Direct Labor. If the printing department had been working overtime, then.the overtime could be reduced or eliminated. But, assuming no overtime, how much would the saving be? It seems unlikely .thata printer could be fired or even put on less.than'a 40~hourwork week. Thus, although there might be a $228 saving, it is I11uchmoreJjkelythat.therewill be no reduction in direct labor. 2. Materials and Supplif?s. There would be a $294 saving ill materials aIldsupplies. ~ ~ 3. Allocated Overhead Costs: TherewiU be no reduction" inthe prirlting.dePartm~nt's monthly $5000 overhead, for there will be 110reduction in departmentf:lOOl:space. (Actually, of course, there may be a slight reduction in the AfII1:S pO~er~,g§ls!fth~l?ri1l,Wt~d~l?artInent does less work.) . j The firm will save $294 in materials and.supplies ancll)1ayorillay not save $228 ill ciiI:ect labor if the printing department no longer does the sl:1ippingdepartInellt work. }'lJ.el))'axl.mUW saving would be $294 + 228... $A2&:.~u!i~Jl!e$lril?Bing d~(>'lftIDeIJfis~pe~tt€d ~9 obtain .." its printing from the outside printer, the fifJ;llmust pay $688.50 alnonth. 17b.esavillgf,1:'omn9t doing the shipping departInent work in the printing departInerit wouldllot.exceed .$5:42,aridit. probably would be only $294, The result would bea net increase in CQstto ~heAI1).1..For this reason, the shipping department should be discouraged from sending itsprintirig to the outside printer. Gathering cost data presents other difficulties. One way to look at the financial consequences--costs and benefits-of various alternatives is as follows. . Market Consequences. These consequences have an established price in the marketplace. We can quickly determine raw material prices, machinery costs, labor costs, and so forth. . Extra-Market Consequences. There are other items that are not directly priced in the marketplace. But by indirect means, a price may be assigned to these items. (Economists call these prices shadow prices.) Examples might be the cost of an employee injury or the value to employees of going from a 5-day to a 4-day, 40-hour week. . Intangible Consequences. Numerical economic analysis probably never fully de- scribes the real differences between alternatives. The tendency to leave out consequences that do not have a significant impact on the analysis itself, or on the conversion of the finaldecision into actual money, is difficultto resolve or eliminate. How does one evaluate the potential loss of workers' jobs due to automation? What is the value of landscaping around a factory? These and a variety of other consequences may be left out of the numerical calculations, but they should be considered in conjunction with the numerical results in reaching a decision. j ~< :>--:..'-~~£~~ . '\~.~,;::--J:~t ~~,f/!~~Y~7_~~~~~~~"~~!'~~~'T-.: ~~'"-"t~~\ 't.~1~~~~:+:?:~_'-lt';:f~~P-1':(-:1~~~#)sJr\'?;;'--'- / _ _. I --, -: . 1 . . .0- - . I~;.;.'&\>..:~-~j).f.~:~>_,-:,!~';';;;i \ C!~;~<~:-:'"'""3i,~ ",-j,];.,". .:S c._i,~_>,:.,:~...~--,i;';'- ~ . -- . . ..... ~,;,...,";::J:,A'"+'. - :.! The Decision-Making Process 11 4. Identify Feasible Alternatives One must keep in mind that unless the best alternative is considered, the result will always be suboptimal.1Two types of alternatives are sometimes ignored. First, in many situations a do-nothing alternative is fe3,$ible.This may be the "Let's keep doing what we are now doing," or the "Let's not spend any money on that problem" alternative. Second, there are often feasible (but unglamorous) alternatives, such as "Patch it up and keep it running for another year before replacing it." There is no way to ensure that the best alternative is among the alternatives being considered. One should try to be certain that all conventional alternatives have been listed and then make a serious effort to suggest innovative solutions. Sometimes a group of people considering alternatives in an innovative atmosphere-brainstorming--can be helpful. Even impractical alternatives may lead to a better possibility. The payoff from a new, innovative alternative can far exceed the value of carefully selecting between the existing alternatives. Any good listing of alternatives will produce both practical and impractical alternatives. It would be of little use, however,to seriously consider an alternativethat cannot be adopted. An alternative may be infeasible for a variety of reasons. For example, it might violate fundamental laws of science, require resources or materials that cannot be obtained, or it might not be available in time. Only the feasible alternatives are retained for further analysis. 5. Select the Criterion to Determine the Best Alternative The central task of decision making is choosing from among alternatives.How is the choice made? Logically, to choose the best alternative, we must define what we mean by best. There must be a criterion, or set of criteria, to judge which alternative is best. Now, we recognize that best is a relative adjective on one end of the following relative subjective judgment: Worst Good Better relative subjective judgment spectrum Since we are dealing in relative terms, rather than absolute values, the selection will be the alternative that is relatively the most desirable. Consider a driver found guilty of speeding and given the alternatives of a $175 fine or 3 days in jail. In absolute terms, neither alternative is good. But on a relative basis, one simply makes the best of a bad situation. There may be an~nlimited number of ways that one mightjudge the various alternatives. Several possible criteria are: . Create the least disturbance to the environment. . Improve the distribution of wealth among people. 1A group of techniques called value analysis is sometimes used to examine past decisions. With the goal of identifying a better solution and, hence, improving decision making, value analysis reexamines the entire process that led to a decision viewed as somehow inadequate. 12 MAKING ECONOMIC DECISIONS ·· · · · Minimize the expenditure of money. Ensure that the benefits to those who gain from the decision are greater than the losses of those who are harmed by the decision.2 Minimize the time to accomplish the goal or objective. Minimize unemployment. .. . Maximize profit. Selecting the criterion for choosing the best alternative will not be easy if different groups support different criteria and desire different alternatives. The criteria may conflict. For example, minimizing unemployment may require increasing the expenditure of money. Or minimizing environmental disturbance may conflict with minimizing time to complete the project. The disagreement between management and labor in collective bargaining (concerning wages and conditions of employment) reflectsa disagreement over the objective and the criterion for selecting the best alternative. The last criterion-maximize profit-is the one normally selected in engineering decision making. When this criterion is used, all problems fall into one of three categories: fixed input, fixed output, or neither input nor output fixed. Fixed Input. The amount of money or other input resources (like labor, materials, or equipment) are fixed. The objective is to effectively utilize them. Examples: A project engineer has a budget of $350,000 to overhaul a portion of a petroleum refinery. You have $300 to buy clothes for the start of school. · · For economic efficiency, the appropriate criterion is to maximize the benefits or other outputs. Fixed Output. There is a fixed task (or other output objectives or results) to be accomplished. Examples: · A civil engineering firm has been given the job of surveying a tract of land and preparing a "record of survey" map. You wish to purchase a new car with no optional equipment. · The economically efficient criterion for a situation of fixed output is to minimize the costs or other inputs. Neither Input nor Output Fixed. The third category is the general situation, in which the amount of money or other inputs is not fixed, nor is the amount of benefits or other outputs. Examples: · A consulting engineering firm has more work available than it can handle. It is considering paying the staff for working evenings to increase the amount of design work it can perform. 2This is the Kaldor criterion. r: ~ r ~' Lc~",",..~...~,;;::,,,-",~, '~'.' . l :,~~~- _.:'~:i~':",..::';:~c. ";::';~::'~';':~,:::.,,,:.':';<-",,,.,,i;.,._,,,,-.,.._,...,;;~ ..~:.;;".~_>", '_'Ec~"'__"..c ---- -.,-.~~>J 14 MAKING ECONOMIC DECISIONS neighbors unhappy, the environment more polluted, and one's savings account smaller. But, to avoid unnecessary complications, we assume that decision making is based on a single criterion for measuring the relative attractiveness of the various alternatives. If necessary, one could devise a single composite criterion that is the weighted average of severaldifferent choice criteria. . ." .. . . To choose the best alternative, the outcomes for each alternative must be stated in a comparable way. Usually the consequences of each alternativeare stated in terms of money, that is, in the form of costs and benefits. This resolution of consequences is done with all monetary and nonmonetary consequences. The consequences can also be categorized as follows: Market consequences-where there are established market prices available Extra-market consequences-no direct market prices, so priced indirectly Intangible consequences-valued by judgment not monetary prices. In the initial problems we will examine, the costs and benefits occur over a short time period and can be considered as occurring at the same time. In other situations the various costs and benefits take place in a longer time period. The result may be costs at one point in time followed by periodic benefits. We will resolve these in the next chapter into a cash flow diagram to show the timing of the various costs and benefits. For these longer-term problems, the most common error is to assume that the current situation will be unchanged for the do-nothing alternative. For example, current profits will shrink or vanish as a result of the actions of competitors and the expectations of customers; and trafficcongestion normally increases overthe years as the number of vehicles increases-doing nothing does not imply that the situation will not change. 8. Choosing the Best Alternative . Earlier we indicated that choosing the best alternative may be simply a matter of determining which alternative best meets the selection criterion. But the solutions to most problems in economics have market consequences, extra-market consequences, and intangible consequences. Since the intangible consequences of possible alternatives are left out of the numerical calculations, they should be introduced into the decision-making process at this point. The alternative to be chosen is the one that best meets the choice criterion after considering both the numerical consequences and the consequences not included in the monetary analysis. During the decision-makingprocess certain feasible alternatives are eliminated because they are dominated by other, better alternatives. For example, shopping for a computer on-line may allow you to buy a custom-configured computer for less money than a stock computer in a local store. Buying at the local store is feasible, but dominated. While elimi- . nating dominated alternatives makes the decision-making process more efficient, there are dangers. Having examined the structure of the decision-making process, it is appropriate to ask, When is a decision made, and who makes it? If one person performs all the steps in decision making, then he is the decision maker. When he makes the decision is less clear. I j I 1~~;.~:J<.~#"~{~;iit~",-!!:,~i~..::~~.-:;.."",.-i~'~.'.:~-~";'2.:i;;~.;;;i,,~~~":;',~,4:~~J.f.{t~~~~ " . .--'-- ----.-.......- Engineering Decision Making for Current Costs 15 The selection of the feasible alternatives may be the key item, with the rest of the analysis a methodical process leading to the inevitable decision. We can see that the decision may be drastically affected, or even predetermined, by the way in which the decision-making process is carried out. This is illustrated by the following example. Liz, a young engineer, was assigned to make an analysis of additional equipment needed for the machine shop. The single criterion for selection was that the equipment should be the most economical, considering both initial costs and future operating costs. A little investigation by Liz revealed three practical alternatives: 1. A new specialized lathe 2. A new general-purpose lathe 3. A rebuilt lathe available from a used-equipment dealer A preliminary analysis indicated that the rebuilt lathe would be the most economical. Liz did not like the idea of buying a rebuilt lathe, so she decided to discard that alternative. She prepared a two-alternative analysis that showed that the general-purpose lathe was more economical than the specialized lathe. She presented this completed analysis to her manager. The manager assumed that the two alternatives presented were the best of all feasible alternatives, and he approved Liz's recommendation. At this point we should ask: Who was the decision maker, Liz or her n'Ianager? Although the manager signed his name at the bottom of the economic analysis worksheets to authorize purchasing the general-purpose lathe, he was merely authorizing what already had been made inevitable, and thus he was not the decision maker. Rather Liz had made the key decision when she decided to discard the most economical alternative from further consideration. The result was a decision to buy the better of the two less economically desirable alternatives. 9. Audit the Results An audit of the results is a comparison of what happened against the predictions. Do the results of a decision analysis reasonably agree with its projections? If a new machine tool was purchased to save labor and improve quality,did it? If so, the economic analysis seems to be accurate. If the savings are not being obtained, what was overlooked? The audit may help ensure that projected operating advantages are ultimately obtained. On the other hand, the economic analysis projections may have been unduly optimistic. We want to know this, too, so that the mistakes that led to the inaccurate projection are not repeated. Finally, an effective way to promote realistic economic analysis calculations is for all people involved to know that there will be an audit of the results! ENGINEERING DECISION MAKING FOR CURRENT COSTS Some of the easiest forms of engineering decision making deal with problems of alternate designs, methods, or materials. If results of the decision occur in a very short period of time, one can quickly add up the costs and benefits for each alternative. Then, using the suitable economic criterion, the best alternative can be identified. Three example problems illustrate these situations. 16 MAKING ECONOMIC DECISIONS A concrete aggregate mix is required to contain at least 31% sand by volume for proper batching. One source of material, which has 25% ~_a.nd an.~_75% coarse aggregate, sells for $3 per cubic meter (m3). Another source, which has 40% sand and 60% coarse aggregate, sells for $4AO/m3. Determine the least cost per cubic meter of blended aggregates. The least cost of blended aggregates will result from maximum us~ of the lower-cost material. The higher-cost material will be used to increase the proportion of sand up to the minimumlevel (31%) specified. .. Let x = Portion 1- x = Portion of blended aggregates froIll $4AO/m3source of blended aggregates from $3.DO/Ill3.source .'" Sand Balance x(0.25) + (1 = x) (0040)=0.31 0.25x + 0040- OAOx - 0.31 x- 0.31- 0040 · .--. -0.09 - 0.25- 0040- -0.15 =Q.60 Thus the blended aggregates will contain 60% of $3.00/Ill3Illaterial 40% of $4.40/m3 material . I The least cost per cubic meter of blended.aggregates.is I ;; 0.60($3.00) -I-0040($4.40) -1.~O-l-l.76 .'. i I ...$3.S6/m3 A machine part is manufactured at a unit cost of 40,i for material and 15,i for direct labor. . An investment of $500,000 in tooling is required. The order calls for 3 million pieces. Halfway , .0. throughthe order,a new1.11ethod of manufacturecan.be put into.effectthatwi11~reduce fue UIiit 1 costs to~4,i for materi('ilandlO,i forcdifect labor-=-but"'it::wiJl"require,:$~OQ;00Q'for~ddiijopa.I=-->1, tooling. This tooling will not be usefUlfor future orders.Oth~r costs are allocated at 2.5 tiIDesthe · direcflabor cost. What, if'<'lPything,should be done? _ __ _oIao ",,,,, ____ Jt.. ..~, ..' r ~'"--~~~ .~,:.;.;~~~k~i~~~~iL.:..;':':':-.~~;~j:_~'_,L.._~~k~~"'_d'!::~K<~~~{'...:..;:'.=1:~~i:~ Engineering Decision Making for Current Costs 17 Since there is only one way to handle first 1.5 million pieces, our problem concerns only the second half of the order. ..0 '" '... , Alternative A: Continue with Present Method 1,500,000 pieces x 0040 = Material cost Direct labor cost Other costs $600,000 225,000 562,500 1,500,000pieces x 0.15= 2.50 x direct labor cost = -, $1,387,500 Cost for remaining 1,500,000 pieces Alternative B: Change the Manufacturing Method Additional tooling cost Material cost 1,500,000 pieces x 0.34 = $100,000 510,000 Directlaborcost 1,500,000pieces x 0.10 __ 150,000 Other costs 2.50 x direct labor cost - 375,000 Cost forremaining 1,500,000 pieces $1,135,000 Before making a final decision, one should closely examine the Other costs to see that they do, in fact, vary as the Direct labor cost varies.Assuming they do, the decision would be to change the manufacturing method. ...... In the design of a cold-storage warehouse, the specifications call for a maximum heat transfer through the warehouse walls of 30,000 joules per hour per square meter of wall when there is a 30°C temperature difference between the inside surface and the outside surface of the insulation. The two insulation materials being considered are as follows: Insulation Material Rock wool Foamed insulation Cost per Cubic Meter $12.50 14.00 Conductivity (J-m/m2-oC-hr) 140 110 The basic equation for heat conduction through a wall is: Q where Q = heat = K(b.T) L transfer, in J/hr/m2 of wall K = cpnduc,tiyityin J-mlm2_oC.,hr b.T ~ clj.fferencein temperature between the two surfaces, in ~C L = thickness of insulating material, in meters Which insulation material should be selected? _ _ __ , - ... -- IiIII - .. - .. - -- - I L~ ~-~- -:ce._,""_. ,.:-",~,Lc.d~i.- ",<,.:--~~~l;;;~;,:;:d:f.i~~k:- ".f...';Si~~~::"">._i,:-,..i .'--,_"_''.';-~,~." .-,' _:.'_:"~;"h.~. .:"-i:--.~" ~ 18 MAKING ECONOMIC DECISIONS There are two steps required to solve th~_pr.()ble~:First, the required thickness of each of the alternate materials must be calculated. Then, since the problem is one of providing a fixed.output (heat transfer through the wall limited to a fixed maximum amount), the criterion is to minimize the input (cost). Required Insulation Thickness Rock wool Foamed insulation 30000 , 30,000 _ 140(30) L L = 110(30) L L-O.llm O.14m . ~ Cost of Insulation per Square Meter of Wall Unit cost~ COSt/Ill3x IlJ,sulati()lJ, thickness, in..il1e.t~J:S Rock wool Unit cost. $12.50 x 0.14Ill~ $1.75/il12 Foamedinsulation Unit cost .. $14.00 X 0.11IIi-$154/IIl2 -~ = ......... ! The foamed insulation is the lesser cost alternative.However, tl1ete.lsalvilJ,talJ,gible cOn.sttailJ,t tl1at must be considered. How thick is the available wall space? EngineeplJ,geconomy and.tljet.iil1e valUeof money are neededto decide what the maX1.mUmheattransfershould be. Whatis the cost of more insulation versus the cost of cooling the warehouse oVerits life? SUMMARY Classifying Problems Many problems are simple and thus easy to solve. Others are of intermediate difficulty and need considerable thought and/or calculation to properly evaluate. These intermediate problems tend to have a substantial economic component, hence are good candidates for economic analysis. Complex problems, on the other hand, often contain people elements, along with political and economic components. Economic analysis is still very important, but the best alternative must be selected considering all criteria-not just economics. The Decision-Making Process Rational decision.making uses a logical method of analysis to select the best alternativefrom among the feasible alternatives.The following nine steps can be followed sequentially,but decision makers often repeat some steps, undertake some simultaneously, and skip others altogether. . 1. Recognize the problem. 2. Define the goal or objective: What is the task? 3. Assemble relevant data: What are the facts? Are more data needed, and is it worth more than the cost to obtain it? 4. Identify feasible alternatives. _..-..- Problems 19 5. Select the criterion for choosing the best alternative: possible criteria include political, economic, environmental, and humanitarian. The single criterion may be a composite of several different criteria. 6. Mathematically model the various interrelationships. 7. Predict the outcomes for each ilIiemative. 8. Choose the best alternative. 9. Audit the results. Engineering decision making refers to solving substantial engineering problems in which economic aspects dominate and economic efficiency is the criterion for choosing from among possible alternatives. It is a particular case of the general decision-making process. Some of the unusual aspects of engineering decision making are as follows: 1. Cost-accounting systems, while an important source of cost data, contain allocations of indirect costs that may be inappropriate for use in economic analysis. 2. The various consequences--costs and benefits---of an alternative may be of three types: (a) Market consequences-there are established market prices (b) Extra-market consequences-there are no dii-ectmarket prices, but prices can be assigned by indirect means (c) Intangible consequences-valued by judgment, not by monetary prices 3. The economic criteria for judging alternatives can be reduced to three cases: (a) For fixed input: maximize benefits or other outputs. (b) For fixed output: minimize costs or other inputs. (c) When neither input nor output is fixed:maximizethe difference between benefits and costs or, more simply stated, maximize profit. The third case states the general rule from which both the first and second cases may be derived. 4. To choose among the alternatives, the market consequences and extra-market consequences are organized into a cash flow diagram. We will see in Chapter 3 that engineering economic calculations can be used to compare differing cash flows. These outcomes are compared against the selection criterion. From this comparison plus the consequences not included in the monetary analysis, the best alternative is selected. 5. An essential part of engineering decision making is the postaudit of results. This step helps to ensure that projected benefits are obtained and to encourage realistic estimates in analyses. PROBLEMS 1-1 Think back to your first hour after awakening this morning. List 15 decision-making opportunities that existed during that hour. After you have done that, mark the decision-making opportunities that you actuallyrecognized this morning and upon which you made a conscious decision. 1-2 Some of the followingproblemswouldbe suitablefor solution by engineering economic analysis. Which ones are they? (a) Would it be better to buy an automobile with a diesel engine or a gasoline engine? 20 MAKING ECONOMIC DECISIONS portion of all its military installations throughout the United States. While many people agreed that saving money was a desirable goal, areas potentially affected by selection to close soon reacted negatively. Congress finally selected a panel of people whose task was to develop a list of installations to close, with the legislation specifying that Congress could not alter the list. Since the goal was to save money, why was this problem so hard to solve? (b) Should an automatic machine be purchased to replace three workers now doing a task by hand? (c) Would it be wise to enroll for an early morning class so you could avoid traveling during the morningtrafficrushhours? 1-3 . .... (d) Would you be better off if you changed your major? (e) One of the people you might marry has a job that pays very little money, while another one has a professionaljob with an excellent salary.Which one should you marry? Which one of the followingproblems is most suitable for analysis by engineering economic analysis? Some 45jt candy bars are on sale for 12 bars for $3. Sandy, who eats a couple of candy bars a week, must decide whether to buy a dozen at the lower price. (b) A woman has $150,000 in a bank checking account that pays no interest. She can either invest it immediately at a desirable interest rate or wait a week and know that she will be able to obtain an interest rate that is 0.15% higher. (c) Joe backed his car into a tree, damaging the fender. He has automobile insurance that will pay for the fender repair. But if he files a claim for payment, they may change his "good driver" rating downward and charge him more for car insurance in the future. (a) 1-4 If you have $300 and could make the right decisions, how long would it take you to become a millionaire? Explain briefly what you would do. 1-5 Many people write books explaining how to make money in the stock market. Apparently the authors plan to make their money selling books telling other people how to profit from the stock market. Why don't these authors forget about the books and make their money in the stock market? 1-6 The owner of a small machine shop has just lost one of his larger customers. The solution to his problem, he says, is to fire three machinists to balance his workforce with his current level of business. The owner says it is a simple problem with a simple solution. The three machinists disagree. Why? 1-7 Every college student had the problem of selecting the college or university to attend. Was this a simple, intermediate, or complex problem for you? Explain. 1-8 Toward the end of the twentieth century, the U.S. government wanted to save money by closing a small 1-9 The college bookstore has put pads of engineering computation paper on sale at half price. What is the minimum and maximum number of pads you might buy during the sale? Explain. 1-10 Consider the seven situations described. Which one situation seems most suitable for solution by engineering economic analysis? (a) Jane has met two college students that interest her. Bill is a music major who is lots of fun to be with. Alex, on the other hand, is a fellow engineering student, but he does not like to dance. Jane wonders what to do. (b) You drive periodically to the post office to pick up your mail. The parking meters require 10jt for 6 minutes-about twice the time required to get from your car to the post office and back. If parking fines cost $8, do you put money in the meter or not? (c) At the local market, candy bars are 45 jt each or three for $1. Should you buy them three at a time? (d) The cost of automobile insurance varies widely from insurance company to insurance company. Should you check with several companies when your insurance comes up for renewal? (e) There is a special local sales tax ("sin tax") on a variety of things that the town council would like to remove from local distribution. As a result a store has opened up just outside the town and offers an abundance of these specific items at prices about 30% less than is charged in town. Should you shop there? (f) Your mother reminds you that she wants you to attend the annual family picnic. That same Saturday you already have a date with a person you have. been trying to date for months. (g) One of your professors mentioned that you have a poor attendance record in her class. You wonder whether to drop the course now or wait to see how you do on the first midterm exam. Unfortunately, the course is required for graduation. ... -------.. Problems 1-11 An automobile manufacturer is considering locating an automobile assembly plant in your region. List two simple, two intermediate, and two complex problems associated with this proposal. 1-12 Consider the following situations. Which ones appear to represent rational decision making? Explain. (a) Joe's best friend has decided to become a civil engineer, so Joe has decided that he, too, will become a civil engineer. (b) Jill needs to get to the university from her home. She bought a car and now drives to the university each day. When Jim asks her why she didn't buy a bicycle instead, she replies, "Gee, I never thought of that." (c) Don needed a wrench to replace the spark plugs in his car. He went to the local automobile supply store and bought the cheapest one they had. It broke before he had finished replacing all the spark plugs in his car. 1-13 Identify possible objectives for NASA. For your favorite of these, how should alternative plans to achieve the objective be evaluated? 1-14 Suppose you have just 2 hours to answer the question, How many people in your home town would be interested in buying a pair of left-handed scissors? Give a step-by-step outline of how you would seek to answer this question within two hours. 1-15 A college student determines that he will have only $50 per month available for his housing for the coming year. He is determined to continue in the university, so he has decided to list all feasible alternatives for his housing. To help him, list five feasible alternatives. 1-10 Describe a situation where a poor alternative was selected, because there was a poor search for better alternatives. 1-17 If there are only two alternatives available and both are unpleasant and undesirable, what should yo~ do? 1-18 The three economic criteria for choosing the best alternative are minimize input, maximize output, and maximize the difference between output and input. For each of the following situations, what is the appropriate economic criterion? (a) A manufacturer of plastic drafting triangles can sell all the triangles he can produce at a fixed price. As he increases production, his unit costs increase as a result of overtime pay and so forth. The manufacturer's criterion should be _' 21 (b) An architectural and engineering firm has been awarded the contract to design a wharf for a petroleumcompanyfor a fixedsumof money.The engineering firm's criterion should be_. '(c) A book publisher is about to set the list price (retail price) on a textbook. The choice of a low list price wouldmean less advertisingthan would be used for a higher list price. The amount of advertising will affect the number of copies sold. The publisher's criterion should be _' (d) At an auctionof antiques,a bidder for a particular porcelain statue would be trying to _' 1-19 As in Problem 1-18, state the appropriate economic criterion for each of the following situations. (a) The engineering school held a raffle of an automobile with tickets selling for'50 ~ each or three for $1. When the students were selling tickets, they notedthat manypeople had troubledeciding whetherto buy one or three tickets. Thisindicates the buyers' criterion was _' (b) A student organization bought a soft-drink machine for use in a student area. There was considerablediscussionoverwhetherthey should set the machine to charge SOt, 75~, or $1 per drink. The organizationrecognized that the number of soft drinks sold would depend on the price charged. Eventually the decision was made to charge 75~. Their criterion was_. (c) In many cities, grocery stores find that their sales are much greater on days when they have advertised their special bargains. However,the advertised special prices do not appear to increase the total physical volumeof groceries soldby a store. This leads us to conclude that many shoppers' criterion is _' (d) A recently graduated engineer has decided to return to school in the evenings to obtain a master's degree. He feels it should be accomplished in a manner that will allow him the maximum amount of time for his regular day job plus time for recreation. In working for the degree, he will _' 1-20 Seven criteria are given in the chapter for judging' which is the best alternative. After reviewing the list, devise three additional criteria that might be used. 1-21 Suppose you are assigned the task of determining the route of a new highway through an older section of town. The highway will require that many older homes must be either relocated or torn down. Two 22 MAKINGECONOMIC DECISIONS possible criteria that might be used in deciding exactly where to locate the highway are: (a) Ensure that there are benefits to those who gain from the decision and that no one is harmed by the decision. . .- neither too wildly colored nor too extreme in tailoring. Bill found three types of sport shirts in the local department store: · Rather somber shirts that Bill's father would want (b) Ensure that the benefits to those who gain from the decision are greater than the losses of those who are harmed by the decision. Which criterion will you select to use in determining the route of the highway? Explain. · him to buy Good-looking shirts that appealed to Bill Weird shirts that were even too much for Bill 1-22 Identify benefits and costs for Problem 1-21. 1-23 In the fall, Jay Thompson decided to live in a university dormitory. He signed a dorm contract under which he was obligated to pay the room rent for the full college year. One clause stated that if he moved out during the year, he could sell his dorm contract to another student who would move into the dormitory as his replacement. The dorm cost was $2000 for the two semesters, which Jay had already paid. A month after he moved into the dorm, he decided he would prefer to live in an apartment. That week, after some searching for a replacement to fulfill his dorm contract, Jay had two offers. One student offered to move in immediately and to pay Jay $100 per month for the eight remaining months of the school year. A second student offered to move in the second' semester and pay $700 to Jay. Jay estimates his food cost per month is $300 if he lives in the dorm and $250 if he lives in an apartment with three other students. His share of the apartment rent and utilities will be $200 per month. Assume each semester is 41/2months long. Disregard the small differences in the timing of the disbursements or receipts. (a) What are the three alternatives available to Jay? (b) Evaluate the cost for each of the alternatives. (c) What do you recommend that Jay do? 1-24 In decision making we talk about the construction of a model. What kind of model is meant? 1-25 An electric motor on a conveyor burned out. The foreman told the plant manager that the motor had to be replaced. The foreman indicated that there were no alternatives and asked for authorization to order the replacement. In this situation, is any decision making taking place? If so, who is making the decision(s)? 1-26 Bill Jones's parents insisted that Bill buy himself a new sport shirt. Bill's father gave specific instructions, saying the shirt must be in "good taste,"that is, · He wanted a good-looking shirt but wondered how to convince his father to let him keep it. The clerk suggestedthat Bill take home tWoshirts for his father to see and return the one he did not like. Bill.selected a good-looking blue shirt he liked, and also a weird lavender shirt. His father took one look and insisted that Bill keep the blue shirt and return the lavender one. Bill did as hisfather instructed.What was the key decision in this decision process, and who made it? 1-27 A farmer must decide what combination of seed, water, fertilizer, and pest control will be most profitable for the coming year. The local agricultural college did a study of this farmer's situation and prepared the following table. Plan A B C D Cost/acre $ 600 1500 1800 2100 Income/acre $ 800 1900 2250 2500 The last page of the college's study was torn off, and hence the farmer is not sure. which plan the agricultural college recommends. Which plan should the farmer adopt? Explain. 1-28 Identify the alternatives, outcomes, criteria, and process for the selection of your college major? Did you make the best choice for you? 1-29 Describe a major problem you must address in the next two years. Use the techniques of this chapter to . structure 1-30 the problem and recommend a decision. One strategy for solving a complex problem is to break the problem into a group of less complex problems and then find solutions to the smaller problems. The resultis the solutionof the complexproblem.. Give an example in which this strategy will work. Then give anotherexample in which this strategywill not work. 1-31 On her firstengineeringjob, Joy Hayes was giventhe responsibility of determining the production rate for a new product. She has assembled data as indicated Problems possible to achieve her boss's criterion. She asks your advice. What would you tell her? on the two graphs: $2000 ----------------------$1900 $1800 $1700 $1600 $1500 $1400 -----------------$1300 1-32 Willie Lohmann travels from city to city in the conduct of his business. Every other year he buys a used . ~ $1200 $1100 $1000 §' car for about $12,000. The auto dealer allows about $8000 as a trade-in allowance with the result that the salesman spends $4000 every other year for a car. Willie keeps accurate records, whi.ch show that all other expenses on his car amount to 22.3 ~ per mile for each mile he drives. Willie's employer has two plans by which salesmen are reimbursed for their car expenses: (a) Willie will receive all his operating expenses, and in addition will receive $2000 each year for the decline in value of the automobile. (b) Willie will receive 32~ per mile but no operating expenses and no depreciation allowance. G't ~ 23 $900 ,. $800 $700 $600 $500 $400 $300 $200 $100 , If Willie travels 18,000 miles per year, which method of computation gives him the larger reimbursement? At what annual mileage do the two methods give the same reimbursement? o 50 100 150 200 250 Output (unitslhr) 1-33 Maria, a college student, is getting ready for three final examinations at the end of the school year. Between now and the start of exams, she has 15 hours of study time available. She would like to get as high a grade average as possible in her math, physics, and engineering economy classes. She feels she must study at least 2 hours for each course and, if necessary, will settie for the low grade that the limited study would yield. How much time should Maria devote to each class if $16 $15 $14 ..... '8 $13 ~ c.. '" $12 she estimates her grade in each subject as follows: lS '0 Q $11 as u ifbI) Mathematics .S ~ tI) Study Hours I $10 ---;---I I I I $9 t I I I I I I J____ I I I 2 I I I- -- - ~-- -- ~-- -- ~-- -- $8 I I I I I I I I I $71 ~ ~ ~ o 50 I I I I I I 3 4 5 6 7 8 I I I ~---I I 100 150 200 250 Output (unitslhr) (a) Select an appropriate economic criterion and estimate.the production rate based upon it. (b) Joy's boss told Joy: "I want you to maximize output with minimum input." Joy wonders if it is r Engineering Economy Physics Grade Study Hours Grade Study Hours Grade 25 35 44 52 59 65 70 2 3 4 5 6 7 8 35 41 49 59 68 77 85 2 3 4 5 6 7 8 50 61 71 79 86 92 96 1-34 Two manufacturing companies, located in cities 90 miles apart, have discovered that they both send , their trucks four times a week to the other city full of cargo and return empty. Each company pays its driver I ~. i;, c.. ~,._~_ . <>;,~,;~;.;:" '¥"~~~;';' ;, "d;.7Ji;~._'i.:.;h,,"~;:.: ~''''~,~...''''''.J-i-i.:,., h~ ";0 ~ --;,,:,'-."-~ ; - ~;>~ --:',.";~- _~';'~.c =,.:c"ii:~~ 24 MAKING ECONOMIC DECISIONS vegetable section of the stores. The product is perishable and any remaining unsold after one week in the store is discarded. The supplier will deliver the packages to the stores, arrange them in the display space, and remove and dispose of any old packages. The price the supplierwill charge the stores depends on the size of the total weekly order for all the stores. $185 a day (the round trip takes all day) and have truck operatingcosts (excludingthe driver)of 60fC a mile. How much could each company save each week if they shared the task, with eachs-ending its truck twice a week and hauling the other comparifs cargo on the return trip? 1-35 A city is in need of increasing its rubbish disposal facilities. There is a choice of two rubbish disposal areas, as follows. Weekly Order Less than 1000packages 1000-1499 1500-1999 2000 or more Area A: A gravel pit with a capacity of 16 million cubic meters. Owing to the possibility of high ground water, however, the Regional Water Pollution-Control Board has restricted the lower 2 million cubic meters of fill to inert material only (earth, concrete, asphalt, paving, brick, etc.). The inert material, principally clean earth, must be purchased and hauled to this area for the bottom fill. The vegetable buyer estimates the quantity that can be soldper week, at various selling prices,as follows: Selling Price 60fC 45 40 33 26 Area B: Capacity is 14 million cubic meters. The entire capacity may be used for general rubbish disposal. This area will require an average increase in a round-trip haul of 5 miles for 60% of the city, a decreased haul of 2 miles for 20% of the city. For the remaining 20% of the city, the haul is the same distance as for Area A. · 1-38 Cost of inert material placed in Area A will be · · ·· . $2.35 1m3. Average speed of trucks from last pickup to disposal site is 15 miles per hour. The rubbish truck and a two-man crew will cost $35 per hour. Truck capacity of 41/2tons per load or 20 m3. Sufficient cover material is available at all areas; however, inert material for the bottom fill in Area A must be hauled in. Which of the sites do you recommend? Area B) (Answer: 1-36 An oil company is considering adding an additional grade of fuel at its service stations. To do this, an additional3000-gallon tank must be buried at each station. Discussions with tank fabricators indicate that the least expensive tank would be cylindrical with minimum surface area. What size tank should be ordered? (Answer: 8 it diameter by 8 ft length) 1-37 The vegetable buyer for a group of grocery stores has decided to sell packages of sprouted grain in the Packages Sold per Week 300 600 1200 1700 2300 The sprouted grain will be sold at the same price in all the grocery stores. How many packages should be purchased per week, and at which of the five prices listed above should they be sold? . Assume the following conditions: Price per Package 35fC 28 25 20 Cathy Gwynn, a recently graduated engineer, decided to invest some of her money in a "Quick Shop" grocery store. The store emphasizes quick service, a limited assortment of grocery items, and rather high prices. Cathy wants to study the business to see if the store hours (currently 0600 to 0100) can be changed to make the store more profitable. Cathy assembled the following information. Daily Sales in the Time Period $ 20 40 60 200 180 300 400 100 30 60 20 Time Period 0600-0700 0700-0800 0800-0900 0900-1200 1200-1500 1500-1800 1800-2100 2100-2200 2200-2300 2300-2400 2400-0100 The cost of the groceries sold averages 70% of sales. The incremental cost to keep the store open, including I I~~~..~ I c~~.c:;x.~,,, ~'''''''~'_': 'c :."'~-j~ .",:" ..< .,.,"'..;, :.~., -4~ ...- Problems the clerk's wage and other incremental operating costs, is $10 per hour. To maximize profit, when should the store be opened, and when should it be closed? 1-39 Jim Jones, a motel owner, noticed that just down the street the "Motel 36" advertises a $36-per-night room rental rate on its sign. As a result, this competitor rents all 80 rooms every day by late afternoon. Jim, on the other hand, does not advertise his rate, which is $54 per night, and he averages only a 68% occupancy of his 50 rooms. There are a lot of other motels nearby, but only Motel 36 advertises its rate on its sign. (Rates at the other motels vary from $48 to $80 per night.) Jim estimates that his actual incremental cost per night for each room rented, rather than remaining vacant, is $12. This $12 pays for all the cleaning, laundering, maintenance, utilities, and so on. Jim believes his eight alternatives are: Resulting Occupancy Rate Alternative Advertise 1 2 3 4 and Charge $35 per night 42 per night 48 per night 54 per night Do Not Advertise 100% 94 80 66 5 6 7 8 and Charge $48 per night 54 per night 62 per night 68 per night 70% 68 66 56 What should Jim do? Show how you reached your conclusion. 1-40 A firm is planning to manufacture a new product. The sales department estimates that the quantity that can be sold depends on the sellingprice. As the selling price is increased, the quantity that can be sold decreases. Numerically they estimate: P = $35.00 - 0.02Q where P = selling price per unit Q = quantity sold per year On the other hand, the management estimates that the average cost of manufacturing and selling the product will decrease as the quantity sold increases. They estimate C where C = cost = $4.00Q + $8000 to produce and sell Q per year The firm's management wishes to produce and sell the product at the rate that will maximize profit, that is, where income minus cost will be a maximum. What quantity should the decision makers plan to produce and sell each year? (Answer: 775 units) 1-41 A manufacturing firm has received a contract to assemble 1000units of test equipment in the next year. The firm must decide how to organize its assembly operation. Skilled workers, at $22 per hour each, could be assigned to individually assemble the test equipment. Each worker would do all the assembly steps, and it would take 2.6 hours to complete one unit. An alternate approach would be to set up teams of four less skilledworkers (at $13 per hour each) and organize the assembly tasks so that each workerdoes part of the assembly. The four-man team would be able to assemble a unit in one hour. Which approach would result in more economically assembly? 1-42 A grower estimates that if he picks his apple crop now, he will obtain 1000 boxes of apples, which he can sell at $3 per box. However, he thinks his crop will increase by 120 boxes of apples for each week he delays picking, but that theprice will drop at a rate of l5~ per box per week; in addition, he estimates approximately 20 boxes per week will spoil for each week he delayspicking. Whenshould he pick hiscrop to obtain the largest total cash return? How much will he receive for his crop at that time? II f" ;. ~ I i-. ~i:.~,~ 25 1 ~ -"~~'O"'c:;~ --1- l.>: ~s;t~~ ,~J<~.",z"..~-,.";'.~"'~.';;.~:~:d~ -.;.."";... :.,;,.:: "/ci._~..;' ;,'-'. ~:. :~."">, <>','-.c,.~"-'. , ,.. ". i~...:.: {;,j After Completing This Chapter... The student should be able to: ·· · · · ·· · Define various cost concepts. Provide specificexamples of how and why these engineering cost concepts areimportant. Define engineering cost estimating. Explain the three types of engineering estimate, as well as common difficulties encountered in making engineering cost estimates. Use several common mathematical estimating models in cost estimating. Discuss the impact of the learning curve on cost estimates. . State the relationship between cost estimating and estimating project benefits. Draw cashflow diagrams to show project costs and benefits. QUESTIONS TO CONSIDER ;;. 1. By investing heavily in warehouses and other infrastructure, Webvan incurred large "fixed costs" that it would have to pay regardless of whether it attracted customers. By contrast, Tesco invested a more modest sum up front and hired employees only when customer orders increased enough to warrant it. How might these choices have affected the financial fates of the two companies? 2. In most cases, businesses that are seeking 'financing for start-up or expansion must develop detailed estimates of their likely costs and future earnings. But Webvan convinced investors that it was operating in a "new world" of Internet COlTIlTIerce, to which the old rules did not apply. How did this affect investors' willingness to accept Webvan's estima.tesof its financial prospects? 3. Generally, businesses view cost considerations as a constraint. Inthis case, however,Jhe dotcom boom of the 1990s stood that rule on its head: the more Webvan spent, the more money investors seemed willing to give the company-at least until the boom ran its course. Would Webvan have been better off with investors who asked more questions and imposed more limits? Why or why not? ' '. ..,- .",.~ c_-.:__-,-.-.-~-,,,, _ 4"'@OO~lifIt~ Engineeri ng Costs and Cost Estimating - Webvan Hits the Skids Webvan, an on-line supermarket, aimed to revolutionize the humdrum business of selling groceries. Consumers could order their weekly provisions with a few clicks and have the goods delivered right to their door. It sounded like a great business plan, and the company had no trouble attracting capital during the dotcom boom of the late 1990s.Eager investors happily poured hundreds of millions into the company. With that kind of money to spend, Webvan invested lavishly in building infrastructure, including large warehouses capable of filling 8000 orders a day. The firm rapidly expanded to serve multiple cities nationwide and even acquired a competing on-line company, Home Grocer. But the hoped-for volume of customers never materialized. By early 2000, Internet grocers had managed to capture only a small part of the food sales market-far short of the 20% they had anticipated. When the dotcom boom went bust, Webvan suddenly looked much less attractiveto investors,who quickly snapped their wallets shut. Without new money coming in, Webvan suddenly had to face an uncomfortable fact: it was spending far more than it was earning. Finally, in 2001, Webvan went bankrupt. A rival on-line grocer, Peapod, narrowly escaped the same fate-but only because a Dutch retailer was willing to buy the company and continue pumping money into it. Interestingly, at the same time Webvan was burning through millions in dotcom cash, a bricks-and-mortar supermarket chain in Britain called Tesco also decided to get into the on-line grocery business. Tesco invested around $56 million in a computerized processing system and, instead of building warehouses, had employees in each store walk the aisles filling orders. Unlike Webvan,Tesco made a profit. II I \ .t~i' ;~~.JiL~~;,,~ ,.;;;~~.~.. .'--;-~i~,,_;,~~~~~~~~,:.~." . ~.:"~::'.~~~,"';~-F...);.> ~~ ;.,~~ ;""..;" ,,~.-- .-: .>,,' -. ~.~ - : -~: ."~~ ':', " ~.-"~.;'-:~j 28 ENGINEERINGCOSTSAND COST ESTIMATING This chapter defines fundamental cost concepts. These include fixed and variable costs, marginal and average costs, sunk and opportunity costs, recurring and nonrecurring costs, incremental cash costs, book costs, and life-cycle costs. We then describe the various types of estimates and difficulties sometimes encountered. The models that are described include unit factor, segmenting, cost indexes";power sizing, triangulation, and learning curves. The chapter discusses estimating benefits, developing cash flow diagrams, and drawing these diagrams with spreadsheets. Understanding engineering costs is fundamental to the engineering economic analysis process, and therefore this chapter addresses an important question: Where do the numbers come from? ENGINEERING COSTS Evaluating a set of feasible alternatives requires that many costs be analyzed. Examples include costs for initial investment, new construction, facility modification, general labor, parts and materials, inspection and quality, contractor and subcontractor labor, training, computer hardware and software, material handling, fixtures and tooling, data management, and technical support, as well as general support costs (overhead).In this section we describe several concepts for classifying and understanding these costs. Fixed, Variable, Marginal, and Average Costs Fixed costs are constant or unchanging regardless of the level of output or activity. In contrast, variable costs depend on the level of output or activity. A marginal cost is the variable cost for one more unit, while the average cost is the total cost divided by the number of units. For example, in a production environment fixed costs, such as those for factory floor space and equipment, remain the same even though production quantity, number of employees, and level of work-in-process may vary.Labor costs are classified as a variable cost because they depend on the number of employees in the factory. Thusfixed costs are level or constant re~ardless of output or activity, and variable costs are changing and related to the level of output or activity. As another example, many universities charge full-time students a fixed cost for 12 to 18 hours and a cost per credit hour for each credit hour over 18. Thus for full-time students who are taking an overload (> 18 hours), there is a variable cost that depends on the level of activity. This example can also be used to distinguish between marginal and average costs. A marginal cost is the cost of one more unit. This will depend on how many credit hours the student is taking. If currently enrolled for 12 to 17 hours, adding one more is free. The marginal cost of an a~ditional credit hour is $0. However,if the student is taking 18 or more . hours, then the marginal cost equals the variable cost of one more hour. To illustrate average costs, the fixed and variable costs need to be specified. Suppose the cost of 12 to 18 hours is $1800 per term and overload credits are $120/hour.If a student takes 12 hours, the average cost is $1800/12 = $150 per credit hour. If the student were to take 18 hours, the average cost decreases to $1800/18 = $100 per credit hour. If the student takes 21 hours, the average cost is $102.86 per credit hour [$1800 + (3 x $120) /21]. r- I ~-. ~ .. ~ . - ':- '. ,. -. ~ '-. ...~ . -, ~¥~f~~~~~.~~~:,-;)";!-.~j1J;:Ji2~~~~;~~~.~{;;;.l~~,~£.~~.~:.::{~j~'r)~~~'~>".i;-~ ---~- . r Engineering Costs 29 Average cost is thus calculated by dividing the total cost for all units by the total number of units. Decision makers use average cost to attain an overall cost picture of the investment on a per unit basis. Marginal cost is used to decide w~~therthe additional unit should be made, purchased, or enrolled in. For the full-time'studentat our exampleuniversity,the marginalcost of another credit is $0 or $120 depending on how many credits the student has already signed up for. An entrepreneur named DK was considering the money-making potential of chartering a bus to take people from his hometown to an event in a larger city. DK planned to provide transportation, tickets to the event, and refreshments on the bus for his customers.He gathered data and categorized the predicted expenses as either fixed or variable. DK's Fixed Costs Bus rental Gas expense Other fuels Bus driver DK's Variable Costs Event ticket Refreshments $80 75 20 50 $12.50 per person 7.50 per person Develop an expression of DK's total fixed and total variable costs for chartering this trip. , SOlUT.ION DK's fixed costs will be incurred regardless of how many people sign up for the trip (even if only one person signs up!). These costs include bus rental, gas and fuel expense, apd the cost to hire a driver: Total fixed costs = 80 + 75 + 20 + 50 = $225 DK's variable costs depend on how many people sign up for the charter, which is the level of activity. Thus for event tickets and refreshments, we would write = - - From Example 2-1 we see how it is possible to calculate total fixed and total variable costs. Furthennore, these values can be combined into a single total cost equation as follows: Total cost = Total fixed cost + Total variable cost (2-1) The relationship between total cost and fixedand variable costSare shownin Figure 2-1. The fixed-cost portion of $3000 is the same across the entire range of the output variable x. Often, the variable costs are linear (y equals a constant times x); however, the variable costs can be nonlinear. For example, employees are often paid at 150% of their hourly rate for overtime hours, so that production levels requiring overtime have higher variable costs. .~ , I I I \ { .~~~~ _: I. ...:.~:<,,:.<..-~~ ...:~:-:~~ -'L '~~ :.,.. ~_i£~. ...z~-:: ", ,.~-~.:.;~,: :;~:; .~_co, j-~~~ ~.:-. - ,;~...~..~,~.;. : ~:.-~...,~;'.::..~ _..:."- . '~, ~-, :;,;,~.:..~;..-}~.; ~~~:.Jj 30 ENGINEERING COSTS AND COST ESTIMATING $8000 FIGURE 2-1 Fixed, variable, and total costs. Total Cost $6000 ... en 8 Variable Cost $4000 $2000 Fixed Cost 5 15 10 Volume Totalcost in Figure2-1 is a fixedcost of $3000plus a variablecost of $200 per unit for straight-time production of up to 10 units and $300 per unit for overtime production of up to 5 moreunits. . Figure 2-1 can also be used to illustrate marginal and average costs. At a volume of 5 units the marginal cost is $200 per unit, while at a volume of 12 units the marginal cost is $300 per unit. The respective average costs are $800 per unit, or (3000 + 200 x 5)/5, and $467 per unit, or (3000 + 200 x 10 + 300 x 2)/12. In Example 2-1, DK develop~d ap overall total cost equation for his business expenses. Now he wants to evaluate the potential to make money from this chartered bus trip. SOLlJTION ! ' ~. We use Equation 2-1 to find DK's total cost equation: Total cost"_. total fixed'cost +~otal'variable'cost = $225 + - ;;; ($20)(number of people on the trip) where number of people on the trip ='X. Thus, ::: Total cost = 225 + 20x Using this relationship, DK can calculate the total cost for any number of people-:-up to the fcapacity of the bus. What he lacks is a revenue equatiQnto offset his costs. DK's total revenue :; from this trip can.b~expressed as: i'! = Total revenue:::: (Charter ticketprice)(Number ~- -- ;;;;:: = ~ of people on the trip) - - - = (Ticket price) (x) -- -- - - -... 11';' ~: \i.. / 1.~~._;'$~~k~~~~"~~,_~;;jl.,=:,.~~~:.,,.~.,,:;'.~~_.,.;:,_':~~2.iiL'f~~:~ Engineering Costs 31 DK believes that he could attract 30 people at a charter ticket price of $35. Thus Total profit = (Total revenue) - (Tot~ costs) .':"-- (35x) - (225 + 20x)- 15x - 225 .. ',,;;~'" :-:. Atx - 30, Total profit = 35 x 30 '"- (225 + 20 x 30) = $225 I'! So, if3P people take the charter,DKwill net acprofit of $225. Th§ ,§ojIle}Vh~tsimpA$tic.analysis ignores the value of DK's time-he would have to "pay himself" out of his $?25 profit. In Examples 2-1 and 2-2 DK developed totalcost and total revenueequations to describe the charter bus proposal. These equations can be used to create what is called a profit-loss breakeven chart (see Figure 2-2). Both the costs and revenuesassociated with various levels of output (activity) are placed on the same set of x-y axes. This allows one to illustrate. a breakevenpoint (in terms of costs and revenue) and regions of profit and loss for some business activity. These terms can be defined as follows. Breakeven point: The level of business activity at which the total costs to provide the product, good, or service are equal to the revenue (or savings) generated by providing the service. This is the level at which one "just breaks even." Profit region: The output levelof the variablex greater than the breakevenpoint, where total revenue is greater than total costs. Loss region: The output level of the variable x less than the breakeven point, where . total costs are greater than total revenue. Notice in Figure 2-2 that the breakeven point for the number of persons on the charter trip is 15 people. For more than 15 people, DK will make a profit. If fewer than 15 sign up FIGURE 2-2 Profit-loss breakeven chart for Examples 2-1 and 2-2. $1200 $1000 $800 .... 8 ~ $600 Totalrevenue y = 35x...... .. .. .. .... .. .. $400 $200 Breakeven point $0 0 5 10 15 20 Customers .. 32 ENGINEERING COSTS AND COST ESTIMATING there will be a net loss. At the breakeven level the total cost to provide the charter equals the revenue received from the 15passengers. Wecan solve for the breakevenpoint by settingthe total costs and total revenue expressions equal to each other and solving for the unknown value of x. From Examples 2-1 and 2-2: Total cost $225 + 20x x = Total revenue = 35x = 15 people Su n k Costs A sunk cost is money already spent as a result of a past decision. Sunk costs should be disregarded in our engineering economic analysis because current decisions cannot change the past. For example, dollars spent last year to purchase new production machinery is money that is sunk: the money allocated to purchase the production machinery has already been spent-there is nothing that can be done now to change that action. As engineering economists we deal with present and future opportunities. Many times it is difficult not to be influenced by sunk costs. Consider 100 shares of stock in XYZ, Inc., purchased for $15 per share last year. The share price has steadily declined over the past 12 months to a price of $10 per share today. Current decisions must focus on the $10 per share that could be attained today (as well as future price potential), not the $15 per share that was paid last year. The $15 per share paid last year is a sunk cost and has no influence on present opportunities. As anotherexample, when Regina was a sophomore, she purchased a newest-generation laptop from the college bookstore for $2000. By the time she graduated, the most anyone would pay her for the computer was $400 because the newest models were faster, cheaper and had more capabilities. For Regina the original purchase price was a sunk cost that has no influence on her present opportunity to sell the laptop at its current market value ($400). Opportu n ity Costs An opportunity cost is associated with using a resource in one activity instead of another. Every time we use a business resource (equipment, dollars, manpower, etc.) in one activity, we give up the opportunity to use the same resources at that time in some other activity. Every day businesses use resources to accomplish various tasks-forklifts are used to transport materials, engineers are used to design products and processes, assembly lines are used to make a product, and parking lots are used to provide parking for employees' vehicles. Each of these resources costs the company money to maintain for those intended purposes. However, that cost is not just made up of the dollar cost, it also includes the opportunity cost. Each resource that a firm owns can feasibly be used in several alternative ways. For instance, the assembly line could produce a different product, and the parking lot could be rented out, used as a building site, or converted into a small airstrip. Each of these alternative uses would provide some benefit to the company. A firm that chooses to use the resource in one way is giving up the benefits that would be derived from using it in those other ways. The benefit that would be derived by using the resource in this "other activity" is the opportunity cost for using it in the chosen activity. Opportunity cost may also be considered a forgone opportunity cost because we are forgoing the benefit that could have been realized. A formal definition of opportunity ~' b .';'~. ' ;c~::'"' ., ",'.. '">"'~";~"",~,,,, ~J''''''i';'''.''~'~",;~;"",~~","",;,;,,"-~ - ---- ---.------ Engineering Costs 33 cost might be: An opportunity cost is the benefit that is forgone by engaging a business resource in a chosen activity instead of engaging that same resource in the forgone activity. As an example, suppose that friends invite a college student to travel through Europe over the summer break. In considering the offer, the student might calculate all the out-ofpocket cash costs that would be incurred. Cost estimates might be made for items such as air travel, lodging, meals, entertainment, and train passes. Suppose this amounts to $3000 for a to-week period. After checking his bank account, the student reports that indeed he can afford the $3000 trip. However, the true cost to the student includes not only his outof-pocket cash costs but also his opportunity cost. By taking the trip, the student is giving up the opportunity to earn $5000 as a summer intern at a local business. The student's total cost will comprise the $3000 cash cost as well as the $5000 opportunity cost (wages forgone)-the total cost to our traveler is thus $8000. A distributor of electricpumps must decide what to do with a "lot" of old electric pumps purchased 3 years ago. Soon after the distributor purchased the lot, technology advances made the old pumps less desirable to customers. The pumps are becoming more obsolescent as they sit in inventory. The pricing manager has the following information. Distributor's purchase price 3 years ago Distributor's storage costs to date Distributor's list price 3 years ago Current list price of the same number of new pumps Amount offered for the old pumps from a buyer 2 years ago Current price the lot of old plimps would bring $ 7,000 1,000 9,500 12,000 5,000 3,000 Looking at the data, the pricing manager has concluded that the price should be set at $8000. This is the money that the firm has "tied up" in the lot of old pumps ($7000 purchase and $tOOO stor~$e), and it ~as re~soned th~t the c..oml?~.yshould at least recqy~r pup f;os}.Fu_rthermo.r~, the pricing manager has argtied that an $8000 price would be $1500 less than the list price from 3 years ago, and it would be $4000 less than what a lot of new pumps would cost ($12,000 $8000). What would be your advice on price? " = - = ~ SOlUTlpN Let's look'more closely at each of the data items. DistrIbutor's purchase price 3 years ago: This is a sunk cost that should not be considered in setting the price today. 1:1 Distributor's storage costs to date: The storage costs for keeping the pumps in inventory a],:esunk costs; that is, they have been paid. Hence they should not influence the pricing decision. ------------- --- - - - __I I 34 ENGINEERING COSTS AND COST ESTIMATING Distributor's listprice 3 years ago: If there have been no willing buyers in the past 3 years at this price, it is unlikely that a buyer will emerge in the future. This past list price should have no influence on the current pricing decision. ~ Current listprice of newerpumps: Newer pumps now include technology and features that have made the older pumps less valuable. Directly comparing the older pumps to those with new technology is misleading. However, the price of the new pumps and the value of the new features help determine the market value of the old pumps. Amount offeredfrom a buyer 2 years ago: This is a forgone opportunity. At the time of the offer, the company chose to keep the lot and thus the $5000 offered became an opportunity cost for keeping the pumps. This amount should not influence the current pricing decision. Current price the lot could bring: The price a willing buyer in the marketplace offers is called the asset's market value. The lot of old pumps in question is believed to have a current market value of $3000. From this analysis, it. is easy to see the flaw in the' pricing manager's reasoning. In an engineering economist analysis we deal only with today's and prospective future opportunities. It is impossible to go back in time and change decisions that have been made. Thus, the pricing manager should recommend to the distributor that the price be set at the current value that a buyer assigns to the item: $3000. . Recurring and Nonrecurring Costs Recurring costs refer to any expense that is known, anticipated, and occurs at regular intervals. Nonrecurring costs are one-of-a-kind expenses that occur at irregular intervals and thus are sometimes difficult to plan for or anticipate from a budgeting perspective. Examples of recurring costs include those for resurfacing a highway and reshingling a roof. Annualexpenses for maintenance and operation are also recurring expenses. Examples of nonrecurring costs include the cost of installing a new machine (including any facility modifications required), the cost of augmenting equipment based on older technology to restore its usefulness, emergency maintenance expenses, and the disposal or close-down costs associated with ending operations. In engineering economic analyses recurring costs are modeled as cash flowsthat occur at regular intervals.(such as every year or every 5 years.) Their magnitude can be estimated, and they can be included in the overall analysis. Nonrecurring costs can be handled easily in our analysis if we are able to anticipate their timing and size. However,this is not always so easy to do. Incremental Costs One of the fundamental principles in engineering economic analysis is that in making a choice among a set of competing alternatives, focus should be placed on the differences between those alternatives. This is the concept of incremental costs. For instance, one ~~- Engineering Costs 35 may be interested in comparing two options to lease a vehicle for personal use. The two lease options may have several specifics for which costs are the same. However, there may be incremental costs associated with one option not required or stipulated by the other. In comparing the two leases, the focus should be on the differences between the alternatives, not on the costs that are the same. Philipis choosingbetweenmodelA (a budgetmodel)and model B (with more featuresand a . higher purchase price). What incremental costs would Philip incur if he chose model B instead of the less expensive model A? Cost Items Model B Model A $10,000 $17,500 Purchase price Installation costs 3,500 5,000 Annual maintenance costs 750 2,500 1,200 2,000 Annual utility expenses 700 500 Disposal costs after useful life . .- j SOLUTION . i. -. We are interested in the incremental or extra costs that are associated with choosing model B instead of model A. To obtain these we subtract model A costs from model B costs for each category (cost item) with the following results. Incremental - Cost Items Purchase price Installation costs Annual maintenance costs Annual utility expenses Disposalcosts afterusefullife = ;; (Model B Cost A Cost) 17,500 - 10,000 Cost of B $7500 5,000 - 3,500 750 -2,500 2,000 - 1,200 1500 -1750/yr 800/yr 500 ~ 700 -200 ~otice that for the cost categooesgiven,.the~incremel1tal costso(model B are both positive 'I:; and negative. Positive incremental costs mean that model B costs more than model A, and negative incremental costs indicate that there would be a savings (reduction in cost) if model B where chosen instead. Because model Bhas more featiife"S', a~d~'tisif)n =wou1~ctlsohTlvelo reflecTconsid~raT1onthe :t:; incremental benefits offered by that model. Cash Costs Versus Book Costs A cash cost requires the cash transaction of dollars "out of one person's pocket" into "the pocket of someone else." When you buy dinner for your friends or make your monthly automobile payment you are incurring a cash cost or cash flow. Cash costs and cash flows are the basis for engineering economic analysis. . I ! . .. .' , -. .'. ~:i..~s<::.:,..:~::::.:~-~_'E:~~!.-b .~, .>;>.~~~~~~..- ~.,~;:..t.:~~,.~:,.-'~-~-~ : : ~.~0~~-'J..~~~ . r ' -__~ . -::'..'L- - ~'. - ..". ~-~~-~~~"";.~.~ -~ 36 ENGINEERING COSTS AND COST ESTIMATING Book costs do not require the transaction of dollars "from one pocket to another." Rather, book costs are cost effects from past decisions that are recorded "in the books" (accounting books) of a firm. In one common book cost, asset depreciation (which we discuss in Chapter 11), the expense paid for a particular business asset is "written off" on a company's accounting sy'stem over a number of periods. Book costs do not ordinarily represent cash flows and thus are not included in engineering economic analysis. One exception to this is the impact of asset depreciation on tax payments-which are cash flows and are included in after-tax analyses. Life-Cycle Costs . The products, goods, and services designed by engineers all progress through a life cycle very much like the human life cycle. People are conceived, go through a growth phase, reach their peak during maturity, and then gradually decline and expire. The same general pattern holds for products, goods, and services. As with humans, the duration of the different phases, the height of the peak at maturity, and the time of the onset of decline and termination all vary depending on the individual product, good, or service. Figure 2-3 illustrates the typical phases that a product, good or service progresses through over its life cycle. Life-cycle costing refers to the concept of designing products, goods, and services with a full and explicit recognition of the associated costs over the various phases of their life cycles. Two key concepts in life-cycle costing are that the later design changes are made, the higher the costs, and that decisions made early in the life cycle tend to "lock in" costs that are incurred later. Figure 2-4 illustrates how costs are committed early in the product Time Beginning Needs COJ;lceptual AS$eSsment and Justification Phe or PreXimjI)ary DeSign Ehase Detailed r>e$igJ:lEhase ."'. or Constnicti!:m Phase ) End ."'- QperalioniJ,l U§eEhase Decline and Retirement Phase ..,,- Requirements Impact Analysis Allocation of Resources Overall Feasibility Proof of Concept Detailed Specifications Conceptual Prototype/ Design Planning Breadboard Development and Testing Component and Supplier Selection Product, Goods Operational Use & Services Built All Supporting Facilities Built Use by Ultimate Customer Operational Use Maintenance 'Planning and Support Production or Construction Phase Processes, Materials and Methods Use Declining Use Phase Out Retirement Responsible Disposal Decline and Retirement Planning Detailed Design Planning FIGURE 2-3 Typical life cycle for products, goods and services. ~~ ~. I 1~:~~~~~.',~~.~.;~~~~~~~.~;~~~~ c;-=-..:+",~~",,~.,.1""<'~~;~&'~.ii~ -;:-j~~.;~~;':.i._"~y;:g,;,j;'~~"'"-!Jj"~:"L~::;;'-' Engineering Costs 37 100% ... CI) o U 80% II) >. 60% U ~ ;.::s 40% "0 I '5 ~ 20% JTime Project Phase FIGURE 2-4 Cumulativelife-cycle costs committedand dollars spent. High II) b/) --------... ...... Ease of ......... Changing Design', !:: o:s ..r:: U '+-< o ... CI) ,, ,, ,, , o U ... o II) CI) o:s ~ Time Project Phase FIGURE 2-5 Life-cycle design change costs and ease of change. life cycle-nearly 70-90% of all costs are set during the design phases. At the same time, as the figure shows, only 10-30% of cumulative life-cycle costs have been spent. Figure 2-5 reinforces these concepts by illustrating that downstream product changes' are more costly and that upstream changes are easier (and less costly) to make. When planners try to save money at an early design stage, the result is often a poor design, calling for change orders during construction and prototype development. These changes, in turn, are more costly than working out a better design would have been. From Figures 2-4 and 2-5 we seethat the time to consider all life-cycle effects, and make design changes, is during the needs and conceptual/preliminary design phases-before a lot of dollars are committed. Some of the life-cycle effects that engineers should consider t~ r.., ' ,I I ..J~~~;1;.~~:;;.~ti_-.,~~.-/,,,i,.;i'h.,;; :;~:£~Ji ,",,.;L: _}.i,,,,..~~,::ji,,,,.c_-,;i;,.;.,,,.-;::;;,:;,.::;. ..< ."I;j:""3,~.,;,;i~ 38 ENGINEERINGCOSTS AND COST ESTIMATING at design time include product costs for liability, production, material, testing and quality assurance, and maintenance and warranty. Other life-cycle effects include product features based on customer input and product disposal effects on the environment. The key point is that engineers who design products and the systems that produce them should consider all life-cycle costs. . .. COST ESTIMATING Engineering economic analysis focuses on the future consequences of current decisions. Because these consequences are in the future, usually they must be estimated and cannot be known with certainty. Examples of the estimates that may be needed in engineering economic analysisincludepurchase costs, annualreven.ue,yearly maintenance, interestrates for investments, annual labor and insurance costs, equipment salvage values, and tax rates. Estimating is the foundation of economic analysis. As is the case in any analysis procedure, the outcome is only as good as the quality of the numbers used to reach the decision. For example, a person who wants to estimate her federal income taxes for a given year could do a very detailed analysis, including social security deductions, retirement savings deductions, itemized personal deductions, exemption calculations, and estimates of likely changes to the tax code. However, this very technical and detailed analysis will be grossly inaccurate if poor data are used to predict the next year's income. Thus, to ensure that an analysis is a reasonable evaluation of future events, it is very important to make careful estimates. Types of Estimate The American poet and novelist Gertrude Stein wrote in The Making of Americans in 1925 that "a rose is a rose is a rose is a rose." However, what holds for roses does not necessarily hold for estimates because "an estimate is not an estimate." Ms. Stein was not suggesting that all roses are the same, but it is true that all estimates are not the same. Rather, we can define three general types of estimate whose purposes, accuracies, and underlying methods are quite different. Rough estimates: Order-of-magnitude estimates used for high-level planning, for determining macrofeasibility, and in a project's initial planning and evaluation phases. Rough estimates tend to involve back-of-the-envelope numbers with little detail or accuracy. The intent is to quantify and consider the order of magnitude of the numbers involved. These estimates require minimum resources to develop, and their accuracy is generally -30% to +60%. Notice the nonsymmetry in the estimating error. This is because decision makers tend to underestimate the magnitude of costs (negative economic effects). Also as Murphy's law predicts, there seem to be more ways for results to be worse than expected than there are for the results to be better than expected. Semidetailed estimates: Used for budgeting purposes at a project's conceptual or preliminary design stages. These estimates are more detailed, and they require additional time and resources to develop. Greater sophistication is used in developing Cost Estimating 39 semidetailed estimates than the rough-order type, and their accuracy is generally -15 to +20%. Detailed estimates: Used during ap!oject' sdetailed design and contractbidding phases. These estimates are made from detailed quantitative models, blueprints, product specification sheets, and vendor quotes. Detailed estimates involve the most time and resources to develop and thus are much more accurate than rough or semidetailed estimates. The accuracy of these estimates is generally -3 to +5%. The upper limits of +60%for rough order, +20%for semi-detailed, and +5%for detailed estimates are based on construction data for plants and infrastructure. Final costs for software, research and development, and new military weapons often correspond to much higher percentages. In considering the three types of estimate it is important to recognize that each has its unique purpose, place, and function in a project's life. Rough estimates are used for general feasibility activities, semidetailed estimates support budgeting and preliminary design decisions, and detailed estimates are used for establishing design details and contracts. As one moves from rough to detailed design, one moves from less to much more accurate estimates. However, this increased accuracy requires added time and resources. Figure 2-6 illustrates the trade-off between accuracy and cost. In engineering economic analysis, the resources spent must be justified by the need for detail in the estimate. As an illustration, during the project feasibility stages we would not want to use our resources (people, time, and money) to develop detailed estimates for unfeasible alternatives that will be quickly eliminated from further consideration. However, regardless of how accurate an estimate is assumed to be, it only an estimate of what the future will be. There will be some error even if ample resources and sophisticated methods are used. Difficulties in Estimation Estimating is difficult because the future is unknown. With few exceptions (such as with legal contracts) it is difficult to anticipate future economic consequences exactly. In this section we discuss several aspects of estimating that make it a difficult task. FIGURE 2-6 Accuracy versus cost trade-off in estimating. High ~e .,::1 '" ~ .... o .... '" o U Low Low Medium High Accuracy of Estimate ~. i(. ~. ' I " .. ~- ~:_};,_ \ ':'_~~' ~-;*.;k~~~~,~~i,-.-,,~, ".:;.;.__ ..~ --."_..'.. .ijj 40 ENGINEERING COSTS AND COST ESTIMATING One-of-a-Kind Estimates Estimated parameters can be for one-of-a-kind or first-runprojects. The first time something is done, it is difficult to estimate costs required to design, produce, and maintain a product over its life cycle. Conside~the. prqjec.tedcost estimates that were developed for the first NASA missions. The U.S. space program initially had no experience with human flight in outer space; thus the development of the cost estimates for design, production, launch, and recovery of the astronauts, flight hardware, and payloads was a "first-time experience."The same is true for any endeavor lacking local or global historical cost data. New products or processes that are unique and fundamentally different make estimating costs difficult. The good news is that there are very few one-of-a-kind estimates to be made in engineering design and analysis. Nearly all new technologies, products, and processes have "close cousins" that have led to their development. The concept of estimation by analogy allows one to use knowledge about well-understood activities to anticipate costs for new activities. Inthe 1950s, at the start of the military missile program, aircraft companies drew on their in-depth knowledge of designing and producing aircraft when they bid on missile contracts. As another example, consider the problem of estimating the production labor requirements for a brand new product, X. A company may use its labor knowledge about Product Y, a similar type product, to build up the estimate for X. Thus, although "first-run" estimates are difficult to make, estimation by analogy can be an effective tool. Time and Effort Available Our ability to develop engineering estimates is constrained by time and person-power availability. In an ideal world, it would cost nothing to use unlimited resources over an extended period of time. However, reality requires the use of limited resources in fixed intervals of time. Thus for a rough estimate only limited effort is used. Constraints on time and person-power can make the overall estimating task more difficult. If the estimate does not require as much detail (such as when a rough estimate is the goal), then time and personnel constraints may not be a factor.When detail is necessary and critical (such as in legal contracts), however, requirements must be anticipated and resource use planned. Estimator Expertise Consider two common phrases: The past is our greatest teacher and knowledge is power. These simple axioms hold true for much of what we encounter during life, and they are true in engineeringestimating as well. The more experiencedand knowledgeable the engineering estimator is, the less difficult the estimating process will be, the more accurate the estimate will be, the less likely it is that a major error will occur, and the more likely it is th~t the estimate will be of high quality. How is experience acquired in industry? One approach is to assign inexperienced engineers relatively smallerjobs, to create expertise and build familiarity with products and processes. Another strategy used is to pair inexperienced engineers with mentors who have vast technical experience. Technical boards and review meetings conducted to ''justify the numbers" also are used to build knowledge and experience. Finally, many firms maintain databases of their past estimates and the costs that were actually incurred. Estimating Models 41 ESTIMATING MODELS This section develops several estimating models that can be used at the rough, semidetailed, or detailed design levels. For rough estimates the models are used with rough data, likewise for detailed design estimates they-areused-withdetailed data. The level of detail will depend upon the accuracy of the model's data. Per-Unit Model The per-unit model uses a "per unit" factor, such as cost per square foot, to develop the estimate desired. This is a very simplistic yet useful technique, especially for developing estimates of the rough or order-of-magnitude type. The per unit model is commonly used in the construction industry. As an example, you may be interested in a new home that is constructed with a certain type of material and has a specificconstruction style. Based on this information a contractor may quote a cost of $65 per square foot for your home. If you are interested in a 2000 square foot floorplan, your cost would thus be: 2000 x 65 = $130,000. Other examples where per unit factors are utilized include ·· ·· · ·· Service cost per customer Safety cost per employee Gasoline cost per mile Cost of defects per batch Maintenance cost per window · Mileage cost per vehicle Utility cost per square foot of floor space Housing cost per student It is important to note that the per-unit model does not make allowances for economies of scale (the fact that higher quantities usually cost less on a per-unit basis). In most cases, however, the model can be effective at getting the decision maker "in the ballpark" of likely costs, and it can be very accurate if accurate data are used. Use the per-unit model to estimate the cost per student that you will incur for hosting 24 foreign exchange students at a local island campground for 10 days. During camp you are planning the following activities: ·· ·· 2 days of canoeing 3 campsite-sponsored day hikes 3 days at the lake beach (swimming, volleyball, etc.) Nightly entertainment After calling the campground and collecting other information, you have accumulated the following data: - - -.... ·· Vanrental from your city to the camp (one way) is $50 per 15 person van plus gas. Camp is 50 miles away, the van gets 10 miles per gallon, and gas is $1 per gallon. - 42 ENGINEERING COSTS AND COST ESTIMATING · ·· ·· · · · Each cabin at the camp holds 4 campers, and rent is $10 per day per cabin. Meals are $10 per day per camper; no outside food is allowed. Boat transportation to the island is $2 per camper (one way). Insurance/grounds fee/overhead is $fpetday.per camper. Canoe rentals are $5 per day per canoe, canoes hold 3 campers. Day hikes are $2.50 per camper (plus the cost for meals). Beach rental is $25 per group per half-day. Nightly entertainment is free. You are asked to use the per unit factpr to estimate the cosfperstudent ontJ:1istrip. Forplanning purppses we assume that there will be 100% participation in all actiyities.We will break the total cost down into categories of transportation, living, and entertainment. Transportation ill' Costs Van travel to andfrom,camp: 2 vans x :2trips x ($50/van + 50 ]:Iplesx 1 gal/lOmiles x ~ $lIgal) = $220 .; ~.~,~ ;; ~~ :::;' ::;: ;; :;= ",". Boat travelto andfrom island: 2 trips...x$2lcamperx 24 c~pers .$96 Transportation costs =~20+ 96 =-$316 Living Costs Mealsfor the lO-day period: 24 campers x $10/cainper/day x 10 days $2400 Cabin rentalfor the lO-day period.~24 campers x 1. capinl4 carIlperSx$10/day/cabin. x 10 days-.$600 lnsurance/Overhad expensefor the lO-dtiyperiod: 24 campers x $1/day/ca1DJ>er x 10 days = $240 - Living costs = 2400 +.600+ 240 = .$.~Z40 Entertfiinment Costs Canoe rental costs:' 2 canoe days x 24calllpeJ,"s xf _ __ ~_ =j8Q_ ~ _ + _... _"'" _ r:C-_ ""__;; -- ." .., - "'" __ _,_i:~:;:': canoe/3 c~per$ r~"-- v ,;-,:~=- x$5/day/canoe =~::J ~~ Beach rental costs: 3 days)( 2 ha.lf-days/dayx$25!b;iU'.,day..l$150 Day hike costs: .24 campers x3 dayl1jJ.(e$x$2.50/C;iIp.per/dayhilte ...$18Q Nightly entertainment: This is free! Can you believe it? _.~_.- F ... '" -- -~ .. h;,..~;_:. ii~~~_-:~.i{~=~J;;;~~':"~-;:;i::~;~;{jL_'''''-''''''''''''''~~_~£+~~;~4i~,.~~~~ .. --..--.-- Estimating Models 43 Total cost Total cost for 10-day period = Transportation co,s!s+ Living costs + Entertainment costs = 316 + 3240 + 410 __ $3966 Thus, the cost per student would be $3966/24 = $165.25. Thus, it would cost you $165.25 per student to host the students at theisland campground for the lO-day period. In this case the per-urtitmodel gives you a very detailed cost estimate (although its accuracy depends on the accuracy of your data and assumptions you've made). Segmenti ng Model The segmenting model can be described as "divide and conquer." An estimate is decomposed into its individual components, estimates are made at those lower levels, and then the estimates are aggregated (added) back together. It is much easier to estimate at the lower levels because they are more readily understood. This approach is common in engineering estimating in many applications and for any level of accuracy needed. In planning the camp trip of Example 2-5, the overall estimate was segmented into the costs for travel, living, and entertainment. The example illustrated the segmenting model (division of the overall estimate into the various categories) together with the unit factor model to make the subestimates for each category. Example 2-6 provides another example of the segmenting approach. Clean Lawn Corp. a manufacturer of yard equipment is planning to introduce a new high-end industrial-use lawn mower called the Grass Grabber. The Grass Grabber is designed as a walkbehind self-propelled mower. Clean Lawn engineers have been asked by the accounting depart'ment to estimate the material costs that will make up the new mower. The material cost estimate will be used, along with estimates for labor and overhead to evaluate the potential of this new model. : :-..;..~ '--t~--t-: 'SOLUTIO~~ . " . . ~-- '-.-'-.--:, -.' .., .' .;,-I' '.-'; ;:. .'.'~ The engineers decide to decompose the design specifications for the GraSs Grabber into its subcomponents, estimate the material costs for each of the subcomponents, and then sum these costs up tQobtain their overall estimate. The engineers are using a segmenting approach to build up their estimate. Mter careful consideration; the engineers have divided the mower into the" following$ajoHmbsystems: chassis, drive tram~'controls,'andcutting/coUectionsystem. Each of these isfqrther divided as appropriate, and unjt material costs were estimated at this lowest of .. - - ~ - , 11;:.; - --- -- -- - - - --. . " 1 -. \ ; : ~~ ~._;,...).s_~.~_~_.-~~~)-- ~ -":" ~ - --=-,~ .::;; 44 ENGINEERINGCOSTS AND COST ESTIMATING levels as follows: Unit Material Cost Estimate Cost Item A. Chassis A.I Deck A.2 Wheels A.3 Axles B. B.I B.2 B.3 B.4 B.5 B.6 $ 7.40. 10.20. 4.85 $22.45 Drive train Engine Starter assembly Transmission ;Drivedisc assembly Clutch linkage ;Beltassemblies $38.50. 5.90. 5.45 10.0.0. 5.15 7.70. Unit Material Cost Estimate Cost Item C. Controls C.1 Handle assembly C.2 Engine linkage C.3 Blade linkage C.4 Speed control linkage C05 Drive control assembly C.6 Cutting height adjuster D. D.1 D.2 D.3 $ 3.85 8.55 4.70. 21.50. 6.70.. 7.40. $52.70. Cutti!lglCollection system Blade assembly Side chute Grass bag and adapter $10.80. 7.0.5 7.75 $72:70. $25.60. The tOtalmaterial cost estimate of $173.45 was calculated by summing up theestimates for each of the four major subsystem levels (chassis, drive train, controls, and cuttinglcollection system). It should be noted that this cost represents only the material portion of the overall cost to produce the mowers. Other costs would include labor and overhead items. ~. - '-~--'~-'~..~ --. In Example 2-6 the engineers at Clean Lawn Corp. decomposed the cost estimation problem into logical elements. The scheme they used of decomposing cost items and numbering the material components (A.I, A.I, A.2, etc.) is known as a work breakdown structure. This technique is commonly used in engineering cost esti(llatingand project management of large products, processes, or projects. A work breakdown structure decomposes a large "work package" into its constituent parts which can then be estimated or managed individually. In Example 2-6 the work breakdown structure of the Grass Grabber has three levels. At the top level is the product itself, at the second level are the four major subsystems, and at the third level are the individual cost items. Imagine what the product work breakdown structure for a Boeing 777 looks like. Then imagine trying to manage the 777's design, engineering, construction, and costing without a tool like the work breakdown structure. Cost Indexes . Cost indexes are numerical values that reflect historical change in engineering (and other) costs. The cost index numbers are dimensionless, and reflect relative price change in either. individual cost items (labor,material, utilities) or groups of costs (consumerprices, producer prices). Indexes can be used to update historical costs with the basic ratio relationship given in Equation 2-2. Cost at time A Cost at time B Index value at time A Index value at time B (2-2) ,. ~. ,1.~ ~> ) K:' - ~..: .~'. . . ~. . .'. '".',~~~.it~.iC..,,~~.~~;.~t~.~." + ~'. ~.~~~~~c~._.4~.j,;~"",~ .." - -. . d~ :...:;;;:.:~'" . ..' .." ,;c..;.:i~&;.:Ji.~:(<!f~:;"~ Estimating Models 45 Equation 2-2 states that the ratio of the cost index numbers at two points in time (A and B) is equivalent to the dollar cost ratio of the item at the same times (see Example 2-7). Miriam isinterested in estimating the annuallabQrand materialcosts for a new production facility. She was able to obtain the following labor and material cost data: Labor costs · · ·· Lapor cost index value was at 124 ten years ago and is 188 today. Annual labor costs for a similarfacility were $575,500 ten years ago. Material Costs -'J!i!~ Material cost index value was at 544 three years ago and is 715 today. AJ;:inualmaterialcosts for a similar facility were $2,455,OPQthJ;eeyears ago. .. , .... _. .. : h ,-,.. Miriam will use Equation 2,.2to develop her cost estimat~s for;annual.labor and material costs. Labor _ Annual cost today " Annual cost 10 years ago Annual cost today II Index value today Index value 10 years ago . 188 124 x $575,500 $871,800 lv.{aterial~ Annual cost today Annual cost 3 years ago Index value today - Index value 3 years ago .. ;:;:;;;:= - - :;;:: ..:::!:!J: " 715 =:;;;; . ;; ..~11lWJll C.Qst..!op~y'; _521-~~J,#~.,2,PQO.-:=...$3,221,-..OOO :;;~ 4 ='=... = J Cost index data are collected and published by several private and public sources in the United States (and world). The U.S. government publishes data through the Bureau of Labor Statistics of the Department of Commerce. The Statistical Abstract of the UnitedStates publishes cost indexes for labor, construction, and materials. Another useful source for engineering cost index data is the Engineering News Record. Power-Sizing Model The power-sizingmodel is usedto estimatethe costs of industrialplants and equipment. The model"scalesup" or "scalesdown"knowncosts,therebyaccountingforeconomiesof - - ---- 46 -_.- - .- - - --- ENGINEERING COSTS AND COST ESTIMATING scale that are common in industrial plant and equipment costs. Consider the cost to build a refinery.Would it cost twice as much to build the same facility with double the capacity? It is unlikely. The power-sizing model uses the exponent (x), called the power-sizing exponent, to reflect economies of scale in the size or capacity: Cost of equipment A Cost of equipment B = Size(capacity) of equipment A Size(capacity) of B ( x ) (2-3) where x is the power-sizing exponent, costs of A and B are at the same point in time (same dollar basis), and size or capacity is in the same physical units for both A and B.' The power-sizing exponent (x) can be 1.0 (indicating a linear cost-versus-size/capacity relationship) or greater than 1.0 (indicating diseconomies of scale), but it is usually less than 1.0 (indicating economies of scale). Generally the ratio should be less than 2, and it should never exceed 5. This model works best in a "middle" range-not very small or very large size. Exponent values for plants and equipment of many types may be found in several sources, including industry reference books, research reports, and technical journals. Such exponent values may be found in Perry's Chemical Engineers' Handbook, Plant Design and Economics for Chemical Engineers, and Preliminary Plant Design in Chemical Engineering. Table 2-1 gives power sizing exponent values for several types of industrial facilities and equipment. The exponent given applies only to equipment within the size range specified. In Equation 2-3 equipment costs for both A and B occur at the same point in time. This equation is useful for scaling equipment costs but not for updating those costs. When the time of the desired cost estimate is different from the time in which the scaling occurs (per Equation 2-3) cost indexes accomplish the time updating. Thus, in cases like Example 2-8 involving both scaling and updating, we use the power sizing model together with.cost indexes. TABLE2-1 Example Power-Sizing Exponent Values Equipment/Facility Size Range Blower, centrifugal Compressor Crystallizer, vacuum batch Dryer, drum, single atmospheric Fan, centrifugal Filter, vacuum rotary drum Lagoon, aerated Motor 10,000-100,000 Reactor, 300 psi Tank, atmospheric, horizontal Power-Sizing Exponent ft3fmin 200-2100 hp 500-7000 ft2 10-100 ft2 20,000-70,000 ft2fmin 10-1500 ft2 0.05-20 million gal/day 5-20 hp 100-1000 gal 100-40,000 gal 0.59 0.32 0.37 0.40 1.17 0.48 1.13 0.69 0.56 0.57 Estimating Models 47 Based on her work in Example 2-7, Miriam has been asked to estimate the cost today of a 25.0.0ft2 heat exchange system for the new plant being-analyzed.'She has the following data. · · · Her company paid $5.0,.0.0.0 for a 10.0.0ft2 heat exchanger 5 years ago. Heat exchangers within this range of capacity have a power sizing exponent (x) of .0.55. Five years ago the Heat Exchanger Cost Index (HECI) was 13.06;it is 1487 today. mmJ:imI Miriam will first use Equation 2-3 to scale up the cost of the 10.0.0ft2 exchanger to one that is 25.0.0ft2using the .0.55power-sizing exponent. Cost of 25.0.0ft2 equipment Cost of 10.0.0ft2 equipment 25.0.0ft2 equipment - ( ( ) 1.0.0.0ft2 equipIIlent 25.0.0 Cost of 25.0.0ft2 equipment - ~ 10.0.0 0.55 ) 0.55 x 5.0,.0.0.0= $82,8.0.0 Miriam knows that the $82,8.0.0reflects only the scaling up of the cost of the 10.0.0ft2 model to a 25.0.0ft2 model. Now she will use Equation 2-2 and the HECI data to estimate the cost of a 25.0.0ft2 exchanger today. Miriam's cost esti)]late would be: Equipment cost today _Index value today Equipment cost 5 years ago - Index Value5 years ago 1487 Equipment cost today = 13.06 x $82,8.0.0 '-0..$94,3.0.0 Triangulation Triangulationis used in engineering surveying. A geographical area is divided into triangles from which the surveyor is able to map points within that region by using three fixed points and horizontal angular distances to locate fixed points of interest (e.g., property line referencepoints). Since any point can be located with two lines, the third line represents an extra perspective and check. We will not use trigonometry to arrive at our cost estimates, but we can utilize the concept of triangulation. We should approach our economic estimate from different perspectives because such varied perspectives add richness, confidence, and quality to the estimate. Triangulation in cost estimating might involve using different sourcesof data or using different quantitative models to arrive at the value being estimated. As decision makers we should always seek out varied perspectives. Improvement and the Learning Curve One commonphenomenon observed, regardless of the task being performed, is that as the numberof repetitions increases, performance becomes faster and more accurate. This is the ..- - 48 .- --. -.. - -- .- ENGINEERING COSTS AND COST ESTIMATING conceptof learningandimprovementin the activitiesthatpeopleperfonn.Fromour own experiencewe all knowthat our fiftiethrepetitionis completedin muchlesstime than we neededto accomplishthe taskthe firsttime. The learning curve captures the relationship between task perfonnance and task repetition. In general. as output doubles the-unit production time will be reduced to some fixed percentage. the learning curve percentage or learning curve rate. For example. it may take 300 minutes to produce the third unit in a production run involving a task with a 95% learning time curve. In this case the sixth (2 x 3) unit will take 300(0.95) = 285 minutes to produce. Sometimes the learning curve is also known as the progress curve. improvement curve. experience curve. or manufacturing progress function. Equation2-4 gives an expressionthat can be used for time estimatingin repetitive tasks. (2-4) = time requirement for the Nth unit of production Tinitial= time requirement for the first (initial) unit of production where TN N b =number of completed units (cumulative production) = learning curve exponent (slope of the learning curve on a log-log plot) Asjust given.a learning curve is often referred to by itspercentage.learning slope.Thus. a curve with b = -0.074 is a 95% learning curve because 2-0.074 = 0.95. This equation uses 2 because the learning curve percentage applies for doubling cumulative production. The learning curve exponent is calculated using Equation 2-5. b = log (learning curve expressed as a decimal) log 2.0 (2-5) Calculate the time required to produce the hundredth unit of a production run if the first unit took 32.0 minutes to produce and the learning curve rate for production is 80%. !"" "' -,.,," "-,:,,,,",'-' -SOL UTlp~+ Two = T1 X 100Iog0.80/log2.0 Two = 32.0 X 100-0.3219 .. Two = 7.27 '" minutes It is particularly important to account for the learning-curve effect if the production run involves a small number of units instead of a large number. When thousands or even millions of units are being produced. early inefficiencies tend to be "averaged out" because of the larger batch sizes. However. in the short run. inefficiencies of the same magnitude can lead to rather poor estimates of production time requirements. and thus production cost estimatesmay be understated.ConsiderExample2-10 and the resultsthat might be observedif thelearning-curveeffect is ignored.Noticein this examplethata "steadystate" Estimating Models 49 time is given. Steady state is the time at which the physical constraints of performing the task prevent the achievement of any more learning or improvement. Estimate the overall labor cost portion due to a task that has a learning-curve rate of 85% and reaches a steady state value after 16 units of 5.0 minutes per unit. Labor and benefits are $22 per hour, and the task requires two skilled workers. The overall production run is 20 units. SOlUTlQN Because we know the time required for the 16thunit, we can use Equation 2-4 to calculate the time required to produce the first unit. Tl6 = Tl X 16Iog0.85/log2.0 5.0 = TI X 16-0.2345 TI - 9.6 minutes ;; Now we use Equation 2-4 to calculate the time requirements for each unit in the production run as well as the total production time required. TN , Unit Number, N 1 2 3 4 5 6 7 '8 ':; 9 10 = Time (min) to produce Nth Unit 9.6 8.2 7.4 6.9 6.6 6.3 6.1 Cumulative Time from ltoN 9.6 17.8 24.2 32.1 38.7 45.0 51.1 5'.9 57.0 5.7 5.6 62.7 68.3 9.6 :;;;: X N-0.2345 Unit Number, N 11 12 13 14 15 16 17 = 18 19 20 Time (min) to produce Nth Unit 5.5 5.4 5.3 5.2 5.1 5.0 50 5.0 5.0 5.0 ;; Cumulative Time from ltoN 74.0 79.2 84.5 89.7 94.8 99.8 lQ1.8" 109.8 114.8 119.8 ;: =The foral ctTmulativetime oftKe proctuctioniim is~119.8minutes (2.01ours). Thus the total labor cost'~stimate would be: ,< 2.0 hours x $22/hour per worker x 2 workers = $88 If we ignot~ the learning-cu~v~effect and c~cul'!t~the lilQprcost portion based only,on tbe steady statelabotH:ate,tbe estip.1atewould be 0.083 hours/unit x 20 units x $22/hour per worker x 2 workers = $73.04 This estiIIl'£lteis understated by about 20% from what the true cost would be. . I L I , ~ . - . ,s:~~~~Jjc"",~~~~-~ I - ~?..J:~~~i-i:~~':';:_~~;"'-"'~-:"'~£;~i..i'i:';~~~j.~.'~;.,.~-. ~.'- ::-.~- __000._- .~~~ -.- - 50 -_. __A._.-_. ENGINEERING COSTS AND COST ESTIMATING ESTIMATING BENEFITS This chapter has focused on cost tenns and cost estimating. However, engineering economists must often also estimate benefits. Example benefits include sales of products, revenues from bridge tolls--and-electric power sales, cost reductions from reduced material or labor costs, reduced time spent in trafficjams, and reduced risk of flooding. Many engineering projects are undertaken precisely to secure these benefits. The cost concepts and cost estimating models can also be applied to economic benefits. Fixed and variable benefits, recurring and nonrecurring benefits, incremental benefits, and life-cycle benefits all have meaning. Also, issues regarding the type of estimate (rough, semidetailed, and detailed) as well as difficulties in estimation (one of a kind, time and effort, and estimator expertise) all apply directly to estimating benefits. Last, per unit, segmented, and indexed models are used to estimate benefits. The concept of triangulation is particularly important for estimating benefits. The uncertainty in benefit estimates is also typically asymmetric, with a broader limit for negative outcomes. Benefits are more likely to be overestimated than underestimated, so an example set of limits might be (-50%, +20%). One difference between cost and benefit estimation is that many costs of engineering projects occur in the near future (for design and construction), but the benefits are further in the future. Because benefits are often further in the future, they are more difficult to estimate accurately, and more uncertainty is typical. The estimation of economic benefits for inclusion in our analysis is an important step that should not be overlooked. Many of the models, concepts, and issues that apply in the estimation of costs also apply in the estimation of economic benefits. CASH FLOW DIAGRAMS The costs and benefits of engineering projects occur over time and are summarized on a cash flow diagram (CFD). Specifically,a CFD illustrates the size, sign, and timing of individual cash flows. In this way the CFD is the basis for engineering economic analysis. A cash flow diagram is created by first drawing a segmented time-based horizontal line, divided into appropriate time units. The time units on the CFD can be years, months, quarters, or any other consistent time unit. Then at each time at which a cash flow will occur, a vertical arrow is added-pointing down for costs and up for revenues or benefits. These cash flows are drawn to relative scale. The cash flows are assumed to occur at time 0 or at the end of each period. Consider Figure 2-7, the ~FD for a specific investment opportunity whose cash flows are described as follows: Timing of Cash Flow At time zero (now or today) 1 time period from today 2 time periods from today 3 time periods from today 4 time periods from today 5 time periods from today Size of Cash Flow A positive cash flow of $100 A negative cash flow of $100 A positive cash flow of $100 A negative cash flow of $150 A negative cash flow of $150 A positive cash flow of $50 r. ~':. t ". ~~2i~~~j~;;t~i;-i.¥b~~...~.c.~-;,;'::~~-;;-';'.L:_,-i'."';~":::~~",;,~~i;.,.iJ--,-i~,;,:~. ."-,-.-_..~;:c'_,.;;-;;;s;c~ FIGURE 2-7 An example cash flow diagram (CFD). $100 (+) + (-) ~ tD~":~~ $100 ~nd of Period 2 IS also Beginning 1 i of Period 3. /0-1-2-3-4-J Time0 ! ("today") $100 $50 j j $150 $150 Categories of Cash Flows The expenses and receipts due to engineering projects usually fall into one of the following categories. First cost = expense to build or to buy and install Operations and maintenance (O&M) = annual expense, such as electricity, labor, and minor repairs Salvage value = receipt at project termination for sale or transfer of the equipment (can be a salvage cost) Revenues = annual receipts due to sale of products or services Overhaul = major capital expenditure that occurs during the asset's life Individual projects will often have specific costs, revenues, or user benefits. For example, annual operations and maintenance (O&M) expenses on an assembly line might be divided into direct labor, power, and other. Similarly, a public-sector dam project might have its annual benefits divided into flood control, agricultural irrigation, and recreation. Drawing a Cash Flow Diagram The cash flow diagram shows when all cash flows occur. Look at Figure 2-7 and the $100 positive cash flow at the end of period 2. From the time line one can see that this cash flow can also be described as occurring at the beginning of period 3. Thus, in a CFD the end of period t is the saI}letime as the beginning of period t + 1. Beginning-of-period cash flows (such as rent, lease, and insurance payments) are thus easy to handle: just draw your CFD and put them in where they occur. Thus O&M, salvages, revenues, and overhauls are assumed to be end-of-period cash flows. The choice of time 0 is arbitrary. For example, it can be when a project is analyzed, when funding is approved, or when construction begins. When construction periods are assumed to be short; first costs are assumed to occur at time 0, and the first annual revenues and costs start at the end of the first period. When construction periods are long, time 0 is usually the date of commissioning-when the facility comes on stream. Perspective is also important when one is drawing a CFD. Consider the simple trans- action of paying $5000 for some equipment. To the firm buying the equipment, the cash fJ,owis a cost and hence negative in sign. To the firm selling the equipment, the cash flow is a revenue and positive in sign. This simple example shows that a consistent perspective is required when one is using a CFD to model the cash flows of a problem. One person's cash outflow is another person's inflow. Often two or more cash flows occur in the same year, such as an overhaul and an O&M expense or the salvage value and the last year's O&M expense. Combining these into one -. .------- 52 -.---.... ENGINEERING COSTS AND COST ESTIMATING total cash flowper year would simplify the cash flowdiagram. However, it is better to show each individually, to ensure a clear connection from the problem statement to each cash flow in the diagram. Drawing Cash Flow 'Diagrams with a Spreadsheet One simple way to draw cash flow diagrams with "arrows" proportional to the size of the cash flows is to use a spreadsheet to draw a stacked bar chart. The data for the cash flows is entered, as shown in the table part of Figure 2-8. To make a quick graph, select cells B1 to D8, which are the three columns of the cash flow.Then select the graph menu and choose column chart and select the stack option. Except for labeling axes (using the cells for year o to year 6), choosing the scale for the y axis, and adding titles, the cash flow diagram is done. Refer to the appendix for a review of basic spreadsheet use. (Note: a bar chart labels periods rather than using an x axis with arrows at times 0, 1,2 ) '''''':'_''''",'i:."c-::;,'-;z;: ,>.----'..7'""''':''-.''.''''::-'5' . "'"".'":'"'7:''-', .--------,---, A B C D 1 Year Capital Costs O&M Overhaul 2 0 1 2 3 4 5 6 - 80000 :3 4 5 6 7 8 9 :~io 11 , ''''1''i' . 13 "14 15 16 '11=' 18 19 "'20 I 21 ~ ; 22 -12000 -12000 -12000 -12000 -12000 -12000 10000 I F E -25000 $10 $0 ~ -$10 § gj o -$20 .s '-' -$30 ~ -$40 o Ii:: -$50 ..c:: ~ -$60 U -$70 -$80 o 1 2 3 Year 4 5 6 FIGURE 2-8 Example of cash flow diagram in spreadsheets. SUMMARY This chapter has introduced the following cost concepts: fixed and variable, marginal and average, sunk, opportunity, recurring and nonrecurring, incremental, cash and book, and life-cycle. Fixed costs are constant and unchanging as volumes change, while variable Summary 53 costs change as output changes. Fixed and variable costs are used to find a breakeven value between costs and revenues, as well as the regions of net profit and loss. A marginal cost is for one more unit, while the average cost is the total cost divided by the number of units. . Sunk costs result from past decis.io~sand shouldnot influenceour attitudetoward current and future opportunities. Remember, "sunk costs are sunk." Opportunity costs involve the benefit that is forgone when we choose to use a resource in one activity instead of another. Recurring costs can be planned and anticipated expenses; nonrecurring costs are one-of-a-kind costs that are often more difficult to anticipate. Incremental costs areeconomic consequences associatedwith the differencesbetween two choices of action. Cash costs are also known as out-of-pocket costs that represent actual cash flows. Book costs do not result in the exchange of money, but rather are costs listed in a firm's accounting books. Life-cycle costs are all costs that are incurred over the life of a product, process, or service. Thus engineering designers must consider life-cycle costs when choosing materials and components, tolerances, processes, testing, safety,service and warranty, and disposal. Cost estimating is the process of "developing the numbers" for engineering economic analysis. Unlike a textbook, the real world does not present its challenges with neat problem statements that provide all the data. Rough estimates give us order-of-magnitude numbers and are useful for high-level and initial planning as well asjudging the feasibility of alternatives. Semidetailed estimates are more accurate than rough-order estimates, thus requiring more resources (people, time, and money) to develop. These estimates are used in preliminary design and budgeting activities. Detailed estimates generally have an accuracy of ::1::3-5%.They are used during the detailed design and contract bidding phases of a project. Difficulties are common in developing estimates. One-of-a-kind estimates will have no basis in earlier work, but this disadvantage can be addressed through estimation by analogy. Lack of time availableis best addressedby planning and by matching the estimate's detail to the purpose-one should not spend money developing a detailed estimate when only a rough estimate is needed. Estimator expertise must be developed through work experiences and mentors. Several general models and techniques for developing cost estimates were discussed. The per-unit and segmenting models use different levels of detail and costs per squarefoot or other unit. Cost index data are useful for updating historical costs to formulate current estimates. The power-sizing model is useful for scalingup or down a known cost quantityto account for economies of scale, with different power-sizing exponents for industrial plants and equipment of different types. Triangulation suggests that one should seek varying perspectives when developing cost estimates. Different information sources, databases, and analytical models can all be used to create unique perspectives. As the number of task repetitions increases, efficiencyimproves because of learning or improvement. This is summarizedin thelearning-curvepercentage, wheredoublingthe cumulativeproductionreduces the time to complete the task, which equals the learning-curvepercentage timesthe current production time. Cash flow estimation must include project benefits. These include labor cost savings, avoided quality costs, direct revenue from sales, reduced catastrophic risks, improvedtraffic flow, and cheaper power supplies. Cash flow diagrams are used to model the positive and negative cash flows of potential investment opportunities. These diagrams provide a consistent view of the problem (and the alternatives)to support economic analysis. ... 54 .-.-..- ENGINEERINGCOSTS AND COST ESTIMATING PROBLEMS 2-1 Custom-designed home Stock-plan home Next 50 kw-hr per hp of connected load at 6.6~ per kw-hr Next 150 kw-hr per hp of connected load at 4.0~ per kw-hr All electricity over 250 kw-hr per hp of connected load at 3.7 ~ per kw-hr The shop uses 2800 kw-hr per month. (a) Calculate the monthly bill for this shop. What are the marginal and average costs per kilowatt-hour? (b) Suppose Jennifer, the proprietor of the shop, has the chance to secure additional business that will require her to operate her existing equipment more hours per day. This will use an extra 1200 kw-hr per month. What is the lowest figure that she might reasonably consider to be the "cost" of this additional energy? What is this per kilowatt-hour? $128,000 128,500 Bob w~s willing to pay the extra $500 for it. Bob's wife, however, felt they should go ahead with the custom-designed home, for, as she put it, "We can't afford to throwaway a set of plans that cost $4000." Bob agreed, but he disliked the thought of building a home that is less desirable than the stock plan home. Then he asked your advice. Which house would you advise him to build? Explain. ~-2 Venus Computer can produce 23,000 personal com- 2-3 First 50 kw-hr per hp of connected load at 8.6~ per kw-hr Bob Johnson decided to purchase a new home. After looking at tracts of new homes, he decided .that a custom-built home was preferable. He hired an architect to prepare the drawings. In due time, the architect completed the drawings and submitted them. Bob liked the plans; he was less pleased that he had to pay the architect a fee of $4000 to design the house. Bob asked a building contractor to provide a bid to construct the home on a lot Bob already owned. While the contractor was working to assemble the bid, Bob came across a book of standard house plans. In the book was a home that he and his wife liked better than the one designed for them by the architect. Bob paid $75 and obtained a complete set of plans for this other house. Bob then asked the contractor to provide a bid to construct this "stock plan" home. In this way Bob felt he could compare the costs and make a decision. The building contractor submitted the following bids: puters a year on its daytime shift. The fixed manufacturing costs per year are $2 million and the total labor cost is $9,109,000. To increase its production to 46,000 computers per year, Venus is considering adding a second shift. The unit labor cost for the second shift would be 25% higher than the day shift, but the total fixed manufacturing costs would increase only to $2.4 million from $2 million. (a) Compute the unit manufacturing cost for the daytime shift. (b) Wouldadding a second shift increase or decrease the unit manufacturing cost at the plant? A small machine shop, with 30 hp of connected load, purchases electricity under the following monthly rates (assume any demand charge is included in this schedule): (c) She contemplates installing certain new machines that will reduce the labor time required on certain operations. These will increase the connected load by 10 hp, but since they will operate only on certain special jobs, will add only 100 kw-hr per month. In a study to determine the economy of installing these new machines, what should be considered as the "cost" of this energy? What is this per kilowatt-hour? 2-4 1\vo automatic systems for dispensing maps are being compared by the state highway department. The accompanying breakeven chart of the comparison of these systems (System I vs System II) shows total yearly costs for the number of maps dispensed per year for both alternatives. Answer the following questions. (a) What is the fixed cost for System I? (b) What is the fixed cost for System II? (c) What is the variable cost per map dispensed for System I? (d) What is the variable cost per map dispensed for System II? (e) What is the breakeven point in terms of maps dispensed at which the two systems have equal annual costs? (f) For what range of annual number of maps dispensed is System I recommended? , .. ---- Problems 2-7 $10 Totalcost, System I: y = 0.9Ox+ 1.0 $8 -;;;"0 !a rI.I $6 6 @, rI.I Total cost, System II: y = O.lOx + 5.0 $4 0 () $2 $0 0 5 Maps Dispensed per year (thousands) 10 Two new rides are being compared by a local amusement park in terms of their annual operating costs. The two rides are assumed to be able to generate the same level of revenue (and thus the focus on costs). The Thmmy Thgger has fixed costs of $10,000 per year and variable costs of $2.50 per visitor. The Head Buzzer has fixed costs of $4000 per year, and variable costs of $4 per visitor. Provide answers to the following questions so the amusement park can make the needed comparison. (a) Mathematically determine the breakeven number of visitors per year for the two rides to have equal annual costs. (b) Develop a graph that illustrates the following: (Note: Put visitors per year on the horizontal axis and costs on the vertical axis.) · · (g) For what range of annual number of maps dispensed is System ITrecommended? (h) At 3000maps per year, what arethe marginaland average map costs for each system? 2-5 · Mr. Sam Spade, the president of Ajax, recently read in a report that a competitor named Bendix has the following relationship between cost and production quantity: 2-8 55 Accurate total cost lines for the two alternatives (show line, slopes, and equations). The breakeven point for the two rides in terms of number of visitors. The ranges of visitors per year where each alternativeis preferred. Consider the accompanying breakeven graph for an investment, and answer the following questions as they pertain to the graph. c = $3,000,000 - $18,000Q + $75Q2 $40 where C = total manufacturing cost per year and Q 2-6 = number $35 of units produced per year. " , ..... A newly hired employee, who previously worked for Bendix, tells Mr. Spade that Bendix is now producing 110 units per year. If the selling price remains unchanged, Sam wonders if Bendix is likely to increase the number of units produced per year, in the near future.He asks you to look at the information and tell him what you are able to deduce from it. A privately owned summer camp for youngsters has the followingdata for a 12-weeksession: Charge per camper Fixed costs Variable cost per camper Capacity $120 $48,000 $80 200 ~ Total Revenue,." ,. $30 $25 x '-' $20 E '0 $15 Q $10 ,.., ,. ,.,. ,.,. ,.,. ,. .... ,.,.,." - ",,." ,.,. ,.,.,.,. ,. o 250 $5 $0 500 750 1000 1250 Output (units/year) per week per session per week campers (a) Give the equation to describe total revenue for x units per year. (b) Give the equation to describe total costs for x units per year. (c) What is the "breakeven" level of x in terms of costs and revenues? (d) If you sell 1500 units this year, will you have a profit or loss? How much? (a) Develop the !Ilathematical relationships for total cost and total revenue. (b) What is the total number of campers that will allow the camp to just break even? (c) What is the profit or loss for the 12-week session if the camp operates at 80% capacity? 2-9 Quatro Hermanas, Inc. is investigating implementing some new production machinery as part of its fy, ~ t. O' , ,~!} . '6"" -';/<;i;';":-'"''':4':;J:~:~_''''''' .> ;':';""" ,._;~,~ 56 ENGINEERING COSTS AND COST ESTIMATING operations. Three alternatives have been identified, and they have the following fixed and variablecosts: Alternative A B C Annual Fixed Costs $100,000 200,000 150,000 Annual Variable Costs per Unit $20.00 5.00 7.50 Determine the ranges of production (units produced per year) over which each alternative would be recommended for implementation by Quatro Hermanas. Be exact. (Note: Consider the range of production to be from 0-30,000 units per year.) 2-10 Three alternative designs have been created by Snakisco engineers for a new machine that spreads cheese between the crackers in a Snakisco snack. Each machine design has unique total costs (fixed and variable) based on the annual production rate of boxes of these crackers. The costs for the three designs are given (where x is the annual production rate of boxes of cheese crackers). Design A B C Fixed Cost $100,000 350,000 600,000 Variable Cost ($Ix) 20.5x 1O.5x 8.0x You are asked to do the following. (a) Mathematically determine which of the machine designs would be recommended for different levels of annual production of boxes of snack crackers. Management is interested in the production interval of 0-150,000 boxes of crackers per year. Over what production volume would each design (A or B or C) be chosen? (b) Depict your solution from part (a) graphically, putting x per year on the horizontal axis and $ on the vertical axis, so that management can see more easily the following: i. Accurate total cost lines for each alternative (show line, slopes, and line equations). ii. Any relevant breakeven, or crossover points in terms of costs between the alternatives. iii. Ranges of annual production where each alternative is preferred. iv. Clearly label your axes and include a title for the graph. 2-11 A small company manufactures a certain product. Variable costs are $20 per unit and fixed costs are $10,875. The price-demand relationship for this product is P = -0.25D + 250, where P is the unit sales price of the product and D is the annual demand. Use the data (and helpful hints) that follow to work out answers to parts (a)-(e). ··· Total cost = Fixed cost + Variable cost Revenue = Demand x Price Profit = Revenue - Total cost Set up your graph with dollars on the y axis, (between o and $70,000) and, on the x axis, demand D: (units produced or sold), between 0 and 1000 units. (a) Develop the equations for total cost and total revenue. (b) Find the breakeven quantity (in terms of profit and loss) for the product. (c) What profit would the company obtain by maximizing its total revenue? (d) What is the company's maximum possible profit? (e) Neatly graph the solutions from parts (a), (b), (c), and (d). 2-12 A painting operation is performed by a production worker at a labor cost of $1.40 per unit. A robot spraypainting machine, costing $15,000, would reduce the labor cost to $0.20 per unit. If the device would be valueless at the end of 3 years, what is the minimum number of units that would have to be painted each year tojustify the purchase of the robot machine? 2-13 Company A has fixed expenses of $15,000 per year and each unit of product has a $0.002 variable cost. Company B has fixed expenses of $5000 per year and can produce the same product at a $0.05 variable cost. At what number of units of annual production will Company A have the same overall cost as Company B? 2-14 A firm believes the sales volume (S) of its product depends on its unit selling price (P) and can be determined from the equation P = $100 - S. The cost (C) of producing the product is $1000 + lOS. (a) Draw a graph with the sales volume(S) from 0 to 100 on the x axis, and total cost and total income from 0 to 2500 on the y axis. On the graph -draw the line C = $1000 + lOS. Then plot the curve of total income [which is sales volume (S) x unit selling price ($100 - S)]. Mark the breakeven points on the graph. (b) Determine the breakeven point (lowest sales volume where total sales income just equals total production cost). (Hint: This may be done by trial Problems Paint information: Your house has a surface area of 6000 ft2. One can of paint can cover 300 ft2. You are estimating the cost to put on two coats of paint for the entire house, using the cost per can given. Note the incremental decrease in unit cost per can as you purchase more and more cans. and error or by using the quadratic equation to locate the point at which profit is zero.) (c) Determine the sales volume (8) at which the firm's profit is a maximum. (Hint: Write an equation for profit and solve it by trial and error, or as a minima-maxima calculus problem.) Number of Cans Purchased 2-15 Consider the situation of owning rental properties that local university students rent from you on an academic year basis. Develop a set of costs that you could classify as recurring and others that could be classified as nonrecurring. First 10 cans purchased Second 15 cans purchased Up to next 50 cans purchased Cost per Can $15.00 $10.00 $7.50 Labor information: You plan to hire five painters who will paint for 10 hours per day each. You estimate that the job will require 4.5 days of their painting time. The painter's labor rate is $8.75 per hour. 2-16 Define the difference between a "cash cost" and a "book cost." Is engineering economic analysis concerned with both types of cost? Give an example of each, and provide the context in which it is important. 2--- 57 In your own words, develop a statement of what the authors mean by "life-cycle costs." Is it important for a firm to be aware of life-cycle costs? Explain why. Fixed cost information: There is a fixed cost of $200 per job that the painting company charges to cover travel expenses, clothing, cloths, thinner, administration, and so on. 2-18 In lookiiig at Figures 2-4 and 2-5, restate in your own words what the authors are trying to get across with these figures. Do you agree that this is an important effect for companies? Explain. 2-22 You are interested in having a mountain cabin built for weekend trips, vacations, to host family, and perhaps eventually to retire in. After discussing the project with a local contractor, you receive an estimate that the total construction cost of your 2000 ft2 lodge will be $150,000. Costs within each category include labor, material, and overheaditems. The percentage of costs for each of severalitems (categories) is broken down as follows: 2-19 In the text the authors describe three effects that complicate the process of making estimates to be used in engineering economy analyses. List these three effects and comment on which of these might be most influential. 2-20 Northern Tundra Telephone (NIT) has received a contract to install emergency phones along a new 100-mile section of the Snow-Moose Thrnpike. Fifty emergency phone systems will be installed about 2 miles apart. The material cost of a unit is $125. NTT will need to run underground communication lines that cost NTT $7500 per mile (including labor) to install. There will also be a one-time cost of $10,000 to network these phones into NTT's current communication system. You are asked to develop a cost estimate of the project from NIT's perspective. If NTT adds a profit margin of .35% to its costs, how much will it cost the state to fund the project? Percentage of Cost Items Total Costs 8% Construction permits, legal and title fees 15 Roadway, site clearing, preparation 13 Foundation, concrete, masonry 12 Wallboard, flooring,carpentry 13 Heating, ventilation,air conditioning 10 Electric, plumbing, communications 12 Roofing, flooring 17 Painting, finishing 100 2-21 You and your spouse are planning a second honeymoon to the Cayman Islands this summer and would like to have your house painted while you are away. Estimate the total cost of the paint job from the information given below, where: (a) What is the cost per square foot of the 2000 ft2 lodge? (b) If you are also considering a 4000 ft2 layout option, estimate your construction costs if: i. All cost items (in the table) change proportionately to the size increase. Costtotal = Costpaint + Costlabor + Costfixed ~ ." -- I. . - '.. '. -. - _. j;!I¥;,;\.~w~.,-),;:~~~i,i$~~~~~~'; ' . -.°0'. ~-~ . -. -. . - . .-., ,_._~,,~ '~..":' a..:.~l.. - ~ 58 - ----.--------- ENGINEERING COSTS AND COST ESTIMATING n. The first two cost items do not change at all; all others are proportionate. 2-23 SungSam, Inc. is currently designing a new digital camcorder that is projected to have the following per unit costs to manufacture: Cost Categories Materials costs Labor costs Overhead costs Total Unit Cost . . - . . . Unit Costs $112 85 213 $410 SungSam adds 30% to its manufacturing cost for corporate profit. Answer the following questions: (a) What unit profit would SungSam realize on each camcorder? (b) What is the overall cost to produce a batch of 10,000 camcorders? (c) What would SungSam's profit be on the batch of 10,000 if historical data shows that 1% of product will be scrapped in manufacturing, and 3% of finished product will go unsold, 2% of sold product will be returned for refund? (d) How much can SungSam afford to pay for a contract that would lock in a 50% reduction in the unit material cost previously given? ITSungSam does sign the contract, the sales price will remain the same as before. 2-24 Fifty years ago, Grandma Bell purchased a set of goldplated dinnerware for $55, and last year you inherited it. Unfortunately a house fire at your home destroyed the set. Your insurance company is at a loss to define the replacement cost and has asked your help. You do some research and find that the Aurum Flatware Cost Index (AFCI) for gold-plated dinnerware, which was 112 when Grandma Bell bought her set, is at 2050 today. Use the AFCI to update the cost of Bell's set to today's cost to show to the insurance company. 2-25 Your boss is the director of reporting for the Athens County Construction Agency (ACCA). It has been his job to track the cost of construction in Athens County. 1\venty-five years ago he created the ACCA Cost Index to track these costs. Costs during the first year of the index were $12 per square foot of constructed space (the index value was set at 100 for that first year). This past year a survey of contractors revealed that costs were $72 per square foot. What index number will your boss publish in his report for this year? ITthe index value was 525 last year, what was the cost per square foot last year? 2-26 An refinisher of antiques named Constance has been so successful with her small business that she is planning to expand her shop with all new equipment. She is going to start enlarging her shop by purchasing the following equipment. Cost of PowerNew Original Original Sizing Equipment Size Equipment Size Equipment Exponent Varnishbath $3500 0.80 50 gal 75 gal $250 0.22 Powerscraper 3/4 hp 1.5hp Paint booth 3ft3 12 ft3 $3000 0.6 What would be the net cost to Constance to obtain this equipment-assume that she can trade the old equipment in for 15% of its original cost. Assume also that the relative price to purchase the equipment has not changed over time (that is, there has been no inflation in equipment prices). 2-27 Refer to Problem 2-26 and now assume the prices for the equipment that Constance wants to replace have not been constant. Use the cost index data for each piece of equipment to update the costs to the price that would be paid today. Develop the overall cost for Constance, again assuming the 15% trade-in allowance for the old equipment. Use any necessary data from Problem 2-26. Original Equipment Varnishbath Power scraper Paint booth Cost Index When Originally Purchased 1.54 780 49 Cost Index Today 171 900 76 2-28 Five years ago, when the relevant cost index was 120, a nuclear centrifuge cost $40,000. The centrifuge had a capacity of separating 1500 gallons of ionized solution per hour. Today, it is desired to build a centrifuge with capacity of 4500 gallons per hour, but the cost index now is 300. Assuming a power-sizing exponent to reflect economies of scale, x, of 0.75, use the power-sizing model to determine the approximate cost (expressed in today's dollars) of the new re~ctor. 2-29 Padre works for a trade magazine that publishes. lists of Power-Sizing Exponents (PSE) that reflect economies of scale for developing engineering estimates of various types of equipment. Padre has been unable to find any published data on the VMIC machine and wants to list its PSE value in his next issue. Given the following data (your staff was able to find data regarding costs and sizes of the VMIC Problems machine) calculate the PSE value that Padre should publish. (Note: The VMIC-I00 can handle twice the volume of a VMIC-50.) VMIC equipment index 5 years ago = booklets that offer excuses for missed appointments. His assistant Doc Duckout has collected information that Randy will need in developing his estimate: Cost of direct labor Cost of materials Cost of VMIC-100 today $100,000 Cost of VMIC-50 5 years ago $45,000 VMIC equipment index today = 214 151 2-30 Develop an estimate for each of the following situations. (a) The cost of a 500-mile automobile trip, if gasoline is $1 per gallon, vehicle wear and tear is $0.08 per mile, and our vehicle gets 20 miles per gallon. (b) The total number of hours in the average human life, if the average life is 75 years. (c) The number of days it takes to travel around the equator using a hot air balloon, if the balloon averages 100 miles per day, the diameter of the earth is ""4000 miles. (Note: Circumference = 7r times diameter.) (d) The total area in square miles of the United States of America, if Kansas is an average-sized state. Kansas has an area of 390 miles x 200 miles. 59 Cost of overhead items Desired profit $20 per hour $43.75 per batch of 25 booklets 50% of direct labor cost 20% of total manufacturing cost Doc also finds out that (1) they should use a 75% learning curve for estimating the cost of direct labor, (2) the time to complete the 15tbooklet is estimated at 0.60 hour, and (3) the estimated time to complete the 25th booklet should be used as their standard time for the purpose of determining the unit selling price. What would Randy and Doc's estimate be for the unit selling price? 2-35 Develop a statement that expresses the extent to which cost estimating topics also apply to estimating benefits. Provide examples to illustrate. 2-32 Rose is a project manager at the civil engineering consulting firm of Sands, Gravel, Concrete, and Waters, Inc. She has been collecting data on a project in which concrete pillars were being constructed, however not all the data are available. She has been able to find 2-36 On December 1, AI Smith purchased a car for $18,500. He paid $5000 immediately and agreed to pay three additional payments of $6000 each (which includes principal and interest) at the end of 1, 2, and 3 years. Maintenance for the car is projected at $1000 at the end of the first year and $2000 at the end of each subsequent year. AI expects to sell the car at the end of the fourth year (after paying for the maintenance work) for $7000. Using these facts, prepare a table of cash flows. out that the 10th pillar required 260 person-hours to construct, and that a 75% learning curve applied. She is interested in calculating the time required to construct the 15tand 20th pillars. Compute the values for her. 2-37 Bonka Toys is considering a robot that will cost $20,000 to buy. After 7 years its salvage value will be $2000. An overhaul costing $5000 will be needed in year 4. O&M costs will be $2500 per year. Draw the cash flow diagram. 2-33 Sally Statistics is implementing a system of statistical process control (SPC) charts in her factory in an effort to reduce the overall cost of scrapped product. The current cost of scrap is $X per month. If an 80% learning curve is expected in the use of the SPC charts to reduce the cost of scrap, what would the percentage reduction in monthly scrap cost be after the charts have been implemented for 12 months? (Hint: Model each month as a unit of production.) 2-38 Pine Village needs some additional recreation fields. Construction will cost $225,000 and annual O&M expenses are $85,000. The city council estimates that the value of added youth leagues is about $190,000. annually. In year 6 another $75,000 will be needed to refurbish the fields. The salvage value is estimated to be $100,000 after 10 years. Draw the cash flow diagram. 2-31 If 200 labor hours were required to produce the 15tunit in a production run and 60 labor hours were required to produce the 7th unit, what was the learning-curve rate during production? 2-34 Randy Duckout has been asked to develop an estimate of the per-unit selling price (the price that each unit will be sold for) of a new line of hand-crafted 2-39 Identify your major cash flows for the current school term as first costs, O&M expenses, salvage values, revenues, overhauls, and so on. Using a week as the time period, draw the cash flow diagram. <'",.;;,,,~,ji;~!!!- olio- _ ..- After Completing This Chapter... The student should be able to: ·· ·· Define and provide examples of the time value of money. Distinguish between simple and compound interest, and use compound interest in engineering economic analysis. Explain equivalence of cash flows. Solve problems using the single payment compound interest formulas. QUESTIONS TO CONSIDER ;:;. 1. Tobaccobonds oftenpay interest of as much as 7%, whileconventionalstate government bonds may pay only 5%. On a $1 million bond with a term of 25 years, how much does this interest differential amount to, in total? 2. Assuming a 2% interest differential, over a 25-year period, how much are states losing by opting to take the settlement money "up front" instead of collecting the long-term payout? 3. When themoney they get from sellingbonds runs out, state governmentswillhave to find a new way of closing their budget gaps. Already, some commentators are suggesting that states may soon look to other business sectors as "deep pockets." The tobacco industry was particularly vulnerable because it sold a product that came to be seen as dangerous and socially undesirable. What other industries might face the same fate in years to come? How about fast-food restaurant chains, or sellers of alcohol products? Should a company like Anheuser-Buschbe worried? Should companies in potentially "problematic" industries consider setting aside money for future settlement costs? - - - - - r - -- - - - --- - f Interest and Equivalence t I f r Going Up in Smoke State governmentsthroughoutthe United States agreed to a largefinancialwindfall in 1998. Tobaccocompaniesagreedto pay in perpetuityto settleclaims arisingfromthe health effects of smoking. Payments over the first 25 years are estimated to be nearly $250 billion. State officials announced they would earmark these funds for goals such as health care, education and, of course, antismoking campaigns. But several states involved in the settlement were chronically short of money and were desperate to plug budget deficits. If only they could get their hands on the full value of that tobacco settlementnow, instead of waiting for the payments to dribble in year by year. That's when someonehit on the idea ofraising instant money by issuing "tobacco bonds": the states would sell bonds to investors and pay them interest out of the tobacco settlement payments the states were receiving. State governments could get the whole sale price of the bond up front; investors would get ongoing income fromthe bond. . A growing number of states are now selling these bonds, and pocketingquickbillions. To attract buyers, however, the states have to pay a high rate of interest on their tobacco bonds, since investors view them as riskier than other government-backed securities. Investors reason that tobacco companies could go broke, after all--especially if the no-smoking laws being passed allover the country cause enough people to snuff out their cigarette habit for good. 62 INTERESTAND EQUIVALENCE In the first chapter, we discussed the engineering economic decision process. In Chapter 2 we described models used to estimate the costs and benefits that are summarized in cash flow diagrams. For many of the situations we examined, the economic consequences of an alternativewere immediate,- ~or took in a very short period of time, as in Example 1-2 .' . -- place . (the decision on the design of a concrete aggregate mix) or Example 1-3 (the change of manufacturing method). In such relatively simple situations, we total the various positive and negativeaspects, compare our results, and quickly reach a decision. But can we do the same if the economic consequencesoccur over a considerable period of time? No we cannot, because money has value over time. Would you rather: (1) receive $1000 today or (2) receive $1000 ten years from today? Obviously, the $1000 today has more value. Money's value over time is expressed by an interest rate. In this chapter, we describe two introductory concepts involving the time value of money: interest and cash flow equivalence. COMPUTING CASH FLOWS Installing an expensivepiece of machineryin a plant obviouslyhas economic consequences that occuroveranextendedperiodof time.If the machinerywerebought on credit,the simple process of payingfor it maytake severalyears.What abouttheusefulness of the machinery? Certainly it was purchased because it would be a beneficial addition to the plant. These favorable consequences may last as long as the equipment performs its useful function. In these circumstances,we do not add up the various consequences; instead, we describe each alternativeas cash receipts or disbursements at different points in time. In this way, each alternativeis resolved into a set of cash flows. This is illustrated by Examples 3-1 and 3-2. The manager has decidedto purchase a new $30,000 mixing machine. The machine may be paid for in one of two ways: 1. Pay the full price now minus a 3% discount. 2. Pay $5000 now; at the end of one year, pay $8000; at the end of each of the next four years, pay $6000. List the alternatives in theform of a table of cash flows. I.. In this problem the two alternatives represent different ways to pay for the mixing machine. While the first plan represents a~lumpsum of $29,100 now, the second one calls for payments cOhtinuinguntilthe end of the fifthyear.Theproblemis to convertan alternativefnto cash receipts or disbursements and show the tirn,ingof each.receipt or disbursement.The result is called a cash flQwtable or, more simply,a set of cashflows. III_ - - - -- - -- - - - .. - - - ...-.. - -.. .. -_. - Computing Cash Flows 63 The cash flowsfor both the alternativesin this problem are very simple. The cash flowtable, with disbursements given negative signs, is as follows: End of Year o (now) 1 2 3 4 5 Pay in FuUNow-$29,100 o o o o o Pay Over 5 Years ---$5000 -8000 -6000 -6000 .-6000 -6000 ~..~. _..~. .~ = . A man borrowed $1000from a bank at 8% interest.He agreedto repay the loan in twoend-of-year payments. At the end of the first year, he will repay half of the $1000 principal amount plus the interest that is due. At the end of the second year,he will repay the remaining half of the principal amount plus the interest for the second year. Compute the borrower's cash flow. . SOL '.TI()N In engineering economic analysis, we normally refer to the beginningof the first year as "time 0." At this point the man receives$1000 from the bank. (A positivesignrepresents a receipt of money and a negative sign, a disbursement.)Thus, at time 0, the cash flow is +$1000. At the end of the first year, the man pays 8% interest for the use of $1000 for one year. The interest is 0.08 x $1000 = $80. In addition, he repays half the $1000 loan, or $500. Therefore, the end-of-year-I cash flow is -$580. At the end of the second year, the payment is 8% for the use of the balance of the principal ($500) for the one-year period, or 0.08 x 500 = $40. The $500 principal is also repaid for a total end-of.;.year-2cash flow at -$540. The cash flow is: End of Year Cash Flow +$1000 o (now) -580 1 2 -540 l In this chapter, we will demonstrate techniques for comparing the value of money at differentdates, an ability that is essential to engineeringeconomicanalysis.Wemust be able to compare, for example, a low-costmotor with a higher-cost motor. If there were no other consequences, we would obviously prefer the low-cost one. But if the higher-cost motor were more efficient and thereby reduced the annual electric power cost, we would want to consider whether to spend more money now on the motor to reduce power costs in the 64 INTERESTAND EQUIVALENCE future. This chapter will provide the methods for comparing the alternatives to determine which motor is preferred. TIME VALUE OF MONEY We often find that the monetary consequences of any alternative occur over a substantial period of time-say, a year or more. When monetary consequencesoccur in a short period of time, we simply add up the various sums of money and obtain a net result. But can we treat money this way when the time span is greater? Which would you prefer, $100 cash today or the assurance of receiving $100 a year from now? Youmight decide you wouldprefer the $100 now because that is one way to be certain of receiving it. But suppose you were convinced that you would receive the $100 one year hence. Now what wouldbe your answer? A little thought should convinceyou that it still would be more desirable to receive the $100 now.If you had the money now,rather than a year hence, you wouldhave the use of it for an extra year. And if you had no current use for $100, you could let someone else use it. Money is quite a valuable asset-so valuable that people are willing to pay to have money available for their use. Money can be rented in roughly the same way one rents an apartment, only with money,the charge is called interest instead of rent. The importance of interest is demonstratedby banks and savings institutions continuously offering to pay for the use of people's money,to pay interest. If the current interest rate is 9% per year, and you put $100 into the bank for one year, how much will you receive back at the end of the year? You will receive your original $100 together with $9 interest, for a total of $109. This example demonstrates the time preference for money: we would rather have $100 today than the assured promise of $100 one year hence; but we might well considerleaving the $100 in a bank if we knew it would be worth $109 one year hence. This is because there is a time value of money in the form of the willingness of banks, businesses, and people to pay interest for the use of various sums. Simple Interest Simple interest is interest that is computed only on the original sum and not on accrued interest. Thus if you were to loan a present sum of money P to someone at a simple annual interest rate i (stated as a decimal)for a period of n years, the amount of interest you would receive from the loan wouldbe: Totalinterest earned = P x i x n = Pin (3-1) At the end of n years the amountof money due you, F, would equal the amount of the loan P plus the total interest earned. That is, the amount of money due at the end of the loan would be F or F = PO + in). = P + Pin (3-2) . r- - Time Value of Money 65 You have agreed to loan a friend $5000 for 5 years at a simple interest rate of 8% per year. How much interest will you receive from the loan'? How much will your friend pay you at the end of 5 years? Total interest earned= Pin = ($5000)(0.08)(5 yr) - $2000 Amount due at end'of loan = P + Pin = 5000 + 2000 = $7000 I ;;.., In Example 3-3 the interest earned at the end of the first year is (5000)(0.08)(1) = $400, but this money is not paid to the lender until the end of the fifth year. As a result, the borrower has the use of the $400 for four years without paying any interest on it. This is how simple interest works, and it is easy to see why lenders seldom agree to make simple interest loans. Compound Interest With simple interest, the amount earned (for invested money) or due (for borrowedmoney) in one period doesnot affectthe principalfor interest calculationsin later periods.However, this is not how interest is normally calculated. In practice, interest is computed using the compound interest method. For a loan, any interest owed but not paid at the end of the year is added to the balance due. Thus, the next year's interest is calculated based on the unpaidbalance due, which includes the unpaid interestfromthe precedingperiod. In this way, compound interest can be thought of as interest on top of interest. This distinguishes compound interest from simple interest. In this section, the remainder of the book, and in practice you should assume that the rate is a compound interest rate. The few exceptions will clearly state use "simple interest." Tohighlight the differencebetween simple and compoundinterest, rework Example 3-3 using an interest rate of 8% per year compound interest. How will this change affect the amount that your friend pays you at the end of 5 years? ~ ~ . . ..-: . . Original loan amount (original principal) = $5000 Loan term = 5 years Int~rest rate charged 1.________ 8%per year cOlnpound interest - ..- - - - - - - --- - - - - - ..- - - -- -- - -- - --- -- -- 66 ------...- INTERESTAND EQUIVALENCE In the following table we calculate on a year-to-year basis the total dollar amount due at the end of each year. Notice that this amount becomes the principal upon which interest is calculated in the next year (this is the compoundingeff~?t). ._ . ___ Unpaid Total Principal Total Amount Due at the End of Year n, New Total Principal for Year n + 1 $5000 x 0.08 = 400 $5000 + 400 = 5400 5400 5400 x 0.08 = 432 5400 + 432 = 5832 3 5832 5832 x 0.08 = 467 5832 + 467 = 6299 4 6299 6299 x 0.08 = 504 6299 + 504 = 6803 5 6803 6803 x 0.08 = 544 6803 + 544 = 7347 Year Total Principal (P) on Which Interest is Calculated in Year n 1 $5000 2 Interest (I) Owed at the End of Year n from Year n's , The total amount due at the end of the fifth year, $7347, is the amount that your friend will give you to repay the original loan. Notice that this amount is $347 more than the amount youreceived for loaning the same amount, for the same period, at simple interest. This, of course, is because of the effect of interest being earned (by you) on top of interest. Repaying a Debt To better understand the mechanics of interest, let us say that $5000 is owed and is to be repaid in 5 years, together with 8% annual interest. There are a great many ways in which debts are repaid; for simplicity,we have selected four specificways for our example. Table 3-1 tabulates the four plans. In Plan 1, $1000 will be paid at the end of each year plus the interest due at the end of the year for the use of moneyto that point. Thus, at the end of the first year, we will have had the use of $5000. The interest owed is 8% x $5000 = $400. The end-of-year payment is, therefore, $1000principalplus $400 interest, for a total payment of $1400.At the end of the second year, another $1000 principal plus interest will be repaid on the money owed during the year.This time the amountowedhas declinedfrom $5000 to $4000because of the $1000 principal payment at the end of the first year. The interest payment is 8% x $4000 = $320, making the end-of-year payment a total of $1320. As indicated in Table 3-1, the series of payments' continues each year until the loan is fully repaid at the end of the fifth year. Plan 2 is another way to repay $5000 in 5 years with interest at 8%. This time the . end-of-year payment is limited to the interest due, with no principal payment. Instead, the $5000 owed is repaid in a lump sum at the end of the fifth year. The end-of-year payment in each of the first four years of Plan 2 is 8% x $5000 = $400. The fifth year, the payment is $400 interestplus the $5000 principal, for a total of $5400. Plan 3 calls for five equal end-of-yearpayments of $1252 each. At this point, we have not shown how the figure of $1252 was computed (see later: Example 4-3). However, it is Time Value of Money TABLE 67 3-1 Four Plans for Repayment of $5000 in 5 Years with Interest at 8% (a) Year (b) Amount Owed (c) Interest Owed for (d) Total Owed at (e) at Beginning of Year That-Year, 8% x (b) End of Year, Principal Payment (b) + (c) Plan 1: At end of each year pay $1000 principalplus interestdue. 1 $5000 $ 400 $5400 2 4000 320 4320 3 3000 240 3240 4 2000 160 2160 5 1000 80 1080 $1200 (f) Total End-of-Year Payment $1000 1000 1000 1000 1000 $1400 1320 1240 1160 1080 $5000 $6200 Plan 2: Pay interest due at end of each year and principalat end of 5 years. 1 $5000 $ 400 $5400 $ o 2 5000 400 5400 o o 3 5000 400 5400 o 4 5000 400 5400 5000 5 5000 400 5400 $ 400 400 400 400 5400 $5000 $7000 $ 852 921 994 1074 1159 $1252* 1252 1252 1252 1252 $5000 $6260 $ $ $2000 Plan 3: Pay in five equal end-of-yearpayments. 1 $5000 $ 400 2 4148 331 3 3227 258 4 2233 178 5 1159 93 $5400 4479 3485 2411 1252 $1260 Plan 4: Pay principal and interest in one payment at end of 5 years. 1 $5000 $ 400 $5400 2 5400 432 5832 3 5832 467 6299 4 6299 504 6803 5 6803 544 7347 $2347 0 o o o 5000 $5000 0 o o o 7347 $7347 *The exact value is $1252.28, which has been rounded to an even dollar amount. clear that there is some equal end-of-year amount that would repay the loan. By following the computations in Table 3-1, we see that this series of five payments of $1252 repays a $5000 debt in 5 years with interest at 8%. Plan 4 is still another method of repaying the $5000 debt. In this plan, no payment is made until the end of the fifth year when the loan is completely repaid. Note what happens at the end of the first year: the interest due for the first year-8% x $5000 = $400-is not paid; instead, it is added to the debt. At the second year, then, the debt has increased to $5400. The second year interest is thus 8% x $5400 = $432. This amount, again unpaid, is ---- -- - . - -- ._--- -- - --- 68 INTERESTAND EQUIVALENCE added to the debt, increasing it further to $5832. At the end of the fifth year, the total sum due has grown to $7347 and is paid at that time (see Example 3-4). Note that when the $400 interest was not paid at the end of the first year, it was added to the debt and, in the second ye~, tilere was interest charged on this unpaid interest. That is, the $400 of unpaid interest resulted in 8% x $400 = $32 of additional interest charge in the second year. That $32, together with 8% x $5000 = $400 interest on the $5000 original debt, broughtthe total interestchargeat the end of the secondyear to $432. Charginginterest on unpaid interest is called compound interest. We will deal extensivelywith compound interestcalculationslaterin thischapter. . With Table 3-1 we have illustrated four different ways of accomplishing the same task, that is, to repay a debt of $5000 in 5 years with interest at 8%. Having described the alternatives,we will now use them to present the important concept of equivalence. EQUIVALENCE When we are indifferent as to whether we have a quantity of money now or the assurance of some other sum of money in the future, or series of future sums of money, we say that the present sum of money is equivalent to the future sum or series of future sums. If an industrial firm believed 8% was an appropriate interest rate, it would have no particular preference about whether it received $5000 now or was repaid by Plan 1 of Table 3-1. Thus $5000 today is equivalent to the series of five end-of-year payments. In the same fashion, the industrial firm would accept repayment Plan 2 as equivalentto $5000 now. Logic tells us that if Plan 1 is equivalent to $5000 now and Plan 2 is also equivalent to $5000 now, it must follow that Plan 1is equivalent to Plan 2. In fact, allfour repayment plans must be equivalent to each other and to $5000 now. Equivalenceis an essential factor in engineering economic analysis. In Chapter 2, we sawhow an alternativecouldbe representedby a cash flowtable.Howmight two alternatives with different cash flows be compared? For example, consider the cash flows for Plans 1 and 2: Year 1 2 3 4 5 Plan 1 -$1400 -1320 -1240 -1160 -1080 Plan 2 -$400 -400 -400 -400 -5400 -$6200 -$7000 If you were given your choice between the two alternatives,which one would you choose? Obviously the two plans have cash flows that are different. Plan 1 requires that there be . larger payments in the first 4 years, but the total payments are smaller than the sum of Plan 2's payments. To make a decision, the cash flows must be altered so that they can be compared. The technique of equivalence is the way we accomplish this. Using mathematicalmanipulation,we can determine an equivalentvalue at some point in time for Plan 1 and a comparable equivalent value for Plan 2, based on a selected Equivalence 69 interest rate. Then we canjudge the relative attractivenessof the two alternatives,not from their cash flows,but from comparable equivalentvalues. Since Plan 1, like Plan 2, repays a present sum of $5000 with interest at 8%, the plans are equivalentto $5000 now; therefore, the alternatives are equally attractive. This cannot be deduced from the given cash flows alone. It is necessary to learn this by deieimining the equivalentvalues for each alternative at some point in time, which in this case is "the present." Difference in Repayment Plans The four plans computed in Table 3-1 are equivalent in nature but different in structure. Table 3-2 repeats the end-of-year payment schedules from Table 3-1 and also graphs each plan to show the debt still owed at any point in time. Since $5000 was borrowed at the beginning of the first year, all the graphs begin at that point. We see, however, that the four plans result in quite different situations on the amount of money owed at any other point in time. In Plans 1 and 3, the money owed declines as time passes. With Plan 2 the debt remains constant,whilePlan 4 increases the debt until the end of the fifth year.These graphs show an important differenceamong the repaymentplans-the areas under the curvesdiffer greatly. Sincethe axes are Money Owed and Time,the area is their product: Money owed x Time, in years. In the discussion of the time value of money, we saw that the use of money over a time period was valuable, that people are willing to pay interest to have the use of money for periods of time. When people borrow money, they are acquiring the use of money as represented by the area under the curve for Money owed vs Time-.It follows that, at a given interest rate, the amount of interest to be paid will be proportional to the area under the curve. Since in each case the $5000 loan is repaid, the interest for each plan is the total minus the $5000 principal: Plan Total Interest Paid 1 $1200 2 2000 3 1260 4 2347 We can use Table 3-2 and the data from Table 3-1, to compute the area under each of the four curves, that is, the area bounded by the abscissa, the ordinate, and the curve itself. We multiply the ordinate (Money owed) by the abscissa (1 year) for each of the fiveyears, then add: Shaded area = (Money owed in Year 1)(1year) + (Moneyowedin Year2)(1year) +... + (Moneyowedin Year5)(1year) or Shaded area [(Money owed)(Time)] = Dollar-Years .. 70 . _. --- INTERESTAND EQUIVALENCE TABLE 3-2 End-of-Year Payment Schedules and Their Graphs Plan 1: At end of each year pay $1000 principal plus interest due. Year 1 2 3 4 5 End-of- Year Payment $1400 1320 1240 1160 - 1080 $6000 $5000 ] $4000 0 .... us:: $3000 o $2000 $1000 $0 $6200 1 2 3 4 5 Time (years) Plan 2: Pay interest due at end of each year and principal at end of 5 years. $6000 Year 1 2 3 4 5 End-of- Year Payment $ 400 400 400 400 5400 - $5000 "'c:I $4000 0 $3000 s:: o $2000 $1000 $0 $7000 $6000 Plan 3: Pay in five equal end-of-year payments. Year 1 2 3 4 5 End-of- Year Payment $1252 1252 1252 1252 - 1252 $5000 "'c:I $4000 0 $3000' s:: $2000 $1000 $0 $6260 Year 1 2 3 4 5 $7347 1 2 3 4 5 Time (years) $7000 Plan 4: Pay principal and interest in one payment at end of 5 years. End-of- Year Payment 0 $ 0 0 0 7347 - 1 2 3 4 5 Time (years) $6000 $5000 $4000 .... 0 $3000 $2000 $1000 $0 h_. ...-...."., 1 2 3 4 5 TIme(years) Equivalence 71 The dollar-years for the four plans would be as follows: Dollar-Years (Money (Money (Money (Money (Money owed owed owed owed owed in Year 1)(1 year) in Year 2)(1 year) in Year 3)(1 year) in Year 4)(1 year) in Year 5)(1 year) Total dollar-years Plan 1 Plan 2 Plan 3 Plan 4 $ 5,000 4,000 3,000 2,000 1,000 $15,000 $ 5,000 5,000 5,000 5,000 5,000 $25,000 $ 5,000 4,148 3,227 2,233 1,159 $15,767 $ 5,000 5,400 5,832 6,299 6,803 $29,334 With the area under each curve computed in dollar-years, the ratio of total interest paid to area under the curve may be obtained: Total Interest Paid Area Under Curve Plan (dollar-years) Ratio of Total Interest Paid to Area Under Curve 1 2 3 4 $1200 2000 1260 2347 15,000 25,000 15,767 29,334 0.08 0.08 0.08 0.08 We see that the ratio of total interest paid to the area under the curve is constant and equal to 8%. Stated another way,the total interestpaid equals the interestrate times the area under the curve. From our calculations,wemore easily see why therepaymentplansrequire thepayment of different total sums of money,yet are actually equivalentto each other.The key factor is that the four repaymentplans provide the borrower with differentquantities of dollar-years. Since dollar-years times interest rate equals the interest charge, the four plans result in different to~ interest charges. Equivalence Is Dependent on Interest Rate In the example of Plans 1-4, all calculations were made at an 8% interest rate. At this interest rate, it has been shown that all four plans are equivalentto a present sum of $5000. But what would happen if we were to change the problem by changing the interest rate? If the interest rate were increased to 9%, we know that the required interest payment for each plan would increase,and the calculated repaymentschedules (Table3-1, column f) could no longer repay the $5000 debt with the higher interest. Instead, each plan would repay a sum less than the principal of $5000, because more money would have to be used to repay the higher interest rate. By some calculations (to be explained later in this chapter and in Chapter 4), the equivalentpresent sum that each plan will repay at 9% interest is: Plan 1 2 3 4 Repay a Present Sum of $4877 4806 4870 4775 - n 72 ..... INTERESTAND EQUIVALENCE As predicted,at the higher9% interesteachof the repaymentplansof Table3-1 repaysa present sum less than $5000. But they do not repay the same present sum. Plan 1 would repay $4877 with 9% interest, while Plan 2 would repay $4806. Thus, with interest at 9%, Plans 1 and 2 are no longer equivalent, for they will not repay the same present sum. The two series of payments (Plan 1 and Plan 2) were equivalent at 8%, but not at 9%. This leads to the conclusion that equivalence is dependent on the interest rate. Changing the interest rate destroys the equivalencebetween two series of payments. Could we create revised repayment schemes that would be equivalent to $5000 now with interest at 9%? Yes, of course we could: to revise Plan I of Table 3-1, we need to increase the total end-of-year payment in order to pay 9% interest on the outstanding debt. Year 1 2 3 4 5 Amount Owed at Beginning of Year $5000 4000 3000 2000 1000 9% Interest for Year $450 360 270 180 90 Total End-of-Year Payment ($1000 plus interest) $1450 1360 1270 1180 1090 Plan 2 of Table 3-1 is revised for 9% interest by increasing the first four payments to 9% x $5000 = $450 and the finalpayment to $5450.Twoplans that repay $5000 in 5 years with interest at 9% are: Revised Year 1 2 3 4 5 End-of-Year Plan 1 $1450 1360 lTIO 1180 1090 Payments Plan 2 $ 450 450 ~O 450 5450 Wehave determined that RevisedPlan 1is equivalentto a present sumof $5000 and Revised Plan 2 is equivalentto $5000 now;it followsthat at 9% interest, RevisedPlan 1is equivalent to Revised Plan 2. Application of Equivalence Calculations To understand the usefulness of equivalencecalculations,consider the following: Alternative A: Alternative B: Lower Initial Cost, Higher Initial Cost, . Year Higher Operating Cost Lower Operating Cost -$600 -$850 o (now) 1 -115 -80 2 -115 -80 3 -115 -80 10 -115 -80 -- -- ._--- - -. - --- _.- --- .--- - - -- - -- --.-. " --- --------- Single Payment Compound Interest Formulas 73 Is the least-costalternativethe onethat hasthe lower initial cost andhigher operating costs or the one with higher initial cost and lower continuing costs? Because of the time value of money, one cannot add up sums of money at different points in time directly. This means that alternatives cannot be compared in actual dollars at different points in time; instead comparisons must be made in some equivalent comparable sums of money. It is not sufficientto comparethe initial $600 against $850. Inste~d,we must compute a value that represents the entire stream of payments. In other words, we want to determine a sum that is equivalent to Alternative A's cash flow; similarly, we need to compute the equivalent sum for Alternative B. By computing equivalent sums at the same point in time ("now"), we will have values thatmay be validlycompared.The methods for accomplishing this will be presented later in this chapter and Chapter 4. . Thus far we have discussed computingequivalentpresent sums for a cash flow.But the technique of equivalenceis not limitedto a present computation.Instead, we could compute the equivalent sum for a cash flow at any point in time. We could compare alternatives in "Equivalent Year 10" dollars rather than "now" (Year 0) dollars. Further, the equivalence need not be a single sum; it couldbe a series of payments or receipts. In Plan 3 of Table 3-1, the series of equal payments was equivalentto $5000 now. But the equivalencyworks both ways. Supposewe ask the question,What is the equivalentequal annualpayment continuing for 5 years,givena presentsumof $5000and interest at 8%? The answer is $1252. SINGLE PAYMENT COMPOUND INTEREST FORMULAS To facilitate equivalence computations, a series of interest formulas will be derived. To simplify the presentation, we'll use the following notation: i = interest rate per interest period. In the equations the interest rate is stated as a decimal(thatis, 9% interestis 0.09). n = number P = a present F = a future sum of money. The future sum F is an amount, n interest periods from of interest periods. sum of money. the present, that is equivalentto P with interest rate i. Suppose a present sum of money P is invested for one yearl at interest rate i. At the end of theyear,we shouldreceivebackourinitialinvestmentP, togetherwithinterestequalto i P, or a total amount P + i P. Factoring P, the sum at the end of one year is P (1 + i). Let us assumethat, instead of removingour investmentat the end of one year, we agree to let it remain for another year. How much wouldour investmentbe worth at the end of the second year? The end-of-first-yearsum P(1 + i) will draw interest in the second year of. i P(1 + i). This means that, at the end of the second year, the total investmentwill become P(1+i)+iP(1+i) IA more general statement is to specify "one interest period" rather than "one year." Since, however, it is easier to visualize one year, the derivation will assume that one year is the interest period. 74 INTERESTAND EQUIVALENCE This may be rearranged by factoring P(1 + i), which gives us P(1 + i)(l + i) or If the process is continued for a third year, the end-of-the-third-yeartotal amount will be P(1 + i)3; at the end of n years, we will have P(l + i)n. The progression looks like this: Year 1 2 3 n Amount at Beginning of Interest Period P P(1 + i) P(1 + i)2 P(l + i)n-l + Interest for Period +iP +iP(l+i) +iP(1+i)2 + iP(1 + i)n-l = Amount at End of Interest Period =P(1+i) = P(1 +i)2 = P(1 + i)3 = P (1 + i)n In other words, a present sum P increases in n periods to P (1 + i)n. We therefore have a relationship between a present sum P and its equivalentfuture sum, F. Future sum = (Present sum) (1 + i)n F = P(1 + i)n (3-3) This is the single payment compound amount formula and is written in functional notation as (3-4) F=P(FjP,i,n) The notation in parentheses (F j P, i, n) can be read as follows: To find a future sum F, given a present sum, P, at an interest rate i per interest period, and n interest periods hence. . Functional notation is designed so that the compound interest factors may be written in an equation in an algebraically correct form. In Equation 3-4, for example, the functional notation is interpreted as which is dimensionallycorrect.Withoutproceedingfurther,we can see that when we derive. a compound interest factor to find a present sum P, given a future sum F, the factor will be (P j F, i, n); so, the resulting equation would be P=F(PjF,i,n) which is dimensionally correct. Single Payment Compound Interest Formulas 75 If $500 were depositedin a bank savings account, how much would be in the account three years hence if the bank paid 6% interest compoun.~tecJ.@1.1u~Jy? We can draw a diagram of the problem. [Note:To have a consistent notation, we will represent receipts by upward arrows (and positive signs), and disbursements (or payments) will have downward arrows (and negative signs).] .SQlUl1I0N From the viewpointof theperson depositingthe $500, the diagramfor "today" (Time = 0) through Year3 is as follows: F=? Receipts (+$) r 0-1-2-3 Disbursements (-$) t n=3 i = 0.06 P = 500 Weneed to identify the various elementsof the equation. The present sum P is $500. The interest rate per interest period is 6%, and in 3 years there are three interest periods. The future sum F is to be computed from the formula F = P(1 + it = 500(1 + where P = $500,i = 0.06, n 0.06)3 = $595.50 = 3, and F is unknown. Thus if we deposit $500 in the bank now at 6% interest, there will be $595.50 in the account in three years. The equation F = P (I + i)n need not be solved with a hand calculator.Instead, thesinglepayment compound amount factor, (1 + i)n, is readily determined from computed~tables~The factor is written in convenientnotation as (1 + i)n = (F j P, i, n) = ;;:= = = and in functional notation as r, (Fj P, 6%, 3) Knowing n = 3, locate the proper row in the 6% table.2 To find F given P, look in the ~st column, which is headed "Single Payment, Compound1\.mountFactor": forn = 3,we find 1.191. IN ~_ -- - ...-- 2The appendix contains compound interest tables for rates between 1/4 and 60%. 76 INTERESTAND EQUIVALENCE Thus, = 500(F j P, 6%, 3) = 500(1.191) = $595.50 F Before leaving this problem, let's draw another-diagramof it, this time from the bank's point of VIew. P = 500 Receipts(+) t 0-1-2-3 Disbursements (,--) n=3 i ==0.06 1 F=? This indicates the bank receives $500 now and must make a disbursement of F at the end of 3 years. The computation, from the bank's point of view, is F II = 500(F j P, 6%,3) = 500(1.191) =$595.50 This is exactly the Sameas what was computed from the depositor's viewpoint, since this is just the other side of the same transaction. The bank's future disbursement equals the depositor's future receipt. If we take F = P(l + i)n and solve for P, then P = 1 F (1 + i)n = F (1 + i)-n This is the single payment present worth formula. The equation P = F (1 + i)-n (3-5) in our notation becomes P = F(PjF,i,n) (3-6) If you wish to have $800 iri a savings account at the end of 4 years, and 5% interest was paid annually, how much should you put into the savings account now? , C~LUTION F = $800 P = F(1 + i - 0.05 n =4 P = unJpIown i)~n= 800(1 + 0.05)-4 = 800(0.8227) - $658.16 Thus to have $800 in the savings account at the end of 4 years, we must deposit $658.16 now. I' I ..., Single PaymentCompound Interest Formulas 77 Fo;:; 800 .. Receipts (+) Disbursements r (...c.) n=4 p=? P i = 0.05 f (P I }!i',i,n) ... $800(t'IF, 5%,4) 10;'" - EtOhl thGcornpoundinterest tables, ~ iO' II " ~ J ~ (PI F,5%, 4) I .. 0.8227 P '~"$800(0.8227)~ $658.1'6 ~ Here the problem has an exact answer. In many situations, however, the answer is rounded off, recognizing that it can be only as acc,urate as the input information on which it is based. Suppose the bank changed their interest policy in Example 3-5 to "6% interest, compounded quarterly." For this situation, how much money would be in the account at the end of 3 years, assuming a $500 depositnow? First, we must be certain to understand the meaningo( 6% interest, compoUnded quarterly.' 'I'h~t;e are two elements: 6% [pterest: Unless otherwise described,itis customary-to as~umc;that the sta.tc;dP1teresti~ .for a one-year period.ljthe stated interest is for othetthgn a one-yeatperiod, the fimg frame must be clgarly statgd. "' Compounded quarterlY:Tl1isil1dicat~~ th~reare fgl1rint~rest periods pet y-ear; that is, an rillt~restperiodis.31TIonths IOllg. il We kno~ that the 6% interest is an annual rate because ifit were for a different period, it would have be9P stated. Since we are dealing with four interest periods per 'year, if foijows that the '" - - - - ..._~ !..I - - I 78 INTERESTAND EQUIVALENCE For the total3-year duration,there are 12interestperiods. interest rate per interest period is 111z%. Thus P = $500 F = P (1 + it= i = 0.015'''' n''''::'(4 x 3) = 12 F = unknown P (F / P, i, n) = $500(1 + 0.015)12= $500(F / P, Jlh%, 12) = $500(1.196) = $598 A $500 deposit now would yield $598 in 3 years. Consider the following situation: Year Cash Flow o 1 2 3 4 5 +P 0 0 -400 0 -600 Solve for P assuming a 12% interest rate and using the compound interest tables. Recall that receipts have a plus sign and disbursements or payments have a negative sign. Thus, the diagram is: p R,oeip"(+) t0-1-2-3-4-5 Disbursements ( ) ~ ! 400 600 f = P = 400(P / F, 12%,3) + 600(P / F, 12%,5) , I = 400(0.7118) + 600(0.5674) - $625.16 J It i$ important to understandjust what the solution, $625.16,represents. We can say that $625.16 is tbe amount of moneythat would need to be invested at 12% annual interest to allow for the withdraWalof $400 at the end of 3 years and $600 at the 'endof 5 years. ... u. .. _ _ _..u_ Single Payment Compound Interest Formulas 79 Let's examine the computationsfurther. If $625.16 is invested for one year at 12% interest, it will increase to [625.16+ 0.12(625.16)] = $700.18. If for the second year the $700.18 is invested at 12%, it will increase to [700.18 +0.12(700.18)] = $784.20. And if this is repeated for another year, [784.20+ 0.12(784.20)]= $878.30. . We are now at the end of Year 3. The original $625.16 has increased through the addition of interest to $878.30. It is at this point that the $400 is paid out. Deducting $400 from $878.30 leaves $478.30. The $478.30 can be invested at 12%for the fourth year and will increase to [478.30 + 0.12(478.30)] = $535.70.Andifleft at interestfor anotheryear,it will increaseto [535.70+ 0.12(535.70)] = $600. We are now at the end of Year 5; with a $600 payout; there 1s no money remaining in the account. In other words, the $625.16 wasjust enough money, at a 12% interest rate, to exactly provide for a $400 disbursementat the end of Year3 and alsoa $600 disbursementat the end of Year5. Weend up neither short of money nor with money left over: this is an illustration of equivalence.The initial $625.16is equivalent to the combinationof a $400 disbursement at the end of Year 3 and a $600 disbursementat the end of Year 5. There is another way to see what the $625.16 value of P represents. Suppose at Year0 you were offered a piece of paper that guaranteedyou would be paid $400 at the end of 3 years and $600 at the end of 5 years. How much would you be willing to pay for this piece of paper if you wanted your money to produce a 12% interestrate? This.altemate statement of the problem changes the signs in the cash flow and the diagram: 600 400 Receipts (+) t 0-1-2 1 5 ,3-4 Disbursements ( - ) 1 ! I -- . .. = Year Cash Flow -P o o o 1 2 3 4 5 I I I :;, +400 o +600 :Oalis to recover our init!.aliTIvestmentP togeth enY ith 12%interesq~er year, weca n see that P mu~~be less than the total amountto be received in the future (that is, $400 + 600 $1000). We must ~fl1culatethe presellt stun P thatis equivalent, at 12%interest, to an aggregate of $400 SiIl ce t be ~ . . . I .. . . . . . . .. I i~~ . ~' ... ~ears;~Ild . . . . . $600 in 5 years. . . . . . .. . . = . 80 INTERESTAND EQUIVALENCE Since we have already derived the relationship p~-- = (1 +~. --".-. i) -n F - -we write p :- 400(1 + 0.12)-3 + 600(1 + 0.12r5 $625.17 This is virtuallythe same amount computed from the first statement of this example. [The slight diffe;rence is due to the rounding in the compound interest tables. For example, (1 + 0.12)-5 - 0.56'742I! bu~!hecompound in,teresJtable for J2%~hows 0.5674.] !!: ,. = ~ ~ Both problems in Example 3-8 have been solved by computing the value of P that is equivalent to $400 at the end of Year3 and $600 at the end of Year 5. In the first problem, we received + P at Year0 and were obligated to payout the $400 and $600 in later years. In the second (alternate formation) problem, the reverse was true. We paid -P at Year 0 and would receive the $400 and $600 sums in later years. In fact, the two problems could represent the buyer and seller of the same piece of paper. The seller would receive + P at Year 0 while the buyer would pay - P. Thus, while the problems looked different, they could have been one situationexamined first from the viewpointof the seller and then from that of the buyer. Either way, the solution is based on an equivalencecomputation. The second set of cash flowsin Example 3-8 was: Year o 1 2 3 4 5 Cash Flow -P o o +400 o +600 Ata 12% interest rate, P was computed to be $625.17. Suppose the interest rate is increased to 15%. Will the value of P be larger or smaller? --- On~ can consider P as a sum of moneyinvested at 15% from which one is to obtain $400 at the endlof 3 years and $600 at the end of 5 years. At 12%, the required P is $625.17. AtJ5%, P will ---~ -.. - -..- Summary 81 earn more interest each year, indicating that we can begin with a smaller P and still accumulate enough money for the subsequent cash flows.The computationis: P 400(P / F, i5'%,3) + 600(P/ F, 15%,5) = 400(0.6575) +600(0.4972) = $561.32 The value of P is smaller at 15% than at 12% interest. .-. SUMMARY This chapter describes cash flow tables, the time value of money, and equivalence. The single payment compoundinterest formulaswere derived. It is essential that these concepts and the use of the interest formulas be carefully understood, since the remainder of this book and the practice of engineering economy are based on them. Time value of money: The continuing offer of banks to pay interest for the temporary use of other people's money is ample proof that there is a time value of money. Thus, we would always choose to receive $100 today rather than the promise of $100 to be paid at a future date. Equivalence: What sum would a person be willing to accept a year hence instead of $100 today? If a 9% interest rate is considered to be appropriate, he would require $109 a year hence. If $100 today and $109 a year hence are considered equally desirable, we say the two sums of money are equivalent.But, if on further consideration, we decided that a 12%interest rate is applicable, then $109 a year hence wouldno longerbe equivalentto $100 today.This illustratesthatequivalence is dependent on the interest rate. Single Payment Formulas Theseformulasare for compoundinterest,whichis usedin engineeringeconomy. Compound amount Present worth where i F = P(l + i)n = P(F / P, i, n) P = F(1+ i) -n = F (P / F, i, n) = interest rate per interest period (stated as a decimal) n = number of interest periods P a present sum of money = F = a future sum of money; the future sum F is an amount, n interest periods from the present, that is equivalentto P with interest rate i This chapteralso definedsimpleinterest,whereinterestdoes not carry over and become part of the principal in subsequent periods. Unless otherwise specified,all interest rates in this text are compound rates. 82 INTERESTAND EQUIVALENCE PROBLEMS 3-1 In your own words explain the time value of money. From your own life (either now or in a situation -that might occur in your future), provide an example in which the time value of money would be important. 3-8 What sum of money now is equivalentto $8250 two years hence, if interest is 4% per 6-month period? (Answer: $7052) 3-9 The local bank offers to pay 5% interest on savings 3-2 Magdalen, Miriam, and Mary June were asked to consider two different cash flows: $500 that they could receive today and $1000 that would be received 3 years from today. Magdalen wanted the $500 dollars today, Miriam chose to collect $1000 in 3 years, and Mary June was indifferent between these two options. Can you offer an explanation of the choice made by each woman? deposits. In a nearby town, the bank pays 1.25% per 3month period (quarterly). A man who has $3000 to put in a savings account wonders whether the increased interest paid in the nearby town justifies driving his car there to make the deposit. Assuming he will leave all money in the account for 2 years, how much additional interest would he obtain from the out-of-town bank over the local bank? 3-3 A woman borrowed $2000 and agreed to repay it at the end of 3 years, together with 10% simple interest per year. How much will she pay 3 years hence? 3-10 A sum of money invested at 2% per 6-month period (semiannually), will double in amount in approximately how many years? (Answer: 171/zyears) 3-4 A $5000 loan was to be repaid with 8% simple annual interest. A total of $5350 was paid. How long was the loan outstanding? 3-5 Solve the diagram below for the unknown Q assuming a 10% interest rate. 3-11 The Apex Company sold a water softener to Marty Smith. The price of the unit was $350. Marty asked for a deferred payment plan, and a contract was written. Under the contract, the buyer could delay paying for the water softener if he purchased the coarse salt for recharging the softener from Apex. At the end of 2 years, the buyer was to pay for the unit in a lump sum, with interest at a rate of 1.5% per quarter-year. According to the contract, if the customer ceased buying salt from Apex at any time prior to 2 years, the full payment due at the end of 2 years would automatically become due. Six months later, Marty decided to buy salt elsewhere and stopped buying from Apex, whereupon Apex asked for the full payment that was to have been due 18 months hence. Marty was unhappy about this, so Apex offered as an alternative to accept the $350 with interest at 10% per semiannual period for the 6 months that Marty had been buying salt from Apex. Which of these alternatives should Marty accept? Explain. 200 t 0-1-2-3-4 t n (Answer: Q = $136.60) =4 Q 3-6 The followingseries of paymentswill repaya present sum of $5000 at an 8% interest rate. Using single payment factors, what present sum is equivalent to this series of payments at a 10%interest rate? Year 1 2 3 4 5 3-7 End-of-Year PaYIDent $1400 1320 1240 1160 1080 A man wentto his bank and borrowed $750. He agreed to repay the sum at the end of 3 years, together with the interest at 8% per year. How much will he owe the bank at the end of 3 years? (Answer: $945) 3-12 The United States recently purchased $1 billion of 30-year zero-coupon bonds from a struggling foreign nation. The bonds yield 41/z% per year interest. The zero-coupon bonds pay no interest during their 30-year life. Instead, at the end of 30 years, the U.S. government is to receive back its $1 billion. together with interest at 4112%per year. A U.S. senator' objected to the purchase, claiming that the correct interest rate for bonds like this is 5lf4%. The result, he said, was a multimillion dollar gift to the foreign Problems country without the approval of Congress. Assuming the senator's math is correct, how much will the foreign country have saved in interest when it repays the bonds at 41/z% instead of 51/4% at the end of 30 years? 3-13 One thousand dollars is borrowed for one year at an interest rate of 1% per month. If the same sum of money could be borrowed for the same period at an interest rate of 12% per year, how much could be saved in interest charges? 3-14 A sum of money Q will be received 6 years from now. At 5% annual interest. the present worth now of Q is $60. At the same interest rate, what would be the value of Q in 10 years? 3-15 In 1995 an anonymous private collector purchased a painting by Picasso entitled Angel Fernandez de Soto for $29,152,000. The picture depicts Picasso's friend deSoto seated in a Barcelona cafe drinking absinthe. The painting was done in 1903 and valued then at $600. If the painting was owned by the same family until its sale in 1995, what rate of return did they receive on the $600 investment? 3-16 (a) If $100 at Time "0" will be worth $110 a year hence and was $90 a year ago, compute the interest rate for the past year and the interest rate next year. (b) Assume that $90 invested a year ago will return $110 a year from now. What is the annual interest rate in this situation? 3-17 How much must you invest now at 7.9% interest to accumulate $175,000 in 63 years? 83 3-18 We know that a certain piece of equipment will cost $150,000 in 5 years. How much will it cost today using 10% interest? 3-19 The local garbage company charges $6 a month for - - ---- -garbage collection. It had been their practice to send out bills to their 100,000 customers at the end of each 2-month period. Thus, at the end of February it would send a bill to each customer for $12 for garbage collection during January and February. Recently the firm changed its billing date: it now sends out the 2-month bills after one month's service has been performed. Bills for January-February, for example, are sent out at the end of January. The local newspaper points out that the firm is receiving half its money before the garbage collection. This unearned money, the newspaper says, could be temporarily invested for one month at 1% per month interest by the garbage company to earn extra income. Compute how much extra income the garbage company could earn each year if it invests the money as described by the newspaper. (Answer: $36,000) 3-20 Sally Stanford is buying an automobile that costs $12,000. She will pay $2000 immediately and the remaining $10,000 in four annual end-of-year principal payments of $2500 each. In addition to the $2500, she must pay 15% interest on the unpaid balance of the loan each year. Prepare a cash flow table to represent this situation. After Completing This Chapter... The student should be able to: Solve problems modeled by the uniform series compound interest formulas. Use arithmetic and geometric gradients to solve appropriately modeled problems. ·· ·· Apply nominal and effective interest rates. Use discrete and continuous compoundingin appropriate contexts. · Use spreadsheets and financial functions to model and solve engineering economic analysis problems. QUESTIONS TO CONSIDER . .::;, 1. Set aside your calculator and test your intuition for a minute. What annual rate of return do you suppose it would take to turn $5000 into $20 million in 51 years? 100%? 200%? 2. Now get out the calculator. What was Anne Scheiber's actual average annual rate of return during the years she was investing? 3. How does that rate compare with the overallperformance of the stock market during the period from 1944 to 1995? .. . More Interest Formulas Anne Scheiber's Bonanza When Anne Scheiber died in 1995,age 101,she left an estate worth more than $20 million. It all went to YeshivaUniversityin New York.The university's officialswere grateful. But who, they asked, was Anne Scheiber? She wasn't a mysterious heiress or a business tycoon, it turned out. She was a retired IRS auditor who had started investing in 1944, putting $5000-her life savings up to that point-into the stock market. Scheiber's portfolio was not based on getrich-quick companies. In fact, it contained mostly "garden variety" stocks like Coca Cola and Exxon. Scheiber typically researched stock purchases carefully, and then held onto her shares for years, rather than trading them. She also continually reinvestedher dividends. Despite her increasing wealth, Scheiber never indulged in a lavish lifestyle. Acquaintances described her as extremely frugal and reported that she lived as a near recluse in her small apartment. What motivated Scheiber to accumulate so muchmoney? In largepart, it seemsto havebeen a reactionto her life experiences.Scheiberworkedfor over 20 yearsas a tax auditorbut failed to receivepromotions. Sheattributedher lackof professional advancementto discrimination againsther as a Jewish woman.At herrequest, the endowmentshe left to YeshivaUniversity was used to fund scholarships and interest-free loans for women students. 86 MORE INTERESTFORMULAS Chapter 3 presentedthe fundamentalcomponentsof engineeringeconomic analysis,including formulas to compute equivalent single sums of money at different points in time. Most problems we will encounter are much more complex. Thus this chapter developsformulas for payments that are a uniform series or are increasing on an arithmetic or geometric gradient. Later in the chapter;noilliriaIand effective interest are discussed. Finally, equations are derived for situations where interest is continuously compounded. UNIFORM SERIES COMPOUND INTERESTFORMULAS Many times we will find uniform series of receipts or disbursements. Automobile loans, house payments, and many other loans are based on a uniform payment series. It will often be convenient to use tables based on a uniform series of receipts or disbursements. The series A is defined as follows: A = An end-of-periodl cash receipt or disbursementin a uniform series, continuing for n periods, the entire series equivalent to P or F at interest rate i The horizontal line in Figure 4-1 is a representation of time with four interest periods illustrated. Uniform payments A have been placed at the end of each interest period, and there are as many A's as there are interest periods n. (Both these conditions are specifiedin the definition of A.) Figure 4-1 uses January 1 and December 31, but other I-year periods could be used. In Chapter 3's section on single payment formulas, we saw that a sum P at one point in time would increase to a sum F in n periods, according to the equation F = P(1+ it We will use this relationship in our uniform series derivation. A FIGURE 4-1 The general relationship between A andF. Year A A t t 0-1-2-3-4 t I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I = ~~ 0 ~ ~~ 0 ~ ~~ 0 ~ ~ v ~ v ~ n=4 A t F v.... ~ 1In textbooks on economic analysis, it is customary to define A as an end-of-period event rather than a. beginning-of-period or, possibly, middle-of-period event. The derivations that follow are based on this end-of-period assumption. One could, of course, derive other equations based on beginning-of-period or mid period assumptions. Uniform Series Compound Interest Formulas 87 Looking at Figure 4-1, we see that if an amount A is investedat the end of each year for 4 years, the total amount F at the end of 4 years will be the sum of the compound amounts of the individual investments. A A A A A + + + + + ..q'-- A A A -+ + + 0-1-2-3-4=0-1-2-3-4+0-1-2-3-4+0-1-2-3-4+0-1-2-3-4 1 F 1 A(l + i)3 + A(l + i)2 t ~ + A(l + i) + A In the general case for n years, F = A(1 + i)n-l + ... + A(1 + i)3 + A(1 + i)2 + A(1 + i) + A (4-1) Multiplying Equation (4-1) by (1 + i), we have . (1 + i)F = A(1 + i)n + . . . + A(1 + i)4 + A(1 + i)3 + A(1 + i)2 + A(1 + i) (4-2) Factoring out A and subtracting Equation 4-1 gives (4-3) (4-4) Solving Equation 4-4 for F gives (1 + i)n - 1 F=A [ i . ] =A(F/A,z%,n) (4-5) Thus we have an equation for F when A is known. The term inside the brackets is called the uniform series compound amount factor and has the notation (F / A, i, n). A man deposits $500 in a credit union at the end of each year fot 5 years. The credit union pays 5% interest, compounded annually.At the end of 5 years, immediately after the fifth deposit, how much does the man have in his account? -- - ---------- 88 MORE INTERESTFORMULAS .... SbU:JTlON1 c.".,," ...,>~ .... I The diagram on the left shows the situation from the man's point of view; the one on the right, from the credit union's point of view.Either way,-thediagram of the five deposits and the desired computationof the future sum F duplicates the situation for the uniform series compound amount formula F 0-1-2-3-4-5 t A r t t A A t 0 t A A A + + 1--'--2 A A + + 3--'--4---5 A + A 1 F (1 + on - 1. F - A.[ · i... where A = $500, n . ..5, i. ...0.05, Funknown. F . . . ] ;: = A(FIA, z%,n) Filling i.11 fi1eknownvariable$gives .$500(F I A, 5%, 5) $500(5.526) . $2763 Th~re willbe $2763 in the account followingthefifthdeposih If Equation 4-5 is solved for A, we have A-F - i [ (1 + On- I ] = F(AI F, i%, n) (4-6) where is called the uniform series sinking fund2 factor and is written as (AI F, i, n). 2A sinking fund is a separate fund into which one makes a unifonn series of money deposits (A) with the goal of accumulating some desired future sum (F) at a given future point in time. Uniform Series Compound Interest Formulas 89 Jim Hayes read that out west, a parcel of land could be purchased for $1000 cash. Jim decided to save a uniform amount at the end of eacn month.sothat he would have the required $1000 at the end of one year. The local credit union pays 6% interest, compounded monthly. How much would Jim have to deposit each month? In this example, n = 12 F = $1000 A = unknown = A = 1000(AjF, 1/2%,12)= 1000(0.0811) = $81.10 Jim would have to deposit $81.10 each month. If we use the sinking fund formula (Equation 4-6) and substitute for F the single payment compound amount formula (Equation 3-3), we obtain A-F - A i -PI i [ (1 + i)n - 1] (+) = P [ (1i(I+i)n . + l)n - 1] i n [ (1 + i)n - 1] . = P(AjP,I%,n) (4-7) Wenow have an equation for determiningthe value of a seriesof end-of-period paymentsor disbursements-A when the present sum P is known. The portion inside the brackets i (1 + i)n [ (1 + i)n - 1] is called the uniform.series capital recovery factor and has the notation (Aj P, i, n). Consider a situation in which you borrow $5000. You will repay the loan in five equal end-ofthe-year payments. The first payment is due one year after you receive the loan. Interest on the loan is 8%. What is the size of each of the five payments? - - ... -.. - - -.. _ _. J 90 MORE INTERESTFORMULAS p r O-1 t A P = $5000 2~3-4 t A n=5 t 5 t A t A A i__8% A -- I= 4 4 = unknown = ~-~~ Tb;eannualloan paymeQtis $1252. In Example 4-3, with interest at 8%, a present sum of $5000 is equivalent to five equal. end-of-period disbursements of $1252. This is another way of stating Plan 3 of Table 3-1. The method for determining the annual payment that would repay $5000 in 5 years with 8% interest has now been explained. The calculation is simply A = 5000(Aj P, 8%, 5) = 5000(0.2505) = $1252 If the capital recovery formula (Equation 4-7) is solved for the present sum P, we obtain the uniform series present worth formula (1 + i)n - 1 = A(PjA, i%, n) P = A [ i (1 + i)n ] (4-8) and (1 + i)n - 1 (PjA,i%,n)= [ i(1+i)n ] which is the uniform series present worth factor. -- An investor holds a time payment purchase contract on some machine tools. The contract calls for.the payment of $140 at the end of eactlmonth for a5..yearperiod/I'hefirstpaYmetlt is due in one month..fle offers to sell you the cOlltractfor $6800 cash today.lf you otherwise can make 1% per monThon yourmoQey,WqUId. you accept or reject tlJ,ewvestpr's.pffer? - -- ~ - I' 1 Uniform Series Compound Interest Formulas 91 n = 60 i=1% p = 'SOlUfflbN,) " ,I.. .,' In this problem we are being offered a contract that will pay $140 per month for 60 months. We must determine whetherthe contractis worth $6800, if we consider 1% per month to be a suitable interestrate. Using the uniform series pt:esentworth formula, w~\yill <;;.omgutethe present worth of the cOntract. P - A(PjA,i,n) .140(PjA, 1%,60) = 140(44.955) = $6293.70 It is clear that if we pay the $6800 asking price for the contract, we wil).receive less than the 1% per month interest we desire. We wi11,therefore,reject the investor's offer. - .- .~--.- Suppose we decided to pay the $6800 for the time purchase contract in Example 4-4. What monthly rate of return would we obtain on our investment? In this situation, we know P,.A, and n, but we do not know i. The problem may be solved by using either the uniform series present worth formula P=A(PjA,i,n) I or the uniform series capital recovery formula r I A=P(AjP,i,n) Either way, we have one equation with one unknown. P = $6800 A = $140 -.. - n..---- 60 i - unknown - -- --- --..... 92 MORE INTERESTFORMULAS P = A(P/A,i,n) $6800 = $140(P/A, i, 60) . .~.;... .. ... . ._~ .- 6800 (P / A, i, 60) = = 48.571 140 We know the value of the uniform series present worth factor, but we do not know the interest rate i. As a result, we need to look through several compound interest tables and then compute the rate of return i by interpolation. Entering values from the tables in the appendix, we find (PIA, i, 60) 48.571.-.J 5L726:lJ 48.174 d The rate of return, which is between 1/2%and %%, may indeed be computedby a linear interpolatiOh.The interestformulas are not linear, so a linear interpolation will not give an exact solution. To minimize the error, the interpolation should be computed usinginterest rates as close to the correct answer as possible. [Since a/b = e/d, a = bee/d)], we write Rate ofreturn i = 0.5%+ a = 0.5% + bee/d) 51.726 - 48.571 ( 6 (3.552) = = 0.50%+ 0.25%.. 51.72 - 48.174 = 0.50% ~ + 0.25% = 0.72% per month 3.155 ) 0.50% + 0.22% The)monthly rate of return on our investmentwould be 0.72% per month. Using a 15% interest rate, compute the value of F in the following cash flow: Year 1 2 3 4 5 Cash Flow + 100 + 100 +100 0 -F -.. -.... - ' r Uniform Series Compound Interest Formulas 100 100 100 t t t 0-1-2-3-4-5 . 93 1 F .. ... . We see that the cash flow diagram is not the same as the diagram shown in Example 4-1, which is a sinking fund factor diagram. A A A t t t 0-1-2---'---3 1 F Since the diagrams do not agree, the problem is more difficult than those We've discussed so far. The general approach to use in this situation is to convert the cash flow fronl its present form into standard forms, for which we have compoundinterest factors and compound interest tables. On~ way to solve this problem is to consider the cash flow as a series of single paYJ:l1ents P and thet}to compute their sum F. In other words, the cash flow is bJ;okeninto three parts, each IJ, one of Whichwe can solve. , . " 100 100 100 0 t1)-2---,3 t t _w 100 4 5= + 0 . - t 1"--'2.=3~4--.,5 ,.. .... { I I F F W FI + F'z+ F3 .. JOO(FI P ,15%, 4) +lQO(f " II " IP, 15%, 3) + 100(F I P, 15%, 2) = 100(1.749) -I- 100(1.521) + 100(1.322) $459.20 The val~~of Fin theilfustrated cash flow is $459.20. - - - ""... - """'" '" -- - 11II... __ ___ _ ___ 1" I 94 MORE INTERESTFORMULAS IALTERNATESOLUTION. 100 100 t t 0-1-2-3-4-5 100 100 t t - 0-1-2-3-4-5 100 t 100 t 1 F F Looked at this way, we first solve for Fl' FI = 100(F / A, 15%,3) = 100(3.472)= $347.20 Now FI can be considered a present sum P in the diagram 3-4-5 F and so F = FI(F/ P, 15%,2) : = 347.20(1.322) ~ $459.00 ]he slightly different value from the preceding computation is due to rounding in the compound interest tables. This has been a two-step solution: FI = 100(F/ A, 15%,3) I" F-:- FI(F/P, 15%,2) - - .. ---'- -- - ...- - - --. -- ...- -.. -- - -- I u + __ n Uniform Series Compound Interest Formulas One could substitute the value of FI frO!llthe first equation into the second equation and solve for F, without computing Fl, F = $459.00 , ~;;.~ = 100(F lA, 15%, 3)(P I P, 15%,2) = 100(3.472)(1.322) ;;; = =t-~ ;!iI:Ci Consider the following situation: 0~1-2-3-4 t 30 t t i = 15% 1 p The diagram is not in a standardform, indicating that there will be a multiple-stepsolution.There are at least three differentways of computingthe answer.(It is important that you understandhow the three computations are made, so please study all three solutions.) ,:- - -"-- j SOLUfflON 1 . P = PI + P2 + P3 = 20(P I F, 15%,2) ::; ~ + 30(P I F, 15%,3) + 20(P I F, 15%,4) ~:;:; :: ~ = 20(0.7561) + 30(0.6575) + 20(0.5718) - $46.28 30 P= W 0-1-1-3'--4 +0 !; - -- - ---- ,--1-2'--3'--4 ! -- t + 0-1-2-3-1 w ~ ----- - - - -..------ 95 96 MORE INTERESTFORMULAS !SOlUTION 2 30 t. 0-1-2-3-4 t 2; 1 P 1 F The relationship between P and F in the diagram is P = F(PjF, 15%,4) Next we compute the future sums of the three payments, as follows: 30 F t = 0-1-2-3-4 t + 0-1-2-3-4 t F t + 0-1-2-3-4 t 2t = FI + F2+ 20 - 20(Fj P, 15%,2) + 30(Fj P, 15%,1) + 20 Combining the two equations, we have P = [FI + F2+ 20](P j F, 15%,4) = [20(Fj P, 15%,2) + 30(Fj P, 15%, 1) + 20](P j F, 15%,4) = [20(1.322) + 30(1.150) + 20](0.5718) = :..f ;..'~.'.'.._ ",."_",,,"~~"-'_"..',_H"'_' ,:, $46.28 ,.- ~ ...SOlUJ}Q~[3. =' 30 0-1-2-3-4 ~ P P Y t 30 Y t tt ~ PI = PI (PjF, Y 1-2-3-4 Y = Y 1-2-3-4 Y Y + 1-2-3-4 ~ ~ = PI PI = 20(P j A, 15%,3) + lO(Pj F, 15%,2) 15%, 1) -- .. ..- -- - --- --- ---- Relationships Between Compound Interest Factors 97 Combining, we have 15%, 2)](P I F, 15%, 1) P - [20(P I A, 15%,3) + . 10(PIF, . -.. . '~- = [20(2.283) + 10(0.7561)](0.8696) = $46.28 = -- RELATIONSHIPS BETWEEN COMPOUND INTEREST FACTORS From the derivations,we see there are several simple relationshipsbetween the compound interest factors. They are summarized here. Single Payment Compound amount factor = 1 Present worth factor 1 (FIP,i,n) (4-9) = (PIF,i,n) Uniform Series . Capital recovery factor = 1 Present worth factor 1 (AIP,i,n) (4-10) = (PIA,i,n) 1 Compound amount factor = Sinking fund factor 1 (F I A, i, n) = (AI F, i, n) (4-11) The uniform series present worth factor is simply the sum of the n terms of the single payment present worth factor n (PIA,i,n) = L(PIF,i, 1=1 J) (4-12) For example: (P I A, 5%, 4) = (PI F, 5%, 1) + (PI F, 5%, 2) + (PI F, 5%,3) + (PI F, 5%, 4) 3.546 = 0.9524 + 0.9070 + 0.8638 + 0.8227 98 MORE INTERESTFORMULAS TJIeuniform series compound amount factor equals 1plus the sum of (n - 1) terms of the single payment compound amount factor n-l (FfA;i;n) = 1+ L (F I P, i, J) (4-13) J=1 For example, (FIA,5%,4) = 1+(FIP,5%,I)+(FIP,5%,2)+(FIP,5%,3) 4.310 = 1 + 1.050 + 1.102 + 1.158 The uniform series capital recoveryfactor equals the uniform series sinkingfund factor plus i: (AIP,i,n) = (AIF,i,n)+i (4-14) For example, (AI P, 5%, 4) 0.2820 = (AI F, 5%, 4) + 0.05 = 0.2320 + 0.05 This may be proved as follows: (AIP,i,n) = (AIF,i,n)+i 1 i (1 + i)n [ (1 + i)n - 1] = [ . (1 + i)n - 1] + l Multiply by (1 + i)n - 1 to get ARITHMETIC GRADIENT It frequently happens that the cash flowseries is not of constant amount A. Instead, there is a uniformly increasing series as shown: t 0-1-2-3-4-5 t 1 r r r Arithmetic Gradient 99 Cash flows of this form may be resolved into two components: A+4G A + 2GA+ 3G 4G _.- A+G i A t r 1 A 3G -". A A A =r-l-Z-3-4-S p 2G t t t t t . r-l-Z-3-4-S . ~ 0 t , r + 1-I-Z-3-4-S Note that by resolving the problem in this manner, the first cash flow in the arithmetic gradient series becomes zero. This is done so that G is the change from period to period, and because the gradient (G) series normally is used along with a uniform series (A). We already have an equationfor P', and we need to derive an equation for P". In this way, we will be able to write P = P' + p" = A(P/A, i, n) + G(PfG, i, n) Derivation of Arithmetic Gradient Factors The arithmetic gradientis a series of increasing cash flowsas follows: (n - })G (n - 2)G 2G o O-}-2-3~n X t r -}-n ! NOdune~O~ The arithmetic gradientseriesmay be thought of as a series of individual cash flows: (n - })G (n - 2)G 2G 0-I-Z-3-+-n-l-t i t X + 0-I-z-3-+-n-1-1 O-}-2-3-+-n-}-n + O-}-2-3-+-n-}-n + + Fn 1 Fill The value of F for the sum of the cash flows = FI + Fn + . . . + Fill+ FlY, or F = G(1 + i)n-2+ 2G(1+ i)n-3+... + (n - 2)(G)(1 + i) 1+ (n - l)G (4-15) 100 MORE INTERESTFORMULAS Multiply Equation 4-15 by (1 + i) andfactor out G, or (1 + i)F = G[(1 + i)n-l + 2(1+ i)n-2 +.". + (n - 2)(1 + i)2 + (n -1)(1 + i)l] (4-16) Rewrite Equation 4-15 to'show other-terms in the series, F = G[(1+ i)n-2+ ". " + (n - 3)(1 + i)2 + (n - 2)(1 + i)1 + n - 1] (4-17) Subtracting Equation 4-17 from Equation 4-16, we obtain F + iF - F = G[(1 + it-1 + (1 + i)n-2 + ." . + (1 + i)2 + (1 + i)1 + 1]- nG (4-18) In the derivation of Equation 4-5, the terms inside the brackets of Equation 4-18 were shown to equal the series compoundamount factor: [(1 + i)n-l + (1 + i)n-2 + .". + (1+ i)2 + (1+ i)1+ 1] = ( 1 + ")n 1 1.I Thus, Equation 4-18 becomes . IF (1 +i)n -1 =G i [ - nG ] Rearranging and solving for F, we write F = ~ [(1+ir (4-19) - 1 - n] Multiplying Equation 4-19 by the single payment present worth factor gives P = = (PjG,i,n)= G (1 + i)n - 1 i [ i G _n 1 ] [ (1 + i)n ] (1 + i)n - in - 1 i2(1 + i)n [ ] (1+i)n-in-1 i2(1+i)n [ (4-20) ] Equation 4-20 is the arithmetic gradient present worth factor. MultiplyingEquation4-19 by the sinking fund factor, we have A = G (1 + i)n - 1 _ n i [ . (AjG,I,n) i i ] [ (1 + i)n - 1] = G In (1 + i)n - in - 1 = [ i(l +i)n -i ] = [ i - (1 +i)n -1 [ ] Equation 4-21 is the arithmetic gradient uniform series factor. (1 + i)n - in - 1 i (1 + i)n - i ] (4-21) . Arithmetic Gradient A man has purchased a new automobile. He wishes to set aside enough money in a bank account to pay the maintenance on the car for the firstS years-.-Ithas been estimated that the maintenance cost of an automobile is as follows: Year 1 2 3 4 5 Maintenance Cost $120 150 180 210 240 Assume the maintenance costs occur at the end of each year and that the bank pays 5% interest. How much should the car owner deposit in the bank now? SOLU)"ION 240 210 0-1-2-3-4-5 t p The cash flowmay be broken into its two components: 240 210 180 150 120 120 120 120 120 60 If"1 = 0-1-2-3-4-5 r r r 0-1-2~3 r 4-5 r + o 0-1-2 30 ! t ,;:;-3~4-5 G::; 30 t 1 p A~120 1 Both components represent cash flows for which compound interest factors have been derived. The first is a uniform series present worth, and the second is an arithmetic gradient seriespresent worth: P = A(P j A, 5%, 5) + G(I' jG, 5%,5) --- - . 101 102 MORE INTERESTFORMULAS Note that the value of n in the gradient factor is 5, not 4. In deriving the gradient factor, we had (n - 1) terms containing G. Here there are four terms containing G. Thus,(n - 1) = 4, so n = 5. P = 120(P j A, 5%, 5) + 30(P jG, 5%" 5) = 120(4.329) + 30(8.237) = 519+247 = $766 He ~houlddeposit $766 in the bank now. ,."'-.."" =..-.~ On a certain piece of machinery, it is estimated that the maintenance expense will be as follows: Year 1 2 3 4 Maintenance $100 200 300 400 What is the equivalent uniform annual maintenance cost for the machinery if 6% interest is used? SOLUTION 400 It 0-1-2-3-4 = ! Ao T r ! ! AA [ ! A, " The first cash flow in the arithmetic gradient series is zero, hence the diagram is not in proper fOffilJorthe arithmeticgradientequation. Asin.Example 4-8, the cash flpwmust beresolved into -------- --- --- --- I Arithmetic Gradient 103 two components: 400 300 200 t 200 TI ,100 t t 0~1--'-'---'-2---'---3~4 J A = 100 + 100(Aj G, 6%, 4)= 100+ 100(1.427) .'$242.70 . The equiyalent unifonn annualmafntenan,ceCosti$$242.70. -- :.0-_ -. .- :: .;;;, :iiiiii -",.,,-~ ...~:;;~ II!! ~ J , i A textile mill in India installed a number of new looms. It is expected that initial maintenance costs and expenses for repairs will be high but will then decline for several years. The projected cost is: Year 1 2 3 4 Maintenance and Repair Costs (rupees) 24,000 18,000 12,000 6,000 What is the projected equivalent annual maintenance and ,repair cost if interest is 1O%? = The projecte~fcashflowis not in the fonn of the arithmetic gradientfactors.Both factors were derived for an increasing gradient overtime. The factors cannot be uSeddirectly for a decpning ;II . . I 104 MORE INTERESTFORMULAS gradient. Instead, we will subtract an increasing gradient from an assumed uniform series of payments. 24' 000 18,000 12,000 6000 o 0-1-2-3-4 - 0-1-2-3-4 t [ i t 60;0 1 A' 1 A' 1 A' 1 A' = 0-1-2-3-4 A' = 24,000- 6000(AjG, 10%,4) =,24,000 -- 6000(1.38t) = 15,714 rupees The projected equivalent uniform maintenanceand repair cost is 15,714rupees per year. oS Compute the value of P in the diagram. Use a 10% interest rate. 150 100 50 0-1-2-3-4-5~6 o t r p ~.,,_. "'.-; ", SptlJ11()~' '- With the arithmetic gradient series present worth factor, we can compute a present sum J. - -- ----------- - -- - - ,. I Geometric Gradient 105 150 100 50 0 -l-- --- -2-3-4 1 o t r 5 6 ! p J It is important that you closely examine the location of J .Based on the way the factor was derived,therewillbe onezerovalu~in thegradientserjesto therightof J " (If this seems strange or incorr~ct,review the beginning of this section on arithmetic gradients.) J - G(PIG, i, n) = 50(PI G, 10%,4) == 50(4.378) (Note: 3 would be incorrect.) . 218.90 · Then P .. J(P IF, 10%,2) To o1;>tain the present worth of theft.1turesum J, use tbe (P I F,'i, n) factor. Combining, we have P [I = 50(P IG, .. " 10%, 4)(P I}?, 10%,2) 50(4.378)(0.8264)' $180.90 The value of Pis $1.80.90. GEOMETRIC GRADIENT Inthe precedingsection, we saw that the arithmetic gradientis applicablewhere the periodby-periodchangein a cash receipt or paymentis a uniformamount.There areother situations where the period-by-period change is a uniform rate, g. For example, if the maintenance costs for an automobileare $100 the first year and they increaseat a uniformrate, g, of 10% per year, the cash flow for the first 5 years would be as follows: Cash Flow Year $100.00 1 100.00 110.00 2 100.00 + 10%(100.00) = 100(1 + 0.10)1 3 4 5 110.00+ 10%(110.00)= 100(1+ 0.10)2 121.00 + 10%(121.00) = 100(1 + 0.10)3 133.10 + 10%(133.10) = 100(1+ 0.10)4 - 121.00 133.10 146.41 .I 106 MORE INTERESTFORMULAS 0-1-2-3-4-5 ---- _ :.. 00 1.00 - 110.00---_ 121.00----.... 133.10-.... ... 146.41' From the table, we can see that the maintenance cost in any year is $100(1 + gt-1 Stated in a more general form, (4-22) . where g = uniform rate of cash flowincrease/decreasefrom period to period, that is, the geometric gradient Al = value of cash flow at Year 1 ($100 in the example) An = value of cash flow at any Year n Since the present worth Pnof any cash flow An at interest rate i is (4-23) we can ~ubstituteEquation 4-22 into Equation 4-23 to get This may be rewritten as P = A1(1 +0-1 t( x=1 X-l 1 +~ 1+ l ) (4-24) The present worth of the entire gradient series of cash flowsmay be obtainedby expanding Equation 4-24: P = A1(1 +0-1 t( X-l 1 +~ x=1 1+ l ) (4-25) . ,. Geometric Gradient A A 2 -;---r--- 107 __A3 f O-1-2-3-..t---n-l-n : I p In the general case, where i =fg, Equation 4-24 may be written out as follows: P (~::) = Al (1 + i)n-I + Al (1 + i)-I ( ) + . . . + Al(1+ i)-I 1+ : = Al (1 + i)-I and b P (~::) 2 n-I 1+ Let a + Al (1 + i)-I = (1 + g)/(1 (4-26) , + i). Equation 4-26 becomes = a + ab + ab2 + . . . + abn-I (4-27) Multiply Equation 4-27 by b: bP = ab + ab2 + ab3 + . . . + abn-I + abn (4-28) Subtract Equation 4-28 from Equation 4-27: P - bP =a - abn P(1 - b) = a(1 - bn) P= a(1 - bn) I-b Replacing the original values for a and b, we obtain: (4-29) where i =f g. 108 MORE INTERESTFORMULAS The expressionin the brackets of Equation 4-29 is the geometric series present worth factor where i =J g. (PIA, g, i,-it) In the special case of i ... [ 1 ~-(1 +.g)n(1 + i)-n z-g ] where i =Jg (4-30) = g, Equation 4-29 becomes (PIA,g,i,n) = [n(1+ i)-I] (4-31) where i = g The first-yearmaintenance cost for a new automobileis estimated to be $100, and it increases at a uniform rate of 10% per year. Using an 8% interest rate, calculate the present worth of cost of the first 5 years of maintenance. STEP-BY-STEP SOLUTION , Year n 1 2 3 4 5 Maintenance Cost 100.00 100.00+ 110.00+ 121.00+ 133.10+ 10%(100.00) 10%(110.00) 10%(121.00) 10%(133.10) (P / F, 8%, n) = 110.00 x x = = 133.10 x 0.9259 = 0.8573 = 0.7938 = 0.7350 = 146.41 x 0.6806 = 100.00 = 121.00 PWof Maintenance $ 92.59 94.30 96.05 97.83 99.65 $480.42 x = III P = Al . [ 1 - (1 +. g)n(1 + i)-n Z- g ] iiiI ::=;' where i =Jg I = 100.00 1 - (1.10)5(1.08)-5 -0.02 [ ] = $480.42 The present worth of cost of maintenance for the first 5 years is $480.42. I I I I. Nominal and Effective Interest NOMINAL 109 AND EFFECTIVE INTEREST Consider the situationof a person depositing $100 into a bank that pays 5% interest, compounded semiannually.How much would be in the savings account at the end of one year? . SOlU~tO~n Five percent interest, compoundedsemiannually,means that the bank pays 21/2%every6 months. Thus, the initial amount P = $100 wouldbe creditedwith 0.025(100) = $2.50 interest at the end of 6 months, or P --* P + Pi = 100+ 100(0.025) = 100 + 2.50 = $102.50 The $102.50 is left in the savings account; at the end of the second 6-month period, the interest earned is 0.025(102.50) = $2.56, for a total in the account at the end of one year of 102.50 + 2.56 = $105.06, or (P + Pi) --* (P + Pi) + i(P + Pi) = P(1 + i)2 = 100(1+ 0.025)2 = $105.06 Nominal interest rate per year, r, is the annual interest rate without considering the effect of any compounding. In the example, the bank pays 21/2%interest every 6 months. The nominal interest rate per year, r, therefore, is 2 x 21h%= 5%. Effective interest rate per year, ia, is the annual interest rate taking into account the effect of any compoundingduring the year. In Example 4-13 we saw that $100 left in the savings account for one year increased to $105.06, so the interest paid was $5.06. The effective interest rate per year, ia, is $5.06/ $100.00 = 0.0506 = 5.06%. r = Nominal in~erestrate per interest period (usually one year) i = Effective interest rate per interest period = Effectiveinterest rate per year m = Number of compounding subperiods per time period ia Using the method presented in Example 4-13, we can derive the equation for the effective interest rate. If a $1 deposit were made to an account that compounded interest m times 110 MORE INTERESTFORMULAS per year and paid a nominal interest rate per year, r, the interest rate per compounding subperiod would be r/ m, and the total in the account at the end of one year would be . $t(1 +:)m orsimply (1 + :)m If we deduct the $1 principal sum, the expression would be Therefore, Effective interest rate per year where r (4-32) = nominal interest rate per year m =numberof compoundingsubperiodsperyear Or, substitutingthe effectiveinterest rate per compounding subperiod, i = (r / m), Effective interest rate per year where (4-33) i = effectiveinterest rate per compoundingsubperiod m = number of compoundingsubperiods per year Either Equation 4-32 or 4-33 may be used to compute an effectiveinterest rate per year. Oneshouldnote that i was describedin Chapter 3 simply as the interestrate per interest period. We were describing the effective interest rate without maldng any fuss about it. A more precise definition, we now know, is that i is the effective interest rate per interest period. Although it seems more complicated, we are describing the same exact situation, but with more care. The nominalinterestrate r is often givenfor a one-yearperiod (but it couldbe givenfor either a shorter or a longer time period). In the special case of a nominal interest rate that is given per compoundingsubperiod, the effective interest rate per compounding subperiod, i, equalsthe nominal interest rate per subperiod,r. In thetypicaleffectiveinterestcomputation,there aremultiplecompoundingsubperiods (m> 1). The resulting effective interest rate is either the solution to the problem .or an intermediatesolution,whichallowsus to use standardcompoundinterestfactors to proceed to solvethe problem. For continuous compounding (which is described in the next section), Effective interest rate per year (4-34) - - '-,- -_. .. --'.. Nominal and EffectiveInterest 111 If a savings bank pays 11/2%interest every 3 months, what are the nominal and effective interest rates per year? Nominal interest rate per year r =4 x Effective interest rate per year ia = = 11/2%= 6% (1+:) m 0.06 1 + "4 ( - 1 4 ) - 1 = 0.061 =6.1% Alternately, Effective interest rate per year ia = (1 + i)m - = (1 + 1 0.015)4 - 1 = 0.061 =6.1% Table 4-1 tabulates the effective interest rate for a range of compoundingfrequencies andnominalinterestrates. It shouldbe notedthatwhena nominalinterestrate iscompounded I TABLE 4-1 Nominal and Effective Interest Nominal Interest Rate per Year EffectiveInterest Rate per Year,i/l When Nominal Rate Is Compounded r (%) Yearly Semiannually Monthly Daily Continuously 1 2 3 4 5 6 8 10 15 25 1.0000% 2.0000 3.0000 4.0000 5.0000 6.0000 8.0000 10.0000 15.0000 25.0000 1.0025% 2.0100 3.0225 4.0400 5.0625 6.0900 8.1600 10.2500 15.5625 26.5625 1.0046% 2.0184 3.0416 4.0742 5.1162 6.1678 8.3000 10.4713 16.0755 28.0732 1.0050% 2.0201 3.0453 4.0809 5.1268 6.1831 8.3278 10.5156 16.1798 28.3916 1.0050% 2.0201 3.0455 4.0811 5.1271 6.1837 8.3287 10.5171 16.1834 28.4025 -~ 112 ----------------- ------------- MORE INTERESTFORMULAS annually,the nominal interest rate equals the effective interest rate. Also, it will be noted thatincreasingthe frequencyof compounding (for example,from monthly to continuously) hasonly a small impact on the effective interest rate. But if the amount of money is large, evensmall differencesin the effective interest rate canbe significant. A loan sharklendsmoney on the following terms: "If I give you $50 on Monday, you owe me $60 on the following Monday." (a) What nominal interest rate per year (r) is the loan sharkcharging? (b) What effectiveinterest rateper year (ia)is he charging? (c) If the loan shark started with $50 and was able to keep it, as well as all the money he received, out in loans at all times, how much money would he have at the end of one year? F = P(FjP, i, n) 60 = 50(F j P, i, 1) (FjP, i, 1) = 1.2 Therefore, i = 20% per week. Nominalinterest rate per year = 52 weeks x 0.20 = 10.40 = 1040% Effective interest rate per year ia = (1 + ~) m 10.40 ( = 1+ 52 - 1 52 ) - 1 = 13,105- 1 = 13,104- 1,310,400% Or Effectiveinterest rate per year ia = (1 + i)m ~ 1 - (1 + 0.20)52- 1 = 13,104 - 1,310,400% I -- - - - -- .. - ...--- - --- - ..--- II I' I I __ _ ~...J 113 Nominal and Effective Interest F = P (1 + i)n = 50(1 + 0.20)52 = $655,200 With a nominal interest rate of 1040%per year and effectiveinterest rate of 1,310,400%per year, ifhe started with $50, the loan shark would have $655,200 at the end of one year. When the varioustime periods in a problem match, we generally can solve the problem using simple calculations. Thus in Example 4-3, where we had $5000 in an account paying 8% interest, compounded annually, the five equal end-of-year withdrawals are simply computed as follows: A = P(Aj P, 8%, 5) = 5000(0.2505) = $1252 Consider how this simple problem becomes more difficult if the compounding period is changed so that it no longer match the annual withdrawals. f-~It T:::: ... ~ ' ,,-": ... ' .'...'ff"'H. ." ... " ... .. J; "'. 0,. II. " I; .J\'. ..1. ' ~~.. . . '.0 ,. ":l" .. "" vi, On January 1, a woman deposits $5000 in a'credit union that pays 8% nominal annual interest, compoundedquarterly.She wishesto withdrawall the moneyin fiveequal yearly sums,beginning December 31 of the first year. How much should she withdraw each year? SOlUjTlON Since the 8% nominal annual interest rate r is compounded quarterly,we know that the effective interest rate per interest period, i, is 2%; and there are a total of 4 x 5 = 20 interest periods in 5 years. For the equation A = P (A j P, i, n) to be used, there must be as many periodic withdrawals as there'are interest periods, n. In this example we have 5 withdrawalsand 20 interest periods. w w w w w f f f f f 0---1---2---3---4---5---6---7---8---9--10--11--12--13--14--15--16--17--18--19--20 j = 2% per quarter n = 20 quarters 1 $5000 == = To solve the problem, we must adjust it so that it is in one of the standard forms for which we have~;;:. compoundinterest factors.This means we must first comp;pt~ either an equivalentA.{oreach - :;: ~ C! ~ ~ ;; = = a~ -. - .. - -- -- -.. - -~---- _ - - -- 114 - -- ------ - -- - -- ---- MORE INTEREST FORMULAS 3-month interest period or an effective i for each time period between withdrawals. Let's solve the problem both ways. Compute an equivalent A for each 3-month time period. If we had been required to compute the amount that could be withdrawn quarterly,the diagram would be as follows:" AAAAAAAAA A iiiiii A A A A A A A A A A i i i i i i i i i i i i i i 0-1-2-3-4-5-6-7-8-9-10-11-12-13-14-15-16-17-18-19-20 i = 2% per quarter n = 20 quarters 1 $5000 A = P(AI P, i, n) = 5000(AI P, 2%, 20) = 5000(0.0612) = $306 Now, since we know A, we can construct the diagram that relates it to our desired equivalent annualwithdrawal,W: 1111111111111111111 0-1-2-3-4-5-6-7-8-9-10-11-12-13-14-15-16-17-18-19-20 1 1 w 1 w 1 w w r$3~ 1 w Looking at a one-year period, i 0-1~2~3-4 A = $306 i t i i ==2% per quarter n=4 quarters W = A(FIA,i,n) 1 - w- ----- .306(FIA,2%,4) 306(4.122) = ,,= 'II .._ ___ iii _ .. .__ __ _ ----.------ _d_ ~ontinuous Gompound~ng ~TjF~~ Compute an effective i for the time period between withdrawals. Between withdrawals, W, there are four interest periods, hence m = 4 compoundingsubperiods per year. Since the nominal interestrate per year, r, is 8%, we canproceed to compute the effective interest rate per year. ffi r . . E ectivemterestrateper year ia.= m 0.08 4' (1+ m ) - 1 = (1+ 4 ) - 1 = 0.0824 = 8.24% per year Now the problem may be redrawn as follows: w w w w w t t t t t 0-1-2-3-4-5 i = 8.24% per year n = 5 years 1 $5000 This diagram may be directly solved to determine the annual withdrawal W using tl),ecapital recovery factor: W = P(AI =P P, i, n) = 5000(AI P, 8.24%,5) . i (1 + i)n = 5000 0.0824(1 + 0.0824)5 [ (1 + i)n - 1] [ (1 + 0...0824)5- 1 ] = 5000(0.2520) = $1260 The depositor should withdraw $1260 per year. CONTINUOUS COMPOUNDING Two variables we have introduced are: r m = Nominal interest rate per interest period = Number of compounding subperiods per time period Since the interest period is normally one year, the definitionsbecome: r m = Nominal interest rate per year = Number of compounding subperiods per year 115 116 MORE INTERESTFORMULAS - ! - -- ....- r - = Interest rate per interest period m mn = Number of compounding subperiods in n years Single Payment Interest Factors: Continuous Compounding The single payment compound amount formula (Equation 3-3) may be rewritten as If we increasem, the number of compoundingsubperiodsper year, withoutlimit, m becomes very large and approaches infinity,and r / m becomes very small and approaches zero. This is the condition of continuous compounding, that is, where the duration of the interest period decreases from some finite duration D.t to an infinitely small duration dt, and the number of interest periods per year becomes infinite. In this situation of continuous compounding: mn F=P lim m~oo ( ) 1+..c. m . (4-35) An important limit in calculus is: lim(1 + X)l/x = 2.71828 = e x~o (4-36) If we setx = r / m, then mn may be written as (1/x)(rn). Asm becomesinfinite,x becomesO. Equation 4-35 becomes F = P[lim(1 x-+o + x)l/xyn Equation4-36 tellsus the quantity insidethe bracketsequals e. So returningto Equation 3-3, we find that F = P(l P = F(1 + i)-n + it becomes F = Pern (4-37) becomes P = Fe-rn (4-38) and We see that for continuous compounding, or, as shown earlier, Effective interest rate per year (4.,34) _ Continu_ous C~mpounding 117 To find compound amount and present worth for continuous compounding and a single payment, we write: Compoundamount Presentworth F = P(ern) = P[F / P, r, n] (4-39) P = F(e-rn) = F[P/ F, r, n] (4-40) Square brackets around the factors distinguish continuous compounding. If your hand calculator does not have eX, use the table of ern and e-rn, provided at the end of the appendix containing the compound inter~st tables. ~_$~;~~~~r'~" If you were to deposit $2000in a bank that pays 5% nominal interest, compoundedcontinuously, how much wouldbe in the account at the end of 2 years? The single payment compoundamount equation for continuous compounding is where r = nominalinterest rate = 0.05 n = number of years F =2 = 2000e(O.05)(2) = 2000(1.1052) = $2210.40 There would be $2210.40in the account at the end of 2 years. ....... --A bank offers to sell savings certificatesthat will pay the purchaser $5000 at the end of 10 years but will pay nothingto the purchaserin the meantime.If interest is computed at 6%, compounded continuously,at what price is the bank sellingthe certificates? F = $5000 r = 0.06 P= 5000e-O.06xlO= 5000(0.5488) = $2744 TherefQre,the bank is selling the $5000 certificatesfor $2744. = 10 years 1'J 118 MORE INTERESTFORMULAS 9V!).. I.... - -~ .:~ <-. .~ How long will it take for money to double at 10% nominal interest, compounded continuously? F = Pern 2 = 1 eO.IOn eO.IOn = 2 or O.lOn = 102 = 0.693 n = 6.93 years It will take 6.93 years for money to double at 10% noniinal interest, compounded continuously. If the savings bank in Example 4-14 changed its interest policy to 6% interest, compounded continuously, what are the nominal and the effectiveinterest rates? - SOLUTION The nominal interest rate remains at 6% per year. Effectiveinterest rate = er - 1 = eO.06 - 1 = 0.0618 = 6.18% ..:. Uniform Payment Series: Continuous Compounding at Nominal Rate r per Period Let us now substitute the equation i = er - 1 into the equations for end-of-period compounding. Continuous Compounding Sinking Fund er [AI F, r, n] = ern -1 _1 (4-41) Continuous Compounding Capital Recovery [AlP, r,n]= -- ern(er - 1) ern - 1 (4-42) - Continuous Cpmpounding 119 Continuous Compounding Series Compound Amount - 1 ern (4-43) [F/A,r,n] = er-l Continuous Compounding Series Present Worth - 1 ern [P / A, r, n] = ern(er (4-44) _ 1) In Example 4-1, a II.1andeposited $500 per year into a credit union that paid 5% interest, compounded annually. At the end of 5 years, he had $2763 in the credit union. How much would he have if the institution paid 5% nominal interest, compounded continuously? SOLUTION . ,., . -. " ' ! ...~..'- '-", .-~;""~~ A = $500 r = 0.05 n ern ( F=A[F/A,r,n]=A = 5 years _1 er - 1 ) eOo05(5) - =500 ( 1 e 005 0 - 1 ) = $2769.84 He would have $2769.84. In Example 4-2, Jim Hayes wished to save a uniform amounteach month so he wouldhave $1000 at the end of a year. .Based on 6% nominal interest, compounded monthly, he had to deposit $81.10 per month. How much would he have to deposit if his credit union paid 6%.nominal interest, compounded continuously? '. .. ,.. .,.,.... SOLUTlO~;; The deposits are made monthly; hence, there are 12 compounding subperiods in the one-year time period. F = $1000 : .. .'. r= nommal mterest rate/mterest penod n = 12 compounding - - __ _ r........_ = =;;;; 0.06 ~ 5 12 - 0.00 subperiods iQ the one-year geri9PQfJhe PJohl~,n:1. ---- -- - -- ~ = ---- ~ :: - -- 120 - - --- - -- MORE INTERESTFORMULAS - 1 ern 1 ( °.005013 = 1000 ( ) A = F[AI F, r, n] = F 0.061837 er _ ) ( eO.OO5 - 1 ) = 1000 eO.OO5(l2) _ 1 = $81.07 He would have to deposit $81.07 per month. Note that the differencebetween monthly and continuous compounding is just 3 cents per month. Continuous, Uniform Cash Flow (One Period) with Continuous Compounding at Nominal Interest Rate r Equations for a continuous, uniform cash flow during one period only, with contin~ous compounding, can be derived as follows. Let the continuous, uniform cash flowtotaling P be distributed over m subperiods within one period (n = 1). Thus PI m is the cash flQw atthe end of each subperiod. Since the nominal interest rate per period is r, the effective interest rate per subperiod is rim. Substituting these values into the uniform series compound amount equation (Equation 4-5) gives F [1 + (rlm)]m m[ rim =p - 1 ] (4-45) Setting x = rim, we obtain F = P [1 +xY/x - 1 m[ rim =P ] [(1 +X)l/xy [ r -1 (4-46) ] As m increases, x approaches zero. Equation 4-36 says ° lim(1 + X)l/x = e X hence Equation 4-46 for one period becomes p LJ ' ,. ... .. 9 , F .(4-47) --- - - - -r -- Continuous Compounding 121 MultiplyingEquation 4-43 by the singlepayment, continuouscompoundingfactors, we can find F for any future time and P for any present time. For Any Future Time F = 15 ( ; 1) er (er(n-l» = 15 [.(er -r~;(ern)] n F Compound amount [Fj15, r, n] = (er -IHern) rer [ (4-48) ] ForAny Present Time P =F ( ) er r 1. (e-rn)= F er - 1 rern ] [ 1 p Presentworth [PjF,r,n] = [ er - 1 -rern (4-49) ] . A self-service gasoline station has been equipped with an automatic teller machine (ATM).Customers may obtain gasoline sin!ply byins~rting tpeir ATM bank card into tJ:J.emachine and filling their tank with gasoline. When they have finished, theATM unit automatically deducts the gasoline purchaSe from the customer's bank account and credits it to the gas station's bank account. The ga§§Ultionres;ejx~~,,"$40,qOO~et month.;inthisJIlanner,,,,,with,thecash flowing"uniformly throughout -- ---' ;:: I" -..-----..- I 1 --- --- - - -- -"1------ --- - - . ..- -. 122 -....... ...-. MORE INTERESTFORMULAS th~ month. If the bank pays 9% nominal interest, compounded continuously, how much will be in the gasoline station bank account at the end of the month? The exampleproblem may be solvedby either Equation 4-47 or 4-48. Here the latter, the general equation,is used. " .." . . r = N omm al mterest rate per mon th = _ F = P (er [ = 40000 , There _ 1)(ern) rer ] =40,000 12 p = 0.0075 (eO.OO75 _ l)(eo.oo 75(1)) [ 0.0075eo.OO75 (0.0075282)(1.0075282) = $40 150 0.00755646 ' [ ] will be $40,150 in the station's bank account. SPREADSHEETS FOR ECONOMIC 0.09 - ] LJ 9 F ANALYSIS Spreadsheets are used in most real world applications of engineering economy. Common tasks include the following: 1. Constructing tables of cash flows. 2. Using annuity functions to calculate a P, F, A, N, or i. 3. Using a block function to find the present worth or internal rate of return for a table of cash flows. 4. Making graphs for analysis and convincingpresentations. 5. Calculating "what-if' for different assumed values of problem variables. Constructing tables of cash flowsrelies mainly on the spreadsheetbasics that are coveredin Appendix A. These basics include using and naming spreadsheet variabl~s,understanding the difference between absolute and relative addresses when copying a formula, and formatting a cell. Appendix A uses the example of the amortization schedule shown in Table 3-1, Plan 3. This amortization schedule divides each scheduledloan payment into principal and interest and includes the outstandingbalance for each period. . Becausespreadsheetfunctionscan be foundthroughpointingandclickingon menus, those steps are not detailed. In Excel the starting point is the Ix button. Excel functions are used here, but there are only minor syntax differencesfor most other spreadsheetprograms. Spreadsheet Annuity Functions In tables of engineeringeconomyfactors, i is the table, n is the row, and two of P, F, A, and G definea column. The spreadsheetannuityfunctions list four arguments chosenfrom n, A, P, F, and i, and solve for the fifthargument.The Typeargument is optional. If it is omitted or 0, then the A value is assumed to be the end-of-periodcash flow.If the A valuerepresents the beginning-of-periodcash flow,then a value of 1 can be entered for the Type variable. - - - - - - - --- - - - --- - . . -- - --._._-- --- Spreadsheets for Economic Analysis To find the equivalent P To find the equivalent A To find the equivalent F To find n To find i 123 -PV(i,n,A,F,Type) -PMT(i,n,P,F,Type) - FV(i ,n,A,P ,Type) NPER(i ,A,P ,F,Type) RATE(n,A,P ,F,Type,guess) The sign convention for the first three functions seems odd to some students. The PV of $200 per period for 10 periods is negative and the PV of -$200 p.erperiod is positive. So a minus sign is inserted to find the equivalent P, A, or F. Without this minus sign, the calculated value is not equivalent to the four given values. A new engineer wants to save money for down payment on a house. The initial deposit is $685, and $375 is deposited at the end of each month. The savings account earns interest at an annual nominal rate of 6% with monthly compounding.How much is on deposit after 48 months? Because deposits are made monthly,the nominal annual interest rate of 6% must be converteqto = 1/2%per month for the 48 months. Thus we must find F if i = 0.5% = 0.005, n 48, A = 375, and P . 685. Note both the initial and periodic deposits are positive cash flows for the savings .. account. The Excel function is multiplied by-lor $21,156.97. -FV(0.005,48,375,685,0), and the result is A new engineer buys a car with 0% down financing from the dealer. The cost with all taxes, registration, and license fees is $15,732. If each of the 48 monthly paymentsis $398, what is the monthly interest rate? What is the effective annual interest rate? - . . -'. ." . -- .~ SOlUTION~:; .. -.' '-..~ ' ::. I . -~ :.:.","--. .., - ". ".- ..-, ' The RATE function can be used to find the monthly interest rate.. given that n = 48, A = -398,P=15,732, and F=O. The Excel function is RATE(48,-398,15732,0) and the result is 0.822%. The effective annual interest rate is 1.0082212- 1 = 10.33%. Spreadsheet Block Functions Cash flows can be specified period-by-period as a block of values. These cash flows are analyzed by block functions that identify the row or column entries for which a present worth or an internal rate of return should be calculated. In Excel the two functions are NPV(i,values) and IRR(values,guess). - --- T -- - -- --- - 124 MORE INTERESTFORMULAS Economic Criteria Net present value Internal rate of return Values for Periods 1to n Oto n Excel Function NPV(i,values) IRR(values,guess); guess argument is optional Excel's IRR function can be used to find the interest rate for a loan with irregular payments (other applications are covered in Chapter 7). These block functions make different assumptions about the range of years included. For example, NPV(i,values) assumesthat Year0 is NOT included, while IRR(values,guess) assumes that Year 0 is included. These functions require that a cash flow be identifiedfor eachperiod. Youcannotleavecells blankeven if the cash flowis $0. The cash flowsfor 1to n are assumed to be end-of-period flows. All periods are assumed to be the same length of time. . Also the NPV functions returns the present worth equivalent to the cash flows,unlike the PV annuity function, which returns the negative of the equivalent value. For cash flowsinvolvingonly constant valuesof P, F, and A this block approachseems to be inferior to the annuity functions. However,this is a conceptually easy approach for more complicated cash flows,such as arithmetic gradients. Suppose the years {row 1) and the cash flows (row 2) are specifiedin columns B through E. A B C D E F 1 Year 0 1 2 3 4 2 Cash flow -25,000 6000 8000 10,000 12,000 If an interest rate of 8% is assumed, then the present worth of the cas~ .flowscan be calculated as = B2+NPV(.08,C2:F2), which equals $4172.95. This is the present worth equivalent to the five cash flows, rather than the negative of the.present worth equivalent returned by the PV annuityfunction.The internal rate of return calculatedusing IRR(B2:F2) is 14.5%.Notice how the NPV function does not include the Year0 cash flow in B2, while the IRR function does. · PW = B2+NPV(.08,C2:F2) · IRR = IRR(B2:F2) NPV range without Year IRR range with Year0 - q . Using Spreadsheets for Basic Graphing Often we are interested in the relationship between two variables. Examples include the number and size of payments to repay a loan, the present worth of an M.S. degree and how long until we retire, and the interest rate and present worth for a new machine. This kind of two-variable relationship is best shown with a graph. As we will show in Example 4-26, the goal is to place one variable on each axis of the graph and then plot the relationship. Modem spreadsheets automate most steps <;>f drawing a graph, so that it is quite easy. However, there are two very similar types of chart, and we must be careful to choose the xy chart and not the line chart. Both charts measure the y variable, but they treat the x variable differently. The xy chart measures the x variable; thus its x value is measured along the x axis. For the line chart, each x value is placed an equal distance along the x axis. Thus x values of 1, 2,4,8 would be spaced evenly,rather - - . Spreadsheets for EconolT,licAnalysis 125 than doubling each distance. The line chart is really designed to plot y values for different , categories,suchaspricesfor modelsof carsor enrollmentsfor differentuniversities. Drawing an xy plot with Excel is easiest if the table of data lists the x values before the y values. This conventionmakes it easy for Excel to specify one set of x values and several sets of y values. The block ofxy values is selected,and then the chart tool is selected. Then the spreadsheet guides the user through the rest of the steps. Graph the loan payment as a function of the number of payments for a possible auto loan. Let the number of monthly payments vary between 36 and 60. The nominal annual interest rate is 12%, and the amountborrowedis $18,000.. The spreadsheet table shown in Figure 4-2 is constructed first. Cells AS:B 10 are selected, and then the Chartwizard icon is selected.The first step is to select anxy (scatter)plot with smoothed A B 1 12% 2 1% 3 $18,000 D C i 6 # Payments 36 Monthly Payment $597.86 7 42 $526.96 8 48 $474.01 9 54 $433.02 10 60 $400.40 11 12 13 14 15 16 17 18 = PMT($A$2,A6,-$A$3) .. I $600.00 =CI) S j:I. --------------->. $500.00 ---------------------------- :2 = 0 ::E 19 20 F monthly interest r-ate amount borrowed 4 5 E nominal annual interest rate , 48 I $400.0036 42 54 60 Number.of Payments . 21 22 -..- - .. - == --- == = ==- = 'II' = ; .'"I I I -- - - - - - -- 126 MORE INTERESTFORMULAS lines, and without markers as the chart type. (The other choices are no lines, straight lines, and adding markers for the data points.) The second step shows us the graph and allows the option of changing the data cells selected. The third step is for chart options. Here we add titles for the two axes and turn off showing the legend. (Since we have only one line in our graph, the legend is not needed. Deleting it leaves more room for the graph.) In the fourth step we choose where the chart is placed. Because xy plots are normally graphed with the origin set to (0, 0), an a~active plot is best obtained if the minimum and maximum values are changed for each axis. This is done by placing the mouse cursor over the axis and left-clicking. Handles or small black boxes should appear on the axis to show that it has been selected. Right-clicking brings up a menu to select format axis. The scale tab allows us to change the minimum and maximum values. SUMMARY The.compoundinterest formulas describedin this chapter,along withthose in Chapter3~will be referred to throughout the rest of the book. It is very important that the reader understand the concepts presented and how these formulas are used. The following notation i.sused consistently. i n P F = effective interest rate per interest period3(stated as a decimal) = number of interest periods = a present sum of money = a future sum of money: the future sum F is an amount, n interest periods from the present, that is equivalentto P with interest rate i A = an end-of-period cash receipt or disbursementin a uniform series continuing for n periods; the entire series equivalent to P or F at interest rate i G = uniform period-by-period increase or decrease in cash receipts or disbursements; the arithmetic gradient g = uniform rate of cash flow increase or decrease from period to period; the geometric r = nominal gradient. . interest rate per interest period (see footnote 3) ia = effectiveinterestrate per year(annum) m = number of compounding subperiodsper period (see footnote 3) P, F = amount of money flowing continuously and uniformly during a given period Single Payment Formulas (derived in Chapter 3) Compoundamount: F = P(1 +i)n = P(FjP,i,n) 3Normally the interest period is one year, but it could be some other period (e.g., quarter, month, semiannual). ._- - - --- T - - ---- Summary Present worth: P = F (1 + i) -n = F (P I F, i, n) Uniform Series Formulas Compoundamount: (1 + i)n -1 F=A [ i ] = A(FIA, i, n) Sinking fund: Capital recovery: A =P i(1 + i)n [ (1 + i)n 1] _ = P(AI P, i, n) Present worth: (1 + i)n - 1 P =A[ i (1 + i)n ] = A (P I A, i, n) Arithmetic Gradient Formulas Arithmetic gradient present worth: (1 +i)n -in . P = G[ -1 i2(1+ i)n = G(PIG,i,n) ] Arithmetic gradient uniform series: (I+i)n-in-1 A-G - [ i (1 + i)n - i 1 ] -- - G[ i n - G(AIG (1 + i)n - 1] - . l n) , , Geometric Gradient Formulas Geometric series present worth, where i =I- g: 1-(I+g)n(1+i)-n P = Al [ i_g =AI(PIA,g,i,n) ] Geometric series present worth, where i = g: P = AI[n(1 + i)-I] = AI(PIA, g, i, n) = AI(PIA, i, i, n) - - - - ------ 127 128 MORE INTERESTFORMULAS Single Payment Formulas: Continuous at Nominal Rate r per Period Compound amount: Compounding F = P(ern) = P[F I P, r, n] P = F(e-rn) = F[P I F, r, n] Present worth: Note that square brackets around the factors are used to distinguish continuous. compounding. Uniform Payment Series: Continuous Compounding at Nominal Rate r per Period Continuouscompoundingsinkingfund: er-1 A = F ern - 1 = F[AjF,r,n] [ ] COJ,ltinuouscompounding capital recovery: A =P - ern(er = P[AIP,r,n] - 1 ] ern [ 1) Continuous compoundingseries compound amount: - 1 ern F=A . [ er - 1 ] =A[FIA,r,n] . Continuous compounding series present worth: ern P=A - [ ern (er 1 =A[PIA,r,n] - 1)] Continuous, Uniform Cash Flow (One Period) with Continuous Compounding at Nominal Interest Rate r Compoundamount F = P[ (er - 1Hern) rer - - - - - --- ] = F[F IF, . ----- r, n] --- -- --- ---... - _. _.. h __..__ Problems 129 Present worth er - 1 _ P=F [ - _ rern ] _ =F[P/F,r,n] Nominal Interest Rate per Year, r The annual interest rate without considering the effect of any compoun<:fu.1g. Effective Interest Rate per Year, ia The annualinterestrate takingintoaccountthe effectof anycompoundingduringtheyear. Effectiveinterestrateper year(periodiccompounding): or Effective interest rate per year (continuouscompounding): ia. er - 1 PROBLEMS 4-1 Solve diagrams (a)-(c) for the unknowns R, S, and T, assuming a 10% interest rate. 4-2 For diagrams (a)-(d), compute the unknown values: B, C, V, x, respectively, using the minimum number of compound interest factors. 200 200 200 200 100 100 100 f f f r 0-1-2-3-4-5 s R ! (b) 3. Y 0-1-2-3-4-5 ~ T t c (a) 4x 2x 10 10 10 10 t ~ T ~ T ~ T ~ T t t 0-1-2-3-4-5 = $464.10;S = $218.90;T = $54.30) x t + r t 0-1-2-3-4 i=1O% I ! (c) (Answers: R (b) t I r 1 B 120 90 r i= 10% i= 10% n=5 (a) r 0-1-2-3-4 t V 500 r i=1O% x=? (d) (c) , - -- --- -- --- 130 MORE INTERESTFORMULAS (Answers: B x = $66.24) 4-3 = $228.13; C = $634; V = $51.05; 100 For diagrams (a)-(d), compute the unknown values C, A, F, A, respectively. 100 t 1 0-1-2-3 75 1 50 2+ t 1 1 t z y 0-1-2-3-4 ~ (c) 10% (d) i = 10% 4-5 C Compute the value of P in the diagram. (a) 200 A A A A A t t t t t t t 250 A T0-1-2-3 T 0-1-2-3-4-5-6 1 I i = 10% 1 p 500 (b) (Answer: P 120 80 I +0 4-6 t i = 10% ~ A ~ A ~ A 10% = $589.50) 500 TTiT3r ~ A 0-1-2-3-4-5 (d) (Answers: C = $109.45; A A = $60.78) ~ I F (c) i Use a 10% interest rate to compute the value of X iJ the diagram. 0-1-2-3-4 4-4 i = $115; F = $276.37; For diagrams (a)-(d), compute the unknown values W, X, Y, Z, respectively. 1 X 4-7 Use a 15% interest rate to compute the value of P iJ the diagram. 300 100 200 50 21 0-1-2-3-4 t 7f i 0-1-2-3-4-5 i = 10% 1 w if 1 r 5f rrr 0-1-2---=-3-4-5- i = 10% 1 x (b) (a) 1 p 'I i '.~ i! ., L....__ ---,---- - -- 131 Problems 4-8 If i = 12%,what is the value of B in the diagram? 800 800 800 4-13 If i = 12%,compute G in the diagram. 60 50 f 0-1-2-3-4 r r 40 r 20 1 4-9 1 t ! 1.5B X 'f 0-1-2-3-4-5-6 Compute the value of n for the diagram, assuming a ~ 500 10% interest rate. A=1 t t 4-14 t t.t t r ~ 500 Compute the value of D in the diagram. 300 n=? 0-1-2-'-."-' -2-. -1-1 F = 35.95 100 ~ D A= 50 ttttttttt 0-1-2-3-4-5-6-7-8-9-'" - t T 0-1-2-3-4-5-6-7-8 4-10 What is the value of n, for this diagram, based on a 31/2%interestrate? 4-15 tn ~ D ~ D r i = 10% ~ D Compute E for the diagram. 300 300 n=? 200 200 1 1000 4-11 r T If r If 0-1-2-3-4-5-6-7-8 Compute the value of J for the diagram, assuming a 10% interest rate. 100 100 100 i = 10% 4-16 r r r 0-1-2-3-4-5-6-7-8 J ! J If T ! J r 3B ~ C ~ C ~ C 4B -- -------- t Using a 10% interest rate, compute B in the diagram. Compute the value of C for the diagram, assuming a 10% interest rate. 300 0-1-2-3-4-5-6-7-8 r E 400 ! 4-12 t E r - - 132 MORE INTERESTFORMULAS 4-17 Find the value of P for the following cash flow diagram. 3000 4-22 If $200 is deposited in a savings account at the beginning of each of 15 years, and the account draws interest at 7% per year, how much will be in the account at the end of 15 years? 1250 4-23 A company deposits $2000 in a bank at the end of every year for 10 years. The company makes no deposits during the subsequent 5 years. If ~e bank pays 8% interest, how much would be in the account at end of 15 years? rTT 4-24 How much must be deposited now at 5.25% interest to produce $300 at the end of ~very year for 10 years? p=? 4-18 The following cash flow transactions are said to be equivalent in terms of economic desirability at an interest rate of 12% compounded annually. Determine the unknown value A. 750 600 A + 0-1-2-3-4 4-21 Derive an equation to find the end of year future sum F that is equivalent to a series of n beginning-of-year payments B at interest rate i. Then use the equation to determine the future sum F equivalent to six B payments of $100 at 8% interest. (Answer: F = $792.28) A + A + A + 150 + TTI = 0-1-2-3-4-5 Cash Flow 1 Cash Flow 2 4-19 Upon the birth of his first child, Dick Jones decided to establish a savings account to partly pay for his son's education. He plans to deposit $20 per month in the account, beginning when the boy is 13 months old. The savings and loan association has a current interest policy of 6% per annum, compounded monthly, paid quarterly. Assuming no change in the interest rate, how much will be in the savings account when Dick's son becomes 16 years old? . 4-20 Mary Lavor plans to save money at her bank for use in December. She will deposit $30 a month, beginning on March 1 and continuing through November 1. She will withdraw all the money on December 1. If the bank pays 1/2% interest each month, how much money will she receive on December 1? 4-25 A student wants to have $30,000 at graduation 4 years from now to buy a new car. His grandfather gave him $10,000 as a high school graduation present. How much must the student sa,\,e each year if he deposits the $10,000 today and can earn 12% on both the $10,000 and his earnings in a mutual fund his grandfather recommends? 4-26 A city engineer knows that she will need $25 million in 3 years to' replace toll booths on a toll road in the city. Traffic on the road is estimated to be 20 million vehicles per year. How much per vehicle should the toll be to cover the cost of the toll booth replacement project? Interest is 10%. (Simplify your analysis by assuming that the toll receipts are received at the end of each year in a lump sum.) 4-27 Using linear interpolation, determine the value of (P / A, 61/2%,10)from the compoundinteresttables. Compute this same value using the equation. Why do the values differ? 4-28 A man wants to help provide a college education for his young daughter. He can afford to invest $600/yr for the next 4 years, beginning on the girl's fourth birthday. He wishes to give his daughter $4000 on her 18th, 19th, 20th, and 2Pt birthdays, for a total of $16,000. Assuming 5% interest, \yhat uniform annual investment will he have to make on the girl's 8th through 17th birthdays? (Answer: $792.73) 4-29 How many months will it take to payoff a $525 debt, with monthly payments of $15 at the end of -----.. - - -- -- --- --- Problems each month, if the interest rate is 18%, compounded monthly? (~nswer: 50 months) 4-30 For how many months, at an interest rate of 1% per month, does money have to be invested before it will double in value? 4-31 A bank recently announced an "instant cash" plan for holders of its bank credit cards. A cardholder may receive cash from the bank up to a preset limit (about $500). There is a special charge of 4% made at the time the "instant cash" is sent to the cardholders. The debt may be repaid in monthly installments. Each month the bank charges 11/2%on the unpaid balance. The monthly payment, including interest, may be as little as $10. Thus, for $150 of "instant cash," an initial charge of $6 is made and added to the balance due. Assume the cardholder makes a monthly payment of $10 (this includes both principal and interest). How many months are required to repay the debt? If your answer includes a fraction of a month, round up to the next month. 133 paid balance is computed at 11/2% per month, how many payments will there be, and what is the amount of the final payment? (Answers: 16 payments; final payment: $1.99) 4-36 A companybuys a machinefor $12,000,whichit agrees to pay for in five equal annual payments, beginning one year after the date of purchase, at an interest rate of 4% per annum. Immediately after the second payment, the terms of the agreement are changed to allow the balance due to be paid off in a single payment the next year. What is the final single payment? (Answer: $7778) 4-32 A man borrowed $500 from a bank on October 15th. He must repay the loan in 16 equal monthly payments, due on the 15th of each month, beginning November 15th. If interest is computed at 1% per month, how much must he pay each month? (Answer: $33.95) 4-37 Anengineeringstudentboughta car at a localusedcar lot. Including tax and insurance, the total price was $3000. He is to pay for the car in 12 equal monthly payments, beginning with the first payment immediately (in other words, the first payment was the down payment). Nominal interest on the loan is 12%, compounded monthly. After six payments (the down payment plus five additional payments), he decides to sell the car. A buyer agrees to pay a cash amount to payoff the loan in full at the time the next payment is due and also to pay the engineering student $1000. If there are no penalty charges for thiS early payment of the loan, how much will the car cost the new buyer? 4-33 4-38 4-34 In Table 3-1 in the text, four plans were presented for the repayment of $5000 in 5 years with interest at 8%. Still another way to repay the $5000 would be to make four annual end-of-year payments of $1000 each, followed by a final payment at the end of the fifth year. How much would the final payment be? An engineer borrowed $3000 from the bank, payable in six equal end-of-year payments at 8%. The bank agreed to reduce the interest on the loan if interest rates declined in the United States before the loan was fully repaid. At the end of 3 years, at the time of the third payment, the bank agreed to reduce the interest rate from 8% to 7% on the remaining debt. What was the amount of the equal annual end-of-year payments for each of the first 3 years? What was the amount of the equal annual end-of-year payments for each of the last 3 years? 4-35 A $150 bicyclewas purchasedon December 1 with a $15 down payment. The balance is to be paid at the rate of $10 at the end of each month, with the first payment due on December 31. The last payment may be some amount less than $10. If interest on the un- A realtor sold a house on August 31, 1997, for $150,000 to a buyer in which a 20% down payment was made. The buyer took a 15-year mortgage on the property with an effective interest rate of 8% per annum. The buyer intends to payoff the mortgage owed in yearly payments starting on August 31, 1998. (a) How much of the mortgage will still be owed after the payment due on August 31,2004, has been made? (b) Solve the same problem by separating the interest and the principal amounts. 4-39 To provide for a college education for his daugh- ter, a man opened an escrow account in which equal deposits were made. The first deposit was made on January 1, 1981, and the last deposit was made on January 1, 1998. The yearly college expenses including tuition were estimated to be $8000, for each of the 4 years. Assuming the interest rate to be 5.75%, how much did the father have to deposit each year in the escrow account for the daughter to draw $8000 per year for 4 years beginning January 1, 1998? 4-40 Develop a complete amortization table for a loan of $4500, to be paid back in 24 uniform monthly 134 MORE INTERESTFORMULAS installments, based on an interest rate of 6%. The amortization table must include the following colqmn headings: Payment Number, Principal Owed (beginning of period), Interest Owed in Each Period, Total Owed (end of each period), Principal Paid in Each Payment, Uniform Monthly Payment Amount 4-45 Considerthe followingcash flow: Year Cash Flow o -$P 1 + 1000 2 +850 3 +700 4 +550 5 +400 6 You must also show the equations used to calculate each column of the table. You are encouraged to use spreadsheets. The entire table must be shown. 4-41 Using the loan and payment plan developed in Problem 4-40, determine the month that the final payment is due, and the amount of the final payment, if $500 is paid for payment 8 and $280 is paid for payment 10. This problem requires a separate amortization table giving the balance due, principal payment, and interest payment for each period of the loan. 4-42 A couple borrowed $80,000 at 7% to purchase a house. The loan is to be repaid in equal monthly payments over a 30-year period. The first payment is paid exactly at the end of the first month. Calculate the interest and principal in the second payment, if the second payment is made 33 days after the first payment. 4-43 Jim Duggan made an investment of $10,000 in a savings account 10 years ago. This account paid interest of 51/2% for the first 4 years and 61/2% interest for the remaining 6 years. The interest charges were compounded quarterly. How much is this investment worth now? 4-44 Consider the cash flow: Year o 1 2 3 4 5 Cash Flow -$100 +50 +60 +70 +80 +140 Which one of the following is correct for this cash flow? (a) 100 = 50 + lO(AjG, i, 5) + 50(P j F, i, 5) b () 50(PjA, i, 5) + lO(PjG, i, 5) + 50(PjF, i, 5) 100 = 1 (c) 100(Aj P, i, 5) = 50 + lO(AjG, i, 5) (d) None of the equationsare correct. T 7 8 . +400 +400 +400 Alice was asked to compute the value of P for the cash flowat 8% interest. She wrote three equations: (a) P = 1000(PjA, 8%, 8) -150(PjG, 8%, 8) + 150(P jG, 8%, 4)(P j F, 8%,4) (b) P = 400(P j A, 8%, 8) + 6oo(p j A, 8%, 5)'150(P jG, 8%,4) (c) P = 150(P jG, 8%,4) + 850(P j A, 8%,4) + 400(P j A, 8%, 4)(P j F, 8%,4) Which of the equations is correct? 4-46 It is estimated that the maintenance cost on a new car will be $40 the first year. Each subsequent year, this cost is expected to increase by $10. How much would you need to set aside when you bought a new car to pay all future maintenance costs if you planned to keep the vehicle for 7 years? Assume interest is 5% per annum. (Answer: $393.76) 4-47 A young engineer wishes to become a millionaire by the time he is 60 years old. He believes that by careful investment he can obtain a 15% rate of return. He plans to add a uniform sum of money to his investment program each year, beginning on his 20th birthday and continuing through his 59th birthday. How much money must the engineer set aside in this project each year? 4-48 The council members of a small town have decided that the earth levee that protects .the town from a nearby river should be rebuilt and strengthened. The town engineer estimates that the cost of the work at the end of the first year will be $85,000. He estimates that in subsequent years the annual repair costs will decline by $10,000, making the second-year cost $75,000; the third-year $65,000, and s~ forth. The council members want to know what the equivalent present cost is for the first 5 years of repair work if interest is 4%. (Answer: $292,870) 4-49 A company expects to install smog control equipment on the exhaust of a gasoline engine. The local - -----.-- .... . ...-..---.-- Problems smog control district has agreed to pay to the firm a lump sum of money to provide for the first cost of the equipment and maintenance during its lO-year usefullife. At the end of 10 years the equipment, which initially cost $10,000, is valueless. The company and smog control district have agreed that the following are reasonable estimates of the end-of-year maintenance costs: Year 1 2 3 4 5 $500 100 125 150 175 Year 6 7 8 9 10 $200 225 250 275 300 Assuming interest at 6% per year~how much should the smog control district pay to the company now to provide for the first cost of the equipment and its maintenance for 10 years? (Answer: $11,693) 4-50 (b) How much will Mark have at the end of 6 years if his stock market investments increase only at 8% annually? 4-53 The Macintosh Company has an employee savings plan that allows every employee to invest up to 5% of his or her annual salary. The money is invested in company common stock with the company guaranteeing that the annual return 'rill never be less than 8%. Jill was hired at an annual salary of $52,000. She immediately joined the savings plan investing the full 5% of her salary each year. If Jill's salary increases at an 8% uniform rate, and she continues to invest 5% of it each year, what amount of money is she guaranteed. to have at the end of 20 years? 4-54 The football coach at a midwestem university was given a 5-year employment contract that paid $225,000 the first year, and increased at an 8% tmiform rate in each subsequent year. At the end of the first year's football season, the alumni demanded that the coach be fired. The alumni agreed to buy his remaining years on the contract by paying him the. equivalent present sum, computed using a 12% interest rate. How much will the coach receive? 4-55 Traffic at a certain intersection is currently 2000 cars per day. A consultant has told the city that trafficis expected to grow at a continuous rate of 5% per year for the next 4 years. How much traffic will be expected at the end of 2 years? 4-56 A local bank will lend a customer $1000 on a 2-year car loan as follows: A debt of $5000 can be repaid, with interest at 8%, by the following payments. Year 1 2 3 4 5 Payment $ 500 1000 1500 2000 X The payment at the end of the fifth year is shown as X. How much is X? 4-51 A man is purchasing a small garden tractor. There will be no maintenance cost during the first 2 years because the tractor is sold with 2 years free maintenance. For the third year, the maintenance is estimated at $20. In subsequent years the maintenance cost will increase by $20 per year (i.e., fourth-year maintenance will be $40, fifth-year $60, etc.). How much would need to be set aside now at 8% interest to pay the maintenance costs on the tractor for the first 6 years of ownership? 4-52 Mark Johnson saves a fixed percentage of his salary at the end of each year. This year he saved $1500. For the next 5 years, he expects his salary to increase at an 8% annual rate, and he plans to increase his savings at the same 8% annual rate. He invests his money in the stock market. Thus there will be six end-of-year investments (the initial $1500 plus five more). Solve the problem using the geometric gradient factor. (a) How much will the investments be worth at the end of 6 years if they increase in the stock market at a 10% annual rate? 135 Money to pay for car Two years' interest at 7%: 2 x 0.07 x 1000 24 monthlypayments = $1000 140 $1140 1140 - 24 = $47.50 The first payment must be made in 30 days. What is the nominal annual interest rate the bank is receiving? 4-57 A local lending institution advertises the "51-50 Club." A person may borrow $2000 and repay $51 for the next 50 months, beginning 30 days after receiving the money. Compute the nominal annual interest rate for this loan. What is the effective interest rate? 4-58 A loan company has been advertising on television a plan that allows people to borrow $1000 and make a payment of $10.87 per month. This payment is for interest only and includes no payment on - - ---- 136 MORE INTERESTFORMULAS the principal. What is the nominal annual interest rate that, they are charging? 4-59 What effective interest rate per annum corresponds to a nominal rate of 12% compounded monthly? (Answer: 12.7%) 4-60 A woman opened an account in a local store. In the charge account agreement, the store indicated it charges 11/2% each month on the unpaid balance. What nominal annual interest rate is being charged? What is the effective interest rate? 4-61 The Bawl Street Journal costs $206, payable now, for a 2-year subscription. The newspaper is published 252 days per year (5 days per week, except holidays). If a 10% nominal annual interest rate, compounded . quarterly,is used: (a) What is the effective annual interest rate in this problem? (b) Compute the equivalent interest rate per 1/252of a year. (c) What is a subscriber's cost per copy of the newspaper, taking interest into account? 4-62 Pete Sampras borrows $10,000 to purchase a car. He must repay the loan in 48 equal end-of-period monthly payments. Interest is calculated at 1.25% per month. Determine the following: (a) The nominal annual interest rate (b) The effective annual interest rate (c) The amount of the monthly payment 4-63 Picabo Street borrows $1000. Torepay the amount she makes 12 equal monthly payments of $90.30. Determine the following: (a) The effective monthly interest rate (b) The nominal annual interest rate (c) The effective annual interest rate 4-64 At the Central Furniture Company, customers who purchase on credit pay an effective annual interest rate of 16.1%, based on monthly compounding. What is the nominal annual interest rate that they pay? 4-65 What monthly interest rate is equivalent to an effective annual interest rate of 18%? 4-66 A bank advertises it pays 7% annual interest, compounded daily, on savings accounts, provided the money is left in the account for 4 years. What effective annual interest rate do they pay? 4-67 To repay a $1000 loan, a man paid $91.70 at the end of each month for 12 months. Compute the nominal interest rate he paid. ------- 4-68 A student bought a $75 used guitar and agreed to pa: for it with a single $85 payment at the end of 6 months Assuming semiannual (every 6 months) compound ing, what is the nominal annual interest rate? What i: the effective interest rate? 4-69 A firm charges its credit customers 13/4%interest pe month. What is the effective interest rate? 4-70 A thousand dollars is invested for 7 months at at interest rate of 1%. per month. What is the nomi nal interest rate? What is the effective interest rate' (Answers: 12%; 12.7%) 4-71 What interest rate, compounded quarterly, is equiva. lent to a 9.31 % effective interest rate? 4-72 If the nominal annual interest rate is 12% com pounded quarterly, what is the effective annual in. terest rate? 4-73 A contractor wishes to set up a special fund by male. 'ing uniform semiannual end-of-period deposits fo] 20 years. The fund is to provide $10,000 at the enc of each of the last 5 years of the 20-year period. 1.1. interest is 8%, compounded semiannually, what is the required semiannual deposit? 4-74 What amount will be reqUired to pur<?hase,on a man' 40th birthday, an annuity to provide him with 30 equaJ semiannual payments of $1000 each, the first to be received on his 50th birthday, if nominal hiterest i 4% compounded semiannually? 4-75 A man decides to deposit $50 in the bank today and to make 10 additional deposits every 6 months beginning 6 months from now, the first of which will be $50 and increasing $10 per deposit after that. A few minutes after making the last deposit, he decides to withdraw all the money deposited. If the bank pays . 6% nominal interest compoUnded semiannually, how much money will he receive? 4-76 A man makes an investment every 3 months at a nominal annual interest rate of 28%, compounded quarterly. His first investment was $100, followed by investments increasing $20 each 3 months. Thus, the second investment was $120, the third investment $140, and so on. If he continues to make this series of investments for a total of 20 years, what will be the value of the investments at the end of that time? 4-77 A 25-year old engineer is opening an individual retirement account (IRA) at a bank. Her goal is to accwnulate $1 million in the account by the time she retires from work in 40 years. The bank manager estimates Problems she may expect to receive 8% nominal annual interest, compounded quarterly, throughout the 40 years. The engineer believes her income will increase at a 7% annual rate during her career. She wishes to start with as Iowa deposit as possible to her IRA now and increase it at a 7% rate each year. Assuming end-ofyear deposits, how much should she deposit the first year? 4-78 What single amount on April 1, 1998, is equivalent to a series of equal, semiannual cash flows of $1000 that starts with a cash flow on January 1, 1996, and ends with a cash flow on January 1, 2005? The interest rate is 14% and compounding is quarterly. 4-79 Paco's saving account earns 13% compounded weekly and receives quarterly deposits of $38,000. His first deposit occurred on October 1, 1996, and the last deposit is scheduled for April 1, 2012. Tisha's account earns 13% compounded weekly. Semiannual deposits of $18,000 are made into her account, with the first one occurring on July 1, 2006, and the last one occurring on January 1, 2015. What single amount on January 1,2007, is equivalent to the sum of both cash flow series? was paid on money not in the account for the full 3-month period. How much was in Ann's account at the end of 3 years? (Answer: $3912.30) 4-83 annual cash flow will occur on January 1,2010. What is the amount of each semiannual cash flow? Use What is the present worth of a series of equal quarterly payments of $3000 that extends over a period of 8 years if the interest rate is 10% compounded monthly? 4-84 The first of a series of equal seII}iannualcash flows occurs on July 1, 1997, and the last occurs on January 1, 2010. Each cash flow is equal to $128,000. The nominal interest rate is 12% compounded semiannually. What single amount on July 1,2001 is equivalent to this cash flow system? 4-85 A man buys a car for $3000 with no money down. He pays for the car in 30 equal monthly payments with interest at 12% per annum, compounded monthly. What is his monthly loan payment? (Answer: $116.10) 4-86 On January 1, Frank Jenson bought a used car for $4200 and agreed to pay for it as follows: 1/3down payment; the balance to be paid in 36 equal monthly payments; the first payment due February 1;an annual. interest rate of 9%, compounded monthly. (a) What is the amount of Frank's monthly payment? (b) During the summer, Frank made enough money to cause him to decide to payoff the entire balance due on the car as of October 1. How much did Frank owe on October 1? 4-80 The first of a series of equal, monthly cash flows of $2000 occurred on April 1, 1998, and the last of the monthly cash flows occurred on February 1, 2000. This series of monthly cash flows is equivalent to a series of semiannual cash flows. The first semiannual cash flow occurred on July 1,2001, and the last semi- 4-87 pounding on all accounts. A series of monthly cash flows is deposited into an account that earns 12% nominal interest compounded monthly. Each monthly deposit is equal to $2100. The first monthly deposit occurred on June 1, 1998 and the last monthly deposit will be on January 1,2005. The account (the series of monthly deposits, 12% nominal interest, and monthly compounding) also has equivalent quarterly withdrawals from it. The first quarterly withdrawal is equal to $5000 and occurred on Octo- effective annual interest rate? 19.3%) 4-88 (Answers: $278.70; The Rule of 78's is a commoDly used method of computing the amount of interest when the balance of a loan is repaid in advance. Adding the numbers representing 12 months gives ber 1,1998. The last $5000 withdrawalwill occur on ---- " II ;I Iu ;i IIi: !Ir !im If a 12-month loan is repaid at'the ecfofl one month, for example, the interest the borrower would be charged is 12/78 of the year's interest. If the loan is repaid at the 4-82 Ann deposits $100 at the end of each month into her bank savings account. The bank paid 6% nominal interest, compounded and paid quarterly. No interest I ,II ,illl 1+ 2 + 3 + 4 + 5 + .. . + 11+ 12 = 78 January 1, 2005. How much remains in the account after the last withdrawal? -l------- On January 1, Laura Brown borrowed $1000 from the Friendly Finance Company. The loan is to be repaid by four equal payments, which are due at the end of March, June, September, and December. If the finance company charges 18% interest, compounded quarterly, what is the amount of each payment? What is the a nominal interest rate of 12% with monthly com4 1 137 ~-- J I! u_......_ 138 .. ... ... ..- -....- - -- -- MORE INTERESTFORMULAS end of 2 months,the total interest charged ould be (12 + 11)/78, or 23/78 of the year's interest. After 11 months the interest charge would therefore be 77/78 of the total year's interest. (a) What washis monthly loan payment? (b) What must he have paid (in addition to his regular 120th monthly payment) to pay the remaining balance of his loan? (c) Recomputepart (a) using 6% compounded continuously. Helen.Reddy borrowed $10,000 on January 1 at 9% annual interest, compounded monthly. The loan was to be repaid in 12 equal end-of-period payments. Helen made the first two payments and then decided to repay the balance of the loan when she pays the third payment. Thus she will pay the third payment plus an additional sum. You are to calculate the amount of this additional sum (a) Based on the rule of 78s. (b) Based on exact economicanalysis methods. 4-93 An automobile may be purchased with a $3000 down payment now and 60 ,monthly payments of $280. If the interest rate is 12% compounded monthly, what is the price of the automobile? 4-94 A man has $5000 on deposit in a bank that pays 5% interest compounded annually. He wonders how much more advantageous it would be to transfer his funds to another bank whose dividend policy is 5% interest, compounded continuously. Compute how much he would have in his savings account at the end of 3 years under each of these situations. 4-95 A friend was left $50,000 by his uncle. He has decided to put it into a savings account for the next year or so. He finds there are varying interest rates at savings institutions: 43/8% compounded annually, 41/4%compounded quarterly, and 41/8%compounded continuously. He wishes to select the savings institution that will give him the highest return on his money. What interest rate should he select? 4-89 A bank is offeringa loan of $25,000 with a nominal interest rate of 18% compounded monthly, payable in 60 months. (Hint: The loan origination fee of 2% will be taken out from the loan amount.) (a) What is the monthlypayment? (b) If a loan origination fee of 2% is charged at the time of the loan, what is the effective interest rate? 4-90 Our cat, Fred, wants to purchase a new litter box. The cost is $100 and he'll finance it over 2 years at an annual rate of 18% compounded monthly and to be repaid in 24 monthly payments. (a) What is his monthlypayment? (b) At the time of the thirteenth payment, Fred decides to payoff the remainder of the loan. 4-96 One of the local banks indicates that it computes the interest it pays on savings accounts by the continuous compounding method. Suppose you deposited $100 in the bank and they pay 4% per annum, compounded continuously. After 5 years, how much money will there be in the account? Using regular compound interest factors, determine the amount of this last payment. 4-97 A college professor won $85,000 in the state lottery; income taxes will take about half the amount. She plans to spend her sabbatical year on leave from the university on an around-the-world trip with her husband, but she must continue to teach 3 more years first. She estimates the trip will cost $40,000 and they will spend the money as a continuous flow of funds during their year of travel. She will put nough of her lottery winnings in a bank account now to pay for the trip. The bank pays 7% nominal interest, compounded continuously. She asks you to compute how much she should set aside in the account for the 4-91 Our cat, Fred, has convinced me that I should set up an account that will assure him of his Meow Mix for the next 4 years. I will deposit an amount P today that will permit Fred to make end-of-the-month withdrawals of $10 for the next 48. months. Consider an interest rate of 6% compounded monthly and that the account will be emptied with the last withdrawal. (a) What is the value of P that I must deposit today? (b) What is the account balance immediately after the 24th withdrawal has been made? 4-92 When Jerry Garcia was alive he bought a house for $500,000 and made a $100,000 down payment. He obtained a 30-year loan for the remaining amount. Payments were made monthly. The nominal annual interest rate was 9%. After 10 years (120 payments) he decided to pay the remaining balance on the loan. - .- -_. - - - --- r --- trip. 4-98 Michael Jacks deposited $500,000 into a bank for 6 months. At the end of that time, he withdrew the money and received $520,000. If the bank paid interest based on continuous compounding: --- Problems (a) What was the effective annual interest rate? (b) What was,the nominal annual interest rate? 4-99 How long will it take for $10,000, invested at 5% per year, compounded continuously, to triple in value? 4-100 A bank pays 10% nominal annual interest on special three-year certificates. What is the effective annual interest rate if interest is compounded: (a) Every three months? (b) Daily? (c) Continuously? 4-101 Bart Simpson wishes to tour the country with his friends. To do this, he is saving money for a bus. (a) How much money must Bart deposit in a savings account paying 8% nominal annual interest, compounded continuously, in order to have $8000 in 41/2years? (b) A friend offers to repay Bart $8000 in 41/2years if Bart gives him $5000 now. Assuming continuous compounding, what is the nominal annual interest rate of this offer? 4-102 Select the best alternative among the following five alternatives. Assume the investment is for a period of 4 years and P = $10,000. (a) 11.98% interest rate compounded continuously (b) 12.00% interest rate compounded daily (c) 12.01% interest rate compounded monthly (d) 12.02% interest rate compounded quarterly (e) 12.03% interest rate compounded yearly 4-103 What single amount on October 1, 1997, is equal to a series of $1000 quarterly deposits made into an account? The first deposit occurs on October 1, 1997 and the last deposit occurs on January 1, 2011. The account earns 13% compounded continuously. 4 4 You are taking a $2000 loan. You will pay it back in four equal amounts, paid every 6 months starting 3 years from now. The interest rate is 6% compounded semiannually. Calculate: (a) The effective interest rate, based on both semiannual and continuous compounding (b) The amount of each semiannual payment (c) The total interest paid 4-105 If you want a 12% rate ofreturn, continuously compounded, on a project that will yield $6000 at the end of 21/2years, how much must you be willing to invest now? (Answer: $4444.80) 4-106 A department store charges 13/4%interest per month, compounded continuously, on its customer's charge - - - - 139 accounts. What is the nominal annual interest rate? What is the effective interest rate? (Answers: 21 %; -23.4%) 4-107 A bank is offering to sell 6-month certificates of deposit for $9500. At the end of 6 months, the b~ will pay $10,000 to the certificate owner. Based on a 6-month interest period, compute the nominal annual interest rate and the effective annual interest rate. 4-108 Two savings banks are located across the street from each other. The West Bank put a sign in the window saying, "We pay 6.50%, compounded daily." The East Bank put up a sign saying, "We pay 6.50%, compounded continuously." Jean Silva has $10,000 which she will put in a bank for one year. How much additional interest will Jean receive by placing her money in the East Bank rather than the West Bank? 4-109 Sally Struthers wants to have $10,000 in a savings account at the end of 6 months. The bank pays 8%, compounded continuously. How much should Sally deposit now? (Answer: $9608) 4-110 The I've Been Moved Corporation receives a constant flow of funds from its worldwide operations. This money (in the form of checks) is continuously deposited in many banks with the goal of earning as much interest as possible for "IBM." One billion dollars is deposited each month, and the money earns an average of 1/2%interest per month, compounded continuously. Assume all the money remains in the accounts until the end of the month. (a) How much interest does IBM earn each month? (b) How much interest would IBM earn each 1I).0nth if it held the checks and made deposits to its bank accounts just four times a month? 4-111 A group of 10 public-spirited citizens has agreed that they will support the local school hot lunch program. Each year one of the group is to pay the $15,000 cost that occurs continuously and uniformly during the year. Each member of the group is to underwrite the cost for one year. Slips of paper numbered year 1 through year 10 are put in a hat. As one of the group, you draw the slip marked year 6. Assuming an 8% nominal interest rate per year, how much do you need to set aside now to meet your obligation in year 6? 4-112 A forklift truck costs $29,000. A company agrees to purchase such a truck with the understanding that it will make a single payment for the balance due in 3 years. The vendor agrees to the deal and offers two 140 MORE INTERESTFORMULAS different interest schedules. The first schedule uses an annual effective interest rate of 13%. The second schedule uses 12.75% compounded continuously. (a) Which schedule should the company accept? (b) What would be size of the single payment? 4-113 PARC Company has money to invest in an employee benefit plan, and you have been chosen as the plan's trustee. As an employee yourself, you want to maximize the interest earned on this investment and have found an account that pays 14% compounded continuously. PARC is providing you $1200 per month to put into your account for 7 years. What will be the balance in-this account at the end of the 7-year period? 4-114 Barry, a recent engineering graduate, never took engineering economics. When he graduated, he was hired by a prominent architectural firm. The earnings from this job allowed him to deposit $750 each quarter into a savings account. There were two banks offering a savings account in his town (a small town!). The first bank was offering 4.5% interest compounded continuously. The second bank offered 4.6% compounded monthly. Barry decided to deposit in the first bank since it offered continuous compounding. Based on the foregoing information, did he make the right decision? 4-115 A local finance company will loan $10,000 to a homeowner. It is to be repaid in 24 monthly payments of $499 each. The first payment is due 30 days after the $10,000 is received. What interest rate per month are they charging? (Answer: 11/2%) 4-116 Mr. Sansome withdrew $1000 from a savings account and invested it in common stock. At the end of 5 years, he sold the stock and received a check for $1307. If Mr. Sansome had left his $1000 in the savings account, he would have received an interest rate of 5%, compounded quarterly. Mr. Sansome would like to compute a comparable interest rate on his common stock investment. Based on quarterly compounding, what nominal annual interest rate did Mr. Sansome receive on his investment in stock? What effective annual interest rate did he receive? 4-117 The treasurer of a firm noted that many invoices were received with the following terms of payment: "20/010 days, net 30 days". Thus, if he were to pay the bill within 10 days of its date, he could deduct 2%. On the other hand, if he did not promptly pay the bill, the full amount would be due 30 days from the date of the invoice. Assuming a 20-day compounding period, the 2% deduction for prompt payment is equivalent to what effective interest rate per year? 4-118 In 1555, King Henry borrowed money from his bankers on the condition that he pay 5% of the loan at each fair (there were four fairs per year) until he had made 40 payments. At that time the loan would be considered repaid. What effective annual interest did King Henry pay? 4-119 One of the largest automobile dealers in the city advertises a 3-year-old car for sale as follows: Cash price $3575, or a down payment of $375 with 45 monthly payments of $93.41. Susan DeVaux bought the car and made a down payment of $800. The dealer charged her the same interest rate used in his advertised offer. How much will Susan pay each month for 45 months? What effective interest rate is being charged? (Answers: $81.03; 16.1%) 4-120 In 1990 Mrs. John Hay Whitney sold her painting by Renoir, Au Moulin de ZaGaZette, depicting ari openair Parisian dance hall, for $71 million. The buyer also had to pay the auction house commission of 10%, or a total of $78.1 million. Mrs. Whitney purchased the painting in 1929 for $165,000. (a) What rate of return did she receive on her investment? (b) Was the rate ofreturn really as high as you computed in (a)? Explain. 4-121 A woman made 10 annual end-of-year purchases of $1000 worth of common stock. The stock paid no dividends. Then for 4 years she held the stock. At the end of the 4 years she sold all the stock for $28,000. What interest rate did she obtain on her investment? 4-122 For some interest rate i and some number of interest periods n, the uniform series capital recovery factor is 0.1728 and the sinking fund factor is 0.0378. What is the interest rate? . . 4-123 The following beginning-of-month (BOM) and end - of month (EOM) amounts are to be deposited in a savings account that pays interest at 9%, compounded . monthly: Today (BOM 1) EOM2 EOM6 EOM7 BOM 10 -- $400 270 100 180 200 - ur Set up a spreadsheet to calculate the account balance at the end of~e first year (EOM12). The spreadsheet must include the following column headings: Month Number, Deposit BOM, Account Balance at BOM, Interest Earned in Each Month, Deposit EOM, Account Balance at EOM. Also, use the compound interest tables to draw a cash flow diagram of this problem and solve for the account balance at the EOM 12. . $2 million for this year. The net revenue will decrease 15% per year for 5 years, when the assembly plant will be closed (at the end of Year 6). If the firm's interest rate is 10%, calculate the PW of the revenue stream. _ . .. . $10,000 and is increase interest rate 6%. at 8% per year for 4 years? The 4-126 What is the present worth of cash flows that begin at to stay steady; otherwise they will drop 3% per year. What is the present worth of the savings over the next 5 years? Ten years? The interest rate is 10%. _ 4-133 A 30-year mortgage for $120,000 has been issued. . JIlLfuel oil to a natural gas unit. The investment win be $2500 installed. The cost of the natural gas will aver.. age $60 per month over the year, instead of the $145 per month that the fuel oil costs. If the interest rate is 9% per year, how long will it take to recover the initial investment? 4-135 Develop a general-purpose spreadsheet to calculate A out the balance due, principal payment, ~d interest payment for each period of a loan. The user inputs to the spreadsheet will be the loan amount, the number of payments per year, the number of years payments are made, and the nominal interest rate. Submit printouts of your analysis ofa loan in the amount of $15,000 at 8.9% nominal rate for 36 months and for 60 months of payments. 4-128 Calculate and print out the balance due, principal pay- ment, and interest payment for each period of a new car loan. The nominal interest is 9% per year, compounded monthly. Payments are made monthly for 5 years. The original loan is for $17,000. 4-129 For the used car loan of Problem 4-127, graph the JIlmonthly payment. . (a) As a function of the interest rate (5-15%). (b) As a function of the number of payments (24-48). 4-130 For the new car loan of Problem 4-128, graph the JI. monthly payment. (a) As a function of the interest rate (4-14%). (b) As a function of the number of payments (36-84). The interest rate is 10% and payments are made monthly. Print out the balance due, principal payment. and interest payment for each period. 4-134 A homeowner may upgrade a furnace that runs on $30,000 andrate decrease at 15% per year for 6 years? The interest is 10%. payment, and interest payment for each period of a used-car loan. The nominal interest is 12% per year, compounded monthly. Payments are made monthly for 3 years. The original loan is for $11,000. earns 6% interest. Your salary increases by 5% per year. What value does your savings book show after 40 years? .IL per year from current profits of $200,000. Investing in a design change will allo~ the profit per widget 4-127 Calculate and print out the balance due, principal ~ at the end of each year in a savings account that 4-132 The market volume for widgets is increasing by 15% 4-125 What is the present worth of cash flows that begin at . 141 4-131 Your beginning salary is $50,000. You deposit 10% 4-124 Net revenues at an older manufacturing plant will be .II. Problems . 4-136 Use the spreadsheet developed for Problem 4-135 to JI. ments. Analyze a $100,000mortgage loan at a nomanalyze 180-month and 360-month house loan pay- . inal interest rate of 7.5% and submit a graph of the. interest and principal paid over time. You need not submit the printout of the 360 payments because it will not fit on one page. II '! --- 1 After Completing This Chapter... The student should be able to: ·· · . . Define and apply the present worth criteria. Compare two competing alternative choices using present worth (PW). Apply the PW model in cases with equal, unequal, and infinite project lives, Compare multiple alternatives using the PW criteria. Develop and use spreadsheets to make present worth calculations. QUESTIONS. TO CONSIDER. ) 1. So far,Boeinghas not been ableto sellthe Sonic Cruiseridea to airlines.Fastis good, they say,but cheap is better. Why isn't increased speed more attractive to airlines, especially overlong-haulflights?Can't airlineschargemoreforfasterflights? . 2. In many cases, air travelers actually spend as much time on the group.das in the air. Going through security,changing planes, clearing customs, and picking up baggage can collectively add hours to their travel time. How might this affect a traveler's decision about whether to pay more to fly on a faster plane like the Sonic Cruiser? 3. Airlines seem to be more interested in the new Airbusmodel. Can you state somereasons for this apparent preference? ~ - - - --------- ,~ ---- Present Worth Analysis Boeing Versus Airbus For several years, Boeing has been working on development of the Sonic Cruiser, an innovative lightweight plane that can fly near the speed of sound. This is about 15%faster than a conventionaljet, so flying the Sonic Crqiser would allow airlines to cut long-haul flight times. Meanwhile, Airbus, Boeing's rival in the passenger aircraftbusiness, has also been working on its versionof a lighterplane. But Airbusis betting on size rather than speed. Its new model, the A380, is known as the super-jumbo and will ". be the largest passengerplane in the world. Airbus is hoping to use several recently' developed composite materials, with the aim of reducing by 20% both the weight and the manufacturing cost. Airbus estimates that flying the composite plane will allow airlines to cut their operating costs by about 8%. I .' .. ~ ~_. .. J.' ' . '. . .. ..,.,!,," 1 J :--1 = ---- -- - -- ;01 . . - .. ...- 144 PRESENTWORTH ANALYSIS In Chapters 3 and 4 we accomplished two important tasks. First, we presented the concept of equivalence. We are powerless to compare series of cash flows unless we can resolve them into some equivalent arrangement. Second, equivalence, with alteration of cash flows from one series to an equivalent sum or series of cash flows, created the need for compound interest factors. So, we derived a whole series of compound interest factors-some for periodic compounding and some for continuous compounding. This background sets the stage for the chapters that follow. ASSUMPTIONS IN SOLVING ECONOMIC ANALYSIS PROBLEMS One of the difficulties of problem solving is that most problems tend to be very complicated. It becomes apparent that some simplifying assumptions are needed to make such problems mana,geable. The trick, of course, is to solve the simplified problem and still be satisfied that the solution is applicable to the real problem! In the subsections that follow, we will consider six different items and explain the customary assumptions that are made. These assumptions apply to all problems and examples, unless other assumptions are given. End-of- Year Convention As we indicated in Chapter 4, economic analysis textbooks and practice follow the endof-period convention. This makes"A" a seriesof end-of-period receipts or disbursements. (We generally assumein problems that all seriesof receipts or disbursementsoccur at the end of the interest period. This allows us to use values fromour compoundinteresttables without any adjustments.) A cash flow diagram of P, A, and F for the end-of-period conventionis as follows: Year0 Jan 1 End of Year 1 Dee 31 Jan 1 .End of Year 2 Dee 31 ,A t A p F If one were to adopt a middle-of-period convention, the diagram would be: Year0 Jan 1 Middleof Year 1 Jun 30 Dee 31 Jan 1 A p Middle of Year 2 Jun 30 End of Year 2 Dee 31 A F As the diagrams illustrate, only A shifts; P remains at the beginning-of-period and F at the regardless of the convention.The compound interest tables in the Appendix end-of-period, are based on the end-of-period convention. -- - .- -- -- - - -.- ----- -------. Assumptions in Solving Economic Analysis Problems 145 Viewpoint of Economic Analysis Studies When we make economic analysis calculations,we must proceed from a point of reference. Generally, we will want to take the point of view of a total firm when doing industrial economic analyses. Example 1-1 vividly illustrated the problem: a firm's shipping department decided it could save money having its printing work done outside rather than by the in-house printing department. An analysis from the viewpoint of the shipping depar:tment supported this, as it could get for $688.50 the same printing it was paying $793.50 for in house. Further analysis showed, however, that its printing department costs would decline less than using the commercial printer would save. From the viewpoint of the firm, the net result would be an increase in total cost. From Example 1-1 we see it is important that the viewpoint of the study be carefully considered. Selecting a narrow viewpoint,like that of the shipping department, may result in a suboptimal decision from the viewpoint of the firm. For this reason, the viewpoint of the total firm is used in industrial economic analyses. Sunk Costs We know that it is the differences between alternatives that are relevant in econonllc analysis. Events that have occurred in the past really have no bearing.on what we should do in the future.Whenthejudge says,"$200fineor 3 daysinjail,'~the eventsthat led to . these ~nhappy alternatives really are unimportant. It is the current and future differences between the alternatives that are important. Past costs, like past events;have no beari.Iigon deciding between alternatives unless the past costs somehow affect the present or future costs. In general, past costs do not affect the present or the future, so we refer to them as . sunk costs and disregard them. Borrowed Money Viewpoint In most economic analyses, the proposed alternativesinevitably require money to be spent, and so it is natural to ask the source of that money. The source will vary from situation to situation. In fact, there are two aspects of money to determine: one is the financing-the obtaining of money-problem; the other is .the investment-the spending of moneyproblem. Experience has shown that these two concerns should be distinguished. When separated,the problems of obtaining moneyand of spending it are both logical and straightforward. Failure to separate them sometimesproduces confusing results and poor decision making. The conventionalassumptionin economicanalysisis that themoney required to finance alternatives/solutionsin problem solving is considered to be borrowed at interest rate i. Effect of Inflation and Deflation For the present we will assume that prices are stable. This means that a machine that costs $5000 today can be expected to cost the same amount several years hence. Inflation and deflation can be serious problems for after-tax analysis and for cost and revenues whose inflation rates differ from the economy's inflation rates, but we assume stable prices for now. T - -- 146 PRESENTWORTH ANALYSIS Income Taxes This aspect of economic analyses, likeinflationand deflation,mustbe considered to findthe real payoff of a project. However,taxes will often affectalternativessimilarly,allowingus to compare our choices without considering income taxes. So, we will defer our introduction of income taxes into economic analyses until later. ECONOMIC CRITERIA We have shown how to manipulate cash flows in a variety of ways, and in so doing we can now solve many kinds of compound interest problems. But engineering economic analysisis more than simplysolvinginterestproblems.The decisionprocess (seeFigure 1-1) requires that the outcomes of feasible alternativesbe arranged so that they may be judged for economic efficiency in terms of the selectioncriterion. The economic criterion will be one of the following, depending on the situation1: Situation For fixed input For fixed output Neither input nor output fixed Criterion Maximize output Minimize input Maximize (output - input) We will now examine ways to resolve engineeringproblems, so that criteria for economic efficiencycan be applied. Equivalenceprovides the logic by which we may adjust the cash flowfor a given alter-' native into someequivalent sum or series.Toapplythe selectioncriterionto the outcomesof the feasible alternatives,we must first resolve them into comparable units. The question is, How should they be compared? In this chapter we'll learn how analysiscan resolve alternatives into equivalent present consequences,referred to simply as present worth analysis. Chapter 6 will showhow given alternativesare convertedinto an equivalentuniform annual cashflow, and Chapter 7 solves for the interest rate at which favorableconsequences-that is, benefits-are equivalent to unfavorable consequences--or costs. As a general rule, any economic analysis problem may be solved by the methods presented in this and in the two following chapters. This is true because present worth, annual cash flow, and rate of return are exact methods that will always yield the same solutionin selectingthebest alternativefromamonga setof mutuallyexclusivealternatives.2 Some problems, however, may be more easily solved by one method than by another. For" this reason, we now focus on the kinds of problems that are most readily solved by preseJ?-t worth analysis. IThe short table summarizes the discussion on selection of criteria in Chapter 1 (see item 5, select the criterion to determine the best alternative, page 9). 2"Mutually exclusive" means that selecting one alternative precludes selecting any other alternative. For example, constructing a gas station and constructing a drive-in restaurant on a particular piece of vacant land are mutually exclusive alternatives. .. 1 ----- T Applying Present Worth Techniques 147 ApPLYING PRESENT WORTH TECHNIQUES One of the easiest ways to compare mutually exclusive alternatives is to resolve their consequences to the present time. The three criteria for economic efficiencyare restated in terms of present worth analysis in Table 5-1. TABLE5-1 Present Worth Analysis Situation Criterion Fixed input Amount of money or other input resources are fixed Maximize present worth of benefits or other outputs Fixed output There is a fixed task, benefit, or other output to be accomplished Minimize present worth of costs or other inputs Neither input nor output is fixed Neither amount of money, or other inputs, nor amount of benefits, or other output, is fixed Maximize (present worth of benefits minus present worth of costs), that is, maximize net present worth Present worth analysis is most frequently used to determine the present value of future money receipts and disbursements. It would help us, for example, to determine a present worth of income-producing property, like an oil well or an apartment house. If the future income and costs are known, then we can use a suitable interest rate to calculate the present worth of the property.This shouldprovide a good estimate of the price at which the property could be bought or sold. Another application might be determining the valuation of stocks or bonds based on the anticipated future benefits from owning them. In present worth analysis,careful considerationmust be givento the timeperiod covered by the analysis. Usually the task to be accomplishedhas a time period associated with it. In that case, the consequences of each alternative must be considered for this period of time which is usually called the analysis period, or sometimes the planning horizon. The analysis period for an economy study should be determined from the situation. In some industries with rapidly changing technologies, a rather short analysis period or planning horizon might be in order. Industries with more stable technologies (like steelmaking) might use a longer period (say, 10-20 years), while government agencies frequently use analysis periods extending to 50 years or more. Three different analysis-period situations are encountered in economic analysis problems: 1. The useful life of each alternativeequals the analysis period. 2. The alternatives have useful lives different from the analysis period. 3. There is an infinite analysis period, n = 00. 1. Useful Lives Equal the Analysis Period Since different lives and an infinite analysis period present some complications, we will begin with four examples in which the useful life of each alternative equals the analysis period. ....--...- - - - --- .. 148 ~~... - +.. h._.. + + PRESENTWORTH ANALYSIS ,#";ci~~;'~i31~~'~~~It; A firm is considering which of two mechanical devices to install to reduce costs in a particular situation. Both devices cost $1000 and have useful lives of 5 years and no salvage value. Device A can be expected to result in $300 savings annually.Device B will provide cost savings of $400 the first year but will decline $50 annually, making the second-year savings $350, the third-y~ar savings $300, and so forth. With interest at 7%, which device should the firm purchase? SOLUTION The analysisperiod can convenientlybe selectedas the useful life of the devices, or 5 years. Since both devices cost $1000, there is a fixed input (cost) of $1000 regardless of whether A or B is chosen. The appropriatedecision criterion is to choose the alternativethat maximizes the present worth of benefits. DeviceA DeviceB 400 A = 300 r 0-1-2-3-4-5 ! t t t t i 01-2-3-4-5 ! n = 5 years PW of Benefits 350 r 300 250 t t 200 t n = 5 years PW of Benefits PW of benefits A = 300(PjA, 7%,5) = 300(4.100)= $1230 PW of benefits B = 400(PjA, 7%, 5) - 50(PjG, 7%, 5) = 400(4.100) - 50(7.647) = $1257.65 Device B has the larger present worth of benefits and is, therefore, the preferred alternative.It is worth noting that, if we ignore the time value of money,both alternativesprovide $1500 worth of benefits over the 5-year period. Device B provides greater benefits in the first 2 years and smaller benefits in the last 2 years. This more rapid flowof benefits from B, although the total magnitude equals that of A, results in a greater present worth of benefits. Wayne County will build an aqueduct to bring water in from the upper part of the state. I~can be built at a reduced size now for $300 million and be enlarged 25 years hence for an additional $350 million. An alternativeis to construct the full-sized aqueduct now for $400 million. Both alternativeswouldprovide the needed capacity for the 50-year analysis period. MaintenanCecosts are small and~may.beignored. At 6% interest, which alternatiye should be selected? - --- - --- -- -- ---- 149 Applying PresentWorth Techniques SOL~TION This problem illustrates staged construction. The aqueduct may be built in a single stage, or in a smaller first stage followed many years later by a second stage to provide the additional capacity when needed. For the Two-Stage Construction PW of cost = $300 million + 350 million(P / F, 6%, 25) = $300 million + 81.6 million = $381.6 million For the Single-Stage Construction PW of cost = $400 million The two-stage constructionhas a smaller present worth of cost and is the preferred construction plan. A purchasing agent is considering the purchase of some new equipment for the mailroom. Two different manufacturers have provided quotations. An analysis of the quotations indicates the following: Manufacturer Speedy Allied Cost $1500 1600 Useful Life (years) 5 5 End-of-Useful..Life Salvage Value $200 325 The equipmentof both manufacturersis expected to perform at the desired levelof (fixed)output. For a 5-year analysis period, which manufacturer's equipment should be selected? Assume 7% interest and equal maintenance costs. , I" SOLUTION' I ", For fixed output, the criterion is to minimize the present worth of cost. Speedy PW of cost ..:..1500 - 200(Pj F, 7%!,5) = 1500 - 200(0.7130) =:J?Op - J~3=-$l357 --- ... - -... ~ -- ----- - - - -- -- - - -- - 150 PRESENTWORTH ANALYSIS Allied PW of cost = 1600 - 325(P / F, 7%,5) = 1600 - 325(0.7130) = 1600 - 232 = $1368 Since it is only the differences between alternatives that are relevant, maintenance co~ts may be left out of.the economic analysis. Although the PWs of cost for all the alternatives are nearly identical, we would, nevertheless, choose the one with minimum present worth of cost unless there were other tangible or intangible differences that would change the decision.Buy the Speedy equipment. A finn is trying to decide which of two weighing scales it should install to check a package-filling operation in the plant. The ideal scale would allow better control of the filling operation and result in less overfilling. If both scales have lives equal to the 6-year analysis period, which on~ should be selected? Assume an 8% interest rate. Alternatives Atlas scale Tom Thumb scale ~OlUTION Cost $2000 3000 Uniform Annual Benefit $450 600 End-of-Useful-Life Salvage Value $100 700 .'. Atlas Scale PW of benefits - PW of cost = 450(P / A, 8%, 6) + 100(P/ F, 8%, 6) - 2000 = 450(4.623) + 100(0.6302) - 2000 = 2080+ 63 --'-'2000= $143 Tom Thumb Scale PW of benefits - PW'of cost = 600(P / A, 8%, 6) + 700(P /F, 8%"p) = 600(4.623) + 700(0.6302) - 3000 = 2774 + 441 - 3000. $215 I ~ 3000 I I I The salvage value of th~ scale, it .shol}ld be noted,!s ;sjmpLy tre<l,tedas..oanotherbenefit of.:: the alternative. Since the criterion is to maximize tbe"present worth of' benefits minus the present worth of cost, the preferred alternative is the Tom Thumb scale. " -J ~'~""..,. ; ... - -- ----- " -- . Applying Present Worth Techniques 151 In Example 5-4, we compared two alternatives and selectedthe one in which present worth of benefits minus present worth of cost was a maximum. The criterion is called the net present worth criterion and written simply as NPW: Net present worth = Present worth of benefits - Present worth of cost NPW = PW of benefits - PW of cost 2. Useful Lives Different from the Analysis Period In present worth analysis, there always must be an identified analysis period. It follows, then, that each alternativemust be considered for the entire period. In Examples 5-1 to 5-4, the useful life of each alternative was equal to the analysis period. Often we can arrange it this way, but in many situations the alternatives will have useful lives different from the analysis period. This section describes one way to evaluate alternatives with lives different from the study period. . In Example 5-3, suppose that the Allied equipment was expected to have a 1O-year useful life, or twice that of the Speedyequipment.Assuming the Allied salvagevalue would still be $325 in 10 years, which equipment should now be purchased? We can recompute the present worth of cost of the Allied equipment, starting as follows: PW of cost = 1600 - 325(P / F,.7%, 10) = 1600- 325(0.5083) = 1600- 165= $1435 Thepresent worthof cost has increased.This is due, of course,to the more distantrecoveryof the salvage value. More importantly,we now find ourselves attempting to compare Speedy equipment, with its 5-year life, against the Alliedequipment with a 1O-yearlife. Because of the variation in the useful life of the equipment, we no longer have a situation offixed output. Speedy equipment in the mailroom for 5 years is certainly not the same as 10 years of service with Allied equipment. . For present worth calculations, it is important that we select an analysis period and judge the consequences of each of the alternatives during the selected analysis period. As such, it is not a fair comparison to compare the NPW of the Allied equipment over its 1O-yearlife against the NPW of the Speedy equipment over its 5-year life. Not only is the firm and its economic environment important in selecting an analysis period, but also the specific situation being analyzed is important. If the Allied equipment (Example 5-3) has a useful life of 10years, and the Speedy equipment will last 5 years, one methodis to select an analysisperiod that is the least common multiple of their useful lives. Thus we would compare the 1O-yearlife of Allied equipment against an initial purchase of Speedy equipmentplus its replacement with new Speedy equipment in 5 years. The result is to judge the alternatives on the basis of a 1O-yearrequirement in the mailroom. On this basis the economic analysis is as follows. 1 -- -- r - - - --- - 152 PRESENTWORTH ANALYSIS Assuming the replacement Speedy equipment 5 years hence will also cost $1500, r Original Speedy Investment ---1 200 0-1-2-3-4-5-6-7-8-9-10 1 1500 200 t 1 1500 t J . I--- ReplacementSpeedyInvestment PW of cost = 1500+ (1500- 200)(P / F, 7%, 5) - 200(P / F, 7%, 10) = 1500+ 1300(0.7130) - 200(0.5083) = 1500 + 927 - 102 = $2325 For the Allied equipment,on the other hand, we have the following results: . 325 t 0-1-2-3-4-5-6~7-8-9-10 ... 1600 PW of cost = 1600 - 325(P / F, 7%, 10) = 1600 - 325(0.5083) = $1435 For the fixed output of 10 years of service in the mailroom, the Allied equipment, with it: smaller present worth of cost, is preferred. . We have seen that setting the analysis period equal to the least common multiple 0 the lives of the two alternativesseems reasonable in the revised Example 5-3. What woul< one do, however, if the alternatives had useful lives of 7 and 13 years, respectively? Here the least common multiple of lives is 91 years. An analysisperiod of 91 years hardly seem: realistic. Instead, a suitable analysis period should be based on how long the equipment i: likely to be needed. This may require that terminal values be estimated for the alternative: at some point prior to the end of their useful lives. Figure 5-1 graphically represents this concept. As Figure 5-1 indicates, it is not neces. sary for the analysis period to equal the useful life of an alternativeor some multiple of tht useful life. To properly reflect the situation at the end of the analysis period, an estimate i1 required of the market value of the equipment at that time. The calculations might be easie] if everything came out even, but it is not essential. T ----------- Applying Present Worth Techniques Alternative 1 Salvage Value t 0-1-2-3-4-5-6-7-8-9-10-11-12-13-14 Initial 1 Cost TerminalValueat End of 10thYear t Replacement Cost 1 I. ! I I I 7-yearLife Alternative 2 153 7-yearLife I Terminal Value at End of 10th Year t . 0-1-2-3-4-5-6-7-8-9-10-11-12-13 1' . Initial i 1 Cost I I 13-year Life lO-year Analysis Period FIGURE 5.1 Superimposing a lO-year analysis period on 7- and 13-year alternatives. A diesel manufacturer is considering the two alternative production machines graphically depicted in Figure 5-1. Specific data are as follows. Initial cost Estimated salvage value at end of useful life Useful life of equipment, in years Alt. 1 $50,000 $10,000 7 Alt. 2 $75,000 $12,000 13 The manufacturer uses an interest rate of 8% and wants to use the PW method to compare these alternatives over an analysis period of 10 yeats. Estimated market value, end of lO-year.analysis period Alt. 1 $20;000 Alt. 2 $15,000 SOLUTION ~ In this case, the decision maker is setting the analysis period at 10 years rather than acceptiug a common multiple of the lives of the alternatives, or assuming that the period of needed service is infinite (to be discussed in the next section). This is a legitimate approach-perhaps the diesel manuf~cturer will be phasing out this model at the end of the 10-year period. In.any event, we n~cl t<)!fOffiRNyt,e a\!emaJiyesOY,yI; tbe 1O~y.ear.period. ;;;; ~1iII - -----.. 'II' il I I" 'll 1 1' , ~ = ~ -- r u I ,I 1 -- - -- -- - --- -.. 154 ... u .._. -. PRESENTWORTH ANALYSIS As.illustrated in Figure 5-1, we may assumethat Alternative 1will bereplacedby an identical machineafter its 7-year useful life. Alternative 2 hasa 13-yearuseful life. The dieselmanufacturer has provided an estimated market value of the equipment at the time of the analysisperiod. As such we can now comparethe two choices over 10 yearsas follows: PW (Alt. 1) = -50,000 + (10,000--50,000)(P / F, 8%,7)+ 20,000(P/ F, 8%,10) = -50,000 - 40,000(0.5835)+ 20,000(0.4632) = -$64,076 PW (Alt. 2,)= -75,000 + 15,000(P/ F, 8%, 10) - -75,000 +. 15,000(0.4632) = -$69,442 I. To minimize PW of coststhe dieselmanufacturer should selectAlt 1. 3. Infinite Analysis Period: Capitalized Cost Another difficulty in present worth analysis arises when we encounter an infinite analysis period (n = 00). In governmentalanalyses, a service or condition sometimes must be maintained for an infinite period. The need for roads, dams, pipelines, and other components of national, state, or local infrastructure is sometimes considered to be permanent. In these situations a present worth of cost analysis would have an infinite analysis period. We call this particular analysis capitalized cost. Capitalizedcost is thepresent sum of moneythatwouldneed to be set asidenow,at some interest rate, to yield the funds required to provide the service (or whatever)indefinitely.To accomplish this, the money set aside for future expenditures must not decline. The interest received on the money set aside can be spent, but not the principal. When one stops to think about an infinite analysis period (as opposed to something relatively short, like a hundred years), we see that an undiminished principal sum is essential; otherwise one will of necessity run out of money prior to infinity. In Chapter 4 we saw that principal sum + interest for the period = amount at end of J?~!i0d,or P + iP = P + iP If we spend i P, then in the next interest period the principal sum P will again increase to P + i P. Thus, we can again spend i P. This concept may be illustrated by a numerical example. Suppose you deposited $200 in a bankthatpaid 4% interestannually.Howmuchmoneycouldbe withdrawneachyear without reducing the balance in the account below the initial $200? At the end of the first year, the $200 would have earned 4%($200) --- T = $8 interest. If this interest were withdrawn, . Applying Present Worth Techniques 155 the $200 would remain in the account. At the end of the second year, the $200 balance = $8. This $8 could also be withdrawn would again earn 4%($200) and the account would still have $200. This procedure could be continued indefinitelyand the bank account would always contain $200. The year-by-year situation would be depicted like this: Year1: $200 initial P ~ 200 + 8 = Withdrawal i P = - 208 8 = Year 2: $200 ~ 200 + 8 Withdrawal i P 208 =- 8 $200 and so on Thus, for any initial present sum P, there can be an end-of-period withdrawal of A equal to i P each period, and these withdrawals may continue forever without diminishing the initial sum P. This gives us the basic relationship: For n = 00, A = Pi This relationship is the key to capitalized cost calculations. Earlier we defined capitalized cost as the present sum of money that would need to be set aside at some interest rate to yield the funds to provide the desired task or service forever. Capitalized cost is therefore the P in the equation A = i P. It follows that: Capitalized cost A P=- (5-2) i If we can resolve the desired task or service into an equivalent A, the capitalized cost may be computed. The following examples illustrate such computations. How much should one set aside to pay $50 per year for maintenance on a gravesite if interest is assumed to be 4%? For perpetual maintenance,the principal sum must remain undiminished after making the annual disbursement. 'SOLUTION , . Capitalized cost P ~ P= 50' . '. == One ~hould$et aside. $1250. = .~ _ !!IC:'''= == =====-_ = Annual disbursement Interest rate i A . 0.04= $1250 ~ = IiiII:I=' ==;;:::::1:::11< = 111I --- -- --- -- - 156 PRESENT WORTH ANALYSIS A city plans a pipeline to transport water from a distant watershed area to the city. The pipeline will cost $8 million and will have an expected life of 70 years. The city anticipates it will need to keep the water line in service indefinitely.Compute the capitalized cost, assuming 7% interest. -SOLUTION The capitalized cost equation p=- A i is simple to apply when there are end-of-period disbursements A. Here we have renewals of the pipeline every 70 years. To compute the capitalized cost, it is necessary to first compute an end-of-period disbursement A that is equivalent to $8 million every 70 years. $8 million $8 million t o t t 70 years 140 years $8 million $8 million t n=oo 1 Capitalized Cost p The $8 million disbursement at the end of 70 years may be resolved into an equivalent A. $8 million n 70 ;:: A A = F(Aj F, i, n) = $8 rnillion(A/ F, 7%, 70) = $8 million(0.0Q()62) .= $4960 I Each 70-year period is identical to this one and the infinite series is shown in Figure 5-2. . . CapItalIzed cost P ........ '" I - --- . A " . 4960 = $8 millIon+ -:= $8 ID1llion + _ 0 .07 l = $8,071,000 -- -- ---- --- --- ' Applying Present Wortb Techniques 157 $8 million A = 4960 n=oo Capitalized Cost p --. FIGURE 5-2 Using the sinking fund factor to compute an infinite series. AlTE~NATESOLUTION~1J Instead of solving for an equivalent end-of-period payment A based on afuture $8 million disbursement, we could find A, given a present $8 million disbursement. A = P(Aj P, i, n) = $8 million(Aj P, 7%, 70) = $8 milIion(0.0706) = $565,000 On this basis, the infinite series is shown in Figure 5-3. Carefully note the difference between this andFigure5-2.Now: . . . A 565,000 CapItalizedcost P = -i = 0.07 - $8,071,000 FIGURE5-3 Using the capitalrecovery factor to compute an infiniteseries. A = 565,000 n=oo p I I Another way of solving the problem is to assume the interest is 70 years. Compute an equivalent interest rate for the 70-year period. Then the capitalized cost may be computed by using \. I Equation 4-33 for m ~ = 70 i70yr = = (1 + i1yr)70= ~ 1 = (1 + 0.07)70- J = 112.989 ~ ;. . $8 million Capitalized cost = $8 mIlIion +. 112.989 = $8,071,000 . -- -1 . ... - -- - ----- 158 PRESENTWORTH ANALYSIS Multiple Alternatives So far the discussion has been based on examples with only two alternatives.But multipll alternative problems may be solved by exactly the same methods employed for problen with two alternatives. (The only reason for avoiding multiple alternatives was to simpli1 the examples.) Examples 5-8 and 5-9 have multiple alternatives. A contractor has been awarded the contract to construct a 6-mile-Iong tunnel in the mountains. During the 5-year construction period, the contractor will need water from a nearby stream. He will construct a pipeline to convey the water to the main construction yard. An analysis of costs for various pipe sizes is as follows: Pipe Sizes (in.) Installed cost of pipeline and pump Cost per hour for pumping 2 3 4 6 $22,000 $1.20 $23,000 $0.65 $25,000 $0.50 $30,000 $0.40 The pipe and pump will have a salvage value at the end of 5 years, equal to the cost to remove them. The pump will operate 2000 hours per year. The lowest interestrate at which the contractor is willing to investmoney is 7%. (The minimum required interestrate for investedmoney is called the minimum attractive rate of return, or MARR.) Select the alternativewith the least present worth of cost. SdllJTI ON ... -,. -.'.. .-.'.' . We can comptite the present worth of cost for each alternative. For each pipe size, the present worth of cost is equal to the installed cost of the pipeline and pump plus the present worth of 5 years of pumping costs. Pipe Size (in.) II ~I Installed cost of pipeline and pump 1.20 x 2000 hr x (P / A, 7%, 5) 0.65 x 2000 hr x 4.100 0.50 x 2000 hr x 4.100 0.40 x 2000 hr x 4.100 Present worth of cost Select the 3 in. pipe size. 2 6 $22,000 9,840 $30,000 3,280; $33,280 $31,840 - .. :=; ~- =.= , IiIooL_ _____ -- r - - - - --- - Applying Present Worth Techniques 159 An investor paid $8000 to a consulting firm to analyze what he might do with a small parcel of land on the edge of town that can be bought for $30,000.In their report, the consultantssuggested four alternatives: Alternatives A B C D Uniform Net Annual Benefit Total Investment Including Land* $ Do nothing Vegetable market Gas station Small motel 0 50,000 95,000 350,000 $ Terminal Value at End of 20 yr 0 5,100 10,500 36,000 $ 0 30,000 30,000 150,000 *Includes the land and structures but does not include the $8000 fee to the consulting .finn. Assuming 10% is the minimum attractive rate pf retlirn, what should the investor. do? Alternative A represents the "do-nothing" alternative. Generally, one feasible alternativein any situation is to remain in the present status and do nothing. In this problem, the investor could decide that the most attractive alternative is notto purchase the property ~d develop it. Thisis clearly a do-nothing decision. We note, however,that if he does nothing, the total venture wouldnot be a very satisfactory one. Thisis because the investor spent $8000for professional advice"on the possible uses of the property. But because the $8000 is a past cost, it is a sunk cost. The only relevant costs in an economic analysis arepresent andfilJure costs; past events and past, or sunk, costs are gone and cannot be allowedto affectfuture planning. (Past costs may be relevan.tincomputingdepreciation charges and income taxes, but 110whereelse.) The past should not deter the investor from making the best decision now, regardless of the costs that brought him to this situatipn arid point of time. This problemis one of neitherfixed input nor fixedoutput,so our criterionwillbe tomaxiIni:z:e the present worth of benefits minu~ the present worth of cost; Or,simply stated, maxirni:z:enet present Worth. I Alter!lative A, Do Nothing 'I' NPW- 0 li~ I r ~ Alternative B, Vegetable Market f 'f NPW= -50,000 +5100(Pj A, 1O%i4(»+ 30,000(P/F, 10%, 2Q) I I , = """"50,0~0 --I-Q10Z~85~4) + 30,000(<l1~86)j! 1:1 : = ~50,OOO+ 43,420 + 4460 ;;; -~212Q - - "= ..........-..... - _ = ;;:"" ~ !I'==:'=&_ u..__... <=: ==;t =: ~ I 160 PRESENTWORTH ANALYSIS AI(ernative C, Gas Station NPW = -95,000 + 1O,500(P/ A, 10%,20) + 30,000(P / F, 10%,20) = -95,000 + 89,400 + 4460 = -$1140 Alternative D, Small Motel NPW = -350,000 + 36,000(P/ A, 10%,20)+ 150,000(P/ F, 10%,20) = -350,000 + 306,500+ 22,290 = -$21,210 The criterion is to maximize net present worth. In this situation, one alternative has NPW equal to zero, and three alternatives have negative values for NPW. We will select the best of the.four alternatives, namely, the do-nothing Alt. A, with NPW equal to zero. A piece ofland may be purchased for $610,000 to be strip-mined for the underlying coal. Annual net income will be $200,000 per year for 10 years. At the end of the 10 years, the surface of the land will be restored as required by a federal law on strip mining. The reclamation will cost $1.5 million more than the resale value of the land after it is restored. Using a 10% interest rate, determine whether the project is desirable. , . -.-" ;SQlUTION The investment opportunity may be described by the following cash flow: Year o 1-10 10 NPW Cash Flow (thousands) -$610 +200 (per year) -1500 = -610 + 200(P / A, 10%, 10) - 1500(P/ F, 10%, 10) = -610 + 200(6.145) - 1500(0.3855) = -610 + 1229 - 578 = +$41 Since NPW is positive, the project is desirable. (See Appendix 7A for a more complete analysis of this type of problem. At interest rates of 4.07% and i8.29~the NPW= 0.)< I!;!:I ! ! i, I ..L ,I --- - - . Applying Present Wort~ Techniques \:~1{ ~ 161 ~~~~ Two pieces of construction equipment are being analyzed: Year o 1 2 3 4 5 6 7 8 Alt. A -$2000 + 1000 +850 +700 +550 +400 +400 +400 +400 Alt.B -$1500 +700 +300 +300 +300 +300 +400 +500 +600 Based on an 8% interest rate, which alternativeshould be selected? SOUUTION'J . . ".. '". '. -. I', ,,-. . _ ~ Joo; Alternative A l00 ~ --850-r - ---- __- -700 J r---1--==f===~--4r---i9---iF--~ 0-1-2-3-4-5-6-7-8 1 2000 PW of benefits = 400(PlA, 8%,8) + 600(PI A, 8%,4) - 150(PIG, 8%,4) = 400(5.747) + 600(3.312) - 150(4.650) -' 3588.50 PW of cost = 2000 Net present worth = 3588.50 - 2000 = +$1588.50 Alternative B .~ I I I I .1 I r = = Ii:: I 1500 L - -- , I :::: ..... ~- 162 PRESENTWORTH ANALYSIS PW of benefits = 300(P / A, 8%, 8) + (700 - 300)(P / F, 8%, 1) + 100(P/ G, 8%, 4)(P / F, 8%,4) = 300(5.747) + 400(0.9259) + 100(4.650)(0.7350) = 2436.24 PW of cost = 1500 Net present worth = 2436.24 - 1500 = +$936.24 To maximize NPW,choose Alt. A. SPREADSHEETSAND PRESENTWORTH Spreadsheetsmake it easy to build more accurate models with shorter time periods. When using factors, it is common to assume that costs and revenues are uniform for n years. With spreadsheets it is easy to use 120 months instead of 10 years, and the cash flows can be estimated for each month. For example, energy costs for air conditioning peak in the summer, and in many areas there is little construction during the winter. Cash flows that depend on population often increase at x% per' year, such as for electric power and transportationcosts. , In spreadsheets any interest rate is entered exactly-so no interpolation is needed. This makes it easy to calculate the monthly repayment schedule for a car loan or a house mortgage. Examples 5-12 and 5-13 illustrate using spreadsheets to calculate PWs. NLE Construction is bidding on a project whose costs are divided into $30,000 for start-up and $240,000 for the first year. If the interest rate is 1% per month or 12.68% per year, what is the present worth with monthly compounding? .. SOlUTION~ . - - --. -.. -. ._-"~"." -, ~, Figure 5-4 illustratesthe spread,sheetsolutionwith the assumptionthat costs are distributedevenly throughout the year (-20,000 = -240,000/12). __ --- - - i!!! 1 - - - - - -- - := ill;! =" 118888I__........_ --- Spreadsheets and Present Worth FIGURE 5-~ Spreadsheet with monthly cash flows. A B 1% 1 163 D C i 2 -30,000 3 -240,000 annual amount 5 Month Cash Flow 6 0 - 30000 initial cash flow 4 7 1 -20000 8 9 2 3 -20000 -"20000 10 4 -20000 11 5 ...,,20000 12 6 -20000 13 7 -'-20000 14 8 -20000 15 9 -20000 16 10 ,.,....20000 17 11 ""-20000 18 12 ---20000 19 NPV -$255,102 I .. I , -- =NPV(A1,B7:B18)+B6 , Since the costs are uniform, th~factor ~olt1tioriis: PWIjlon .=-"30,OQO+20,000(P I I lA, 1%,12) == --,-$255,102 I I The value of monthly periods.can be'itl'l1strated..bycomputing thePW assuming an annual peqoCt.Theresults differbY4'Il01:e !h.al1$,12,OOO, because$2Q,000 at the el1dofMonth~ 1 through 12 is not the same ?s$240,OOO.at theynd QfMonth 12. The timing of tile ca~htlows makest:lJ,e differe:flce,even thougb the effective.interesttate§'are tbesa,tIl.e. 1 1 PWannual '" = -30,000 240,QOO(P[F,12,68%,1» ,90,000 240,OQ0/1.:1268-$242,993 " I J - Regina Industries has a new product whose sales are expected to be 1.2,3.5, 7, 5, (llld 3 million units per year over the next 5 years; Prpductiol1, distrib4tipn, (llld..ov~rh~adicos~s are stable at $120 per unit The price will be $200 petunitfoI:tbefi,r~t~2 yearS,al1a.thbn$r80~$i6(), aIld~$f4Ufor the next 3 years. The remaining R&D (llldproduction costs are $300 n:;riUion.Jfiis 15%, WJ;1atis b:: thep!e$,eptwor;.t~2';~~~~w!,~~d~ctZ=*,,,,,,~== 11_ "".. 1--- _ _ ___ , J;I ~~=::;a ==~ ;;;;;~~:= 164 PRESENTWORTH ANALYSIS SOLUTION. It is easiest to calculate the yearly net revenue per unit before building the spreadsheet shown in Figure 5-5. Those values are the yearly price minus the $120 of costs, which equals $80, $80, $60, $40, and $20. A 1 2 3 4 5 6 7 8 9 10 B C D E Sales (M) Net Revenue 1.2 3.5 7 5 3 80 80 60 40 20 Cash Flow ($M) -300 96 280 420 200 60 $469 Million 12% i Year 0 1 2 3 4 5 D4+NPV(Al,D5:D9)= FIGURE 5-5 Present worth of a new product. SUMMARY Present worth analysis is suitable for almost any economic analysis problem. But it is particularly desirable when we wish to know the present worth of future costs and benefits. And we frequentlywant to knowthe valuetoday of such things as income-producingassets, stocks, and bonds. For present worth analysis, the proper economic criteria are: Maximize the PW of benefits Fixed input Minimize the PW of costs Fixed output Neither input nor Maximize (PW of benefits - PW of costs) output is fixed or, more simply stated: Maximize NPW Tomake validcomparisons, we need to analyze each alternativein a problem over the same analysis period or planning horizon. If the alternatives do not have equal lives, some technique must be used to achieve a common analysis period. One method is to select an analysis period equal to the least commonmultiple of the alternative lives.Another method is to select an analysis period and then compute end-of-analysis-period salvage values for the alternatives. Capitalized cost is the present worth of cost for an infinite analysis period (n = 00). When n = 00, the fundamental relationship is A = i P. Some form of this equation is used whenever there is a problem with an infinite analysis period. The numerous assumptions routinely made in solving economic analysis problems include the following. -- ---- - --- -._--- Problems 165 1. Present sums P are beginning of period and all series receipts or disbursements A and future sums F occur at the end of the interest period. The compound interest tables were derived on this basis. 2. In industrial economic analyses, the appropriate point of reference from which to compute the consequences of alternatives is the total firm. Taking a narrower view of the consequences can result in suboptimal solutions. 3. Only the differences between the alternatives are relevant.Past costs are sunk costs and generally do not affect present or future costs. For this reason they are ignored. 4. The investment problem is isolated from the financing problem. We generally assume that all required money is borrowed at interest rate i. 5. For now, stable prices are assumed. The inflation-deflation problem is deferred to Chapter 14. Similarly, our discussion of income taxes is deferred to Chapter 12. 6. Often uniform cash flowsor arithmetic gradients arereasonable assumptions.However, spreadsheets simplify the finding of PW in more complicated problems. PROBLEMS Jl Most problems could be solved with a spreadsheet, but calculators and tabulated factors are often easier. The 5-3 What is the valueof P for the situationdiagrammed? icon indicates that a spreadsheet is recommended. 5-1 Compute P for the following diagram. 300 200 200 100 100 t r 5; 0-1-2-3-4 I I 0 i = 12% i = 10% 1 1 p p 5-2 t r 0-1-2-3-4 Compute the value of P that is equivalent to the four cash flows in the following diagram. r 5-4 30 30 t t 0-1-2-3 (Answer: P r Compute the value of Q in the following diagram. 50 i = 15% = $498.50) 50 50 50 50 50 t0-1-2-3-4-5 t t t t t ., i = 12% :Ii II 1 Q 1 p II~ r t, ----- - - - - 166 5-5 PRESENTWORTH ANALYSIS Computethe value of P for the followingdiagram. 5-8 If i = 10%, what is the value of P? B t 0-[-2-3-4-5-6-7-8-9 120 r 120 t r 5; B o-[-2-3-4-5-6-7~~'oo 1 i=lO% P 5-9 j p 5-6 B t t t t 120 r B A stonecutter, assigned to carve the headstone for a well-known engineering economist, began with the following design. Compute the value of P for the following diagram. 5G 4G 3G 120 2G 60 60 60 60 60 t t t t t 0-1-2-3-4-5 G f r 0-1-2-3-4-5 r r i = 10% 1 p p (Answer: P 5-7 = $324.71) He then started the equation as follows: Use a geometric gradient factor to solve the following diagram for P. P = G(PjG,i,6) He realized he had made a mistake. The equation should have been 266.20 P = G(PjG, i, 5) + G(PjA, i, 5) The stonecutter does not want to discard the stone and start over. He asks you to help him with his problem. The right-hand side must be for the first equation to be correct for the carved figure, mul!iplied by one compound interest factor, taking the form: i = 15% . ''Op= G(PjG, p , i, ) Write the complete equation. - T i, 6)( - -- ----- --- --- -- ---------- Problems 5-10 Using 5% nominal interest, compounded continuously, solve for,P. 500 5-15 A manufacturer is considering purchasing equipment which will have the following financial effects: Year 0 1 2 3 4 500 LJ .i .. 1 2 167 3 Disbursements $4400 660 660 440 220 Receipts 0 $ 880 1980 2420 1760 If money is worth 6%, should he invest in the equipment? P 5-11 Find P for the following cash flow diagram. 1000 r 1000 1000 T3t. r TT2+ r 0-1-2-3-4-5-6-7-8-9-1O-11-12-13~oo 1000 r i=8% 1 p 5-12 The annual income from a rented house is $12,000. The annual expenses are $3000. If the house can be sold for $145,000 at the end of 10 years, how much could you afford to pay for it now, if you considered 18% to be a suitable interest rate? (Answer: $68,155) 5-13 Consider the following cash flow.At a 6% interest rate, what is the value of P, at the end of Year 1, that is equivalentto the benefitsat the end of Years2 through 7? 5-16 Jerry Stans, a young industrial engineer, prepared an economic analysis for some equipment to replace one production worker. The analysis showed that the present worth of benefits (of employing one less production worker) just equaled the present worth of the equipment costs, based on a lO-year useful life for the equipment. It was decided not to purchase the equipment. A short time later, the production workers won a new 3-year union contract that granted them an immediate 40~-per-hour wage increase, plus an additional 25 ~-per-hour wage increase in each of the two subsequent years. Assume that in each and every future year, a 25 ~-per-hour wage increase will be granted. Jerry Stans has been asked to r~vise his earlier economic analysis. The present worth of benefits of replacing one production employee will now increase. Assuming an interest rate of 8%, the justifiable cost of the automation equipment (with a lO-year useful life) will increase by how much? Assume the plant operates a single 8-hour shift, 250 days per year. 5-17 . Year Cash Flow 1 -P 2 3 4 5 6 7 +100 +200 +300 +400 +500 +600 5-14 How much would the owner of a building be justified in paying for a sprinkler system that will save $750 a year in insurance premiums if the system has to be replaced every 20 years and has a salvagevalueequal to 10%of its initialcost? Assumemoneyis worth 7%. (Answer: $8156) In a present worth analysis of certain equipment, one alternative has a net present worth of +420, based on a 6-year analysis period that equals the useful life of the alternative. A 10% interest rate was used in the computations. The alternative device is to be replaced at the end of the 6 years by an identical item with the same cost, benefits, and useful life. Based on a 10% interest rate, compute the net present worth of the alternative equipment for the 12-year analysis period. (Answer: NPW = +657.09) 5-18 A project has a net present worth of -140 as of January 1,2000. If a 10% interest rate is used, what is the project NPW as of December 31, 1997? 5-19 On February 1, the Miro Company needs to purchase some office equipment. The company is presently short of cash and expects to be short for several - -- ._- -.-'-" . 'I 11 I J 168 PRESENTWORTH ANALYSIS months. The company treasurer has indicated that he could pay for the equipment as follows: Date April 1 June 1 Aug. 1 Oct. 1 Dec. 1 Payment $150 300 450 600 750 A local office supply firm will agree to sell the equipment to the firm now and accept payment according to the treasurer's schedule. If interest will be charged at 3% every 2 months, with compounding once every 2 months, how much office equipment can the Miro Company buy now? (Answer: $2020) 5-20 By installing some elaborate inspection equipment on its assembly line, the Robot Corp. can avoid hiring an extra worker who would have earned $26,000 a year in wages and an additional $7500 a year in employee benefits. The inspection equipment has a 6-year useful life and no salvage value. Use a nominall8% interest rate in your calculations. How much can Robot afford to pay for the equipment if the wages and worker benefits would have been paid (a) At the end of each year (b) Monthly (c) Continuously (d) Explain why the answers in (b) and (c) are larger than in (a). Assume the compounding matches the way the wages and benefits would have been paid (i.e., annually, monthly, and continuously, respectively). 5-21 Annual maintenance costs for a particular section of highway pavement are $2000. The placement of a new surface would reduce the almual maintenance cost to $500 per year for the first 5 years and to $1000 per year for the next 5 years. After 10 years the annual maintenance would again be $2000. If maintenance costs are the only saving, what investment can be justified for the new surface. Assume interest at 4%. 5-22 An investor is considering buying a 20-year corporate bond. The bond has a face value of $1000 and pays 6% interest per year in two semiannual payments. Thus the purchaser of the bond will receive $30 every 6 months in addition to $1000 at the end of 20 years, along with the last $30 interest payment. If the investor wants to receive 8% interest, compounded semiannually, how much would he or she be willing to pay for the bond? I ! LL 5-23 A road building contractor has received a major highway construction contract that will require 50,000 m3 of crushed stone each year for 5 years. The needed stone can be obtained from a quarry for $5.80/m3. As an alternative, the contractor has decided to try and purchase the quarry. He believes that if he owned the quarry, the stone would only cost him $4.301m3. He thinks he could resell the quarry at-the end of 5 years for $40,000. If the contractor uses a 10% interest rate, how much would he be willing to pay for the quarry? 5-24 A new office building was constructed 5 years ago by a consulting engineering firm. At that time the firm obtained a bank loan for $100,000 with a 12% annual interest rate, compounded quarterly. The terms of the loan call for equal quarterly payments to repay the loan in 10 years. The loan also allows for its prepayment at any time without penalty. As a result of internal changes in the firm, it is now proposed to refinance the loan through an insurance company. The new loan would be for a 20-year term with an interest rate of 8% per year, compounded quarterly. The new equal quarterly payments would repay the loan in the 20-year period~ The insurance company requires the payment of a 5% loan initiation charge (often described as a "5-point loan fee"), - which will be added to the new loan. . (a) What is the balance due on the original mortgage if 20 payments have been made in the last five years? (b) What is the difference between the equal quar-. terly payments on the present bank loan and the proposed insurance company loan? 5-25 ffiP Inc. is considering establishing a new machine to automate a meat packing process. The machine will save $50,000 in labor annually. The machine can be purchased for $200,000 today and will be used for a period of 10 years. It is has a salvage value of $10,000 at the end of its useful life. The new machine will require an annual maintenance cost of $9000. The corporationhas a minimumrate of returnof 10%.Do you recommendautomatingthe process? 5-26 Argentinais consideringconstructinga bridge across the Rio de la Plata to connect its northern coast to the southern coast of Uruguay. If this bridge is constructed, it will reduce the travel time from Buenos Aires, Argentina, to Sao Paulo, Brazil, by over 10 hours and has the potential to significantly improvethe flowof manufacturedgoodsbetweenthe .. - __h '--'-' --- --- Problems two countries. The cost of the new bridge, which will be the longest bridge in the world, spanning over 50 miles, will be $700 million. The bridge will require an annual maintenance of $10 million for repairs and upgrades and is estimated to last 80 years. It is estimated that 550,000 vehicles will use the bridge during the first year of operation, and an additional 50,000 vehicles per year until the tenth year. These data are based on an a toll charge of $90 per vehicle. The annual traffic for the remainder of the life of the bridge life will be 1,000,000 vehicles per year. The Argentine government requires a minimum rate of return of 9% to proceed with the project. (a) Does this project provide sufficient revenues to offset its costs? . (b) What considerations are there besides economic factors in deciding whether to construct the bridge? 5-27 A student has a job that leaves her with $250 per month in disposable income. She decides that she will use the money to buy a car. Before looking for a car, she arranges a 100% loan whose terms are $250 per month for 36 months at 18% annual interest. What is the maximum car purchase price that she can afford with her loan? 5-28 The student in Problem 5-27 finds a car she likes and the dealer offers to arrange financing. His terms are 12% interest for 60 months and no down payment. The car's sticker price is $12,000. Can she afford to purchase this car with her $250 monthly disposable income? 5-29 The student in Problem 5-28 really wants this particular car. She decides to try and negotiate a different interest rate. What is the highest interest rate that she can accept, given a 60-month term and $250 per month payments? 5-30 We know a car costs 60 monthly payments of $199. The car dealer has set us a nominal interest rate of 4.5% compounded daily. What is the purchase price of the car? 5-31 A machine costs $980,000 to purchase and will provide $200,000 a year in benefits. The company plans to use the machine for 13 years and then will sell the machine for scrap, receiving $20,000. The company interest rate is 12%. Should the machine be purchased? 5-32 A corporate bond has aface value of$l 000 with maturity date 20 years from today. The bond pays interest -- 169 semiannually at a rate of 8% per year based on the face value. The interest rate paid on similar corporate bonds has decreased to a current rate of 6%. Determine the market value of the bond. 5-33 Calculate the present worth of a 4.5%, $5000 bond with interest paid semiannually. The bond matures in 10 years, and the investor desires to make 8% per year compounded quarterly on the investment. 5-34 A rather wealthy man decided he would like to arrange for his descendants to be well educated. He would like each child to have $60,000 for his or her education. He plans to set up a perpetual trust fund so that six children will receive this assistance in each generation. He estimates that there will be four generations per century, spaced 25 years apart. He expects the trust to be able to obtain a 4% rate of return, and the first recipients to receive the money 10 years hence. How much money should he now set aside in the trust? (Answer: $389,150) 5-35 The president of the E. L. Echo Corporation thought it would be appropriate for his firm to "endow a chair" in the Department of Industrial Engineering of the local university; that is, he was considering giving the university enough money to pay the salary of one professor forever. That professor, who would be designated the E. L. Echo Professor of Industrial Engineering, would be paid from the fund established by the Echo Corporation. If the professor holding that chair will receive $67,000 per year, and the interest received on the endowment fund is expected to remain at 8%, what lump sum of money will the Echo Corporation need to provide to establish the endowment fund? (Answer: $837,5(0) 5-36 A man who likes cherry blossoms very much would like to have an urn full of them put on his grave once each year forever after he dies. In his will, he intends to leave a certain sum of money in the trust of a local bank to pay the florist's annual bill. How much money should be left for this purpose? Make whatever assumptions you feel are justified by the facts presented. State your assumptions, and compute a solution. 5-37 A home builder must construct a sewage treatm~nt plant and deposit sufficient money in a perpetual trust fund to pay the $5000 per year operating cost and to replace the treatment plant every 40 years. The plant will cost $150,000, and future replacement plants will also cost $150,000 each. If the trust fund earns --- - --- .....- 170 PRESENTWORTH ANALYSIS 8% interest, what is the builder's capitalized cost to const:l1Jctthe plant and future replacements, and to pay the operating costs? 5-38 The local botanical society wants to ensure that the gardens in the town park are properly cared for. They recently spent $100,000 to plant the gardens. They would like to set up a perpetual fund to provide $100,000 for future replantings of the gardens every 10 years. If interest is 5%, how much money would be needed to forever pay the cost of replanting? 5-39 An elderly lady decided to distribute most of her considerable wealth to charity and to retain for herself only enough money to provide for her living. She feels that $1000 a month will amply provide for her needs. She will establish a trust fund at a bank that pays 6% interest, compounded monthly. At the end of each month she will withdraw $1000. She has arranged that, upon her death, the balance in the account is to be paid to her niece, Susan. If she opens the trust fund and deposits enough money to pay herself $1000 a month in interest as long as she lives, how much will Susan receive when her aunt dies? 5-40 What amount of money deposited 50 years ago at 8% interest would provide a perpetual payment of $10,000 a year beginning this year? 5-41 A small dam was constructed for $2 million. The annual maintenance cost is $15,000. If interest is 5%, compute the capitalized cost of the dam, including maintenance. 5-42 A depositor puts $25,000 in a savings account that pays 5% interest, compounded semiannually. Equal annual withdrawals are to be made from the account, beginning one year from now and continuing forever. What is the maximum annual withdrawal? 5-43 A trust fund is to be established for three purposes: (1) to provide $750,000 for the construction and $250,000 for the initial equipment of a small engineering laboratory; (2) to pay the $150,000 per year laboratory operating cost; and (3) to pay for $100,000 of replacement equipment every 4 years, beginning 4 years from now. At 6% interest, how much money is required in the trust fund to provide for the laboratory and equipment and its perpetual operation and equipment replacement? 5-44 The local Audubon Society has just put a new bird feeder in the park at a cost of $500. The feeder has a useful life of 5 years and an annual maintenance cost of $50. Our cat, Fred, was very impressed with the project. He wants to establish a fund that will maintain the feeder in perpetuity (that's forever!). Replacement feeders cost $500 every 5 years. If the fund will earn 5% interest, what amount must he raise for its establishment? Note that both maintenance and replacement costs following the intial investment must be covered. 5-45 We want to donate a marble birdbath to the city park as a memorial to our cat, Fred, while he can still enjoy it. We also want to set up a perpetual care fund to cover future expenses "forever." The initial cost of the bath is $5000. Routine annual operating costs are $200 per year, but every fifth year the cost .will be $500 to cover major cleaning and maintenance as well as operation. (a) What is the capitalized cost of this project if the interest rate is 8%? (b) How much is the present worth of this project if it is to be demolished after 75 years? The final $500 payment in the 75th year will cover the year's operating cost and the site reclamation. 5-46 Dr. Fog E. Professor is retiring and wants to endow a . chair of engineering economics at his university. It is expected that he will need to cover an annual cost of $100,000 forever. What lump sum must.he donate to the university today if the endowment will earn 10% interest? . 5-47 A man had to have the muffler replaced on his 2-yearold car. The repairman offered two alternatives. For $50 he would install a muffler guaranteed for 2 years. But for $65 he would install a muffler guaranteed "for as long as you own the car." Assuming the present owner expects to keep the car for about 3 more years, which muffler would you advise him to have installed if you thought 20% were a suitable interest rate and the less expensive muffler would only last 2 years? 5-48 A consulting engineer has been engaged to advise a town how best to proceed with the construction of a 200,000 m3 water supply reservoir. Since only 120,000 m3 of storage will be required for the next 25 years, an alternative to building the full capac-. ity now is to build the reservoir in two stages. Initially, the reservoir could be built with 120,000 m3 of capacity and then, 25 years hence, the additional 80,000 m3 of capacity could be added by increasing the height of the reservoir. Estimated costs are as follows: - -. "--. Problems in two stages :t stage: 120,000 m3 ~servoir ond stage: Add 80,000 m3 If capacity, additional onstruction and naintenance costs Construction Cost Annual Maintenance Cost $14,200,000 $75,000 $12,600,000 $25,000 $22,400,000 full capacity now ),000 m3 reservoir Option 1 Option 2 Use an 8-year analysis period and a 10% interest rate to determine which alternative should be selected: First cost Uniform annual benefit Useful life, in years A $5300 $1800 4 B $10,700 $2,100 8 Two alternative courses of action have the following schedules of disbursements: Year 0 1 2 3 4 5 A -$1300 0 0 0 0 0 -$1300 B -$100 -200 -300 -400 - -500 -1500 Based on a 6% interest rate, which alternative should be selected? 2 Telefono Mexico is expanding its facilities to serve a new manufacturing plant. The new plant will require 2000 telephone lines this year, and another 2000 lines after expansion in 10 years. The plant will be in Provide one cable now with capacity to serve 4000 lines. The cable cost will be $200,000, and annual maintenance costs will be $15,000. Provide a cable with capacity to serve 2000 lines now and a second cable to serve the other 2000 lines in 10 years. The cost of each cable will be $150,000, and each cable will have an annual maintenance of $10,000. The telephone cables will last at least 30 years, and the cost of removing the cables is offset by their salvage value. If interest is computed at 4%, which construction plan is preferred? . subscription. ) 171 operation for 30 years. The telephone company is evaluating two options to serve the demand. $100,000 A ""'ekly businessmagazineoffersa I-year subscriptil-_ for $58 and a 3-year subscription for $116. If you thoughtyou wouldread the magazinefor at least the next 3 years, and consider 20% as a minimum rate of return, which way would you purchase the magazine:Withthree I-year subscriptionsor a single 3-year subscription? (Answer: Choose the 3-year --- (a) Which alternative should be selected based on a 10% interest rate? (b) Will your answer to (a) change if the demand for additional lines occurs in 5 years instead of 10 years? 5-53 Dick Dickerson Construction Inc. has asked to you help them select a new backhoe. You have a choice between a wheel-mounted version, which costs $50,000 and has an expected life of 5 years and a salvage value of $2000, and a track-mounted one, which costs $80,000, with a 5-year life and an expected salvage value of $10,000. Both machines will achieve the same productivity. Interest is 8%. Which one will you recommend? Use a present worth analysis. 5-54 Walt Wallace Construction Enterprises is investigating the purchase of a new dump truck. Interest is 9%. The cash flows for two likely models are as follows: Annual First Operating Model Cost Cost A $2000 $50,000 B $1000 $80,000 Annual' Salvage Income Value Life $9,000 $10,000 lOyr $12,000 $30,000 lOyr (a) Using present worth analysis, which truck should the firm buy, and why? (b) Before the construction company can close the deal, the dealer sells out of Model B and cannot get any more. What should the firm do now and why? J ' I! l~' ) ..a..- 5-55 Two different companies are offering a punch press for sate. Company A charges $250,000 to deliver and install the device. Company A has estimated that the 'Ir' I 172 ---.---- PRESENTWORTH ANALYSIS machine will have maintenance and operating costs of $4000 a year and will provide an annual benefit of $89,000. Company B charges $205,000 to deliver and install the device. Company B has estimated maintenance and operating costs of the press at $4300 a year, with an annual benefit of $86,000. Both machines will last 5 years and can be sold for $15,000 for the scrap metal. Use an interest rate of 12%. Which machine should your company purchase, based on the forgoing data? 5-56 A battery manufacturing plant has been ordered to cease discharging acidic waste liquids containing mercury into the city sewer system. As a result, the firm must now adjust the pH and remove the mercury from its waste liquids. Three "firmshave provided quotations on the necessary equipment. An analysis of the quotations provided the following table of costs. Bidder Foxhill Instrument Quicksilver Almaden Annual Installed Operating Cost Cost $8000 $ 35,000 40,000 100,000 Annual Income from Mercury Salvage Recovery Value $2000 $20,000 7000 2000 2200 3500 0 0 cost, with all controls, would be least for natural gas at $30,000; for fuel oil it would be $55,000; and for coal it would be $180,000. If natural gas is used rather than fuel oil, the annual fuel cost will increase by $7500. If coal is used rather than fuel oil, the annual fuel cost will be $15,000 per year less. Assuming 8% interest, a 20-year analysis period, and no salvage value, which is the most economical installation? 5-59 Austin General Hospital is evaluating new office equipment offered by three companies. In each case the interest rate is 15% and the useful life of the equipment is 4 years. Use NPW analysis to determine the company from which you should purchase the equipment. First cost Maintenanceand B C $15,000 $1,530 $20,000 $1,890 10 20 20 The salvage value at the end of the useful life of each alternative is zero. At the end of 10 years, Alternative A could be replaced with another A with identical cost and benefits. The maximum attractive rate of return is 6%. Which alternative should be selected? 5-58 A steam boiler is needed as part of the design of a new plant. The boiler can be fired by natural gas, fuel oil, or coal. A decision must be made on which fuel to use. An analysis of the costs shows that the installed 8,000 3,000 13,000 6,000 Company C $20,000 900 11,0 4,500 5-60 The following costs are associated" with three tomato-peeling machines being considered for use in a canning plant. Machine A First cost Mainnance 5-57 A firm is considering three mutually exclusive alternatives as part of a production improvement program. The alternatives are: A $10,000 $1,625 Company B $25,000 400 operating costs Annualbenefit Salvagevalue If the installation can be expected to last 20 years and money is worth 7%, which equipment should be purchased? (Answer: Almaden) Installed cost Uniform annual benefit Useful life, in years Company A $15,000 1,600 Machine B Machine C' and $15,000 $63,000 $9,000 $67,000 $12,000 operating costs Annual benefit Salvage value Usefullife, in years $38,000 $13,000 4 $31,000 $19,000 6 . $37,000 $22,000 12 $52,000 " If the canning company uses an interest rate of 12%, which is the best alternative? Use NPW to make yom decision. (Note: Consider the least common multiple as the study period.) 5-61 A railroad branch line to a missile site is to be con. structed. It is expected that the railroad line will be used for 15 years, after which the missile site will be removed and the land turned back to agricultural use The railroad track and ties will be removed at tha' time. In building the railroad line, either treated OJ untreated wood ties may be used. Treated ties have ar installed cost of $6 and a 1O-year life; untreated tie! are $4.50 with a 6-year life. If at the end of 15 yean - - Problems rate, which alternative should be selected? Each alternative has a six-year useful life. the ties then in place have a remaining useful life of 4 years or more, they will be used by the railroad elsewhere and have an estimated salvage value of $3 each. Any ties that are removed at the end of their service life, or too close to the end of their service life to be used elsewhere, can be sold for $0.50 each. Determine the most economical plan for the initial railroad ties and their replacement for the 15year period. Make a present worth analysis assuming 8% interest. 5-65 The management of an electronics manufacturing firm believes it is desirable to automate its production facility. The automated equipment would have a lO-year life with no salvage value at the end of 10 years. The plant engineering department has surveyed the plant and suggested there are eight mutually exclusive alternatives available. 5-62 Consider A-E, five mutually exclusive alternatives: Initial cost Uniform annual benefits for first ) years F, .ist 5 years A $600 100 B $600 100 C $600 100 D $600 150 E $600 150 50 100 110 0 50 Plan 1 2 3 4 5 6 7 8 The interest rate is 10%. If all the alternatives have a lO-year useful life, and no salvage value, which alternative should be selected? 5-63 An investor has carefully studied a number of companies and their common stock. From his analysis, he has decided that the stocks of six firms are the best of the many he has examined. They represent about the same amount of risk and so he would like to determine one single stock in which to invest. He plans to keep the stock for 4 years and requires a 10% minimum attractive rate of return. Which stock from Table P5-63, if any, should the investor consider purchasing? (Answer: Spartan Products) 173 Net Annual Benefit (thousands) $51 39 26 15 57 23 47 33 Initial Cost (thousands) $265 220 180 100 305 130 245 165 If the firm expects a 10% rate of return, which plan, if any, should it adopt? (Answer: Plan 1) 5-66 A local symphony association offers memberships as follows: Continuing membership, per year Patron lifetime membership $ 15 375 The patron membership has been based on the symphony association's belief that it can obtain a 4% rate of return on its investment. If you believed 4% to be an appropriate rate of return, would you be willing to purchase the patron membership? Explain why or why not. 5-64 Six mutually exclusive alternatives, A-F (see Table P5-64), are being examined. For an 8% interest TABLE P5-63 Common Stock Western House Fine Foods Mobile Motors Spartan Products U.S. Tire Wine Products Price Annual End-of- Year per Share $233/4 45 305/8 12 333/8 521/2 Dividend per Share $1.25 4.50 0 0 2.00 3.00 Estimated Price at End of 4 Years $32 45 42 20 40 60 II I i I TABLEP5-64 Initial cost Uniform annual benefit A $20.00 6.00 B $35.00 9.25 C $55.00 13.38 D $60.00 13.78 E $80.00 24.32 -- --- --- -- F $100.00 24.32 --- -' --- . 174 PRESENTWORTH ANALYSIS 5-67 Using capitalized cost, detennine which type of road surface is preferred on a particular section of highway. Use 12% interest rate. A Initial cost Annual maintenance Periodic resurfacing Resurfacing interval $500,000 35,000 350,000 10 years subcontractors quoted the following prices and ten of payment: B $700,000 25,000 450,000 15 years Paving Co. Quick Price $85,000 Tartan Faultless $82,000 $84,000 Payment Schedule 50% payable immediate 25% payable in 6 month 25% payable at the end of one year Payable immediately 25% payable immediate] 75% payable in 6 month. The building contractor uses a 12% nomine interest rate, compounded monthly, in this type ( bid analysis. Which paving subcontractor should b awarded the paving job? 5-68 A city has developed a plan to provide for future municipal water needs. The plan proposes an aqueduct that passes through 500 feet of tunnel in a nearby mountain. Two alternatives are being considered. The first proposes to build a full-capacity tunnel now for $556,000. The second proposes to build a halfcapacity tunnel now (cost = $402,000), which should be adequate for 20 years, and then to build a second parallel half-capacity tunnel. The maintenance cost of the tunnel lining for the full-capacity tunnel is $40,000 every 10 years, and for each half-capacity tunnel it is $32,000 every 10 years. The friction losses in the half-capacity tunnel will be greater than if the full-capacity tunnel were built. The estimated additional pumping costs in the single half-capacity tunnel will be $2000 per year, and for the two half-capacity tunnels it will be $4000 per year. Based on capitalized cost and a 7% interest rate, which alternative should be selected? 5-71 Given the following data, use present worth analysi to find the best alternative, A, B, or (. A Initial cost Annual benefit Salvage value Usefullife $10,000 6,000 1,000 2 years B 15,000 10,000 - 2,000 3 years C $12,00( 5,00< 3,00( 4 years Use an analysis period of 12 years and 10% interest. 5-72 Consider the following four alternatives. Three are "do something" and one is "do nothing." Cost Net annual benefit 5-69 An engineer has received two bids for an elevator to be installed in a new building. The bids, plus his evaluation of the elevators, are shown in Table P5-69. The engineer will make a present worth analysis using a 10% interest rate. Prepare the analysis and detennine which bid should be accepted. Usefullife, in years ABC $0 $50 $30 o 12 4.5 51010 D $40 6 At the end of the 5-year useful life of B, a replacement is not made. If a 1O-year analysis period and a 10% interest rate are selected, which is the preferred alternative? 5-70 A building contractor obtained bids for some asphalt paving, based on a specification. Three paving TABLEP5-69 Engineer's Estimates Alternatives Bids: Installed Cost Service Life (years) Annual Operating Cost, Including Repairs Salvage Value at End of Service Life Westinghome Itis $45,000 54,000 10 15 $2700/yr 2850/yr $3000 4500 <\I .\ - -- --- --- I Problems 5-73 A cost analysis is to be made to determine what, if anything, shoqld be done in a situation offering three "do-something" and one "do-nothing" alternatives. Estimates of the cost and benefits are as follows. Alternatives 1 2 3 4 Uniform End-ofUseful Annual Useful-Life Life Cost Benefit Salvage Value (years) $500 $135 $ 0 5 600 100 250 5 700 100 180 10 0 0 0 0 Use a 10-year analysis period for the four mutually exclusive alternatives. At the end of 5 years, Alternatives 1 and 2 may be replaced with identical alternatives (with the same cost, benefits, salvage value, and useful life). (a) If an 8% interest rate is used, which alternative should be selected? (b) If a 12% interest rate is used, which alternative should be selected? 5-80 Assume annual mortgage payments of $12,000 for .. JIL 30 years and an interest rate of 6.168% per year. What - initial principal or PW will this repay? 5-81 Why do the values in Problems 5-78, 5-79, and 5-80 differ? 5-82 Ding Bell Imports requires a return of 15% on a,ll .- 4 years and an interest rate of 1/2% per month. What initial principal or PW will this repay? 5-75 Assume annual car payments of $6000 for 4 years and .- _ _ an interest rate of 6% per year. What initial principal or PW will this repay? projects. If Ding is planning an overseas development project with the cash flows shown in Table P5-82, what is the project's net present value? 5-83 Maverick Enterprises is planning a new product. Annual sales, unit costs, and unit revenues are as tab- ulated; the first cost of R&D and setting up the assembly line is $42,000. If i is 10%, what is the PW? Year 1 2 3 4 5 6 5-74 Assume monthly car payments of $500 per month for A- _ 5-77 Why do the values in Problems 5-74, 5-75, and 5-76 differ? 5-78 Assume mortgage payments of $1000 per month for L 30 years and an interest rate of 1/2% per month. What initial principal or PW will this repay? . 5-79 Assume annual mortgage payments of $12,000 for Annual Sales $ 5000 6000 9000 10,000 8000 4000 C.ost/unit $3.50 3.25 3.00 2.75 2.5 2.25 Annual production, unit costs, and unit' revenues are in the table. The first cost of the mine setup is $8 million. If i is 15%, what is the PW? . Year 1 2 3 4 5 6 7 Annual Production (tons) 70,000 90,000 120,000 100,000 80,000 60,000 40,000 A initial principal or PW will this repay? 30 years and an interest rate of 6% per year. What TABLEP5-82 Year 0 1 Net cash ($) 0 -120,000 2 -60,000 Price/unit $6 5.75 5.50 5.25 4.5 3 5-84 Northern Engineering is analyzing a mining project. 5-76 Assume annual car payments of $6000 for 4 years and an interest rate of 6.168% per year. What initial principal or PW will this repay? 175 3 4 5 6 7 20,000 40,000 80,000 100,000 60,000 Cost per ton $25 20 22 24 26 28 30 Price per ton $35 34 33 34 35 36 37 After Completing This Chapter... The student should be able to: · ·· · · Define equivalent uniform annual cost (EUAC) and equivalent uniform annual benefit~ (EUAB ).' Resolve an engineeringeconomic analysis probleminto its annual cash flowequivalent Conduct an annual worth analysis for a single investment. Use EUAC and EUAB to compare alternativeswith equal, common multiple, or continuous lives, or over some fixed study period. Develop and use spreadsheets to analyze loans for purposes of building an amortization table, calculating interest versus principal, finding the balance due, and determining whether to payoff a loan early. QUESTIONS TO CONSIDER ~ 1. Retailers frequently attract buyer interest by advertising low sale prices for print~rs,but they almost never mention the cost of the ink cartridges. In analyzing the likely cost of operating a printer over its useful lifetime, how much weight should the buyer give to the price of the printer itself, and how much to the cost of the ink cartridges? 2. King Camp Gillette, inventor of the safety razor, designed his product to work with a specially crafted blade that did not need sharpeningand could be disposed of after a few uses. Gillette eventuallyannounced,to everyone's as~onishment,that he would be giving 'his razors away for free. Many people couldn't imagine how he could possibly make money this way. In fact, Gillette revolutionized marketing, and his business revenue soared. Can you explain why? . 3. WouldGillette's strategy work with a product such as, say, an automobile?Why or why not? -- ... Annual Cash Flow Analysis Lowest Prices on the Net! Buy Now! Next time you review your e-mail and scroll through the spam, take a look at how many unsolicited messages are offering cut-rate ink cartridges for your printer. Strange, isn't it? Why would spam pests imagine they can capture your attention with ads for cheap printer ink, especially when competing spammersgenerally are advertising far more (shall we say) intriguing products? The answer becomes clear when you look at theprice of an ink cartridgefor a typicalinkjer printer. Cartridges often cost $30 or more, and they need to be replacedfrequently.By contrast, a good quality inkjet printer can often be purchased for under $100. . '. W --"- ----- - T - -- -- - - - __ - ___ - . h_ __ __ . ___ 178 ANNUAL CASH FLOW ANALYSIS This chapter is devotedto annual cash flowanalysis-the second of the three major analysis techniques. As we've said, alternatives must be resolved into a fonn that allows them to be compared. This means we must use the equivalence concept to convert from a cash flow representing the alternativeinto some equivalent sum or equivalent cash flow. With present worth analysis, we resolved an alternative into an equivalent cash sum. This might have been an equivalent present worth of cost, an equivalent present worth of benefit, or an equivalent net present worth. Here we compare alternatives based on their equivalentannualcash flows.Dependingon theparticularsituation,we may wish to compute the equivalent unifonn annual cost (EUAC),the equivalentunifonn annualbenefit (EUAB), or their difference (EUAB - EUAC). To prepare for a discussion of annual cash flow analysis, we will review some annual cash flow calculations, then examine annual cash flow criteria. Following this, we will proceed with annual cash flow analysis. ANNUAL CASH FLOW CALCULATIONS Resolving a Present Cost to an Annual Cost Equivalencetechniqueswere used in prior chaptersto convertmoney,at one point in time,to some equivalent sum or series. In annual cash flowanalysis, the goal is to convertmoney to an equivalentunifonn annual cost or benefit.The simplestcase is to converta present sum P to a seriesof equivalentunifonn end-of-periodcash flows.This is illustratedin Example6-1. A student bought $1000 worth of home furniture. If it is expected to last 10 years, what will the equivalent unifonn annual cost be if interest is 7%? 0-1-2-3-4-5--6-7-8-9-10 1 P = 1000 ! ! ! ! n ! == ! ! 10years ! ! ! i=7% SOLUTION' Equivalent unifonn annual cost = P (A j P, i, n) - 1000(Aj P, 7%, 10) = $142.40 Equivalentunifonn annual cost is $142.40. ........ Treatment of Salvage Value When there is a salvage value, or future value at the end of the usefu1life of an asset, the result is to decrease the equivalent unifonn annual cost. i i LJ_ -- - - --- --- - Annual Cash Flow Calculations 179 The student in Example 6-1 now believes the furniture can be sold at the end of 10 years for $200. Under these circumstances, what is the equivalent uniform annual cost? Resale Value 8=200 t 0-1-2-3-4-5-6-7-8-9 P 1 = 1000 i ,10 i i i i i i i i i For this situation, the problem may be solved by means of three different calculations. EUAC= P(AI P, i, n) - S(AIF, i, n) = 1000(A I P, 7%, 1O).~200(AIF, 7%, 10) - 1000(0.1424) - 200(0.0724) = = $127.92 142.40 - 14.48 (6-1) This method reflects the annual cost of the cash disbursement minus the annual benefit of the future resale value. EquatiOIl 6-1 describes a relationship that maybe modified by an identity presented ill Chaptet:,4: (AI p, i, n). (AI F, i, n) + i Substituting this into Equation 6-1 gives: i EUAC = P(AI F, t, n) + 'Pi =- SCAlF, t, n) I = (P'-" S)(AIF, i.,it) -+ Pi. I = (1000 - 2(0) (AIF, J ,. .800(0.0724) = ~~ ~ 7%, 10) +1000(0.07) +70 ~ 57.92+70 ;;;;:::: ~ ;;;;~ =;'1~~ i I ~ :-.: (6-3) . '. $127.92 ~ .~ ~~s method c()mputes.'th.e"equiyalel).tarmui.lIc?s~~~et()the(utIrecQyete~US~O(}when~the"furnitty"e ~=.J IS sold, and adds annual mtereston the $1000 mveStinent '..:..: ~iiA .: ~=-W:;;;::~J1'-4>.." ,'-',, ~...; .. __ 'UMtltJ - -- .. r LV "~., :;;;:-~~c.ooF ] I -AI 180 ANNUAL CASH FLOW ANALYSIS SOlUTiQ~N:~{ .. lithe value for (AI F, i, n) from Equation 6-2 is substituted into Equation 6-1, we obtain: EUAC= P(AI P, i, n) - S(AI P, i, n) + Si = (P - S)(AI P, i, n) + Si = (1000 - 200)(AI P, 7%,10) + 200(0.07) = 800(0.1424) + 14 = 113.92 + 14 = $127.92 (6-4) This method computes the annual cost of the $800 decline in value during the 10 years, plus interest on the $200 tied up in the furniture as the salvage value. When there is an initial disbursement P followedby a salvage value S, the annu"alcost may be computed in the three different ways introduced in Example 6-2. EUAC = P(AI P, i, n) EUAC = (P EUAC = (P - (6-1) S(AI F, i, n) - S)(Aj F, i, n) + Pi - S)(AI P, i, n) + Si (6-3) (6-4) Each of the three calculations gives the same results. In practice, the first and third methods are most commonlyused. TheEUACcalculatedin Equations6-1, 6-3,and 6-41salsoknown as the capital recovery cost of a project. Bill owned a car for 5 years; One day he wondered what his uniform annual cost for maintenance and repairs had been. He assembled the following data: Year 1 2 3 4 5 Maintenance and Repair Cost for Year $ 45 90 180 135 225 Compute the equivalent uniform annual cost (EUAC) assuming 7% interest and end-of-year disbursements. ~ -- LL . Annual Cash Flow Calculations 181 The EUAC may be computed for this irregular series of payments in two steps: 1. Use single payment present worth factors to compute the present worth of costJor the 5 years. 2. With the PW of cost known, use the capital recovery factor to compute EUAC. PW of cost - 45(P I F, 7%?1) + 90(P IF, 7%,2) + 180(PI F, 7%, 3) + 135(P I F, 7%, 4) + 225(P I F, 7%, 5) .. . 45(0.9346) + 90(0.8734) + 180(0.8163) + 135(0.7629}+ 225(0.7130)$531 EUAC = 53l(AIP, 7%,5)531(0.2439) -- $130 Bill reexamined his calculations and found that in his table he had reversed the maintenance and repair costs for years 3 and 4. The correct table is: Maintenance and Repair Cost for Year $ 45 90 135 180 225 Year 1 2 3 4 5 Recompute the EUAC. Ii,. This time the schedule of disbursements is an arithmetic gradjent series plus a uniform annual cost, as follows: 225 180 180 135 135 .. "'10_ !!II _ _ __ __.,.... _=- ~ _ .....--...- ~~- - - --- -- - - 182 ANNUAL CASH FLOW ANALYSIS EUAC = 45 + 45(A/G, 7%, 5) = 45 + 45(1.865) = $129 Since the timing of the expenditures is different in Examples 6-3 and 6-4, we would not expect to obtain the same EUAC. The examples have shown four essential points concerning cash flow calculations: 1. There is a direct relationship between the present worth of cost and the equivalent unifonn annual cost. It is EUAC = (PW of cost)(A/ P, i, n) 2. In a problem, an expenditure of money increases the EUAC, while a receipt of money (obtained,for example,from selling an item for its salvage value) decreases EUAC. 3. When there are irregular cash disbursements over the analysis period, a convenient method of solution is to first determine the PW of cost; then, using the equation in Item 1 above, the EUAC may be calculated. 4. Where there is an arithmetic gradient, EUAC may be rapidly computed by using the arithmetic gradient unifonn series factor, (A / G, i, n). ANNUAL CASH FLOW ANALYSIS The criteriafor economic efficiencyare presentedin Table6-1.One noticesimmediatelythat the tableisquite similarto Table5-1.In thecase of fixedinput, for example,thepresent worth criterion is maximize PW of benefits, and the annual cost criterion is maximize equivalent uniform annual benefits. It is appar~nt that, if you are maximizing the present worth of benefits, simultaneously you must be maximizing the equivalent unifonn annual benefits. This is illustrated in Example 6-5. TABLE6-1 Annual Cash Flow Analysis Input/Output Situation Criterion Fixed input Amount of money or other input resources is fixed There is a fixed task, benefit, or other output to be accomplished ~axUITrizeequivmentunUorm benefits (maxUITrizeEUAB) Neither amount of money, or other inputs, nor amount of benefits, or other outputs, is fixed ~aximize (EUAB Fixed output Neither input nor output is fixed Minimize equivment unUorm annum cost (minimize EUAC) -- -- ---- -- -- - - - EUAC) Annual Cash Flow Analysis 183 A finn is considering which of two devices to install to reduce costs in a particular situation. Both devices cost $1000 and have useful lives of 5 years with no salvage value. DeviceA can be expected to result in $300 savings annually.Device B will provide cost savings of $400 the first year but will decline $50 annually, making the second year savings $350, the third year savings $300, and so forth. With interest at 7%, which device should the finn purchase? Device A EUAB = $300 Device B EUAB = 400 - 50(AjG, 7%, 5) == 400 - 50(1.865) , = $306.75 . To maximize EUAB, select Device B. I J Example 6-5 was presented earlier, as Example 5-1, where we found: PW of benefits A = 300(PjA, 7%, 5) = 300(4.100) = $1230 This is converted to EUAB by multiplying by the capital recovery factor: EUABA = 1230(Aj P, 7%, 5) = 1230(0.2439) = $300 PW of benefits B = 400(P j A, 7%, 5) - 50(P jG, 7%, 5) = 400(4.100) - 50(7.647) = $1257.65 and, hence, EUABB = 1257.65(Aj P, 7%, 5) = 1257.65(0.2439) = $306.75 We see, therefore, that it is easy to convertthe present worth analysis results into the annual cash flow analysis results. We could go from annual cash flow to present worth just as easily, by using the series present worth factor.And, of course, both methods indicate that Device B is the preferred alternative. -- 184 ANNUAL CASH FLOW ANALYSIS Three alternatives are being considered for improving an operation on the assembly line along with the "do nothing" alternative. Equipment costs vary, as do the annual benefits of each in comparison to the present situation. Each of Plans A, B, and C has a lO-year life and a scrap value equal to 10% of its original cost. Plan A $15,000 14,000 8,000 1,500 Installed cost of equipment Material and labor savings per year Annual operating expenses End-of-usefullife scrap value Plan B $25,000 9,000 6,000 2,500 Plan C $33,000 14,000 6,000 3,300 If interest is 8%, which plan, if any, should be adopted? Since neither installed cost nor output benefits are fixed, the economic criterion is to maximize EUAB - EUAC. PlanA Equivalent uniform annual benefit (EUAB) Material and labor per year Scrap value (AI F, 8%, 10) EUAB = Equivalent uniform annual cost (EUAC) Installed cost (AI P, 8%, 10) Annual operating expenses ....EUAC = (EUAB -,-EUAC) , PlanB PlanC Do "Nothing $14,000 104 $9,000 172 $14,000 228 $0 -0 $14,104 $9,172 $14,228 $0 $ 2,235 8,000 $10,235 $ 3,869 $3,725 6,000 $9,725 -$ .553 $ 4,917 6,000 $10,917 $ 3,311 $0 -0 $0 '$0 Based on our criterion of maximizing EUAB -. EUAC, Plan A is the best of the four alternatives. We note, however, that. since the do-nothing alternative has EUAB desirable alternative than Plan B. - EUAC .,0, it isa D;).ore ANALYSIS PERIOD In Chapter 5, we saw that the analysis period is an important consideration in computing present worth comparisons. In such problems, a common analysis period must be used for each alternative. In annual cash flow comparisons, we again have the analysis period question. Example 6-7 will help in examining the problem. J -- - - - Analysis Period 185 Two pumps are being considered for purchase. If interest is 7%, which pump should be bought? Pump A $7000 1500 12 Initial cost End-of-useful-life salvage value Useful life, in years Pump B $5000 1000 6 The annual cost for 12 years of Pump A can beJound by using Equation 6-4: EUAC = (P - S)(Aj P, i, n) + Si = (7000 - 1500)(Aj P, 7%,12) + 1500(0.07) = 5500(0.1259) + 105 = $797 Now compute the annual cost for 6 years of Pump B: I ~, I' EUAC = (5000 - 1000)(Aj P, 7%, 6) + 1000(0.07) = 4000(0.2098) + 70 = $909 For a common analysis period of 12 years, we need to replace Pump B at the end of its 6-year useful life. If we assume that another pump B' can be obtained, having the same $5000 initial cost, $1000 salvage value, and 6-year life, the cash flow wilt be as follows: 1000 1000 t 0-1-2-3-4"---5-6~7-8--9 1 5000 10 1 5000 I, Pump B 6 years 1 ReplacementPumpB' 6 years .t "11 "12 ' -J For the 12-year analysis period, the annual cost for Pump B is I EUAC = [5000- 1000(PIF, 7%, 6) + 5000(P j F, 7%, 6) I I - I II t. L -" 1000(P j F, 7%, 12)] x (AlP, 7%, 12) [5000 - 1000(0.6663) + 5000(0.6663) - 1000(0.4449,11 x (0.125Q) 1 = (5000 - 666 + 3331 - 444)(0.1259) = (7211)(Q.,!.459)= \11.. __ J ' ii. I 1..I' = $9£)9 _ 11IO _ Ii ___...___ ""'--- .. , 1 186 ANNUAL CASH FLOW ANALYSIS The annual cost of B for the 6-year analysis period is the same as the annual cost for the 12-year analysis period. This is not a surprising conclusion when one recognizes that the annual cost of the first 6-year period is repeated in the second 6-year period. Thus the lengthy calculation of EUAC for 12 year~ of Pump B and B' was not needed. By assuming that the shorter-life equipment is replaced by equipment with identical economic consequences, we have avoided a lot of calculations. Select Pump A. Analysis Period Equal to Alternative Lives In the ideal situation, the analysis period for an economy study coincides with the useful life for each alternative.The economy study is based on this analysis period. Analysis Period a Common Multiple of Alternative Lives When the analysis period is a common multiple of the alternative lives (for example, in Example 6-7, the analysis period was 12 years with 6- and 12-year alternative lives), a "replacement with an identical item with the same costs, performance, and so forth" is frequently assumed. This means that when an alternative has reached the en~ of its useful life, it is assumed to be replaced with an identical item. As shown in E~ampIe 6-7, the result is that the EUAC for Pump B with a 6-year useful life is equal to the EUAC for the entire analysis period based on Pump B plus the replacement unit, Pump B'. . Under these circumstances of identical replacement, it is appropriate to compare the annualcash flowscomputedfor alternativesbased on theirownservicelives.InExample 6-7, the annual cost for Pump A, based on its 12-year servicelife, was compared with the annual cost for Pump B,.based on its 6-year service life. Analysis Period for a Continuing Requirement Many timesaneconomicanalysisis undertakento determinehowto providefor a moreor less continuing requirement. One might need to pump water from a well as a continuing requirement. There is no distinct analysis period. In this situation, the analysis period is assumed to be long but undefined. . If, for example,wehad a continuingrequirementto pumpwaterand alternativePumps A and B had useful lives of 7 and 11 years, respectively,what should we do? The customary assumptionis that Pump A's annual cash flow (based on a 7-year life) may be comparedto Pump B's annual cash flow (based on an ll-year life). This is done without m':1chconcern that the least common multiple of the 7- and ll-year lives is 77 years. This cOJIlparison of "different-life" alternatives assumes identical replacement (with identical costs, performance, etc.) when an alternative reaches the end of its useful life. Example 6-8 illustrates the situation. ~- - - Analysis Period 187 Pump B in Example 6-7 is now believed to have a 9-year useful life. Assuming the same initial cost and salvage value, compare it with Pump A using the same 7% interest rate. If we assume that the need for A or B will exist for some continuing period?the comparison of annual costs for the unequal lives is an acceptable technique. For 12 years of Pump A: EUAC = (7000 - 1500)(AI P, 7%, 12) + 1500(0.07) $797 For 9 years of Pump B: EUAC - (5000 -1000)(AI p, 7%, 9) + 1000(0.07). $684 For minimum EUAC, select Pump B. Infinite Analysis Period At times we have an alternativewith a limited (finite)useful life in an infinite analysisperiod situation. The equivalent uniform annual cost may be computed for the limited life. The assumption of identical replacement (replacementshave identical costs, performance, etc.) is often appropriate.Based on this assumption,the same EUACoccurs for each replacement of the limited-life alternative.The EUAC for the infinite analysis period is therefore equal to the EUAC computed for the limited life. With identical replacement, EUACinfinite analysisperiod = EUACfor limited life n A somewhat different situation occurs when there is an alternative with an infinite life in a problem with an infinite analysis period: EUACnfinite analysisperiod When n = 00, we have A = Pi = P(AI P, i, (0) + Any other annual costs and, hence, (AI P, i, (0) equals i. EUACinfinite analysisperiod = Pi + Any other annual costs In the construction of an aqueduct to expand the water supply of a city, ther~are t~o aJ-ternatives for a particular portion of the aqueduct. Either a tunnel can be constructed tlu:ougha mountain, or a pipeline can be laid to go around themountain. If there is a permanentneed for the aqueduct, , J .------ --1 --- 188 ANNUAL CASH FLOW ANALYSIS should the tunnel or the pipeline be selectedfor this particular portion of the aqueduct? Assume a 6% interest rate. ThnnelThrough Mountain $5.5 million Initial cost Maintenance Usefullife Salvage value Pipeline Around Mountain $5 million o o Permanent 50 years o o Thnnel For the tunnel, with its permanent life, Wewant (AI P, 6%, 00). For an infinite life, the capital recovery is simply interest on the invested capital. SO(AI P, 6%, 00) = i, andwe write EUAC = Pi = $5.5 million(0.06) = $330,000 Pipeline EUAC = $5 million(AIP, 6%, 50) = $5 million(0.0634) = $317,000 For fixed output, minimize EUAC. Select the pipeline. The difference in annual cost between a long life and an infinite life is small unless an unusually low interest-rateis used. In Example 6-9 the tunnel is assumed to be permanent. For comparison, compute the annual cost if an 85-year life is assumed for the tunnel. EUAC = $5.5 million(AI P, 6%, 85) = $5.5 million(0.0604) = $332,000 The difference in time between 85 years and infinity is great indeed, yet the difference in annual costs in Example 6-9 is very small. Some Other Analysis Period The analysis period in a particular problem may be something other than one of the four we have so far described. It may be equal to the life of the shorter-lifealternative,the longer-life alternative, or something entirely different. One must carefully examine the consequences l - --- I Analysis Period 189 of each alternative throughout the analysis period and, in addition, see what differences there might be in salvage values, and so forth, at the end of the analysis period. Suppose that Alternative 1 has a 7-year life and a salvage value at the end of that time. The replacement cost at the end of 7 years may be more or lessthan the originalcost. If the replacement is retired prior to 7 years, it will have a terminal value that exceeds the end-of-life salvage value. Alternative 2 has a 13-year life and a terminal value wheneverit isretired. If the situationindicates that 10 years is the proper analysis period, set up the equations to properly compute the EUAC for each alternative. Alternative 1 Salvage Value TerminalValueat End of 10thYear f f O~1-2~3---:::-4~5~6~1---8---9-1O-11---12-13---14 ,jnitii1 R"Plae<meni.. l Cost Cost I i l II I 7~yearLife 7-yearLife I Altemative2 , Terminal Value at End of 1OthYear O~1-2-3___4~5 ' 6" 7~8 9~10 f 11---12-13 1'titii1 l Cost 13-year Life I II I 10-year Analysis Period --I Altel'native.l '. . EUAC1 I r= [Initial cost + (Replacetnentcost - Salvage value)(P j F, i, 7,) - (Terminalvalue)(PjF, i, lO)](AjP, i, 10) Alterp.ative2 :,.=' I: ~~ f1J1C"2,,.DIJit!~co~t-~~Ji~l£k.!W.J~Ic.P - -- - - ~JPl -- = 190 ANNUAL CASH FLOW ANALYSIS USING SPREADSHEETS TO ANALYZE LOANS Loan and bond payments are made by firms, agencies, and individual engineers. Usually, the payments in each period are constant. Spreadsheetsmake it easy to: · Calculate the loan's amortization schedule · Decide how a payment is to be split between principal and interest · · Find the balance due on a loan Calculate the number of payments remaining on a loan. Building an Amortization Schedule As illustrated in Chapter 4 and Appendix 1,an amortizationschedule lists for each payment period: the loan payment, interest paid, principal paid, and remaining balance. For each period the interest paid equals the interest rate times the balance remaining from the period' before. Then the principal payment equals the payment minus the interest paid. Finally, this principal payment is applied to the balance remaining from the' preceding period to calculate the new remaining balance. As a basis for comparison with sprea~sheet loan functions, Figure 6-1 shows this calculation for Example 6-11. An engineer wanted to celebrate graduating and getting a job by buying $2400 of new furniture. Luckily the store was offering 6-month financing at thelow interest rate of 6% per year nominal' (really 1/2%per month). Calculate the amortization schedule. A 1 B 2400 2 E D Initial balance i 0.50% 3 C 6 N 4 $407.03 ""'.--PM)'(A2,A3,Al) Payment 5 6 ~: . p I . t , ~:; ~ ~ Month 8 0 9 1 10 2 Interest , #Al ==D8--C9 395.03 2004.97 10.02 397.00 1607.97 11 3 8;04 398.99 1208.98 12 4 6.04 400.98 807.99 13 5 4.04 402,99 14 15 6 2.03 405.00 16 I , .,"'" ,"" . . AIDomzati6n _ __ I \. . ....... . '.0 .= ==$A$2*Dl3 * ,,> schedrile:for fUnllture1oah. - .. --- , 240"0.00 12.00 18 _ Payment , 17 '" '1'1~'FIGURE'6:r _ ~"1 _ El11:ling Balance Principal 7 , 405.00 0.00 '",. "k'''' I .... , ,, .' <;;J'!i; == ,",' '. .' . . '. .. " ;;.. :;;;:11:1: ""'==,." ,'--" L - - -- ..,.~. ..;. I j Using Spreadsheets to Analyze Loans 191 The first step is to calculate the monthly payment: A -2400(AjP,lh%, 6) .2400 (0..J.696) ..$407.Q With tlIi~iIlformation to p.al).dtQe engiIieer cap. us~. the $pJ;ea<l$he~tofEiguI:e 6-1 to obtafu the amowzatiQJ.).schedule. '1: = How Much to Interest? How Much to Principal? For a loan with constant payments, we can answer these questions for any period without the full amortization schedule. For a loan with constant payments, the functions IPMT and PPMT directly answer these questions. For simple problems, both functions have four arguments (i, t, n, -P), where t is the time period being calculated. Both functions have optional arguments that permit adding a balloon payment (an F) and changing from endof-period payments to beginning-of-period payments. For example, consider Period 4 of Example 6-11. The spreadsheet formulas give the same answer as shown in Figure 6-1. Interes4 = IPMT(0.5%, 4, 6, -2400) = $6.04 Principalpayment4= PPMT(0.5%, 4,6, -2400) = $400.98 Finding the Balance Due on a Loan An amortizationschedule can be used to calculatethe balance due on a loan. Or more easily the balance due equals the present worth of the remaining payments. Interest is paid in full after each payment, so later payments are simply based on the balance due. A. caris pUJ.'chasedWith a 48-month, 9% nominal loan with an initial balance of $15,000. What is the,balance dtl~ halfway through the 4 years? 'J!l Th,.eW$t~Jep j~JQ;f81q1l1at~ !h.~1J1P!lfuJypay.w.el).t".3-ta mon(hlyinteJ;est;-ate,J>fJ /4%.Jbis ~quals " II ,IPaYIIlel).f' I : = d! R: '..15,QOO(AJP,'0.75%,48) . ~{/15',"6(0){6.iQ2];9) ~;$17330~'" or. ...PMT(0,75%, 48, -15,000) bt J!.1373;28 ~ L -.- - ---- ~W" . 192 ANNUAL CASH FLOW ANALYSIS The next step will use ~e spreadsheet answer, because it is more accurate (there are only three significantdigits in the tabulated factor). After 24 paymentsand with 24 left, theremainingbalance equals (P/ A, i, Nremaining) payment Balance = (P / A, 0.75%, 24)$373.28 = (21.889)(373.28) =$8170.73 or or = PV(0.75%, 24,373.28) . $8170.78 Thus halfway through the repayment schedule, 54.5% of the original balance is still owed. ! . Pay Off Debt Sooner by Increasing Payments Paying off debt can be a good investment because the investmentearns the rate of interest on the loan. For example, this could be 8% for a mortgage, 10% for a car loan, or 19% for a credit card. When making extra payments on a loan, the common question is: How much sooner will the debt be paid off? Until the debt is paid off, any early payments are essentially locked up, since the same payment amount is owed each month. The firstreason thatspreadsheetsare convenientis fractionalinterestrates. For example, an auto loan might be at a nominal rate of 13% with monthly compounding or 1.08333% per month. The second reason is that the function NPER calculates the number of periods remaining on a loan. NPER can be used to calculate how much difference one extra payment makes or how much difference increasing all payments by x% makes. Extra payments are applied entirely to principal, so the interest rate, remaining balance, and payment amounts are all known. Nremaining equals NPER(i, payment, remaining balance) with optional argumentsfor beginning-of-period cash flows and balloon payments. The signs.of the payment and the remaining balance must be different. Maria has a 7.5% mortgage with monthly payments for 30 years. Her original balance was $100,000, and she just made her twelfth payment. Each month she also pays into a reserve account, which the bank uses to pay her fire and liability insurance ($900 annually) and property taxes ($1500 annually). By how much does she shorten the loan if she makes an extra loan payment today? If she makes an extra total payment? If she increases each total payment to 110% of her current total payment? SOLUTION The first step is to calculate Maria's loan payment for the 360 months. Rather than calculating a six.-significantdigit monthly interest rate,it is easier to use 0.075/12 in the spreadsheetformulas. Payment-CPMT(O.01s/12,360, -100000) = $699.21 ...- - - - ...- -- - -- --- Summary 193 The remainjng balance after 12suchpayments is thepresent worthof theremaining 348payments. Balance12 = PV(0.075jI2,348,699.21) = $99,077.53 (after 12 payments, she has paid off $922!) If she pays an extra $699.21, then the number of periods remaining is NPER(0.075jI2,-699.21, 99077.53 - 699.21) = 339.5 This is 8.5 payments less than the 348 periods left before the extra payment. If she makes an extratotal payment, then Total payment = 699.21 +900j12 NPER(0.075j12, + 1500j12 - $899.21jmonth -699.21,99244 -,,-.899.21) = 337.1 or2.4;more paYlllentssaved.If sqemakes an extra 10%paYlllentOnthetotalpaymenfof $899.21, th~ ~ NPER([0.075jI2, ---(1.1)(899.21-200)],99077.53) .246.5 payments or IOl.5 payments saved. IIQ'bte that.$200 of the total paYlllentgoes to payfori:nsnf<Ulceand t~es. SUMMARY Annual cash flowanalysis is the secondof the threemajor methodsof resolving alternatives into comparablevalues. When an alternativehas an initial cost P and salvage value S, there are three ways of computing the equivalent uniform annual cost: · EUAC = P(Aj P, i, n) - S(Aj F, i, n) · EUAC= (P - S)(Aj F, i, n) + Pi · EUAC = (P - (6-1) (6-3) S)(Aj P, i, n) + Si (6-4) . All three equationsgivethe same answer.This quantityis also known as the capital recovery. cost of the project. The relationship between the present worth of cost and the equivalent uniform annual cost is: · EUAC = (PW of cost)(Aj P, i, n) The three annual cash flow criteria are: For fixed input For fixed output Neither input nor output fixed Maximize EUAB Minimize EUAC Maximize EUAB - EUAC In present worth analysis there must be a common analysis period. Annual cash flow analysis, however,allows some flexibilityprovided the necessary assumptions are suitable in the situation being studied. The analysis period may be different from the lives of the --- 194 ANNUAL CASH FLOW ANALYSIS alternatives, and provided the following criteria are met, a valid cash flow analysis may l made. 1. When an alternativehas reached the end of its useful life, it is assumedto be replace by an identical replacement (with the same costs, performance, etc.). 2. The analysis period is a common multiple of the useful lives of the alternatives, ( there is a continuing or perpetual requirement for the selected alternative. If neither condition applies, it is necessary to make a detailed study of the consequence of the various alternatives over the entire analysis period with particular attention to th difference between the alternatives at the end of the analysis period. There is very little numerical differencebetween a long-life alternativeand a perpetua alternative. As the value of n increases, the capital recovery factor approaches i. At tho limit, (AI P, i, (0) = i. One of the most common uniform payment series is the repayment of loans. Spread sheets are useful in analyzing loans (balance due, interest paid, etc.) for several rea sons: they have specialized functions, many periods are easy, and any interest rate can bt used. PROBLEMS 6-1- Compute the value of C for the following diagram, based on "a 10% interest rate. 6-3 Computethe value of E: 60 45 60 t 1 0-1-2-3-4 ~ 0-1-2-3-4 ~ r 'f ! ! E ~ ~ c c r E ! E + ! E i = 12% ~ c 3f c 6-4 If i = 6%, compute the value of D that is equivalent to the two disbursementsshown. (Answer: C 6-2 = $35.72) 200 Compute the value of B for the following diagram: B B B B B t t t t t , 1 ~ 100 D t t t t t "t t D i = 15% D 0-1-2-3-4-5-6 0-1-2-3-4-5 ~ 100 r- 100 ~ .-'~Answer: D = $52.31) D D D Problems 6-5 For the diagram, compute the value of D: 300 300 195 300 1.5D t t 0-1-2-3-4-5 t r i It T r T TT r T ItT r T 0-1-2-3-4-5-6-7-g-9-1O-1l-12 t n = co ! A! A! A! A! A! A! A! A! A! A! A! = 12% A i = 10% 1 500 6-6 6-9 On April 1 a man borrows $100. repaid in three equal semiannual payments. If the annual interest pounded semiannually, how much (Answer: $35.69) 6-10 An electronics firm invested $60,000 in a precision inspection device. It cost $4000 to operate and main~ tain in the first year and $3000 in each of the subsequent years. At the end of 4 years, the firm changed their inspection procedure, eliminating the need for the device. The purchasing agent was very fortunate in being able to sell the inspection device for $60,000, the original price. The plant manager asks you to com~ pute the equivalent uniform annual cost of the device during the 4 years it was used. Assume interest at 10% per year. (Answer: $9287) What is C in the accompanying figure? 40 30 20 i 0-1-2-3-4-5 t r C C 10 ! C 10 ! C i = 10% The loan is to be (every 6 months) rate is 7% comis each payment? 6-11 A firm is about to begin pilot plant operation on a pro6-7 cess it has developed. One item of optional equipment that could be obtained is a heat exchanger unit. The company finds that a unit now available for $30,000 could be used in other company operations. It is estimated that the heat exchanger unit will be worth $35,000 at the end of 8 years. This seemingly high salvage value is due primarily to the fact that the $30,000 purchase price is really a rare bargain. If the firm believes 15% is an appropriate rate of return, what annual benefit is needed to justify the purchase of the heat exchanger unit? (Answer: $4135) If interest is 10%,what is A? ~ i ~ r 0-1-2-3-4-5-6-7 t ~ t r i ~ ~ ~ ~ ~ ~ ~ A A A A A A A 6-12 6-8 The maintenance foreman of a plant in reviewing his records found that a large press had the following maintenance cost record: As shown in the cash flow diagram, there is an annual disbursement of money that varies from year to year from $100 to $300 in a fixed pattern that repeats forever. If interest is 10%, compute the value of A, also continuing forever, that is equivalent to the fluctuating disbursements. -. 5 years ago 4 years ago 3 years ago 2 years ago Last year -~ $ 600 700 800 900 1000 . 196 ANNUAL CASH FLOW ANALYSIS After consulting with a lubrication specialist, he changed the preventive maintenance schedule. He believ~s that this year maintenance will be $900 and will decrease $100 a year in each of the following 4 years. If his estimate of the future is correct, what will be the equivalent uniform annual maintenance cost for the 10-year period? Assume interest at 8%. (Answer: $756) estimated 500 bales a year that would be producec A wastepaper company will pick up the bales at th plant and pay Johnson $2.30 per bale for them. USI an annual cash flow analysis in working this problem (a) If interest is 8%, is it economical to install ant operate the bailer? (b) Would you recommend that the bailer be installed? 6-13 A firm purchased some equipment at a very favorable price of $30,000. The equipment resulted in an annual net saving of $1000 per year during the 8 years it was used. At the end of 8 years, the equipment was sold for $40,000. Assuming interest at 8%, did the equipment purchase prove to be desirable? 6-19 An engineer has a fluctuating future budget for th~ maintenance of a particular machine. During each 01 the first 5 years, $1000 per year will be budgeted. During the second 5 years, the annual budget will be $1500 per year. In addition, $3500 will be budgeted for an overhaul of the machine at the end of the fourth year, and another $3500 for an overhaul at the end of the eighth year. The engineer asks you to compute the uniform annual expenditure that would be equivalent to these fluctuating amounts, assuming interest at 6% per year. 6-14 When he started work on his twenty-second birthday, D. B. Cooper decided to invest money each month with the objective of becoming a millionaire by the time he reaches age 65. If he expects his invesnnents to yield 18% per annum, compounded monthly, how much should he invest each month? (Answer: $6.92 a month.) 6-15 Linda 0' Shay deposited $30,000 in a savings account as a perpetual trust. She believes the account will earn 7% annual interest during the first 10 years and 5% interest thereafter. The trust is to provide a uniform end-of-year scholarship at the university. What uniform amount could be used for the student scholarship each year, beginning at the end of the first year and continuing forever? 6-16 A machine costs $20,000 and has a 5-year useful life. At the end of the 5 years, it can be sold for $4000. If annual interest is 8%, compounded semiannually, what is the equivalent uniform annual cost of the machine? (An exact solution is expected.) 6-17 The average age of engineering students at gradu:' ation is a little over 23 years. This means that the working career of most engineers is almost exactly 500 months. How much would an engineer need to save each month to become a millionaire by the end of his working career? Assume a 15% interest rate, compounded monthly. 6-18 The Johnson Company pays $200 a month to a trucker to haul wastepaper and cardboard to the city dump. The material could be recycled if the company were to buy a $6000 hydraulic press bailer and spend $3000 a year for labor to operate the bailer. The bailer has an estimated useful life of 30 years and no salvage value. Strapping material would cost $200 per year for the 6-20 An engineer wishes to have $5 million by the time he retires in 40 years. Assuming 15% nominal interest, compounded continuously, what annual sum must he set aside? (Answer: $2011) 6-21 Art Arfons, a K-State-educated engineer, has made a considerable fortune. He wishes to start a perpetual sch?larship for engineering students at KState. The scholarship will provide a student with an annual stipend of $2500 for each of 4 years (freshmen through senior), plus an additional $5000 during the senior year to cover entertainment expenses. Assume that students graduate in 4 years, a new award is given every 4 years, and the money is provided at the beginning of each year with the first award at the beginning of year one. The interest rate is 8%. (a) Determine the equivalent uniform annual cost (EUAC) of providing the scholarship. . (b) How much money must Art donate to K-State? 6-22 Jenny McCarthy is an engineer for a municipal power plant. The plant uses natural gas, which is currently provided from an existing pipeline at an annual cost of $10,000 per year. Jenny is considering a project to construct a new pipeline. The initial cost of the new pipeline would be $35,000, but it would reduce the annual cost to $5000 per year. Assume. an analysis period of 20 years and no salvage value for either the existing or new pipeline. The interest rate is 6%. (a) Determine the equivalent uniform annual cost (EUAC) for the new pipeline? (b) Should the new pipeline be constructed? --:-' - -- - - - -------- Problems 6-23 A machine has a first cost of $150,000, an annual at the end of the 3 year period, the car is returned to the leasing company. (c) Lease the car with an option to purchase it at the end of the lease; pay $360 a month for 36 months; at the end of that time, Claude could purchase the car, if he chooses, for $3500. Claude believes he should use a 12% interest rate in determining which alternative to select. If the car could be sold for $4000 at the end of 3 years, which method should he use to obtain it? operation and, maintenance cost of $2500, a life of 10 years, and a salvage value of $30,000. At the end of years 4 and 8, it requires a major service, which costs $20,000 and $10,000, respectively. At the end of year 5, it will need to be overhauled at a cost of $45,000. What is the equivalent uniform annual cost of owning and operating this particular machine? 6-24 Mr. Wiggley wants to buy a new house. It will cost $178,000. The bank will loan 90% of the purchase price at a nominal interest rate of 10.75% compounded weekly, and Mr. Wiggley will make monthly payments. What is the amount of the monthly payments if he intends to pay the house off in 25 years? 6-27 A motorcycle is for sale for $2600. The motorcycle dealer is willing to sell it on the following terms: No down payment; pay $44 at the end of each of the first 4 months; pay $84 at the end of each month after that until the loan 6-25 Steve Lowe must pay his property taxes in two equal installments on December 1 and April 1. The two payments are for taxes for the fiscal year that begins on July I and ends the following June 30. Steve purchased a home on September 1. He estimates the annual property taxes will be $850 per year. Assuming the annual property taxes remain at $850 per year for the next several years, Steve plans to open a savings account and to make uniform monthly deposits the first of each month. The account is to be used to has been paid in full. Based on these terms and a 12% annual interest rate compounded monthly, how many $84 payments will be required? 6-28 When he purchased his home, AI Silva borrowed $80,000 at 10% interest to be repaid in 25 equal annual end-of-year payments. After making 10 payments, AI found he could refinance the balance due on his loan at 9% interest for the remaining 15 years. To refinance the loan, AI must pay the original lender the balance due on the loan, plus a penalty charge of 2% of the balance due; to the new lender he also must pay a $1000 service charge to obtain the loan. The new loan would be made equal to the balance due on the old loan, plus the 2% penalty charge, and the $1000 service charge. Should AI refinance the loan, assuming that he will keep the house for the next 15 years? Use an annual cash flow analysis in working this problem. pay the taxes when they are due. To begin the account, Steve deposits a lump sum equivalent to the monthly payments that will not have been made for the first year's taxes. The savings account pays 9% interest, compounded monthly and payable quarterly (March 31, June 30, September 30, and December 31). How much money should Steve put into the account when he opens it on September 1? What uniform monthly deposit should he make from that time on? (A careful exact solution is expected.) (Answers: Initial deposit $350.28; monthly deposit $69.02) 6-29 A company must decide whether to provide their salesmen with company-owned automobiles or to pay the salesmen a mileage allowance and have them drive their own automobiles. New automobiles would cost about $18,000 each and could be resold 4 years later for about $7000 each. Annual operating costs would be $600 per year plus 12,t per mile. If the sales~en drove their own automobiles, the company probably would pay them 30}!!per mile. Calculate the number of miles each salesman would have to drive each year for it to be economically practical for the company to provide the automobiles. Assume a 10% annual interest rate. Use an annual cash flow analysis. 6-26 Claude James, a salesman, needs a new car for use in his business. He expects to be promoted to a supervisory job at the end of 3 years, and so his concern now is to have a car for the 3 years he expects to be "on the road." The company reimburses salesmen each month at the rate of 251iper mile driven. Claude has decided to drive a low-priced automobile. He finds, however, that there are three different ways of obtaining the automobile: (a) Purchase for cash; the price is $13,000. (b) Lease the car; the monthly charge is $350 on a 36month lease, payable at the end of each month; -- 197 ---- ------- - - 198 ANNUALCASH FLOWANALYSIS 6-30 Your company must make a $500,000 balloon payment on a lease 2 years and 9 months from today. You have been directed to deposit an amount of money quarterly, beginning today, to provide for the $500,000 payment. The account pays 4% per year, compounded quarterly. What is the required quarterly deposit? Note: Lease payments are due at beginning of the quarter. 6-31 A manufacturer is considering replacing a production machine tool. The new machine, costing $3700, would have a life of 4 years and no salvage value, but would save the firm $500 per year in direct labor costs and $200 per year indirect labor costs. The existing machine tool was purchased 4 years ago at a cost of $4000. It will last 4 more years and will have no salvage value at the end of that time. It could be sold now for $1000 cash. Assume that money is worth 8% and that the difference in taxes, insurance, and so forth, for the two alternatives is negligible. Use an annual cash flow analysis to determine whether the new machine should be purchased. 6-32 1\vo possible routes for a power line are under study. Data on the routes are as follows: Length First cost Maintenance Useful life, in years Salvage value Yearly power loss Annual property taxes ! ~ t i !I I i I I ji j II ! II I i I !! i tI I; Under the Lake 5km $25,OOO/km $400/km/yr 15 $5000/km $500/km 2% of first cost If 7% interest is used, should the power line be routed around the lake or under the lake? (Answer: Around the lake.) ! I Around the Lake 15km $5000/km $200/km/yr 15 $3000/km $500/km 2% of first cost 6-33 An oil refinery finds that it is now necessary to subject its waste liquids to a costly treating process before discharging them into a nearby stream. The engineering department estimates that the waste liquid processing will have cost $30,000 by the end of the first year. By making process and plant alterations, it is estimated that the waste treatment cost will decline $3000 each year. As an alternate, a specialized firm, Hydro-Clean, has offered a contract to process the waste liquids for 10 years for a fixed price of$15,000 per year, payable at the end of each year. Either way, there should be no need for waste treatment after 10 years. The refinery manager considers 8% to be a suitable interest rate. Use an annual cash flow analysis to determine whether he should accept the Hydro-Clean offer. 6-34 Bill Anderson buys an automobile every 2 years a! follows: initially he makes a down payment of $600C on a $15,000 car. The balance is paid in 24 equa: monthly payments with annual interest at 12%. Wher he has made the last payment on the lo~n, he trade! in the 2-year-old car for $6000 on a new $15,000 car and the cycle begins over again. Doug Jones decided on a different purchase plan He thought he would be better off if he paid $15,OOC cash for a new car. Then he would make a monthly de. posit in a savings account so that, at the end of2 years he would have $9000 in the account. The $9000 plm the $6000 trade-in value of the car will allow Dou~ to replace his 2-year-old car by paying $9000 for, new one. The bank pays 6% interest, compoundec quarterly. . . (a) What is Bill Anderson's monthly payment to pa) off the loan on the car? (b) After he purchased the new car for cash, ho" much per month should Doug Jones deposit ii his savings account to have sufficient money fo] the next car two years hence? (c) Why is Doug's monthly savings account deposi smaller than Bill's payment? ' 6-35 Alice White has arranged to buy some home record. ing equipment. She estimates that it will have a 5-ye3.J useful life and no salvage value. The dealer, who is , friend, has offered Alice two alternative ways to pa~ for the equipment: (a) Pay $2000 immediately and $500 at the end 0: one year. (b) Pay nothing until the end of 4 years, when a singh payment of $3000 must be made. Alice believes 12% is a suitable interest rate Use an annual cash flow analysis to determine whicl method of payment she should select. (Answer: Select (b» 6-36 Two mutually exclusive alternatives are being con sidered. Year o I 2 3 4 5 A -$3000 +845 +845 +845 +845 +845 B -'$5000 +1400 + 1400 + 1400 +1400 +1400 ;' ,,\. 'l :t LL Problems One of the alternatives must be selected. Using a 15% nominal inteJ:est rate, compounded continuously, determine which one. Solve by annual cash flow analysis. 6-37 6-40 6-41 6-38 A suburban taxi company is considering buying taxis with diesel engines instead of gasoline engines. The cars average 50,000 km a year, with a useful life of 3 years for the taxi with the gas engine and 4 years for the diesel taxi. Other comparative information is as follows: 6-42 Diesel Gasoline $13,000 $12,000 48~ 51~ Fuel cost per liter 35 28 Mileage, in km/liter Annual repairs $ 300 $ 200 500 500 Annual insurance premil1;In. . .. End-of-useful-life resai.; v~""" 2,000 3,000 Vehicle cost A company must decide whether to buy Machine A or Machine B: Machine B $20,000 10 $10,000 1,000 $10,000 $50 15 10 $180 60 5 Consider the following two mutually exclusive alternatives: A B Cost $100 $150 Uniform annual benefit 16 24 Useful life, in years 00 20 Some equipment will be installed in a warehouse that a firm has leased for 7 years. There are two alternatives: A B Cost $100 $150 Uniform annual benefit 55 61 Useful life, in years 3 4 At any time after the equipment is installed, it has no salvage value. Assume that Alternatives A and B will be replaced at the end of their useful lives by identical equipment with the same costs and benefits. For a 7-year analysis period and a 10% interest rate,: use an annual cash flow analysis to determine which alternative should be selected. Use an annual cash flow analysis to determine the more economical choice if interest is 6%. Machine A $10,000 4 B Alternative B may be replaced with an identical item every 20 years at the same $150 cost and will have the same $24 uniform annual benefit. Using a 10% interest rate, and an annual cash flow analysis, determine which alternative should be selected. Which machine would you choose? Base your answer on annualcost. (Answers:X = $1252; Y = $1106) Initial cost Usefullife, in years End-of-useful-life salvage value Annual maintenance A The analysis period is 10 years, but there will be ~o replacement for Alternative B at the end of 5 years. Based on a 15% interest rate, determine which alternative should be selected. Use an annual cash flow analysis in working this problem. Machine X Machine Y First cost $8000 $5000 12 5 Estimated life, in years o $2000 Salvage value Annual maintenance cost 150 o 6-39 Consider the following alternatives: Cost Uniform annual benefit Useful life, in years A certain industrial firm desires an economic analysis to determine which of two different machines should be purchased. Each machine is capable of performing the same task in a given amount of time. Assume the minimum attractive return is 8%. The following data are to be used in this analysis: 199 6-43 A pump is required for 10 years at a remote location. The pump can be driven by an electric motor if a power line is extended to the site. Otherwise, a gasoline engine will be used. Use an annual cash flow analysis to determine, based on the following data and a 10% interest rate, how the pump should be powered. Gasoline Electric First cost $2400 $6000 Annual operating cost 1200 750 Annual maintenance 300 50 Salvage value 300 600 Life, in years 5 10 o At a 10% interest rate, which machine should be installed?Use an annual cash flowanalysisin working this problem. (Answer:Machine A) - -- -- 200 ANNUAL CASH FLOW ANALYSIS 6-44 The town of Dry Gulch needs an additional supply of water from Pine Creek. The town engineer has selected two plans for comparison: a gravity plan (divert water at a point 10 miles up Pine Creek and carry it through a pipeline by gravity to the town) and a pumping plan (divert water at a point closer to the town and pump it to the town). The pumping plant would be built in two stages, with half-capacity installed initially and the other half installed 10 years later. 6-47 Carp, Inc. wants to evaluate two methods of shippinJ their products. The following cash flows are associ ated with the alternatives: First cost Maintenance and operating costs + Cost gradient (begin Year 1) Annual benefit Salvage value Useful life, in years An analysis will assume a 40-year life, 10% interest, and no salvage value. Costs are as follows: Initial investment Additional investment in 10th year Operation and maintenance Power cost Average first 10 years Average next 30 years . . Gravity Pumping $2,800,000 $1,400,000 None 200,000 1O,OOO/yr None None 25,000/yr First cost Annual maintenance Salvage value Service life, in years Masonite $250 20 10 4 Brick $1000 10 100 20 Tire Warranty (months) 12 24 36 48 First cost Maintenance and operating costs Annual benefit Salvage value Useful life, in years Machine B $25,000 400 13,000 6,000 10 Assume an interest rate of 12%. Use annual cash flow analysisto determine which machine should be selected. 154,000 142,000 10 .303,000 210,000 20 Price per Tire $39.95 59.95 69.95 90.00 determine whichtire he should buy. . 6-49 Consider the following three mutually exclusiv( alternatives: Cost Uniform annual benefit Useful life, in years A $100 10 00 B $150.00 17.62 20 C $200.00 55.48 5 Assuming that Alternatives B and C are replaced wid: identical replacements at the end of their useful lives. and an 8% interest rate, which alternative should b( selected? Use an annual cash flow analysis in workin~ this problem. (Answer: Select C) 6-50 A new car is purchased for $12,000 with a 0% down. .IlL 8,000 3,000 7 +750/yr The student feels that the warranty period is a gooc estimate of the tire life and that a 10% interest ran is appropriat~. Use an annual cash flow analysis tc 6-46 The manager in a canned food processing plant is trying to decide between two labeling machines. Their respective costs and benefits are as follows: Machine A $15,000 1,600 +900/yr 6-48 A college student has been looking for a new tire fo his car and has located the following alternatives: 50,000/yr 100,000/yr Uncle Elmo needs to replace the family privy. The local sanitary engineering firm has submitted two alternative structural proposals with respective cost estimates as shown. Which construction should Uncle Elmo choose if his minimum attractive rate of return is 6%. Use both a present worth and annual cost approach in your comparison. B $1,700,000 29,000 Use an interest rate of 15% and annual cash flow anal ysis to decide which is the most desirable alternative Use an annual cash flow analysis to determine the more economical plan. 6-45 A $700,000 18,000 9% loan. The loan's length is 4 years. After making 30 payments, the owner desires to payoff the loan'~ remaining balance. How much is owed? 6-51 A year after buying her car, Anita has been offered JIL ajob in Europe. Her for car60 loan is for If$15,000 a 9% nominal interest rate months. she can atsell the car for $12,000, how much does she get to keep after paying off the loan? __n.. - . - ........ .. ... Problems 201 6-52 A $78,000 mortgage has a 30-year term and a 9% 6-55 A 30-year mortgage for $145,000 is issued at a 6% (a) What is the monthly payment? (b) After the first year of payments, what is the outstanding balance? (c) How much interest is paid in month 13? How much principal? -,.. (a) (b) How long does it take to payoff the mortgage,if $1000 per month is paid? (c) How long does it take to payoff the mortgage, if 20% extra is paid each month? A nominal interest rate. A nominal interest rate. What is the monthly payment? 6-53 A $92,000 mortgage has a 30-year term and a 9% 6-56 Solve Problem 6-32 for the breakeven first cost per -. nominal interestrate. (a) What is the monthly payment? (b) After the first year of payments, what fraction of the loan has been repaid? (c) After the first 10 years of payments, what is the outstanding balance? . (d) How much interest is paid in Month 25? How much principal? 1 A 30-year mort~~~e for $95,000 is issued at a 9% JI. . . nominal interest rate. (a) What is the monthly payment? (b) How long does i!~e to pay off the mortgage, if $1000 per month is,paid? (c) How long does it take to pay off the mortgage, if double payments are made? AIL kilometer of going under the lake. 6-57 Redo Problem 6-38 to calculate the EUAW of the _ as a function of miles driven per year to - alternatives see if there is a crossover point in the decision process. .. Graph your results. 6-58 Set up Problem 6-18 on a spreadsheet to make all the A input variable andbailer determine various scenarios which data would make the economical. 6-59 Develop a spreadsheet to solve Problem 6-44. What A is the breakeven cost of the additional primping investment in year 10? ~"--- ----------- -- - -" I I' 'I I l,.. ~ _/:>~~;J~~> After Completing This Chapter... The student should be able to: Evaluate project cash flowsusing the internal rate of return (IRR) methodology. .. .. Plot a project's present worth (PW) against the interest rate. Use an incremental rate of return analysis to evaluate competing alternat:ives. Develop and use spreadsheets to make IRR and incremental rate of ret~ calculatio~ .QUESTIONSTO CONSIDER 1. Imagine that you are a manufacturer who is consideringusing RFID chips to track Y01 products. What cost andbenefit factorswouldyou considerin decidingw.hetherto embe chips in your pr9ducts, and how much to invest in the necessary techD.ology? 2. Now assume that you are a retailer who is also considering the RFID technology.Ho' wouldyour cost and benefit considerations differ from those of a product manufacture} What specificfactors would you consider? What if the manufacturers you buy producl from have already adopted the technology? How would this affect your decision? 3. Since October 2000, the U.S. military has been using RFID technology in its Joint Tou AssetVisibility(ITAV) program, whichtracksequipmentand supplies.ITAVhas reduce. duplicate orders, helped speed up troop deployment, and allowed mi~tary planners to accomplish missions with fewer ships, planes, and ground vehicles. How is the ITA' program likely to affect the pace ofRFID's adoptionby priva~ industry? 4. Some consumers have already expressed concern about RFID chips, calling them aJ invasion of privacy.What ethical and social concerns are raised by RFID chips? 5. In response to consumer worries, RFID chip manufacturers are developing "kill" tech nologiesthat can allowconsumersto disable RFIDchips after they havepurchasedgood: containing these miniature tracking devices. How might this affect consumer attitudes', I_ -.:0 --- - -- --- -- -- - - Rate of Return Analysis Big Brother Is Tracking You II Do you ever get the feeling that your closet is spying on you? Well, you may be right. Manufacturers of products ranging from jeans to shampoo are trying out a new technology called radio-frequencyidentification (RFID), which embeds tiny tracking chips in merchandise. The chips can follow goods from the time they leave the plant until you buy them at your local shopping mall-and even after you take them home. RFID chips offer a wealth of advantages to product makers and retailers. They can tell shippers when trucks are delayed, and they can alert store security when products are being stolen. They can also tell managers when stocks of popular products are running low. The chips still suffer from some technical drawbacks, including vulnerability to interference from cell phones and other electronic devices. A bigger drawback,however,is their cost. As of early 2003,RFID chips cost about30 cents each, which makes them too expensive to embed in low-cost products like soap and shaving cream. Moreover,the scanner~required to read the chips currently cost about $1000 each. And companies that want to use the technology must also investin the electronicequipmentnecessary to keep track of the chips. The cost hurdle may be falling soon, though. A major manufacturer of RFID chips says it can produce them for 5 cents each if the customer orders a batch of at least a billion. Even more substantialcost reductions arelikely when the technology becomes more widely adopted. I --- ------ --- - - --- ----- ---- -- ----- 204 RATEOF RETURN ANALYSIS In this chapter we will examine three aspects of rate of return, the third of the three major analysis methods. First, we describe the meaning of "rate of return"; then, the calculation of rate of return is illustrated; finally,rate of return analysis problems are presented. In an appendix to this chapter, we describe difficulties sometimes encountered in attempting to compute an interest rate for cash flows of certain kinds. INTERNAL RATE OF RETURN . i In Chapter 3 we examined four plans to repay $5000 in 5 years with interest at 8% (Table 3-1). In each of the four plans the amount loaned ($5000) and the loan duration (5 years) were the same. Yet the total interest paid to the lender varied from $1200 to $2347, depending on the loan repayment plan. We saw, however, that the lender received 8% interest each year on the amount of money actually owed. And, at the end of 5 years, the principal and interest payments exactly repaid the $5000 debt with interest at 8%. We say the lender received an "8% rate of return." Internal rate of return is defined as the interest rate paid on the unpaid balance of a loan such that the payment schedule makes the unpaid loan balance equal to zero when the final payment is made. Instead of lending money, we might invest $5000 in a machine tool wi~ a 5-year usefullife and an equivalent uniform annual benefit of $1252. An appr<?priatequestion is, What rat~ of return would we receive on this investment? The cash flow would be as follows: Year o 1 2 3 4 5 Cash Flow -$5000 +1252 +1252 + 1252 +1252 +1252 We recognize the cash flow as Plan 3 of Table 3-1. We know that five payments of $1252 are equivalentto a present sum of $5000 when interest is 8%. Therefore, the rate of return on this investment is 8%. Stated in terms of an investment, we may define internal rate of return as follows: Internal rate of return is the interest rate earned on the unrecovered investment such that the payment schedule makes the unrecovered investment equal to zero at the end of the life of the investment. It must be understoodthat the 8% rate of return does not mean an annual return of 8% on the $5000 investment, or $400 in each of the 5 years with $5000 returned at the end of Year5. Instead, each $1252 payment represents an 8% return on the unrecovered investmentplus - --- 205 Calculating Rateof Return the partial return of the investment.This may be tabulated as follows: Year 0 1 2 3 4 5 Cash Flow -$5000 +1252 +1252 +1252 +1252 +1252 Unrecovered Investment at Beginning of Year 8% Return on Unrecovered Investment Investment Repayment at End of Year Unrecovered Investment at End of Year $5000 4148 3227 2233 1159 $ 400 331 258 178 - 93 $ 852 921 994 1074 - 1159 $5000 $4148 3227 2233 1159 0 $1260 This cash flow represents a $5000 investmentwith benefitsthat produce an 8% rate of returnon the unrecoveredinvestment. . . Althoughthetwodefinitionsofinternalrateofreturnarestateddifferently,oneinterms . of a loan and the other in terms of an investment, there is only one fundamental concept being described. It is that the int~rnal rate of return is the interest rate at which the benefits are equivalent to the costs or the present worth (PW)is O.Since we are describing situations of funds that remain within the investment throughout its life, the resulting rate of return is described as the internal rate of return, i. CALCULATING RATE OF RETURN To calculate a rate of return on an investment, we must convert the various consequences of the investment into a cash flow. Then we will solve the cash flow for the unknown value of the internal rate of return (IRR). Five forms of the cash flow equation are as follows: PW of benefits - PW of costs = 0 (7-1) =1 (7-2) Net present worth = 0 (7-3) =0 (7-4) = PW of benefits (7-5) PW of benefits PW of costs EUAB - EUAC PW of costs t The five equations represent the same concept in different forms. They can relate costs and benefits with the IRR as the only unknown. The calculation of rate of return is illustrated by the following examples. - -- .- - -- -- 206 RATEOF RETURN ANALYSIS An $8200 investment returned $2000 per year over a 5-year useful life. What was the rate of return on the investment? Using Equation 7-2, we write PW of benefits =1 PW of costs 2000(P I A, i, 5) 8200 =1 Rewriting, we see that 8200 (PIA, i, 5) = 2000 = 4.1 Then look at the compound interest tables for the value of i where (PIA, i, 5) = 4.1; if no tabulated value of i gives this value, we will then find values on either side of the desired value (4.1) and interpolate to find thelM. From interest tables we find: i (P/A,i,S) 6% 4.212 7% 4.100 8% 3.993 In this example, no interpolation is needed because the internal rate of return is exactly 7%. An investmentresulted in the following cash flow.Compute the rate,of return. . Year Cash Flow o 1 2 3 4 -$700 + 100 +175 + 250 +325 EUAB - EUAC = 0 100 + 75(AI G,i, 4) -- 700(AI P, i, 4) =0 In this situation, we navetwo differentinterestfactors in the equation.We wilLnotbe able to solve '"" i it as easily as ExaIllple 7-1. Since there is no convenient direct method of so!ution, we will solve !"! "'... .a . . -- --- -- Calculating Rate of Return the equatiop by trial and error. Try i 207 = 5% first: EUAB - EUAC = 0 100+ 75(AjG, 5%, 4) -700(Aj P, 5%, 4) = 0 100 + 75(1.439) - 700(0.2820) =0 At i = 5%, EUAB- EUAC = 208 -197 = +11 The EUAC is too low.If the interest rate is increased,EUAC will increase. Try i = 8%: EUAB - EUAC = 0 100+ 75(AjG, 8%,4) -700(Aj P, 8%,4) = 0 100 + 75(1.404) - 700(0.3019) Ati --I =0 = 8%, EUAB - EUAC = 205 - 211 =-6 -I' This time the EUAC is too large. We see that the true rate of return is between 5% and 8%. Try i =7%: EUAB - EUAC = 0 100+ 75(AjG, 7%, 4) -700(Aj P, 7%, 4) = 0 =0 100 + 75(1.416) - 700(0.2952) At i = 7%, EUAB -EUAC -.:...206 - 206 =0 The IRR is 7%. Calculate the rate of return on the investment on the following cash flow. Year o 1 2 3 4 5 Cash Flow -$100 +20 +30 +20 +40 +40 = - - -- - -- -~- b iiii..._ - ---- - -- --- - -- 208 RATEOF RETURN ANALYSIS Using a net present worth of 0, try i NPW = -100 = 10%: + 20(P / F, 10%, 1) + 30(P / F, 10%,2) + 20(P / F, 10%,3) + 40(P / F, 10%,4) + 40(P / F, 10%,5) = -100 + 20(0.9091) + 30(0.8264) + 20(0.7513) + 40(0.6830) + 40(0.6209) = -100 + 18.18 + 24.79 + 15.03 + 27.32 + 24.84 = -100 + 110.16 = +10.16 The trial interest rate i is too low.Select a second trial, i NPW = = 15%: -100 + 20(0.8696) + 30(0.7561) + 20(0.6575) + 40(0.5718) + 40(0.4972) = -100 + = -100 + = -4.02 17.39 + 22.68 + 13.15 + 22.87 + 19.89 95.98 FIGURE 7-1 Plot of NPW versus interest rate i. 15 5% These 10 and 15% points areplotted in Figure 7-1. By linear interpolation we compute the rate of return as follows: . IRR = 10% + (15% - 10%) ( 10.16 .. 10.16 + 4.02 )- 13.5% We can prove that the rate of return is very close to 131/2%by showing that the unrecoveredinvestment is very close to zero at the end of the life of the investment. f' - - - - - -- - - - Problems 6-52 A $78,000 mortgage has a 30-year term and a 9% -. nominal intere.strate. (a) What is the monthly payment? 6-53 A $92,000 mortgage has a 30-year term and a 9% nominal interest rate. (a) What is the monthly payment? (b) After the first year of payments, what fraction of the loan has been repaid? (c) After the first 10 years of payments, what is the outstandingbalance? 6-55 A 30-year mortgage for $145,000 is issued at a 6% . (b) After the first year of payments, what is the outstanding balance? (c) How much interest is paid in month 13? How much principal? JI. _ 201 .. . nominal interest rate. (a) What is the monthly payment? (b) How long does it take to payoff the mortgage, if $1000 per month is paid? (c) How long does it take to payoff the mortgage, if 20% extra is paid each month? 6-56 Solve Problem 6-32 for the breakeven first cost per kilometer of going under the lake. &:IiiiIiiii8 6-57 Redo Problem 6-38 to calculate the EUAW of the all. alternativesas a function of miles driven per year to see if there is a crossover point in the decision process. Graph your results. . (d) How much interest is paid in Month 25? How much principal? 1 A 30-year mortgage for $95,000 is issued at a 9% 6-58 Set up Problem 6-18 on a spreadsheet to make all the . A input variable andbailer detennine various scenarios which data would make the economical. 6-59 Develop a spreadsheet to solve Problem 6-44. What JII.. nominal interest rate. (a) What is the monthly payment? (b) How long does it take to pay off the mortgage, if $1000 per month is paid? (c) How long does it take to payoff the mortgage, if double payments are made? is the breakeven cost of the additional plimping investment in year 10? = _.--- --- --- L 'i~;'''' ,1; II .:W. -", _.,.",,~~<,.-~-_. ::z.."~'" ---- After Completing This Chapter... The student should be able to: Evaluate project cash flowsusing the internal rate of return (IRR) methodology. Plot a project's present worth (PW) against the interest rate. ·. .. Usean incrementalrateof returnanalysisto evaluatecompetingalternatives. Develop and use spreadsheets to make IRR and incremental rate of ret~ calculatio~ QUESTIONS TO CONSIDER 1. Imagine that you are a manufacturer who is consideringusing RFID chips to track YOI products.What cost and benefitfactors wouldyouconsiderin decidingwhetherto embt chips in your products, and how much to invest in the necessary technology? 2. Now assume that you are a retailer who is also considering the RFID technology.Ho' would your cost and benefit considerations differfrom those of a product manufactureI What specific factors would you consider? What if the manufacturersyou buy product' from have already adopted the technology? How would this affect your decision? 3. Since October 2000, the U.S. military has been using RFID technology in its Joint Tou AssetVisibility(ITAV) program, whichtracksequipmentand supplies.ITAVhas reduce duplicate orders, helped speed up troop deployment, and allowed mi~tary planners t, accomplish missions with fewer ships, planes, and ground vehicles. How is the ITA' program likely to affect the pace qf RFID's adoptionby priva~ industry? 4. Some consumers have already expressed concern about RFID chips, calling them aJ invasion of privacy.What ethical and social concerns are raised by RFID chips? 5. In response to consumer worries, RFID chip manufacturers are developing "kill" tech nologiesthat can allowconsumersto disableRFID chips afterthey havepurchasedgood: containing these miniature tracking devices. How might this affect consumer attitudes~ ~ """"" I - --- ~_ ,_ok_-' - ._.,- -- Rate of Return Analysis Big Brother Is Tracking You " Do you ever get the feeling that your closet is spying on you? Well, you may be right. Manufacturers of products ranging from jeans to shampoo are trying out a new technology called radio-frequency identification (RFID), which embeds tiny tracking chips in merchandise. The chips can follow goods from the time they leave the plant until you buy them at your local shopping mall-and even after you take them home. RFID chips offer a wealth of advantagesto product makers and retailers. They can tell shippers when trucks are delayed, and they can alert store security when products are being stolen. They can also tell managers when stocks of popular products are running low. The chips still suffer from some technical drawbacks, including vulnerability to interference from cell phones and other electronic devices. A bigger drawback,however,is their cost. As of early 2003,RFID chips cost about30 cents each, which makes them too expensive to embed in low-cost products like soap and shaving cream. Moreover,the scanner~required to read "'1 '. the chipscurrentlycost about$1000each.And companies that want to use the technology must also investin the electronicequipmentnecessary to keep track of the chips. The cost hurdle may be falling soon, though. A major manufacturer of RFID chips says it can produce them for 5 cents each if the customer orders a batch of at least a billion. Even more substantialcost reductions arelikely when the technology becomes more widely adopted. - - -- - - --- - -- --- -- -- 204 RATEOF RETURN ANALYSIS In this chapter we will examine three aspects of rate of return, the third of the three major analysis methods. First, we describe the meaning of "rate of return"; then, the calculation of rate of return is illustrated; finally, rate of return analysis problems are presented. In an appendix to this chapter, we describe difficulties sometimes encountered in attempting to compute an interest rate for cash flows of certain kinds. INTERNAL RATE OF RETURN In Chapter 3 we examined four plans to repay $5000 in 5 years with interest at 8% (Table 3-1). In each of the four plans the amount loaned ($5000) and the loan duration (5 years) were the same. Yet the total interest paid to the lender varied from $1200 to $2347, depending on the loan repayment plan. We saw, however, that the lender received 8% interest each year on the amount of money actually owed. And, at the end of 5 years, the principal and interest payments exactly repaid the $5000 debt with interest at 8%. We say the lender received an "8% rate of return." Internal rate of return is defined as the interest rate paid on the unpaid balance of a loan such that the payment schedule makes the unpaid loan balance equal to zero when the final payment is made. Insteadof lendingmoney,we might invest$5000in a machinetool wit;ha 5-yearuse- . , ; , fullife and an equivalent uniform annual benefit of $1252. An apprqpriate question is, What rate of return would we receive on this investment? The cash flow would be as follows: Year o 1 2 3 4 5 Cash Flow -$5000 +1252 +1252 +1252 +1252 +1252 We recognize the cash flow as Plan 3 of Table 3-1. We know that five payments of $1252 are equivalentto a present sum of $5000 when interest is 8%. Therefore, the rate of return on this investment is 8%. Stated in terms of an investment, we may define internal rate of return as follows: Internal rate of return is the interest rate eamed on the unrecovered investment such that the payment schedule makes the unrecovered investment equal to zero at the end of the life of the investment. It must be understood that the 8% rate of return doesnot mean an annualreturn of 8% on the $5000 investment, or $400 in each of the 5 years with $5000 returned at the end of Year5. Instead, each $1252 payment represents an 8% return on the unrecovered investmentplus 205 CalculatingRateof Return the partial return of the investment.This may be tabulated as follows: Year 0 1 2 3 4 5 Cash Flow -$5000 +1252 + 1252 + 1252 +1252 +1252 Unrecovered Investment at Beginning of Year 8% Return on Unrecovered Investment Investment Repayment at End of Year Unrecovered Investment at End of Year $5000 4148 3227 2233 1159 $ 400 331 258 178 - 93 $1260 $ 852 921 994 1074 - 1159 $5000 $4148 3227 2233 1159 0 This cash flow represents a $5000 investmentwith benefits that produce an 8% rate of returnon the unrecoveredinvestment. . Althoughthetwodefinitionsofinternalrateofreturnarestateddifferently,oneinterms . of a loan and the other in terms of an investment, there is only one fundamental concept being described. It is that the int~rnal rate of return is the interest rate at which the benefits are equivalent to the costs or the present worth (PW)is O.Sincewe are describing situations of funds that remain within the investment throughout its life, the resulting rate of return is described as the internal rate of return, i. CALCULATING RATE OF RETURN To calculate a rate of return on an investment, we must convert the various consequences of the investment into a cash flow. Then we will solve the cash flow for the unknown value of the internal rate of return (IRR). Five forms of the cash flow equation are as follows: PW of benefits - PW of costs = 0 PW of benefits (7-1) =1 (7-2) =0 (7-3) EUAB - EUAC = 0 (7-4) PWof costs Net present worth PW of costs = PW of benefits (7-5) The five equations represent the same concept in different forms.They can relate costs and benefits with the IRR as the only unknown. The calculation of rate of return is illustrated by the following examples. - -- - - -- - - - -- 206 j RATEOF RETURN ANALYSIS I I An $8200 investment returned $2000 per year over a 5-year useful life. What was the rate of return on the investment? J .. . ScpLUTION Using Equation 7-2, we write PW of benefits =1 PW of costs 2000(P I A, i, 5) 8200 =1 Rewriting, we see that 8200 (PIA, i, 5) = -2000 = 4.1 Then look at the compound interest tables for the value of i where (PIA, i, 5) = 4.1; if no tabulated value of i gives this value, we will then find values on either side of the desired value (4.1) and interpolate to find the IRR. From interest tables we find: i 6% 7% 8% (PIA, i, s) 4.212 4.100 3.993 In this example, no interpolation is needed because the internal rate of return is exactly 7%. . An investmentresulted in the following cash flow.Compute the rate.of return. Year Cash Flow o 1 2 3 4 -$700 + 100 +175 + 250 +325 ..~OLUTION EUAB - EUAC = 0 100+ 75(AIG, i, 4) -700(AI P, i, 4) = 0 In this situation, we have two different interestfactors in the equation. WewilLnotbe able to solve it as easily as Example 7-1. Since there is no convenientdirect method of solution, we will solve --- - - - -- Calculating Rate of Return the equatiotl by trial and error. Try i 207 = 5% first: EUAB - EUAC =0 100 + 75(AfG, 5%, 4) -700(Af P, 5%, 4) = 0 100 + 75(1.439) -700(0.2820) =0 Ati = 5%, EUAB - EUAC = 208 -197 = +11 The EUAC is too low.If the interest rate is increased, EUAC will increase. Try i = 8%: EUAB - EUAC = 0 100+ 75(AfG, 8%,4) -700(Af P, 8%,4) = 0 100 + 75(1.404) - 700(0.3019) = 0 Ati = 8%, EUAB - EUAC = 205 - 211 = -6 This time the EUAC is too large. We see that the true rate of return is between 5% and 8%. Try i =7%: EUAB - EUAC = 0 100 + 75(AfG, 7%, 4) -700(Af P, 7%, 4) = 0 100 + 75(1.416) - 700(0.2952)= 0 At i = 7%, EUAB- EUAC- 206- 206 = 0 The IRR is 7%. ~ Calculate the rate of return on the investmenton the following cash flow. Year o 1 2 3 4 5 Cash Flow -$100 +20 +30 +20 +40 +40 ,.....- - - - . - -. - -- -- 208 RATEOF RETURN ANALYSIS Using a net present worth of 0, try i NPW = = 10%: -100 + 20(P / F, 10%, 1) + 30(P / F, 10%,2) + 20(P / F, 10%,3) + 40(P / F, 10%,4) + 40(P / F, 10%,5) = -100 + 20(0.9091) + 30(0.8264) + 20(0.7513) + 40(0.6830) + 40(0.6209) = -100 + 18.18 + 24.79 + 15.03 + 27.32 + 24.84 = -100 + 110.16 = +10.16 The trial interest rate i is too low. Select a second trial, i NPW = -100 = 15%: + 20(0.8696) + 30(0.7561) + 20(0.6575) + 40(0.5718) + 40(0.4972) = -100 + 17.39 + 22.68 + 13.15 + 22.87 + 19.89 = -100 + 95.98 = -4.02 FIGURE 7-1 Plot of NPW versus interest rate i. IS '€ 10 ~ fi '" S ~ Q. V Z 0 S% ...:..S These 10 and 15% points are plotted in Figure 7-1. By linear interpolation we compute the rate of return as follows: , i IRR = 10% + (15% -10%) - .. . 10.16 10.1 + 4.02 ( .. 6.. ) -13.5% We can prove that the rate of return is very close to 131/2%by showing that the unrecoveredinvestment is very close to zero at the end of the life of the investment. -- 209 Calculating Rate of Return Year Cash Flow Unrecovered Investment at Beginning of Year 0 1 2 3 4 5 -$100 +20 +30 +20 +40 +40 $100.0 93.5 76.1 66.4 35.3 13% % Return on Unrecovered Investment Investment Repayment at End of Year Unrecovered Investment at End of Year $13.5 12.6 10.3 8.9 4.8 $ 6.5 17.4 9.7 31.1 35.2 $93.5 76.1 66.4 . 35.3 0.1* *This small unrecovered investment indicates that the rate ofreturn is slightly less than 131/2%. If in Figure 7-1 net present worth (NPW) had been computed for a broader range of values of i, Figure 7-2 would have been obtained. From this figure it is apparent that the error resulting from linear interpolation increases as the interpolationwidth increases. +50 '€ cu '" 0 £- -50 FIGURE 7-2 Replot ofNPW versus interest rate i over a larger range of values. Plot of NPW versus Interest Rate i Figure 7-2-the second.plot of NPW versus interest rate i-is an important source of information. For.a cash flow representing an investment followed by benefits from the investment, the plot ofNPW versus i (we will call it an NPW plot for convenience)would have the form of Figure 7-3. Year 0 1 2 3 4 Cash Flow -p +Benefit A +A +A +A + i=: 0 .. .~ FIGURE 7-3 NPW plot for a typical investment. - - - - -- - -- -- r, "'" 210 RATEOF RETURN ANALYSIS If, on the other hand, borrowed money was involved,the NPW plot would appear as in Figure 7-4. This form of cash flowtypically results when one is a borrower of money. In such a case, the usual pattern is a receipt of borrowed money early in the time period with a later repayment of an equal sum, plus payment of interest on the borrowed money. In all cases in which interest is charged, the NPW at 0% will be negative. Cash Flow +p - Repayment A -A -A -A Year 0 1 2 3 4 + 0 FIGURE 7-4 TYPicalNPW plot for borrowed money. How do we determine the interestrate paid by the borrower in this situation'l~ically we would write an equation, such as PW of income = PW of disbursements,aild solve for the unknown IRR. Is the resulting IRR positive or negative from the borrower's point of view? If the lender said he was receiving, say, + 11% on the debt, it might seem reasonable to state that the borrower is faced with -11 % interest. Yet this is not the way interest is calculated. A banker says he pays 5% interest on savings accounts and charges 11% on personal loans. Both rates are positive. Thus, we implicitly recognize interest as a charge for the use of someone else's money and a receipt for letting others use our money.In determining the interest rate in a particular situation, we solve for a single unsigned value of it. We then view this value of i in the customary way, that is, as either a charge for borrowing money or a receipt for lending money. A new corporate bond was initially sold by a stockbroker to an investor for $1000. The issuing corporation promised to pay the bopdholder $40 interest on the $1000 face value of the bond every 6 months, and to repay the $1000 at the end of 10 years. After one year the bond was sold by the original buyer for $950. (a) ~at.rate of.,J:etl1lJ1d,!dJ!1e or!ginaJ.buy~rreceiv~ on pisjnve§.tment? (b) What rate ofreturn can the newbuyer (paying $950) expect to receive if he keeps the bond for its remaining 9-year life? ii..E.__ a - -- -- __ _ _ - - -- -- - -- - - -- -- - - -- _ :1 a...~ .,. . Calculating Rate of Return 211 950 40J 0-1-2 t t 1 1000 Since $40 is received each 6 months, we will solve the problem using a 6-month interest period. Let PWof cost = PW of benefits, and.write 1000 = 40(P / A, i, 2) + 950(P / F, i, 2) Try i = 11/2%: 1000 ~ 40(1.956) + 950(0.9707) = 78.24 + 922.17 ~ 1000.41 Theinterestrateper6 months,~ mon, is verycloseto I1h%.Thismeansthenominal(annual) interest rate is 2 x 1.5% = 3%. The effective (annual) interest rate = IRR = (1 + 0.015)2 1 = 3.02%. SOUhION'TO - b 1000 A =40 t t t t t t t t t t t t. t t t t I-l-2-3-4-5 J t t 6--7-8--:-:::-IO~II-12-\3~14-15~16-17~18 950 Given the same $40 semiannual interest payments, for 6-month interest periods we write 950 = 40(P / A, i, 18) + IOOO(P/ F, i, 18) Try i =5%: 950 ~ 40(11.690) + 1000(0.4155) = 467.60+415.50 ~883.10 The PW of benefits is too low. Try a lower interest rate, say, i -.". '950 ~40(12.659) ~ 999.96 - - = 4%: 4- 1000(0.4936)~'''506.3<1+~93.60~ - == ~ -- - --.... - ---------- 212 RATEOF RETURNANALYSIS The valUeof the IRR is between 4 and 5%. By interpolation, IRR 4% t (1%) !he nominal interestrate is 2 x 4.43 % ' 9.05%. 999.96 -950.00 999.96 -883.10 ( ) .. . 4.43% 8.86%. Tbe ef(ective interestra.te is (1+ 0.0443)L.,. 1 . RATE OF RETURN ANALYSIS Rate of return analysis is probably the most frequently used exact analysis technique' in industry. Although problems in computing rate of return sometimes occur, its major advantageoutweighs the occasionaldifficulty.The major advantageis that we can compute a single figure of merit that is readily understood. Consider these statements: ·· · The net present worth on the project is $32,000. The equivalent uniform annual net benefit is $2800. The project will produce a 23% rate of return. While none of these statements tells ~e complete story, the third one gives a measure of desirability of the project in terms that are wide.1yunderstood. It is this acceptance by engineers and business leaders alike of rate of return that has promoted its more frequent use than present worth or annual cash flow methods. There is another advantage to rate of return analysis. In both present worth and annual cash flow calculations, one must select an interest rate for use in the calculations-and this may be a difficult and controversialitem. In rate of return analysis, no interest rate is introducedinto the calculations(exceptasdescribedin Appendix7A).Instead,we computea rate of return (more accurately called internal rate of return)from the cash flow.To decide how,to proceed, the calculated rate of return is compared with a preselected minimum attractive rate of return, or simply MARR. This is the same value of i used for present worth and annual cash flow analysis. When there are two alternatives, rate of return analysis is performed by computing the incremental rate of return-~IRR-on the difference between the alternatives. Since we want to look at increments of investment,the cash flow for the difference between the alternativesis computed by taking the higher initial-cost alternativeminus the lower initialcost alternative. If ~IRR is the same or greater than the MARR, choose the higher-cost alternative.If .6.IRRis less than the MARR, choose the lower-costalternative. Two-Alternative Situation .6.IRR~ MARR ~IRR < MARR Decision Choose the higher-cost alternative Choose the lower-costalternative Rate of return and incremental rate of return analysis are illustrated by Examples 7-5 through 7-8.' ------ --- - - - - -. -- ... - -. . - - _.. -- - -- - - _._- Rate of Return Analysis 213 If an electromagnet is installed on the input conveyor of a coal-processing plant, it will pick up scrap metal in the coal. The removal of this metal will save an estimated $1200 per year in costs associated with machinery damage due to metal. The electromagnetic equipment has an estimated useful life of 5 years and no salvage value. Two suppliers have been contacted: Leaseco will provide the equipment in return for three beginning-of-year annual payments of $1000 each; Saleco will provide the equipment for $2783. If the MARR is 10%, which supplier should be selected? SOLUTION Since both suppliers will provide equipment with the same useful life and benefits, this is a fixedoutput situation. In rate of return analysis, the method of solution is to examine the differences between the alternatives.By taking Saleco - Leaseco, we obtain an increment of inves-tment.. Year 0 1 2 3 4 5 Compute Leaseco -$1000 1200 { +-1000 1200 {+ -1000 + 1200 +1200 + 1200 Saleco -$2783 Difference Between Alternatives: Saleco Leaseco -$1783 - + 1200 +1000 + 1200 + 1000 . 0 0 0 + 1200 + 1200 +1200 the NPW at various interest rates on the increment of investment represented by the difference between the alternatives. ,. ..' pw* Year n CashFlow: Saleco-Leaseco At0%. At8% At 20% At 00% 0 1 2 3 4 5 -$1783 + 1000 + 1000 0 0 0 -$1783 + 1000 + 1000 0 0 0 -$173 +926 +857 0 0 0 -$1783 +833 +694 0 0 0 -$1783 0 0 0 0 0 -256 -1783 NPW= +217 *Each year the cash flow is multiplied by (P / F, i, n). 0 .,., -' .........,.g " At O%:(P / F, 0%, n) = 1 for all values of n At 00%: (P / F, 00%,0) = 1 (Pj F, 00%, n) = 0 for all other values of n ---.- - --- - - - - ---'- ---- - - ----- " 214 RATEOF RETURN ANALYSIS From the plot of these data in Figure 7-5, we see that NPW = 0 at i = 8%. 217 '€ ~ i:$ G) '" ~ Q.., ... G) Z -1783 FIGURE 7-5 NPW plot for Example 7-5. Thus, the incrementalrate of return-b.IRR--of is less than the 10% MARR. Select Leaseco. selectingSalecorather than Leaseco is 8%.This Youmust select one of two mutually exclusive alternatives.(Note:Engineering economists often use the term "mutually exclusivealternatives" to emphasizethat selecting one precludes selecting any other alternative.)The alternatives are as follows: Year o 1 Alt. 1 -$10 +15 Alt. 2 -$20 +28 Any money not invested here may be invested elsewhere at the MARR of 6%. If you can choose only one alternativeone time, which one would you select using the internal rate of return (lRR) analysis method? :.~9!-qTION Using the IRR analysis method, we will select the lesser-cost alternative(Alt. 1), unless we find that the additional cost of Alt. 2 produces enough additionalbenefitsto make it preferable instead. If we consider Alt. 2 in relation to Alt. 1, then Higher-cost [ Alt. 2 ] = LOWer-cost [Alt. 1 Differencesbetween + Alt. 1 and Alt. 2 ] [ ] or LOWer-cost Differe~cesbetwe;n ~t. 1 and Alt. 2 = Higher-cost _ [Alt. 2 ] [Alt. 1 ] - - - ..-- -- - - - Rate of Return Analysis 215 The choic~ between the two alternatives reduces to an examination of the differences between them. We can compute the rate of return on the differences between the alternatives.Writing the alternatives again, Year o 1 Alt. 1 -$10 '+15 Alt. 2 - Alt. 1 -$20 - (-$10) = -$10 +28 - (+15) = +13 Alt. 2 -$20 +28 PW of cost of differences (Alt. 2 - Alt. 1) = PW of benefit of differences (Alt. 2 - Alt. 1) 10 = 13(PjF, i, 1) Thus, 10 (PjF,i, 1) = -13 =0.7692 One can see that if $10 increases to $13 in one year, the interestrate must be 30%.The compound interest tables confirm this conclusion. The 30% rate of return on the difference between the alternatives is far higher than the 6% MARR. The additional $10 investment to obtain Alt. 2 is superior to investingthe $10 elsewhere at 6%. To obtain this desirable increment of investment, with its 30% rate of return, Alt. 2 is selected. Tounderstandmore about Example 7-6, compute the rate of return for each alternative. Alternative 7 PW of cost of Alt. 1 = PW of benefit of Alt. 1 $10 = $15(P j F, i, 1) Thus, 10 (PjF, i, 1) = - = 0.6667 15 From the compound interest tables: rate of return = 50%. Alternative 2 PW of cost of Alt. 2 $20 = PW = of benefit $28(P of Alt.2 j F, i, 1) I~ - --- - - - - - - -- _. ----- 216 RATEOF RETURN ANALYSIS Thus, . (PjF, 1,1) 20 = -28 = 0.7143 From the compound interest tables: rate of return = 40%. One may be tempted to select Alt. 1, based on these rate of return computations. We have already seen, however, that this is not the correct solution. Solve the problem again, this time using present worth analysis. Present Worth Analysis Alternative 1 NPW= -10+ 15(PjF,6%, 1) = -10+ 15(0.9434)=+$4.15 Alternative 2 NPW = -20 + 28(P j F, 6%, 1) = -20 + 28(0.9434)= +$6.42 Alternative 1 has a 50% rate of return and an NPW (at the 6% MARR) of +$4.15. Alternative 2 has a 40% rate of return on a larger investment,with the result that its NPW (at the 6% MARR) is +$6.42. Our economic criterion is to maximize the return, rather than the rate of return. To maximize NPW, select Alt. 2. This agrees with therate of return analysis on the differencesbetween the alternatives. If the computations for Example 7-6 do not convince you, and you still think Alternative 1 would be preferable, try this problem. You have $20 in your wallet and two alternativeways of lending Bill some money. i. (a) Lend Bill $10 with his promise of a 50% return. That is, he will pay you back $15 at the agreed time. (b) Lend Bill $20 with his promise of a 40% return. He will pay you back $28 at the same agreed time. You can select whether to lend Bill $10 or $20. This is a one-time situation, and any money not lent to Bill will remain in your wallet. Which alternativedo you choose? SOLUTION So you see that a 50% return on the smaller sum is less rewardingfo you tnan 40% on th"elarger sum? Since you would prefer to have $28 than $25 ($15 from Bill plus $10 remaining in your wallet) after the loan is paid, lend Bill $20. - - ;, Rate of Return Analysis 217 :! .... !1'~.~ Solve Example 7-6 again,but this time computethe interestrate on theincrement (Alt. 1 - Alt. 2). How do you interpret the results? This time the problem is being viewed as follows: Alt. 1 = A1t.2 + [Alt. 1 - Alt. 2] Year Alt. 1 -$10 +15 o 1 Alt. 2 -$20 +28 [Alt. 1 - Alt. 2] -$10 - (-$20) = +$10 +15 - (+28) = -13 We can write one equation in one unknown: NPW = PW of benefit of differences - PW of cost of differences = 0 +lO-13(PIF,i, 1) =0 Thus, 10 (PIF, i, 1) = -13 = 0.7692 Once again the interest rate is found to be 30%. The critical question is, what does the 30% represent? Looking at the increment again: Year Alt. 1 - Alt. 2 o 1 1..,..r ." +$10 -13 - The cash flowdoes not representan investment;instead, it represents a loan. It is as if we borrowed $10 in Year 0 (+$10 represents a receipt of money) and repaid it in Year 1 (-$13 represents a disbursement). The 30% interest rate means this is the amount we would pay for the use of the $10 borrowed in Year0 and repaid inYear 1. . Is this a desirable borrowing? Since the MARR on investments is 6%, it is reasonable to assume our maximum interest rate on borrowing would also be 6%. Here the interest rate is 30%, which means the borrowing is undesirable. Since Alt. 1 = Alt. 2 + (Alt. 1 -- Alt. 2), and we do not like the (Alt. 1 - Alt. 2) increment, we should reject Alternative 1, which contains the undesirable increment. This means we should select Alternative 2-the same conclusion reached in EXan1ple7-6. Example 7-8 illustrated that one can analyze either increments of investment or increments of borrowing. When looking at increments of investment, we accept the increment when the incremental rate of return equals or exceeds the minimum attractive rate ''''~ a..II - --- ,.,., --- - - - --- - -- ---- ----------- 218 RATEOF RETURN ANALYSIS of return (~IRR ~ MARR). When looking at increments of borrowing, we accept the increment when the incremental interest rate is less than or equal to the minimum attractive rate of return (~IRR ~ MARR). One way to avoid much of the possible confusion is to organize the solution to any problem so that one is examining increments of investment. This is illustrated in the next example. A firmis considering which of two devicesto install to reduce costs in a particular situation. Both devices cost $1000, and both have useful lives of 5 years and no salvage value. Device A can be expected to result in $300 savings annually. Device B will provide cost savings of $400 the first year but will decline $50 annually,making the second-year savings $350, the third-year savings $300, and so forth. For a 7% MARR, which device should the firm purchase? This problem has been solved by present worth analysis (Example5-1) and annual cost analysis (Example 6-5). This time we will use rate of return analysis.The examplehas fixedinput ($1000) and differing outputs (savings). In determiningwhether to use an (A - B) or (B - A) differencebetween the alternatives,we seek an increment of investment.By looking at both (A - B) and (B - A), we find that (A - B) is the one that represents an increment of investment. Year o 1 2 3 4 5 Device A -$1000 +300 +300 +300 +300 +300 Device B -$1000 +400 +350 +300 +250 +200 Difference Between Alternatives: Device A Device B $0 -100 -50 - o +50 +100 For the difference between the alternatives,write a single equation with i as the only unknown. EUAC = EUAB [100(Pj F, i, 1) + 50(P j F, i, 2)](Aj P, i, 5) [50(Fj P, i, 1) + 100](Aj F, i, 5) The equation is cumbersome, but need not be solved. Instead, we observe that the sum of the costs (-100 and -50) equals the sum of the benefits (+50 and + 100). This indicates that 0% is ili~ IRRA-B on the A - B increment of investment. This is less than the 7% MARR; therefore, the increment is undesirable. Reject Device A and choose Device B. As describedin Example7-8,if theincrel11ent eXal1:).ined is (B ~ A),'the interestratewould again be 0%, indicating a desirable borrowing situation. Wewould choose Device B. , :; =::z' ==;; :::I ... .-- --.-------- Rate of Return Analysis 219 Analysis Period , In discussing present worth analysis and annual cash flow analysis, an important consideration is the analysis period. This is also true in rate of return analysis. The method of solution for two alternativesis to examine the differencesbetween the alternatives. Clearly, the examination must cover the selected analysisperiod. For now, we can only suggest that the assumptionsmade shouldreflectone's perceptionof the future as accuratelyas possible, In Example 7-10 the analysis period is a common multiple of the alternative service lives and identical replacement is assumed. This problem illustrates an analysis of the differences between the alternatives over the analysis period. Two machines are being considered for purchase. If the MARR is 10%, which machine should be bought? Use an IRR analysis comparison. Initial cost Uniform annual benefit End-of-useful-lifesalvage value Usefullife, in years Machine X $200 95 50 6 Machine Y $700 120 150 12 SOU!JTION The solution is based on a 12-year analysisperiod and a replacement machine X that is identical to the present machineX. The cash flowfor the differencesbetween the alternatives,is as follows: Year MachineX Machine Y Machine Y - Machine X 0 1 2 3 4 5 -$200 +95 +95 +95 +95 +95 -$700 +120 +120 +120 +120 +120 -$500 +25 +25 +25 +25 +25 +120 +25 6 I l ii:I;=:; '='= ;;: ::,;; 1:; == I -. +50 +95 +95 9 +95 10 +95 11 +95 .... ...--- +150 { -200 +95' 7 8 12 1:1 - +50 {+95 +120 +25 +25 +25 +25 =-+25 +120 +150 +25 +100 +120 +120 +IW +120 : -,. II; II! I I !II! --- ..- 0; I --- - - - -- - ---------- - - --- 220 RATEOF RETURNANALYSIS PW of cost of differences = PW of benefits of differences 500 = 25(P / A, i, 12) + 150(P/ F, i, 6) + 100(P/ F, i, 12) The sum of the benefits over the 12 years is $550, which is only a little greater than the $500 additional cost. This indicates that the rate of return is quite low. Try i = 1%. 500 ~ 25(11.255) + 150(0.942) + 100(0.887) ~ 281 + 141 + 89 =511 The interest rate is too low.Try i = 11/2%: 500 '!/::25(10.908) + 150(0.914) + 100(0.836) '!/::273 + 137 + 84 = 494 The internal rate of return on the Y - X increment, IRRy-X, is about 1.3%, far below the 19% minimum attractive rate of return. The additional investment to obtain Machine Y yields an. . . " unsatisfactory rate of return, therefore X is the preferred alternative. SPREADSHEETS AND RATE OF RETURN ANALYSIS The spreadsheetfunctions covered in earlier chapters are particularly useful in caIculatiDg internal rates of returns (IRRs). If a cash flow diagram can be reduced to at most"one P, one A, and/or one F, then the RATE investmentfunction can be used. Otherwise the IRR blockfunction is used with a cash flow in each period. . The Excel investmentfunction is RATE(n,A,P ,F ,type,guess).The A, P, and F cannot all be the same sign. The F, type, and guess are optional arguments. The "type" is end or beginning of period cash flows (for A, but not F), and the "guess" is the starting value in the search for the IRR. . ConsideringExample7-1,where P = -8200, A = 2000,and n = 5, the RATEfunction would be: RATE(5,2000,-8200) which gives an answer of 7.00%, which matches that found in Example 7-1. For Example 7-2, where P = -700, A = 100, G = 75, and n = 4, the RATE function cannot be used, since it has no provisions for the arithmetic gradient, G. Suppose the years (row 1) and the cash flows (row 2) are specified in columns B through E. The internal rate of retum calculated using IRR(B2:F2) is 6.91%. A B D E F C 1 Year 0 1 2 3 4 2 Cash flow -700 100 175 250 325 Figure 7-6 illustrates using a spreadsheet to graph the present worth of a cash flow series versus the interest rate. The interest rate with present worth equal to 0 is the IRR. The Summary 1 2 3 4 , 5 6 ; .-.Q'.....,. _.._C-. .'.'__._...r A Year 0 1 2 3 B Cash flow -700 100 175 250 4 325 . . . ._.._---- C D i 0% 2% 4% 6% 7 221 .. E PW 150.0 102.1 58.0 17.4 8% -20.0 10% -54.7 F =$B$2 + NPV(D2,$B$3:$B$6) 8 150.0 : 9 10 '11 '€ 100.0 12 0 13 50.0 C QJ 14 £ 15 '" 0.0 16 : 17 I -50.0 2% 0% 4% 6% Interest rate 18 19 I I I 10% 8% I FIGURE 7-6 Graphing present worth versus i. y axis on this graph has been modified so that the x axis intersects at a present worth of -50 rather than at O.To do this click on the y axis, then right-click to bring up the "format axis" option. Selectthis and then selectthe tab for the scale of the axis.This has a selectionfor the intersection of the x axis. This process ensures that the x-axis labels are outside the graph. SUMMARY Rate of return may be defined as the interest rate paid on the unpaid balance of a loan such that the loan is exactlyrepaid if the scheduleof paymentsis followed.On an investment,rate of return is the interest rate earned on the unrecovered investment such that the payment schedule makes the unrecovered investment equal to zero at the end of the life of the investment.Although the two definitionsof rate of return are stated differently,there is only one fundamental concept being described.It is that the rate of return is the interest rate i at which the benefits are equivalent to the costs. There are a variety of ways of writing the cash flowequation in which the rate of return i may be the single unknown. Five of them are as follows: PW of benefits - PW of costs = 0 PWof benefits =1 PWof costs NPW= 0 ; ,. II, 11 ,[ 1'1 EUAB- EUAC = 0 PW of costs = PW of benefits --- ------------ --- -~ - 222 RATEOF RETURN ANALYSIS Rate of return analysis: Rate of return analysis is the most frequently used method iJ industry, as the resulting rate of return is readily understood. Also, the difficultie in selecting a suitable interest rate to use in present worth and annual cash flov analysis are avoided. Criteria Two Alternatives Computethe incrementalrateof return-~IRR-on the alternatives.Then, the incrementof investmentbetweel ·· if ~IRR 2: MARR,choosethehigher-costalternative,or, if ~IRR < MARR, choose the lower-cost alternative When an increment of borrowing is examined, where ~IRR is the incremental interes. rate, · · if ~IRR ::::MARR, the increment is acceptable, or if ~IRR > MARR, the increment is not acceptable Three or More Alternatives Incremental analysis is needed, which is described in Chapter 8. Looki ng Ahead Rate of return is further described in Appendix 7A. This material.concentrates on the difficulties that occur with some cash flows that yield more than one root for the rate of return equation. , ) PROBLEMS 7-1 For the following diagram, compute the IRR to within 112%. . 7-2 For the following diagram, compute the interest rate at which the costs are equivalent to the benefits. i 80 80 80 80 11' 200 80 80 t t t t t t 0-1-2-3-4-5-6 j ( , ! , ) 200 : ! 1 200 50 60 (Answer: 50%) ) L -- - - - -- Problems 7-3 For the following diagram, compute the rate of return. 25 l 0-1-2-3-4-5 Compute the rate of return on the $1000 investment to within 0.1%. (Answer: 5.4%) 7-8 15 Consider the following cash flow: 10 Year t o i I 2f 1 2 3 4 5 1 42.55 7-4 i 'to i i 'to 'to i ... n = 00 7-9 o 1-10 1 3810 .to 7-6 7-7 Cash Flow -$100 +27 (Answer: 23.9%) For the following diagram, compute the rate of return. A = 1000 oI I I I I I I I o +200 +150 +100 +50 Compute the rate of return for the following cash flow to within 112%. Year 7-5 Cash Flow -$400 Write one equation, with i as the only unknown, for the cash flow. In the equation you are not to use more than two single payment compound interest factors. (You may use as many other factors as you wish.) Then solve your equation for i. For the following diagram, compute the rate of return on the $3810 investment. 500 500 500 500 0-1-2-3-4-5-6-7- 223 11Oli.ii.i.ii1... n = 7-10 Solve the following cash flow for the rate of return to within an 112%. Year 00 o 1 2 3 4 5 1 5000 Considerthe followingcash flow: Year Cash Flow o -$500 1 +200 2 +150 3 +100 4 +50 Compute the rate of return represented by the cash flow. Considerthe followingcash flow: Year Cash Flow o -$1000 1 0 2 +300 3 +300 4 +300 5 +300 7-11 Compute the rate of return for the following cash flow. Year 1-5 6-10 7-12 Cash Flow -$500 -100 +300 +300 +400 +500 Cash Flow -$233 +1000 Compute the rate of return for the following cash flow to within 0.5%. Year o 1 2 3 4 5 Cash Flow -$640 o 100 200 300 300 (Answer: 9.3%) ,, ( , -- -- - 224 RATEOF RETURN ANALYSIS graduate by $3000 per year. During the subsequent 10 years, she projects an annual income that is $6000 per year higher. During the last 20 years of employment, she estimates an annual salary that is $12,000 above the level of the non-college graduate. If her estimates are correct, what rate of return will she receive as a result of her investment in a college education? 7-13 Compute the rate of return for the following cash flow. Year Cash Flow o -$500 1-3 4 0 +4500 7-14 A woman went to the Beneficial Loan Company and borrowed $3000. She must pay $119.67 at the end of each month for the next 30 months. (a) Calculate the nominal annual interest rate she is paying to within ::1::0.15%. (b) What effective annual interest rate is she paying? 7-20 An investor purchased a one-acre lot on the outskirts of a city for $9000 cash. Each year he paid $80 of property taxes. At the end of 4 years, he sold the lot. After deducting his selling expenses, the investor received $15,000. What rate ofreturn did he receive on his investment? (Answer: 12.92%) 7-15 Helen is buying a $12,375 automobile with a $3000 down payment, followed by 36 monthly payments of $325 each. The down payment is' paid immediately, and the monthly payments are due at the end of each month. What nominal annual interest rate is Helen paying? What effective interest rate? (Answers: 15%; 16.08%) 7-21 A popular reader's digest offers a lifetime subscrip-' tion to the magazine for $200. Such a subscnption may be given as a gift to an infant at birth (the parents can read it in those early years), or taken but by an individual for himself. Normally, the magazine. costs $12.90 per year. Knowledgeable people say it probably will continue indefinitely at this $12.90 rate. . What rate of return would be obtained if a life subscription were purchased for an infant, rather than paying $12.90 per year beginning immediately? You may make any reasonable assumption!;, but the compound interest factors must be correctly used. 7-16 Peter Minuit bought an island from the Manhattoes Indians in 1626 for $24 worth of glass beads and trinkets. The 1991 estimate of the value of land on this island was $12 billion. What rate of return would the Indians have received if they had retained title to the island rather than selling it for $24? 7-17 A man buys a corporate bond from a bond brokerage house for $925. The bond has a face value of $1000 and pays 4% of its face value each year. If the bond will be paid off at the end of 10 years, what rate of return will the man receive? (Answer: 4.97%) 7-18 A well-known industrial firm has issued $1000 bonds that carry a 4% nominal annual interest, rate paid semiannually. The bonds mature 20 years from now, at which time the industrial firm will redeem them for $1000 plus the terminal semiannual interest payment. From the financial pages of your newspaper you learn that the bonds may be purchased for $715 each ($710 for the bond plus a $5 sales commission). What nominal annual rate of return would you receive if you purchased the bond now and held it to maturity 20 years hence? (Answer: 6.6%) 7-22 On Apri12, 1988, an engineer bought a $1000 bond of an American airline for $875. The bond paid 6% on its principal amount of $1000, half in each of its April 1 and October 1 semiannual payments; it repaid the $1000 principal sum on October 1, 2001. What nominal rate of return did the engineer receive from the bond if he held it to its maturity (on October 1, 2001)? (Answer: 7.5%) 7-23 The cash price of a machine tool is $3500. The dealer is willing to accept a $1200 down payment and 24 end-of-month monthly payments of $110 each. At what effective interest rate are these terms equivalent? (Answer: 14.4%) 7-24 A local bank makes automobile loans. It charges 4% per year in the following manner: if$3600 is borrowed to be repaid over a 3-year period, the bank interest charge is $3600 x 0.04 x 3 years = $432. The bank deducts the $432 of interest from the $3600 loan and gives the customer $3168 in cash. The customer must repay the loan by paying 1136of $3600, or $100, at the end of each month for 36 months. What nominal annual interest rate is the bank actually charging for this loan? 7-19 One aspect of obtaining a college education is the prospect of improved future earnings in comparison to non-college graduates. Sharon Shay estimates that a college education has a $28,000 equivalent cost at graduation. She believes the benefits of her education will occur throughout 40 years of employment. She thinks that during the first 10 years out of college, her income will be high~r than that of a non-college -. ~ , - -- Problems 7-25 Upon graduation, every engineer must decide whether to go on to graduate school. Estimate the costs of going full time to the university to obtain a master of science degree. Then estimate the resulting costs and benefits. Combine the various consequences into a cash flow table and compute the rate of return. Nonfinancial benefits are probably relevant here too. 7-26 A table saw costs $175 at a local store. You may either pay cash for it or pay $35 now and $12.64 a month for 12 months beginning 30 days hence. If you choose the time payment plan, what nominal annual interest rate will you be charged? (Answer: 15%) 7-27 An investment of $5000 in Biotech common stock proved to be very profitable. At the end of 3 years the stock was sold for $25,000. What was the rate of return on the investment? 7 -, The Southern Guru Copper Company operates a large J!!! National mine in a Assembly South American legislator in the the said incountry. a speechAthat most of capital for the mining operation was provided by loans from the Word Bank; in fact, Southern Guru has only $500,000 of its own money actually invested in the property. The cash flow for the mine is: Year o 1 2 3 4 5 6 7 Cash Flow $0.5 million investment 3.5 million profit 0.9 million profit 3.9 millionprofit 8.6 million profit 4.3 million profit 3.1 millionprofit 6.1 million profit The legislator divided the $30.4 million total profit by the $0.5 million investment. This produced, he said, a 6080% rate of return on the investment. Southern Guru claims the actual rate of return is much lower. They ask you to compute their rate of return. 7-29 An insurance company is offering to sell an annuity for $20,000 cash. In return the firm will guarantee to pay the purchaser 20 annual end-of-year payments, with the first payment amounting to $1100. Subsequent payments will increase at a uniform 10% rate each year (second payment is $1210; third payment is $1331, etc.). What rate of return will the purchaser receive if he buys the annuity? 7-30 A bank proudly announces that it has changed its interest computation method to continuous compound- - --- 225 ing. Now $2000 left in the bank for 9 years will double to $4000. (a) What nominal interest rate, compounded continuously, is the bank paying? (b) What effective interest rate is it paying? 7-31 Fifteen families live in Willow Canyon. Although several water wells have been drilled, none has produced water. The residents take turns driving a water truck to a fire hydrant in a nearby town. They fill the truck with water and then haul it to a storage tank in Willow Canyon. Last year truck fuel and maintenance cost $3180. This year the residents are seriously considering spending $100,000 to install a pipeline from the nearby town to their storage tank. What rate of return would the Willow Canyon residents receive on their new water supply pipeline if the pipeline is considered to last (a) Forever? (b) 100 years? (c) 50 years? (d) Would you recommend that the pipeline be installed? Explain. 7-32 Ian purchased 100 shares of Peach Computer stock for $18 per share, plus a $45 brokerage commission. Every 6 months she received a dividend from Peach of 50 cents per share. At the end of 2 years, just after receiving the fourth dividend, she sold the stock for $23 per share and paid a $58 brokerage commission from the proceeds. What annual rate of return did she receive on her investment? 7-33 The Diagonal Stamp Company, which sells used postage stamps to collectors, advertises that its average price has increased from $1 to $5 in the last 5 years. Thus, management states, investors who had purchased stamps from. Diagonal 5 years ago would have received a 100% rate of return each year. (a) To check their calculations, compute the annual rate of return. (b) Why is your computed rate of return less than 100%? 7-34 An investor purchased 100 shares of Omega common stock for $9000. He held the stock for 9 years. For the first 4 years he received annual end-of-year di-vidends of $800. For the next 4 years he received annual dividends of $400. He received no dividend for the ninth year. At the end of the ninth year he sold his stock for $6000. What rate of return did he receive on his investment? - - ------- .....---------- 226 .- RATEOF RETURNANALYSIS 7-35 You spend $1000 and in return receive two payments of $1094.60--:<>neat the end of 3 years and the other at the end of 6 years. Calculate the resulting rate of return. 7-36 A mine is for sale for $240,000. It is believed the mine will produce a profit of $65,000 the first year, but the profit will decline $5000 a year after that, eventually reaching zero, whereupon the mine will be worthless. What rate of return would this $240,000 investment produce for the purchaser of the mine? 7-37 Fred, our cat, just won the local feline lottery to the tune of 3000 cans of "9-Lives" cat food (assorted flavors). A local grocer offers to take the 3000 cans and in return, supply Fred with 30 cans a month for the next 10 years. What rate of return, in terms of nominal annual rate, will Fred realize on this deal? (Compute to nearest 0.01 %.) 7-38 An apartment building in your neighborhood is for sale for $140,000. The building has four units, which are rented at $500 per month each. The tenants have long-term leases that expire in 5 years. Maintenance and other expenses for care and upkeep are $8000 annually. A new university is being built in the vicinity and it is expected that the building could be sold for $160,000 after 5 years. (q) What is the internal rate of return for this investment? (b) Should this investment be accepted if the your other options have a rate of return of 12%? 7-39 A new machine can be purchased today for $300,000. The annual revenue from the machine is calculated to be $67,000, and the equipment will last 10 years. Expect the maintenance and operating costs to be $3000 a year and to increase $600 per. year. The salvage value of the machine will be $20,000. What is the rate of return for this machine? 7-40 AI Larson asked a bank to lend him money on January 1, based on the following repayment plan: the first payment would be $2 on February 28, with subsequent monthly payments increasing by $2 a month ,. be willing to lend AI on the proposed repayment plan ? 7-41 The following advertisement appeared in the Wall Street Journal on Thursday, February 9, 1995. ''There's nothing quite like the Seville SmartLease. Seville SLS $0 down, $599 a monthj36 months." First month's lease payment of $599 plus $625 refundable security deposit and a consumer down payment of $0 for a total of $1224 due at lease signing. Monthly payment is based on a net capitalized cost of $39,264 for total monthly payments of $21,564. Payment examples based on a 1995 Seville SLS: $43,658 MSRP including destination charge. Tax, license,' title fees, and insurance extra. Option to purcha.se at lease end for $27,854. Mileage charge of $0.15 per mile over 36,000 miles. (a) Set up the cash flows. (b) Determine the interest rate (nominal and effective) for the lease. 7-42 After 15 years of working for one employer, you transfer to a new job. During these years your employer contributed (that is, she diverted from your salary) $1500 each year to an account for your retirement (a fringe benefit), and you contributed a matching amount each year. The whole fund was invested at 5% during that time and the valu of the account now stands at $30,000. You are now faced with two alternatives. (1) You may leave both contributions in the fund until retirement in 35 years, during which you will get the future value of this amount at 5% interest per year. (2) You may take out the total value of "your" contributions, which is $15,000 (one-half of the total $30,000). You can do as you wish with the money you take out, but the other half will be lost as far as you are concerned. In other words, you can give up $15,000 today for the sake of getting now the other $15,000. Otherwise, you must wait 35 years on an arithmetic gradient. (The March 31 payment wouldbe $4; the April 30 paymentwouldbe $6, etc.) more to get the accumulatedvalue of the entire fund. Which alternative is more attractive? Explain your The payments are to continue for 11 years, making a total of 132 payments. (a) Compute the total amount of money AI will pay the bank, based on the proposed repayment plan. (b) If the bank charges interest at 12% nominal per year, compounded monthly, how much would it 7-43 A finance company is using the following "Money by Mail" offer. Calculate the yearly nominal IRR received by the company if a customer chooses the loan of $2000 and accepts the credit insurance (Life and Dis.). i I I choice. ] j I - Problems Non-Negotiable For the Amount of $3000 or $2000 or $1000 Pay to the Order of I Feel Rich Limited Time Offer For the Amount of $3000 Dollars Pay to the Order of I Feel Rich Total of Payments $4,280.40 Number of Monthly Payments 36 Months Amount Financed $3,246.25 For the Amount of $1000 Dollars Pay to the Order of I Feel Rich Total of Payments $1,300.80 Number of Monthly Payments 24 Months Amount Financed $1055.26 INE 1/96 Dollars To borrow $3000, $2000, or $1000 APR 18.95% Finance Charge $1034.29 Monthly Payment $118.90 Credit Line Premium $83.46* Credit Disabilit Premium $162.65* $3000 loan terms APR 19.95% Finance Charge $594.25 For the Amount of $2000 Dollars Pay to the Order of I Feel Rich Total of Payments $2,731.50 Number of Monthly Payments 30 Months Amount Financed $2,137.25 227 Monthly Payment $91.05 Credit Line Premium $44.38* Credit Disabilit Premium $92.87* $2000 loan terms APR 20.95% Finance Charge $245.54 Monthly Payment $54.20 Credit Line Premium $16.91 * Credit Disabilit Premium $39.02* $1000 loan terms *Credit insurance. If selected, premium will be paid from amount financed. If not selected, cash advance is total amount financed. 7-44 A finance company is using the "Money Mail" offer shown in Problem 7-43. Calculate yearly nominal IRR received by the company customer chooses the $3000 loan but declines credit insurance. 7-45 In his will, Frank's uncle has given Frank the choice between two alternatives: Alternative 1 Alternative 2 by the if a the $2000 cash $150 cash now plus $100 per month for 20 months beginning the first day of next month 7-46 The owner of a comer lot wants to find a use that will yield a desirable return on his investment. After much study and calculation, the owner decides that the two best alternatives are: Build Soft Build Ice Cream Stand Gas Station First cost $120,000 $80,000 5,000 3,000 Annual property taxes Annual income 16,000 11,000 20 ,20 Life of building, in years Salvage value (a) At what rate of return are the two alternatives equivalent? (b) If Frank thinks the rate ofretum in (a) is too low, which alternative should he select? o If the owner wants a minimum attractive rate of return on his investment of 6%, which of the two alternatives would you recommend? -- --- o 228 RATEOF RETURN ANALYSIS 7-52 Two mutually exclusive alternatives are being 7-47 Two alternatives are as follows: considered. Year A -$2000 +800 +800 +800 o 1 2 3 B -$2800 +1100 + 1100 +1100 Year 0 1 2 3 4 5 If 5% is considered the minimum attractive rate of return, which alternative should be selected? A -$2500 +746 +746 +746 +746 +746 B -$6000 + 1664 +1664 + 1664 + 1664 + 1664 7-48 Consider two mutually exclusive alternatives: X -$100 35 35 35 35 Year o 1 2 3 4 If the minimum attractive rate of return is 8%, which alternative should be selected? Solve the problem by (a) Present worth analysis (b) Annual cash flow analysis (c) Rate of return analysis y -$50.0 "16.5 16.5 16.5 16.5 " If the minimum attractive rate of return is 10%, which alternative should be selected? 7-49 Consider these two mutually exclusive alternatives: Year o 1 2 3 4 A -$50 17 17 17 17 B -$53 17 17 17 17 7-53 A contractor is considering whether to purchase or lease a new machine for his layout site work. Purchasing a new machine will cost $12,000 with a salvage value of $1200 at the end of the machine's useful life of 8 years. On the other hand, leasing requires an annual lease payment of $3000. Assuming that the MARR is 15% and on the basis of an internal rate of return analysis, which alternative should the contractor be advised to accept. The cash flows are ~ follows: Year (n) o At a MARR of 10%, which alternative should be selected? (Answer: A) 7-50 1\\'0 mutually exclusive alternatives are being considered. Both have a 10-year useful life. If the MARR is 8%, which alternative is preferred? Initial cost Uniform annual benefit A $100.00 19.93 B $50.00 11.93 7-51 Consider two mutually exclusive alternatives: X -$5000 -3000 +4000 +4000 +4000 Year o 1 2 3 4 y -$5000 +2000 +2000 +2000 +2000 If the MARR is 8%, which alternative should be selected? - -- Alt. A (purchase) -$12,000 1 2 3 4 5 6 7 8 Alt. B (lease) o -$3000 -3000 -3000 - 3000 -3000 - 3000 -3000 -3000 +1200 7-54 Two hazardous environment facilities are being evaluated, with the projected life of each facility being 10 years. The caSh flows are as follows: First cost Maintenanceand operating cost Annual benefitS Salvage value Alt. A $615,000 10,000 -Alt.B $300,000 25,000 158,000 65,000 92,000 -5,000 - - - --- Difficulties in Solving for an Interest Rate The company uses a MARR of 15%. Using rate of return analysis, wpich alternative should be selected? 7-55 Two alternatives are being considered: Two alternatives are being considered: First cost Uniform annual benefit Useful life, in years A $9200 $1850 8 Initial cost Uniform annual benefit Usefullife, in years B $5000 $1750' 4 If the minimum attractive rate of return is 7%, which alternative should be selected? 7-56 7-57 A $9200 1850 8 B $5000 1750 4 Base your computations on a MARR of 7% and an 8-year analysis period. If identical replacement is assumed, which alternative should be selected? 7-58 Two investmentopportunitiesare as follows: Jean has decided it is time to purchase a new battery for her car. Her choices are: First cost Guarantee period, in months 229 Zappo Kicko $56 $90 12 24 Jean believes the batteries can be expected to last only for the guarantee period. She does not want to invest extra money in a battery unless she can expect a 50% rate of return. If she plans to keep her present car another 2 years, which battery should she buy? First cost Uniform annual benefit End-of-useful-life salvage value Usefullife, in years A $150 25 20 15 B $100 22.25 o 10 At the end of 10 years, Alt. B is not replaced. 'Thus, the comparison is 15 years of A versus 10 years of B. If the MARR is 10%, which alternative should be selected? APPENDIX 7A Difficulties in Solving for an Interest Rate Mter completing this chapter appendix, students should be able to: . · . Describe why some project's cash flows cannot be solved for a single positive interest rate. Identify when multiple roots can occur. . Evaluate how many potential roots exist for a particular project. Use the modified internal rate of return (MIRR) methodology in multiple-root cases. Example 7A-I illustrates the situation. 230 1 RATEOF RETURN ANALYSIS The Going Aircraft Company has an opportunity to supply a large airplane to Interair, a foreign airline. Interair will pay $19 million when the contract is signed and $10 million one year later. Going estimates its second- and third-year costs at $50 million each. Interair will take delivery of the airplane during Year 4 and agrees to pay $20 million at the end of that year and the $60 million balance at the end of Year 5. Compute the rate ofretum on this project. SplUTION The PW of each cash flowcan be computed at various interest rates. For example, for Year2 and i = 10%: PW = -50(P / F, 10%,2) = -50(0.826) = -41.3. Year 0 1 2 Cash Flow +$19 +10 -50 -50 +20 +60 "3 4 5 0% +$19 +10 -50 -50 +20 - +60 10% +$19 +9.1 -41.3 -37.6 +13.7 +37.3 20% +$19 +8.3 -34.7 -28.9 +9.6 +24.1 40% +$19 +7.1 -25.5 -18.2 +5.2 +11.2 50% +$19 +6.7 -22.2 -14.8 +3.9 +7.9 +$9 +$0.1 -$2.6 -$1.2 +$0.5 PW= 10.00 8.00 6.00 ~ Q., . 4.00 2.00 0.00 -2.00 -4.00 0% 10% 20% 30% 40% 50% FIGURE 7A-l PW plot. The PW plot-for this cash flow is represented in Figure 7A- L We see theftthis cash flow produces twopointsat whichPW = 0: one at 10.24%and the otherat 47.30%. --------- r - Difficulties in Solving for an Interest Rate 231 Why Multiple Solutions Can Occur Example7A-l producedunexpectedresults.This couldhappenbecausethere weretwo , changes in the signs of the cash flows. Years 0 and 1 were positive, Years 2 and 3 were negative, and Years 4 and 5 were positive. The cash flow series went from positive cash flowsto negative cash flowsto positive cash flows. Most cash flow series have only one shift in sign. Investmentsstart with one or more years of negative cash flows followed by many years of positive cash flows. Loans begin with a positive cash flow that is repaid with later negative cash flows. These problems have a unique rate of return because they have a single change in the sign of the cash flows. Havingmore than one sign change in the cash flowseries can produce multiple points or roots at which the PW equals O.To see how many roots are possible, we link solving an economic analysis problem to solving a mathematical equation. A project's cash flows are the values from CFoto CFn. Year 0 1 2 Cash Flow CFo CFI CFz n CFn The equation to find the internal rate of return, where PW = 0, is written as follows: PW = 0 -:-CFo + CFI(1 + i)-I + CFz(1 + i)-z +... (7A-l) + CFn(1+ i)-n If we let x = (1 + i) -I, then Equation 7A-I may be written (7A-2) Equation 7A-2 is an nth-order polynomial, and Descartes' rule describes the number of positive roots for x. The rule is: If a polynomial with real coefficients has m sign changes, then the number of positive roots will be m - 2k, where k is an integer between 0 and m12. A sign change exists when successivenonzero terms, written according to ascending powers of x, have different signs. If x is greater than zero, then the number of sign changes in the cash flowsequals the number in the equation. Descartes' rule means that the number of positive roots (values of x) of the polynomial cannot exceed m, the number of sign changes. The number of positive roots for x must either be equal to m or less by an even integer. - - -- ------ III 232 RATEOF RETURN ANALYSIS Thus, Descartes' rule for polynomials gives the following: Number of Sign Changes, m Number of Positive Values of x o o 1 20rO 3 or 1 4, 2, or 0 1 2 3 4 N~ber of Positive Values of i o 10rO 2, 1, or 0 3, 2, 1, or 0 4, 3, 2, 1, or 0 If x is greater than I, the corresponding value of i is negative. If there is only one root and it is negative, then it is a valid IRR. One example of a valid negative IRR comes from a bad outcome, such as the IRR for a failed research and development project. Another example would be a privately sponsored student loan set up so that if a student volunteers . for the Peace Corps, only half of the principal need be repaid. If a project has a negative and a positive root for i, then for most projects only the positive root of i is used. . Projects with Multiple Sign Changes Example 7A-I is representativeof projects for which there are initial payments (e.g., when the order for the ship, airplane, or building is signed), then the bulk of the costs occur, and then there are more payments on completion. . Projectsof other types often have two or more sign changesin their cash flows.Projects with a salvage cost typically have two sign changes. This salvage cost can be large for environmentalrestoration at termination. Examples include pipelines, open.pit mines, and nuclear power plants. The following cash flow diagram is representative. $50K t t 0-1-2-3-4-5-6-7 t t t j t ~tK $180K Many enhancement projects for existing mines and deposits have a pattern of two sign changes. Example 7A-2 describes an oil well in an existing field. The initial investment recovers more of the resource and speeds recovery of resources that would have been recovered eventually.The resources shifted for earlier recovery can lead to :twosign changes. In Example 7A-3, we consider staged construction, where three sign changes are common. --- Difficulties 233 in Solving for an Interest Rate Other examples can be found in incremental comparisons of alternatives with unequal lives. In the next section we learn how to determine whether each of these examples has multiple roots. Adding an oil well to an existing field costs $4 million (4M). It will increase recovered oil by $3.5M, and it shifts $4.5M worth of production from Years 5, 6, and 7 to earlier years. Thus, the cash flows for Years 1 through 4 total$8M and Years 5 through 7 total -$4.5M. If the well is justified, one reason is that the oil is recovered sooner. How many roots for the PW equation are possible? . SOLUTION. The first step is to draw the cash flow diagram and count the number of sign changes. The following pattern is representative, although most wells have a longer life. ~$4M = There are two sign changes, thus there may be 0, 1, or 2 positive roots for the PW 0 equation. The additio}Jalrecovery corresponds to an investment, and the shifting of (ecovery to earlier years corresponds to a 10aJ;). (positive cash flow now and negative later). Thus, the oil wells are neither anirlVe$hnent }JOtaloan; they are a combination of both. ;I I '11., .'! , I II II A projecthas a first cost of $120,000. Net revenues begin at $30,000 in Year 1 and then increase * ~ by $2000 pefyear. In Year 5 the facility is expanded at a cost of $60,000 so that demand can confiJrueto expandat $2000peryear.HowfilanYrootsfor the PWequationarepossible? _~ . i"' ':'" I~ - (. - -- ~ - -~l - - - - 234 "1 RATEOF RETURN ANALYSIS SOLUTION. The first step is to draw the cash flow diagram. Then counting the three sign changes is easy. $30K $32K t $34K t $36K t $42K t t 0-1-2-3-4-5-6-7-8-9-10 ! $40K -$221{ ~ JoK ~ $44K t t $46K t $48K t $38K -$120K With three sign changes, there may be 0, 1, 2, or 3 positive roots for the PW = 0 equation. Evaluati ng How Many Roots There Are The number of sign changes tells us how many roots are possible-not how many roots there are. Rather than covering the many mathematical approaches that may tell us if the root will be unique, it is more useful to employ the power of the spreadsheet.A spreadsheet can show us if a root is unique and the value of each root that exists. The approach is simple. For any set of cash flows,graph the PW as a function of the interest rate. We are interested in positive interest rates, so the graph usually starts at i = O. Since Descartes' rule is based on x > 0, and x = (1+ i) -1, sometimes the graph is started close to i = -1. This will identify any negative values of i that solve the equation. If the root is unique, we're done. We've found the internal rate of return. If there are multiple roots, then we know the project's PW at all interest rates-including the one our organizationuses. Wecan alsouse the approachof the next sectionto finda modifiedinternal rate of return. The easiest way to build the spreadsheetis to use the NPV(i ,values) function in Excel. Remember that this function applies to cash flowsin periods 1 to N, so that the cash flow at time 0 must be added in separately.The following examplesuse a spreadsheetto answer the question of how many roots each cash flow diagram in the last section has. This project is representative of ones with a salvage cost. How many roots for the PW equation exist? $50K t t 0-1~2-3-4-5-6-7 j - t t t t '. .. _-- .$180K = -= $1 - - -- : ----- .. :;:: -- Difficulties in Solving for an Interest Rate Figure 7A-2 shows the spreadsheet calculations and the graph ofPW versus i. A B 1 Year Cash Flow 2 0 3 1 50 -30% 219.99 4 2 50 -20% 189.89 5 3 50 -10% 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 4 5 6 7 50 50 50 -70 IRR root D E i PW -40% -126.39 C -180 250.00 200,00 150.00 100.00 50.00 c.. 0.00 -50.00 -100.00 -150.00 -40% G =$B$2 + NPV(D2,$B$3:$B$9) 114.49 50.00 1.84 -33.26 -59.02 -78.24 -92.88 ; 0% 10% 20% 30% 40% 50% 10.45% - 38.29% F f I I I -20% 0% i 20% 40% FIGURE 7A-2 PW versus i for project with salvage cost. In this case, there is I positive root of .10.45%.The value can be used as an IRR. There is also a negative root of i = -"38.29%. This root is not useful. With two sign changes, other similar diagrams may have 0, I, or 2 positive roots for tlie PW = 0 equation. The larger the value of the final salvage cost, the more likely 0 or 2 positive roots are to occur. . -- _-~IIIIIIIII!!'!!I!!!!!I~ --- ------- 235 .... I 236 RATEOF RETURN ANALYSIS $3.5M $2.5M t 1 $1.5M t I 0-1-2-3-4-5-6-7-8 $01M -$J.5M ~ -$I.5M 1 -$2.5M j -$4M SOLUTION Figure 7A-3.shows the spreadsheetcalculations and the graph of PW .versusi. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 A Year 0 1 2 3 4 5 6 7 B. Cash Flow -4.00 3.50 2.50 1.50 0.50 -0.50 - 1.50 -2.50 root root 4.73% 37.20% C D i 0% 5% 10% 15% 20% 25% 30% 35% 40% 45% 50% E PW -0.50 0.02 0.28 0.37 0.36 0.29 0.19 0.06 -0.08 -0.22 -0.36 F G =$B$2 + NPV(D2,$B$3:$B$9) 0040 0.20 0.00 c..-0.20 -0040 I -0.60 0% / I I 10% 20% I 30% 40% 50% i FIGURE7A-3PW versusi foroilwell. = --~ - 41this case, there are 2 positiye rpots..at4.73and 37.20%. These roots ~ not "seful.1}ris proj7ct", is a combination of an investment and a loan, so we don't even know if we want a high rate or a low rate. If our interest rate is about 20%, then the project has a positive PW. However, small = 1:1I:I:;; == iii!'==-= changes~inthe,datacan mak.efor~largechangesin.these,JesuJts. 11;':;;1;;:;;'1:10= . - It is useful to apply the modified internal rate of return described in the next section. --- - --- Difficulties in Solving for an Interest Rate 237 A project has a first cost of $120,000.Net revenuesbegin at $30,000in Year1,and then increaseby $2000 per year. In Year 5 the facility is expanded at a cost of $60,000 so that demand can continue to expand at $2000 per year. How many roots for the PW equation have a positive value for i? $30K $32K $34K $36K t t t t 0-1-2-3-4-5-6-7-8-9 ! -$120K $40K $42K t $44K t t $46K t $48K t ,.10 -$22K ~ $~K - $38K Figure 7A-4 shows the spreadsheet calculations and the graph ofPW versus i. 1 2 3 4 A Year 0 1 2 B Cash Flow -120.00 30.00 32.00 5 3 34.00 6 4 7 5 8 6 9 10 11 12 7 8 9 10 I C D i -80% -60% I -40% I I -22.00 73 40.00 20% 7 42.00 44.00 46.00 48.00 I 30% i 40% IRR 21.69% 15 ! I I 48 I 62 I ! 60% -71 !{ 70% -78 II ,83 87 90% I , i I 80% > 16 17 150 19 p., 23 " -- 24 =,':::I . - --,50 -100 ," ____ve"'"" -- I I :: ::II I" 10% ....,._ - --- ">I'" I '20% =, I I 30% 40% ,<c "'''' 1- ;,::;........ i forp«>jec,WltJ, staged eOl)J;toJefjOo. .. 50% i iIGt1RW 7A4f!W -- .:. 0% 26 rL__,_· I 0 25 . i 50 -' , 22 - I 100 21 F= I "" i1. , 20 I ,, I I 200 18 It i r -28 50% i =$B$2 +NJ>V(D2,$B$3:$B$12) 210 10% I G 1460 0% . F 17651 .....30% 36.00 13 14 E pw 579646330 735723 .~== ~-. . -. -. - .... .~.- ~._~. 1 238 RATEOF RETURN ANALYSIS In this case, there is 1positive root of 21.69%. The value can be used as an IRR, and the project i~ very attractive. ~ - . When there are two or more sign changes in the cash flow, we know that there are several possibilities concerning the number of positive values of i. Probably the greatest danger in this situation is to fail to recognize the multiple possibilities and to solve for a value of i. The approach of constructing the PW plot both establishes whether there are multiple roots and what their values are. This may be tedious to do by hand but is very easy with a spreadsheet (or a graphing calculator). If there is a single positive value of i, we have no problem. On the other hand, if there is no positive value of i, or if there are multiple positive values, the situation may be attractive,unattractive, or confusing. Where there are multiple positive values of i, none of them should be considered a suitable measure of the rate of return or attractiveness of the cash flow. Modified Internal Rate of Return (MIRR) Two external rates of return can be used to ensure that the resulting equation is solvable for a unique rate of return-the MIRR. The MIRR is a measure of the attractivenessof the cash flows,but it is also a function of the two external rates of return. The rates that are external to the project's cash flows are (1) the rate at which the organization normally invests and (2) the rate at which it normally borrows. These are external rates for investing, einv,and for financing, efin.Because profitable firms invest at higher rates than they borrow at, the rate for investingis generally higher than the rate for financing.Sometimesa singleexternalrate is used for both,but thisrequires thequestionable assumption that investing and financinghappen at the same rat~. The approach is: 1. 2. 3. 4. Combine cash flowsin each period into a single net receipt, R" or net expense, Et. Find the present worth of the expenses with the financingrate. Find the future worth of the receipts with the investingrate. Find the MIRR which makes the present and future worths equivalent. The result is Equation 7A-3. This equation will have a unique root, since it has a single negativepresent worth and a single positive future worth. There is only one sign change in the resulting series. (F / P, MIRR, n) L Et(P t / F, efin,t) = L RtCF/ P, einv,n t t) (7A-3) There are other external rates of return, but the MIRR has historically been the most clearly defined. All of the external rates of return are affectedby the assumed values for the investing and financing rates, so none are a true rate of return on the project's cash flow. The MIRR also has an Excel function, so it now can very easily be used. Example 7A-7 illustrates the calculation, which is also summarized in Figure 7A-5. J r Difficulties in Solving for an Interest Rate 239 External Investing Rate -$0.5M -$1.5M - $2.5M External Financing Rate -$4M P(1 + MIRR)n = F FIGURE 7A-5 MIRR for the oil well. Adding an oil well to an existing field had the cash flows summarized in Figure 7A-5. If the firm normally bOlTowsmoney at 8% and invests at 15%,find the modifiedinternal rate of return (MIRR). i~OlUTlO~ Figure 7A-6 shows the spreadsheet calculations. " A 1 2 C I D 3 Year . 0 1 4 Cash Flow -4.00 3.50 13.64% ,... I I E F G H I I 8% externalfinancingrate 15% externalinvestingrate 5 -.' B I 2 3 4 5 6 7 1.50 0.50 -0.50 -1.50 -2.50 I Cell A5 contains =MIRR(A4:14,AI,A2) 2.50 , FIGURE7A-6MIRR for oil well. Itis also possible to calculate the MIRR by hand. While more work, the proceSsdoes clarify what the MIRRfunction.is doing. 1. Each period's cash flow is already a single net receipt or,y~Rendittlre. 2. Find the present.worth of the expenses with the financingrate. ::: :: 1:1 IIIIi '...'"'" -. ~ -- -- RW ~ --4M -' O.5M(PIF, 8%, 5) -1.5M(P I F, 8%,6) -- 2.5M(P IF, 8%, 7) II !II= = ~ == '.', ~iM. -- Qft~M,(Q~q~.Q§} --1~jO.63,02) -' 2.5M4o.5835)..;:;~.(j,,;z44M#k1'~ -- . 'I I 240 .- RATEOF RETURN ANALYSIS 3., Find the future worth of the receipts with the investingrate. FW = 3.5M(FjP, 15%,6) +2.5M(FjP, 15%,5) + 1.5M(FjP, 15%,4) +O.5M(FjP, 15%,3) = 3.5M(2.313) + 2.5M(2.011) + 1.5M(1.749) + O.5M(1.521) = 16.507M 4. Find the MIRR that makes the present and future worths equivalent. 0= (1 +MIRRt .PW +FW 0= (1 +MIRR)7(-6.744M) + 16.507M (1 + MIRR)7 = 16.507Mj6.744M = 2.448 (1 + MIRR) = 2.4481/7 = 1.1364 MIRR = 13.64% The MIRR does allow calculation of a rate of return for any set of cash flows.However, the result is only as realistic as the externalrates that are used. The MIRR value can depend as much on the external rates that are used, as it does on the cash flowsthat it is describing. SUMMARY In cash flows with more than one sign change, we find that solving the cash flow equation can result in more than one positive rate of return. Typicalsituations include a new oil well in an existing field, a project with a significantsalvage cost, and staged construction. In a sign change, successivenonzero values in the cash flow have different signs (that is, they change from + to -, or vice versa). Zero sign changes indicates there is no rate of return, as the cash flow is either all disbursements or all receipts. One sign change is the usual situation, and a single positive rate of return generally results. There will be a negative rate of return whenever loan repayments are less than the loan or an investmentfails to return benefits at least equal to the investment. Multiple sign changesmay result in multiplepositiveroots for i. When they occur,none of the positive multiple roots are a suitable measure of the project's economic desirability. If multiple roots are identified by graphing the present worth versus the interest rate, then the modified internal rate of return can be used to evaluate the project. Graphing the present worth versus the interest rate ensures that the analyst recognizes that the cash flowhas multiple sign changes. Otherwisea rate could be found and used that is not in fact a meaningful descriptor of the project. The modified internal rate of return (MIRR)relies on rates for investingand borrowing that are external to the project. The number of sign changes are reduced to one, ensuring that the MIRR can be found. PROBLEMS Unless the problem asks a different question or provides different data: (1) determine how many roots are possible, and (2) graph the PW versus the inter- --------- est rate to see whether multiple roots occur. If the root is a unique IRR, it is the project's rate of return. If there are multiple roots, then use an external -- - - -- Problems investing rate of 12% and an external borrowing rate of 6%. Compute and use the MIRR as the project's rate of return. 7A-1 7A-2 7A-3 Then determine a quarterlyrate of return, a nominal annual rate, and an effectiveannualrate. 7A-4 Find the rate of return for the following cash flow: Year Cash Flow o -$15,000 1 2 3 4 10,000 -8,000 11,000 13,000 7A-5 Giventhe followingcash flow,determine the rate of return on the project. (Answer: 11.3%MIRR) Year Cash Flow o -$500 1 +2000 2 -1200 3 - 300 Given the following cash flow, determine the rate of (Answer: 21.2% IRR) return on the investment. A group of businessmen formed a partnership to buy and race an Indianapolis-type racing car. They agreed to pay an individual $50,000 for the car and associated equipment. The payment was to be in a lump sum at the end of the year. In what must have been "beginner's luck," the group won a major race the first week and $80,000. The rest of the first year, however, was not as good: at the end of the first year, the group had to payout $35,000 for expenses plus the $50,000 for the car and equipment. The second year was a poor one: the group had to pay $70,000 just t()clear up the racing debts at the end of the year. During the third and fourth years, racing income just equaled costs. When the group was approached by a prospective buyer for the car, they readily accepted $80,000 cash, which was paid at the end of the fourth year. What rate of return did the businessmen obtain from their racing venture? (Answer: 9.6% MIRR) IRR) A student organization, at the beginning of the fall quarter, purchased and operated a soft-drink vending machine as a means of helping finance its activities. The vending machine cost $75 and was installed at a gasoline station near the university. The student organization pays $75 every 3 months to the station owner for the right to keep the vending machine at the station. During the year the student organization owned the machine, they received the following quarterly income from it, before making the $75 quarterly payment to the station owner: Quarter Fall Winter Spring Summer Income $150 25 125 150 At the end of one year, the student group resold the machine for $50. Determine the quarterly cash flow. - --r 241 Year o 1 2 3 (Answer: IRR = 21.1 % Cash Flow -$500 200 -500 1200 7A-6 Giventhe followingcash flow,determinethe rate of return on the investment. Year Cash Flow o -$100 1 360 2 -570 3 360 7A-7 Compute the rate of return on the investment characterizedby the followingcash flow. Year Cash Flow o -$110 1 -500 2 300 3 -100 4 400 5 500 7A-8 Compute the rate of return on the investmentchar~ acterizedby the followingcash flow. Year Cash Flow o -$50.0 1 +20.0 2 -40 3 +36.8 4 +36.8 5 +36.8 (Answer: IRR 7A-9 = 15.4%) A firminvested$15,000in a projectthat appearedto haveexcellentpotential.Unfortunately,a lengthylabor dispute in Year3 resulted in costs that exceeded 242 RATEOF RETURN ANALYSIS benefits by $8000. The cash flow for the project is as follows: Year Cash Flow o -$15,000 1 +10,000 2 +6,000 3 -8,000 4 +4,000 5 +4,000 6 +4,000 Compute the rate of return for the project. 7A-10 The followingcashflowhas nopositiveinterestrate. The project, which had a projected life of 5 years, was terminated early. Cash Flow Year o 1 2 -$50 +20 +20 There is, however, a negative interest rate. Compute its value. (Answer: -13.7%) 7A-11 For the following cash flow, compute the rate of return. Year Cash Flow o 1 2 3 4 5 -$20 o -10 +20 -10 +100 7A-12 Given the following cash flow: Year Cash Flow o -$800 1 2 3 4 5 +500 +500 -300 +400 +275 What is the rate of return on the investment? (Answer: 26.55% IRR) 7A-13 Consider the following cash flow. Year o 1 2 Cash Flow -$100 240 -143 If the minimum attractive rate of return is 12%, should the project be undertaken? - - - --- 7A-14 Refer to the strip-mining project in Example 5-10. Compute the rate of return for the project. 7A-15 Consider the following cash flow. Year o 1 2 3 Cash Flow -$500 800 170 -550 Compute the rate of return on the investment. 7A-16 Determine the rate of return on the investment for the following cash flow. . Year o 1 2 3 Cash Flow -$100 +360 -428 168 7A-17 Compute the rate of return on the investment on the following cash flow. Year o 1 2 3 4 5 6 Cash Flow -$1200 +358 +358 +358 +358 +358 -394 7A-18 Determine the rate of return on the investment on the following cash flow. Year o 1-3 4 5-8 Cash Flow -$3570 + 1000 -3170 +1500 7A-19 Bill purchased a vacation lot he saw advertised on television for an $800 down and monthly payments of $55. When he visited the lot he had purchased, he found it was not something he wanted to own. After 40 months he was finally able to sell the lot. The new purchaser assumed the balance of the loan on the lot and paid Bill $2500. What rate of return did Bill receive on his investment? - -- - -- Problems Year 0 1 2 3 4 5 7A-20 Computethe rate of return on an investmenthaving the followingcash flow. Year 0 0 2-9 10 Cash Flow -$850 +600 +200 -1800 7A-21 Assumethat the followingcash flowsare associated with a project. Year 0 1 2 3 4 5 Cash Flow -$16,000 -8,000 11,000 13,000 -7,000 8,950 Compute the rate of return for this project. 7A-22 Compute the rate of return for the following cash flow. Year 0 1 2 3 4 5 6 7 8 Cash Flow -$200 +100 +100 +100 -300 +100 +200 +200 -124.5 7A-25 Consider the following situation. Year Cash Flow What is the rate of return associated with this project? 7A-24 A project has been in operationfor 5 years, yielding the followingannual cash flows: - - -- 1--- o -$200 1 2 +400 -100 What is the rate of return? 7A-26 An investor is considering two mutually exclusive projects. He can obtain a 6% before-tax rate of return on external investments, but he requires a minimum attractive rate of return of7% for these projects. Use a 10 year analysis period to compute the incremental rate of return from investing in Project A rather than Project B. Initial capital investment Net uniform annual income Salvage value 10 years hence Computedrate of return !3 Followingarethe annualcostdata fora tomatopress. Cash Flow -$210,000 88,000 68,000 62,000 -31,000 30,000 55,000 65,000 Cash Flow -$103,000 102,700 -87,000 94,500 -8,300 38,500 Calculate the rate of return and state whether it has been an acceptable rate of return. (Answer: 20%) Year 0 1 2 3 4 5 6 7 243 Project A: Build Drive-Up Photo Shop $58,500 Project B: Buy Land in Hawaii $ 48,500 6,648 0 30,000 138,000 8% 11% 7A-27 In January, 2003, an investor purchased a convertible debenture bond issued by the XLA Corporation. The bond cost $1000 and paid $60 per year interest in annual payments on December 31. Under the convertible feature of the bond, it could be converted into 20 shares of common stock by tendering the. bond, together with $400 cash. The day after the investor received the December 31, 2005, interest payment, he submitted the bond together with $4;00 to the XLA Corporation. In return, he received the 20 shares of common stock. The common stock paid no dividends. On December 31, 2007, the investor . sold the stock for $1740, terminating his 5-year investment in XLA Corporation. What rate of return did he receive? ~~-' '~":'~-~$-.~1 ~. .:.:..~ ~ ..~ f~ . ...--- ~~--:-:-~~~. . ~~ . ~'~ ~k:t?~jt~-~___,;.,~-,~~~;~~~~~.'- ',.&~~- ~ . #_'_:~ 15 ----..,-po . After Completing This Chapter... The student should be able to: Define incremental analysis and differentiate it from a standard present worth, annual worth, and internal rate of return analyses. Use a graphical incrementaltechnique to visualize and solve problems involvingmutually exclusive choices. Use the generalized incremental approach with the internal rate of return method for comparing alternatives. · lliustrate how the present worth of costs versus benefits can be used incrementally. Use spreadsheets to solve incremental analysisproblems. · · · · QUESTIONSTO CONSIDER ;. 1. In today's more environmentally aware decision-making climate, costly features such as the proposed light-rail lanes may have to be included in any proposal for new bridge construction just to get the project off the drawing board. How does this affect the "threshold" cost of public works projects? 2. Any bridge that is built would have to undergo extensive environmental impact review and a public comment period prior to approval. This is likely to result in pressure for additional environmentally friendly components to be built into the project, including methods for limiting emissions from cars that are waiting to access the bridge. How would this affect the ultimate cost of the project? r-._ - - -- - ;,. -------- 1, . I l Incremental Analysis Tapping into Controversy When New York's Tappan Zee Bridge opened in 1955,it seemed big enough to handle any amount of traffic that was likely to come its way.The bridge spanned the Hudson River at a crossing in Westchester County, well north of New York City, near.towns seemingly so removed from Manhattan that they were referred to as "satellite communities." Invoking visions of westward expansion and unpopulated vistas, state government officialsproudJ.Y told voters that the $81 million bridge project would open up "new frontiers." During its first full year of operation, the bridge' carried only 18,000 vehicles per day, far below its design capacity of 100,000.By 2000, the satellite communities were called "suburbs," and the TappanZee Bridge was being used by 135,000 vehicles every day. That number is expected to increase to 175,000 over the next two decades. Clearly, something must be done to ease the growing congestion on ~e bridge. But what? The bridge's usable space has already been expandedonce by convertinga large median strip into a reversibletrafficlane. And the bridge itself is nearing the end of its useful life, which originally was intendedto be 50 years. Upgrading the existing bridge would likely cost over $1 billion and would not adequately alleviate trafficcongestion. Building a new bridge could cost as much as $4 billion dollars-quite a jump from the original price tag. Why so costly? Inflation explains part of it, of course. The original Tappan Zee Bridge would cost almost $600 million in today's dollars. The new bridge would also be bigger. It would have eight lanes instead of six, a 33% increase. But much of the cost would come from additional expenses that could not have been contemplated 50 years ago. Backers of the original bridge happily predicted that it would increase property values and attract new homebuyers to "outlying areas." And so it did. But nowadays, this effect is called suburban sprawl, and most taxpayers want less of it, not more. Many local residents fear that enlarging the bridge would simply encourage even more traffic-soon leading back to the same quandary they are now in. To address these concerns, some government officials have suggested incorporatinglight-rail tracks and bus lanes into the new structure, in an attempt to encourage greater use of mass transit. Others have suggestedinstallation of a more sophisticatedtollcollection system that would give discounts to off-peaktravelers. Any such additionsto the bridge would considerably drive up the cost of construction: ----- I I ,I II Ii II ! l' II II II -------- --------.----. -- I - - -- - -- 246 INCREMENTAL ANALYSIS We now see how to solve problems by each of three major methods, with one exception: for three or more alternatives,no rate of return solutionwas given. The reason is that under these circumstances, incremental analysis is required and it has not been djscussed. This chapter will show how to solve that problem. Incremental analysis can be defined as the examination of the differences between alternatives.By emphasizing alternatives,we are really deciding whether or not differential costs arejustified by differential benefits. In retrospect, we see that the simplest form of incremental analysis was presented in Chapter 7. We did incremental analysis by the rate of return evaluation of the differences between two alternatives.Werecognized that the two alternatives could be rela.tedas follows: Higher-~ost [alternatIve ] Differences Lower-~ost + [alternatIve ] [between them ] = Wewill see that incrementalanalysiscan be examinedeither graphically or numerically.We will first look at graphicalrepresentations of problems, proceed with numerical'solutionsof rate of return problems, and see that a graphical representationmay be useful in examining problemswhetherusingincrementalanalysisor not. " GRAPHICAL SOLUTIONS In the last chapter, we examined problems with two alternatives. O~ method of solution represented a form of incremental analysis. A graphical review of that situation will help to introduceincrementalanalysis. " This is a review of Example 7-6, featuring two mutually exclusive alternatives: Year AIt.1 Alt. 2 o -$10 -$20 1 +15 +28 If 6% interest is assumed, which alternative should be selected? For this problem, we will plot the two alternatives on a graph showingPW of benefits versusPW of cost. Alternative 1 PW of cost PW of benefits = $15(P / F, 6%, 1) = =:::I".. $14.15 ---.. T = $10 - - 11II .._ ___ -- - ~ = II' 15(0.9434) =11'I:II ill:a Graphical Solutions FIGURE 8-t Graph ofPW of benefitsversus PW of cost. 247 $30 Alternative 2 $27 -------------------------- $24 $6 $3 o $3 $6 $9 $12 $15 $18$21 $24 $27$30 PresentWorthof Cost Alternative 2 PW of cost - $20 PW of benefits = $28(PjF, 6%,1) = 28(0.9434) = $26.42 The alternatives are plotted in Figure 8-1, which looks very simple yet tells us a great deal about the situation.' On a graph of PW of benefits versus PW of cost (for convenience,we willcall it a benefit..., cost graph), there will be a line where NPW O.Where the scales used on the two axes are identical, as in this case, the resulting line will be at a 45° angle. If unequal scales are used~the line will be at some other angle. For the chosen interest rate (6% in this example), thisNPW= 0 line divides the graph iIlto areas of desirable and undesirable alternatives.To the left (or above) the line is "de~irable,"while to the right (or below) the line is "undesirable." Wesee that to the left of the liIle,PW of bynefits exceeds thePW of cost or, we could say, NPW is positive. To the right of the lU1e,in the shaded area, PW of benefits is less than PW of cost; thus, NPW is negative. In this example,both alternativesare to the left of the NPW=0 line. Therefore, both alternatives will have a rate of return greater than the 6% interest rate used in COIlstructingthe graph. In fact, other rate of return lines could also be computed and plotted on the graphfor this special case ofa one-year analysis period: We must emphasize ai the outs~t that the additioIlal rate of return lines shownin Figure 8-2 can be plotted only for this special situation. For:analysis period~ greater than one y~ar,the NPW . . 0 line is the only line that caIl be accurately cir"aWn. The graphicalresultS in Figure 8-2 agree with the calculations.made in Example 7-6, Where the ' I . I , . =' ~rates,ofretQrnJor'the'two alternatives were"50%' tind 40%, respectively. :: I' ' il l:1 .. Ii ~ Figure 8..2showsthat1;beslope of a line on the graph represents a p~!cular rate ofrettirnfor thisspecialcase?fa oIle-rearan~ysispe~od.Bet:veenthe2riginapdAlt.1,theslqp~!~We~~ a'=5Q%~Tat~DfY'eti.Ibi':wIfile fr15m'the o~n.gihToKft~ 2: the sTope represeIlts a'"40% raie~fretlllJJ. - -- - - - am .. .,"'".... <., - - - 248 INCREMENTALANALYSIS Rate of Return $30 $30 $27 $27. Alternative 2 $24 '" CI) $24 . Difference ~c:: $21 tE G.> $21 c:: Betweenthe Alternatives' <U ~ E $18 o $18 '5 Alternative 1 '€ o $15 'E $15 o ~c:: $12 ~o ~o /. / 0 ~J'// / I,i / / ~~~o~o ;v ~\:J ~ 11)/\~ 4- .h .b~o'.;; /\:J~o 1 ~ $12 5 G.> ~ Alternative '8 ~ $9 p.., p.., $9 $6 $6 $3 $3 $3 $6 $3 $9 $12. $15 $18 $21 .$24 $27 $30 PresentWorth of Cost FIGURE 8-2 Benefit-cost gra.ph for Example 8-1 with a one-yearanalysisperiod. . $6 $9 $12 $15 $18 $21 $24 $27 $30 PresentWorth of Cost FIGURE 8-3 Benefit-cost graph for Alternatives 1 and 2 with a one-year analysis period. Since [Higher-cost Alt. 2] = [Lower-costAlt. 1] + [Differencebetween them] the difference between the alternatives can be represented by a line shown in Figure 8-3. Viewed in this manner, we clearly see that Alt. 2 may be considered to be two' separate increments of investment. The first increment is Alt. 1 and the second one is the difference between the alternatives.Thus we will select Alt. 2 if the difference between the alternativesis a desirable increment of investment. Since the slope of the line represents a rate of return, we see that the increment is desirable if the slope of the increment is greater than the slope of the 6% line that corresponds to NPW = O. We can see that the slope is greater; hence, the increment of investment is, attractive. In fact, a careful examination shows that the "Differencebetween the alternatives"line has the same slope as.the 30% rate of .return line. We can say, therefore, that the incremental rate of return from selecting Alt. 2 rather than Alt. 1 is 30%. This is the same as was computedin Example 7-6. We conclude that Alt. 2 is the preferred alternative. Solve Example 7-10 by means of a benefit-cost graph. Two machines are being considered for purchase. If the minimum attractive rate of return (MARR) is 10%, which machine should be .pought? - -- .. -- Graphical Solutions Machine X $200 95 50 6 Initial cost Uniform annual benefit End-of-useful-life-salvage value Usefullife, in years ,. . -. 249 Machine Y $700 120 150 12 .'. SOLUTION Noting that for both Machine X and Machine Y, the PW of cost reflectsthe conventionof treating salvage value as a reduction in cost rather than a benefit, and using a 12-yearanalysis period, we have Machine X PW of cost:;::: 200 + (200 -- 50)(P I F, 10%,6) - 50(P / F, 10%,12) = 200+ 150(0.5645)- 50(0.3186) = 269 PW of benefit :;:::95(PIA,1O%,12) = 95(6.814):;::: 647 Machine Y fW of cost:;::: 700 - 150(P IF, 10%,12) = 700 - PW of benefit = 120(P I A,10%,12)-120(6.814) 150(0.3186)==652 I :;:::818 I Wben the two alternativesplotted (Figure 8-4), we see that the incrementY-X has a slope much less than the 10%rate of return line. The rate of return on the increment of investmentis less than 10%;hence, the increment is undesirable. This means that Machine X should be selected rather than Machine Y. EIGURE8-4 Benefit'-{;ostgraph. I. .t $900 y $800 '" $700 q:: g0) $600 'I Ii J:Q 'S" $500 II € ,\ 'I ~i: $400 I] Ii' I ~ $300 J: $2QO "' ,,$Joo := = - " $100 $200 $300 $400 $500 $600 $700 $800 Present Worth o£?€ost==f==- 'III~ "" - 250 INCREMENTALANALYSIS FIGURE 8-5 Solving multiple-alternative problems by successive two-alternative analyses. Three Alternatives ~1 B A~Y'i' Bestof3 tLI C Alternatives Four Alternatives 1\vo-Alternative Analysis J K Best of 4 Alternatives L M Five Alternatives 1\vo-Alternative Analysis p Q R S T Best of 5 Altemlttives The two example problems show us some aspects of incremental analysis. We now will examine some multiple-alternativeproblems. We can solve multiple-alternativeproblems by means of present worth and annual cash flow analysis without any difficulties.Rate of return analysis requires that the differences between two alternatives must be examined to see whether either is desirable. Now, if we can choose between two alternatives,then by a successiveexamination we can choose from multiple alternatives.Figure 8-5 illustrates the method. Consider three mutually exclusive alternatives: Initial cost Unifofriiannual bem5fit A $2000 ~41O ~ C $5000 700" B $4000 639 Each alternative has a 20-year life and no salvage value. If the MARR is 6%, which alternative II shouldhe,selected? = 1S; -- -- - - : ----- - - Graphical Solutions 251 At 6%, (PW of benefits) = (Uniform annual benefit) x (Series present worth factor). PW of benefits = (Uniform annual benefit)(P / A, 6%,20) PW of benefits for Alt. A = $410(11.470) = $4703 = $639(11.470) = $7329 Alt. C = $700(11.470) - $8029 Alt. B Figure 8-6 is a plot of the situation; we see that the slope of the line from the origin to A is greater than the 6% line (NPW = 0). Thus therate ofretumfor A is greater than 6%. For the increment of additional cost of B over A, the slope of line B-A is greater than the 6% line. This indicates that the rate of return on theincrement of investmentalso exceeds 6%.But J:4eslope of increment C-B indicates its rate of return is less than 6%, b,ence,undesirable. Weconclude that the Ainvestment is satisfactory as well as the B-,-Aincrement; therefore, B is satisfactory.The C-B increment is unsatisfactory; so C is undesirable ip comparison to B. Our decision is to select Alternative B, E'IGURE 8-6 Benefit--cost graph for Example 8-3. $9000 $8000 I 1 I '" ilii 1' $3000 $5000 $7000 PresentWorthof Cost 'Ii 1 . 1' .1I' ,I ' ,ql I I ;1 I \ .'I! :11 If' Further study of the three alternativesof Ex:ample8-3 reveals that Alt. A's uniform annualbenefit was overstated.It is now projected to be $122 rather than $410. Replotthebep.efit-cost graph for ~ this cbanged,situ,<i.tiop. '" ~ L_ -_,_ _ ,_____ -- '= ~ _ - ; em' III~= ~ IB ~,; '" ~ i:i: ""'.>"iI 252 INCREMENTALANALYSIS Alt. A': PW of benefits = 122(P / A, 6%, 20) = 122(11.470) = 1399 The revised plot of the three alternatives (Figure 8-7) shows that the revised Alt. A' is no longer desirable. We see that it has a rate of return less than 6%. FIGURE 8-7 Benefit-cost graph for Example 8-4. $9000 $8000 $7000 -'" f.,t:; g(\) $6000 ~ '"0 $5000 ..Q 1:: $4000 ~- = ~ $3000 (\) '" ~ $2000 $1000 o $1000 $3000 $5000 $7000 PresentWorthof Cost $9000 Now we wish to examineAlt. B. Shouldwe compareit with thedo-nothingalternative(wIiich is representedby the origin) or as a B-A' increment over A'? Graphically;shouldwe examine line O-B or B-A'? Since A' is an undesirable alternative, it should be discarded and not considered further. Thus ignoring A', we should compare B with the do-nothing alternative, which is line O-B. Alternative B is preferred over the do-nothing alternative because it has a rate of return greater than 6%. Then increment C--B is examined and, as we saw,it is an undesirable increment of investment.The decision to select B has not changed,which shouldbe no surprise:if an inferior A had become an even less attractive A', we still would select the superior alternative, B. The graphical solution of the four exampleproblems has helped us visualize the mechanics of incremental analysis. While problems could be solved this way, in practice they are solved mathematically rather than graphically. We will now proceed to solve problems mathematically by incremental rate of return analysis. INCREMENTAL RATE OF RETURN ANALYSIS To illustrate incremental rate of return analysis, we will solve three of the examples again by mathematical, rather than graphical, methods. -- - - --- -- Incremental Rate of Return Analysis Solve Example 8-1 mathematically.With two mutually exclusive alternatives and a 6% MARR, which alternative should be selected? Year o 1 Alt. 2 -$20 +28 Alt. 1 -$10 +15 Examine the differences between the alternatives. Alt. 2 - Alt. 1 Year = -10 o -20 - (-10) 1 +28 - (+15) -+13 Ip.crementalinternal rate of return (~IRR), 10 = 13(PIF, i, 1) 10 (P I F, i, 1) = --,.. 13 = 0.7692 LlIRR. 30% The Ll:IBRis greater than the MARR; hence, we will select the alteroative that gives this increment. [Higher-cost All. 2] = [Lower-costAlt. 1] + [mcremel1tbetween theIll] . Select Alternative 2. Recompute Example 8-3 for MARR = 6%. Each alternative has a 20-year life and no salvage value. Initial cost Uniform annual benefit B $4000 639 A $2000 410 C $5000 700 A practical first step is to compute the rate of return for each alt~rnative. Alt~matiye A 410(PI A, t, 20) - -"-"11 .A . ""' . .-"2000 (p'. I.l;1,l, ""L. 0) = "'="'-="''-t. 878. 2000 . = . 410 ''''~ ,' . -- nT/ r-" "20 -/o'~ w ..------ -- - r--I 253 1 254 INCREMENTALANALYSIS Alternative B 4000 = 639(P I A, i, 20) _4000 = 6.259 i (PIA, i, 20) = 639 = 15% Alternative C = 700(P I A, i, 20) 5000 . (PIA, i, 20) 5000 = -700 = 7.143 The rate of return is between 12 and 15%: i = 12% + 7.469 -7.143 7.469 - 6.259 ( ) (3%) = 12.8% At this point, we would reject any alternative that fails to meet the MARR criterion of 6%. All three alternatives exceed the MARR in this example. Next, we arrange the alternatives in order of increasing initial cost. Then we can examine the increments between the alternatives. A $2000 410 20% Initial cost Uniform annual benefit Rate of return Incremental cost Incremental uniform annual benefit B $4000 639 15% C $5000 700 12.8% Increment B-A $2000 229 Incremental rate of return: 2000 (PIA,i,20) = 229(P I A, = 2000 229 i, 20) ~IRR 9.6% The B-A increment is satisfactory; therefore, B is preferred over A. Because B is preferred to A, the next incremental comparison is of C with B. (If A was preferred to B, then the next incremental comparison wouldhave been of C with A.) - ..-- Incremental cost JncremenJal'l,.uMofIIJ, an,nualb~nefi! - ...-- - - ...--- Increment C-B $1000 pI - --- - ----- _ ,_ _"_,, 'U' ,__ Incremental Rate of Return Analysis 1000 = 61(P IA, i, 20) 1000 (PIA, i, 20) 255 ~IRR = 61 2.0% The C- B increment has an unsatisfactory 2% rate of return; therefore, B is preferred over C. (Answer: Alt. B) Solve Example 8.4 mathematically.Alternative A in Example 8-6 was believed to have an overstated benefit. The new situation for A (we will again call it A') is a uniform annual benefit of 122. Compute the rate of return for A'. SOLUTION 2000 (PIA, i, 20) = 122(P I A, i, 20) 2000 = -122 = 16.39 i -.2% This time Alt. A' has a rate of return less than the MARR of 6%. AlternativeA' is rejected, and the problem now becomes selecting the better of B and C. In Example 8-6.we saw that the increment C-B had a ~IRR of 2% and it, too, was undesirable. Thus, we again select Alternative B. The following information is provided for five mutually exclusive alternatives that have 20-year useful lives. If the minimum attractiverate of return is 6%, which alternativeshould be selected? Alternatives A Cost Uniform annual benefit PW of benefit* C B $4000 639 7330 $2000 410 4700 ; = 20q;h= w: Rate of return 15% *PW of benefit = (uniform annual benefit)(P I A, 6%,20)= val1,leswill be,used later to plot a curve'f()r,P~f . - -- - ...... $6000 761 8730 :I: 'ii% D $1000 117 - -1340 10% E $9000 785 9000 < . iI 6% 11.470 (uniform annual benefit). These c()stversus'PWofbenBftt,1I' ---- II I ;; I I - -- ! 256 INCREMENTAL ANALYSIS SOLUTION We see that the rate of return for each alternative equals or exceeds the MARR, therefore, no alternatives are rejected at this point. Next, we rearrange the alternativesto put them in order of increasing cost: Alternatives Cost Unifonn annual benefit Rate of return D B A C f $1000 117 10% $2000 410 20% $4000 639 15% $6000 761 11% $9000 785 6% Increments L\Cost L\Annual benefit L\Rate of return B-D A-B C-A $1000 293 29% $2000 229 10% $2000 122 2% Beginning with the analysis of increment B-D, we compute..anIRRB-D of29%. Alternativ~ B is thus preferred to Alt. D, and D may be discarded at this point. The IRRA-B for A-B is also satisfactory,so A is retained and B is now discarded:The C-A incrementhas a rate of return less than the MARR. Therefore, C is discarded and A continues to be retained. . At this point, we have examined four alternatives-D; B, A, C-arid hav.eretained A after discarding the other three. Now we must decide whether A or E is the superior alternative. The increment we will examine is E-A. (Note:Increment E-C would have no particular meaning for we have already discarded C.) ~Cost ~Annual benefit Increment E-A $5000 146 Over the 20-year useful life, the total benefits (20 x 146 - 2920) are less than the cost. Thus, the rate of return on this incrementis negative.This is an unsatisfactoryincrement, so E is discarded. Alternative A is the best of the five alternatives. The benefit-cost graph (Figure 8-8) of this example illustrates an interesting situation. All five alternatives have rates of return equal to or greater than the MARR of 6%. Yet, on detailed examination, we see that Alternatives C and E contain increments of i~vestment that are unsatisfactory. Even though C has an 11% rate of return, it is unsatisfactory in comparison to Alt. A. Also noteworthy, the project with the greatest rate of return-Alternative B-is not the best alternative, since the proper economic criterion in this situation is to accept all separable increments of investmentthat have a rate of return greater than the 6% MARR. --- - Elements in Incremental Rateof Return Analysis 257 FIGURE 8-8 Benefit-cost graph for Example 8-8 data. .6% $9000 $8000 J!J / A I '. $7000 $6000 !.;:1 j:Q 'c; $5000 "€ $4000 -= $3000 $2000 $1000 o $1000 $3000 $5000 $7000 $9000 Present Worth of Cost We found a desirable A-B increment with a 10% IRRA-B. For the relationship between Alternatives A and B, and the computed rates of return, we write: Higher-cost Alt. A: [ 15% rate of return ] = Alt. B: 20% rate [ ofreturn ] + Differencesbetween A and B: [ 10% rate ofreturn ] By selecting Alt. A we have, in effect, acquired a 20% rate of return on $2000'and a 10% rate of return on an additional $2000, both desirable outcomes. Taken together as A, the result is a 15%rate of return on a $4000 investment.This is economically preferable to a 20% rate of return on a $2000 investment, assuming we seek to invest money whenever we can identify an investmentopportunitythat meets our 6% MARR criterion. This implies that we have sufficientmoney to accept all investmentopportunities for which the MARR is exceeded that come to our attention. This assumption of an abundarit supply of money is considered appropriate in most industrial analyses, but it is not likely to be valid for individuals. The selection of an appropriate MARR is discussed in Chapter 15. ELEMENTS IN INCREMENTAL RATE OF RETURN ANALYSIS 1. Be sure all the alternatives are identified. In textbook problems the alternatives will be well defined,but industrial problems may be less clear.Before proceeding, one must have all the mutually exclusive alternatives tabulated, including the "do-nothing" or "k~ep doing the same thing" alternative, if appropriate. 2. (Optional) Compute the rate of return for each alternative. If one or more alternatives has a rate of return equal to or greater than the minimum attractive rate of return (IRR 2: MARR), then any other alternativeswith an IRR less than the MARR may be immediatelyrejected. This optional step requirescalculationsthatmayor may not eliminate I) . - -- -r - ----- 258 INCREMENTALANALYSIS alternatives.In an exam, this step probably shouldbe skipped,but in lesspressing situations, these are logical computations. 3. Arrange the remaining alternatives in ascending order of investment. The goal is to organizethe alternativesso that the incrementalanalysiswillbe of separableincrements of investmentwhen we analyze the higher-cost alternativeminus the lower-costalternative. In textbook problems, this is usually easy-but there are lots of potential difficulties in following this simple rule. The ordering of alternatives is not a critical element in incremental analysis; in fact, the differences between any two alternatives,like X and Y, can be examined either as an X-Y increment or a Y-X increment. If this is done in a random fashion, however, we will be looking sometimes at an increment of borrowing and sometimes at an increment of investment.It can get a little difficultto keep all this straight;the basic goal is to restrict the incremental analysis, where possible, to increments of investment. 4. Make a two-alternative analysis of the first two alternatives. If the increment examined is Y - X, which represents an increment of investment. Higher-cost [ Alt. Y ] _ LOWer-cost [ Alt. X Differences between ] + [them (Y - X) ] Compute the ~IRR on the increment of investment. The criterion is: ·· · If ~IRR ~ MARR, retain the higher-cost Alt. Y. If ~IRR < MARR, retain the lower-costAlt. X. Reject the other alternativeused in the analysis. Sometimes the two alternatives being examined cannot be described as "higher cost" and "lower cost." In Example 7-8, we encounteredtwo alternatives,A and B, with equal investments. There we selected the A-B increment because it was an increment of investment. On the other hand, when we are examihingan incrementof borrowing, the criteria are as follows: · · If ~IRR ::::MARR, the increment is acceptable. If ~IRR > MARR, the increment is not acceptable. 5. Take the preferred alternative from Step 4, and the next alternative from the list created in Step 3. Proceed with another two-alternativecomparison. 6. Continue until all alternatives have been examined and the best of the multiple alternatives has been identified. Incremental Analysis with Unlimited Alternatives At times the possible alternatives are a more or less continuous function. For example, in an analysis to determine the economical height of a dam, the number of alternatives could be infinite. If the alternatives were limited, however, to heights in even feet, the number of alternatives would still be large and would have many of the qualities of a continuous function, as in the following example. - -- --- Elements in Incremental Rate of Return Analysis 259 A careful analysis has been made of the consequences of constructinga dam in the Blue Canyon. It would be feasible to construct a dam at this site with a height anywhere from 200 to 500 feet. Using a 4% MARR and a 75-year life, the various data have been used to construct Figure 8-9. Note particularly that the dam heights are plotted on the x axis along with the associated PWof cost. What height of dam should be constructed? FIGURE 8-9 Benefit-costgraph for Example 8-9. 11 :1 II 1' I1 ~ 1' -c: 4) CIJ e i:l.o I1 "I ,I , 1 Ii II !! Ii 'II 1 2 3 4 5 6 7 8 9 10 11 PresentWorth of Cost (millionsof dollars) I 200 I I 250300 I I 400 I 500 Ii j PaIn Height (ft) Five points have been labeled on Figure 8-9 to aid in the discussion. Damheignts below Point A have a PW of cost> PW of benefit; hence, the rate of return is le.ssthanMARR.. and we woulq not build a dam of these heights..In the region of Point B, an increment of additional PW of cost-.6.C-produces a larger increment of PW ofbenefit-AB. These are,tI1erefore,desirable increments of additional investment and, hence, dam height. At poirit D and also atPoint E, tI1e reverse is true. An increment of additionalJnvestment-.6.C-p],"oduces."a smalle.rincreIIientof PW of benefit-.6.B; this is undesirable.We do not want these increments, so.tI1e~aIIlshould not be bUiltto these heights. At Point C, we see that fJ.B= .6.C.LoWer dam heights have desirable increments of.in.vestmeIltand higher dam heights have unfavorable increments of investri;lent.1;'heoptimal dam height, th.erefore, is where.6.B = .6.C.In Figure 8..9,this corresponds II I to a heighto£app],"oximately 250 feet. Another way of definingthe poin.tatwhich fJ.B= AC is to describe it as tI1epointwnere the slope of the curve equals the slope of the NPW 0 line, il The techniquesfor solvingdiscrete alternativesor continuousfunction alternativesarereally the same. We proceed by adding increments whenever .6.IRR2:MARR and discarding increments when, fJ.IRR< MARR. ii, "~ ", -- 260 INCREMENTALANALYSIS PRESENT WORTH ANALYSIS WITH BENEFIT-COST GRAPHS Any of the exampleproblems presented so far in this chaptercould be solved by the presel worth method. The benefit-cost graphs we introduced earlier can be used to graphicall solve problems by present worth analysis. In present worth analysis,where neither input nor outputis fixed,the economiccriterio is to maximize NPW. In Example 8-1, we had two alternatives and the MARR equc to 6%: These points are plotted in Figure 8-10, which has the NPW = 0 line at 450,since identical scales were used on both axes. The point for Alt. 2 is plotted at the coordinates (PW of cost, PW of benefits). We drop a vertical line from Alt. 2 to the diagonal NPW = 0 line. The coordinates of any point on the graph are (PW of cost, PW of benefits), but along,the NPW = 0 line (450line), the x and y coordinates are equal. Thus, the coordinates of Point4 are also (pW of cost, PW of cost). Since NPW = PW of benefits - PW of cost the vertical distance from Point 4 to Alt. 2 represents NPW.Similarly, the vertical distance between Point 3 and Alt. 1presents NPW for Alt. 1.Sincethe criterionis to maximizeNPW, FIGURE 8-10 Benefit-cost graph for PW analysis. 2 $27 $24 '" $21 tE ~ $18 ~ r:Q 'C3 $15 '€ ~ $12 C ~ £ $9 $6 $3 $3 .. --'-- - $6 $9 $12 $15 $18 $21 PresentWorth of Cost Choosing an Analysis Method 261 c $9000 $8000 A '" $7000 ..... I.t: g cu j:Q $6000 B 'C; $5000 I- € $4000 ..... Q £ $3000 $2000 $1000 $1000 $3000 $5000 $7000 PresentWorthof Cost $9000 FIGURE 8-11 Present worth analysis of Example 8-8 using a benefit-cost plot. we select Alt. 2 with larger NPW. The same technique is used in situations where there are multiple alternatives or continuous alternatives. In Example 8-8 there were five alternatives; Figure 8-11 plots these five alternatives. We can see that A has the greatest NPW and, therefore, is the preferred alternative. CHOOSING AN ANALYSIS METHOD At this point, we have examined in detail the three major economic analysis techniques: present worth analysis, annual cash flow analysis, and rate of return analysis. A practical question is, which method should be used for a particular problem? While the obvious answer is to use the method requiring the least computations, a number of factors may affect the decision. . 1. UnlesstheMARR-minimum attractiverate ofreturn (orminimum requiredinterest rate for investedmoney)-is known, neither present worth analysis nor annual cash flow analysis is possible. 2. Present worth analysis and annual cash flow analysis often require far less computation than rate of return analysis. 3. In some situations, a rate of return analysis is easier to explain to people unfamiliar with economic analysis. At other times, an annual cash flow analysis may be easier to explain. 4. Business enterprises generally adopt one, or at most two, analysis techniques for broad categories of problems. If you work for a coq>orationand the policy manual specifies rate of return analysis, you would appear to have no choice in the matter. Since one may not always be able to choose the analysis technique computationally best suited to the problem, this book illustrates how to use each of the three methods in all -- . 262 INCREMENTALANALYSIS feasible situations. Ironically,the most difficultmethod-rate of return analysis-is the one most frequently used by engineers in industry! SPREADSHEETS AND INCREMENTAL ANALYSIS An incremental analysis of two alternativesis easily done with the RATEor IRR functions when the lives of the alternatives are the same. The problem is more difficult, however, when the lives are different.As discussedin Chapters5 and 6, when comparing alternatives having lives of different lengths, the usual approach is to assume that the alternatives are repeated until the least commonmultiple of their lives.This can be done with a spreadsheet, but Excel supports an easier approach. Excel has a tool called GOAL SEEKthat identifies a formula cell, a target value, and a variablecell. This tool causes the variablecell to be changed automaticallyuntil the formula cell equals the target value. To find an IRR for an incremental analysis, the formula cell can be the difference between two equivalent annual worths with a target value of O.Then if the variable cell is the interest rate, GOAL SEEK will find the IRR. In Excel this tool is accessed by selecting T(ools) on the main toolbar or menu and G(oal seek) on the submenu.As shown in Example 8-10, the variable cell (with the interest rate) must somehow affectthe formula cell (differencein equivalentuniform annual worths or costs (EUAWsor EUACs», although it need not appear directly in the formula cell. IiI Figure 8-12a and 8-12b the interest rate (cell AI) appears in the EUAC formulas (cells D3 and D4), but not in the formula cell (cell D5). Two different asphalt mixes can be used on a highway. The good mix will last 6 years, and it will cost $600,000 to buy and lay down. The better mix will last 10 years, and it will cost $800,000 to buy and lay down. Find the incremental IRR for using the more expensivemix. SOLUTION This example would be difficult to solve without the GOAL SEEK tool. The least common multiple of 6 and 10 is 30 years, which is the comparison period. With the GOAL SEEK tool a very simple spreadsheet does the job. In Figure 8-12a the spreadsheeUs shown before GOAL SEEK. Figure 8-12b shows the result after the goal (D5) has been set .0 and A1selected as the variable cell to change. . -. A B -. C .. D E F ;...---,.,'-...,.".,.-",.."..,.,-",-. 1 8.00% Interest rate 2 Alternative Cost Life EUAC 3 4 5 Good Better 600000 6 129,789 ! 800000 " 10 difference= ..119,2 . to,5§ .. ,:: -PMT($A$I,C3,B3) -- -- :I'::S'-PMT($A$1C4,B4)' ~ -....... . ,0=;D3 -D4 FIGURE 8-12a Spreadsheetbefore,GOAI--SEp:rC.. -.. ----- -T ~ Summary A , 1 2 3 4 5 B 14.52% Interestrate Alternative Cost Good 600000 Better 800000 C D Life 6 10 EUAC 156,503 156,503 0 difference = E 263 F = -PMT($A$1,C3,B3) = -PMT($A$1,C4,B4) =,D3-D4 FIGURE 8-12b Spreadsheet after GOAL SEEK. The incremental IRR found by GOAL SEEK is 14.52%. .... SUMMARY A graph of the PW of benefits versus PW of cost (called a benefit-cost graph) can be an effective way to examine two alternatives by incremental analysis. And since multiplealternative incremental analysis is done by the successive analysis of two alternatives, benefit-cost graphs can be used to solve multiple-alternative and continuous-function alternatives as easily as two alternativeproblems. The important steps in incremental rate of return analysis are as follows: 1. Check to see that all the alternatives in the problem are identified. 2. (Optional)Computethe rate of return for each alternative.If one or more alternatives has IRR :::MARR, reject any alternatives with IRR < MARR. ,3. Arrange the remaining alternatives in ascending order of investment. 4. Make a two-alternative analysis of the first two alternatives. 5. Take the preferred alternative from Step 4, and the next alternativefrom the list in Step 3. Proceed with another two-alternativecomparison. 6. Continue until all alternatives have been examined and the best of the multiple alternatives has been identified. Decision Criteria for Increments of Investment ·. · If LlIRR ::: MARR, retain the higher-cost alternative. If LlIRR < MARR, retain the lower-cost alternative. Reject the other alternative used in the analysis. Decision Criteria for Increments of Bo"owing · · If, LlIRR ~ MARR, the increment is acceptable. If, LlIRR > MARR, the increment is not acceptable. Benefit-cost graphs, which plot the PW of benefits versus the PW of cost, can also be used in present worth analysis to graphically show the NPW for each alternative. - J --- 264 INCREMENTALANALYSIS PROBLEMS Unless otherwise noted, all problems for Chapter 8 should be solved by rate of return analysis. 8-1 A firm is considering moving its manufacturing plant from Chicago to a new location. The industrial engineering department was asked to identify the various alternatives together with the costs to relocate the plant, and the benefits. The engineers examined six likely sites, together with the do-nothing alternative of keeping the plant at its present location. Their findings are summarized as follows: Plant Location Denver Dallas San Antonio Los Angeles Cleveland Atlanta Chicago First Uniform Annual Cost ($OOOs) Benefit ($OOOs) $ 52 $300 137 550 117 450 167 750 18 150 49 200 8-3 Alternative E F 8-4 The annual benefits are expected to be constant over the 8-year analysis period. IT the firm uses 10% annual interest in its economic analysis, where should the manufacturing plant be located? (Answer: Dallas) 8-2 Alternative A B C D Annual Benefit $100 at end of first year;increasing$30 per year thereafter $10 at end of first year;increasing$50 per year thereafter Annual end-of-year benefit= $260 $450at end of first year; declining$50 per year thereafter Useful Life (years) 10 10 Calculated Rate of Return 10.0% 11.2% The owner of a downtown parking lot has employed a civil engineering consulting firm to advise him on the economic feasibility of constructing an office building on the site. Bill Samuels, a newly hired civil engineer, has been assigned to make the analysis. He has assembled the following data: Alternative In a particular situation, four mutually exclusive alternatives are being considered. Each of the alternatives costs $1300 and has no end-of-useful-life salvage value. Useful Life (years) 10 Cost $3000 5850 Annual End-ofYears Benefit $ 488 1000 IT the MARR remains at 8%, which one of the six alternatives should be selected? Neither Alt. E nor F has any end-of-useful-life salvage value. (Answer: Alt. F) o o A more detailed examination of the situation in Problem 8-2 reveals that there are two additional mutually exclusive alternatives to be considered. Both cost more than the $1300 for the four original alternatives. Sell parking lot Keep parking lot Build I-story building Build 2-story building Build 3-story building Build 4-story building Build 5-story building Calculated Rate of Return 10.0% Total Investment$. 0 200,000 400,000 555,000 750,000 875,000 1,000,000 Total Net Annual Revenue from Property $ 0 22,000 60,000 72,000 100,000 105,000 120,000 *Includes the value of the land. 10 8.8% 10 15.0% 10 18.1% IT the MARR is 8%, which alternative should be selected? (Answer: Alt. C) The analysis period is to be 15 years. For all alternatives, the property has an estimated resale (salvage) value at the end of 15 years equal to the present totaJ investment. If the MARR is 10%, whatrecommendation should Bill make? 8-5 An oil company plans to purchase a piece of vacan1 land on the comer of two busy streets for $70,000. On properties of this type, the company installs businesses of four different types. Problems Cost of Plan Improyements. A $ 75,000 B Type of Business Conventional gas station with service facilities for lubrication, oil changes, etc. Automatic carwash facil- 230,000 ity with gasoline pump island in front c 30,000 D 130,000 First cost Uniform annual benefit Usefullife, in years End-of-useful-life salvage value Computed rate of return Discount gas station (no service bays) Gas station with low-cost, quick-carwash facility 8-8 D $85 17 10 o o o o 16.8% 20.2% 15.1% 15.1% A $200 59.7 B $300 77.1 C $600 165.2 5 5 5 o o o 15% 9% 11.7% _ Net Annual Income $23,300 44,300 10,000 27,500 MARR.:::_%." 8-9 A firm is considering three mutually exclusive alternatives as part of a production improvement program. The alternatives are as follows: A B C $10,000 1,625 10 $15,000 1,625 20 $20,000 1,890 20 8-10 Consider four mutually exclusive alternatives,each having an 8-year useful life: A C B D First cost $1000 $800 $600 $500 Uniform annual 122 120 97 122 benefit 750 500 500 0 Salvagevalue Three mutually exclusive projects are being consid.ered: First cost Uniform annual benefit Salvagevalue Usefullife, in years A $1000 150 B $2000 150 C $3000 0 1000 5 2700 6 5600 . 7 When each project reaches the end of its useful life, it would be sold for its salvage value and there would be no replacement. If 8% is the desired rate of return, which project should be selected? Given the following four mutually exclusive alternative,s and using 8% for the MARR, which alternative should be selected? --------. --- . If the minimum attractive rate of return is 8%, which alternative should be selected? For each alternative, the salvage value at the endof-useful-life is zero. At the end of 10 years, Alt. A could be replaced by another A with identical cost and benefits. The MARR is 6%. If the analysis period is 20 years, which alternative should be selected? 8-7 C $50 10 10 For what range of values of MARR is Alt. C the preferred alternative? Put your answer in the follow% .::: ing form: "Alternative C is preferred when If the oil company expects a 10% rate of return on its investments, which plan (if any) should be selected? Installed cost Uniform annual benefit Usefullife, in years B $50 12 10 Consider the following three mutually exclusive alternatives: First cost .Uniform annual benefit Useful life, in years End-of-useful-life salvage value Computed rate of return In each case, the estimated useful life of the improvements is 15 years. The salvage value for each is estimated to be the $70,000 cost of the land. The net annuat income, after paying all operating expenses, is projected as follows: 8-6 A $75 16 10 (Answer: A) *Cost of improvements does not include the $70,000 cost of land. Plan A B C D 265 -- ..- --_.---- -- -- 266 INCREMENTAL ANALYSIS 8-11 Consider three mutually exclusive alternatives: Year o 1 2 3 4 Buy X -$100.0 +31.5 +31.5 +31.5 +31.5 BuyY -$50.0 +16.5 +16.5 +16.5 +16.5 8-15 Given the following: Year Do Nothing o 1 o o o o o = = 8-12 Consider the three alternatives: 0 1 2 3 4 5 A -$100 +30 +30 +30 +30 +30 B -$150 +43 +43 +43 +43 +43 Initial cost Annual benefitin each of first 5 years Annual benefit in each of subsequent5 years Which alternative should be selected: (a) IfMARR = 6%? (b) IfMARR = 8%? (c) If MARR = 1O%? 8-13 A firm is considering two alternatives: . A B $10,700 $5500 2,100 1800 o 8 o 4 At the end of 4 years, another B may be purchased with the same cost, benefits, and so forth. If the MARR is 10%, which alternative should be selected? 8-14 Consider the following alternatives: Initial cost Uniform annual benefits A $300 41 8-16 Consider four mutually exclusive alternatives: A B C 'D Initial cost $770.00 $1406.30 $2563.30 0 Uniform annual 420.00 420.00 420.00 0 benefit 2 4 8 0 Useful life, in years 6.0% 7.5% 6.4% 0 Computed rate of return The analysis period is 8 years. At the end of 2, 4, and 6 years, Alt. A will have an identical replacement. Alternative B will have a single identical replacement at the end of 4 years. Over what range of values of MARR is Alt. B the preferred alternative? 8-17 Consider the three alternatives: Do Nothing 0 0 0 0 0 0 Initial cost Uniform annual benefits Salvage value at end of useful life Useful life, in years Y -$20 +28 Over what range of values ofMARR is Y the preferred alternative? Which alternative should be selected: (a) If the minimum attractive rate of return equals 6%? 9%? (b) If MARR (c) If MARR = 10%? (d) IfMARR 14%? (Answers: (a) X; (b) Y; (c) Y; (d) Do nothing) Year X -$10 +15 B $600 98 C $200 35 Each alternative has a lO-year useful life and no salvage value. If the MARR is 8%, which alternative should be selected? A $1500 250 B $1000 250 C $2035 650 450 250 145 Each alternative has a lO-year useful life and lio salvage value. Based on a MARR of 15%, which alternative should be selected? Where appropriate, use an external interest rate of 10% to transform a cash flow to one sign change before proceeding with rate of return analysis. 8-18 A new 1O,000~square-meter warehouse next door to the Tyre Corporation is for sale for $450,000. The terms offered are $100,000 down with the balance being paid in 60 equal monthly payments based on 15% interest. It is estimated that the warehouse would have a resale value of $600,000 at the end of 5 years. Tyre has the needed cash available and could buy the warehouse but does not need all the warehouse space at this time. The Johnson Company has offered to lease half the new warehouse for $2500 a month. Tyre presently rents and utilizes 7000 square meters of warehouse space for $2700 a month. It has the option of reducing the rented space to 2000 square -- - -- ;' Problems meters, in which case the monthly rent would be $1000 a mont;h. Further, Tyre could cease renting warehouse space entirely. Tom Clay, the Tyre Corp. plant engineer, is considering three alternatives: 1. Buy the new warehouse and lease the Johnson Company half the space. In turn, the Tyre-rented space would be reduced to 2000 square meters. 2. Buy the new warehouse and cease renting any warehouse space. 3. Continue as is, with 7000 square meters of rented warehouse space. Based on a 20% minimum attractive rate of return, which alternative should be selected? 8-19 Consider the following alternatives: Initial cost Uniform annual benefit Useful life, in years A $100.00 10.00 B $150.00 17.62 C $200.00 55.48 Infinite 20 5 Use present worth analysis, an 8% interest rate, and an infinite analysis period. Which alternative should be selected in each of the two following situations? 1. Alternatives B and C are replaced at the end of their useful lives with identical replacements. 2. Alternatives Band C are replaced at the end of their useful lives with alternatives that provide an 8% rate of return. 8-20 A problem often discussed in the engineering economy literature is the "oil-well pump problem."} Pump 1 is a small pump; Pump 2 is a larger pump that costs more, will produce slightly more oil, and will produce it more rapidly. If the MARR is 20%, which pump should be selected? Assume that any temporary external investment of money earns 10% per year and that any temporary financing is done at 6%. } One of the more interesting exchanges of opinion about this problem is in Prof. Martin Wohl's "Common Misunderstandings About the Internal Rate of Return and Net Present Value Economic Analysis Methods;" and the associated discussion by Professors Winfrey, Leavenworth, Steiner, and Bergmann, published in Evaluating Transportation Proposals, Transportation Research Record 731, Transportation Research Board, Washington, D.C. See also Appendix 7A in Chapter 7. Pump 1 ($OOOs) -$100 +70 +70 Year 0 1 2 267 Pump 2 ($OOOs) -$110 +115 +30 8-21 Three mutually exclusive alternatives are being studied. If the MARR is 12%, which alternative should be selected? Year o 1 2 3 4 A -$20,000 +10,000 +5,000 +10,000 +6,000 c - $20,000 B -$20,000 +10,000 +10,000 +10,000 +5,000 +5,000 +5,000 +15,000 o 8-22 The South End bookstore has an annual profit of $170,000. The owner is considering opening a second bookstore on the north side of the campus. He can lease an existing building for 5 years with an option to continue the lease for a second 5-year period. If he opens the second bookstore, he expects the existing store will lose some business that will he gained by "The North End," the new bookstore.. It will take $500,000 of store fixtures and inventory to open The North End. He believes that the two stores will have a combined profit of $260,000 a year after all the expenses of both stores have been paid. The owner's economic analysis is based on a 5-yearperiod. He will be able to recover this $500,000 investment at the end of 5 years by selling the store fixtures and inventory. The owner will not open The North End unless he can expect a 15% rate of return. What should he do? Show computations to justify your decision. . 8-23 A paper mill is considering two types of pollution control equipment. Initial cost Annual chemical cost Salvage value Usefullife, III years Neutralization $700,000 40,000 175,000 5 Precipitation $500,000 110,000 125,000 5 The firm wants a 12% rate ofreturn on any avoidable increments of investment. Which equipment should be purchased? 8-24 A stockbroker has proposed two investments in lowrated corporate bonds paying high interest rates and ..~ T L 268 INCREMENTALANALYSIS selling below their stated value (in other words, junk bonds), Both bonds are rated as equally risky. Which, if any, of the bonds should you buy if your MARR is 25%? Annual Current Market Stated Interest Price, Including Bond Value Payment BuyingCommission Maturity. Bond Gen Dev $1000 RJR 1000 $480 630 $ 94 140 15 years 15 First cost Annual upkeep Service life, in years *At maturity the bondholder receives the last interest payment plus the bond stated value. 8-25 Three mutually exclusive alternatives are being considered. A Initial investment Annual net income Computed rate of return C B $22,000 2,077 7% $50,000 5,093 8% $15,000 1,643 9% Each alternative has a 20-year useful life with no salvage value. If the minimum attractive rate of return is 7%, which alternative should be selected? 8-26 A firm is considering the following alternatives, as well as a fifth choice: do nothing. Initial cost Uniform annual net income ($1000s) Computed rate of return 1 2 3 4 $100,000 26.38 $130,000 38.78 $200,000 47.48 $330,000 91.55 10% 15% 6% 12% Each alternative has a 5-year useful life. The firm's minimqm attractive rate of return is 8%. Which alternative should be selected? Alternatives 8-27 Initial cost Annual benefit Salvage value Life, in years MARR required A B C D $2000 800 2000 5 6% 5000 500 1500 6 6% 4000 400 1400 7 6% 3000 1300 3000 4 6% Find the best alternative using incremental IRR analysis. r 8-28 Our cat Fred's summer kitty-cottage needs a new roof. He's considering the following two proposals and feels a 15-year analysis period is in line with his remaining lives. Which roof should he choose if his MARR is 12%? What is the actual value of the IRR on the incremental cost? (There is no salvage value for old roofs.) Thatch $20 5 3 Slate $40 2 5 8-29 Don Garlits is a landscaper. He is considering the purchase of a new commercial lawn mower, either, the Atlas or the Zippy. The minimum attractive rate of return is 8%, and the table provides all the necessary information for the two machines. Initial cost Annual operation and maintenance cost Annual benefit Salvage value Usefullife, in years Atlas $6700 1500 Zippy $16,900 1,200 4000 1000 3 4,500 . 3,500 6 (a) Determine the rate of return on the Atlas mower (to the nearest 1%). (b) Does the rate ofreturn on the Zippy mower exceed the MARR? (c) Use incremental rate of return analysis to decide which machine to purchase. 8-30 QZY, Inc. is evaluating new widget machines offered by three companies. The machines have the following . characteristics: First cost Maintenance and operating Annual benefit Salvage value Usefullife, in years Company A Company B Company C $15,000 1,600 $25,000 400 $20,000 900 8,000 3,000 4 13,000 6,000 4 9,000 4,500 4 MARR = 15%. Using rate of return analysis, from which company, if any, should you purchase the widget machine? ---TOO- .-- --. - .-. Problems 8-31 The Croc Co. is considering a new milling machine from among three alternatives: Alt. A B C Alternative L. Initial Cost Annual Profit $100,000 $30,000 66,000 300,000 500,000 80,000 269 Profit Rate 30% 22% 16% Deluxe Regular Economy Which alternative should be selected? Use a rate of First cost Annual benefit Maintenance and $220,000 79,000 38,000 $125,000 43,000 13,000 $75,000 28,000 8,000 return analysis. operating costs Salvage value 16,000 6,900 3,000 8-34 The New England Soap Company is considering adding some processing equipment to the plant to aidin the removal of impurities from some raw materials. By adding the processing equipment, the firm can purchase lower-grade raw material at reduced cost and upgrade it for use in its products. Four different pieces of processing equipment are being considered: All machines have a life of 10 years, and MARR = 15%. Using incremental rate of return analysis, which alternative, if any, should the company choose? 8-32 Wayward Airfreight, Inc. has asked you to recommend a new automatic parcel sorter. You have obtained the following bids: First cost Salvage value Annual benefit Yearly maintenance and operating cost Useful life, in years SHIP-R SORT-Of U-SORT-M $184,000 38,300 75,300 21,000 $235,000 44,000 89,000 21,000 $180,000 14,400 68,000 12,000 7 7 7 Initial investment Annual saving in materials costs Annual operating cost A B C D $10,000 4,000 $18,000 6,000 $25,000 7,500 $30,000 9,000 2,000 3,000 3,000 4,000 The company can obtain a 15% annual return on its' investment in other projects and is willing to invest money on the processing equipment only as long as_ it can obtain 15% annual return on each increment of money invested. Which one, if any, of the alternatives should be selected? Use a rate of return analysis. Using an MARR of 15% and a rate of return analysis, which alternative, if any, should be selected? 8-33 A firm must decide which of three alternatives to adopt to expand its capacity. The firm wishes a minimum annual profit of 20% of the initial cost of each separable increment of investment.. Any money not invested in capacity expansion can be invested elsewhere for an annual yield of 20% of initial cost. 8-35 Build a spreadsheet to find the EAC of each roof in .II. Problem 8-28.ofUse GOAL SEEK tool of Excel to find the IRR the the incremental investment. 8-36 Build a spreadsheet to find the EAW of each lawn- JI. mower Problem 8-29. the GOAL SEEK tool of Excel toinfind the IRR of Use the incrementalinvestment. . --;0 _.............-. - --- -- -- -- -- After Completing This Chapter... The student should be able to: . . . . . . . Use future worth, benefit-cost ratio, payback period, and sensitivity analysis methods. to solve engineering economy problems. . Link the use of thefuture worth analysisto the present worth and annual worth methods developedearlier. . Mathematically develop the benefit-cost ratio, and use thistnodel to select alternatives and make economic choices. Understand the concept of thepayback period of an investment,and be able t~ calculate this quantity for prospectiveprojects. Demonstrate a basic level understanding of sensitivity and breakevenanalyses and the use of these tools in an engineering economic analysis. Use a spreadsheet to perform sensitivity and breakeven analyses. QUESTIONS TO CONSIDER ) 1. Officeleases frequently require building owners, rather than tenants, to pay heating and cooling costs. What effect might this have on the decision making of potential tenants for green buildings?' 2. The green building movement has had more success among developers who hold onto their buildings for years and rent them out, rather than selling them as soon as they are constructed. What factors might influence their views? 3. Manyenvironmentallyfriendlybuildingsare architecturallydistinctiveand featurebetter quality materials and workmanshipthan traditionalcommercialstructures.Environmental advocateshopethese characteristicswillhelp greenbuiJ,.dings attracta rent "premium." How might these features make the buildings more attractive to tenants? .. Other Analysis Techniques Clean, Green, and.Far Between Designers and engineers have made great progress in developingenvironmentallyfriendly construction materials and building techniques. And "green" office buildings offer numerous . i . advantages. They use energy more efficiently, provide a more natural environment for. workers, and can even reduce indoor air pollution. The only problem is, no one wants to build them. In the commercial sector, barely 1% of officeconstructionprojects have bothered to applyfor certification as green buildings. Commercial developers have found th~t environmentallyfriendly features add to the expenseof construction.Eventhoughadditionalcostsmayincrease I.. overall construction expenses by only 1 or 2%, they can make it harder for:the builder to recoup its investment and break even on the project. Moreover, the extra expenditures apparently do little to attract potential tenants. Renters may like the idea of being cleaner and greener-but they don't want to pay extra for it. .1' d i'j l '1 = =-- ..... ---- .: 1 ,j .~ ., --. -. .. - . - . - ... ... 272 .. ..... OTHER ANALYSISTECHNIQUES Chapter 9 examines four topics: · Future worth analysis · Benefit-cost ratio analysis ·· Sensitivity and breakeven analysis Payback period Future worth analysis is very much like present worth analysis, dealing with then (future worth) rather than with now (present worth) situations. Previously,we have written economic analysis relationships based on either: PW of cost = PW of benefit or EUAC = EUAB Instead of writing it in this form, we could define these relationships as PW of benefit PW of cost =1 or EUAB =1 EUAC When economic analysis is based on these ratios, the calculations are called benefit-cost ratio analysis. Payback period is an approximate analysis technique, generally defined as the time required for cumulativebenefits to equal cumulativecosts. Sensitivitydescribes the relative magnitude of a particular variation in one or more elements of a problem that is sufficient to change a particular decision. Closely related is breakeven analysis, which determines the conditions where two alternatives are equivalent.Thus, breakeven analysis is a form of sensitivity analysis. FUTURE WORTH ANALYSIS Wehaveseenhow economic analysistechniquesresolvedalternativesinto comparableunits. In present worth analysis, the comparison was made in terms of the present consequences of taking the feasible courses of action. In annual cash flow analysis, the comparison was in terms of equivalent uniform annual costs (or benefits). We saw that we could easily convertfrom present worth to annualcash flow,and vice versa. But the concept of resolving alternatives into comparable units is not restricted to a present or annual comparison. The comparison may be made at any point in time. In many situations we would like to know what thefuture situation will be, if we take some particular course of action now. This is called future worth analysis. Ron Jamison, a 20-year-old college student, consumes about a carton of cigarettes a week. He wonders how much money h~ could accumulate by age 65 if he quit smoking now and put his cigarett~moneyinto~asavings account. Cig"aretiescost $35 per carton. Ron expects:that a savings account would earn 5% interest, compounded semiannually. Compute Ron's future worth at age 65. -. ;,."..'::1: .... Future Worth Analysis 273 SOUJTIONi Semiannual saving $35lcarton x 26 weeks . . $910 Future worth (FW)- A(FjA,,2'l2%, 90) = 910(329.2) . . $299,57f An east coast firm has decided to establish a second plant in Kansas City. There is a factory for sale for $850,000 that, with extensiveremodeling, could be used. As an alternative,the company could buy vacant land for $85,000 and have a new plant constructed there. Either way, it will be 3 years before the company will be able to get a plant into production. The timing and cost of the various components for the factory are given in the following cash flowtable. Year Construct New Plant Remodel Available F~ctory o Buy land $ 85,000 Purchase factory $850,000 1 Design and initial 200,000 Design and remodeling 250,000 construction costs costs 2 Balanceof . 1,200,000 Additional remodeling 250,000 construction costs costs 3 Setup of production 200,000 Setup of production 250,000 equipment equipment If interest is 8%, which alternative results in the lower equivalent cost when th~,fiim begins production at the end of the third year? New Plant Future worth of cost (FW) - 85,000(Fj P,8%, 3) + 200,000(FjA, 8%,3) + 1,000,OOO(FjP, 8%,1)- $1,836,000 Remodel Available Factory Future worth of cost (FW) iii . 850,OOO(Fj P, 8%, 3) + 250,000(FjA., 8%, 3) . 274 OTHER ANALYSISTECHNIQUES The total cost of remodeling the available factory ($1,600,000) is smaller than the total cost of a new plant ($1,685,000). The timing of the expenditures, however, is less favorable than building the new plant. The new plant is projected to have the smaller future worth of cost and thus is the preferred altefl1ative.. BENEFIT-COST RATIO ANALYSIS At a given minimum attractive rate of return (MARR), we would consider an alternative acceptable, provided or PW of benefits - PW of costs ~ 0 EUAB - EUAC 2: 0 These could also be stated as a ratio of benefits to costs, or B ; Benefit-cost ratio C ; , , ' = _ PW of benefit EUAB > 1 PW of costs - EUAC - Rather than using present worth or annual cash flow analysis to solve problems, we can base the calculations on the benefit-cost ratio, BIe. The criteria are presentedin Table9-1. We will illustrate "B/C analysis" by solving the same example problems worked by other economic analysis methods. ' , , ' TABLE9-1 Benefit-Cost Ratio Analysis Fixed input Fixed output Neither input nor output fixed Situation Criterion Amount of money or other input resources are fixed Fixed task, benefit, or other output to be accomplished Maximize BIC Neither amount of money or other inputs nor amount of benefits or other outputs are fixed Two alternatives: Compute incremental benefit-cost ratio (~BI ilC) on the increment of investment between the alternatives. If ilBI ilC ::::1, choose higher-cost alternative; otherwise, choose lower~cost alternative. , Three or more alternatives: Solve by benefit-cost ratio incremental analysis Maximize B IC , Benefit-Cost Ratio Analysis 275 A finn is trying to decide which of two devices to install to reduce costs in a particular situation. Both devices cost $1000 and have useful lives of 5 years and no salvage value. Device A can be expected to result in $300 savings annually. Device B will provide cost savings of $400 the first year, but savings will decline by $50 annually, making the second...yearsavings $350, the third-year savings $300, and so forth. With interest at 7%, which device should the firm purchase? jSOtuTION Wehave used three types of analysis thus far to solve this problem: present worthin Example 5-1, annual cash flowin Example 6-5, and rate of return in Example 7-9. Device A = $1000 PW of cost PW of benefits = 300(PjA, 7%,5) - -BC 300(4.100) = $J230 PW of benefit PW of costs · 1230 --;---= 1000 1.23 Device B PW of cost -- $1000 PW of benefit = 400(P j A, 7%, 5) - 50(P fG-, 7%, 5) = 400(4.100).--' 50(7.647)= r 1640 --' 382= 1258 BPW of benefit 1258 --- = . .' ... ., = --- --1.26 CPWof costs 1000 I . To ma)!:imizethe benefit-cost ratio, select Device B,. r Twomachines are being considered for purchase. Assuming 10%interest, which machine should be bought? Machine X Machine Y Initial cost $200 $700 95 ::I;;;: 120 Unifonn, annual benefit 50 150 End-of-useful-lifesalvage value 6 =~ :=.ef2'i1''I: ~= .,,, UsefuUife,.in years - --~ 1: ---- -.. , ~ 1 ·-- .d .. ( , -- - - - 276 OTHER ANALYSISTECHNIQUES SOLUTION Assuming a 12-year analysis period, the cash flowtable is: Year o 1-5 Machine X -$200 +95 Machine Y -$700 +120 +95 -200 { +50 +95 +95 { +50 6 7-11 12 +120 +120 +120 +150 We will solve the problem using B C EUAB EUAC and considering the salvage value of the machines to be reductions in cost, rather than increases in benefits. This choice affects the ratio value, but not the decision. Machine X EUAC = 200(Aj P, 10%, 6) - 50(Aj F, 10%,6) = 200(0.2296) - 50(0.1296) = 46 - 6 = $40 EUAB = $95 Note that this assumes the replacement for the last 6 years has identical costs. Under these circumstances, the EUAC for the fi~st6 years equals the EUACfor all 12 years. Machine Y EUAC = 700(Aj P, 10%, 12) - 150(Aj F, 10%, 12) - 700(0.1468) - 150(0.0468) = 103 -7 = $96 EUAB = $120 Machine Y - Machine X dB = ;; = 12095 _ 96 - 40 LSC= 25 = 0.45 56 = Since the incremental benefit-costratio isless than 1, it represents an undesirable increment of investmenf! We-=thereforechoosethe lower-cost alternative-Machine Xi' If we had computed -- -- ---- - -- - -- - r Benefit-Cost Ratio Analysis 277 benefit-co~t ratios for each machine, they would have been: Machine X B 95 - = - = 2.38 C 40 Machine Y B 120 C =. 96 = 1.25 Although BIC = 1.25 for Machine Y (the higher-cost alternative), we must not use this fact as the basis for selecting that the more expensive alternative. The incremental.benefibcost ratio, ~B I b.C, clearly shows that Y is a less desirable alternative than X. Also, we must not jump to the conclusion that the best alternative is always the one with the largest B/Cratio. This, too, may lead to incorrect decisions-as we shall see when we examine problems with three orJIlor~ alternatives. I I . Consider the fivemutually exclusive alternativesfrom Example 8-8 plus an additionalalternative, F. They have 20-year useful lives and no salvage value. If the minimum attractiverate of return is 6%, which alternative should be selected? A B D E F C Cost $1000 $9000 $4000 $2000 $6000 $10,000 8730 1340 9000 PW of benefit 7330 4700 9,500 B - PW of benefits 1.34 1.83 2.35 1.46 1.00 0.95 C PW of cost ',-' -.". SOLUTION. Incremental analysisis needed to solve the problem.The steps in the solution are the same as the ones presented for incremental rate of return, except here the criterion is b.B/ b.C, and the cutoff is 1,rather than b.IRR with a cutoff of MARR. 1. Be sure all the alternatives are identified. 2. (Optional) Compute the BIC ratio for each alternative. Since there are alternatives for which B/C ::::1, we will discard any withB/C < 1. Discard Alt. F. 3. Arrange the remaining alternativesin ascending order of investment. D B A E C $1000 $2000 $6000 $9000 $4000 Cost (= PW of cost) 4700 8730 9000 PW of benefits 1340 7330 1.34 2.35 1.83 1.46 1.00 B/C ~ b.Cost ~Benefit b.BI b.C Increment B-D $1060 ~ 3360 3.36 Increment A-B $20062630 J..32 ; Increment C-A : $2000 1400 " =~ il Q.2J).c "_II _ ...__ - --- - 278 OTHER ANALYSISTECHNIQUES 4. Examine each separable increment of investment. If ~BI ~c < 1, the increment is not attractive. If D.B I ~C 2: 1, the increment of investment is desirable. The increments B-D and A-B are desirable. Thus, of the first three alternatives (D, B, and A), Alt. A is the preferred alternative.Increment C-Ais not attractiveas ~BI ~C = 0.70,whichindicates that of the first four alternatives (D, B, A, and C), A continues as the best of the four. Now we want to decide between A andE, which we'll do by examining the increment of investmentthat represents the difference between these alternatives. Increment E-A D.Cost $500b 1670 ~Benefit 0.33 D.BI4C The increment is undesirable. We choose Alt. A as the best of.the six alternatives. One should note that the best alternativein this example does not have the highest B/C ratio. I, ' Benefit-cost ratio analysis may be graphically represented. Figure 9.1 is a graph of Example 9-5. We see that F has a B/C < 1 and can be discarded. Alternative D is the starting point for examining the separable increments of investment. The slope of line B-D indicates a ~BI D.Cratio of > 1. This is also true for line A-B. Increment C-A has a slope much flatter than BIC = 1, indicating an undesirable increment of investment.Alternative C is therefore discarded and A retained. Increment E-A is similarlyunattractive. Alternative A is therefore the best of the six alternatives. Note particularly two additional things about Figure 9-1: fin;t, even if alternatives with B/C ratio < 1 had not been initially excluded, they would have been systematically eliminated in the incremental analysis. Since this is the case, it is not essential to compute the B/C ratio for each alternativeas an initial step in incremental analysis. Nevertheless,it seems like an orderly and logical way to approach a multiple-alternativeproblem. Second, Alt. B had the highestB/C ratio (B/C = 2.35), but Itis not thebest ofthe sixaltematives.We saw the same situation in rate of return analysis of three or more alternatives.The reason is the same in both analysis situations.We seek to maximizethe total profit, not the profitrate. FIGURE 9-1 Benefit-cost ratio graph of Example9-5. $10,000 B/C=2.0 B/C=1.5 B/C=1.0 $9000 ~ $8000 I+:: ~ ~ $7000 Eo $6000 '€ o $5000 ~ , $4000 -= ~ $3000 a !!:! ~ $2000 , , " " $1000 o 4000 6000 8000 PresentWorthof Cost 10,600 . Benefit-Cost Ratio Analysis Continuous 279 Alternatives At times the feasiblealternativesare a continuous function.When we considered the height of a dam in Chapter 8 (Example 8-9), we found that it was possible to build the dam anywhere from 200 to 500 feet high. In many situations, the projected capacity of an industrial plant can be varied continuously over some feasible range. In these cases, we seek to add increments of investmel).t where /1Bj /1C 2: 1 and avoid increments where /1Bj /1C < 1. The optimal size of such a project is where /1Bj /1C = 1.Figure 9-2a shows the line of feasible alternatives with their Maximum Present Worth ",of Benefits Obtainable at Each Level of Cost I . rIC=O.5 I o 'H'. '"_~+ . . _ _.' '-. ~." .~.n<_,,_._. I Present Worth of Cost I II (a) I I I-I I I I I I OptimalValuefor PW of Cost Present Worth of Cost (b) Present Worth of Cost (c) FIGURE 9-2 Selecting optimal size of project: (a) feasible alternatives, (b) changes in ~B / ~C, and (c) total NPW plotted versus size of the project. --- - 280 - . -- .. -- ... ..-- OTHER ANALYSISTECHNIQUES costs and benefits. This may represent a lot of calculations to locate points through which the line passes. Figure9-2b showshowthe incrementalbenefit-costratio (L\.Bj L\.C) changesas one moves along the line of feasible alternatives.Figure 9-2b also plots the ratio of incremental netpresent worthto incrementalcost (L\.NPWj L\.C). As expected, we are addingincrements of NPW as long as L\.BjL\.C> 1. Finally, in Figure 9-2c, we see the plot of (total) NPW versus the size of the project. This three-part figure demonstrates that present worth analysis and benefit-cost ratio analysis lead to the same optimal decision. We saw in Chapter 8 that rate of return and present worth analysis led to identical decisions. Any of the exact analysis methodspresent worth, annual cash flow,rate of return, or benefit-cost ratio-will lead to the same decision. Benefit-cost ratio analysis is extensivelyused in economic analysis at all levels of government. PAYBACK PERIOD Payback period is the period of time required for the profit or other benefits from an investment to equal the cost of the investment. This is the general definition for payback period, but there are other definitions. Others consider depreciation of the investment, interest, and income taxes; they, too, are simply called "payback period." For now, we will limit our discussion to the simplest form. Payback period is the period of time required for the profit or other benefits of an investment to equal the cost of the investment. The criterionin all situationsis to minimize thepaybackperiod. The computationofpayback period is illustrated in Examples 9-6 and 9-7. The cash flowsfor two alternatives are as follows: Year o 1 2 3 4 5 A -$1000 +200 +200 +1200 +1200 +1200 B -$2783 +1200 +1200 + 1200 + 1200 +1200 You may assume the benefits occur throughout the year rather than just at the end of the year. Based ong~.!>a~kperiod, ,,;:hichaltern~tiveshould be selected? L ---- - - - - - - -- -- - ... --- - - - r .r _.u.._ --. --.-.-... Payback Period 281 Alternative A Payback period is the period of time required for the profit or other benefits of an investmentto equal the costof the investment.In the first 2 years,only $400ofthe$1000costis recovered.The remaining $600 cost is recovered in the first half of Year3. Thus the payback period for Alt. A is 2.5 years. Alternative B Since the annual benefits are uniform, the payback period is simply $2783/$1200 per year - 2.3 years To minimize the payback period, choose Alt. B. A firm is trying to decide which of two weighing scalesit shouldinstall to check a package-filling operation in the plant. If both scales have a 6-year life, which one should be selected? Assume an 8% interest rate. . Alternative Atlas scale Tom Thumb scale Uniform Annual Benefit $450 600 Cost $2000 3000 End-of-Useful-Life Salvage Value $100 700 ..SQkl)T10~... Atlas Scale Cost Payback period = Uniform annual benefit 2000 = 4.4 years = 450 Tom Thumb Scale Cost Payback period ~ Uniform annual _3000 -II 600 !I ... 5 years =-= ..- ------ - benefir = ::! Ii: 0; - - ..- .._- - 282 OTHER ANALYSISTECHNIQUES $3000 $3000 $2000 $2000 $1000 $1000 o 12341 5 o Cost = $3000 1 2 3 4 5 4.4 FIGURE Year Year (a) (b) 9-3 Payback period plots for Example 9-7: (a) Atlas scale and (b) Tom Thumb scale. Figure 9-3 illustrates the situation. To minimize payback period, select the Atlas scale. There are four important points to be understood about payback period calculations: 1. This is an approximate, rather than an exact, economic analysis calculation. 2. All costs and all profits, or savings of the investment,prior to payback are included without considering differences in their timing. . 3. All the economic consequences beyond the payback period are completely ignored. 4. Being an approximate calculation, payback period mayor may not select the correct alternative. This is, the payback period calculations may select an alternative different from that found by exact economic analysis techniques. . This last point-that payback period may select the wrong alternative-was illustrated by Example 9-7. When payback period is used, the Atlas scale appears to be the more attractive alternative. Yet, when the same problem was solved earlier by the present worth method . (Example 5-4), the Tom Thumb scale was the chosen alternative. A review of the problem reveals the reason for the differentconclusions.The $700 salvagevalue at the end of 6 years for the TomThumb scale is a significantbenefit.The salvage value occurs after the payback period, so it was ignored in the payback calculation. It was considered in the present worth analysis, with the result that the Tom Thumb scale was in fact more desirable. But if payback period calculations are approximateand may lead to the selectionof the wrong alternative, why is the method used at all? There are two primary answers: first, the calculations can be readily made by people unfamiliar with economic analysis. One does not need to know how to use gradient factors, or even to have a set of compound interest tables. Second, payback period is a readily understood concept. Earlier we pointed out that this was also an advantage to rate of return. Moreover,payback period does giveus a useful measure,telling us how long it will take for the cost of the investmentto be recoveredfrom thebenefitsof the investment.Businesses and industrial firms are often very interested in this time period: a rapid return of invested J .u~I.:.. ~ r .. --- --+_. ... ......-.-. Payback Period 283 capital means that it can be reused sooner for other purposes. But one must not confuse the , speedof the returnof the investment,as measuredby the paybackperiod,witheconomic efficiency.They are two distinctly separate concepts. The former emphasizes the quickness with which invested funds return to a firm; the latter considers the overall profitability of the investment. We can create another situation to illustrate how selecting between alternativesby the payback period criterion may result in an unwise decision. A :firmis purchasing production equipment for a new plant. Two alternativemachines are being considered for a particular operation. . Dura Machine $35,000 $1000 the first year, increasing $3000 per year thereafter 8 Tempo Machine $30,000 $12,000 the first year, declining $3000 per year thereafter 4 Installed cost Net annual benefit after all annual expenses have been deducted Useful life, in years Neither machine has anysalvage value. Compute the payback period for each of the alternatives. SOLUTION BASED ON PAYBACKPERIOD Cost $35,000 $35,000 = $35,000 Cost = $30,000 $30,000 $30,000 ./T .eo $25,000 !/ I- .<q'lJ $20,000I$15,000 .... S'lJY I-0PI I $25,000 I $20,000 I $15,000 $10,000 $10,000 $5,000 $5,000 0 1 2 3 Year 4 5 (a) FIGURE = II ~ == 0 II I' :ill II 1 2 3 Year 4 5 (b) 9-4 Payback period plots for Example 9-7: (a) Tempo machine and (b) Dura machine. The Tempo machine has a declining annual benefit, while the Dura has an increasing annual benefit. Figure 9-4 shows the Tempo has a 4-year payback period and the Dura has a 5-year. Ril~b<!p!5;p~lii>4. lJ> .Q1ir}jmize.thepayback period, theTempo is selected.' - - ......... - - - - - -- -- .. --- ~ ~ ==1 i ----- - 284 OTHER ANALYSISTECHNIQUES Now, as a check on the payback period analysis, compute the rate of return for each alternative." Assume the minimum attractive rate of return is 10%. The cash flowsfor the two alternatives are as follows: Year o 1 2 3 4 5 6 7 8 TempoMachine -$30,000 + 12,000 +9,000 +6,000 +3,000 o o o o o Dura Machine -$35,000 +1,000 +4,000 +7,000 +10,000 + 13,000 +16,000 +19,000 +22,000 +57,000 Tempo Machine Since the sum of the cash flows for the Tempo machine is zero, we see immediately that the $30,000 investment just equals the subsequent benefits. The resulting rate of return is 0%. Dura Machine 35,000 = 1000(PjA, i, 8) +3000(PjG, i, 8) Try i = 20%: 35,000 ~ 1000(3.837) + 3000(9.883) ~3837 + 29,649 = 33,486 The 20% interest rate is too high. Try i = 15%: 35,000 ~ 1000(4.487) + 3000(12.481) ~4487 + 37,443 = 41,930 II .. This time, the interest rate is too low. Linear interpolation would show that the rate of return is approximately 19%. Using an exact calculation-rate of return--;-,-itis clear that the Tempo is not very attJ.'active economiq.l}y. Y~t it ~s thi~ aJle!ll~t:tve,and nq.t."the.,QurawacJ:rine, that w,~ preferred based On the payback period calculations. On the other hand, the shorter payback period for the Tempo does give a measure of the speed of the return of the investment not found in the Dura. The conclusion to be drawn~iSJhatJiqfiidity;ffild,profitability may be two quite different criteria. \1 ..,~..~ -- - - ~ Sensitivity and Breakeven Analysis 285 From the discussionandthe examples,we see thatpaybackperiodcanbe helpful in providing a measure of the speed of the return of the investment. This might be quite important, for example, for a company that is short of working capital or for a firm in an industry experiencing rapid changes in technology.Calculation of payback period alone, however, must not be confused with a careful economic analysis.Wehave shown that a short payback period does not always mean that the associated investment is desirable. Thus, payback period should not be considered a suitable replacement for accurate economic analysis calculations. SENSITIVITY AND BREAKEVEN ANALYSIS Sincemany datagathere~in solvinga problemrepresentprojectionsof futureconsequences; there may be considerableuncertaintyregardingthe data's accuracy.Sincethe desiredresult of the analysis is decision making, an appropriate question is, To what extent do variations in the data affect my decision? When small variationsin a particular estimate would change selection of the alternative, the decision is said to be sensitive to the estimate. To better evaluate the impact of any particular estimate, we compute "the variation to a particular . estimate that would be necessary to change a particular decision."This is called sensitivity analysis. An analysisof the sensitivityof a problem's decisionto its variousparametershighlights the important and significantaspects of that problem. For example, one might be concerned that the estimates for annual maintenance and future salvage value in a particular problem may vary substantially. Sensitivity analysis might indicate that the decision is insensitive to the salvage-value estimate over the full range of possible values. But, at the same time, we might find that the decision is sensitiveto changes in the annual maintenance estimate. Under these circumstances, one should place greater emphasis on improving the annual maintenance estimate and less on the salvage-valueestimate. As indicated at the beginning of this chapter, breakeven analysis is a form of sensitivity analysis. To illustrate the sensitivity of a decision between alternatives to particular estimates, breakeven analysis is often presented as a breakeven chart. . Sensitivityand breakevenanalysisfrequently areuseful in engineeringproblems called stage construction. .Should a facility be constructed now to meet its future full-scale requirement, Or should it be constructed in stages as the need for the increased capacity arises? Three examples of this situation are: · · · Should we install a cable with 400 circuits now or a 200-circuit cable now and another 200-circuit cable later? A 10 cm water main is needed to serve a new area of homes. Should it be installed now, or should a 15 cm main be installed to ensure an adequate water supply to adjoining areas later, when other homes have been built? An industrial firm needs a new warehouse now and estimates that it will need to double its size in 4 years. The firm could have a warehouse built now and later enlarged, or have the warehouse with capacity for expanded operations built right away. Examples 9-9 and 9-10 illustrate sensitivityand breakevenanalysis. --~ -- -- -- --- 286 OTHER ANALYSISTECHNIQUES Consider a project that may be constructed to full capacity now or may be constructed in two stages. Construction Costs Two-stageconstruction Construct first stage now Construct second stage n years from now Full-capacity construction Construct full capacity now $100,000 120,000 140,000 Other Factors 1. All facilities will last for 40 years regardless of when they are installed; after 40 years, they will have zero salvage value. 2. The annual cost of operation and maintenance is the same for both two-stage construction and full-capacity construction. 3. Assume an 8% interest rate. Plot "age when second stage is constructed" versus "costs for both alternatives." Mark the breakevenpoint on your graph. What is the sensitivityof the decisionto second-stageconstruction 16 or more years in the future? ' [Sp'LOTION' ~.. ~'-- ~".".",.~,,: '.. ...."..',.~""..- ~. " Since we are dealing with a common analysis period, the calculations J;Ilaybe either annual cost or present worth. Present worth calculations appear simpler and are used here. Construct Full Capacity Now PW of cost = $140,000 Two-Stage Construction First stage constructed now and the second stage to be constructedn years hence...Compute the PW of cost for severalvalues of n (years). PW of cost n=5 n = 10 n -20 n=30 PW = 100,000 + 120,000(P IF, 8%, n) 100,000+ 120,000(0.6806) -- $181,700 PW = 100,000+ 120,000(0.4632)' J55,600 PW 100,900 + 120,000(0.+145) --: 125,700 pw" .100,00.0-h;120,.000(0..0994,). _ -u- '_.~.'M1!l't ____ '."-"_.~. <po ,', "'-"-'-"-"~'''' ~~ .. '," '--r:,: -..J.ll.,2iliL -""---',.: :..,.".,..."--~ -' ._ C!IIIii These data are plotted in the form of a breakevell chart in Figure 9..5. - -.. - -r -- - ~ ~ Sensitivity and Breakeven Analysis FIGURE 9-5- Breakeven chart for Example 9-9. 287 $250,000 $200,000 ... '" o u 'Ci $150,000 -E ~i:: $100,000 '" Q) £ $50,000 o 5 10 15 Year 20 25 30 Age When SecondStage C0E-structed Figure 9-5 portrays the PW of cost for the two alternatives. The x-axis variable is the time when the second stage is constructed. We see that the PW of cost for two-stage construction naturally decreases as the time for the second stage is deferred. The one-stage construction (full capacity now) is unaffected by the x-axis variable and, hence, is a horizontal line on the graph. The breakeven point on the graph is the point at which both alternatives have equivalent costs. We see that if, in two-stage construction, the second stage is deferred for 15 years, then the PWof cost of two-stage construction is equal to one-stage construction; Year 15 is the breakeven point. The graph also shows that if the second stage were to be needed prior to Year 15, then one-stage construction, with its smaller PW of cost, would be preferred. On the other hand, if the second stage would not be required until after 15 years, two-stage construction is preferred. The decision on how to construct the project is sensitive to the age at which the second stage is needed only if the range, of estimates includes 15 years. For example, if one estimated that the second stage capacity would be needed between 5 and 10 years hence, the decision is insensitive t9 that estimate. For any value within that range, the decision does not change. The more eQollomical thing to do is to build the full capacity now. But, if the second-stage capacity were to be needed sometime between, say, 12 and 18 years, the deciSion would be sensitive to the estimate of when the full capacity would be needed. One question posed by Example 9-9 is how sensitive the decision is to the need for the second stage at 16 years or beyond. The graph shows that the decision is insensitive. In all cases for construction on or after 16 years, two-stage construction has a lower PW of cost. =! .~ -- I --- ---- :tu.. 288 OTHER ANALYSISTECHNIQUES --. Example 8-3 posed the following situation. Three mutually exclusivealternatives are given,each with a 20-year life and no salvage value. The minimum attractive rate of return is 6%. Initial cost Uniform annual benefit A $2000 410 B $4000 639 C $5000 700 In Example 8-3 we found that Alt. B was the preferred alternative.Here we would like to know how sensitive the decision is to our estimate of the initial cost of B. If B is preferred at an initial cost of $4000, it will continue to be preferred at any smaller initial cost. But how much higher than $4000 can the initial cost be and still have B the preferred alternative? The computations may be done several different ways. With neither input nor output fixed,maximizing net present worth is a suitable criterion. Alternative A NPW = PW of benefit - PW of cost = 41O(P j A, 6%, 20) = 410(11.470) - 2000 2000 = $2703 Alternative B Let x = initial cost of B. NPW = 639(P j A, 6%, 20) - x = 639(11.470)- x = 7329- x Alternative C NPW = 700(PjA, 6%,20) - 5000 = 700(11.470) - 5000 = $3029 For the three alternatives,we see that B will only maximize NPW as long as its NPW is greater than 3029. 3029 = 7329 - x x :: - - -- = 7329 - 3029 = $4300 Therefore, B is tbe preferred alternativeif its initial cost does not exceed $4300. Figure 9-6 is a breakeven chart for the three alternatives.Here the criterion is to maximize NPW; as a result, the graph shows that B is preferred if its initial cost is less than $4300. At an - - --- - f ---- Graphing with Spreadsheetsfor Sensitivity and Breakeven Analysis FIGURE 9~6 Breakeven chart for Example 9-10. $6000 Alternative B Preferred $5000 289 IAlternative C II I Preferred '€ $4000 ~ 5'" $3000 £ ~ $2000 $1000 o $1000 $2000 $3000 $4000 $5000 $6000 Initial Cost of Alternative B initial cost above $4300, C is preferred. We have a breakeven point at $4300. When B has an ipitial cost of $4300, B and C are equally desirable. Sensitivity analysis and breakeven point calculations can be very useful in identifying how different estimates affect the calculations. It must be recognized that these calculations assume that all parameters except one are held constant, and the serisitivityof the decision to that one variable is evaluated. Later we will look further at the impact of parameter estimates on decision making. GRAPHING WITH SPREADSHEETS FOR SENSITIVITY AND BREAKEVEN ANALYSIS Chapter 4 introduced drawing xy plots with spreadsheets, and Chapter 7 reviewed this procedure for plotting present worth versus i. The Chapter 7 plot is an example of breakeven analysis, as it is used to determine at what interest rate does the project break even or have a present worth of O. This section will present some of the spreadsheet tools and options that can make the xy plots more effective and attractive. The spreadsheet tools and options can be used to: · · Modify the x or y axes Specify the minimum or maximum value Specify at what value the other axis intersects (default is 0) Match line types to data Use line types to distinguish one curve from another Use markers to show real data Use lines without markers to plot curves (straight segments or smooth curves) - - -- - -- -- - 292 OTHER ANALYSISTECHNIQUES "shoWlegend" off. Similarly,we can turn the x-axis gridlines on. The line style for these gridlines should be changed to match the y-axis gridlines. This allows us to see that the breakeventime is between 14 and 15 years. To make the graph less busy, change the scale on the x axis so that the interval is 5 years rather than automatic.Also eliminate the gridlines for the y axis (by selecting the chart area, chart options, and gridlinestabs). The graph size can be increased for easier reading, as well. This may require specifying art interval of 10,000for the scale of the y axis. Finally to addthe labels for thefull-capacitycurveand the two-stagecurve, findthe toolbarfor graphics, which is open when the chart is selected(probably alongthe bottom of the spreadsheet). Select the text box icon, and click on a location close to the two-stage chart. Type in the label for two-stage construction.Notice how including a return and a few spaces can shape the label to fit the slanted line. Add the label for full construction. Figure 9-8 is the result. A B C Full Capacity 140,000 Two Stage 100,000 1 Time 2 0 3 n D 120,000 4 5 life 6 i 40 8% 7 Present worth of 8 n 9 Full Capacity 140,000 Two Stage 0 10 5 140,000 181,670 11 10 140,000 155,583 12 20 140,000 125,746 13 30 140,000 111,925 14 15 16 17 18 19 20 21 22 23 tj0 155,000 1"---... ..... .. Two Stage ......... 0 "€ 145,000 I- f'" 135,000 p.. 125,000 10 - I , I ......... Construction U =$C$2+PV($B$6,A1O,0, -$C$3) - I - , ......... ...... ...... ,, I , I '.', ' I Full Capacity Construction - .... ...... .., : ...... ...... ...... - ...... '" 20 15 Years until Capacity Needed 24 FIGURE 9-8 Spreadsheet of Figure 9-7 with improved graph. -- - - Problems 293 SUMMARY In this chapter, we have looked at four new analysis techniques. Future worth: When the comparison between alternatives will be made in the future, the calculation is called future worth. This is very similar to present worth, which is based on the present, rather than a future point in time. Benefit-cost ratio analysis: This technique is based on the ratio of benefi,tsto costs using either present worth or annual cash flow calculations.The method is graphically similar to present worth analysis. When neither input nor output is fixed, incremental benefit-cost ratios (~B/ ~C) are required. The method is similar in this respect to rate of return analysis. Benefit-cost ratio analysis is often used at the various levels of government. Paybackperiod: Here we definepayback as the period of time required for the prQfit or other benefits of an investmentto equal the cost of the investment. Although simple to use and simple to under~tand,payback is a poor analysis technique for ranking alternatives. While it provides a measure of the speed of the return of the investment, it is not an accurate measure of the profitability of an. investment. Sensitivity and breakeven analysis: These techniques are used to see how sensitive a decision is to estimates for the various parameters. Breakeven analysis is dOJ?e to locate conditions under which the alternatives are equivalent. This is often presented in the form of breakevencharts. Sensitivity analysis is an examination of a range of values for some parameters to determine their effect on a particular decision. PROBLEMS 9-1 For a 12% interest rate, compute the value of F in the following diagram. 9-2 Compute F for the followingdiagram. 200 f r0-1 -2-3-t-' 100 100 100 t t t I 0-1-2-3-4-5-6 t 100 1 i=lO% F + 100 + 100 + F -- - - --- - - -- 294 OTHER ANALYSISTECHNIQUES 9-3 Compute F for the followingdiagram. 300 8~nTG 250 ~~FOO 7f 200 0-1-2-"'-8-9-10 P i = 12% 150 100 t 0-1-2-3-4-5 1 I i = 12% 9.4 = $1199) 9-9 You have an opportunity to purchase a piece of vacant land for $30,000 cash. If you buy the property, you plan to hold it for 15 years and then sell it at a profit. During this period, you would have to pay annual property taxes of $600. You would have no income from the property. Assuming that you would want a 10% rate of return from the investment, at what net price would you have to sell it 15 years hence? (Answer: $144,373) 9.10 An individual's salary is now $32,000 per year and he anticipates retiring in 30 years. If his salary is increased by $600 each year and he deposits 10% of his yearly salaryinto a fund that earns7% interest compoundedannually,what is the future worthof the amountaccumulatedat the time of his retirement? 2x 0-1-2-3-4 x . i = 15% 1 F 9.5 For the following diagram, compute F. 30 30 30 30 30 30 25 10 i t f r I 0-1-2-3-4-5-6-7-8-9-10-11 i = 10% 1 F 9-6 System 2 A 20-year-old student decided to set aside $100 on his 2pt birthday for investment. Each subsequent year through his 55th birthday, he plans to increase the sum for investment on a $100 arithmetic gradient. He will not set aside additional money after his 55th birthday. If the student can achieve a 12% rate of return, what is the future worth of the investments on his 65th birthday? (Answer: $1,160,700) 3x t Calculate the present worth and the future worth of a series of 10 annual cash flows with the first cash flow equal to $15,000 and each successive cash flow increasing by $1,200. The interest rate is 12%. The total cash flow series is a combination of Systems 1 and 2. F 9.8 4x t i = 12% The interest rate is 16% per year and there are 48 compounding periods per year. The principal is $50,000. What is the future worth in 5 years? For the following diagram, compute F. 1 ~ P 9-7 1 (Answer: F I F System 1 F i 0-1-2-"'-8-9-10 9-11 Stamp collecting has become an increasingly popular-and expensive-hobby. One favorite method is to save plate blocks (usually four stamps with the printingplate numberin the margin)of each new stamp as it is issued by the post office.But with the rising postage rates and increased numbers of new stamps being issued, this collecting plan costs more each year. Stamps, however, may have been a good place to invest money over the last 10 years, as the demand for stamps previously issued has caused resale prices to increase 18% each year. Suppose a collector .------------- Problems purchased $100 worth of stamps 10 years ago, and increased his pqrchases by $50 per year in each subsequent year. After 10 years of stamp collecting, what is the future worth of the stamp collection? 9-12 9-13 Sally deposited $100 a month in her savings account for 24 months. For the next 5 years she made no deposits. What is the future worth in Sally's savings account at the end of the 7 years, if the account earned 6% annual interest, compounded monthly? (Answer: $3430.78) In the early 1980s, planners were examining alternate sites for a new airport to serve London. In their economic analysis, they computed the value of the structures that would need to be removed from various airport sites. At one airport site, the twelfth-century Norman church of St. Michaels, in the village of Stewkley, would be demolished. The planners used the value of the fire insurance policy on the churcha few thousand pounds sterling-as the value of the structure. An outraged antiquarian wrote to the London Times that an equally plausible computation would be to assume that the original cost of the church (estimated at 100 pounds sterling) be increased at the rate of 10% per year for 800 years. Based on his proposal, what would be the future worth ofSt. Michaels? (Note: There was great public objection to tearing down the church, and it was spared.) 9-14 9-17 9-16 Don Ball is a 55-year-old engineer. According to mortality tables, a male at age 55 has an average life expectancy of 21 more years. In prior years, Don has accumulated $48,500 including interest, toward his retirement. He is now adding $5000 per year to his retirement fund. The fund earns 12% interest. Don's goal is to retire when he can obtain an annual income from his retirement fund of $20,000 per year, assuming he lives to age 76. He will make no provision for a retirement income after age 76. What is the youngest age at which Don can retire, based on his criteria? Jean invests $100 in Year 1 and doubles the amount each year after that (so the investment is $100, 200, 400,800, . . .). If she continuesto do this for 10years, and the investment pays 10% annual interest, what is the future worth of her investment at the end of 10 years? 9-18 If you invested $2500 in a bank 24-month certificate of deposit paying 8.65%, compounded monthly, what is the future worth of the certificate of deposit when it matures in 2 years? 9-19 After receiving an inheritance of $25,000 on her 21st birthday, Ayn Rand deposited the inheritance"in a savings account with an effective annual interest rate of 6%. She decided that she would make regular deposits on each future birthday, beginning with $1000 on her 22nd birthday and then increasing the amount by $200 in each following year (i.e., $1200 on her 23rd birthday, $1400 on her 24th birthday, etc.). What was the future worth of Ayn's deposits on her 56th birthday? 9-20 The Association of General Contractors (AGe) wished to establish an endowment fund of $1 million in 10 years for the Construction Engineering Technology Program at Grambling State University in Grambling, Louisiana. In doing so, the AGC established an escrow account in which 10 equal end-ofyear deposits that earn 7% compound interest were made. After seven deposits, the Louisiana legislature revised laws relating to the licensing fees AGe can charge its members, with the result that there was no deposit at the end of Year 8. What must the amount of the remaining equal end-of-year deposits be sure that the $1 million is available to Grambling State for its Construction Engineering Technology Program? Bill made a budget and planned to deposit $150 a month in a savings account, beginning September 1. He did this, but on the following January 1, he reduced the monthly deposits to $100 a month. In all he made 18 deposits, four at $150 and 14 at $100. If the savings account paid 6% interest, compounded monthly, what was the future worth of his savings account immediately after he made the last deposit? (Answer: $2094.42) 9-15 A company deposits $1000 in a bank at the beginning of each year for 6 years. The account earns 8% interest, compounded every 6 months. What is the future worth of the account at the end of 6 years? Make a careful, accurate computation. 295 9-21 J 1 On her birthday, a 25-year-old engineer is considering investing in an individual retirement account (IRA). After some research, she finds a mutual fund wi~ an average return of 10% per year. What is the future worth of her IRA at age 65 if she makes annual investments of $2000 into the fund beginning on her 25th birthday? Assume that the fund continues to earn an annual return of 10%. J 1 -- --- -"" 296 --- - ---- OTHER ANALYSISTECHNIQUES 9-22 IPS COrp. will upgrade its package-labeling mac~nery. It costs $150,000 to buy the machinery and have it installed. Operation and maintenance costs are $1500 per year for the first 3 years and increase by $500 per year for the remaining years of the machine's 1O-year life. The machinery has a salvage value of 5% of its initial cost. Interest is 10%. What is the future worth of cost of the machinery? 9-23 A company is considering buying a new bottlecapping machine. The initial cost of the machine is $325,000 and it has a 1O-year life. Monthly maintenance costs are expected to be $1200 per month for the first 7 years and $2000 per month for the remaining years. The machine requires a major overhaul costing $55,000 at the end of the fifth year'of service. Assume that all these costs occur at the end of the appropriate period. What is the future value of all the costs associated with owning and operating this machine if the nominal interest rate is 7.2%? 9-24 A family starts an education fund for their son Patrick when he is 8 years old, investing $150 on his eighth birthday increasing the yearly investment by $150 per year until Patrick is 18 years old. The fund pays 9% annual interest. What is the future worth of the fund when Patrick is 18? 9-25 A bank account pays 19.2% interest with monthly compounding. A series of deposits started with a deposit of $5000 on January 1, 1997. Deposits in the series were to occur each 6 months. Each deposit in the series is for $150 less than the one before it. The last deposit in the series will be due on January 1,2012. What is the future worth of the account on July 1, 2014, if the balance was zero before the first deposit and no withdrawals are made? 9-26 Let's assume that a late-twentieth century college graduate got a good job and began a savings account. He is paid monthly and authorized the bank to automatically withdraw $75 each month. The bank made the first withdrawal on July 1, 1997 and is instructed to make the last withdrawal on January 1,2015. The bank pays a nominal interest rate of 4.5% and compounds twice a month. What is the future worth of the account on January 1, 2015? 9-27 Bob, an engineer, decided to start a college fund for his son. Bob will deposit a series of equal, semiannual cash flows with each deposit equal to $1500. Bob made the first deposit on July 1, 1998 and will make the last deposit on July 1,2018. Joe, a friend of Bob's, received an inheritance on April 1, 2003, and has decided to begin a college fund for his daughter. Joe wants to send his daughter to the same college as Bob's son. Therefore, Joe needs to accumulate the same amount of money on July 1,2018, as Bob will have accumulated from his semiannual deposits. Joe never took engineering economics ~d had no idea how to determine the amount that should be deposited. He decided to deposit $40,000 on July 1, 2003. Will Joe's deposit be sufficient? If not, how much should he have put in? Use a nominal interest of7% with semiannual compounding on all accounts. 9-28 A business executive is offered a management job at Generous Electric Company, which offers him a 5-year contract that calls for a salary of $62,000 per year, plus 600 shares of GE stock at the end of the 5 years. This executive is currently employed by Fearless Bus Company, which also has offered him a 5-year contract. It calls for a salary of $65,000, plus 100 shares of Fearless stock each year. The Fearless stock is currently worth $60 per share and pays an annual dividend of $2 per share. Assume end-of-year payments of salary and stock. Stock dividends begin one year after the stock is received. The executive believes that the value of the stock and the dividend will remain constant. If the executive c.onsiders 9% a suitable rate of return in this situation, what must the Generous Electric stock be worth per share to make the two offers equally attractive? Use the future worth analysis method in your comparison. (Answer: $83.76) 9-29 A project will cost $50,000. The benefits at the end of the first year are estimated to be $10,000, increasing at a 10% uniform.rate in subsequent years. Using an 8-year analysis period and a 10% interest rate, compute the benefit-cost ratio. 9-30 Each of the three alternatives shown has a 5-year use- fullife. If the MARR is 10%, which alternative should be selected? Solve the problem by benefit-cost ratio analysis. C A B Cost $600.0 $500.0 $200.0 58.3 Uniform annual benefit 158.3 138.7 (Answer: B) 9-31 Consider three alternatives, each with a 10-year usefullife. If the MARR is 10%, which alternative should be selected? Solve the problem by benefit-cost ratio analysis. = - -- Problems Cost . Uniform annual benefit A $800 142 B $300 60 C $150 33.5 9-32 An investor is considering buying some land for $100,000 and constructing an office building on it. Three different buildings are being analyzed. Building Height Cost of building (excluding cost of land) Resale value* of land and building at end 20-year analysis period Annual rental income after all operating expenses have been deducted 2 Stories 5 Stories 10 Stories $400,000 $800,000 $2,100,000 200,000 70,000 400,000 300,000 256,000 105,000 *Resale value to be considered a reduction in cost, rather than a benefit. Using benefit-cost ratio analysis and an 8% MARR, determine which alternative, if any, should be selected. 9-33 Using benefit-cost ratio analysis, determine which one of the three mutually exclusive alternatives should be selected. First cost Uniform annual benefit Salvage value A $560 140 40 B $340 100 C $120 40 o o Each alternative has a 6-year useful life. Assume a 10% MARR. 9-34 Consider four alternatives, each of which has an 8-year useful life; ABC D Cost $100.0 $80.0 $60.0 $50.0 Uniform annual benefit 12.2 12.0 9.7 12.2 Salvage value 75.0 50.0 50.0 0 297 If the MARR is 8%, which alternative should be selected? Solve the problem by benefit-cost ratio analysis. 9-35 Usingbenefit-cost ratio analysis,a 5-yearusefullife, and a 15%MARR, determinewhic~of the following alternativesshouldbe selected. Cost Uniform annual benefit A B $100 $200 37 69 C $300 83 D E$400 $500 126 150 9-36 Five mutually exclusive investment alternatives have been proposed. Based on benefit-cost ratio analysis, and a MARR of 15%, which alternative should be selected? ~M ABC D E F o -$200 -$125 -$100 -$125 -$150 -$225 1-5 +68 +40 +25 +42 +52 +68 9-37 Able Plastics, an injection-molding firm, has negoti- . ated a contract with a national chain of department. stores. Plastic pencil boxes are to be produced for a 2-year period. AJ:>lePlastics has never produced the item before and, therefore, requires all new dies. If the firm invests $67,000 for speciai removal equipment to unload the completed pencil boxes from' the molding machine, one machine operator can be eliminated. This would save $26,000 per year. The removal equipment has no salvage value and is not expected to be used after the 2-y~ar production contract is completed. The equipment, although useless, would be serviceable for about 15 years. You have been asked to do a payback period analysis on whether to purchase the special removal equipment. What is the payback period? Should Able Plastics buy the removal equipment? 9-38 A cannery is considering installing an automatic casesealing machine to replace current hand methods. If they purchase the machine for $3800 in June, at the beginning of the canning season, they will save $400 per month for the 4 months each year that the plant is in operation. Maintenance costs of the casesealing machine are expected to be negligible. The case-sealing machine is expected to be useful for five annual canning seasons and will have no salvage value at the end of that time. What is the payback period? Calculate the nominal annual rate of return based on the estimates. ... --- 298 9-39 - OTHER ANALYSISTECHNIQUES A project has the followingcosts and benefits.What is the payba~kperiod? Benefits Year Costs o $1400 1 500 2 300 $400 3-10 300 in each year 9-40 A car dealer presently leases a small computer with software for $5000 per year. As an alternative he could purchase the computer for $7000 and lease the software for $3500 per year. Any time he would decide to switch to some other computer system he could cancel the software lease and sell the computer for $500. If he purchases the computer and leases the software, (a) What is the payback period? (b) If he kept the computer and software for 6 years, what would be the benefit-cost ratio, based on a 10% interest rate? 9-41 A large project requires an investment of $200 millions. The construction will take 3 years: $30 million will be spent during the first year, $100 million during the second year and $70 million during the third year of construction. Two project operation periods are being considered: 10 years with the expected net profit of $40 million per year and 20 years with the expected net profit of $32.5 million per year. For simplicity of calculations it is assumed that all cash flows occur at end of year. The company minimum required return on investment is 10%. Calculate for each alternative: (a) The payback periods (b) The total equivalent investment cost at the end of the construction period (c) The equivalent uniform annual worth of the project (use the operation period of each alternative) Make your recommendations based on the foregoing economic parameters. 9-42 Two alternatives are being considered: Initial cost Uniform annual cost Useful life, in years A $500 200 8 B $800 150 8 Both alternatives provide an identical benefit. (a) Compute the payback period if Alt. B is purchased rather than Alt. A. l (b) Use a MARR of 12% and benefit-cost ratio analysis to identify the alternative that should be selected. 9-43 Tom Sewel has gathered data on the relative costs of a solar water heater system and a ~onventional electric water heater. The data are based on statistics for a mid-American city and assume that during cloudy days an electric heating element in the solar heating system will provide the necessary heat. The installed cost of a conventional electric water tank and heater is $200. A family of four uses an average of 300 liters of hot water a day, which takes $230 of electricity per year. The glass-lined tank has a 20-year guarantee. This is probably a reasonable estimate of its actual useful life. The installed cost 'of two solar panels, a small electric pump, and a storage tank with auxiliary electric heating element is $1400. It will cost $60-a year for electricity to run the pump and heat water on cloudy days. The solar system will require $180 of maintenance work every 4 years. Neither theconventional electric water heater nor the solar water heater will have any salvage value at the end of their useful lives. (a) Using Tom's data. what is the payback period if the solar water heater system is installed, rather than the conventional electric water heater? (b) Chris Cook studied the same situation and de~ cided that the solar system will not require the $180 of maintenance every 4 years. Chris believes future replacements of either the conventional electric water heater, or the solar water heater system can be made at the same costs and useful lives as the initial installation. Based on a 10% interest rate, what must be the useful life of the solar system to make it no more expensive than the electric water heater system? 9-44 Consider four mutually exclusive alternatives: ABC D Cost $75.0 $50.0 $15.0 $90.0 4.5 23.8 Uniform annual benefit 18.8 13.9 Each alternativehas a 5-year useful life and .no salvage value. The MARR is 10%. Which alternative should be selected, based on (a) Future worth analysis (b) Benefit-cost ratio analysis (c) The payback period I J -- JIll""""'"" . Problems 9-45 Consider three alternatives: 9-48 299 Three mutually exclusive alternatives are being considered: A First cost Uniform annual benefit UsefUl life, in years* Computed rate of return $50 28.8 2 10% 8 C $150 39.6 6 15% $110 39.6 4 16.4% *At the end of its usefUllife, an identical alternative (with the same cost, benefits,and usefUllife) may be installed. All the alternatives have no salvage value. If the MARR is 12%,which alternativeshouldbe selected? (a) Solve the problem by future worth analysis. (b) Solve the problem by benefit-cost ratio analysis. (c) Solve the problem by payback period. (d) If the answers in parts (a), (b), and (c) differ, explain why this is the case. 9-46 Consider three mutually exclusive alternatives.The MARR is 10%. Year 0 I 2 3 4 X -$100 25 25 25 25 Y -$50 16 16 16 16 Z -$50 21 21 21 21 (a) For Alt. X, computethe benefit-cost ratio. (b) Based on the payback period, which alternative should be selected? (c) Determine the preferred alternativebased on an exact economicanalysismethod. 7 The cash flowsfor three alternativesare as follows: Year 0 1 2 3 4 5 6 A -$500 -400 200 250 300 350 400 B -$600 -300 350 300 250 200 150 C -$900 0 200 200 200 200 200 (a) Based on payback period, which alternative should be selected? (b) Using fUtureworth analysis, and a 12% interest rate, determine which alternative should be selected. Initial cost Benefit at end of the first A $500 200 B $400 200 C $300 200 year Uniform benefit at end of 100 125 100 6 5 4 subsequent years UsefUl life, in years At the end of its usefUl life, an alternative is not replaced. If the MARR is 10%, which alternative should be selected: (a) Based on the payback period? (b) Based on benefit-cost ratio analysis? Year E F G H 9-49 o -$90 1 2 3 4 5 6 20 20 20 20 20 20 -$110 35 35 35 35 0 0 -$100 -$120 0 10 20 30 40 50 0 0 00 0 180 (a) Based on fUture worth analysis, which of the four alternatives is preferred at 6% interest? (b) Based on fUture worth analysis, which alt~rnative is preferred at 15% interest? (c) Based on the payback period, which alternative is preferred? (d) At 7% interest, what is the benefit-cost ratio for Alt. G? _ 9-50 Tom Jackson is preparing to buy a new car. He knows it represents a large expenditure of money, so he wants to do an analysis to see which of two cars is more economical. Alternative A is an American-built compact car. It has an initial cost of $8900 and operating costs of 9,i1km, excluding depreciation. Tom checked automobile resale statistics. From them he estimates the American automobile can be resold at the end of 3 years for $1700. Alternative B is a foreign-built Fiasco. Its initial cost is $8000, the operating cost, also excluding depreciation, is 8,i1km. How low could the resale value of the Fiasco be to provide equally economical transportation? Assume Tom will drive 12,000 km/year and considers 8% as an appropriate interest rate. (Answer: $175) 9-51 A newspaper is considering purchasing locked vending machines to replace open newspaper racks for the sale of its newspapers in the downtown area. The T -- - 300 OTHERANALYSIS TECHNIQUES newspaper vending machines cost $45 each. It is expected that the annual revenue from selling the same quantity of newspapers will increase $12 per vending machine. The useful life of the vending machine is unknown. (a) To determine the sensitivity of rate of return to useful life, prepare a graph for rate of return versus useful life for lives up to 8 years. (b) If the newspaper requires a 12% rate of return, what minimum useful life must it obtain from the vending machines? (c) What would be the rate of return if the vending machines were to last indefinitely? 9-52 Data for two alternatives are as follows: Cost Uniform annual benefit Usefullife, in years A $800 230 5 B $1000 230 X If the MARR is 12%, compute the value of X that makes the two alternatives equally desirable. 9-53 What is the cost of Alt. B that will make it at the breakeven point with Alt. A, assuming a 12% interest rate? Cost Uniform annual benefit Salvage value Usefullife, in years A $150 40 100 6 B $X 65 200 6 9-54 Consider two alternatives: Cost Uniform annual benefit ~sefullife, in years A $500 75 Infinity B $300 75 X Assume that Alt. B is not replaced at the end of its useful life. If the MARR is 10%, what must be the useful life of B to make Alternatives A and B equally desirable? 9-55 Jane Chang is making plans for a summer vacation. She will take $1000 with her in the form of traveler's checks. From the newspaper, she finds that if she purchases the checks by May 31, she will not have to pay a servicecharge.That is, she will obtain $1000worth of traveler's checks for $1000. But if she waits and buys the checks until just before starting her summer - - - trip, she must pay a 1% service charge. (It will cost her $1010 for $1000 of traveler's checks.) Jane can obtain a 13% interest rate, compounded weekly, on her money. To help with her planning, Jane needs to know how many weeks after May 31 she can begin her trip and still justify buying the traveler's checks on May 31. She asks you to make the computations for her. What is the answer? 9-56 Fence posts for a particular job cost $10.50 each to install, including the labor cost. They will last 10 years. If the posts are treated with a wood preservative they can be expected to have a 15-year life. Assuming a 10% interest rate, how much could one afford to pay for the woodpreservativetreatment? . 9-57 A piece of property is purchased for $10,000 and yields a $1000 yearly net profit. The property is sold after 5 years. What is its minimum price to breakeven with interest at 1O%? . 9-58 Rental equipment is for sale for $110,000. A prospective buyer estimates he would keep the equipment for 12 years and spend $6000 a year on maintaining the . equipment. Estimated annual net receipts from equip- . ment rentals would be $14,400. It is estimated rental . equipment could be sold for $80,000 at the end of 12 years. If the buyer wants a 7% rate of return on his investment, what is the maximum price he should pay for the equipment? 9-59 A motor with a.200-horsepower output is needed in the factory for intermittent use. A Graybar motor costs $7000, and has an electrical efficiency of 89%. A B1ueball motor costs $6000 and has an 85% ~fficiency. Neither motor would have any salvage value, as the cost to remove it would equal its scrap value. The annual maintenance cost for either motor is estimated at. $300 per year. Electric power costs 0.746 kW). If a 10% $O.072/kilowatt-hour (1 hp interest rate is used in the calculations, what is the minimum number of hours the higher initial cost Graybar motor must be used each year to justify its purchase? = 9-60 Plan A requires a $100,000 investment now. Plan B requires a~ $80,000 investment now and an additional $40,000 investment at a later time. At 8% interest, compute the breakeven point for the timing of the $40,000 investment. 9-61 A low-carbon-steel machine part, operating in a corrosive atmosphere, lasts 6 years, and costs $350 installed. If the part is treated for corrosion resistance. - - ---- .-...- Problems it will cost $500 installed. How long must the treated part last to be the preferred alternative, assuming 10% interest? At what useful life are the machines equivalent if (a) 10% interest is used in the computations? (b) 0% interest is used in the computations? 9-62 Neither of the following machines has any net salvage value. A B Original cost Annual expenses Operation Maintenance Taxes and insurance $55,000 9-63 A machine costs $5240 and produces benefits of $1000 at the end of each year for eight years. Assume continuous compounding and a nominal annual interest rate of 10%. $75,000 9,500 5,000 1,700 301 (a) What is the payback period (in years)? (b) What is the breakeven point (in years)? (c) Since the answers in (a) and (b) are different, which one is "correct"? 7,200 3,000 2,250 ~'! - - -- - .~,,:<.;~;n _< . --- 4~i:-. <. i'ii/i'7'!#"' After Completing This Chapter... The student should be able to: Use a range of estimated variables to evaluate a project. · Describe possible outcomes with probability distributions. · · · ·· Useeconomicdecisiontreesto describeand solvemorecomplexproblems. Combine probability distributions for individual variables into joint probability distributions. Use expected values for economic decision making. · Measure and consider risk when making economic decisions. Understand how simulation can be used to evaluate economic decisions. QUESTIONS TO CONSIDER ~ 1. How can a company quantify the risk of an event such as discovery of TCE on its property? 2. What are some ways a. company might estimate the cost of such an event,.should.it occur? 3. Priorto the mid-1970s,it wascommon (and legal)for manufacturersto landfillchemicals and other potentially hazardous materials. When Superfund was adopted in 1980, i~ required companies that had disposed of these materials to clean up the landfill sites, even if disposal had taken place years before, at a time when the practice was not against the law. How can companies anticipate and prepare for future laws that might penalize activitieswhich arelegal today,but may become illegal (or at least sociallyunacceptable) in the future? -~ , -- '- -- - - --- - - -- r . - - - -- - U ncertai nty Future Events They Only Thought They Were Done Since the early 1980s, the federal law known as "Superfund" has required companies to clean up property sites that are contaminatedby hazardous waste. So far, hundreds of sites have been cleaned up (or "remediated," in Superfund terminology), at a collective cost that runs to the billions of dollars; hundreds of other sites are still in the process of being decontaminated. Superfund cleanups usually take years to complete, and often involve expensive legal wrangling among the parties involved. A large number of Superfund sites-including some where cleanup has been completed-have been found to contain trichloroethylene (TCE), a widely used solvent. To complicate mattersfurther, the United StatesEnvironmentalProtection Agency (BPA)has now discovered that TCE may be far more toxic than originally thought. Moreover,the cleanuptechniquesused in the past to remove the substance from groundwater may actually have made the problem worseby causingTCE to be emittedin vapor form.This vapor can concentrate in buildings and other enclosed spaces, increasingthepotentialtoxicityevenfurther. . EPA is now studying whether previously cleaned Superfund sites should be "reopened" for additional remediation of TCE. The decision is made more difficultby the fact that some previouslycleaned sites have been sold since the Superfund remediationwas finished,andmay havehad buildings constructed on them. . - -- -- 304 UNCERTAINTY IN FUTURE EVENTS An assembly line is built after the engineering economic analysishas shown that the anticipated product demand will generateprofits.A new motor,heatexchanger,or filtrationunit is installed after analysis has shown that future cost savings will economicallyjustify current costs. A new road.,school, or other public facility is built after analysis has shown that the future demand and benefitsjustify the present cost to build. However,future performance of the assembly line, motor, and so on is uncertain, and demand for the product or public facility is more uncertain. Engineering economic analysis is used to evaluate projects with long-term consequences when the time value of money matters. Thus, it must concern itself with future consequences;but describing thefuture accuratelyisnot easy.In this chapterwe considerthe problem of evaluatingthefuture.The easiestwayto beginis to make a careful estimate.Then we examine the possibility of predicting a range of possible outcomes. Finally,we consider whathappens whenthe probabilitiesofthe variousoutcomesareknownor may be estimated. We will show that the tools of probability are quite useful for economic decision making. ESTIMATES AND THEIR USE IN ECONOMIC ANALYSIS Economic analysis requires evaluating the future consequences of an alternative. In practically every chapter of this book, there are cash flow tables and diagrams that describe precisely the costs and benefits for future years. We don't really believe that we can exactly foretell a future cost or benefit. Instead, our goal is to select a single value representing the best estimate that can be made. We recognize that estimated future consequences are not precise and that the actual values will be somewhat different from our estimates. Even so, it is likely we have made the tacit assumption that these estimates are correct. We know estimates will not always turn out to be correct; yet we treat them like facts once they are in the economic analysis. We do the analysis as though the values are exact. This can lead to trouble. If actual costs and benefits are different from the estimates, an undesirable alternative may be selected. This is because the variability of future consequences is concealed .by assuming that the best estimates will actually occur. The problem is illustrated by Example 10-1. Two alternatives are being considered. The best estimates for the various consequences are as follows: Cost Net annual benefit Usefullife, in years = End-of-useful-life~salvage value A $1000 $150 10 B $2000 $250 10 $100 ..$400 I If interest is 3Y2%, which alternative has the better net present worth (NI>W)? .. __ _.. _ _ ..Iii -- ...... -r J - - -- Estimates and Their Use in Economic Analysis 305 SOLUTION Alternative A NPW = -1000+ = -1000 = 150(PjA, 3Y2%,10) + 100(PjF, 3Y2%,10) + 150(8.317) + 100(0.7089) -1000 + 1248 + 71 = +$319 Alternative B NPW = -2000 = = + 250(P j A, 3Y2%, 10) + 400(P j F, 3Y2%, 10) -2000 + 250(8.317) + 400(0.7089) -2000 + 2079 + 284 = +$363 Alternative B, with its larger NPW, would be selected. Alternate Formation of Example 10-1 Suppose that at the end of 10 years, the actual salvage value for B were $300 instead of the $400 best estimate. If all the other estimates were correct, is B still the preferred alternative? Revised B NPW = -2000 = -2000 = -2000 + 2079 + 213 = +$292 -+ A is now the preferred alternative. + 250(P j A, 3Y2%, 10) + 300(P j F, 3Y2%, 10) + 250(8.317) + 300(0.7089) Example 10-1 shows that the change in the salvage value of Alternative B actually results in a change of preferredalternative.Thus, a more thorough analysisof Example 10-1 wouldconsider(1) which valuesare uncertain, (2) whethertheuncertaintyis :1:5%or ~50 to +80%, and (3) whichuncertain valuesleadto differentdecisions.A more thoroughanalysis, which is done using the tools of this chapter, determines which decision is better over the range of possibilities.Explicitly considering uncertainty lets us make better decisions. The tool of breakeven analysis is illustrated in Example 10-2. iU~ 'Iii I - --- 306 UNCERTAINTY IN FUTURE EVENTS Use the data of Example 10-1 to compute the sensitivity of the decision to the Alt. B salvage value by computing the breakevenvalue. For Alt. A, NPW = +319. For breakeven between the alternatives, NPW A= NPW B +319 = -2000 + 250(P / A, 312%, 10) + Salvage valueB(P/ F, 312%, 10) = -2000 + 250(8.317) + Salvage valueB(0.7089) At the breakeven point Salvage valueB = 319+ 2000- 2079 = 240 0.7089 0.7089 = $339 When Alt. B salvage value >$339, B is preferred; when <$339, A is preferred. Breakeven analysis, as shown in Example 10-2, provides one means of examininJ the impact of the variability of some estimate on the outcome. It helps by answering thl question, How much variability can a parameter have before the decision will be affected' While the preferred decision depends on whether the salvage value is above or below the breakevenvalue,the economicdifferencebetween the alternativesis small when the salvage value is "close" to breakeven. Breakeven analysis does not answer the basic problem 0 how to take the inherent variability of parameters into account in an economic analysis This will be considered next. A RANGE OF ESTIMATES It is usuallymore realistic to describe parameterswith a range of possible values,rather thar a single value. A range could include an optimistic estimate, the most likely estinlMe7-ant a pessimistic estimate. Then, the economic analysis can determine whether the decisioni~ sensitive to the range of projected values. I L A firmis considering an investment.The most likely data values were found during the feasjbility study. Analyzing past data of similar projects shows that optimistic values for the first cost and the annual benefit are 5% better than most likely values. Pessimistic values are 15% worse. The firm's most experienced project analyst has estimated the values for the useful life and salvage value. ----- A Range of Estimates Cost Net annual benefit Usefullife, in years Salvage value Optimistic $950 $210 12 $100 Most Likely $1000 $200 10 $0 307 Pessimistic $1150 $175 8 $0 Compute the rate of return for each estimate. If a 10% before-tax minimum attractive rate of return is required, is the investmentjustified under all three estimates? If it is only justified under some estimates, how can these results be used. , SQLQjJQ~J Optimistic Estimate PW of cost $950 = PW of benefit = 21O(P / A, IRRopt>12) + 100(P / F, IRRopt, 12) IRRopt = 19.8% Most Likely Estimate $1000 (P / A, IRRmostlikely, 10) = 200(P / A, IRRmostlikely,10) = 1000/200 = 5 ~ IRRmostlikely = 15.1% Pessimistic Estimate $1150 - 170(P/A, IRRpess,8) (P / A, IRRpess,8) = 1150/170 . 6.76 ~ IRRpess= 3.9% From the calculations we conclude that the rate of return for this investmentis most likely to be 15.1%, but might range from 3.9%to 19.8%.The investmentmeets the 10%MARR criterion for two of the estimates. These estimates can be considered to be scenarios of what may happen with this project. Since one scenario indicates that the project is not attractive,we need t9 have a method of weighting the scenarios or consideringhow likely each is. Example 10-3 made separate calculations for the sets of optimistic, most likely, and pessimistic values. The range of scenarios is useful. However,if there are more than a few uncertain variables it is unlikelythat all will prove to be optimistic (best case) or most likely or pessimistic (worstcase).It is more likely that manyparameters are the most likely values, while some are optimistic and some are pessimistic. This can be addressed by using average or mean values for each parameter, based on Equation 10-1. Equation 10-1 puts four times the weight on the most likely value - --- -- - ~il 308 UNCERTAINTY IN FUTURE EVENTS than on the other two. This equation was developed as an approximation with the beta distribution. Mean value = Optimistic value + 4(Most likely value) + Pessimistic value 6 (10-1) This approach is illustrated in Example 10-4. Solve Example 10-3by using Equation 10-1. Compute the resulting mean rate of return. . SOLUTION Compute the mean for each parameter: Mean cost = [950 + 4 x 1000 + 1150]/6 = 1016.7 Mean net annual benefit - [210 + 4 x 200 + 170]/6 = [12 + 4 x 10 + Mean salvage life = 100/6 = 16.7 Mean useful life 8]/6 = 196.7 = 10.0 Compute the mean rate of return: PW of cost $1016.7 = PW of benefit = 196.7(j> / A, IRRbeta,10) + 16.7(j>I F,IRRbeta, 10) IRRbeta= 14.2% Iioo Example 10-3 gave a most likely rate of return (15.1%) that differed from the mean rate of return (14.2%) computed in Example 10-4. These values are different because the former is based exclusively on the most likely values and the latter takes into account the variability of the parameters. In examiningthe data, we see thatthe pessimistic valuesare further awayfrom the most likely values than are the optimistic values. This is a common occurrence. For example, a savings of 10 to 20% may be the maximum possible, but a cost overrun can be 50%, 100%, or even more. This causes the resulting weighted mean values to be less favorable than the most likely values. As a result, the mean rate of return, in this example, is less than the rate of return based on the most likely values. ' PROBABILITY We all have used probabilities. For example, what is the probability of getting a "head" when flipping a coin? Using a model that assumes that the coin is fair, both the head and tail outcomes occur with a probability of 50%, or Y2'This probability is the likelihood {' - -------- Probability 309 an event in a single trial. It also describes the long-run relative frequency of occurrence of getting heads in many trials (out of 50 coin flips, we expect to average25 heads). Probabilities can also be based on data, expert judgment, or a combination of both. Past data on weather and climate, on project completion times and costs, and on highway traffic is combined with expertjudgment to forecast future events. These examples can be important in engineering economy. Another example based on long-run relative frequency is the PW of a flood-protection dam that depends on the probabilities of different-sizedfloodsover manyyears. This would be based on data about past floods and would include many years of potential flooding. An example of a single event that may be estimated by expert judgment is the probability of a successful outcome for a research and development project, which will determine its PW. All the data in an engineering economy problem may have some level of uncertainty. However,smalluncertaintiesmaybe ignored,so that moreanalysiscan be donewiththe . large uncertainties. For example, the price of an off-the-shelfpiece of equipmentmay vary by only ::1::5%. The price could be treated as a known or deterministic value. On the other hand, demandover the next 20 yearswill havemore uncertainty.Demandshouldbe 3l\alyzed as a random or stochastic variable. We should establish probabilities for different values of demand. There are also logical or mathematicalrules for probabilities. If an outcome can never happen, then the probability is O.If an outcome will certainly happen, then the probability is 1, or 100%.This means that probabilities cannot be negative or greater than 1; in other words, they must be within the interval [0, 1], as indicated shortly in Equation 10-2. Probabilities are defined so that the sum of probabilities for all possible outcomes is 1 or 100% (Equation 10-3). Summing the probability of 0.5 for a head and 0.5 for a tail leads to a total of 1for the possible outcomes from the coin flip. An explorationwell drilled in a potential oil field will have three outcomes (dry hole, noncommercial quantities, or commercial quantities) whose probabilities will sum to one. Equations 10-2 and 10-3 can be used to check that probabilities are valid. If the probabilities for all but one outcome are known, the equations can be used to find the unknown probability for that outcome (see Example 10-5). o ~ Probability ~ 1 L P(outcomej) = 1, (10-2) where there are Koutcomes (10-3) j=l to K L r l L - - - - - - In a probability course many probability distributions, such as the normal, uniform, and beta are presented. These continuous distributions describe a large population of data. However,for engineering economyit is more common to use 2 to 5 outcomes with discrete probabilities--even though the 2 to 5 outcomes only represent or approximatethe range of possibilities. This is done for two reasons. First, the data often are estimated by expert judgment, so that using 7 to 10 outcomes would be false accuracy.Second, each outcome requires more analysis.In most cases the2 to 5 outcomesrepresents thebest trade-offbetweenrepresenting the range of possibilities and the amount of calculation required. Example 10-5 illustrates these calculations. - - -- --- -- I -- 310 - -_.-- - .- - . " - -- --. - --- UNCERTAINTY IN FUTURE EVENTS What are the probability distributionsfor the annual benefit and life for the following project? The annual benefit's most likely value is $8000 with a probability of 60%. There is a 30% probability that it will be $5000, and the highest value that is likely is $10,000. A life of 6 years is twice as likely as a life of 9 years. sOluTION .. Probabilities are given for only two of the possible outcomes for the annual benefit. The third value is found using the fact that the probabilities for the three outcomes must sum to 1, (Equation 10-3). 1 = P(Benefit is $5000) + P(Benefit is $8000) + P(Benefit is $10,000) P (Benefitis $10,000) = 1 - 0.6 - 0.3 = 0.1 The probabilitydistributioncan thenbe summarizedin a table.Thehistogram orrelativefrequency diagram is Figure 10-1. Annual benefit Probability $5000 0.3 FIGURE 10-1 Probabilitydistribution for annual benefit. $8000 0.6 $10,000 0.1 0.7 0.6 0.5 .~ := 0.4 ~ ~ £ 0.3 0.2 0.1 o $5,000 $8,000 AnnualBenefit $10,000 The problem statement tells us: P(life is 6 years) - - = 2P(life is 9 years) Equation 10-3 can be applied to write a second equation for the two unknown probabilities: P(6) + P£9)= 1 - - - --- .... Joint Probability Distributions 311 CombiningJhese, we write 2P(9) + P(9) __ t P(9) 1/3 P(6) = 2/3 The probability distribution for thelifejs P(6) ... 6,6.7%and P(9) - 33.3%. JOINT PROBABILITY DISTRIBUTIONS Example 10-5 constructed the probability distributions for a project's annual benefit and life. These examples show how likely each value is for the input data of the problem. .We would like to construct a similar probability distribution for the project's present worth. This is the distribution that we can use to evaluate the project. That present worth depends on both input probability distributions, so we need to construct thejoint probability distribution for the different combinations of their values. For this introductory text, we assume that two random variables such as the annual benefit and life are unrelated or statistically independent. This means that the joint probability of a combined event (Event A defined on the first variable and Event B on the second variable) is the product of the probabilities for the two events. This is Equation 10-4: If A and B are independent, then peA and B) = peA) x PCB) (10-4) For example, flipping a coin and rolling a die are statisticallyindependent. Thus, the prob- abilityof {flippinga head and rollinga 4} equalsthe probabilityof a {heads}= Y2times the probability of a {4} = 1/6, for a joint probability = 1/12. The number of outconies in the joint distribution is the product of the number of outcomes in each variable's distribution. Thus, for the coin and the die, there are 2 times 6 or 12 combinations. Each of the 2 outcomes for the coin is combined with each of the 6 outcomes for the die. . Some variables are not statisticallyindependent, and the calculationof theirjoint probability distribution is more complex. For example, a project with low revenues may be terminated early and one with high revenues may be kept operating as long as possible. In these cases annual cash flow and project life are not independent. While this type of relationshipcan sometimesbe modeledwith economicdecisiontrees (coveredlater:in this chapter),we will limit our coveragein this text to the simplercase of independent variables. Example 10-6uses the three values and probabilities for the annual benefit and the two values and probabilities for the life to construct the six possible combinations. Then the values and probabilities are constructed for the project's PW. -- J I II -- --- 312 UNCERTAINTY IN FUTURE EVENTS The project described in Example 10-5 has a first cost of $25,000. The finn uses an interest rate of 10%. Assume that the probability distributions for annual benefit and life are unrelated or statistically independent. Calculate the probability distribution for the PW. 5.QLUTION Since there are three outcomes for the annual benefit and two outcomes for the life, there are six combinations. The first four columns of the following table show the six combinations of life and annual benefit. The probabilities in columns 2 and 4 are multiplied to calculate the joint probabilities in column 5. For example, the probability of a low annual benefit and a short life is 0.3 x 2/3, which equals 0.2 or 20%. The PW values include the $25,000 first cost and the results of each pair of annual benefit and life. For example, the PW for the combination of high benefit and long life is: PW$IO,OOO,9= Annual Benefit $ 5,000 8,000 10,000 5,000 8,000 10,000 -25,000 + 1O,000(P/ A, 10%,9) = -25,000 + 10,000(5.759) = $32,590 Probability 30% 60 10 30 60 10 Life 6 6 6 9 9 9 Joint Probability 20.0% 40.0 6.7 10.0 20.0 3.3 100.0% Probability 66.7% 66.7 66.7 33.3 33.3 33.3 PW -$ 3,224 9,842 18,553 3,795 2.1,072 32,590 Figure 10-2 shows the probabilities for the PW. This is called the histogram, relative frequency distribution, or probability distribution function. 40 30 ,....... '-' g :g .c 8 p., 20 10 o -$3,224 I La $3,795 $9,842 $18,553 $21,072 PresentWorth $32,590 FIGURE 10-2 Probability distribution function::for PW. - - - - - - -- - - 313 Expected Value This prQbability distribution function shows that there is a 20% chance of having a negative PW. It also shows that there is a small 3.3% chance of the PW being $32,590. The three values used to describe possible annual benefits for the project and the two values for life have been combined to describe the uncertainty in the project's PW. Creating a distribution, as in Example 10-6, gives us a much better understanding of the possible PW values along with their probabilities. The three possibilities for the annual benefit and the two for the life are representative of the much broader set of possibilities that really exist. Optimistic, most likely, and pessimistic values are a good way to represent the uncertainty about a variable. Similarly the six val1.lesfor the PW represent the much broader set of possibilities. The 20% probability of a negativePW is one measure of risk that we will talk about later in the chapter. Some problems, such as Examples 10-3 and 10-4,have so many variables or different outcomes that constructing the joint probability distribution is arithmeticallyburdensome. If the valuesin Equation 10-1are treated as a discrete probability distribution function, the probabilities are 1/6, 2/3, 1/6. With an optimistic, most likely, and pessimistic outcome for each of 4 variables, there are 34 = 81 combinations.In Examples10-3and 10-4, the salvage value has only two distinct values, so there are still 3 x 3 x 3 x 2 = 54 combinations. When the problem is important enough, the effort to construct the joint probability distribution is worthwhile. It gives the analyst and the decision maker a better un- . derstanding of what may happen. It is also needed to calculate measures of a project's risk. While spreadsheets can automate the arithmetic, simulation (described at the end of the chapter) can be a better choice when there are a large number of variables and combinations. EXPECTEDVALUE For any probability distribution we can compute the expected value (EV) or arithmetic average (mean). To calculate the EV, each outcome is weighted by its probability, and the results are summed. This is NOT the simple average or unweighted mean. When the class average on a test is computed, this is an unweighted mean. Each student's test has the same weight. This simple "average" is the one that is shown by the button x on many calculators. The expected value is a weighted average,like a student's GPAor gradepoint average. For a GPA the grade in each class is weighted by the number of credits. For the expected value of a probability distribution, the weights are the probabilities. . This is described in Equation 10-5.We sawin Example 10-4,that these expected values can be used to compute a rate of return. They can also be used to calculate a present worth as in Example 10-7. Expected value = OutcomeA x peA) + OutcomeBx PCB)+... (10-5) .-- -- ... - 314 UNCERTAINTY IN FUTURE EVENTS The first cost of the project in Example 10-5 is $25,000. Use the expected values for annual benefits and life to estimate the present worth. Use an interest rate of 10%. = 5000(0.3) + 8000(0.6) + 10,000(0.1) = $7300 EVlife = 6(2/3) + 9(1/3) = 7 years EVbenefit The PW using these values is PW(EV) = -25,000 + 7300(P / A, 10%,7) = -25,000 +6500(4.868) = $10,536 [Note: This is the present worth of the expected values, PW(EV), not the expected value of the present worth, EV(PW). It is an easy value to calculate that approximates the EV(PW), which will be computed using the joint probability distributioI).found in Example 10.6.] Example 10-7 is a simple way to approximate the project's expected PW. But the true expected value of the PW is somewhatdifferent.Tofindit, we must use thejoint probability distribution for benefit and life, and the resulting probability distribution function for PW that was derived in Example 10-6. Example 10-8 shows the expected value of the PW or the EV(PW). Use the probability distributionfunction of the PW that was derivedin Example 10-6to calculate the EV(PW). Does this indicate an attractive project? . ';" The table from Example 10-6can be reused with one additional column for the weighted values of the PW (= PW x probability). Then, the expected value of the PW is calculated by summing the column of present worth values that have been weightedby their probabilities. PW x Joint Annual Life Joint PW Probability Benefit Probability (years) Probability Probability $ 5,000 8,000 10,000 5,000 8,00000 10,000 30% 60 10 30 10 ,. -- 6 6 6 9 9 9 66.7% 66.7 66.7 33.3 33.3 33.3 20.0% 40.0 6.7 10.0 20.0 - 3.3 100.0% -$ 3,224 9,842 18,553 3,795 21,072 32,590 EV(PW) -$ 645 3,937 1,237 380 4,214 1,086 = $10,209 Expected Value 315 With an expected PW of $10,209, this is an attractive project. While there is a 20% chance of a negative PW, the possible positive outcomes are larger and more likely. Having analyzed the project under uncertainty,we are much more knowledgeable about the potential result of the decision to proceed. The$~0,209 value is more aCClli"ate than the approximate value calculated in Example10-7. The values differ.because PW is a nonlinear function of the life. The more accurate value of $~0,209 is lower because the annual benefit values for the longyrlife are discounted by 1/(1 + i) formOry ye~. In Examples 10-7 and 10-8, the question was whether the project had a positive PW. With two or more alternatives, the criterion would have been to maximize the PW. . With equivalent uniform annual costs (EUACs)the goal is to minimize the EUAC. Example 10-9 uses the criterion of minimizing the EV of the EUACto choose the best height for a dam. . A dam is being considered to reduce river flooding. But if a dam is built, what height should it be? Increasing the dam's height will (1) reduce a flood's probability, (2) reduce the damage when floods OCClli", and (3) cost more. Which dam height minimizes the expected total annual cost? The state uses an interest rate of 5% for flood protection projects, and all the dams should last 50 years. Dam Annual Damages If First Cost Flood Occurs Height (ft) P (flood) > Height No dam $ 0 0.25 $800,000 20 0.05 700,000 500,000 30 0.01 800,000 300,000 40 0.002 900,000 200,000 The easiest way to solve this problem is to c400se the dam height with the lowest equivalent unifonn annual cost (EUAC). Calcu.lating the EUAC of the first cost requires multiplying the first coSt by (4jP, 5%, 50). For example, for the da:m20ft high. this is 700,000(Aj P,5%, 50) .. .. $38,344. Calculating the annual expected.floodd.a:rp.age co~tfo1"eachalternativeis simplifiedbecause =- ~~~!~~ : ~ ~or~:."PJ~~ tt~~~tdr~! ~~t'?'2~~usethed.ama~eSi~no~ood,!!:~ ~~4Thus~e=ne~dto ~ ~~ J GaJclllateonfy ffietermIorflo03111g.'I'lilSIS ~~5YmartiPIymg IDeP(IIoOd)tU1lesthe damages 1 if,a floodhappens. For example, the expected annual flood da:mageco~twith no leveeis 0.25 x I t\ ~ $80'p,000, ~i $2g0!OQQ~; y- -_ f.I~: ~= =n ~ =- ~;::; ~;r,; ~ ; '" """"'- I J 316 UNCERTAINTY IN FUTURE EVENTS Then the EUAC of the first cost and the expected annual flood damage are added together to find the total EUAC for each height. The 30 ft dam is somewhat cheaper than the 40 ft dam. Dam Height (ft) No dam 20 30 40 EUAC of First Cost 0 $ 38,344 43,821 49,299 Expected Annual Flood Damages $200,000 25,000 3000 400 Total Expected EUAC $200,000 63,344 46,821 49,699 ECONOMIC DECISION TREES Some engineeringprojects are more complex, and evaluatingthem properly is correspondingly more complex. For example, consider a new product with potential sales.volumes ranging from low to high. If the sales volume is low, then the product may be discontinued early in its potential life. On the other hand, if sales volume is high, additional capacity may be added to the assembly line and new product variations may be added. This can be modeled with a decision tree. The following symbols are used to model decisions with decision trees: Decision node D /-'\ Dl D2: Decision maker chooses 1 of the available paths. Dx ;;;:C1 Chance node 0~C2: Represents a probabilistic (chance) event. Each possible ~. Cy outcome (CI, Cz, ..., Cy) has a probability (PI, pz, ..., py) associated with it. Outcome node --0: Shows result for a partic~lar path through the decision tree. Pruned branch ---H--: The double hash mark indicates that a branch has been pruned because another branch has been chosen. This can happen only at decision nodes, not at chance nodes.The term "pruned" is chosen to correspondwith the gardener's practice of trimming or pruning off branches to make a tree or bush healthier. o Figure 10-3 illustrates how decision nodes D, chance nodes 0, and outcome nodes can be used to describe the problem's structure. Details such as the probabilities and costs can be added on the branches that link the nodes. With the branches from decision and chance nodes the model becomes a decision tree. ----- --- . Economic Decision Trees 317 7: Net Revenue ="$0 4: Net Revenue Year!:;:: $100K 8: Revenue $100Klyear 5: Net Revenlle Year 1 "".$200K Year 2 ... n 9: Net Revenue =. $600K/year 1: Build New Product 6: Net Revenue Year 1 == $400K 10: Net Revenue == $400K/year t == 0 t == 1 t ==2, ..., ex> FIGURE 10-3 Economic decision tree for new product. Figure 10-3illustrates thatdecision trees describetheproblemby startingat the decision that mustbe made and then adding chance and decisionnodesin theproper logical sequence. Thus describing the problem starts at the first step and goes forward in time with sequences of decision and chance nodes. To make the decision, calculations begin with the final nodes in the tree. Since they are the final nodes, enough infonnation is availableto evaluate them. At decision nodes the criterion is either to maximize PW or to minimize EUAC. At chance nodes an expected value for PW or EUAC is calculated. Once all nodes that branch from a node have been evaluated,the originating node <;:an be evaluated. If the originating node is a decision node, choose the branch with the best PW or EUAC and place that value in the node. If the originating node is a chance node, calculate the expected value and place that value in the node. This process "rolls back'" values from the terminal nodes in the tree to the initial decision. Example 10-10 illustrates this. I"Ii fI ".!_ .J; -- 318 UNCERTAINTY IN FUTURE EVENTS What decision should be made on the new product summarized in Figure 1O-3?What is the expected value of the product's PW? The firm uses an interest rate of 10% to evaluate projects. If the product is terminated after oneYear,the capital equipmenthas a salvage value of $550,000 for use with other new products. If the equipment is used for 8 years, the salvage valut? is $0. Evaluating decision trees is done by starting with the end outcome nodes and the decisions that lead to them. In this case the decisions are whetherto terminate after 1year if sales volume is low and whether to expand after 1 year if sales volume is high. The decision to terminate the product depends on which is more valuable, the equipment's salvage value of $550,000or the revenue of $100,000per year for 7 more years. The worth (PW1) of the salvage value is $550,000. The worth (PW1)of the revenue stream at the end of Year 1 shown in node 8 is: PW1 for node 8 = 100,000(PjA, 10%,7) - 100,000(4.868) = $486,800 Thus, terminating the product and using the equipmentfor other products is better. We enter the two "present worth" values at the end of Year1in nodes 7 and 8. We make the arc-tonode 7 bold, and use a double hash mark to show that we're pruning the arc to node 8. The decision to expand at node 6 could be based on whether the $800,000 first cost for expansion can be justified based on increasingannualrevenues for 7 years by $200,000 per year. However,this is difficultto show on the tree. It is easier to calculate the "present worth" values at the end of Year 1 for each of the two choices. The worth (PW1)of node 9 (expand) is: .~ PW1for node 9 - -800,000 + 600,000(P j A, 10%,7) ;::::; -800,000 + 600,000(4.868) .$2,120,800 The value of node 10 (continue withoutexpanding) is: :PW1 for pode 10 . 400,000(Pj A, 10%,7) .4OQ,0()O(4. 868) --'=$1~947,200 This is $173,600less.thaI}tIleexpaJ1sioJ).node,sothee)(pansion shouldhappen if volumeis high, Figure 10=4suni1nanz~swjiat,w~iW.ff~~HiU"S~geofthepro~ess. '" ~ __ - - ...". ~ II _.. .. .. - - ---- ~ Economic Decision Trees 319 7: Net Revenue = $0 4: Net Revenue Year 1 = $looK ~\~ PW = $486,800 I ~ 5: Net Revenue Year 1 = $200K 8: Revenue $ lOOK/year Year 2 ... n 9: Net Revenue III 1: Build New Product 6: Net Revenue Year 1 = $400K =$600K/year ~.PW 1 = $2,120,800 ~ First cost = $800K /g~ ~ ~ t= 0 t =1 10: Net Revenue = $4ooK/year t = 2, ...,00 FIGURE 10-4 Partially solved decision tree for new product. The next step is to calculate the PW at nodes 4,5, and 6. PW at node 4 = (100,000 + 550,000)(P / F, 10%,1) = 650,000(0.9091)- $590,915 PW at node 5 = (200,000)(P / A, 10%,8) - 20Q,000(5.335)= $1,067,000 PW at node 6 = [400,000 - 800,000 + 600,000(P / A, 10%, 7)](P / F, 10%, 1) = [-400,000 + 600,000(4.868)](0.9091) = $2,291,660 Now the expected value at node 2 can be calculated EV at node 2 - 0.3(590,915) + p.6(1,067,002) + ~1(2,~~~l,~669~~$\,04~,640 The cost of selecting node 2 is $1,000,000, so proceeding witb.the prpduct.has an.expect~d PW of $46,640.This is greater than the $0 for not building the project. So the decision is to build. Figure 10-5 is the decision tree anhe finaJ.stage. =' ---- --- 320 UNCERTAINTY IN FUTURE EVENTS 7: Net Revenue = $0 PW '"" $590,915 4: Net Revenue Year 1 '""$100K 8: Revenue $100Klyear PW == $1,067,000 5: Net Revenue Year 1 ~ $200K Year 2 ...n 9: Net Revenue '1:17'11 ~ $600K/year I:: PW = $46,640 PW= $2,291,660 #1 pW1 = $2,120,800 fl:J First cost = $800K 6: Net Revenue Year 1= $4QOK 1: Build New Product 10: Net Revenue = $400K/year ,I : I . I I (=0 (= 2, ...,00 FIGURE 10-5 Solved decision tree for new product. J I I, Example 10-10 is representative of many problems in engineering economy. The main criteria is maximizing PW or minimizing EUAC. However, as shown in Example 10-11, other criteria, such as risk, are used in addition to expected value. Consider the economic evaluation of collision and comprehensive(fire, theft, etc.) insurance for a car. This insurance is typically required by lenders, but once the car has been paid for, this ihsurance is not required. (Liabilityinsurance is'typically a legalrequirement.) Figure 10-6 begins with a decision node with two alternativesfor the next year. Insurance ~il1 ~ost$800 per year with.a $500 d~~~ctibleif a loss occurs.The other option i&to self.-insure, i , i I i. L ----- - whichmeans to go without buying cQ;u.1s1on ana comprebensivein&urance.Tl1enif alosroc~ut=r;= L - ......- .. ...- - - _III!I _III__~~' l' . Economic Decision Trees 321 $0 $300 «$500 deductible) $500 $0 $300 $13,000 FIGURE 10-6 Decision tree for buying auto collision insurance. the owner must replace the vehicle with money from savings or a loan, or do without a vehicle until the owner can afford to replace it. Three accident severities are used to represent the range of possibilities: a 90% chance of no accident, a 7% chance of a small accident (at$300, which is less than the deductible), and a 3% chance of totaling the $13,000 vehicle. Since our driving habits are likely to be the same with and without insurance, the accident probabilities are the Same for both chance nodes. Even though this is a text on engineering economy, we have simplified the problem and ignored the difference in timing of the cash flows. Insurance payments are made at the beginning of the covered period, and accident costs occur during the covered period. Since car insurance is usually paid semiannually, the results of the economic analysis are not changed significantly by the simplification. We focus on the new concepts of expected value, economic decision trees, and risk. What are the expected values for each alternative, and what decision is recommended? .. - - - '. '. -. 'SQhUI10N The expected values are computed by using Equation 10-5.Ifinsured, the maximum cost equals ~ . I the deductible"' of $500.If selr-insure'O, the costis the cosfof tileaccident. EVaccident w/ins. = (0.9 '="= ~ EVaccident w/o ins. X =: 0) + (0.07 1:1I =;; X 300) + (0.03 = X 500) = $36 = = (0.9 X 0) + (0.07 X 300)+ (0.03 X 13,000)= $411 -- '. - - - 322 UNCERTAINTY IN FUTURE EVENTS Thus, buying insurance lowersthe expectedcost of an accidentby $375. Toevaluatewhether we should buy insurance, we must also accountfor the cost of the insurance.Thus, these expected costs are combined with the $0 for self-insuring (total $411) and the $800 for insuring (total') $836). Thus self-insuring has an expefted value cos,tthat is $425 less per year;($836 - $411). This is not surprising, since the premiums collected must cover both.the costs of operating the insurance company and the expected value of the payouts. This is also an example ofexpectedv(llues alone not determining the decision. Buying iIlsuranc,ehas an expectedcosttb,atis $425perye~ highet, but tb,atinsunll1ce}imitsthe maxiInutn loss to $500 rather thaI)$13,000.:rbe$425may be worth spendi,Ilgto avoidthatpsk. RISK Risk can be thought of as the chance of getting an outcomeother than the expectedvaluewith an emphasis on something negative. One common measure of risk is the probability of a loss (see Example 10-6).The other common measure is the standard deviation (a), which measures the dispersion of outcomes about the expected value. For example, many students have used the normal distribution in other classes. The normal distribution has 68% of its probable outcomes within ::1:1stand~d deviation of the mean and 95% within ::1:2standard deviations of the mean. Mathematically, the standard deviation is defined as the square root of the variance. This term is definedas the weightedaverageof the squareddifferencebetweenthe outc.omes of the random variable X and its mean. Thus the larger the difference,between the mean and the values, the larger are the standard deviationand the variance.This is Equation 10-6: Standard deviation (a) = V[EV(X - mean)2] (10-6) Squaring the differences between individual outcomes roWthe EV ensures that positive . -.' and negativedeviationsreceivepositiveweights.Consequently,negativevaluesfor the standard deviation are impossible,and they instantly indicatearithmeticmistakes.The standard deviation equals.0 if only one outcome is possible. Otherwise, the standard deviation is positive. This is not the standard deviation formula built into most calculators, just as the weighted average is not the simple average built into most calculators. The calculator formulas are for N equally likely data points from a randomlydrawn sample, so that each probability is 1/ N. In economic analysis we will use a weighted average for the squared deviations since the outcomes are not equally likely. The second difference is that for calculations (by hand or the calculator), it is easier to use Equation 10-7, which is shown to be equivalent to Equation 10-6 in introductory probability and statistics texts. Standard deviation (a) - = V{EV(X2) - {Outcomei. x peA) + Outcome~ x PCB)+... [EV(X)]2} - expected value2} (10-7) (10-7') --- .- .. --- - - --- 323 Risk This equation is the square root of the difference between the average of the squares and the square of the average.The standarddeviationis used insteadof the variance because the standard deviation is measured in the same units as the expected value. The variance is measured in "squared dollars"-whatever they are. The calculation of a standard deviationby itself is only a descriptivestatistic of limited value. However,as shown in the next section on risk/return trade-offs,it is useful when the standard deviation of each alternativeis calculated and these are compared. But first, some examples of calculating the standard deviation. Consider the economicevaluationof collision andcomprehensive(fire,theft, etc.) insurancefor an auto. One example was described in Figure 10-6.The probabilitiesand outcomes are summarized in the calculation of the expected values, which was done using Equation 10-5. EVaccident w/ins.= (0.9 x 0) + (0.07 x 300) + (0.03 x 500) = $36 EVaccident w/oins.= (0.9 x 0) + (0.07 x 300) + (0.03 x 13,000) = $411 Calculate the standard deviations for insuring and not insuring. 'SOLutiON: . - - -.. - --. The first step is to calculate the EV(outcome2) for each. EV~ident w/ins. EV;ccident w/o ins. = (0.9 X 02) + (0.07 x 3002) + (0.03 x 5002) = $13,800 = (0.9 X 02) + (0.07 x 3002) + (0.03 x 13,0002) = $5,076,300 Then the standard deviationscan be calculated. = -/(13,800 O"w/ins." O"w/oins. - 362) = JI2,504= $112 = -/(5,076,300 -4112) = -/4,907,379 =$2215 As described in Example 10-11, the expected value cost of insuring is $836 (= $36 + $800) and the expected value cost of self-insuring is $411. Thus the expected cost of not insuring is about half the cost of insuring. But the standard deviation of self-insuring is 20 times larger.It is clearlyr,iskier. Which choice is preferred depends on how much risk yoti are comfortable with. As stated before, this is an example of expected values alone not determining the decision. Buying insurance ,has an expected cost:::that ~is-$425=-peryear higher, but that "insurancelimits the maximum loss to $500 rather than $13,000. The $425 may be wortb spendiJ:!gto avoid that risk. : f '" I :I 324 UNCERTAINTY IN FUTURE EVENTS Using the probability distribution for the PW from Example 10-6, calculate the PW's standard deviation. The following table adds a column for (PW) (probability) to calculate the EV(PW2). Annual Benefit Probability $ 5,000 30% 8,000 60 10,000 10 5,000 30 8,000 60 10,000 10 Life (years) Probability 6 66.7% 666.7 6 66.7 9 33.3 9 33.3 9 33.3 Standard deviation Joint Probability 20.0% 40.0 6.7 10,0 20.0 3.3 PW ~$ 3,224 9,842 18,553 3,795 21,072 32,590 EV PW x Probability -$ 645 3,937 1,237 380 4,2!4 1,086 $10,209 PW2x Probability $ 2,079,480 38,j47,9~4 22,950,061 1,442,100 88,797,408 35,392,740 $189,409,745 V{EV(X--:;;~ [EV(X)]2} (J. '. J{l89,405,745~ [l0,209]2}'- y'85, 182,064 = $9229 For those with stronger backgrounds in probability than this chapter assumes, let us consider how the standard deviation in Example 10-13depends on the assumptionof independence between the variables. While exceptions exist, a positive statistical dependence between variables often increases the PW's $tandarddeviation. Similarly,a negativestatistical dependence between variables often decreases the PW's standard deviation. RISK VERSUS RETURN A graph of risk versus return is one way to consider these items together. Figure 10-7 in Example 10-14 illustrates the most common format. Risk measured by standard deviation is placed on the x axis, and return measuredby expected value is placed on the y axis. This is usually done with the internal rates of return of the alternativesor projects. t A large firmis discontinuing an older product, so some facilities are becoming availablefor other uses. The followingtable sunmiarizeseighfnew projects that woulduse the facilities.Coriside~ng expected return and risk, wbjch projects are good candidates? The .firmbelieves it can earn 4% on a risk-free investmentin: govel11p1ent PJ;Qject = ~ !!.~ securities ::: = == (labeled ::;=. as == ~~, ~F). ~- - -- -- - - I Risk Versus Return Project 1 2 3 4 5 6 7 8 F IRR 13.1% 12.0 7.5 6.5 9.4 16.3 15.1 15.3 4.0 325 Standard Deviation 6.5% 3.9 1.5 3.5 8.0 10.0 7.0 9.4 0.0 SOLUTION Answering the question is far easier if we use Figure 10-7.Since a larger expectedreturn is better, we want to select projects that are as "high up" as possible. Since a lower risk is better, we.want to select project~ that are as "far left" as possible. The graph lets us examine the trade-off of accepting more risk for a higher return. FIGURE 10-7 Risk-versus-retum graph. 20 6 7 . . 8 . . 5 5 4 ., F o 246 8 10 Standard Deviation of 1M (%) First, we can eliminate Projects 4 and 5.They are dominated projects. Dominated alternatives are no better than another alternative on all measures and inferior on at least one measure. Project 4. is dominated by Project 3, which has a higher expected return and a lower risk. Project 5 is dominated by Projects 1, 2, and 7. All three have a higher expected return and a lower risk. Second, we look at the efficient frontier. This is the blue line in Figure 10-7 that connects Proj~ctE,P' 3, 2~..7,and 6. I)epencijng, on the trade"of( that~we want to make betweell risk and return, any of these could be the best choice. Project 1 appears to be inferior to Projects 2 and 7. Project 8 appears to be inferior to Projects ::: ., 7"and6. Pmjects 1 and 8 are inside and not on the efficient frontier."== ;;:; ==z:=:= = ~ - ... ---~ -- -- - ------ ---- , -" -- - -. 326 - UNCERTAINTY IN FUTURE EVENTS , There are models of risk and return that can allow us to choose between Projects F, 3, 2, 7, and 6; but thosemodels are beyond what.is coveredhere. - SIMULATION Simulation is a more advanced approach to consideringrisk in engineering economy problems. As such, the followingdiscussionfocuses on what it is. As the examples show,spreadsheet functions and add-in packages make simulation easier to use for economic analysis. Economic simulation uses random samplingfrom the probability distributionsof one or more variables to analyze an economic model for many iterations. For each iteration, all variables with a probability distribution are randomly sampled. These values are used to calculate the PW, IRR, or EUAC.Then the results of all iterations are combined to create a probability distribution for the PW,IRR, or EUAC. . Simulation can be done by hand, using a table of random numbers-if there are only a few random variables and iterations. However,results are more reliable as the number of iterations increases, so in practice this is usually computerized. This can be done in Excel using the RANDO function to generate random numbers, as shown in Example 10-15. Because we were analyzingeach possible outcome,the probabilitydistributionsearlier in this chapter (and in the end-of-chapter problems) used two or three discrete outcomes. This limited the number of combinations that we needed to consider. Simulation makes it easy to use continuous probability distributionslike the uniform, normal, exponential, log normal, binomial, and triangular.Examples 10-15and 10-16use the normal and the discrete . uniform distributions. ShipM4U is considering installing a new, more accurate scale, which will reduce the errorjn computingpostagechargesandsave$250ayear.Thescale'susefullifeisbelievedtobeuniformly distributedover 12, 13, 14, 15,and 16years.The initialcost of the scale is estimatedto be normally distributed with a mean of $1500 and a standarddeviation of $150. Use Excel to simulate 25 random samples of the problem and compute the rate of return for each sample. Construct a graph of rate of return versus frequencyof occurrence. SOLUTION This problem is simple enough to construct a table with each iteration's values of the life and the first cost. From these values and the annual savings of $250, the IRR for eacb iteration can be calculated using the RATE function. These are shown in Figure 10-8. The IRR values. are summarized in a relative frequency diagram in Figure~.. 10-9. ,'" .--'. -«:;11 _. "" (Note: Each time Excel recaJ,cul<Itesthe spreadsheet, different values for all tlie ranaoIIl numbers are generated. Thus the results depend on the set of random numbers, and your results will be different if you create this spreadsheet.) in c: 'Ii = :r::::r:::I:: ~!: ~ I:!: == = _ I . t - -- - - --- --- ..--- ..c_ -- -- -- ---- = ..... r Simulation FIGURE 10-8 Excel spreadsheet for simulation (N = 25). B A 1 250 C 327 D Annual Savings Life First Cost 2 3 Min 12 1500 Mean 4 Max 16 150 Std dev 5 6 Iteration IRR 7 1 12 1277 16.4% 8 2 15 1546 13.9% 9 3 12 1523 12.4% 10 4 16 1628 13.3% 11 5 14 1401 15.5% 12 6 12 1341 13 7 12 1683 15.2% 10.2% 14 8 14 1193 19.2% 15 9 15 1728 11.7% 16 10 12 1500 12.7% 17 11 16 1415 16.0% I . 18 12 12 1610 11.2% 19 13 15 1434 15.4% 20 14 12 1335 15.4% 21 15 14 1468 14.5% 22 16 13 23 17 1469 1409 13.9% 14 15.3% 24 18 15 1484 14.7% 25 19 14 20 15 1594 1342 12.8% 26 27 17.0% 16.8% 21 14 1309 28 22 12 1541 12.1% 29 23 16 1564 14.0% 30 24 13 1590 12.2% 31 25 . 16 1311 17.7% 33 Mean 14 1468 14.4% 34 Std dev 2 135 2.2% 32 Note for students who have had a course in probability and statistics: Creating the random values for life and first cost is done as follows. Select a random number in [0, 1) using Excel's RAND function:,This is tlte Val~eof the ~ug1Ulati,ye di~tributionfunctionfor .thevariable.Convert this to the variable's value by using an inverse function from Excel, or build the inverse function. For the discreteuniform life the function is = .minlife +INT(tange* RANDO).For the normally "distributed first C()st the,functions -- ------ is<-NORMINY(RANDO, mean, standard deviation). - - -.. - - .. - . - - ---- - -- ~ --- - -- 328 UNCERTAINTY IN FUTURE EVENTS 6 5 a I 4 ;>. u s::: Q) 6- 3 r.x.. 2 o 8 FJGURE .10-9 Graph of 1M values. Stand-alone simulation programs and commercialspreadsheet add-in packages such as @Risk and Crystal Ball provide probabilitydistributionfunctions to use for each input variable. In Example 10-16 the functions RiskUniformand RiskNormal are used. The packages also collect values for the output variables,such as the IRR for Example 10-16. In other problems the PW or EUAC could be collected. These values form a probability distribution for the PW, IRR, or EUAC. Fromthis distribution the simulationpackage can calculate the expected return, P(1oss),and thestandarddeviation of the return. Example 10-16 uses @Risk to simulate 1000 iterations of PW for the data in Example 10-15. A simulation package makes it easy to do more iterations. More important still, since it is much easier to use different probabilitydistributions and parameters, more accurate models can be built. Because the modelsare easier to build, they are less likely to contain errors. Consider the scale described in Example 10-15. Generate 1000iterations and construct a frequency distribution for the scale's rate of return. The first IRR (cell A8) of 14.0t% tgat t~ computed in Figure 10-10 is based on the average bYe and the average first cost. TheB~cop:d''1RR(cell 1\1.1)bfi.~.Ol%'is comptited by @Risk using the average of each distributjoJ).. Th.e cell content i§ the RATE fOfJ).1ulawith its l~is1<:UmfofJ).1and ."..f!I. --- :; Simulation A 1 2 3 4 5 6 7 8 9 10 11 I B I IRR for each simulation iteration 14.01% = RA TE(RiskUniform(A3,A4),A5,RiskNorma1(Al,A2)) I 14.15% = average 2.17% = standard value of All deviation of All (IRR) from 1000 iterations DistributionforIRR (cellAll) 16 14 19 12 --- a 22 24 25 26 27 28 (IRR) from 1000 iterations I 17 18 23 F IRR computedusing averages 14.01% = RATE((A3+A4)j2,A5,Al) 16 21 E I 15 20 D -1500 averagefirst cost 150 standarddeviationof first cost 12 minimumvalueof life 16 maximumvalue of life 250 annualbenefit 12 13 14 c 329 .0 0 ------- -- 1O - - 6 2 0 7 - - 8 4 - r- "n, l,n,n.nIr-1I 10 13 14.5 16 17.5 8.5 11.5 19 20.5 IRR(%) ------- - - FIGURE 10-10 Simulation spreadsheetfor Example 10-15 and 10-16. RiskNonnal function, however spreadsheets with @Riskfunctionsdisplay by default the results of using average values. The RATEfunction contains two @Riskfunctions: RiskUnifonn and RiskNonnal. The unifonn distribution has the minimum and maximum values as parameters.The nonnal distribution has the average and standard deviation as parameters. The third IRR (cell A13) is the average for 1000 iterations. It will change each time the =simulationis done.=I'hegraph in Figure 10-10 withl 000'"iterations is much §mootIier""than The graph from Example 10-15,where 25 iterations were done. J I I I . ~ I ~ I I i ="'iF "---- ~ 330 UNCERTAINTY IN FUTURE EVENTS SUMMARY Estimating the future is required for economic analysis, and there are several ways to do this. Precise estimates will not ordinarily be exactly correct, but they are considered to be the best single values to represent what we think will happen. A simple way to representuncertainty is through a range of estimates for each variable, such as optimistic, most likely,and pessimistic. The full range of prospectiveresults.maybe examinedby using the optimisticvalues to solve the problem and then using the pessimistic values. Solving the problem with the most likely values is a good single value estimate. However,the extremes with all optimisticvalues or all pessimistic values are less likely-it is more likely that a mix of optimistic, most likely, and pessimistic values will occur. One approach uses weightedvalues instead of a range of estimates. One set of weights suggestedis: Estimate Relative Weight 1 Optimistic 4 Most likely 1 Pessimistic The most commonly used approach for decision making relies on expected values. Here, known or estimated probabilities for future events are used as weights for the corresponding outcomes. Expected value = OutcomeAx ProbabilityA+ OutcomeBx ProbabilityB+ .. . Expected value is the most useful and the most frequently used technique for estimating a . project's attractiveness. However, risk as measured by standard deviation and the probability of a loss is alsQ important in evaluating projects. Since projects with higher expected returns also frequently have higher risk, evaluating the trade-off's between risk and return is useful in decision making. More complicated problems can be summarized and analyzed by using decision trees, . which allow logical evaluation of problems with sequential chance, decision, and outcome nodes. Where the elements of an economic analysis are stated in terms of probability distributions, a repetitive analysis of a random sample is often done. This simulation-based approachrelies on the premise that a random samplingof increasing size becomes a better and better estimate of the possible outcomes. The largenumber of computationsmeans that simulationis usually computerized. PROBLEMS 10-1 Telephone poles exemplify items that have varying useful lives. Telephone poles, once installed in a location, remain in useful service until one of a variety of events occur. (a) Name three reasons why a telephone pole might be removed from useful service at a particular location. (b) You are to estimate the total useful life of telephone poles. If the pole is removedfrom an original location while it is still serviceable, it will r . .. Problems 331 10-8 The construction time for a bridge depends on weather conditions. The project is expected to take 250 days if the weather is dry and the temperature is hot. If the weather is damp and cool, the project is expected to take 350 days. Otherwise, it is expected to take 300 days. Historical data suggest that the probability of cool, damp weather is 30% and that of dry, hot weather is 20%. Find the project's probability distribution and expected completion time. be installed elsewhere. Estimate the optimistic life, most likely life, and pessimistic life for telephone poles. What percentage of all telephone poles would you expect to have a total useful life greater than your estimated optimistic life? 10-2 The purchase of a used pickup for $9000 is being considered. Records for other vehicles show that costs for oil, tires, and repairs about equal the cost for fuel. Fuel costs are $990 per year if the truck is driven 10,000 miles. The salvage value after 5 years of use drops about $0.08 per mile. Find the equivalent uniform annual cost, if the interest rate is 8%. How much does this change if the annual mileage is 15,000? 5000? 10-9 You recently had an auto accident that was your fault. If you have another accident or receive a another moving violation within the next 3 years, you will become part of the "assigned risk" pool, and you will pay an extra $600 per year for insurance. If the probability of an accident or moving violation is 20% per year, what is the probability distribution of your "extra" insurance payments over the next 4 years? Assume that insurance is purchased annually and that violations register at the end of the year-just in time to affect next year's insurance premium. 10-3 A heat exchanger is being installed as part of a plant modernization program. It costs $80,000, including installation, and is expected to reduce the overall plant fuel cost by $20,000 per year. Estimates of the usefullife of the heat exchanger range from an optimistic 12 years to a pessimistic 4 years. The most likely value is 5 years. Using the range of estimates to compute the mean life, determine the estimated before-tax rate of return. Assume the heat exchanger has no salvage value at the end of its useful life. 10-10 Two instructors announced that they "grade on the curve," that is, give a fixed percentage of each of the various letter grades to each of their classes. Their curves are as follows: 10-4 For the data in Problem 10-2 assume that the 5000, 10,000, and 15,000 mileage values are, respectively, pessimistic, most likely, and optimistic estimates. Use a weighted estimate to calculate the equivalent annual cost. Instructor A 10% 15 45 15 15 Grade A B C D F 10-5 When a pair of dice are tossed, the results may be any whole number from 2 through 12. In the game of craps one can win by tossing either a 7 or an lion the first roll. What is the probability of doing this? (Hint: There are 36 ways that a pair of six-sided dice can be tossed. What portion of them result in either a 7 or an II?) (Answer: 8/36) Instructor B 15% 15 30 20 20 If a random student came to you and said that his object was to enroll in the class in which he could expect the higher grade point average, which instructor would you recommend? (Answer: GPAB =. 1.95, Instructor A) 10-11 A man wants to determine whether or not to invest $1000 in a friend's speculative venture. He will do so if he thinks he can get his money back in one year. He believes the probabilities of the various outcomes at the end of one year are as follows: 10-6 Annual savings due to an energy efficiency project have a most likely value of $30,000. The high estimate of $40,000 has a probability of 0.2, and the low estimate of $20,000 has a probability of 0.30. What is the expected value for the annual savings? (Answer: $29,000) Result 10-7 Over the last 10 years, the hurdle or discount rate for projects from the firm's research and development division been 10% twice, 15% three times, and 20% the rest of the time. There is no recognizable pattern. Calculate the probability distribution and the expected value for next year's discount rate. $2000 (double his money) 1500 1000 500 o (lose everything) Probability 0.3 0.1 0.2 0.3 0.1 -.' - - - --- -- 332 - .UNCERTAINTY IN FUTURE EVENTS What would be his expected outcome if he invests the $1000? ' 10-12 The MSU football team has 10 games scheduled for next season. The business manager wishes to estimate how much money the team can be expected to have left over after paying the season's expenses, including any postseason "bowl game" expenses. From records for the past season and estimates by informed people, the business manager has assembled the following data: Situation Probability Regular season Win 3 games Win 4 games Win 5 games Win 6 games Win 7 games Win 8 games Win 9 games Win 10 games 0.10 0.15 0.20 0.15 0.15 0.10 0.07 0.03 Postseason Bowl game 0.10 Situation . Regular season Win 5 or fewer games Win6to8 games Win 9 or 10 games Postseason Bowl game Net Income $250,000 400,000 600,000 Additional income of $100,000 Based on the business manager's data, what is the expected net income for the team next season? (Answer: $355,000) 10-13 In the New Jersey and Nevada gaming casinos, craps is a popular gambling game. One of the many bets available is the "Hard-way 8." A $1 bet in this fashion will win the player $4 if in the game the pair of dice come up 4 and 4 prior to one of the other ways of totaling 8. For a $1 bet, what is the expected result? (Answer: 80jt) 10-14 If your interest rate is 8%, what is the expected value of the present worth of the "extra" insurance payments in Problem 1O-9? (Answer: $528.7) 10-15 A decision has been made to perform certain repairs on the outlet works of a small dam. For a particular 36-inch gate valve, there are three available alternatives: (a) Leave the valve as it is. (b) Repair the valve. (c) Replace the valve. If the valve is left as it is, the probability of a failure of the valve seats, over the life of the project, is 60%; the probability of failure of the valve stem is 50%; and of failure of the valve body is 40%. If the valve is repaired, the probability of a failure of the seats, over the life of the project, is 40%; of failure of the stem is 30%; and of failure of the body is 20%. If the valve is replaced, the probability of a failure of the seats, over the life of the project, is 30%; of failure of the stem is 20%; and offailure of the body is 10%. The present worth of cost of future repairs and service disruption of a failure of the seats is $10,000; the present worth of cost of a failure of the stem.is $20,000; the present worth of cost of a failure of the body is $30,000. The cost of repairing the valve now is $10,000; and of replacing it is $20,000. If the criterion is to minimize expected costs, which 'alternative is best? 10-16 A man went to Atlantic City with $500 and placed 100 bets of $5 each, one after another, on the same number on the roulette wheel. There are 38 numbers on the wheel and the gaming casino pays 35 times the amount bet if the ball drops into the bettor's numbered slot in the roulette wheel. In addition, the bettor receives back the original $5 bet. Estimate how much money the man is expected to win or lose in Atlantic City. 10-17 A factory building is located in an area subject to occasional flooding by a nearby river. You have been brought in as a consultant to determine whether floodproofing of the building is economically justified. The alternatives are as follows: (a) Do nothing. Damage in a moderate flood is $10,000 and in a severe flood, $25,000. (b) Alter the factory building at a cost of $15,000 to withstand moderate flooding without damage and to withstand severe flooding with $10,000 damages. (c) Alter the factory building at a cost of $20,000 to withstand a severe flood without damage. In any year the probability of flooding is as follows: 0.70, no flooding of the river; 0.20, moderate flooding; and 0.10, severe flooding. If inter.est is 15% and a 15-year analysis period is used, what do you recommend? 10-18 An industrial park is being planned for a tract of land near the river. To prevent flood damage to the industrial buildings that will be built on this low-lying - .- -- -, - - - - -_. Problems land, an earthen embankment can be constructed. The height of tlIe embankment will be determined by an economic analysis of the costs and benefits. The following data have been gathered. Embankment Height Above Roadway (m) 2.0 2.5 3.0 3.5 4.0 Flood Level Above Roadway (m) 2.0 2.5 3.0 3.5 4.0 Initial Cost $100,000 165,000 300,000 400,000 550,000 Average Frequency That Flood Level Will Exceed Height in Col. 1 Once in 3 years Once in 8 years Once in 25 years Once in 50 years Once in 100 years The embankment can be expected to last 50 years and will require no maintenance. Whenever the flood water flows over the embankment, $300,000 of damage occurs. Should the embankment be built? ITso, to which of the five heights above the roadway? A 12% rate of return is required. 10-19 Should the following project be undertaken if its life is 10 years and it has no salvage value? The firm uses an interest rate of 12% to evaluate engineering projects. First Cost $300,000 400,000 600,000 P 0.2 0.5 0.3 Net Revenue $ 70,000 90,000 100,000 P 0.3 0.5 0.2 (Answer: $45,900, yes) 10-20 A robot has just been installed at a cost of $81,000. It will have no salvage value at the end of its useful life. Given the following estimates and probabilities for the yearly savings and useful life, determine the expected rate of return. Savings per year Probability $18,000 20,000 22,000 0.2 0.7 0.1 333 Useful Life (years) Probability 12 1/6 5 2/3 4 1/6 10-21 Five years ago a dam was constructed to impoun~ irrigation water and to provide flood protection for the area below the dam. Last winter a 100-year flood caused extensive damage both to the dam and to the surrounding area. This was not surprising, since the dam was designed for a 50-year flood. The cost to repair the dam now will be $250,000. Damage in the valley below amounts to $750,000. IT the spillway is redesigned at a cost of $250,000 and the dam is repaired for another $250,000, the dam may be expected to withstand a 100-year flood without sustaining damage. However, the storage capaCity of the dam will not be increased and the probability of damage to the surrounding area below the dam will be unchanged. A second dam can be constructed up the river from the existing dam for $1 million. The capacity of the second dam would be more that adequate to provide the desired flood protection. If the second dam is built, redesign of the existing dam spillway will not be necessary, but the $250,000 of repairs must be done. The development in the area below the dam is expected to be complete in 10 years. A new 100year flood in the meantime will cause a $1 million loss. After 10 years the loss would be $2 million. In addition, there would be $250,000 of spillway damage if the spillway is not redesigned. A 50-year flood is also likely to cause about $200,000 of damage, but the spillway would be adequate. Similarly, a 25-year flood would cause about $50,000 of damage. There are three alternatives: (1) repair the existing dam for $250,000 but make no other alterations, (2) repair the existing dam ($250,000) and redesign the spillway to take a 100-year flood ($250,000), and (3) repair the existing dam ($250,000) and build the second dam ($1 million). Based on an expected annual cash flow analysis, and a 7% interest rate, which alternative should be selected? Draw a decision tree to clearly describe the problem. 10-22 A new product's chief uncertainty is its annual net JI.. . revenue. So far, $35,000 has been spent on development, but an additional $30,000 is required finish development. The firm's interest rate is10%. - -- 334 - UNCERTAINTYIN FUTURE EVENTS (a) What is the expected PW for deciding whether to prQceed? (b) Find the P(loss) and the standard deviation for proceeding. P Annual savings State . Probability Net revenue Life, in years Bad OK Great 0.3 -$15,000 5 0.5 $15,000 5 0.2 $20,000 10 10-23 (a) In Problem 10-22 how much is it worth to the firm to terminate the product after 1 year if the net revenues are negative? (b) How much does the ability to terminate early change the P (loss) and the standard deviation? 10-24 Find the probability distribution and the expected PW JI. to modify an assembly line. The first cost is $80,000, and its salvage value is $0. The firm's interest rate is 9%. The savings shown in the table depend on whether the assembly line runs one, two, or three shifts, and on whether the product is made for 3 or 5 years. Useful Life Shifts! Savings! (years) Probability year Probability day 3 0.6 0.3 1 $15,000 0.4 5 2 0.5 30,000 0.2 3 45,000 10-25 In Problem 10-24, how much is it worth to the firm .- 10-28 A new machine will cost $25,000. The machine is expected to last 4 years and have no salvage value. If the interest rate is 12%, determine the return and the risk associated with the purchase. to be able to extend the product's life by 3 ye~~,.at a cost of $50,000, at the end of the product's Imtial useful life? 10-26 Al took a midterm examination in physics and received a score of 65. The mean was 60 and the standard deviation was 20. Bill received a score of 14 in mathematics, where the exam mean was 12 and the standard deviation was 4. Which student ranked higher in his class? Explain. 10-27 The Graham Telephone Company may invest in new switching equipment. There are three possible outcomes, having net present worth of $6570, $8590, and $9730. The probability of each outcome is 0.3, 0.5, and 0.2, respectively. Calculate the expected return and risk associated with this proposal. (Answer: Epw = $8212, Uf'W= $1158) 0.3 $7000 0.4 $8500 0.3 $9500 10-29 What is your risk associated with Problem 10-14? 10-30 Measure the risk for Problem 10-19 using the P(loss), JIL range PWs. of PW values, and standard deviation of the (Answer: Uf'W= $127,900) 10-31 (a) In Problem 10-24, describe the risk using the P(1oss) and standard deviation of the PWs. . (b) How much do the answers change if the possible life extension in Problem 10-25 is allowed? 10-32 An engineer decided to make a careful analysis of the cost of fire insurance for his $200,000 home. From a fire rating bureau he found the following risk of fire loss in any year. Outcome No fire loss $ 10,000 fire loss 40,00Q fire loss 200,000 fire loss Probability 0.986 0.010 0.003 0.001 (a) Compute his expected fire loss in any year. (b) He finds that the expected fire loss in any year is less than the $550 annual cost of fire insurance. In fact, an insurance agent "explains that this is always true. Nevertheless, the engineer buys fire insurance. Explain why this is or IS not a logical decision. 10-33 A firm wants to select one new research and development project. The following table summarizes six possibilities. Considering expected return and risk, which projects are good candidates? The firm believes it can earn 5% on a risk-free investment in government securities (labeled as Project F). Project 1 2 3 4 5 6 F IRR 15.8% 12.0 10.4 12.1 14.2 18.5 5.0 Standard Deviation 6.5% 4.1, 6.3 5.1 8.0 10.0 0.0 r I Problems 10-34 A firm is choosing a new product. The following table summaIjzes six new potential products. Considering expected return and risk, which products are good candidates? The firm believes it can earn 4% on a risk-free investment in government securities (labeled as Product F). Product 1 2 3 4 5 6 F IRR 10.4% 9.8 6.0 12.1 12.2 13.8 4.0 335 Standard Deviation 3.2% 2.3 1.6 3.6 8.0 6.5 0.0 I I. ! . Jl - - - .0_."'" After Completing This Chapter... The student should be able to: · Describe depreciation, deterioration, and obsolescence. Distinguishvarioustypes of depreciableproperty and differentiatebetween depreciation expenses and other business expenses. . · Use historical depreciation methods to calculate the annual depreciation charge and book value over the asset's life. · · Explain the differences between the historical depreciation methods and the modified . · · · economicanalysisproblems. · accelerated cost recovery system (MACRS). Use MACRS to calculate allowable annual depreciation charge and book value over the asset's life for various cost bases, placed in service combinations, property classes, and recovery periods. Fully account for capital gainsllosses, ordinary losses, and depreciation recapture due to the disposal of a depreciated business asset. Use the units of production and depletion depreciation methods as needed in engineering Use spreadsheetsto calculate depreciation. QUESTIONS TO CONSIDER ) 1. Whendeterminingtheir cash flow,ownersof baseballteams dedu~tstadiumdepreciation, which can amount to as much as $5 million per year. How would this affect their calculations? 2. Even though Barry Bonds is in peak.health and hit a record 73 home runs in 2002, the IRS considers him a "wasting asset" and allows the owners of his baseball club to depreciate a large portion of his contract. Explain. 1 - - - -- - ---- - - { I Depreciation Is Baseball Going Broke? In 2001, major league baseball brought in record revenues totaling over $3.5 billion. Great business, you say? Not according to Bud Selig, baseball's commissioner. In December of that year, Selig went before Congress to plead poverty, asserting that baseball had suffered a loss of over $200 million for the year. Many journalists (and even more fans) weren't buying it. A few months later, Forbes magazine published an article stating that baseball had in fact had an operating profit (that is, earnings before interest, taxes, and depreciation) of $75 million for the year. Selig angrily denounced the Forbes article as "pure fiction." In response, Forbes suggested that perhaps Selig was "afraid of being called before Congress and explaining his figures in more detail." r e L I JJ' ~.-o..:.:-"_ -"- --- -- -- -- - - --- - 338 DEPRECIATION We have so far dealt with a variety of economic analysis problems and many techniques for their solution. In the process we have avoided income taxes, which are an imponant element of many private-sector economic analyses. Now,we can move to more realistic_ and,unfonunately,morecomplex-situations. . Our governmenttaxes individualsand businessesto support its processes-lawmaking, domestic and foreign economic policymaking, even the making and issuing of money itself. The omnipresenceof taxes requires that they be included in economic analyses, which means that we must understand something about the way taxes are imposed. For capital equipment,depreciation is required to compute income taxes. Chapter 11 examines depreciation, and Chapter 12 illustrates how depreciation is used in income tax computations. The goal is to support decision making on engineering projects, not to support final tax calculations. BASIC ASPECTS OF DEPRECIATION The word depreciation is definedas a "decreasein value."This is not an entirely satisfactory definition,for value has several meanings. In the context of economic analysis, value may refer either to market value-that is, the monetary value others place on property--or value to the owner. For example, an assembly line is far more valuable to the manufacturing firm that it was designed for, than it is to a used equipment market. Thus, we now have two definitions of depreciation: a decrease in value to the market or a decrease to the . owner. Deterioration and Obsolescence A machine may depreciate because it is deteriorating or wearing out and no longer performingits function as well as when it was new.Many kinds of machineryrequire increased maintenanceas they age, reflectinga slowbut continuingfailure of individualparts. In other types of equipment, the quality of output may decline owing to wear on components and resultingpoorer mating of parts. Anyone who hasworked to maintain artauto has observed deterioration due to failure of individual parts (such as fan belts, mufflers, and batteries) and the wear on components (such as bearings, piston rings, and alternatorbrushes). Depreciationis also caused by obsolescence. A mach~neis described as obsolete when it is no longer needed or usefuL A machine that is in excellent working condition may still be obsolete.In the 1970s,mechanicalbusinesscalculatorswith hundredsof gears and levers became obsolete. The advance of integrated circuits resulted in a completely different and far superior approach to calculator design. Thus, mechanical calculators rapidly declined or depreciatedin value. If your auto depreciated in the last year, that means it has declined in market value. It has less value to potential buyers. On the other hand, a manager who indicates {;hata piece of machineryhas depreciated may be describing a machine that has deterioratedbecause of use and/orbecause it has become obsolete comparedwith newer machinery.Both situations indicate a decline in value to the owner. The accountingprofession defines depreciationin yet another way.Although in everyday conversationwe are likely to use "depreciation" to mean a decline in market value or - ..... Basic Aspects of Depreciation 339 value to the owner, accountants define depreciation as allocating an asset's cost over its useful or depreciable life. Thus, we now have three distinct definitionsof depreciation: 1. Decline in market value of an asset. 2. Decline in value of an asset to its owner. 3. Systematicallocation of an asset's cost over its depreciablelife. Depreciation and Expenses It is this third (accountant's) definition that is used to compute depreciation for business assets. Business costs are generally either expensed or depreciated. Expensed items, such as labor, utilities, materials, and insurance, are part of regular business operations and are "consumed" over short periods of time (sometimesrecurring).These costs do not lose value gradually over time. For tax purposes they are subtractedfrom businessrevenueswhen they occur. Expensed costs reduce income taxes because businesses are able to write off ~eir full amount when they occur. In contrast, business costs due to depreciated assets are not fully written off when they occur. A depreciated asset does lose value gradually and must be written off over an extended period. For instance, consider a plastic injection-molding machine used to produce the beverage cups found at sporting events. The plastic pellets melted into the cup shape lose their value as raw material directly after manufacturing. The raw material cost for production material (plastic pellets) is written off, or expensed, immediately. On the other hand, the plastic injection mold machine itself will lose value over time, and thus its cost (purchase price and installation expenses) are written off (or depreciated) over a period of years. The number of years over which the machine is depreciated is called its depreciable life or recovery period, which is often different from the asset's useful or most economic life. Depreciable life is determined by the depreciation method used to spread out the cost-depreciated assets of many types operate well beyond their depreciable life. Depreciation is a noncash cost that requires no exchange of dollars. Companies do notwritea checkto someoneto pay theirdepreciationexpenses.Rather,thesearebusiness . expenses that are allowed by the government to offset the loss in value of business assets. Remember, the company has already paid for assets up front; depreciationis simply a way to claim these "business expenses" over time. Depreciation is important to the engineering economiston an after-tax basis because, even though it is a noncash cost, it does change the cash flows due to taxes. Depreciation deductions reduce the taxable income of husine"sses and thus reduce the amount of taxes paid. In general business assets can be depreciated only if they meet the following basic requirements: 1. The property must be used for business purposes to produce income. 2. The property must have a useful life that can be determined, and this life must be longer than one year. 3. The propertymustbe an assetthatdecays, getsused up, wears out,becomes obsolete, or loses value to the owner from natural causes. - -- -r - II '-'. 340 .......... DEPRECIATION Consider the costs that areincurred by a local pizza business.Identify each cost as either expensed or depreciated and describe why. ·· Costto pay wagesforjanitor ·· Cost for pizza dough and toppings Cost of a new baking oven Cost of new delivery van Cost of furnishings in dining room · · Utilitycostsfor sodarefrigerator ..SOLUTION. Cost Item Pizza dough and toppings New delivery van Wages for janitor Furnishings in dining room New baking oven Utilities for soda refrigerator Type of Cost Expensed Depreciated Expensed Depreciated Depreciated Expensed Why Life < t year; lose value immediately Meets 3 requirements for depreciation . Life < 1 year; lose value immediately Meet 3 requirements for depreciation Meets 3 requirements for depreciation Life < 1 year; lose value immediately _ ~.."~~.r.. - J Types of Property The rules for depreciation are linked to the classification of business property as either tangible or intangible. Tangibleproperty is further classified as either real or personal. Tangible property can be seen, touched, and felt. Real property includes land, buildings, and all things growing on, built upon, constructed on, or attached to the land. Personal property includes equipment, furnishings, vehicles, office machinery, and anything that is tangible excluding assets defined as realproperty. Intangible property is all property that has value to the owned but cannot be directly seen or touched. Examples include patents, copyrights, trademarks, trade names, and franchises. Many different types of property that wear out, decay, or lose value can be depreciated as business assets. This wide range includes copy machines, helicopters,buildings, interior furnishings, production equipment, and computer networks. Almost all tangible property can be depreciated. One important and notable exception is land, which is never depreciated..Land does not wear out, lose value, or have a determinable useful life and thus does not qualify as a depreciable property. Consider the aspect of loss in value. Rather than decreasing in value, most land becomes more valuable as time passes. In addition to the land itself, expenses for clearing, grading, preparing, planting, and landscaping are not generally depreciated because they have no fixed useful life. Other tangible property that cannot be depreciatet includes factory inventory, containers considered as inventory, equipment used to bui!.: -- - ---- - --- --- --- r Basic Aspects of Depreciation 341 capital improvements, and leased property. The leased property exception highlights the fact that only the owner of property may claim depreciation expenses. Tangible properties used in both business and personal activities, such as a vehicle used in a consulting engineering firm that is also used to take one's kids to school, can be depreciated.However,in suchcases one can take depreciationdeductionsonly in proportion to the use for businesspurposes. Accurate records indicating the portion of use for business and personal activities are required. Depreciation Calculation Fundamentals Tounderstandthe complexitiesof depreciation,the first step is to examine the fundamentals of depreciation calculations. Figure 11-1 illustrates the general depreciation problem of allocating the total depreciation charges over the asset's life. The vertical axis is labeled book value, and the curve of asset cost or basis minus depreciationcharges made starts at the first cost and declines to the salvage value, Book value = Asset cost - Depreciation charges made to date Looked at another way, book value is the asset's remaining unallocated cost. In Figure 11-1,book value goes from a value of B at time zero in the recovery period to a value of S at the end of Year5. Thus, book value is a dynamic variable that changes over an asset's recoveryperiod. The equationused to calculate an asset's book value overtime is: t BV, L = Cost basis - (11-1) dj j=l where = book value of the depreciated asset at the end of time t Cost basis = B = dollar amount that is being depreciated; this includes the asset's B \l, purchase price as well as any other costs necessary to make the asset "ready for use" t . L j=l dj = sum of depreciation deductions taken from time 0 to time t, where dj is the depreciation deduction in Year j FIGURE 11-1 General depreciation. Cost B ~" , T TotalDepreciation ", ,, , Charges " ", CurveValuesDependon ,DepreciationMethod ,, ,, ,, '......... S ~ Salvage Value I o I - - - - 1 234 UsefulLife (years) -- 5 1 - --.. . 342 -._- DEPRECIATION r Equation 11-1 shows that year-to-year depreciation charges reduce an asset's book value over its life. The following section describes methods that are or have been allowed under federal tax law for quantifying these yearly depreciation deductions. HISTORICAL DEPRECIATION METHODS Allowing businesses to deduct the cost for capital expenditures over time (that i$, to depreciate business assets) has long been part of the tax code. However, over time several versions of various depreciation methods have been used to calculate these deductions.In general, accounting depreciation methods can be categorized as follows. Pre-1981 historical methods: These methods include the straight-line, sum-oj-theyears' -digits, and declining balance methods. Each method required estimates.of an asset's useful life and salvagevalue. Firms could elect to use any of these methods for assets, and thus there was little uniformity in how depreciation expenses were reported. 1981-1986 method: With the Economic RecoveryTaxAct (ERTA)of 1981,Congress created the accelerated cost recovery system (ACRS). The ACRS method had three key features: (1) property class lives were created and all depreciated assets assigned to one particular category, (2) the need to estimate salvage values was eliminated because all assets werefully depreciated over their recovery period, and (3) shorter recovery periods were used to calculate annual depreciation,which accelerated the write-off of capital costs more quickly than did the historical methods-thus the name. 1986 to present: The modified accelerated cost recovery system (MACRS) has been in effect since the Tax Reform Act of 1986 (TRA-86). The MACRS method is similar to the ACRS system except that (1) the number of property classes was expanded and (2) the annual depreciationpercentages were modified to include a half-year conventionfor the first and final years. In this chapter, our primary focus is to describe the MACRS depreciation method. However,it is useful to first describe three historical depreciationmethods. These methods are used in some countries, and MACRSis based on two of them. Straight-Line Depreciation The simplest and best known depreciationmethod is straight-line depreciation.To calculate the constant annual depreciation charge, the total'amount to be depreciated, B - S, is divided by the depreciable life, in years, N: 1 Annual depreciation charge 1N B-S N = dt = - is used for the depreciation period because it may be shorter than n, the horizon. -- - ---- (11-2) Historical Depreciation Methods 343 I:"~.:~ l.\~\ '~- ~.~ Consider the following: $900 5 $70 Cost of the asset, B Depreciable life, in years, N Salvage value, S Compute the straight-line depreciation schedule. Annual depreciation charge B- S N = dt = - Year, Depreciation for Year t, Sum of Depreciation Charges Up to Year t, t dt 1 $166 Ldj j=l $166 2 3 4 5 166 166 166 166 332 498 664 830 = 900 - 70 _ = $166 Book Value at the End of Year t, t t BVt=B- Ldj j=l 900 - 166 = 734 - 900 900 900 332 498 664 900 - 830 = 568 = 402 = 236 = 70 = S This situation is illustrated in Figure 11-2.Notice the constant $166 dt each year for 5 years, and that the asset has been depreciated down to a book value of $70, which was the estimated salvage value. FIGURE 11-2 Straight-line depreciation. $900 Straight-Line Annual Depreciation Charge _R . 166 J 734 u ::s 0 0 j:CI 568 402 236 166 70 I 0 I 'S I - 1 1 2 3 4 Time,DepreciableLife = N (years) .......... 5 -~! 1 ,.. 344 DEPRECIATION The straight-line (SL) method is often used for intangible property. Veronica's finn bought a patent in April that was not acquired as part of acquiring a business. She paid $6800 for this patent and must use the straight-line method to depreciate it over 17 years with no salvage value. The annual depreciation is $400 (=$6800/17). Since the patent Was purchased in April, the deduction must be prorated over the 9 months of ownership. This year the deduction is $300 (=$400 x 9/12), and then next year she can begin taking the full $400 per year. Sum-of-Years'-Digits Depreciation Another method for allocating an asset's cost minus salvage value over its depreciable life is called sum-of-years'-digits (SOYD) depreciation. This method results in larger-thanstraight-line depreciation charges during an asset's early years and smaller charges as the asset nears the end of its depreciablelife. Each year, thedepreciationchargeequals a fraction of the total amount to be depreciated (B - S). The denominator of the fraction is the sumof the years' digits. For example, if the depreciable life is 5 years, 1+ 2 + 3 + 4 + 5 = 15 = SOYD. Then 5/15,4/15,3/15,2/15, and 1/15 are the fractions from Year 1 to Year 5. Each year the depreciation charge shrinks by 1/15 of B - S. Because this change is the same every year, SOYD depreciation can be modeled as an arithmetic gradient, G. The equations can also be written as: Remaining depreciable life Sum-of-years' -digits ( depreciation charge for any year ) dt where dt = ( ( -_ N ) ) at beginning of year Sum of years' digits for total depreciable life (Total amount depreciated) - t + 1(B _ S) (11-3) ~~~~ = depreciationcharge in any year t N = number of years in depreciable life SOYD = sum of years' digits, calculated as N(N + 1)/2 B = cost of the asset made ready for use S = estimated salvage value after depreciable life = SOYD Compute the SOYD depreciation schedule for the situation in Example 11.2. Cost of the asset, B Depreciable life, in years, N Salvage value, S $900 5 $70 ..-- SOlUTIQ,~; 'sOYD _ ~5 >&6 ~=15 -- - - -- - Historical Depreciation Methods 345 Thus, dl = 5 - 1 + 1(900 _ 70) = 277 15 d2 = . 5-2+1 15 (900 - 70) = 221 d3 = 5 - 3 + 1(900 _ 70) = 166 15 d4 = ds= Year, Depreciation for Year t, t dt 5-4+1 (900-70) = 111 15 5-5+1 ._ (900 - 70) = 55 Book Value at End of Year t, Sum of Depreciation Charges Up to Year t, t t BVt=B-"Ldj "Ldj j=l 1 2 3 4 5 j=l 900 - 277 - 623 $277 498 664 775 830 $277 221 166 111 55 = 402 900 - 498 900 - 664 900 - 775 900 - 830 = 236 = 125 = 70 = S These data are plotted in Figure 11-3. FIGURE 11-3 Sum~of-years' -digits depreciation. SOYD Depreciation Charges $900 .B 277 623 0,) ::I «i > 0 402 0 j:Q 236 = = L125 70 := . S I 0 I I I - - -- -- I 1t I 1 234 Time,DepreciableLife = N (years) 5 ~.,..,-""--- --- I ~ '. -- ..,.~ .--~ - J - - --~-- - -- ~.- '" 346 - ~- - DEPRECIATION Declining Balance Depreciation Declining balance depreciation applies a constantdepreciation rate to the property's declining book value. Tworates were commonlyusedbefore the 1981and 1986tax revisions, and they areused to computeMACRSdepreciationpercentages.These are 150and 200%of the straight-linerate. Since200% is twice the straight-linerate, it is calleddouble declining balance, or DDB; the general equation is 2 Double declining balance dt = -(Book valuet_I) N (11-4a) Since book value equals cost minus depreciationcharges to date, 2 .. DDB dt = -(Cost - DeprecIationcharges to date) N or (ll-4b) Compute the DDB depreciation schedule for the situationsin Examples 11-2 and 11-3. Cost of the asset, B Depreciable life, in years, N Salvage value, S $900 5 $70 SOLUTION Year, Depreciation for Year t Using Equation 11-4a, Sum of Depreciation Charges Up to Year t, dt Ldj j=I $360 576 706 784 830 t 1 2 3 4 5 t (2/5)900 = 360 (2/5)540 (2/5)324 = 216 = 130 ~7/5l194= 78 (2/5)116= 46 = Book Value at End of Year t BVt=B- - Ldj j=I 900 - 360 - = 540 900 - 576 - 324 900- 706 - 194 900 -784= 116 ::i: Ii == 900 - 830 - 70 - S ~Fjgure 11-;1,illl!strates tl;1p~itufltiO!h -....... t, 'iiiiiii' --- ---- i' - --- - - -- -- - - Modified Accelerated Cost Recovery System (MACRS) FIGURE 1\-4 Decliningbalance depreciation. 347 Declining Balance Depreciation Charges $900 .B 360 I Q) 540 r ::s C; > 0 j:Q 334 194 116 70, 0 I I I I 1 2 3 4 S I 5 Time, Depreciable Life = N (years) The final salvage value of $70 for Examples 11-2, 11-3,and 11-4was chosen to match the ending value for the double-declining balance method. This does not normally happen. If the final salvage value of Example 11-4 had not been $70, the double declining balance method wouldhave had to be modified.One modificationstops further depreciationonce the book value comes to equal the salvage value-this prevents taking too much depreciation. The other modificationwould switch from declining balance depreciationto straightlinethis ensures taking enough depreciation. These modifications are not detailed here because (1) MACRS has been the legally appropriate system since 1986 and (2) as will be shown, MACRS incorporates the shift from declining balance to straight-line depreciation. MODIFIED ACCELERATED COST RECOVERY SYSTEM (MACRS) The modified accelerated cost recovery system (MACRS)depreciation method, introduced by the Tax Reform Act of 1986,has been continued with the Taxpayer Relief Act of 1997. Three major advantages of MACRS are that (1) the "property class lives" are less than the "actual useful lives," (2) salvage values are assumed to be zero, and (3) tables of annual percentages simplify computations. The definition of the MACRS classes of depreciable property is based on work by the U.S. Treasury Department. In 1971 the Treasury published guidelines for about 100 broad asset classes. For each class there was a lower limit, a midpoint, and an upper limit of useful life, called the asset depreciation range (ADR). The ADR midpoint lives were somewhat shorter than the actual average useful lives. These guidelines have been incorporated into MACRS so that the property class lives are again shorter than the ADR midpoint lives. Use of MACRS focuses on the general depreciation system (GDS), which is based on declining balance with switch to straight-line depreciation. The alternative depreciation system (ADS) provides for a longer period of recovery and uses straight-line - 348 .- -- -."-"."... - DEPRECIATION depreciation. Thus it is muchless economicallyattractive.Underlaw,ADSmustbe used for (1) any tangible property used primarily outside the United States, (2) any propertythat is tax exempt or financed by tax-exempt bonds, and (3) farming property placed in service when uniform capitalization rules are not applied. The ADS may also be elected for property that can be depreciated using the GDS system. However, once ADS has been elected for an asset, it is not possible to switch back to the GDS system. Because the ADS makes the depreciation deductions less valuable, unless ADS is specifically mentioned, subsequent discussion assumes the GDS system when reference is made to MACRS. Once a property has been determined to be eligible for depreciation,the next step is to calculate its depreciation deductions over its life. The following information is required to calculate these deductions: ·· · The cost basis of the property. The property class and recoveryperiod of the asset. The asset's placed-in-service date. Cost Basis and Placed-in-Service Date The cost basis, B, is the cost to obtain and place the asset in service fit for use. However, for real property the basis may also include certain fees and charges that the buyer pays as part of the purchase. Examples of such fees include legal and recording fees, abstract fees, surveycharges, transfer taxes, title insurance, and amountsthat the seller owes that you pay (back taxes, interest, sales commissions, etc.). Depreciationfor a businessassetbeginswhenthe assetisplaced in servicefor abusiness purpose. If an asset is purchased and used in a personal context, depreciation may not be taken. If that asset is later used in business for income-producingactivity,depreciationmay begin with the change in usage. Property Class and Recovery Period Each depreciated asset is placed in a MACRS property class, which defines the recovery period and the depreciation percentage for each year. Historically the IRS.assigned each type of depreciable asset a class life or an asset depreciation range. With MACRS, asset class lives have been pooled together in theproperty classes. Table 11-1lists the class lives and GDS and ADS property classes for severalexample depreciable assets. Table 11-2lists the MACRS GDS property classes. The MACRS GDS property classes are described in more detail in Table 11-2. The proper MACRS property class.can be found several different ways. In the list below, the first approach that works should be used. 1. 2. 3. 4. 5. Property class given in problem. Asset is named in Table 11-2. IRS tables or Table 11-1. Class life. Seven-yearproperty for "all other property not assigned to another class." --- - r 349 Modified Accelerated Cost Recovery System (MACRS) TABLE 11-1 Example Class Lives and MACRS Property Classes IRSAsset Class Asset Description 00.11 00.12 00.22 00.241 00.25 00.40 01.11 01.21 13.00 13.30 15.00 20.10 20.20 24.10 32.20 20.1 48.10 48.2 49.12 49.13 49.23 50.00 80.00 Class Life (years) ADR Office furniture, fixtures, and equipment Information systems: computers/peripheral Automobiles, taxis Light general-purpose trucks Railroad cars and locomotives Industrial steam and electric distribution Cotton gin assets Cattle, breeding or dairy Offshore drilling assets Petroleum refining assets Construction assets Manufacture of grain and grain mill products Manufacture of yarn, thread, and woven fabric Cutting of timber Manufacture of cement Manufacture of motor vehicles Telephone distribution plant Radio and television broadcasting equipment Electric utility nuclear production plant Electric utility steam production plant Natural gas production plant Municipal wastewater treatment plant Theme and amusement park assets 10 6 3 4 15 22 10 7 7.5 16 6 17 11 6 20 12 24 6 20 28 14 24 12.5 MACRSProperty Class (years) GDS ADS 7 5 5 5 7 I 15 7 5 5 10 5 10 7 10 6 6 6 15 22 10 7 7.5 16 6 17 11 6 20 12 24 6 20 28 14 24 12.5 5 I 15 7 15 I I 5 15 120 7 I 15 7 Once the MACRS property class is known, as well as the placed-in-service date and cost basis, the year-to-year depreciation deductions can be calculated for GDS assets over their depreciable life using dt where dt B rt = B x rt (11-5) = depreciationdeduction in year t = cost basis being depreciated = appropriateMACRS percentage rate Percentage Tables ~ The IRS has prepared tables to assist in calculating depreciation charges when MACRS GDS depreciation is used. Table 11-3 gives the yearly depreciation percentages (rt) that fife used for the six personal property classes (3-, 5-, 7-, 10-, 15-,and 20-year property classes), and Table 11-4givesthe percentages for nonresidentialreal property.Notice that the values are given in percentages-thus, for example, the value of 33.33% (given in Table 11-3for Ii : ~: I~ J Year 1 for a 3-year MACRS GDS property) is 0.3333. -- ~ - -. 350 -._- - DEPRECIATION TABLE 11-2 MACRSCDS Property Classes Property Class Personal Property (all property except real estate) 3-year property Special handling devices for food and beverage manufacture Special tools for the manufacture of finished plastic products, fabricated metal products, and motor vehicles Property with ADR class life of 4 years or less 5-year property Automobiles* and trucks Aircraft (of non-air-transport companies) Equipment used in research and experimentation Computers Petroleum drilling equipment Property with ADR class life of more than 4 years and less than 10 years 7-year property All other property not assigned to another class Office furniture, fixtures, and equipment Property with ADR class life of 10 years or more and less than 16 ye3!s 10-year property Assets used in petroleum refining and certain food products Vessels and water transportation equipment Property with ADR class life of 16 years or more and less than 20 years IS-year property Telephone distribution plants Municipal sewage treatment plants Property with ADR class life of 20 years or more and less than 25 years 20-yearproperty Municipal sewers Property with ADR class life of 25 years or more Property Class Real Property (real estate) 27.5 years Residential rental property (does not include hotels and motels) 39 years Nonresidential real property *The depreciation deduction for automobiles is limited to $7660 (maximum) the first tax year, $4900 the second year, $2950 the third year, and $1775 per year in subsequent years. Source: U.S. Department of the Treasury, Internal Revenue Service Publication 946, How to Depreciate Property. Washington, DC: U.S. Government Printing Office. . Notice in Table 11-3 that the depreciation percentages continue for one year beyond the property class life. For example, a MACRS lO-year property has an rt value of 3.28% in Year 11. This is due to the half-year convention that also halves the percentage for the first year. The half-year convention assumes that all assets are placed in service at the midpoint of the first year. . Another characteristic of the MACRS percentage tables is that the rt values in any column sum to 100%. This means that assets depreciated using MACRS are fully depreciated at the end of the recovery period. This assumes a salvage value of zero. This is --- -- Modified Accelerated Cost Recovery System (MACRS) 351 for Personal Property: Half-Year Convention TABLE11-3 MACRS Depreciation ApplicablePercentagefor PropertyClass Recovery Year 3-Year Property 5-Year Property 7-Year Property 1 2 3 33.33 44.45 14.81* 20.00 32.00 19.20 14.29 24.49 17.49 10.00 18.00 14.40 5.00 9.50 8.55 3.750 7.219 6.677 4 5 6 7.41 11.52* 11.52 5.76 12.49 8.93* 8.92 11.52 9.22 7.37 7.70 6.93 6.23 6.177 5.713 5.285 6.55* 6.55 6.56 5.90* 5.90 5.91 4.888 4.522 4.462* 6.55 3.28 5.90 5.91 5.90 4.461 4.462 4.461 13 14 15 5.91 5.90 5.91 4.462 4.461 4.462 16 17 18 2.95 4.461 4.462 4.461 7 8 9 8.93 4.46 10 11 12 10-Year Property 15-Year Property 19 20 21 20-Year Property 4.462 4.461 2.231 Computation method · ·· · The 3-, 5-, 7-, and 1O-year classes use 200% and the 15- and 20-year classes use 150% declining balance depreciation. All classes convert to straight-line depreciation in the optimal year, shown with asterisk (*). A half-year of depreciation is allowed in the first and last recovery years. If more than 40% of the year's MACRS property is placed in service in the last 3 months, then a mid quarter convention must be used with depreciation tables that are not shown here. a departure from the pre-1981 historical methods, where an estimated salvage value was considered. Where MACRSPercentage Rates (rt) Come From This section describes the connection between historical depreciation methods and the MACRS percentages that are shown in Table 11-3. Before ACRS and MACRS, the most common depreciation method was declining balance with a switch to straight line. That combined method is used for MACRSwith three further assumptions. --- --------- -- - - - -- - -- 352 - DEPRECIATION TABLE 11-4 MACRS Depreciation Recovery Year 1 2-39 40 for Real Property (real estate)* Recovery Percentages for Nonresidential Real Property (month placed in service) 1 2 3 4 5 6 7 8 9 10 11 12 2.461 2.247 2.033 1.819 1.605 1.391 1.177 0.963 0.749 0.535 0.321 0.107 2.564 2.564 2.564 2.564 2.564 2.564 2.564 2.564 2.564 2:564 2.564' 2.564 0.107 0.321 0.535 0.749 0.963 1.177 1.391 1.605 1.819 2.033 2.247 2.461 *The useful life is 39 years for nonresidential real property. Depreciation is straight line using the midmonth convention. Thus a property placed in service in January would be allowed 11'l2 months depreciation for recovery Year 1. 1. Salvage values are assumed to be zero for all assets. 2. The first and last years of the recovery period are each assumed to be half-year. 3. The declining balance rate is 200% for 3-,5-, 7-, and 1O-yearproperty and 150% for 15- and 20-year property. As shown in Example 11-5, the MACRS percentage rates can be derived from these rules and the declining balance and straight-line methods. However,it is obviously much easier to simply use the rt values from Tables 11-3 and 11-4. Consider a 5-year MACRSproperty asset with an installed and "made ready for use" cost basis of $100. Developthe MACRSpercentage rates (rt) for the asset based on the underlyingdepreciation methods. SOLUTION To develop the 5-year MACRS property percentage rates, we use the 200% declining balance method, switching over to straight line at the optimal point. Since the assumed salvage value is zero, the entire cost basis of $100 is depreciated. Also the $100 basis mimics the 100% that is used in Table 11-3. Let's explain the accompanyingtable year by year. In Year 1 the basis is $100 - 0, aQdthe dt values are halved for the initial half-year assumption. Double declining balance has the larger value, so it is chosen. Since N = 5, the rest of the decliningbalancecomputationsare simply 200%/5 times the basis minus the cumulative depreciation. In Year2 there are 4.5 years remainingfor straightline, so 4.5 is the denominator for dividing the remaining $80 in book value. Similarly in Year 3 there are 3.5 years remaining. In Year"4 the two calculations happen to be identical, so the switc"hfrom DDB to SL can be dOllein either Year 4 or Year 5. Once we know that the SL depreciation is 11.52 at the switch point, then the ,. -:' ~ ." _ _ ',- ,-.,.- smly furt!ler cal,cuiatioll is to halve that for the lastJetg". -- - - - ..- -- ...----- - -- - - -- - - ---------- - liE'" ~_. ,r; "'''' ~ . .-- '..,----- Modified Accelerated Cost Recovery System (MACRS) 353 Notice,that the DDB calculations get smaller every year, so that at some point the straight-line calculations lead to faster depreciation. This point is the optimal switch point, and it is built into Table 11-3 for MACRS. MACRS Tt Year 1 2 3 4 5 6 DDB Calculation 1/2(2/5)(100 - 0) =20.00 (2/5)(100 - 20.00) (2/5)(100 SL Calculation (%) Rates 20.00 (DDB) 1/2(100 - 0)/5 = 10.00 (100 - 20)/4.5 17.78 32.00 (DDB) (100 - 52)/3.5 = 13.71 19.20 (DDB) (100 - 71.20) /2.5 = 11.52 11.52 (either) 11.52 11.52 (SL) (112)(11.52) = 5.76 5.76 (SL) = = 32.00 - 52.00)= 19.20 (2/5)(100 - 71.20) = 11.52 Cumulative Depreciation (%) 20.00 52.00 71.20 82.72 94.24 100.00 . The values given in this example match the rt percentage rates given in Table 11-3 for a 5-year MACRS property. -'-"~-',---~-.~ MACRS Method Examples Remember the key questions in using MACRS: (1) What type of asset do you have, and does it qualify as depreciable property? (2) What amount are you depreciating [costbasis]? and (3) When are you placing the asset in service? Let's look at several examples of using MACRS to calculate both depreciation deductions and book values. Use the MACRS GDS method to calculate the yearly depreciation allowances and book values for a firm that has purchased $150,000 worth of office equipment that qualifies as depreciable property. This office furiritureis estimated to have.a salvage (market) value of $30,000 (20% of the original cost) after the end of its depreciable life. i . SqLUTION 1. 2. 3. 4. 5. The assets qualify as depreciable property. The cost basis is given as $150,000. The assets are being placed in service in Year 1 of our analysis. MACRS GDS applies. The salvage value is not used with MACRS to calculate depreciation or book value. . _ " .-;< _ _,_ _ .J Office equipmetItis listed in Table 11-2 as a 7-year property.We now use the MACRS <1bS- - ~-I 7-year property percentages from Tab ation allowances. We use Equation 11-1 to calculate the book value of the asset. . le 11-3 and Eq. 11_5to calculate I - ..-- .. -. - .. ...- po the year-to-year -- <!erred~ . J -- - --- ---- - _.... ....... --- 1I 354 DEPRECIATION Year, t 1 2 3 4 5 6 7 8 MACRS, Tt 14.29% x 24.49 17.49 12.49 8.93 8.92 8.93 4.46 Cost Basis Cumulative dt dt $150,000 $ 21,435 $ 21,435 150,000 58,170 36,735 150,000 84,405 26,235 150,000 18,735 103,140 150,000 13,395 116,535 129,915 150,000 13,380 150,000 13,395 143,310 150,000 6,690 150,000 100.00% B Vt = B - Cum. dt $128,565 91,830 65,595 46,860 33,465 20,085 6,690 0 $150,000 Notice in this example several aspects of the MACRS depreciation method: (1) the sum of the rt values is 100.00%, (2) this 7-year MACRS GDS property is depreciated over 8 years (-- property class life + 1), and (3) the book value after 8 years is $0. Investors in the JMJ Group purchased a hotel resort in April. The group paid $2.0 million for the hotel resort and $500,000 for the grounds surrounding the resort. The group sold the resort 5 years later in August. Calculate the depreciation deductions for Years 1 through 6. What was the book value at the time the resort was sold. Hotels are nonresidential real property and are depreciated over a 39-year life. Table 11-4 lists the percentages for each year. In this case the cost basis is $2.0 million, and the $500,000 paid for the land is not depreciated. JMJ's depreciation is calculated as fo!1ows: Year 1 (obtained in April) Year 2 Year 3 Year4 Year5 Year 6 (disposed of in August) d1 = 2,000,000(1.819%)= $36,380 dz = d3 = d4 = ds = d6 = 2,000,000(2.564%) 2,000,000(2.564%) 2,000,000(2.564%) 2,000,000(2.564%) 2,000,000(1.605%) Thus the hotel's book value when it was sold was: BV6 -- B - (d1 + dz + d3 + d4 + ds + d6) .~2,OOO,000:£..:: (273,600) ~ $1,'726,400 The value of the land has not challged in terms of book value. --- - -- -- = = = = = 51,280 51,280 51,280 51,280 32,100 ,"I Modified Accelerated Cost Recovery System (MACRS) 355 Comparing MACRSand Historical Methods In Examples 11-2 through 11-4 we used the straight-line, sum-of-the-years'-digits,and declining balance depreciation methods to illustrate how the book value of an assetthat cost $900 and had a salvage value of $70 changed over its 5-year depreciationlife.Figures 11-2through 11-4provided a graphicalviewof book valueoverthe 5-yeardepreciation period using these methods.Example 11-8comparestheMACRSODSdepreciationmethod directly against the historical methods. Consider the furniture that was purchased in Example 11-6. Calculate the asset's depreciation deductions and book values over its depreciable life for MACRS and the historical methods. SOLUTION Table 11-5andFigure 11-5compareMACRSandhistoricaldepreciationmethods.MACRSdepreciation is the most acceleratedor fastestdepreciationmethod-remember its nameis the modified accelerated cost recovery system. The book value drops fastest and furthest with MACRS,thus the present worth is the largest for the MACRS depreciation deductions. Depreciation deductions benefit a firm after taxes because they reduce taxable income and taxes.The time value of moneyensures that it is better to takethese deductions as soon aspossible. In general, MACRS, which allocates larger deductions earlier in the depreciation life, provides more economic benefits than historical methods. I TABLE 11-5 Comparison of MACRSand Historical Methods for Asset I in Example "-6 I MACRS Straight line Year, dt Declining Digits BYt dt BVt 12,000 12,000 12,000 138,000 126,000 114,000 30,000 24,000 19,200 120,000 96,000 76,800 21,818 19,636 17,455 128,182 108,545 91,091 12,000 102,000 12,000 90,000 12,000 78,000 15,360 12,288 9,830 61,440 49,152 39,322 15,273 13,091 10,909 75,818 62,727 51,818 7,864 1,457 0 31,457 30,000 30,000 8,727 6,545 4,364 43,091 36,545 32,182 30,OOO 128,565 91,830 65,595 4 5 6 18,735 13,395 13,380 46,860 33,465 20,085 7 8 9 10 13,395 6,690 0 0 6,690 0 0 0 12,000 12,000 12,000 =1000 $108,217 .-.....--..- - - .. 66,000 54,000 42,000 ",30p =: ;;;:;; $73,734 - $89,918 . -. -- <1 = o:3q,gOp = ii:;;;'" -- -- -- .---.. ------- dt 4" 182 $84,118 -- '" BYt dt 21,435 36,735 26,235 ........--- Sum-of- Years' BYt 1 2 3 PW (10% ) Double -- I I l III I :ili 1 356 DEPRECIATION $150,000 $125,000 «) $100,000 :::I ~ ~ o o $75,000 '..'. ...... .... I:Q $50,000 $25,000 o '."'" 2 .., 4 , ". .,." 6 ,... 8 10 Year, t FIGURE 11-5 Comparing MACRS ( ) and historical depreciation methods: double declining balance ( 4. ), sum of the years' digits (_ _ _ _ _), and straight line ( ; ) DEPRECIAliON AND ASSET DISPOSAL When a depreciated asset is disposed of, the key question is, Which is larger, the asset's book value, BV (what we show in our accounting records after applying the rules set by the government)or the asset's market value, MV (what a willing buyer pays)? If the bookvalue is lower than the market value, then too much depreciationhas been deducted from taxable income. On the other hand, if the book value is higher than the market value, there is a loss on the disposal. In either case, the current level of taxes owed changes. Depreciation recapture(ordinarygains): Depreciationrecapture, also called ordinary gains, is necessarywhen an assetis soldfor morethan anasset's currentbook value. If more than the original cost basis is received, only the amount up to the original cost basis is recaptured depreciation. Ordinary gains are also called depreciation recapturebecause the amount of gain represents the overexpense in depreciation that has been claimed.In other words, we've taken too muchexpensefor the asset's "loss in value."Since MACRSassumes S = 0 for its annual calculations,MACRS often has recaptured depreciation at disposal. Losses: A loss occurs when less than book value is received for a depreciated asset. In the accounting records we've disposed of an asset for a dollar amount less than its book value, which is a loss. In this case a company has not claimed enough depreciation expense. Capital gains: Capital gains occur when more than the asset's original cost basis is received for it. The excess over the original cost basis is the capital gain. As described in Chapter 12, the tax rate on such gains is sometimes lower than the rate on ordinary income, but this depends on how long the investment has been held ("short," ::: 18months; "long," ~ 18 months). In most engineering economic - - - --- -- - - - -- - - Depreciation and Asset Disposal 357 analyses capital gains are very uncommonbecause business andproduction equipment and facilitiesalmost alwayslose value overtime.Capitalgains aremuch more likely to occur for nondepreciated assets like stocks, bonds, real estate, jewelary, art, and collectibles. The relationship between depreciation recapture, loss, and capital gain is illustrated in Figure 11-6. $12,000 $10,000 $12,000 $10,000 $8,000 $10,000 - - - - -$5000-1 . !; $4,000 .. -- :ciatiOn Recapture $6,000 $5000 $4,000 $2,000 $0 ' $8,000 $7000 $6,000 I $10 000 $2000 $2,000 Cost Basis Market Value $0 Book Value I $3000 Loss . Cost Basis Market Value Book Value (b) Loss (a) Depreciation Recapture (ordinary gain) $16,000 $14,000 $14,000 $12,000 $10,000 $10,000 $4000 Capital $8,000 Gain $5000 Depreciation Recapture $6,000 $5000 $4,000 $2,000 $0 Cost Basis Market Value Book Value (c) Capital Gain and Depreciation Recapture FIGURE 11-6 Recaptured depreciation, loss on sale, and capital gain. Consider an asset with a cost basis of $10,000 that has been depreciated using the MACRS method~ This asset is a 3-year MACRS property. What is the gain or loss if the asset is disposed of after 5 years of operation for (a) $7000, (b) $0, and (c) a cost of $2000. ::;; Tq find gain or loss at disposal we compare market and book value. Since MACRS depreciates toa salvage value of 0, and 5 years is greater than the recovery period, the book value equals $0. 11:I I!!I...~iICIi:I L - - -- - - - --- ~ 358 ..--..---- - DEPRECIATION (a) Recaptured depreciation = $7000, since the book value is $7000 higher. (b) Since market value equals salvage value, there is no recaptured depreciation or loss. (c) Since the money is paid for disposal, this is less than the book value, and there is a loss. This generalmethodfor calculatingrecaptureddepreciationor loss appliesto all ofthedepreciation methods described in this chapter. If the asset is in the middle of its depreciable life, then recaptured depreciation and losses are calculated in a similar manner--compare the market and book values at the time of disposal. However,in computingthe book value with MACRSdepreciationa specialrule must be applied for assets disposed of before the end of the recovery period. The rule is to take one half of the allowable depreciation deductionfor that year. This rule assumes th&t disposals take place on averagehalfway through the year. Thus for a 5-year asset disposed of in the middle of Year4, the rate allowed for MACRS depreciation is half of 11.52%or 5.76%. If the asset is disposedof in Year6, it is alreadypast the recovery period, and a halfyear assumptionhas alreadybeen built into the MACRSschedule.Thus, the full r6is taken. Example 11-10illustrates severalcases of disposal before the asset is fully depreciated. Consider the $10,000 asset in Example 11-9.Do the following: 1. Calculate the effect of disposal if this asset is sold during Year2 for $2500. 2. Calculate the effect of disposal if the asset is sold during Year3 for $2500. 3. Calculate the effect of disposal if the asset is sold after Year 3 for $4000 if straight-line depreciationis used over the asset's 5-year life, and a salvage value of $5000 is assumed. SOLUTION 1. Marketvalu~ = $2500 Bookvaluez= 10,000- 1O,009[rl + rzl2] = 10,000- 10,000[0.3333+ 0.4445/2] = $4444.5 Loss = $1944.5 (= 4444.5 - 2500) (Loss, since BV greater than MV) (Note: If the full rather than half deduction were taken in Year 2, then the book value would be $2222.50 less and the loss would become a gain of $278. Since depreciation would increase by that $2222.50, the total of (depreciation) + (loss or gain) would be $4167 in either calculation.) 2. Market value3 = $2500 BoolC'value~~, 10,000 10,000[rp.-/-:= rz *' r3/2] = 10,000 -10,000[0.3333 + 0.4445 + 0.1481/2] Recapturea[fepi'eaatimf --- - 2500~ 1481.50;-1$1018.50 -- ...- -- -- - = $1481.50 (Recaptured, since.MV > BY) Unit-oF-Production Depreciation 3. Straight-linerate = (B - S)/ N Market value4 = (10,000 - = $1000 = $4000 Book value4 = 10,000 - (4)(1000) Loss 5000)/5 359 = 6000 - 4000 = $6000 = $2000 If the asset were disposed of during the year rather than at year's end, then the straightline depreciation deduction would have to be prorated for the number of months during the year that the asset was in service. There is no half-year conventionwith the historical depreciation methods. For example, if disposal occurred on September 30 of Year 4, d4 = (9/12)(1000) = $750. (Note: In Case 1 it was shown that the required half-year convention did not affect the total deductions from taxable income. This is true for the other cases as well, since both recaptured depreciation and losses are treated as ordinary income.) UNIT-Of-PRODUCTION DEPRECIATION At times the recovery of depreciation on a particular asset is more closely related to use than to time. In these few situations (and they are rare), the unit-of-production (UOP) depreciation in any year is .. . Production for year . (B - S) (11-6) Total lIfetime production for asset This method might be useful for machinery that processes natural resources if the resources will be exhausted before the machinery wears out. Historically,this method was sometimes used for constructionequipment that had very heavy use in some years and very light use in others. It is not considered an acceptablemethod for generaluse in depreciating industrial equipment. UOP deprecIatIOn III any year = .. For numerical sjrnilarity with previous examples, assume that equipment costing $900 has been purchased for use in a sand and gravel pit. The pit will operate for 5 years, while a nearby airport is being reconstructed and paved. Then the pit will be shut down, and the equipment removed and sold for $70. Compute the unit-of-production (UOP) depreciation schedule if the airport reconstruction schedule calls for 40,000 m3of sand and gravel as follows.: Year 1 2 3 4 5 Required Sand and Gravel (m3) 4,000 8,000 16,000 8,000 4,000 - : ::I = -- ....-- ------- .-. 360 hh_ ___. DEPRECIATION SO~UTION The cost basis, B, is $900. The salvagevalue, S, is $70. The total lifetimeproduction for the asset is 40,000 m3 of sand and gravel. From the airport reconstruction schedule, the first-year UOP depreciation would be: . . 4000 m3 FIrst-yearUOP deprecIation = '> ($900 - $70 ) 40,000 m = $83 Similar calculations for the subsequent 4 years give the complete depreciation schedule: Year 1 2 3 4 5 UOP Depreciation $ 83 166 332 166 83 $830 It should be noted that the actual unit-of-production depreciation charge in any year is based on the actual production for the year rather than the scheduled production. DEPLETION Depletion is the exhaustion of natural resources as a result of theirremoval. Since depletion covers such things as mineral properties, oil and gas wells, and standing timber, removal may take the form of digging up metallic or nonmetallic minerals, producing petroleum or . natural gas from wells, or cutting down trees. . . Depletion is recognized for income taxes for the same reason depreciation is--capital investment is being consumed or used up. Thus a portion of the gross income should be considered to be a return of the capital investment. The calculation of the depletion allowanceis differentfrom depreciationbecause there are two distinctmethods of calculating depletion: cost depletion and percentage depletion. Except for standing timber and most oil and gas wells, depletion is calculated by both methods and the larger value is taken as depletion for the year.For standing timber and most oil and gas wells, only cost depletionis permissible. Cost Depletion Depreciation relied on an asset's cost, depreciable life, and salvage value to apportion the cost minus salvage value over the depreciable life. In some cases, where the asset is used at fluctuating rates, we might use the unit-of-production (UOP) method of depreciation. For mines, oil wells, and standing timber, fluctuating production rates are the usual situation. - ---- -- -- - Depletion 361 Thus, cost depletion is computed like unit-of-productiondepreciation using: 1. Property cost, less cost for land. 2. Estimated number of recoverableunits (tons of ore, cubic meters of gravel, barrels of oil, million cubic feet of natural gas, thousand board-feet of timber, etc.). 3. Salvage value, if any, of the property. A small lumber company bought a tract of timber for $35,000, of which $5000 was the land's value and $30,000 was the value of the estimated 1Y2million board-feet of standing timber. The first year, the company cut 100,000 board-feet of standing timber. What was the year's depletion allowance? SOLUTION: I, c' ,r.,.. Depletion allowanceper 1000board-ft = $35,000- $5000 1500board-ft = $20 per 1000 board-ft The depletion allowance for the year would be 100,000 board-ft x $20 per 1000board-ft = $2000 - ~. . Percentage Depletion Percentage depletion is an alternatemethod for mineral property and some oil or gas wells. The allowance is a certain percentage of the property's gross income during the year. This is an entirely different concept from depreciation. Unlike depreciation, which allocates cost over useful life, the percentage depletion allowanceis based on the property's gross income. Since percentage depletion is computed on the income rather than the property's cost, the total depletion may exceed the cost of the property. In computing the allowable percentage depletion on a property in any year, the percentage depletion allowance cannot exceed 50% of the property's taxable income computed without the depletion deduction. The percentage depletion calculations are illustrated by Example 11-13. TABLE11-6 PercentageDepletion Allowance for SelectedItems Typeof Deposit Depletion Allowance Lead, zinc, nickel, sulfur, uranium Oil and gas (small producers only) Gold, silver, copper, iron ore Coal and sodium chloride Sand, gravel, stone, clam and oyster shells, brick, tile clay Most other minerals and metallic ores --- --- 22% 15 15 10 5 14 -----------.------ 362 - DEPRECIATION A coal mine has a gross income of $250,000 for the year. Mining expenses equal $210,000. Compute the allowable percentage depletion deduction. !.SOU..J.tION' From Table 11-6, coal has a 10% depletion allowance. The percentage depletion deduction is computed from gross mining income. Then the taxable income must be computed.The allowable percentage depletion deduction is limited to the computedpercentage depletion or 50% of taxable income, whichever is smaller. Computed PercentageDepletion Gross income from mine Depletion percentage $250,000 x 10% Computed percentage depletion $ 25,000 TaxableIncome Limitation Gross income from mine Less: Expenses other than depletion $250,000 -210,000 Taxable income from mine Deduction limitation x Taxableincome limitation 40,000 50% $20,000 Since the taxable income limitation ($20,000) is less than the computed percentage depletion ($25,000), the allowablepercentage depletion deduction is $20,000. As previously stated, on mineral property and some oil and gas wells (small producers only), the depletion deduction can be based on either cost or percentage depletion. Each year, depletion is computed by both methods, and the allowable depletion deduction is the larger of the two amounts. . ' SPREADSHEETS AND DEPRECIATION The spreadsheet functions for straight-line, double declining balance, and sum-of-years'digits depreciation are listed in Table 11-7. Because these techniques are simple and were TABLE 11-7 Spreadsheet Functions for Depreciation Depreciation Technique Excel Straight line Declining balance Sum of years' digits MACRS SLN(cost, DDB(cost, SYD(cost, VDB(cost, salvage, life) salvage, life, period, factor) salvage, life, period) salvage, life, start-period, end_period, factor, no-switch) r Spreadsheets and Depreciation 363 replacedby MACRSin 1986,they are not coveredin detail here. All three functions include parameters for cost (initialbook value), salvage (final salvagevalue), and life (depreciation period). Both DDB and SYD change depreciation amounts every year, so they include a parameter to pick the period (year). Finally, DDB includes afactor. The default value is 2 for 200% or double declining balance, but another commonly used value is 1.5 for 150%. Using VDB for MACRS . The Excel function VDB is a flexible or variable declining balance method. It includes the ability to specify the starting and ending periods, rather than simply a year. It also includes an optional nO-8witchfor problems where a switch from declining balance to straight-line depreciation is NOT desired. To use VDB to calculate MACRS depreciation, the following are true. 1. Salvage= 0, since MACRS assumes no salvage value. 2. Life = recovery period of 3,5, 7, 10, 15, or 20 years. 3. First period runs from 0 to 0.5, 2ndperiod from 0.5 to 1.5, 3rdfrom 1.5 to 2.5, tth from t - 1.5 to t - 0.5, and last period from life -0.5 to life. 4. Factor = 2 for recovery periods of 3,5, 7, or 10 years and = 1.5 forrecovery periods of 15 or 20 years. 5. Since MACRS includes a switch to straight line, no_switchcan be omitted. The starLperiod and end_periodargumentsare from t - 1.5to t - 0.5, because MACRS uses a half-year convention for the first year. Thus the first year has 0 to 0.5 year of depreciation, and the second year starts where the first year stops. When one is writing the Excel function, either the first and last periods must be edited individually,or start_period must be definedwith a minimum of 0 and end_periodwith a maximum of life. This prevents the calculation of depreciation from -0.5 to 0 and from life to life +0.5. The results of using the VDB function match Table 11-4, except that the VDB function has more significant digits rather than being rounded to 2 decimals. Example 11-14 illustrates the use of the VDB function. Return to the data of Example 11-5 which had $150,000 of office equipment, which is 7-year MACRS property.Use VDB to compute the depreciation amounts. SOLUTION. -- The spreadsheetin Figure 11-7de:finest:!lestart...periodwith a minimum of Q¥1d tpe end_pefiog ~lthfnfaximum of life.Thus thisformula couldbe usedfor anyyear of anyrecoveryschedule.Notice that the VDBformula uses the value0 for thesalvage value,rather than referring to the data cell for thes~v~g~ v'tlue.~A~RSoa§sumesca§ill,v~gy.value ot:zer~nomatter what the value truly is: ---- --- -- --- -.. .. -- -.. - ..-,- ........ 364 - DEPRECIATION , A B First Cost 150,000 0 Salvage 7 Life 200% Factor 1 2 3 4 5 6 7 8 9 Period 1 2 3 10 4 $18,742.19 11 5 $13,387.28 12 6 $13,387.28 13 7 $13,387.28 14 8 $6,693.64 15 C D E F G Depreciation $21,428.57 = VDB($A$1,0,$A$3,MAX(0,A7-1.5),MIN($A$3,A7-O.5),$A$4) $36,734.69 or (cost, salvage,life, max(O,(-1.5), min (life,(- .5), factor) $26,239.07 j $150,000 I I = Sum FIGURE 11-7 Using VDB to calculate MACRS depreciation. SUMMARY Depreciation is the foundation for including income taxes in economic analysis. There are three distinct definitionsof depreciation: 1. Decline in asset's market value. 2. Decline in asset's value to its owner. 3. Allocating the asset's cost less its salvage value over its recovery period or depreciable life. While the first two definitions are used in everyday discussions, it is the third, or accountant's, definition that is used in tax computations and in this chapter. Book value is the remaining unallocated cost of an asset, or Book value = Asset cost - Depreciation charges made to date This chapter describes how depreciable assets are written off (or claimed as a business expense) over a period of years instead of expensedin a single period (like wages,material costs, etc.). The depreciation methods described include the historical pre-1981 methods: straight line, sum of the years' digits, and declining balance. These methods required estimatingtheasset'ssalvagevalueanddepreciablelife. . The current tax law specifies use of the modified accelerated capital recovery system (MACRS).This chapterhas focusedon the general depreciationsystem (GDS) with limited discussion of the less attractive alternative depreciation system (ADS). MACRS (GDS) specifiesfaster recoveryperiods and a salvagevalue of zero, so it is generally economically more attractivethan the historical methods. -- .. - .- - -- .- -- -. -- - --. -.. A Problems _.___________ 365 The MACRS system is the current tax law, and it assumes a salvagevalue of zero. This , is in contrast with historical methods, which ensured the final book value would equal the predicted salvage value. Thus, when one is using MACRSit is often necessary to consider recaptured depreciation.This is the excess of salvagevalue over book value, and it is taxed as ordinary income. Similarly, losses on sale or disposal are taxed as ordinary income. Integrating depreciation schedules with cash flows often involves a lot of arithmetic. Thus, the tool of spreadsheetscan be quite helpful.The functionsfor the historicalmethods, straight line, sum of the years' digits, and declining balance, are straightforward. Rather than individuallyentering MACRSpercentagesinto the spreadsheet,the function VDB can be used to calculate MACRS depreciation percentages. Unit-of-production (UOP) depreciation relies on usage to quantify the loss in value. UOP is appropriate for assets that lose value based on the number of units produced, the tons of gravel moved, and so on (vs number of years in service). However,this method is not considered to be acceptable for most business assets. Depletion is the exhaustion of natural resources like minerals, oil and gas wells, and standing timber.The owners of the natural resources are consumingtheir investmentsas the natural resources are removed and sold. Cost depletionis computedbased on the fraction of the resource that is removed or sold. For minerals and some oil and gas wells, an alternate calculationcalled percentagedepletion is allowed.Percentagedepletionis based on income, so the total allowable depletion deductions may exceed the investedcost. PROBLEMS 11-1 A depreciable asset costs $10,000 and has an estimated salvage value of $1600 at the end of its 6-year depreciable life. Compute the depreciation schedule for this asset by both SOYD depreciation and DDB depreciation. (d) MACRS depreciation. 11-5 The company treasurer is uncertain which of four depreciation methods the firm should use for office furniture that costs $50,000, and has a zero salvage value at the end of a 1O-yeardepreciable life. Compute the depreciation schedule for the office furniture using the methods listed: (a) Straight line. (b) Double declining balance. (c) Sum-of-years' -digits. (d) Modified accelerated cost recovery system. 11-2 A million-dollar oil drilling rig has a 6-year depreciable life and a $75,000 salvage value at the end of that time. Determine which one of the following methods provides the preferred depreciation schedule: DDB or SOYD. Show the depreciation schedule for the preferred method. 11-3 A new machine tool is being purchased for $16,000 and is expected to have a zero salvage value at the end of its 5-year useful life. Compute the DDB depreciation schedule for this capital asset. Assume any remaining depreciation is claimed in the last year. 11-6 The RX Drug Company has just purchased a capsulating machine for $76,000. The plant engineer estimates the machine has a useful life of 5 years and little or no salvage value. He will use zero salvage value in the computations. Compute the depreciation schedule for the machine using: (a) Straight-line depreciation. (b) Sum-of-years'-digits depreciation. (c) Double declining balance depreciation. Assume any remaining depreciation is claimed in the last year. 11-4 Some special handling devices can be obtained for $12,000. At the end of 4 years, they can be sold for $600. Compute the depreciation schedule for the devices using the following methods: (a) Straight-line depreciation. (b) Sum-of-years' -digits depreciation. (c) Double declining balance depreciation. -- - -- -- HI II 366 DEPRECIATION 11-7 The Acme Chemical Company purchased $45,000 of research equipment, which it believes will have zero salvage value at the end of its 5-year life. Compute the depreciation schedule for the equipment by each of the following methods: . (a) Straight line. (b) Sum-of-years'-digits. (c) Double declining balance. (d) Modified accelerated cost recovery system. 11-8 Consider a $6500 piece of machinery, with a 5-year depreciable life and an estimated $1200 salvage value. The projected utilization of the machinery when it was purchased, and its actual production to date, are shown below. Year 1 2 3 4 5 Projected Production (tons) 3500 4000 4500 5000 5500 Actual Production (tons) 3000 5000 [Not yet known] Compute the machinery depreciation schedule by each of the following methods: (a) Straight line. (b) Sum-of-years' -digits. (c) Double declining balance. (d) Unit of production (for first 2 years only). (e) Modified accelerated cost recovery system. 11-9 A large profitable corporation purchased a small jet plane for use by the firm's executives in January. The plane cost $1.5 million and, for depreciationpurposes, is assumed to have a zero salvage value at the end of 5 years. Compute the MACRS depreciation schedule. 11-10 For an asset that fits into the MACRS "All property not assigned to another class" designation, show in a table the depreciation and book value over the asset's lO-year life of use. The cost basis of the asset is $10,000. 11-11 A company that manufactures food and beverages in the vending industry has purchased some handling equipment that cost $75,000 and will be depreciated using MACRS GDS. The class life of the asset is 4 years, show in a table the yearly depreciation amount and book value of the asset over its depreciation life. ---- -- ---- 11-12 Consider five depreciationschedules: Year A B C D E 1 $45.00 $35.00 $29.00 $58.00 $43.50 2 36.00 20.00 46.40 34.80 30.45 27.00 30.00 27.84 20.88 21.32 3 4 18.00 30.00 16.70 12.53 14.92 5 9.00 20.00 16.70 7.52 10.44 8.36 6 They are based on the same initial cost, useful life, and salvage value. Identify each schedule as one of the following . .. .. . Straight-line depreciation. Sum-of-years' -digits depreciation. 150% declining balance depreciation. Double declining balance depreciation. Unit-of-production depreciation. Modified accelerated cost recovery system. . 11-13 The depreciation schedule for an asset, with a salvage value of $90 at the end of the recoveryperiod, has been computed by severalmethods. Identifythe depreciationmethodused for each schedule. Year ABC D E. $lO7.0 1 $323.3 $212.0 $424.0 $194.0 2 258.7 339.2 254.4 194.0 216.0 3 194.0 203.5 152.6 194.0 324.0 4 129.3 122.1 91.6 194.0 216.0 5 64.7 122.1 47.4 194.0. . 107.0 6 61.1 ----970.0 lO60.0 970.0 970.0 970.0 j i I I J I 11-14 The depreciatib.n schedule for a microcomputer has been arrived at by several methods. The estimated salvage value of the equipment at the end of its 6-year useful life is $600. Identify the resulting depreciation schedules. Year ABC D 1 $2114 $2000 $1600 $1233 2 1762 1500 2560 1233 3 14lO 1125 1536 1233 4 1057 844 922 1233 5 705 633 922 1233 6 352 475 460 1233 11-15 TELCO Corp has leased some industrial land near its plant. It is building a small warehouse on the site at a cost of $250,000. The building will be ready for use January 1. The lease will expire 15 years after the building is occupied. The warehouse will belong at that time to the landowner, with the result thaI J 1 . ... .. - . Problems 367 there will be no salvage value to TELCO. The warehouse is to be depreciated either by MACRS or SOYD depreciation. If 10% interest is appropriate, which depreciation method should be selected? 11-20 The MACRS depreciation percentages for lO-year personal property are given in Table 11-3. Make the necessary computations to determine if the percentages shown are correct. 11-16 A profitable company making earthmoving equipment is considering an investment of $100,000 on equipment that will have 5-year useful life and a $20,000 salvage value. If money is worth 10%, which one of the following three methods of depreciation would be preferable? (a) Straight-line method. (b) Double declining balance method. (c) MACRS method. 11-21 Use MACRS GDS depreciation for each of the assets, 1-3, to calculate the following items, (a)-(c). 1. A light general-purpose truck used by a delivery 11-17 The White Swan Talc Company purchased $120,000 of mining equipment for a small talc mine. The mining engineer's report indicates the mine contains 40,000 cubic meters of commercial quality talc. The company plans to mine all the talc in the next 5 years as follows: Year 1 2 3 4 5 Talc Production (m3) 15,000 11,000 4,000 6,000 4,000 At the end of 5 years, the mine will be exhausted and the mining equipment will be worthless. The company accountant must now decide whether to use sumof-years' -digits depreciation or unit-of-production depreciation. The company considers 8% to be an appropriate time value of money. Compute the depreciation schedule for each of the two methods. Which method would you reco~end that the company adopt? Show the computations to justify your decision. 11-18 For its fabricated metal products, the Able Corp. is buying $10,000 of special tools that have a 4-year useful life and no salvage value. Compute the depreciation charge for the second year by each of the following methods: (a) DDB. (b) Sum-of -years' -digits. (c) Modified accelerated cost recovery system. 11-19 The MACRS depreciation percentages for 7-year personal property are given in Table 11-3. Make the necessary computations to determine if the percentages shown are correct. business, cost = $17,000. 2. Production equipment used by a Detroit automaker to produce vehicles, cost = $30,000. 3. Cement production facilities used by a construction firm, cost $130,000. (a) The MACRS GDS property class. (b) The depreciation deduction for Year 3. (c) The book value of the asset after 6 years. 11-22 On July 1,Nancy Regan paid $600,000 foracommercial building and an additional $150,000 for the land on which it stands. Four years later, also on July 1, sbe sold the property for $850,000. Compute the modified accelerated cost recovery system depreciation for each of the five calendar years during which she had the property. 11-23 A group of investors has formed Trump Corporation to purchase a small hotel. The asking price is $150,000 for the land and $850,000 for the hotel building. If the purchase takes place in June, compute the MACRS depreciation for the first three calendar years. Then assume the hotel is sold in June of the fourth year, and compute the MACRS depreciation in that year also. 11-24 Mr. Donald Spade purchased a computer in January to keep records on all the property he owns. The computer cost $70,000 and is to be depreciated using MACRS. Donald's accountant pointed out that under a special tax rule (the rule applies when the value of property placed in service in the last 3 months of the tax year exceeds 40% of all the property placed in service during the tax year), the computer and all property that year would be subject to the midquarter convention. The midquarter convention assumes that all property placed in service in any quarter-year is placed in service at the midpoint of the quarter. Use the mid-quarter convention to compute Donald's MACRS depreciation for the first year. 11-25 A company is considering buying a new piece of machinery. A 10% interest rate will be used in the computations. Two models of the machine are available. - - - - 368 - DEPRECIATION Initial cost End-of-useful-life salvage value, S Annual operating cost Usefullife, in years MachineI Machinen $80,000 $100,000 20,000 25,000 18,000 20 ment. All activity would take place during a single summer. The farmer hired an engineering student for $2500 to count the truckloads of material hauled away. The farmer estimated that 2 acres of streambed had been stripped of the sand and gravel. The 640-acre farm had cost him $300 per acre, and the farmer felt the property had not changed in value. He knew that there had been no use for the sand and gravel prior to the construction of the highway, and he could foresee no future use for any of the remaining .50,000 cubic meters of sand and gravel. Determine the fanner's depletion allowance. (Answer: $1462.50) 15,000 first 10 years 20,000 thereafter 25 (a) Determine which machine should be purchased, based on equivalent uniform annual cost. (b) What is the capitalized cost of Machine I? (c) Machine I is purchased and a fund is set up to replace Machine I at the end of 20 years. Compute the required uniform annual deposit. (d) Machine I will produce an annual saving of material of $28,000. What is the rate of return if Machine I is installed? (e) What will be the book value of Machine I after 2 years, based on sum-of- years' -digits depreciation? if) What will be the book value of Machine n after 3 years, based on double declining balance depreciation? (g) Assuming that Machine n is in the 7-year property class, what would be the MACRS depreciation in the third year? 11-26 Equipment costing $20,000 that is a MACRS 3-year property is disposed of during the second year for $14,000. Calculate any depreciation recapture, ordinary losses, or capital gains associated with disposal of the equipment. 11-29 Mr. H. Salt purchased a ys interest in a producing oil well for $45,000. Recoverable oil reserves for the well were estimated at that time at 15,000 barrels ys of which represented Mr. Salt's share of the re~ serves. During the subsequent year, Mr. Smt received $12,000 as his ysshare of the gross income from the sale of 1000 barrels of oil. From this amount, he had to pay $3000 as his share of the expense of producing the oil. Compute Mr. Salt's depletion allowance for the year. (Answer: $3000) 11-30 A heavy construction firm has been awarded a contract to build a large concrete dam. It is expected that a total of 8 years will be required to complete the work. The firm will buy $600,000 worth of special equipment for the job. During the preparation of the job cost estimate, the following utilization schedule was computed for the special equipment: 11-27 An asset with a 8-year ADR class life costs $50,000 and was purchased on January I, 2001. Calculate any depreciation recapture, ordinary losses, or capital gains associated with selling the equipment on December 31, 2003, for $15,000, $25,000, and $60,000. Consider two cases of depreciation for the problem: if MACRS GDS is used, and if straight-line depreciation over the ADR class life is used with a $10,000 salvage value. 11-28 When a major highway was to be constructed nearby, a farmer realized that a dry streambed running through his property might a valuable source of sand and gravel. He shipped samples to a testing laboratory and learned that the material met the requirements for certain low-grade fill material. The farmer contacted the highway construction contractor, who offered 65/ c per cubic meter for 45,000 cubic meters of sand and gravel. The contractor would build a haul road and would use his own equip- Year 1 2 3 4 Utilization (hr/yr) 6000 4000 4000 1600 Year 5 6 7 8 Utilization (hr/yr) 800 800 2200 2200 At the end of the job, it is estimated that the equipment . can be sold at auction for $60,000. (a) Compute the sum-of-years-digits' depreciation schedule. (b) Compute the unit-of-prQduction depreciation schedule. 11-31 Some equipment costs $1000, has a 5-year deprecia.II. ble life, and an estimated $50 salvage value at the end of that time. You have been assigned the problem to determine whether to use straight-line or SOYD depreciation. If a 10% interest rate is appropriate, which is the preferred depreciation method for this profit.!""" l I Problems . corporation? Use a spreadsheet to show your computations of the Qifference in present worths. 11-32 The FOURX Corp. has purchased $12,000 of experimental equipment. The anticipated salvage value is ~ $400 at the end of its 5-year depreciable life. This profitable corporation is considering two methods of depreciation: sum of years' digits and double declining balance. If it uses 7% interest in its comparison, which method do you recommend? Show computations to support your recommendation. Use a spreadsheet to develop your solution. 11-33 Given the data in Problem 11-16, use a spreadsheet ala function to compute the MACRS depreciation schedule. Show the total depreciation taken (=sum( » as --------- . 369 well as the PW of the depreciation charges discounted at the MARR %. 11-34 Office equipment whose initial cost is $100,000 has ~ an estimated actual life of 6 years, with an estimated salvage value of $10,000. Prepare tables listing the annual costs of depreciation and the book value at the end of each 6 years, based on the straight-line, sum-of-years' -digits, and MACRS depreciation. Use spreadsheet functions for the depreciation methods. 11-35 You are equipping an office. The total office equip- Ala ment will have a first cost of $1,750,000 and a salvage value of $200,000. You expect the equipnient will last 10 years. Use a spreadsheet function to compute the MACRS depreciation schedule. . -.-....- After Completing This Chapter... The student should be able to: Calculate taxes due or taxes owed for both individuals and corporations. Understand the incremental nature of the individual and corporate tax rates used for calculating taxes on income. Calculate a combined income tax rate for state and federal income taxes and select an appropriate tax rate for engineering economic analyses. Utilize an after-tax tax table to find the after-taxcash flowsfor a prospective investment project. Calculate after-tax measures of merit, such as present worth, annual worth, payback period, internalrate of return, and benefit-cost ratio,from developedafter-taxcash flows. Evaluate investment alternativeson an after-tax basis including asset disposal. Use spreadsheets for solving after-taxeconomic analysis problems. .. . . . .. QUESTIONS TO CONSIDER ;;. 1. The wind energy production tax credit has become controversial among some commentators, who see it as "corporate welfare." These critics contend that the government should not be in the business of encouraging wind energy-or any other type of energy source, for that matter.Instead, they argue, the market should determine whether alternative energy sources such as wind power succeed or fail. Are these critics right? 2. Proponents of the wind production tax credit counter that the energy market has never been genuinely "free." They point out that other types of energy production (notably coal and oil) have obtained large government subsidies over the years in the form of depletion allowances and other tax provisions. In fact, these proponents argue, carbonbased sources enjoy a tremendous "hidden subsidy" because the environmental and health effects of these more-polluting energy sources are not paid for by coal and oil producers, but are borne by society at large. Are these proponents right? 3. Developing a wind power project takes many years and requires the commitment of large sums of investmentcapital before the project begins to return a profit. What is the effect on investmentwhen the wind power production tax credit is allowed to expire, or is extended for periods of only a few years? 4. The federal income tax was introduced in the United States in 1913.Using the Internet, can you determinehow taxrates changed throughoutthe course of the twentieth century? How has this affected the value of tax credits to industry? ------------ Income Taxes' On with the Wind For at least three decades now, environmental activists and community leaders have bemoaned American dependenceon foreign oil. The war in Iraq again highlighted the precariousness of relying on an unstable region of the world to producea major part of our energy requirements. One solutionto this dilemmais to rely more on renewablesources of energy,such as solarpower and wind.The technologyfor suchalternative energy sources has been around for many years, and most voters seem to favor more reliance on renewables. If good intentions were all it took, we'd be getting much of our electricity from windmills. But, of course, making the transition to greater use of wind power requires a significantinvestmentin infrastructure,especially costly wind turbines. And few investors are willing to plunk downtheir money unless they have a solid expectation of earning a competitivereturn. Until fairly recently, cost factors kept wind energy from becoming an attractive investment. As recently as the late 1980s, wind-generated power cost roughlytwice as muchto produceas energyfromconventional sources. In the past few years, however, wind energy has decreased dramatically in price. The American Wind Energy Association (AWEA) reports that many modem, state-of-the-art windplantscan nowproducepowerfor less than 5 centsper kilowatt~hour, makingthem . - competitive with conventional sources. Unsurprisingly,investmentin wind power production has also increased substantially. How did this happen? In part, it was driven by advances in wind turbine technology. But it was helped significantlyby a provision of the federal tax law contained in the Energy Policy Act of 1992. This statute allowed utilities and other electricity suppliers a "production tax credit" of 1.5 cents per kilowatt-hour (later adjusted to 1.7 cents to account for the effects of inflation). This was a key incentive to the wind power industry which, like all energy producers, must expend large sums on capital assets. The tax credit had a dramatic effect on the wind energy market. During 2001, for instance, energy producers added almost 1700megawatts of wind-generating capacity-enough to power nearly half a million homes. But what the government giveth, the government can take away.When the production tax credit briefly expired at the end of 2001, an estimated $3 billion worth of wind projects were suspended, and hundreds of workers were laid off. I . I t -~ t1 f ----- ~. 372 ~... ~..- -..~-- ... .....-. -, - INCOME TAXES Fortunately for the industry, the credit was subsequently extended to the end of 2003, andis expectedto be extendedfurther. As a result, the AWEA anticipatesthat another 1500 to 1800megawattsof wind-:generatingcapacity will come on line during 2003. - As Benjamin Franklin said, two things are inevitable: death and taxes. In this chapter We will examinethe structureof taxesin the United States.There is, of course,a wide :variety of taxesranging from salestaxes, gasoline taxes, and property taxesto state and federal income taxes,and so forth. Here we will concentrateour attention on federal incometaxes. Income taxesare part of most real problems and often havea substantialimpactthatmust be considered. First, we must understand the way in which taxes are imposed. Chapter 11concerning depreciation is an integral part of this analysis, so the principles covered there must be well understood. Then; having understood the mechanism, we will see how federal income taxes affect our economic analysis. The various analysis techniques will be used in examples of after-tax calculations. A PARTNER IN THE BUSINESS Probably the most straightforward way to understand the role of federal income taxes is to considerthe U.S. governmentas a partner in everybusinessactivity.As a partner,the government shares in the profits from every successful venture. And in a somewhat more complex way, the government shares in the losses of unprofitable ventures. The tax laws are complex, and it is not our purpose to fully explain them. 1Instead, we will examine the fundamental concepts of the federal income tax laws-and we emphasize at the start that thereareexceptionsandvariationsto almosteverystatementwe shallmake! CALCULATION OF TAXABLE INCOME At the mention of income taxes, one can visualize dozens of elaborate and complex calculations. And there is some truth to that vision, for there can be complexities of all sorts in the computation of income taxes. Yet some of the difficulty is removed when one defines incomes taxes as just another type of disbursement. Our economic analysis calculations in prior chapters have dealt with all sorts of disbursements: operating costs, maintenance, labor and materials, and so forth. Now we simply add one more prospective disbursement to the list-income taxes. Taxable Income of Individuals The amount of federal income taxes to be paid depends on taxable income and the income tax rates. Therefore, our first concern is the definition of taxable income. To begin, one must compute his or her gross income: Gross income = Wages, salary, etc. + Interest income + Dividends+ Capi~ gains + Unemploymentcompensation + Other income 1Manygovernmentandprivatesourcesexistthat describedetailedtaxationinformation.Theseinclude the Internal Revenue Service (www.irs.gov), U.S. Treasury (www.treasury.gov), Commerce Clean::, House (www.cch.com), and Research Institute of America (www.riahome.com). ----- - - Calculation of Taxable Income 373 From gross income, we subtract any allowable retirement plan contributions and other ,adjustments. The result is adjusted gross income (AGI). From adjusted gross income, individuals may deduct the following items:2 1. Personal Exemptions. One exemption ($3050 for 2003) is provided for each person who depends on the gross income for his or her living.3 2. Itemized Deductions. Some of these are: (a) Excessive medical and dental expenses (exceeding 71/2%of adjusted gross income) (b) State and local income, property and personal property tax (c) Home mortgage interest (d) Charitable contributions (e) Casualty and theft losses (exceeding $100 + 10% of adjusted gross income) (I) Miscellaneous deductions (some categories must exceed 2% of adjusted gross income) (g) Car and other business expenses (h) Tax benefits for work-relatededucation 3. Standard Deduction. Each taxpayer may either itemize his or her deductions, or instead take a standard deduction as follows: .. (a) Single taxpayers, $4750 (for 2003) (b) Married taxpayers filing a joint return, $7950 (for 2003) The result is taxable income. For individuals, taxable income is computed as follows: Adjusted gross income Taxable income = Gross income - Adjustments = Adjusted gross income - Personal exemption(s) - Itemized deductions or Standard deduction (12-1) Classification of Business Expenditures When an individual or a firm operates a business, there are three distinct types of business expenditure: 1. For depreciable assets. 2. For nondepreciable assets. 3. All other business expenditures. 2The 2003 itemized deduction are limited if adjusted gross income is more than $139,500 ($69,750 if married filing separately.) 3Personal exemptions for high-income taxpayers are being phased out. In 2003, for taxpayers who are married, filing jointly, the personal exemption is phased out beginning at an AGI of $209,250. The phaseout is complete at an AGI of $331,750. For single taxpayers, the beginning and complete phaseout numbers are $139,500 and $262,000, respectively. -- ------ .. 374 - INCOME TAXES Expendituresfor depreciable assets: When facilities or productive equipment with useful lives in excess of one year are acquired, the taxpayer will recover his investment through depreciation charges.4Chapter 11 examined in great detail the several ways in which the cost of the asset could be allocated over its useful life. Expendituresfor nondepreciable assets: Land is considered to be a nondepreciable asset, for there is no finite life associated with it. Other nondepreciable ~sets are properties not used either in a trade, in a business, or for the production of income.An individual'shome and automobileare generallynondepreciableassets. The final category of nondepreciable assets are those subject to depletion, rather than depreciation. Since business firms generally acquire assets for use in the business, their only nondepreciableassets normally are land and assets subjectto depletion. All other business expenditures: This category is probably the largest of all, for it includes all the ordinary and necessary expendituresof operating a business. Labor costs, materials, all direct and indirect costs, and facilities and produ~tiveequipment with a useful life of one year or less are part of the routine expenditures. They are charged as a business expense-expensed-when they occur. Business expenditures in the first two categories-that is, for either depreciable or nondepreciable assets-are called capital expenditures. In the accounting records of the firm, they are capitalized; all ordinary and necessaryexpenditures in the third categoryare expensed. Taxable Income of Business Firms The starting point in computing a firm's taxable income is gross income. All ordinary and necessary expenses to conduct the business-except capital expenditures-are deducted from gross income. Capital expendituresmay not be deducted from gross income. Except for land, business capita! expenditures are charged to expense period by period througb; depreciation or depletion charges. For business firms, taxable income is computed as follows: Taxable income = Gross - income All expenditures except capital expenditures - Depreciation and depletion charges (12-2) Because of the treatment of capital expenditures for tax purposes, the taxable income of a firm may be quite different from the actual cash results. 4There is an exception. In 2003, businesses may immediately deduct (expense) up to $25,000 of business equipment (via the Section 179 Business Deduction) in a year, provided their total equipment expenditure for the year does not exceed $200,000. Between $200,000 and $224,000 in qualifying 179 Property the dollar limit is reduced by the amount over $200,000. Above $224,000 a Section 179 deduction cannot be taken. Income Tax Rates 375 _. During a 3-year period, a finn had the following results (in millions of dollars): Gross income from sales Purchase of special tooling (useful life: 3 years) All other expenditures Cash results for the year Year 1 $200 -60 Year 2 $200 -140 $ 0 -140 $ 60 o Year3 $200 o -140 $ 60 Compute the taxable income for each of the 3 years. .SOl~TlON' The cash results for each year would suggest that Year 1 was a poor one, while Years 2 and 3 were very profitable. A closer look reveals that the finn's cash results were adversely affected in Year 1 by the purchase of special tooling. Since the special tooling has a 3-year useful life, it is a capital expenditure with its cost allocated over the useful life. For straight-line depreciation and no salvage value, we use Equation 11-2 to find the annual charge: B-S Annualdepreciationcharge= - N = 60-0 3 = $20million Applying Equation 12-2,we write Taxableincome = 200 - 140 - 20 = $40 million In each of the 3 years, the taxable income is $40 million. An examination of the cash results and the taxable income in Example 12-1indicates th~t taxable income is a better indicator of the annual perfonnance of the finn. INCOME TAX RATES Incometaxrates for individualschangedmanytimesbetween 1960and 1995,asillustratedin Figure 12-1.Recent movementshave been less dramatic.From 1995to 2000 the maximum rate was 39.6%; it fell to 39.1% in 2001 and was 38.6% in 2002 and 2003. Individual Tax Rates There arefour schedulesof federal incometax rates forindividuals. Singletaxpayersuse the Table 12-1schedule.Marriedtaxpayersfilingajointretum use the Table 12-2schedule.1\vo other schedules (not shown here) are applicable to unmarriea individuals with dependent relatives ("head of household"), and married taxpayers filing separately. 376 - INCOME TAXES 100 90 80 '-" £ 70 60 50 e S 40 .>( 30 20 10 0 1960 1965 1970 1975 1980 1985 1990 Year FIGURE 12-1 Maximum federal income tax rates for individuals. TABLE12-1 2003 Tax Rates for Single Taxpayers Tax Taxable Income Over $ But Not Over 0 $ 6,000 6,000 28,400 28,400 68,800 68,800 143,500 143,500 311 ,950 Over 311,950 Base Tax $ 0.00 600.00 3,960.00 14,868.00 37,278.00 96,235.50 Plus On Income Over 10% 15% 27% 30% 35% 38.6% $ . 0 6,000 28,400 68,800 143,500 311,950 TABLE12-2 2003 Tax Rates for Married Individuals FilingJointly '.Taxable Income Over $ 0 12,000 47,450 114,650 174,700 Tax But Not Over $ 17,000 47,450 114,650 174,700 311,950 Over 311,950 ------ Base Tax Plus 0.00 1,200.00 6,517.50 24,661.50 42,676.50 90,714.00 10% 15% 27% 30% 35% 38.6% $ On Income Over 0 12,000 47,45 114,650 174,700 311,950 $ . ..... ..> . --, - Income Tax Rates 377 An unmarried student earned $8000 in the summer plus another $2000 during the rest of the year. When he files an income tax return, he will be allowed one exemption(for himself). He estimates he spent $1000 on allowable itemized deductions. How much income tax will he pay? SqLUTION Adjusted gross income Taxable income = $8000 + 2000 = $10,000 = Adjusted gross income - Deduction for one exemption ($3050) - Standard deduction ($4750) = 10,000- 3050- 4750= $2200 Federal income tax = 10% (2200) - $220 Corporate Tax Rates Income tax for corporations is computed in a manner similar to that for individuals. The data in Figure 12-2can be presented in the schedule form which follows it. 45 '-' 39% 40 38% <;j 35 a::: 35% 34% 34% Q) 30 Q) e0 0 25 -";js:: ... 25% 20 .g 15 '5 s:: Q) e 10 -0s:: 5 0 115% $ 10M $100,000 $200,000 $300,000 $400,000 TaxableIncome $15M :Ii FIGURE 12-2 Corporation federal income tax rates (2003) rates. - - - - ---- $18,333,333 -- -- - - 378 - INCOME TAXES I Tax Rate Taxable Income 15% 25% 34% 39%* 34% 35% 38% 35% Not over $50,000 $50,000-75,000 $75,000-100,000 $100,000-335,000 $335,000-10 million $10 million-15 million $15 million-18,333,333 :::$18,333,333 Corporate Income Tax 15% over $0 $7,500 + 25% over $50,000 $13,750 + 34% over $75,000 $22,250 + 39% over $100,000 $113,900 + 34% over $335,000 $3,400,000 + 35% over $10 million $5,159,000 + 38% over $15 million $6,425,667 + 35% over $18,333,333 *The extra 5% from $100,000 to $335,000 was chosen so that firms in the $335,000 to $10 million bracket pay a flat 34% tax rate. [(0.39-0.34)](335,000-100,000) (0.34 - 0.15)(50,000) + (0.34 - 0.25)(75,000 - 50,000)] so tax = 0.34 (tax income) ~n $335,000 to $10 million bracket. Similarly, for corporate incomes over $18,333,333 the tax rate schedule is equivalent to a flat tax rate of 35%. = The French ChemicalCorporationwas fOnnedto producehousehold bleach.The firmbought land , for $220,000, had a $900,000 factory building erected, and installed $650,000worth of chemical and packaging equipment. The plant was completed and operations begun on April 1. The gross income for the calendar year was $450,000. Supplies and all operating expenses, excluding the capital expenditures,were $100,000.The firm will use modifiedacceleratedcost recovery system (MACRS) depreciation. (a) What is the first-year depreciationcharge? (b) Whatis the first-yeartaxableincome? (c) How much will the corporationpay in federal.income taxes for the year? lSO'LUTION (a) MACRS depreciation: Chemical eqt1ipm~ntigpersonal propeIty:I3romTable 11-2, iUs probably in the "7-year, all other property" class. I3irst-yeardepreciation(equipment) .,,$650,000x 14.29% -<$92,885 The building is in the 39-yearreal propertyclass.Sm(;e it was placed in serVjceon April 1, the.first-year depreciationis: First-year depre(;iatioIl(buildiIlg). $90Q,000x 1.819%' ..$16,37,1 (see Tablet 1-4) .. ... -- --- I ,- I ~ . ~ ., .' '~II ---- i Income Tax Rates (b) Taxable income 379 = Gross income - All expenditures except capital expenditures - Depreciation and depletion charges (c) Federal = $450,000 - 100,000 - 109,256 = $240,744 income tax = $22,250 + 39%(240,744 - 100,000) = $77,140 Combined Federal and State Income Taxes In addition to federal income taxes, most individuals and corporations pay state income taxes. It would be convenientif we could derive a single tax rate to represent both the state and federal incremental tax rates. In the computation of taxable income for federal taxes, the amount of state taxes paid is one of the allowable itemized deductions. Federal income taxes are not, however, generally deductible in the computation of state taxable income. Therefore, the state income tax is applied to a larger taxable income than is the federal income tax rate. As a result, the combined incremental tax rate will not be the sum of two tax rates. For an increment of income (Lllncome) and tax rate on incremental income (LlTax rate): State income taxes Federal taxable income Federalincometaxes = (LlState tax rate) (Lllncome) = (~lncome)(1 - ~State tax rate) = (~Federaltax rate)(Lllncome)x (1 - b.Statetax rate) The total of state and federal income taxes is [LlStatetax rate + (LlFederaltax rate)(1 - ~State tax rate)](~Income) The tenn in the brackets gives the combined incremental tax rate. Combined incremental tax rate = LlStatetax rate + (LlFederaltax rate)(1 - ~State tax rate) An engineer has an income that puts him in the 27% federal income tax bracket and at the 10% state incremental tax rate. He has an opportunityto earn an extra $500 by doing a small consulting job. What will be his combined state and federal income tax rate on the additional income? :; _._ ...;...;._.:--:.L...;._.__________ 59~~IIOr~Jii I I ; ! ;; = Use Equation 12-3 to find the combined incremental tax rate: 0.10 + 0.27(1"-'-'.0.10)= 34.3%. (12-3) "II I 380 - INCOME TAXES Selecting an Income Tax Rate for Economy Studies Since income tax rates vary with the level of taxable income for both individuals and corporations, one must decide which tax rate to use in a particular situation. The simple answer is that the tax rate to use is the incremental tax rate that applies to the change in taxable income projected in the economic analysis. If a married couple filingjointly has a taxable income of $40,000 and can increase their incomeby $2000, what tax rate shouldbe used for the $2000 of incremental income? From Table 12-2,we see the $2000 falls within the 15% tax bracket. Now suppose this couple could increase their $40,000 income by $10,000. In this situation, Table 12-2 shows that the 15% incremental tax rate should be applied to the first $7,450 and a 27% incremental tax rate to the last $2,550 of extra income. The appropriate incremental tax rate for corporations is equally easy to determine. For larger corporations, the federal incremental tax rate is 35%. In addition, there may be up to a 12-15% state tax. ECONOMIC ANALYSIS TAKING INCOME TAXES INTO ACCOUNT An importantstep in economicanalysis has been to resolve the consequencesof alternatives into a cash flow.Because income taxes have been ignored, the result has been a before-tax cashflow. This before-taxcash flowis an essentialcomponentin economic analysisthat also considers the consequences of income tax. The principal elements in an after-tax analysis are as follows: · Before-tax cash flow ·· Taxableincome(Before-taxcashflow- Depreciation) ·· Depreciation Income taxes (Taxableincome x Incremental tax rate) After-tax cash flow (Before-tax cash flow - Income taxes) These elements are usually arranged to form a cash flow table. This is illustrated by Example 12-5. A IDedium-sizedprofitable corporation is considering the purchase of a $3000 used pickup truck for use by the shipping and .receiving department. puring the trucJ<:'s5-year useful life, it is estimated the .finn will save $800 per year after all the costs of oWniJ:J.g and operati.ngthe truck have been paid. Truck salvage value is estimated at$750. :I I p L' (a) Wbatistfie befo~e-ta:~ ;~te ~lretOtn?~ (b) What is th~after-taxrate !' _ ::I Ii ofretutIl depreciation. ~..,.. ---- ,:.,... ..e>-' ,..,..,',',,';; =;11: - - = =:=: =- ~;; =~ ~=: = = =:e ;;;;;:;;; '= 911tb,is capit<l! expellaitur~? AssuIDe straight4ine t!tiI ---- =tI!!I:.:=: =;;'~ == -- ;:; - a::IICIi I:!i f!Ii!i Economic Analysis Taking Income Taxes into Account 381 For a before-tax rate of return, we must first compute the before-tax cash flow. Year Before- Tax Cash Flow o 1 2 3 4 -$3000 +800 +WO +800 +800 +800 +750 { 5 Solve for the before-tax rate of return, IRRBT: = 800(P / A, i, 5) + 750(P / F, i, 5) 3000 Try i = 15%: 3000 ~ 800(3.352) + 750(0.4972) ~ 2682 + 373 = 3055 Sincei is slightlylow,try i = 18%: 3000 ~ 800(3.127) + 750(0.4371) :fl= 2502 + 328 = 2830 3055 - 3000 IRRBT = 15%+ 3% = 15%+ 3%(0.15)= 15.7% 3055 - 2830 ( . SOUJTI()N,tOEXAM~lE12~5(b) ; ) . . -. . .' -, -. -,. ~.' , .-' "". > , -, For an after-tax rate of return, we must set up the cash flow table (Table 12-3). The starting point is the before-tax cash flow. Then we will need the depreciation schedule for the truck: .. .. B- S Stnnght-ltne depreCIatIOn= Taxable ;;;- = 3000 - 750 = $450 per year ~ income is the before-tax cash flow minus depreciation. For this medium-sizedpro:(itable corporation, let's assume an incremental federal income tax rate of 34%. Therefore income taxes are 34% of taxable income. Finally, the after-tax cash flow equals the before-tax cash flow minus income taxes. These data are used to compute Table 12-3. The after-tax cash flow may be solved to find the after-tax rate of return, IRRAT.Try i = 10%: 3000 :fl=681(P I!I' i: I, ii i / A, 10%,5) + 750(P/ F, 10%,5) ~ 681(3.791) + 750(0.6209) :- 3047 Since I i is slightly low, try i= 12%: 3000 ~ 681(3:605)+150(0.5674)= 3047 ( - 3000 11III _.__ 2881 ., ) IRRAT= 10% + 2% 3047 .. - 2~81 .. .' ~1O.6% - - -- =1 , I. iI. --- ---- ]1 ~ 382 INCOME TAXES JABLE 12-3 Cash Flow Table for Example 12-5 Year (a) Before-Tax Cash Flow 0 1 2 3 4 -$3000 800 800 800 800 5 750 {800 (b) Straight-Line Depreciation (c) A(Taxable Income) (a) - (b) (d) 34% Income Taxes -0.34 (c)* $450 450 450 450 $350 350 350 350 -$119 -119 -119 -119 450 350 -119 (e) After-Tax Cash Flow (a) + (d)t -$3000 681 681 681 681 { 750 681 *Sign convention for income taxes: a minus (-) represents a disbursement of money to pay income taxes; a plus (+) represents the receipt of money by a.decrease in the tax liability. tThe after-tax cash flow is the before-tax cash flow minus income taxes. Based on the income tax sign convention, this is accomplished by adding columns (a) and (d). The calculations required to compute the after-tax rate of return in Example 12-5 were certainly more elaborate than those for the before-taxrate of return. It must be emphasized, however, that often the after-tax rate of return is the key value, since income taxes are a major disbursement that cannot be ignored. An analysis of a firm's sales activitiesindicates that a number of profitable sales are lost each year because the firm cannot deliver some of its products quickly enough. By investing an additional $20,000 in inventoryit is believedthat the firm will realize $1000 more in before-taxprofits in the first year. In the second year, before-tax extra profit will be $1500. Profits for subsequent years are expected to continue to increase on a $500-per-yeargradient. The investmentin the additional inventory may be recovered at the end of a 4-year analysis period simply by selling it and not replenishing the inventory.Compute: (a) The before-tax rate of return. (b) The after-tax rate of return assuming an incremental tax rate of 39%. Inventory is not considered to be a depreciable asset. Therefore, the investment in additional Itlinventory is not depreciated. The cash flow table for the problem is presented in Table 12-4.CI! ~ --- Economic Analysis Taking Income Taxes into Account 383 TABLE 12-4 Cash Flow Table for Example 12-6 Year (a) Before-Tax CashFlow 0 1 2 3 -$20,000 1,000 1,500 2,000 4 (c) A(Taxable Income) (a)- (b) (d) 39% Income Taxes - $1000 1500 2000 -$390 -585 -780 - 2500 -975 (b) Depreciation { 20,000 2,500 (c) x (e) After-Tax CashFlow -0.39 (a) + (d) -$20,000 610 915 1,220 { 20,000 1,525 ..~(t.O . .EXAMPlE12=6(a) SOlW :TlO .. . . .' ~. . ..,..' . Use the following equation to calculate the before-tax rate of return: 20,000 Try i = 1000(P / A, i, 4) + 500(P / G, i, 4) + 20,OOO(PjF, i, 4) = 8%: 20,000 #=1000(3.312) + 500(4.650) + 20,000(0.7350) #=3312 + 2325 + 14,700 = 20,337 Since i is too low, try i = 10%: 20,000 #=1000(3.170) + 500(4.378) + 20,000(0.6830) #=3170 + 2189 + 13,660 = 19,019 20,337 - 20,000 Before-tax rate of return =8% + 2% 20,337 - 19,019 ( .' )= 8.5% For a before-tax cash flow gradie!lt of $500, the resulting after-tax cash flow gradient is (1 - 0.39)(500) = $305. . 20,000 Try i = 61O(Pj A, i,4) + 305(P jG, i, 4) +.20,000(P / F, i, 4) = 5%: 20,000 *610(3.546) + 305(5.103) + 20,000(0.8227) #=20, 173 Since i is too low,try i = 6%: \ 20;000 $olO(3~465r+ 3ot>(4:4945)+ 20,000(0.7921) i: 19,304 After-tax. rate ofietnrn -,~~ ... :::;.5%, + J . r:!t. 20, 173 ( ..'%'h 20,173 ~ _ 20,000 " 04 19,3 ) ~ 5.2% iii II I d 384 INCOME TAXES CAPITAL GAINS AND LOSSES FOR NON DEPRECIATEDASSETS When a nondepreciatedcapital asset is sold or exchanged,there must be entries in the firm's accounting records to reflect the change. If the sellingprice of the capital asset exceeds the original cost basis, the excess is called a capital gain. If the selling price is less than the original cost basis, the difference is a capital loss. Examples of nondepreciated assets include stocks, land, art, and collectibles. Capital [~~:] = Selling price - Original cost basis There have been quite elaborate rules in the past for the tax treatment of capital gains and losses. For example, nondepreciated capital assets held for more than 6 months produced. long-term gains or losses. Assets held less than 6 months produced short-ierm gains and losses. Current tax law sets the net capital gains tax at 20% for most assets held by individuals more than one year. This is in 'contrast to recaptured depreciation, which is taxed at the same rate as other (ordinary) income. The tax treatment of capital gains and losses for nondepreciatedassets is shown in Table 12-5. TABLE12-5 Tax Treatment of Capital Gains and Losses For Individuals Capital gain For assets held for less than I year, taxed as ordinary income. For assets held for more than 1 year, taxed at 20% tax rate. * Capital loss Subtract capital losses from any capital gains; balance may be deducted from ordinary income, but not more than $3000 per year. Excess capital losses may be carried into future taxable years indefinitely. For Corporations Capital gain Capital loss Taxed as ordinary income. Corporations may deduct capital losses only to the extent of capital gains. Any capital loss in the current year that exceeds capital gains can be carried back 2 years, and, if not completely absorbed, is then carried forward for up to 20 years. *Depending on tax bracket, type of asset, and duration of ownership the capital gain tax rate can range from 8 to 28%. In addition, a special 0% capital gain tax rate applies to homeowners who treat their home as a primary residence for at least 2 years before selling. In such cases, single taxpayers are allowed $250,000 in tax-free gain ($500,000 for married couples). Investment Tax Credit I When the economy slows down and unemploymentrises, the U.S. government frequently alters its tax laws to promote greater industrial activity. One technique used to stimulate capital investments has been the investment tax credit. Businesses were able to deduct from 4 to 8% of their new business equipmentpurchases as a tax credit.This meant that the finn's net cost of the equipment was reduced by the amount of the investment tax credit. J I After-TaxCash Flows and Spreadsheets 385 Depending on the specific investment tax credit provisions, the credit might be subtracted ,from the basis for depreciation, or the basis for computing depreciation might remain the full cost of the equipment. The Tax Reform Act of 1986 eliminated the investment tax credit. It is likely, however, that it will reappear at some future time. ESTIMATING THE AFTER-TAX RATE OF RETURN There is no shortcut method for computing the after-tax rate of return from the before-tax rate of return. One possible exception to this statement is in the situation of nondepreciable assets. In this special case, we have After-tax rate of return = (1 - Incremental tax rate) x (Before-tax rate of return) For Example 12-6,we could estimate the after-tax rate of return from the before-taxrate of return as follows: After-tax rate ofreturn = (1 - 0.39)(8.5%)= 5.2% This value agrees with the value computed in Example 12-6(b). This relationship may be helpful for selecting a trial after-tax rate of return when the before-tax rate of return is known. It must be emphasized, however, this relationship 'is only a rough approximation in almost all situations. AFTER-TAXCASH FLOWS AND SPREADSHEETS The starting point for after-tax analysis is to calculate before-tax cash flows (BTCF) and taxable income. Before-tax cash flows may include first costs and principal payments on loans, which are not tax deductible. Taxable income deducts depreciation-which is not a cash flow.Once these principles are understood, spreadsheetscan help with the arithmetic. Taxes are computed based on taxable income, and after-tax cash flows are computed by subtracting taxes from BTCF. Realistic after-tax analysis requires spreadsheets. Even if costs and revenues are the same every year, MACRS depreciation percentages are not. The steps for calculating'an after-taxinternal rate of return are illustrated in Example 12-7.Because some cash flowsare taxed and some are not, the spreadsheetis easier to build if these two classes are separated. Spreadsheet construction is easier, as well, if recaptured depreciationor other gainlloss on disposal or sale is tabulated separately. Taxes are considered even if only the costs of a project are known. The firm that d~es an engineering project must generate profits-or go out of business. Even if a firm has an unprofitable year, the tax law includes carry forward and backward provisions to transfer deductions to profitable years. The depreciation and revenues in Example 12-7 result in a negative taxable income for Year 2. Thus the positive cashflow due to taxes that is shown in Year 2 of Example 12-7 really represents tax savings for the firm. --- 386 INCOME TAXES Return to the data of Example 12-6,where the used truck had a firstcost of $3000, a salvagevalue after 5 years of $750, and savings of $800 per year. Use MACRS depreciation and calculate the after-tax rate of return. Under MACRS, vehicles have a 5-year recovery period. Thus the MACRS depreciation can be cal-. culated by using a VDB function (see "Spreadsheets and Depreciation" at the end of Chapter 11) or by lookup in Table 11-3. The depreciation in Year 5 has been halved, since it is the year of disposal. A 1 2 3 4 5 B C D E F G H MACRS Recaptured Depreciation Tax Income Tax ATCF 3000 First Cost 800 Annual Benefit 5 Recovery Period 750 Salvage Value 0.34 Tax Rate 6 7 Year Untaxed BTCF 8 0 -3000 9 1 10 Taxed BTCF I -3000.0 t 800 600.0 2 800 960.0 11 3 800 12 4 800 13 5 800 172.8 Cum. Depr."'" 2654.4 14 .. 750 200.0 576.0 -76.2 345.6 454.4 -154.5 i 1031.6 7 7 7 15 16 17 54.4 (160.0) 224.0 404.4 / I =SaJvage / - 732.0 -68.0 I 854.4 . 723.8 645.5 .....,350.7 1199.3 11.24% IRR:::; .- ""'.}'aJl:ed .BTCF - ;MACRS + Recapt. Boo.kValUe FIGURE 12-3 Spreadsheet for after-taxJRR calculation. Note .from Figtire 12-3 that usingMACRS rather than straigl1Mine depreciation increases the after-tax 1RR from 10.6% fo 11..24%. This is due sQ!ely to the faster wriJe-offthat is allowed Ul1derMACRS. SUMMARY Since income taxes are part of most problems, no realistic economie-~ysis can ignore their consequences. Income taxes make the U.S. government a partner in every business venture. Thus the government benefits from all profitable ventures and shares in the losses of unprofitableventures. -- Problems 387 The first step in computing individual income taxes is to tabulate gross income. Any ~djustments-for example,allowabletaxpayer contributionsto a retirement fund-are subtracted to yield adjusted gross income. Personal exemptionsand either itemized deductions or the standard deduction are subtracted to find taxable income. This is used, together with a tax rate table, to compute the income tax liability for the year. For corporations,taxable income equals gross income minus all ordinary and necessary expenditures (except capital expenditures) and depreciation and depletion charges. The. income tax computation (whether for an individual or a corporation) is relatively simple, with rates ranging from 10 to 39.6%. The proper rate to use in an economic analysis is the incremental tax rate applicable to the increment of taxable income being considered. Most individuals and corporations pay state income taxes in addition to federal income taxes. Since state income taxes are an allowable deduction in computing federal taxable income,it follows that the taxable income for the federal computation is lower than the state taxable income. . Combined state and federal incremental tax rate = L\Statetaxrate + (L\Federaltax rate)(1 - L\State tax rate) Tointroduce the effect of income taxes into an economic analysis, the startingpoint is a before-tax cash flow.Then the depreciation schedule is deducted from appropriateparts of the before-taxcash flowto obtain taxable income. Income taxes are obtained by multiplying taxable income by the proper tax rate. Before-tax cash flow less income taxes equals the after-tax cash flow. Current tax law has decreased long-term capital gains on most nondepreciated assets for individuals to 20% when held for more than 1 year and provided an exclusion on the gain of the principal residence held for more than 2 years. When dealing with nondepreciable assets, there is a nominal relationship between before-tax and after-taxrate of return. It is After-tax rate of return = (1 - L\Tax rate)(Before-tax rate of return) There is no simple relationship between before-tax and after-tax rate of return in the more usual case of investmentsinvolving depreciable assets. PROBLEMS _ ~ These can be solved by hand, but most will be solved much more easily with a spreadsheet. 12-1 An unmarried taxpayer with no dependents expects an adjusted gross income of $48,000 in a given year. His nonbusiness deductions are expected to be $3400. (a) What will his federal income tax be? (b) He is considering an additional activity expected to increase his adjusted gross income. If this increase should be $16,000 and there should be no change in nonbusiness deductions or exemptions, what will be the increase in his federal income tax? 12-2 John Adams has a $50,000 adjusted gross income from Apple Corp. and allowable itemized deductions of $5000. Mary Eve has a $45,000 adjusted gross in~ come and $2000 of allowable itemized deductions. Compute the total tax they would pay as unmarried individuals. Then compute their tax as a married couple filing a joint return. (Answers: $7618.50 + 6336 = $13,954.50; $15,562.50) ~ -- - 388 INCOME TAXES 12-3 Bill Jackson worked during school and during the first 2 months of pis summer vacation. After factoring in his deductions and personal exemption as a single man, Bill found that he had a total taxable income of $1800. Bill's employer wants him to work another month during the summer, but Bill had planned to spend the month hiking. If an additional month's work would increase Bill's taxable income by $1600, how much more money would he have after paying the income tax? (Answer: $1440) 12-4 A married couple filing jointly have a combined total adjusted gross income of $75,000. They have computed that their allowable itemized deductions are $4000. Compute their federal income tax. (Answer: $10,162.50) . 12-5 Jane Shay operates a management consulting business. The business has been successful and now produces a taxable income of $65,000 per year after all "ordinary and necessary" expenses and depreciation have been deducted. At present the business is operated as a proprietorship; that is, Jane pays personal federal income tax on the entire $65,000. For tax purposes, it is as if she had ajob that pays her a $65,000 salary per year. As an alternative, Jane is considering incorporating the business. If she does, she will pay herself a salary of $22,000 a year from the corporation. The corporation will then pay taxes on the remaining $43,000 and retain the balance of the money as a corporate asset. Thus Jane's two alternatives are to operate the business as a proprietorship or as a corporation. Jane is single and has $2500 of itemized personal deductions. Which alternative will result in a smaller total payment of taxes to the government? (Answer: Incorporation, $8280 versus $11,736) 12-6 Bill Alexander and his wife Valerie are both employed. Bill will have an adjusted gross income this year of $70,000. Valerie has an adjusted gross income of $2000 a month. Bill and Valerie have agreed that Valerie should continue working only until the federal income tax on their joint income tax return becomes $11,500. On what date should Valerie quit her job? 12-7 A company wants to set up a new office in a country where the corporate tax rate is asJollows: 15% of first $50,000 profits, 25% of next $25,000, 34% of next $25,000, and 39% of everything over $100,000. Executives estimate that they will have gross revenues of $500,000, total costs of $300,000, $30,000 in allow- --- able tax deductions, and a one time business start-up credit of $8000. What is taxable income for the first year and how much should the company expect to pay in taxes? 12-8 ARKO oil company purchased two large compressors for $125,000 each. One compressor was installed in the firm's Texas refinery and is being depreciated by MACRS depreciation. The other compressor was placed in the Oklahoma refinery, where it is being de- . preciated by sum-of-years' -digits depreciation with zero salvage value. Assume the company pays federal income taxes each year and the tax rate is Constant. The corporate accounting department noted that the two compressors are being depreciated differently and wonders whether the corporation will wind up paying more income taxes over the life of the equipment as a result of this. What do you tell them? 12-9 Sole Brother Inc. is a shoe outlet to a major shoe manufacturing industry located in Chicago. Sole Brother uses accounts payable as one of its financing sources. Shoes are delivered to Sole Brother with a 3% discount if payment on the invoice is received within 10 days of delivery. By paying after the lO-day period, Sole is borrowing money and paying (giving up) the 3% discount. Although Sole Brother is not required to pay interest on delayed payments, the shoe manufacturers require that payments not be delayed beyond 45 days after the invoice date. To be sure of paying within 10 days, Sole Brothers decides to pay on the fifth day. Sole has a marginal corporate income tax of 40% (combined state and federal). By paying within the 10-day period, Sole is avoiding paying a fairly high price to retain the money owed shoe manufacturers. What would have been the effective annual after-tax interest rate? 12-10 To increase its market share, Sole Brother Inc. decided to borrow $5000 from its banker for the purchase of newspaper advertising for its shoe retail line. The loan is to be paid in four equal annual payments with 15% interest. The loan is discounted 6 points. The 6 "points" is an additional interest charge of 6% of the loan, deducted immediately. This additional interest $300 means the actual amount received 6%(5000) from the $5000 loan is $4700. The $300 additional interest may be deducted as four $75 additional annual interest payments. What is the after-tax interest rate on this loan? = 12-11 A major industrialized state has a state corporate tax rate of 9.6% of taxable income. If a corporation has ---- Problems a state taxable income of $150,000, what is the total state and f~deral income tax it must pay? Also, compute its combined incremental state and federal income tax rate. (Answers: $50,534; 43.93%) 12-12 An unmarried individual in California with a taxable income of about $80,000 has a federal incremental tax rate of 30% and a state incremental tax rate of 9.3%. What is his combined incremental tax rate? 12-13 The Lynch Bull investment company suggests that Steven Comstock, a wealthy New York City investor (his incremental income tax rate is 38.6%), consider the following investment. Buy corporate bonds on the New York Stock Exchange with a face value (par value) of $100,000 and a 5% coupon rate (the bonds pay 5% of $100,000, which equals $5000 interest per year). These bonds can be purchased at their present market value of $75,000. At the end of each year, Steve will receive the $5000 interest, and at the end of 5 years, when the bonds mature, he will receive $100,000 plus the last $5000 of interest. Steve will pay for the bonds by borrowing $50,000 at 10% interest for 5 years. The $5000 interest paid on the loan each year will equal the $5000 of interest income from the bonds. As a result Steve will have no net taxable income during the five years due to this bond purchase and borrowing money scheme. At the end of 5 years, Steve will receive $100,000 plus $5000 interest from the bonds and will repay the $50,000 loan and pay the last $5000 interest. The net result is that he will have a $25,000 capital gain; that is, he will receive $100,000 from a $75,000 investment. (Note: This situation represents an actual recommendation of a brokerage firm.) (a) Compute Steve's after-tax rate of return on this dual bond-plus-loan investment package. (b) What would be Steve's after-tax rate ofreturn if he purchased the bonds for $75,000 cash and did not borrow the $50,000? 12-14 Albert Chan decided to buy an old duplex as an investment. After looking for several months, he found a desirable duplex that could be bought for $93,000 cash. He decided that he would rent both sides of the duplex, and determined that the total expected income would be $800 per month. The total annual expenses for property taxes, repairs, gardening, and so forth are estimated at $600 per year. For tax purposes, Al plans to depreciate the building by the sum-of-years' -digits method, assuming that the building has a 20-year 389 remaining life and no salvage value. Of the total $93,000 cost of the property, $84,000 represents the value of the building and $9000 is the value of the lot. Assume that Al is in the 38% incremental income tax bracket (combined state and federal taxes) throughout the 20 years. In this analysis Al estimates that the income an.d expenses will remain constant at their present levels. If he buys and holds the property for 20 years, what after-tax rate of return can he expect to receive on his investment, using the following assumptions? (a) Al believes the building and the lot can be sold at the end of 20 years for the $9000 estimated value of the lot. (b) A more optimistic estimate of the future value of the building and the lot is that the property can be sold for $100,000 at the end of 20 years. 12-15 Zeon, a large, profitable corporation, is considering adding some automatic equipment to its production facilities. An investment of $120,000 will produce an initial annual benefit of $29,000, but the benefits are expected to decline $3000 per year, making secondyear benefits $26,000, third-year benefits $23,000, and so forth. If the firm uses sum-of-years' -digits depreciation, an 8-year useful life, and $12,000 salvage value, will it obtain the desired 6% after-tax rate of return? Assume that the equipment can be sold for its $12,000 salvage value at the end of the 8 years. Also assume a 46% income tax rate for state and federal taxes combined. 12-16 A group of businessmen formed a corporation to lease for 5 years a piece of land at the intersection of two busy streets. The corporation has invested $50,000 in car-washing equipment. They will depreciate the equipment by suin-of-years'-digits depreciation, assuming a $5000 salvage value at the end of the 5 year useful life. The corporation is expected to have a before-tax cash flow, after meeting all expenses of operation (except depreciation), of $20,000 the first year, declining $3000 per year in future years (second year = $17,000, third year = $14,000, etc.). The corporation has other income, so it is taxed at a combined corporate tax rate of 20%. If the projected income is correct, and the equipment can be sold for $50oo.at the end of 5 years, what after-tax rate of return would the corporation receive from this venture? (Answer: 14%) 12-17 The effective combined tax rate in an owner-managed corporation is 40%. An outlay of $20,000 for certain -, 390 INCOME TAXES new assets is under consideration. It is estimated that for the next 8 years, these assets will be responsible for annual receipts of $9000 and annual disbursements (other than for income taxes) of $4000. Mter this time, they will be used only for stand-by purposes, and no future excess of receipts over disbursements is estimated. (a) What is the prospective rate ofretum before income taxes? (b) What is the prospective rate of return after taxes if straight-line depreciation can be used to write off these assets for tax purposes in 8 years? (c) What is the prospective rate of return after taxes if it is assumed that these assets must be written off for tax purposes over the next 20 years, using straight-line depreciation? 12-18 A firm is considering the following investment project: Year o 1 2 3 4 5 Before-Tax Cash Flow (thousands) -$1000 +500 +340 +244 +100 +100 { + 125 Salvage value } The project has a 5-year useful life with a $125,000 salvage value, as shown. Double declining balance depreciation will be used, assuming the $125,000 , salvage value. The income, tax rate is 34%. If the firm requires a 10% after-tax rate of return, should the project be undertaken? 12-19 The Shellout Corp. owns a piece of petroleum drilling equipment that costs $100,000 and will be depreciated in 10 years by double declining balance depreciation, with conversion to straight-line depreciation at the optimal point. Assume no salvage value in the depreciation computation and a 34% tax rate. Shellout will lease the equipment to others and each year receive $30,000 in rent. At the end of 5 years, the firm will sell the equipment for $35,000. (Note that this is different from the zero-salvage-value assumption used in computing the depreciation.) What is the after-tax rate of return Shellout will receive from this equipment investment? ------- 12-20 A mining corporation purchased $120,000 of production machinery and depreciated it using SOYD depreciation, a 5-year depreciable life, and zero salvage value. The corporation is a profitable one that has a 34% incremental tax rate. At the end of 5 years the mining company changed its method of operation and sold the production machinery for $40,000. During the 5' years the machinery was used, it reduced mine operating costs by $32,000 a year, before taxes. If the company MARR is 12% after taxes, was the investment in the machinery a satisfactory one? 12-21 An automobile manufacturer is buying some special tools for $100,000. The tools are being'depreciated' by double declining balance depreciation using a 4year depreciable life and a $6250 salvage value. It is expected the tools will actually be kept in service for 6 years and then sold for $6250. The before-tax benefit of owning the tools is as follows: Before-Tax Year Cash Flow' 1 $30,000 2 30,000 3 35,000 4 40,000 5' 10,000 6 10,000 6,250 Selling price Compute the after-tax rate of return for this investment situation, assuming a 46% incremental tax rate. (Answer: 11.6%) 12-22 This is the continuation of Problem 12-21. Instead of paying $100,000 cash for the tools, the corporation will pay $20,000 ~ow and borrow the remaining $80,000. The depreciation schedule will remain unchanged. The loan will be repaid by 4 equal end-ofyear payments of $25,240. Prepare an expanded cash flow table that takes into account both the special tools and the loan. (a) Compute the after-tax rate of return for the tools, taking into account the $80,000 loan. (b) Explain why the rate of return obtained in part (a) is different from the rate of return obtained in Problem 12-21. Hints: 1. Interest on the loan is 10%, $25,240 = 80,000 (AlP, 10%,4). Each payment is made up of part interest and part principal. Interest portion for any year is 10% of balance due at the beginning of the year. Problems 2. Interest payments are tax deductible (i.e., they re,duce taxable income and thus taxes paid). Principal payments are not. Separate each $25,240 payment into interest and principal portions. 3. The Year-Ocash flow is -$20,000 (100,000 - 80,000). 4. After-tax cash flow will be before-tax cash flow - interest payment - principal payment - taxes. 12-23 A project will require the investment of $108,000 in equipment (sum-of-years' -digits depreciation with a depreciable life of 8 years and zero salvage value) and $25,000 in raw materials (not depreciable). The annual project income after all expenses except depreciation have been paid is projected to be $24,000. At the end of 8 years the project will be discontinued and the $25,000 investment in raw materials will be recovered. Assume a 34% income tax rate for this corporation. The corporation wants a 15% after-tax rate of return on its investments. Determine by present worth analysis whether this project should be undertaken. 12-24 A profitable incorporated business is considering an investment in equipment having the following beforetax cash flow. The equipment will be depreciated by double declining balance depreciation with conversion, if appropriate, to straight-line depreciation at the preferred time. For depreciation purposes a $700 salvage value at the end of 6 years is assumed. But the actual value is thought to be $1000 and it is this sum that is shown in the before-tax cash flow. Year o 1 2 3 4 5 6 Before-Tax Cash Flow $12,000 1,727 2,414 2,872 3,177 3,358 1,997 1,000Salvage value If the firm wants a 9% after-tax rate of return and its incrementalincome tax rate is 34%, determineby annual cash flow analysis whether the investmentis desirable. 12-25A salad oil bottlingplant can either purchasecaps for the glass bottles at 5 cents each or install $500,000 worth of plastic molding equipmentand manufacture 391 the caps at the plant. The manufacturing engineer estimates the material, labor, and other costs would be 3 cents per cap. (a) If 12 million caps per year are needed and the molding equipment is installed, what is the payback period? (b) The plastic molding equipment would be depre-. ciated by straight-line depreciation using a 5-year useful life and no salvage value. Assuming a 40% income tax rate, what is the after-tax payback period, and what is the after-tax rate of return? 12-26 A firm has invested $14,000 in machinery with a 7year useful life. The machinery will have no salvage value, as the cost to remove it will equal its scrap value. The uniform annual benefits from the machinery are $3600. For a 47% income tax rate, and sum-ofyears' -digits depreciation, compute the after-tax rat~ of return. 12-27 A firm manufactures padded shipping bags. One hundred bags are packed in a cardboard carton. At present, machine operators fill the cardboard cartons by eye: that is, when the cardboard carton looks full, it is assumed to contain 100 shipping bags. Actual inspection reveals that the cardboard carton may contain anywhere from 98 to 123 bags with an average quantity of 105.5 bags. The management has never received complaints from its customers about cartons containing fewer than 100 bags. Nevertheless, management realizes that they are giving away 51/2% of their output by overfilling the cartons. One solution would be to count the shipping bags to ensure that 100 are packed in each carton. Another solution would be to weigh each filled shipping carton. Underweight cartons would have additional shipping bags added, and ov~ht cartons would have some shipping bags removed. This would not be a perfect solution because the actual weight of the shipping bags varies slightly. If the weighing is done, it is believed that the average quantity of bags per carton could be reduced to 102, with almost no cartons containing fewer than 100 bags. The weighing equipment would cost $18,600. The equipment would be depreciated by straight-line depreciation using a lO-year depreciable life and a $3600 salvage value at the end of 10 years. The $18,600 worth of equipment qualifies for a 10% investment tax credit. One person, hired at a cost of $16,000 per year, would be required to operate the weighing equipment and to add or remove padded , . l-'~~h. ----- 392 - INCOME TAXES bags from the cardboard cartons. 200,000 cartons will be ,checked on the weighing equipment each year, with an average removal of 3.5 padded bags per carton with a manufacturing cost of 3 cents per bag. This large profitable corporation has a 50% combined federal-plus-state incremental tax rate. Assume a lO-year study period for the analysis and an after-tax MARR of 20%. Compute: (a) The after-tax present worth of this investment. (b) The after-tax internal rate of return of this investment. (c) The after-tax simple payback period of this investment. 12-30 A corporation with a 34% income tax rate is consider_ ing the following investment in research equipment and has projected the benefits as follows: ' 12-28 Mr. Sam K. Jones, a successful businessman, is considering erecting a small building on a commercial lot he owns very close to the center of town. A local furniture company is willing to lease the building for $9000 per year, paid at the end of each year. It is a net lease, which means the furniture company must also pay the property taxes, fire insurance, and all other annual costs. The furniture company will require a 5-year lease with an option to buy the building and land on which it stands for $125,000 at the end of the 5 years. Mr. Jones could have the building constructed for $82,000. He could sell the commercial lot now for $30,000, the same price he paid for it. Mr. Jones files a joint return and has an annual taxable income from other sources of $63,900. He would depreciate the commercial building by modified accelerated cost recovery system (MACRS) depreciation. Mr. Jones believes that at the end of the 5-year lease he could easily sell the property for $125,000. What is the after-tax present worth of this 5-year venture if Mr. Jones uses a 10% after-taxMARR? Prepare a cash flow table to determine the yearby-year after-tax cash flow assuming MACRS depreciation. (a) What is the after-tax rate.ofreturn? . (b) What is the before-tax rate of return? 12-29 One January Gerald Adair bought a small house and lot for $99,700. He estimated that $9700 of this amount represented the value of the land. He rented the house for $6500 a year during the 4 years he owned the house. Expenses for property taxes, maintenance, and so forth were $500 per year. For tax purposes the house was depreciated by MACRS depreciation (27 .5-year straight-line depreciation with a midmonth convention is used for rental property). At the end of 4 years the property was sold for $105,000. Gerald is married and works as an engineer. He estimates that his incremental state and federal combined tax rate is 24%. What after-tax rate of return did Gerald obtain on his investment in the property? Year o 1 2 3 4 5 6 Before-Tax Cash Flow -$50,000 +2,000 +8,000 +17,600 + 13,760 +5,760 +2,880 12-31 An engineer is working on the layout of a new research and experimentation facility. Two plant operators will be required. If, however, an. additional $100,000 of instrumentation and remote controls were added, the plant could be run by a single oper~tor. The total before-tax cost of each plant operator is projected to be $35,000 per year~The instrumentation and controls will be depreciated by means of the modified accelerated cost recovery system (MACRS). If this corporation (34% corporate tax rate) invests in the additional instrumentation and controls, how long will it take for the after-tax benefits to equal the $100,000 cost? In other words, what is the aftertax payback period? (Answer: 3.24 years). 12-32 A special power tool for plastic products costs $400, has a 4-year useful life, no salvage value, and a 2-year before-tax payback period. Assume uniform annual end-of-year benefits. (a) Compute the before-tax rate of return. (b) Compute the after-tax rate of return, based on MACRS depreciation and a 34% corporate income tax rate. 12-33 The Ogi Corporation, a construction company, purchased a pickup truck for $14,000 and used MACRS depreciation in the income tax return. During the time the company had the truck, they estimated that it saved $5000 a year. At the end of 4 years, Ogi sold the truck for $3000. The combined federal and state income tax rate for Ogi is 45%. Compute the after-tax rate of return for the truck. (Answer: 12.5%) -- --- Problems 12-34 A profitable wood products corporation is considering buying a parcel of land for $50,000, building a small factory building at a cost of $200,000, and equipping it with $150,000 of MACRS 5-year class machinery. If the project is undertaken, MACRS depreciation will be used. Assume the plant is put in service October 1. The before-tax net annual benefit from the project is estimated at $70,000 per year. The analysis period is to be 5 years, and planners assume the sale of the total property (land, building, and machinery) at the end of 5 years, also on October 1, for $328,000. Compute the after-tax cash flow based on a 34% income tax rate. If the corporation's criterion is a 15% after-tax rate of return, should it proceed- with the project? 12-35 A small vessel was purchased by a chemical company for $55,000 and is to be depreciated by MACRS depreciation. When its requirements changed suddenly, the chemical company leased the vessel to an oil company for 6 years at $10,000 per year. The lease also provided that the vessel could be purchased at the end of 6 years by the oil company for $35,000. At the end of the 6 years, the oil company exercised its option and bought the vessel. The chemical company has a 34% incremental tax rate. Compute its after-tax rate of return on the vessel. (Answer: 9.86%) 12-36 Xon, a small oil company, purchased a new petroleum drilling rig for $1,800,000. Xon will depreciate the drilling rig using MACRS depreciation. The drilling rig has been leased to a drilling company, which will pay Xon $450,000 per year for 8 years. At the end of 8 years the drilling rig will belong to the drilling company. If Xon has a 34% incremental tax rate and a 10% after-tax MARR, does the investment appear to be satisfactory? 12-37 The profitable Palmer Golf Cart Corp. is considering investing $300,000 in special tools for some of the plastic golf cart components. Executives of the company believe the present golf cart model will continue to be manufactured and sold for 5 years, after which a new cart design will be needed, together with a different set of special tools. The saving in manufacturing costs, owing to the special tools, is estimated to be $150,000 per year for 5 years. Assume MACRS depreciation for the special tools and a 39% income tax rate. (a) What is the after-tax payback period for this investment? 393 (b) If the company wants a 12% after-tax rate ofreturn, is this a desirable investment? 12-38 Uncle Elmo is contemplating a $10,000 investment in a methane gas generator. He estimates his gross income would be $2000 the first year and increase by $200 each year over the next 10 years. His expenses of $200 the first year would increase by $200 each year over the next 10 years. He would depreciate the generator by MACRS depreciation, assuming a 7-year property class. A lO-year-old methane generator has no market value. The income tax rate is 40%. (Remember that recaptured depreciation is taxed at the same 40% rate). (a) Construct the after-tax cash flow for the 10 year project life. (b) Determine the after-tax rate ofreturn on this investment. Uncle Elmo thinks it should be at least 8%. (c) If Uncle Elmo could sell the generator for $7000 at the end of the fifth year, would his rate of return be better than if he kept the generator for 10 years? You don't have to actually find the rate of return, Just do enough calculations to see whether it is higher than that of part (b). 12-39 Granny's Butter and Egg Business is such that she pays an effective tax rate of 40%. Granny is considering the purchase of a new Thrbo Churn for $25,000. This churn is a special handling device for food manufacture and has an estimated life of 4 years and a salvage value of $5000. The new churn is expected to increase net income by $8000 per year for each of the 4 years of use. If Granny works with an aftertax MARR of 10% and uses MACRS depreciation, should she buy the churn? 12-40 Eric Heiden has a house aDd lot for sale for $70,000. It is estimated that $10,000 is the value of the land and $60,000 is the value of the house. Bonnie Blair is purchasing the house on January 1 to rent and plans to own the house for 5 years. After 5 years, it is expected that the house and land can be sold on December 31 for $80,000. Total annual expenses (maintenance, property taxes, insurance, etc.) are expected to be $3000 per year. The house would be depreciated by MACRS depreciation using a 27.5-year straightline rate with midmonth convention for rental property. For depreciation, a salvage value of zero was used. Bonnie wants a 15% after-tax rate of return Oij her investment. You may assume that Bonnie has an incremental income tax rate of 27% in each ofthe " - - -- .., 394 INCOME TAXES 5 years. Capital gains are taxed at 20%. Detennine the following: (a) The annual depreciation. (b) The capital gain (loss) resulting from the sale of the house. (c) The annual rent Bonnie must charge to produce an after-tax rate of return of 15%. (Hint: Write an algebraic equation to solve for rent.) cars. Joe estimates that the new parking would increase his business and produce an additional beforeincome-tax profit of $7000 per year. It would cost $2500 to demolish the old building. Mr. Lang's accountant advised that both costs (the property and demolishing the old building) would be considered to comprise the total value of the land for tax purposes, and it would not be depreciable. Mr. Lang would spend an additional $3000 right away to put a light gravel surface on the lot. This expenditure, he believes, may be charged as an operating expense immediately and need not be capitalized. To compute the tax consequences of adding the parking lot, Joe estimates that his combined state and federal incre.mental income tax rate will average 40%. If Joe wants a 15% after-tax rate of return from this project, how much could he pay to purchase the adjoining land with the old building? Assume that the analysis period is 10 years and that the parking lot could always be sold to recover the costs of buying the property and demolishing the old building. (Answer: $23,100) 12-41 Bill Cavitt owns a data processing company. He plans to buy an additional computer for $20,000, use the computer for 3 years, and sell it for $10,000. He expects that use of the computer will produce a net income of $8000 per year. The combined federal and state incremental tax rate is 45%. Using MACRS depreciation, complete Table P12-41 to detennine the net present worth of the after tax cash flow using an interest rate of 12%. 12-42 Refer to Problem 12-33. To help pay for the pickup truck the Ogi Corp. obtained a $10,000 loan from the truck dealer, payable in four end-of-year payments of $2500 plus 10% interest on the loan balance each year. (a) Compute the after-tax rate of return for the truck together with the loan. Note that the interest on the loan is tax deductible, but the $2500 principal payments are not. (b) Why is the after-tax rate of return computed in part (a) so much different from the 12.5% obtained in Problem 12-33? 12-43 A store owner, Joe Lang, believes his business has suffered from the lack of adequate customer parking space. Thus, when he was offered an opportunity to buy an old building and lot next to his store, he was interested. He would demolish the old building and make off-street parking for 20 customers' Year I Cash Flow 0 -$20,000 1 +8,000 2 +8,000 3 +8,000 12-44 The management of a private hospital is considering the installation of an automatic telephone switchboard, which would replace a manual switchboard and eliminate the attendant operator's position. The class of service provided by the new equipment is estimated to be at least equal to the present method of operation. To provide telephone service, five operators will work three shifts per day, 365 days per year. Each operator earns $14,000 per year. Company-paid benefits and overhead are 25% of wages. Money co'sts 8% after income taxes. Combined federal and state income taxes are 40%. Annual property taxes and maintenance are 21;2 and 4% of investment, respectively. Depreciation is I5-year straight line. Disregarding After-Tax Cash Flow Tax (45%) Depreciation - +10,000 I I Net Present worth = -- Present Worth (12%) Problems inflation, how large an investment in the new equipment can be eQonomically justified by savings obtained by eliminating the present equipment and labor costs? The existing equipment has zero salvage value. 12-45 A contractor has to choose one of the following alternatives in performing earth-moving contracts: (a) Purchase a heavy-duty truck for $13,000. Salvage value is expected to be $3000 at the end of the vehicle's 7-year depreciable life. Maintenance is $1100 per year. Daily operating expenses are $35. (b) Hire a similar unit for $83 per day. Based on a 10% after-tax rate of return, how many days per year must the truck be used to justify its purchase? Base your calculations on straight-line depreciation and a 50% income tax rate. (Answer: 91 Y2 days) 1" ") The Able Corporationis consideringthe installation of a small electronic testing device for use in conjunction with a government contract the firm has just won. The testing device will cost $20,000, and have an estimated salvage value of $5000 in 5 years when the government contract is finished. The firm will depreciate the instrument by the sum-of-years' -digits method, using 5 years as the useful life and a $5000 salvage value. Assume that Able pays 50% federal and state corporate income taxes and uses 8% aftertax in economic analysis. What minimum equal annual benefit must Able obtain before taxes in each of the 5 years to justify purchasing the electronic testing device? (Answer: $5150) 12-47 A house and lot are for sale for $155,000. It is estimated that $45,000 is the value of the land and $110,000 is the value of the house. If purchased, the house can be rented to provide a net income of $12,000 per year after taking all expenses, except depreciation, into account. The house would be depreciated by straight line depreciation using a 27.5-year depreciable life and zero salvage value. Mary Silva. the prospective purchaser, wants a 10% after-tax rate of return on her investment after considering both annual income taxes and a capital gain when she sells the house and lot. At what price would she have to sell the house at the end of 10 years to achieve her objective? You may assume that Mary has an incremental income tax rate of 27% in each of the 10 years. 12-48 A corporation is considering buying a medium-sized computer that will eliminate a task that must be performed three shifts per day, 7 days per week, except 395 for one 8-hour shift per week when the operation is shut down for maintenance. At present four people are needed to perform the day and night tasks. Thus the computer will replace four employees. Each employee costs the company $32,000 per year ($24,000 in direct wages plus $8000 per year in other company employee costs). It will cost $18,000 pe~ year to maintain and operate the computer. The computer will be depreciated by sum-of-years' -digits depreciation using a 6-year depreciable life, at which time it will be assumed to have zero salvage value. The corporation has a combined federal and state incremental tax rate of 50%. If the firm wants a 15% rate of return, after considering both state and federal income taxes, how much can it afford to pay for the computer? 12-49 A sales engineer has the following alternatives .to consider in touring his sales territory. (a) Buy a new car for $14,500. Salvage value is expected to be about $5000 after 3 years. Maintenance and insurance cost is $1000 in the first year and increases at the rate of $500/year in subsequent years. Daily operating expenses are $50/day. (b) Rent a similar car for $80/day. Based on a 12% after-tax rate of return, how many days per year must he use the car to justify its purchase? You may assume that this sales engineer is in the 30% incremental tax bracket. Use MACRS depreciation. 12-50 A large profitable company, in the 40% federal/state tax bracket, is considering the purchase of a new piece of equipment. The new equipment will yield benefits of $10,000 in Year 1, $15,000 in Year 2, $20,000 in. Year 3, and $20,000 in Year4. The equipment is to be depreciated using 5-year MACRS depreciation starting in the year of purchase (Year 0). It is expected that the equipment will be sold at the end of fourth year at 20% of its purchase price. What is the maximum equipment purchase price the company can pay if iis after-tax MARR is 1O%? 12-51 A prosperous businessman is considering two alternative investments in bonds. In both cases the first interest payment would be received at the end of the first year. If his personal taxable income is fixed at $40,000 and he is single, which investment produces the greater after-tax rate of return? Compute the after-tax rate of return for each bond to within 4 of 1 percent. y - - - - H _ 81' 1I:"_ 'I 396 INCOME TAXES Ann Arbor Municipal Bonds: A bond with a face value of $1000 pays $60 per annum. At the end of 15 years, the bond becomes due ("matures"), at which time the owner of the bond will receive $1000 plus the final $60 annual payment. The bond may be purchased for $800. Since it is a municipal bond, the annual interest is not subject to federal income tax. The difference between what the businessman would pay for the bond ($800) and the $1000 face value he would receive at the end of 15 years must be included in taxable income when the $1000is received. Southern Coal Corporation Bonds: A thousanddollar bond yields $100 per year in annual interest payments. When the bonds mature at the end of 20 years, the bondholder will receive $1000 plus the final $100 interest. The bonds may be purchased now for $1000. The income from corporation bonds must be included in federal taxable income. 12-52 A small-business corporation is considering whether to replace some equipment in the plant. An analysis indicates there are five alternatives in addition to the do-nothing option, Alt. A. The alternatives have a 5-year useful life with no salvage value. Straight-line depreciation would be used. Alternatives A B C D E F Cost (thousands) $ 0 25 10 5 15 30 Before-Tax Uniform Annual Benefits (thousands) $0 7.5 3 1.7 5 8.7 The corporation has a combined federal and state income tax rate of 20%. If the corporation expects a 10% after-tax rate of return for any new investments, which alternative should be selected? 12-53 A corporation with $7 million in annual taxable income is considering two alternatives: Before-TaxCash Flow Year Alt. 1 Alt. 2 o 1-10 11-20 -$10,000 4,500 -$20;000 4,500 4,500 o Both alternatives will be depreciated by straight-line depreciation assuming a lO-year depreciable life and no salvage value. Neither alternative is to be replaced at the end of its useful life. If the corporation has a minimum attractive rate of return of 10% after taxes, which alternative should it choose? Solve the problem by: (a) Present worth analysis (b) Annual cash flow analysis (c) Rate of return analysis (d) Future worth analysis (e) Benefit-cost ratio analysis (f) Any other method you choose 12-54 Two mutually exclusive alternatives are being Con'sidered by a profitable corporation with an annual taxable income between $5 million and $10 million. Before-Tax Cash Flow Year Alt. A Alt. B 0 1 2 3 4 5 -$3000 1000 1000 1000 1000 1000 -$5000 1000 1200 1400 2600 2800 Both alternatives have a 5-yearuseful and depreciable life and no salvage value. Alternative A would be depreciated by sum-of-years' -digits depre,ciation, and Alt. B by straight-line depreciation. If the MARR is 10% after taxes, 'Yhich alternative should"be selected? (Answer: Alt. B) 12-55 A large profitable corporation is considering two mutually exclusive capital investments: Initial cost Uniform annual benefit End-of-depreciable-life salvage value Depreciation method End-of -useful-life salvage value obtained Depreciable life, in years Useful life, in years Alt. A $11,000 3,000 2,000 Alt. B $33,000 9,000 3,000 SL 2,000 SOYD 5,000 3 5 4 5 ~ If the firm's after-tax minimum attractive rate of re- turn is 12% and its incremental income tax rate is J .... 34%, which project should be selected?~, - -- Problems 12-56 Assume that you are debating whether to lease a car or buy one to U$e exclusively in business. The estimated mileage in both cases will be 12,000/year. Insurance costs will be the same in either case at $600 per year. You want to examine the tax impacts of the two options to see which is preferred. Use MACRS depreciation. Assume that you will have leased or purchased the car on June 30, 2003, and that your intend to keep the car for 36 months. List any assumptions that you make in your analysis. Use a 12% MARR, a 40% tax rate, and assume end of year payments. Lease option: $2250 down and $369 a month for 36 months. Purchase option: The car that you are interested in lists for $29,188 and the dealer finance plan calls for 30 monthly payments of $973 with no money down and 0% interest. Estimated value after 36 months $15,200. Remember that the tax law limits the depreciation on cars according to the following rule: - -- - ---- 397 Depreciation Limitation on Automobiles Purchased During 2003 TaxYear Amount First Second Third Fourth and later $7660 4900 2950 1775 12-57 A plant can be purchased for $1,000,000 or it can be leased for $200,000 per year. The annual income is expected to be $800,000 with the annual operating cost of $200,000. The resale value of the plant is estimated to be $400,000 at the end of its lO-year life. The company's combined federal and state income tax rate is 40%. A straight-line depreciation can be used over the 10 years with the full first-year depr~ciation rate. (a) If the company uses ~e after-tax minimum attractive rate of return of 10%, should it lease or purchase the plant? (b) What is the breakeven rate of return of purchase versus lease? 1 J I ;,~~-~... ,, ;..-~': -.,=-~-. - J;.A- :~. .~ ~ ...u...~ After Completing This Chapter... The student should be able to: · Recast an equipment reinvestment decision as a challenger versus defender analysis. · Use the replacementanalysis decision map to select the appropriate economic.analysis technique to apply. . Calculate the minimum cost life of economic challengers. ·· · · "Incorporate concepts such as repeatability assumption for replacement analysis and marginal cost datafor the defender to select the appropriate economic analysis techniques. Perform replacement problems on an after-tax basis, utilizing the defender sign change procedure when appropriate. Use spreadsheets for solving before-tax and after-taxreplacement analysis problems. QUESTIONS TO CONSIDER ;:0, 1. Using the Internet, can you determine what it would cost Intel to build a new fab from scratch? 2. Will Intel need to scrap all its current production assets when it does the upgrade? 3. How might Intel determine what to keep in operation and what to purchase new? , I j ,- :, - -- - -- -- -- ... Replacement Analysis I l The $2 Billion Upgrade In February 2003, Intel. CorPoration announced it was planning to spend $2 billion to modernize and update its silicon wafer manufacturing plant in Chandler, Arizona. The upgradewill allowIntel to manufacture300-millimeterchips,rather than the 200-:millimeter. size it had been producing at the plant, known as "Fab 12."The project is also expectedto double manufacturing capacity at the facility, while lowering overall operating costs. Upgrade work will probably take over a year to be completed. It will include remodeling the interior of the plant, and will require Intel to buy new wafer fabrication tools. Despite the seeminglyhuge price tag for the project,Intel statesthat it will be saving money by upgrading the existing fabrication facilityinstead of building a newone. The companyalso noted that, by deciding to remain in its currentlocation,it wouldbe ableto retainits currentworkforce,which is highly skilled. Moreover,the new,larger wafer will accommodatemore chips, which should lower production costs at the plant. The upgraded plant will also use a newer process that allows chips to hold smaller and fastertransistors. - . I.. ... - -- , 400 REPLACEMENTANALYSIS Up to this point in our economic analysis we have considered the evaluation and selection of new alternatives. Which new car or washing machine should we purchase? What new material handling system or ceramic grinder should we install? However,a choice between new alternatives is not always what we must consider--economic analysis weighs more frequently existing versus new facilities. For most engineers the problem is less likely to be one of building a new plant; rather the goal is more often keeping a present plant operating economically.We are not choosing between new ways to perform the desired task. Instead, we have equipmentperforming the task, and the questionis, Should the existing equipment be retained or replaced? This adversarial situation has given rise to the terms defender and challenger. The defender is the existing equipment;the challenger is the best available replacement equipment.An economicevaluationof the existingdefender and the challenger replacement is the domain of replacement analysis. THE REPLACEMENT PROBLEM The replacement of an existing asset may be appropriate in various situations including obsolescence, depletion, and deterioration due to aging. In each of these cases, the ability of a previously implemented business asset to produce a desired output is challenged.For cases of obsolescence, depletion, and aging, it may be economical to replace the existing asset. We define each of these situations. Obsolescence: Occurs when the technology of an asset is surpassed by newer and/or different technologies. Changes in technology cause subsequent changes in the market demand for older assets. As an example,today's personal computers(PCs) with more RAM, faster clock speeds, largerhard drives,and more powerfulcentral processorshave made older,lesspowerful PCs obsolete.Thus, obsolete assets may need to be replaced with newer, more technologically advanced ones. . Depletion: The gradual loss of market value of an asset as it is being consumed or exhausted. Oil wells and timber tracts are examples of such assets. In most cases the asset will be used until it is depleted, at which time a replacement asset will be obtained. Depletion was treated in Chapter 11.. . Deteriorationdue to aging: The general conditionofloss in value of some asset due to the aging process. Production machinery and other business assets that were once new eventually become aged. To compensate for a loss in functionality due to the agingprocess, additionaloperatingand maintenanceexpenses are usually incurred to maintain the asset at its operating efficiency. Aging equipment often has a greater risk of breakdowns.Planned replaceII}entscan be scheduled to minimize the time and cost of disruptions. Unplanned replacements can be very costly or even, as with an airplane engine, potentially catastrophic. In industry, as in government, expenditures are normally monitored by means of annual budgets. One important facet of a budget is the allocation of money for new capital expenditures, either new facilities or replacement and upgrading of existing facilities. - --- .. ..- ....-......... What Is the Basic Comparison? 401 Replacement analysis may, therefore, produce a recommendation that certain equipment be replaced and money for the rephicement be included in the capital expenditures budget. Even if there is no recommendation to replace the equipment at the current time, such a recommendation may be made the following year or subsequently.At some point, the existing equipmentwill bereplaced, either when it is no longer necessary or when better equipment is available.Thus, the question is not if the defender will be replaced, but whl!n it will be replaced. This leads us to the first aspect of the defender-challenger comparison: Shall we replace the defender REPLACEMENT ANALYSIS DECISION no}1l, or shall we keep it for one or more additional years? MAP Figure 13-1 is a basic decision map for conducting a replacement analysis. Identify Participants of COI11parison Defender Best Challenger Available Not Available No Find Lowest EUAC for Defender Find EUAC Over Given Life ""-0. Analysis Technique 3 Lacking defender rnarg!palcost data. compare the EUAC of defender over its remaining useful life,with the EUAC of challenger at its minimum cost life. Analysis Technique 2 Compare the lowest EUAC of the defender with the EUAC of the challenger at its minimum cosf'ijfe; Analysis Technique. I i Compare the next-year I:marginal cost of the defender with the EUAC of the challenger. FIGURE 13-1 Replacement analysis decision map. Looking at the map, we can see.there are three replacement analysis techniques that are appropriate under different circumstances.The appropriatereplacement analysis technique to use in making a replacement comparisonof old versus new asset is a function of the data available for the alternativesand how the data behave over time. . WHAT Is THE BASICCOMPARISON? By looking at the replacement analysis map, we see that the first step is to identify the basic participants of the economic comparison. Again, in replacement analysis we are interested - -- ---- ._-- --- 402 REPLACEMENTANALYSIS Mutually Exclusive Set of Challengers "' ~ ~ ~ ...'<cc,. ...~~ JJi."<,<:!t'<;.::)@;;,:;., ..c-",:-,~_:__,~,,_,_ Challenger-l "' ,. ..~,v "",,,,ctik:>;f}J~H Challenger-2 «. .2ft "h~W' ~"',._O~"-:---,-;:;.?:/:;".",.:<;. "Current Champion" Asset (previously implemented) Challenger-3 vs "Wmner" of Challengers Comparison FIGURE 13-2 Defender-challenger comparison. in comparing our previously implemented asset (the defender) against the best current available challenger. If the defender proves more economical, it will be retained. If the challenger proves more economical, it will be installed. In this comparisonthe challengerbeing evaluatedagainst a defender has been selected from a mutuallyexclusive set of competing challengers.Figure 13-2illustrates this concept in the context of a drag race between the 4~X~der and the challenger. Notice that the challenger that is competing against the defender has emerged from an earlier competition among a set of potential challengers. Any of the methods for evaluating sets of mutually exclusivealternatives,previouslydiscussed in this text, could be used to identifythe "best" challenger to race against the defender.However,it is important to note that the comparison of these potential challenger alternatives should be made at each alternative's respective minimum cost life. This concept is discussed next. Minimum Cost Life of the Challenger The minimum cost life of any new (or existing) asset is the number of years at which the equivalentuniform annual cost (EUAC)of ownershipis minimized.This minimum cost life is often shorter than either the physical or useful life of the asset due to increasing operating and maintenance costs in the later years.of asset ownership. The challenger asset selected .~--+._-- - -- --- ---- - - -. What Is the Basic Comparison? , 403 to "race" against the defender (in Figure 13-2) is the one having the lowest minimum cost life of all the competing mutually exclusivechallengers. To calculate the minimum cost life of an asset, determine the EUAC that results if the asset is kept for each possible life less than or equal to its useful life. As is illustrated in Example 13-1, the EUAC tends to be high if the asset is kept only a few years, then decreases to some minimum EUAC and increases again as the asset ages. By identifying the number of years at which the EUAC is a minimum and then keeping the asset for that number of years, we are minimi7.ingthe yearly cost of ownership.Example 13-1 illustrates how minimum cost life is calculated for a new asset. A piece of machinerycosts $7500and has no salvagevalueafter it is installed.Themanufacturer's warranty will pay the first year's maintenance and repair costs. In the second year, maintenance costs will be $900, and this item will increase on a $900 arithmetic gradient in subsequent years, Also, operating expenses for the machinery will be $500 'the first year and will increase on a $400 arithmetic gradient in the following years. If interest is 8%, compute the useful life of th~ machinery that results in a minimum EUAC. That is, find its minimum cost life. SOLUTION If Retired at the End of Year n Year, n EUACof Capital Recovery Costs: $7500(AjP, 8%, n) EUACof Maintenance and Repair Costs: $900(A/G,8%, n) $8100 4206 g 1238 1410 4779 5081 , r 1440 2425 1578 5443 $ 1305 1200 1117 1050 995 948 909 2789 3142 3484 3816 4136 4446 4746 5035 1740 1896 .5834 6239 6650 7063 7470' 7871 8265 8648 13. 14 1$1 87q '...... "" . " 5331 4644 4589+- 1661 2048 11. 121 " ." $8600 1878 1622 10 :=&i:-.~' $ 500 854 1264 9, - 0 433 EUACTotal 2910 2264 5 '- Operating Costs: $500 + $400(A/G, 8%, n) 692 880 1062 . - $ EUACof ._".. _ _ .' II; c: .~ ',' : :::11I ~ 2048 2196 2338 2476 2609 2738 =II:I!C;;::I ;; ~1Iii,!! _ '==. ;II --:- .;!I;.. =:ii~, I t .','. 11le tot.~lEUACdata are plotted in Figure 13-3. From either the tabulation gr tbefigure, we see = thattb~lminimum cost life of the machinery is 4 yeaJ."S,with a II1it:).imUJ.nEUAC of $4589£or each ==,.gtJ;hQ.~,!i1a¥e~ === -.== ;;, ==11::II' "III= 1;;1=_ II '11''11'11 ~ i::IIi: II 11;11III:==:m:>"m===- --- --- iiii 404 l REPLACEMENTANALYSIS $1'1,000 $10,000 EUAC of Capital Recovery EUAC 9fMllintenance $9,000 and Repair #EUAC ofpperating $8,000 --t- -- $,7,000 ;;r 0 U r; } Total EUAC ,,: $6,000 ~ $5,000 I $4,000 r". >, r'J!! I ~ -$3,Q90 il fl $4,OOO .. ',i $1,000 0 " id 15 " I~IGURE13-3 Plot ofcostS.fot Example 13.:1. If - a . Looking at Figure 13-3 a bit more closely, we see the effects of each of the individual cost components on total EUAC (capital recovery,maintenance/repair,and operating expense EUACs) and how they behave over time. The total EUAC curve of most assets tends to follow this concave shape-high at the beginningdue to capital recovery costs, and high at the end due to increased maintenance/repairand operating expenses.The minimum EUAC occurs somewherebetween these high points. Use of Marginal Cost Data Once the basic participantsin the defender-challengercomparisonhavebeen identified(see Figure 13-1),two specificquestionsregardingmarginalcosts mustbe answered:Do we have marginal cost datafor the defender? and Are the defender's marginal costs increasingon a year-to-year basis? Let us first define marginalcost and then discuss why is it important to answer these two questions. Marginal costs, as opposed to an EUAC, are the year-by-year costs associated with keeping an asset. Therefore, the "period" of any yearly marginal cost associated with own-. ership is always1year.Onthe other hand, anEUACcan applyto any number of consecutive years. Thus, the marginalcost of ownershipfor anyyear in the life of an asset is the cost for thatyear only. In replacementproblems, the total marginalcost for any year can include the capital recovery cost (loss in market value and lost interest for the year), yearly operating and maintenancecosts, yearly taxes and insurance,and any other expensethat occurs during that year. To calculate the yearly marginal cost of ownership of an asset it is necessary to have estimates of an asset's market value on a year-to-year basis over its useful life, as well as ordinary yearly expenses. Example 13-2illustrates how total marginal cost can be calculated for an asset. r What Is the Basic Comparison? 405 A new piece of production machinery has the following costs. Investment cost . = $25,000 Annual operating and maintenance cost = $2000 the first year and increasing at $500 per year thereafter Annual cost for risk of breakdown = $5000per year for 3ye~s, then increasing by $1500 per year thereafter = 7 years Usefullife MARR = 15%per year Calculate the marginal cost of keeping this ass~tover its useful life. SOLUTION From the problem data we call easily find tbe marginal costs for O&M and risk of breakdowns. However, to calculate the marginal capital recovery cost, we need estimates of the year-to-year market value: the prices of the production machinery would bring over its 7-year useful life. Market value estimates were made as follows: . Year I 2 3 I I t 4 I I I I .Market Value $18,000 13,000 9,000 5 6 6,000 4,000 3,000 7 2,500 I , II We can now calculate the marginal cost (year-to-year cost o,f ownership) of the production machinery over its 7-year useful life. Cost of Total O&M Breakdown Marginal Loss in Market Forgone Interest Costin Risk Cost Year,n Value in Year n in Y~ar n Year n in Year n in Yearn 1 2,5,000 - 18,000 = $7000 25,000(0.15) = $3750 $2000' $5,000 $1.7;750 2 18,000 - 13,000= 5000 18,000(0.15) == 2700 2500 5,000 15,200 3 d3,ooQ - 9,000 = 4000 13,000(0.15) = 1950 3000 5,000 13,950 4 9,000 - 6,000 = 3000 9,000(0.15) == 1350 3500 6,500 14,350 5 6,000 -'- 4,000 = 2000 6,000(0.15) = 900 4000 8,000 14,'9006 4,000- 3,000= 1000 4,000(0.15).~. 600 45QO 9,500 15,~OQ 7 3,000- 2,500== 500 3,000(0,15)' 450 5000 11,odb 16,950 'NpticeC'thflfthe fatalrn1irgiilaf cosrJdfea~ Y'eaf'.j$ mad~~~p=o!, fuss1'0.l11arketvalue, fot:golle _ intere~~,O&M cost, and cost for risk of breakdowns. AsaIIe~fu1J.ple,theYear-5.:marginalcostof $14,900 calculated as 2,000+ 900 + 4,000 +-=.8,000. '='= = = is=:!!I 101 "'= _:llt_ ~I!!I::Ii= -= "= 'll'a= ' ~ ,.-- it 406 -l REPLACEMENT ANALYSIS Do We Have Marginal Cost Data for the Defender? Our decision map indicates that it is necessary to know whether marginal cost data are available for the defender asset to detennine the appropriate replacement technique to use. Usually in engineering economic problems annual savings and expenses are given for all alternatives.However,as in Example 13-2, it is also necessary to have year-to-year salvage value estimates to calculate total marginal costs. If the total marginal costs for the defender can be calculated, and if the data are increasing on a year-to-year basis, then it is appropriate to use replacement analysis technique 1 for comparing the defender to the challenger. Are These Marginal Costs Increasing? We have seen that it is important to know whether the marginal cost for the defender is increasing on a year-to-yearbasis. This is detennined by inspecting the total marginalcost of ownershipof the defender over its remaininglife. Example 13-3illustrates the calculation of the total marginal cost for the defender asset. "An ~sset purchased 5 years agoror $75,000 Canbe sold today f9r$15,OOO.Operating expenses in the pastp.ave been $10,000 per year, but these are estimated to increase in the future by $1500 per year each year. lEis estiplated tbafthe marlcetvalue or the .old a$~etwi.1ldecrease by $1000 per year over the next 5 years. If theMARR usedpyth~ company is 15%, calculate lithetotal marginal cost of ownership of tbjs old asset (thatjs,~the defender) for eacQof the next 5 years. II II , SOLUTION __ ;;Wecalculate the total marginal cost of majntaining the qld asset for the n~xt ?-::yearperioq as ::tollows; I, Qpef~Jing ~! ~I 1 - U 2 II 3 II 4 5 _ , "-C.--';' '.-.:.- .. ',:-- _ ..' Cost .]j'orgoneInterest in:)~earn .' ,," ,'C'.".. ',:/!::-':"_:;:~ $IQ,QOO 1'1,500 J3 000 - _-J9iSO 12,OOO(0~~t5');"'-N ~180t> It, 000(0. rs},. "'~,:1J~~Q' n 'co"~, -:;~.. ',:2':: Year11 i)) Yearn 15~000(0.q)C$~250 J,A,900(Q.};DJ," 2.lPD 13,OOO(o.l~Y M8t~at Cost'in -:;,,,~:,-".~:-." "'i.;...l .:'I:!io 14~500 - ,-",-"'-,.:; 16000 ';"':';'-"~';;"i~ $13250 " .."., i4,600 15,,9SQ !T300 ~ lear, n Loss in Mar~et Value in Year-n 15,000 - 14,000$1000" 14,OQO- 13,000 1000 13,000 - 12,000 - 1000 12,000 ..,... 11,000 . tOQQ 11,000 -- 10,000' . 1000_ ',. . .'...,," .,~,"". 18656" :-~." ".,t,, ' ,,:L..., We can see that the marginal costs increase in each subsequent year of ownership. When the conditionof increasingmarginal costs for the defenderhas been met, then the defenderchallenger comparison is appropriatelymade by using replacementanalysis technique 1. -- -~- -- - -.. .----.....-.. What Is the .. - ... .-. -. .- .--- Basic Comparison? 407 Replacement Analysis Technique 1: Defender Marginal Costs Can Be Computed and Are Increasing When our firstmethodof analyzingthe defender asset againstthebest availablechallengeris used, the basic comparisoninvolvesthe marginalcost data of thedefender and the minimum cost life data of the challenger, In the case of the marginal cost of the defender increasing on a year-to-year basis, we will maintain that defender as long as the marginal cost of keeping it one more year is less than the minimum EUAC of the challenger. Thus our decision rule is as follows: .- Maintain the defender as long as the marginal cost of ownership for one more year is less than the minimum EUAC of the challenger. When the marginal cost of the defender becomes greater than the minimum EUAC of the challenger, then replace the defender with the challenger. One can see that this technique assumes that the current best challenger, with its minimum EUAC, will be available and unchanged in the future. However,it is easy to update a replacement analysiswhen marginalcosts for the defender change or when there is a change in the cost and/or performance of availablechallengers. Example 13-4illustrates the use of this technique for comparing defender and challenger assets. . I~ I Takip.gthe machinery in Example 13-2 as the challenger and the m~chineryin Example 13-3 as the liefender,use replacement analysis technique 1 to determine when, if at all, a replacement decision should be made. SOLUTION " - Replacement analysis technique 1 is appropriate only in the condition of increasing marginal'1 costs for the defender. Since these marginal costs are increasing for the defender (from, Example 13-3), we can proceed by comparing defender marginal costs against the minimum EUACof the challenger asset. In Example 13-2we calculated only the marginalcost&of the challeng~r;thus it is necessary to calculate the challenger's minimum EUAC. The EUAC of keeping this asset for each year of its useful life is worked out as follows. " c;hallenger Total Marginal EUAC of Challenger Ownership Year, 1I <;]08tip Year" If Kept 'l'hrough Year n a 1 $17,750 [17,750(P/F, 15%, 1)](A/P, 15%, 1)=$17,750 2 15,200 [17,750(P/ F, 15%,1) +15,200(P/ F, 15%, 2)](A/P, 15%,2)16,560 3 13,950 [17,756(P/F,15%,1)+... + 13,95Q(P/F,J5%,3)](A/P,IJ%,3) 15,81.0 4 14,350 [17,750(P/P, 15%, 1) + ... + 14,350(P/F,15%,4)](A./J-\15%, 4) 15,520'. 5 !i: = 14,900 -= [17,;950(P7F,15%~1) f.. ::'¥'14;900(P7F;I3"%, 5)](~/J';"i'5%,"'5r ~. "15,43U 6 " 15,600 [17,750(P/F,15%, 1) +... + 15,60Q(P/F,15%, 6)](4/11,1~%,.~ == 15.450 7 16,950 __[!2.,750Ct!! . .~1~,2~Pi~J~, .wr0~7J~ === ~.5.8 = ..' ~~ ~t ---- - 408 REPLACEMENTANALYSIS A minimum EUAGof $15,430is attainedfor the challengerat Year5, whichis the challenger's minimum cost life. Weproceed by comparingthis value against the marginalcosts of the defender from Example 13-3: Defender Total Marginal Cost. in Year n ~ ~13?250 CbatIenger M1nmuInlEtJAC $15,430 2 14,600 15,430 3 15,950 Year,n 1 COlIlpariso~ Result and Recolnnlcndanon Since $13,250 is ressthqn $15,430, keep def~nder. Since$t4,600is.less than $15,430, keep defender. Since$J5,950 is greatel; t1Jqn$15,4~O, II . ;;; II replace defender. Qi . i"1 I I a One may ask, Why can't replacementanalysis technique 1 be used when the marginai costs of the defender do not increase? To answer this question we must understand that for this technique to be valid, the following basic assumptions must be valid: the best challenger will be available "with the same minimum EUAC" at any time in the future; and the period of needed service in our business is indefinitelylong. In other words, we assume that once the decision has been made to replace, there will be an indefiilite replacement of the defender, with continuing "cycles" of the current best challenger asset. These two assumptionstogether are much like the repeatability assumptionsthat allowed us to use the annual cost method from earlier chapters to compare competing alternatives with different useful lives. Taken together, we call these the replacement repeatability assumptions. They allow us to greatly simplify the comparison of the defender and the challenger. We state these assumptions formally below. Replacement Repeatability Assumptions The two assumptions are as follows: 1. The currently available best challenger will continue to be available in subsequent years and will be unchanged in its economic costs. When the defender is ultimately replaced, it will be replaced with this challenger. Any challengers put into service will also be replaced with the same currently available challenger. 2. The period of needed service of the asset is indefinitely long. Thus the challenger asset, once put into service, will continuouslyreplace itself in repeating, unchanged cycles. .. . ... . .-. -.-.......- What Is the Basic Comparison? 409 Given that the defender will ultimately be replaced with the current best challenger,we wouldnot want to everincur a defendermarginalcost greaterthan the challenger's minimum EUAC. And because the defender's marginal costs are increasing, we can be assured that once the marginal cost of the defender has become greater than the challenger's minimum EUAC,it will continue to be so in the future. However,if the marginalcosts do not increase, we have no guaranteethat replacementanalysis technique1 will producethe alternativethat is of the greatest economic advantage.One may ask,Are there ordinaryconditions in whiah the marginalcosts are not increasing?The answer to this questionis yes. Consider the new assetin Example 13-2.This new assethas marginalcosts thatbeginat a high of $17,750,then decrease over the next years to a low of $13,950, and then increasethereafter to $16,950 in Year7. If this asset were implemented and then evaluatedone year after implementationas a defender asset, it would not have increasing marginal costs. Thus, defenders in the early stages of their respective implementations would not fit the requirements of replacement analysis technique 1. In.the situation graphedin Figure 13-3,such defender assets would be in the downward slope of a concave marginal cost curve. Example 13-5illustrates the error that can be introduced when replacementanalysis technique 1 is applied when defen~ers do not have consistently increasing marginal cost curves. ~ I t..~. -- ..,~_.'. ,,_v ... - . . ,. .~, . -, LetuslookagainatthedefenderandchallengerassetsinExample13-4.Thistimeletus arbitrarily ~ change the defender's marginal costs for its 5-year useful life. Now when, if at all, should the defender be replaced with the challenger? Year, n 1 2 3 4 5 II Defender Total Marginal Cost in Year n $16,000 14,000 13,500 15,300 17,500 SOLUTION IIIthi$case the total marginalcosts of the defender arenot consistentlyincreasingfromyear to year. :a()w~ver,ifwe ignore this fact and apply replacementanalysis technique 1, the recoIIlI)1endati9n woulg be to replace the defender now, because the marginal cost of the defender for the first year t$J6,OOO)is greater than.th,eminimUITlEUAC of the challenger ($15,430). Let usteview this d~ci$ion.One can see that' the first-year marginal cost of the defender is greater than.the miniJji1iID. aUA.C: of the c:hallenger, bittin the secondthroughfourthyearsthe marginalcoStsare less.Wus marginalcostsdecrease4e.refor3 yearsto a minimumapdthenincreasethefollowing~, = : ~ ~ = == == ;;. ;; -~~= . = a ~ ~:; = of k~epiIi15 the defender asset each of its remaining 5 years, at xit Uscalculate the EUAC 2 ye~; .. ~ ~ .. ~ . =; . i ...:.lft%." " --- - --------- - --- . 410 -t. i i j - REPLACEMENTANALYSIS Year, n 1 2 3 4 5 EUAC of Defender Ownership H Kept n Years $16,000 15,070 14,618 14,754 15,162 Challenger Minimum EUAC ='$15,430 = 16,000(P IF, 15%, I)(AI P, 15%, 1) = [16,000(P I F, 15%, 1) + 14,000(P I F, 15%, 2)](AI P, 15%,2) = [16,000(P IF, 15%, 1) + ... + 13,500(P IF, 15%, 3)](AI P, 15%,3) = [16,000(PI F, 15%,1)+... + 15,300(PI F, 15%, 4)](AI P, 15%,4) = [16,000(PI F, 15%, 1) +... + 17,5OO(PI F, 15%, 5)](AI P, 15%,5) The minimum EUAC of the defender for 3 years is $14,618, which is less than that of the challenger. But this comparison alone is not sufficient to indicate that we should keep the defender for 3 years and then replace it. In looking at a study period of 5 years, we really have six options. Option 1: :Implementthe challenger today (at $15,430/year). Option 2: Keep the defender for 1 year (at $16,000) and the challenger for 4 years (at $15,430/year). Option 3: Keep the defender for 2 years (at $15,070/year)and the challenger for 3 years (at $15,430/year). Option4: Keep the defender for 3 more years (at $14,618/year)and the challenger for 2 years (at $15,430/year). Option 5: Keep the defender for 4 years (at $14,754/year)and the challenger for 1 year (at $15,430). Option 6: Keep the defender for 5 years (at $15,162/year). The EUAC of each of these options at 15% is: Option 1: EUAC is $15,430. Option 2: EUAC is [16,000 + 15,430(P I A, 15%,4)](P I F, 15%, I)(AI P, 15%,5) = $15,578. . Option 3: EUAC is [15,070(PlA, 15%,2) + 15,430(P lA, 15%, 3)(P I F, 15%, 2)] x (AI P, 15%,~) = $15,255. Option 4: EUAC is [14,618(PlA, 15%,3) + 15,430(P I A, 15%, 2)(P IF, 15%,3)] x (AI P, 15%,5)= $14,877. Option 5: EUAC is [14, 754(P I A, 15%,4) + 15,430(P I F, 15%, 5)](AI P, 15%, 5) = $14,855. Option 6: EUAC is $15,162. I I '1 One can see that 0I?tion5 produces the nrinimum.EUAC.Thus we should keep the defen~er for 4 more years and then reevaluate the defender-'Challengerdecision. Notice that if we replace ;;; ~~the defenderno\V~witlf -- -- tfie"'C1ffi:neii@r'1>eFteplaZ'1fm~ni7lect~8n rutel:ll, w~ wlfrlid ifdr adYieft:: a minimum EUAC over this 5-year period. Notice also that the lowest cost life of 3 years for the defender is NOT we='!II!CI! sJiouldkeep the defender.In Year 5, the defender's marginal 13 how-long ~ II:; =-I: :c 11:I11 II =' ,. .. ---- - l r 4 -- l' What Is the Basic Comparison? -. 411 co~t again expeeds t}lechall~nger's minimum EUAC. We emphasize that this analysis is valid only 1?ecauseof thereplacement repeatabilityassumptions. These assumptionshold that after the 5-yeat;pepod, all altemativ~shave the exact same yearly cash flowof $15,430 (thuswe can ignore theIil)~and these identical cash flowscontillue indefinitely. Example 13-5demonstratesthat when the marginalcosts of the defenderarenot consistently increasing from year to year, it is necessary to calculate the lowest EUAC of the defender. In the next section we describe this calculation. L r Lowest EUACof the Defender How long can a defender asset be kept operating? Anyone who has seen or heard old machinery in operation, whether it is a 50-year-oldautomobileor 20-year-oldpiece of production equipment,has realized that almost any machine can be kept operating indefinitely, provided it receives proper maintenance and repair. However, even though one might be able to keep a defender going indefinitely, the cost may prove excessive. So, rather than asking what remaining operating life,the defender may have, we really want to ask what is the minimum cost life of the asset. The minimum cost life of the defender is defined as the number of years of ownership in the future that results in a minimum EUAC for the defender and the challenger combined, as in Example 13-5. However,for decision-making now, the answer to a simpler calculation is enough. Is the defender's lowest or minimum EUAC less than the challenger's minimum EUAC? An l1-year-old piece of equipment is being considered for replacement. It can be sold for $2000 now,and it is believedthat the same salvagevalue can also be obtainedin futureyears. The current maintenance cost is $500 per year and is expected to increase $100 per year in future years. If the equipment is retained in service, compute the minimum EUAC, based on 10% interest. il jt' III i l Ifill ! SOLUTION II' Here the salvage value is not expecied to decline fromits present $2000. The annual cost of this invested capital is Si = 2000(0.10) = $200. The maintenance is represented by $500 + $100G. A year-by-year computation of EUAC is as follows: Year, n 1 2 == .. Age of Equipment (years) 11 12 EUAC of Invested Capital (= Si) $200 200 ;;;;;:0.3 = ::II = :==;;:'1 3 ' ;:; ;;; : -- -. -14 4 "" 5 15 = =' ; II: EUAC of Maintenance [= 500 + 100(AIG, 10%, n)] Total EUAC $500 548 $700 748 I -- - = -_. -- - - --.. -2()0 --- -.. ----- ~:>~q- ----- -=;::;I ,9'4,- 200 200 II 638 :;; z ~8111' = II au = ill -------- ---- --- 8~8 z=~Jl I I II:!, I'" II '. p , I; 412 REPLACEMENTANALYSIS , . j FIGURE 13-4 EUAGvalues for different remaining lives. 1 ! $900 $800 iI j j ! $600 j ! i "" $500 "" o~ :! u i I I $400 j I $300 I $200 ! I " , i I EIJACOf ti $LOO 11 12 " Ii I~ 13 14 . 1'5 3 Yearn 4 5 I!! Age QfEqUipment " J '1 II).vested CapiW 1 2 II \1 The lowest EUAC is $700. That nUJ;Ilber can be compared with the challenger's best EUAC to determinewhethertbe equipmentshouldbereplacednow. "- These data are plotted "jp'Fig. 13-4. We see that the aIWualcost of continuing to u§e the equipment is jncreasing. Jt ISreasonable to assume tqat if the equipment is not replaf:ed110W, . it will be reviewed again next year. I:fthe def~llder'sand challeilgefs cost,data do not chapge, we can cOl1,1pute the'defender's marginal co~t toyxten.d seiYice ~ach year to decide wheI1;. to ~ ~ replace. Example 13-6 represents a common situation. The salvage value is stable, but the maintenance cost is increasing.The marginal cost to extendthe defender's life and its total EUAC will continue to increase as time passes, which means that the defender's lowest EUAC will be based on retainingthe defender for 1 more year. This is not always the case, as is shown in Example 13-7. ! '3:year-okirpacliine, whose Currentrpatketvalue is $5000, is beIng analyzedlo deternlin~its ecoIioIQiclife il).a reI>!ac~mentatlalysis. Compute its lowest EUi\C using a 10% mter:estf(,ite. i~.g~l\l$,-~aJWe~a~c~..£.s,Uteae..~i~Il!~~~q~:.ta~= ~;< . - -- -- -- - - ~ - ..-L_1:-'=_ ,.','__ 'II,' " Ij.':.' r What Is the Basic Comparison? 413 If Retired at Endof Year n Years of Remaining Life, n 0 1 2 3 4 5 6 7.1 8 9 10 11 Estimated Estimated Salvage Maintenance Value (5) at Cost for Year End of Year n p (P EUACof EUACof Capital Recovery Maintenance S) x (AI P, 10%, n) + Si 100(AI G, 10%, n)Total "EUAC = $5000 4000 3500 3000 2500 2000 2000 2000 2000 2000 2000 2000 .$ 0 100 200 300 400 500 600 700 800 900 1000 $1100 + 864 + 804 + 789 + 791 + 689 + 400 350 300 250 200 200 616 +- 200 562 + 521 + 488 + 462 + 200 200 200 200 $ 0 48 94 138 181 222 262 300 337 372 406 $l5,QQ 1262 1198 1177 1172 1111 1078 1062 1058+1060 1068 SOLUTION A mWimumEUACof $1,058is computed at Year 9,forthe:existing machiI.le. Looking again at Examples 13-6and 13-7,we find that the twocasesrepresent the same machine being examined at different points in its life. (In Year6 and later in Example 13-7 the costs match those of Example 13-6.)Example 13-7indicatesthatthe 5-year-oldmachine has a minimum EUAC at a life of 9 years. For older equipment with a negligible or stable salvage value, it is likely that the operating and maintenance costs are increasing. Under these circumstances,the useful life at which EUAC is a minimum is 1 year. It is necessary to calculate the defender's minimum EUAC when marginal costs are not consistently increasing. Having made this calculation, we can now use replacement analysis technique 2 to compare the defender against the challenger. Replacement Analysis Technique 2: Defender Marginal Cost Can Be Computed and Is Not Increasing Using our second method of analyzing the defender asset against the best available challenger the basic comparison involvesthe lowest EUAC of the defender and the EUAC of the challengerat its minimum cost life. We calculate the minimum EUAC of the defender and compare this directly against the minimum EUAC of the challenger. Remember that the replacement repeatabilityassumptions allow us to do this. In this comparison we then choose the alternative with !he lowest EUAC. In Example 13-5 the comparison involved the defender with the lowest EUAC = $14,618at a life of 3 years and the challenger with an economic life of 5 years and EUAC = $15,430. Here we would recommend that the defender be retained for 4 more years, as its marginal costs for later years will be increasing above the challenger's EUAC. Then, replacementanalysis technique 1 would apply. Consider Example 13-8. ~ J - .-- -.. -- ..,. --..------- I 414 .I REPLACEMENTANALYSIS An economic analysis is to be made to detennine whether existing (defender) equipment in an industrial plant shouldbe replaced. A $4000 overhaul must be done now if the equipmentis to be retained in service. Maintenanceis estimatedat $1800 in each of the next 2 years, after which it is expected to increaseby $1000each year. The defender has no present or future salvagevalue. The equipment describedin Example 13-1is the challenger.Make a replacementanalysisto determiIie whether to retain the defender or replace it by the challenger if 8% interest is used. SOLUTION The first step is to detennine the lowest EUAC of the defender. The pattern of overhaul and maintenance costs (Figure 13-5) suggests that if the overhaul is done, the equipment should be kept for several years. The computation is as follows: If Retired at End of Year n EUAC of Overhaul Year,n $4000(A/I', 8%, n) EUACof Maintenance $1800 + $1000 Gradient from Year3 on 1 2 3 4 5 $4320 2243 1552 1208 1002 $1800 1800 1800 + 308* 1800 + 683t 1800 + 1079 Total EUAC $6120 4043 3660+3691 3881 . *For the first 3' years, the maintenance is $1800, $1800, and $2800. Thus, EUAC = 1800 + 1000(Aj F, 8%, 3) = 1800 + 308. tEUAC = 1800 + loo0(P j G, 8%, 3)(P j F, 8%, 1)(Aj P, 8%,4) = 1800 + 683. FIGURE 13-5 Overhaul and maintenance costs for the defender in Example 13-8. 4800 4000 3800 2800 r 1800 1800 t t f r 0~1~2~3~4~5 Year The lowestEUACof thedefenderis $3660.In Example 13~1,we determinedthattheminimum cost life of the challenger is 4 years and that the resulting EUAC is $4589. If we assume the equipment is needed for at least 4 years, the EUACof the defender ($3660) is less than the EtJAC of the challenger ($4589). In this sifuation the defender should not be replaced yet. If the defender's and challenger's cost data do not change, we can use replacementanalysis technique 1 ,todetennine- when=the,defehder~hould b-M'epJ.a'COO. W~Kii5W; nom I"the.1hifi.ii'fihm : EUAC calculation that the defender should be kept at least 3 years. Is this the best life? The following table computes the marginal cost to answer this question. ~ .. " _ 1IftI:III!In:I = ~;;r:~ II 1111; ~ '" 1:1" '.I; ..-............. --- -- - ------- I I .... .., I ~ rn What Is the Basic Comparisori? 415 I Year,n o 1 :.2 3 4 5 r Overhaul Cost $4000 o o o o o Maintenance Cost $ 0 1800 1800 2800 3800 4800 Margin81 Cost to Extend Service $6120 = 4000(1.08) + 1800 1800 2800 3800 4800 Year 5 is the first year after the year with the lowest EUAC in which the $4800 marginal cost for the defender exceeds the $4589 minimum EUAC for the challenger.Thus, the defender should be kept 4 more years if :::; costs do -' " not change. (Note that if the defender can be overhauled again after 3 years, that might be an even better choice.) t r I - No Defender Marginal Cost Data Available Earlier we described replacement analysis techniques for comparing the defender and challenger if marginal cost data are available for the old asset. You may recall that yearly salvage value estimates were necessary to calculate these marginal costs. Sometimes,however, it is difficultto obtain end-of-the-year salvage value estimates for the defender asset. A defender based on aging technology with a shrinking market might be such an asset. Or, from a student's problem-solving perspective,perhaps salvage value data are not given as part of the problem data. Without these data, it is impossible to calculate the marginal cost of the defender,and thus impossible to use marginal cost data to compare the defender against the challenger.How should we proceed? Given the replacement repeatability assumptions, it is possible for us to calculate the EUAC of the defender over its remaining useful life. Then, knowing that both the defender and challenger will be replaced with indefinite cycles of the challenger, we may use replacementanalysis technique 3 to compare defender and challenger. Replacement Analysis Technique 3: . When Defender Marginal Cost Data Are Not Available When our third method of analyzingthe defender asset againstthe best availablechallenger is used, the basic comparison involvesthe EUAC of the defender over its stated useful life, and the minimum EUAC of the challenger. We will calculate the EUAC of the defender asset over its remaining useful life and compare this directly against the EUACof the challenger at its minimum cost life, and then choose the lesser of these two values. However,in making this basic comparison an often complicating factor is deciding what first cost to assign to the challenger and the defender assets. Defining Defender and Challenger FirstCosts Becausethe defenderis alreadyin service,analystsoftenmisunderstandwhatfirstcostto assignit. Example13-9demonstratesthisproblem. .. --! /II "'. I : -- --- ---- t.~.~ . 416 l REPLACEMENTANALYSIS A laptop word processor model SK-30 was purchased 2 years ago for $1600; it has been depreciated by straight-line depreciation using a 4-year life and zero salvage value. Because of recent innovationsin wordprocessors,the currentprice of t4e SK-30 laptophas been reduced from $1600 to $995. An officeequipment supply finn has offered a trade-in allowance of $350 for the SK-30 on new $1200 model EL-40 laptop. Some discussionrevealed that without a trade-in, the EL-40 can be purchased for $1050, indicating the originally quoted price of the EL-40 was overstated to allow a larger trade-in allowance.The true current market value of the SK-30 is probably only $200. In a replacement analysis, what value should be assigned to the SK-30 laptop? a SOLUTION I " In the example, five different dollar amounts relating to the SK-30 laptop have been outlined: 1. Original cost: The laptop cost.$1600 2 years ago. 2. Present cost: The laptop noW'.sellsfor $995. 3..Book value; The originalcost less2 yearsof depreciationis 1600-%(1600 ~O) 4. Trade-invalue.~'The offer was $350. 5. Market value: The estimate>was$200. = $800. We know that an economic analy§i§is based on the CUrrentsituation, not 0)1the past. We,refer tP past costs as sunk costs to emphasize tbat, since these costs caUJ10tbe a1tet;~d,they are not r.elevant.(There is one exception: past costs may affect present or futUreincome taxes.) In the analysis we want to use actual,cashftows for each altemativ~.a~re the questioI;lis, What value should be used in an economic analysisfor ,theSK~30?Tbe relevant costis the present market Kaluefor the equipment.Neither the ori~inalcost, thepresent cost,the book value,)lorthetrade..in )!'alueis relevant. >" rill. ..'c... , ',' . . .,' . .:"',. . .',. ,'. . ..' ", ,_.-,," At first glance, the trade-in value of an asset would appear to be a suitable present value for the equipment. Often the trade-in price is inflatedalong with the price for the new item. (This practice is so common in new-car showroomsthat the term overtrade is used to describe the excessive portion of the trade-in allowance.The purchaser is also quoted a higher price for the new car.) Distortion of the present value of the defender, or a distorted price for the challenger, can be serious because these distortions do not cancel out in an economic analysis. Example 13-9illustrated that of the severaldifferentvaluesthat can be assigned to the defender, the most appropriate is the present market value. If a trade-in value is obtained, care should be taken to ensure that it actually represents a fair market value. Determining the value for the installed cost of the challenger asset should be less difficult. In such cases the first cost is usually made up of purchase price, sales tax, installation costs, and other items that occur initially on a one-time basis owing to the selection of the challenger. These values are usually rather straightforward to obtain if a thorough analysis is conducted. One aspect to consider in assigninga first cost to the challenger is the potential disposition (or market or salvage)valueof the defender.One must not I --- ........-..-. What Is the Basic Comparison? 417 arbitrarily subtract the disposition cost of the defender from the first cost of the challenger , asset,forthispracticecan leadto an incorrectanalysis. As described in Example 13-9, the correct first cost to assign to the defender SK-30 laptop is its $200 current market value. This value represents the present economic benefit that we would be forgoing to keep the defender. This can be called our opportunityfirst cost. If, instead of assuming that this is an opportunity cost to the defender, we assume it is a cash benefit to the challenger, a potential error arises. Consider the following case involving the SK-30 and EL-40 laptops. Assume the following data: Market value Remaining life SK-30 $200 3 years First cost Useful life EL-40 $1050 3 years In this case the remaining life of the defender (SK-30) is 3 years, and so is the useful life of the challenger (EL-40). If we use an opportunitycost perspective,then the calculated capital recovery effect of first cost using an annual cost comparison is: Annualized first coStSK-30= $200(Aj P, 10%,3) = $80 Annualized first costEL-40 = $1050(Aj P, 10%, 3) = $422 The difference in annualized first cost between the SK-30 and EL-40 is: AFCEL-40 - AFCSK-30 = $422 - $80 = $342 Now use a cashflow perspectiveto look at the first costs of the defender and challenger. In this case we use the actual cash that changes hands when each alternative is selected. A first cost of zero ($0) cash would be assigned to the defender and a first cost of $850 to the challenger(-$1 050 purchaseprice of the challengerand +$200 in salvagevalue from defender). We calculate the differencedue to first cost between the SK-30 and EL-40 to be: Annualized first coStSK-30 = $O(Aj P, 10%,3) = = $0 Annualiied first costEL-40 ($1050 - 200)(Aj P, 10%,3) AFCEL-40- AFCSK-30= $342 - $0 ~ - ---- I~ l = $342 When both the remaining life of the defender and the useful life of the challenger are the same, 3 years in this case, the analysis of the first cost yields an identical (and correct) result. Both the opportunity cost and cashflow perspectivesfor considering first cost of the defender and challenger result in a difference of $342 between the two alternatives on an annual cost basis. Now see whathappens when the remaining life of the defender is Q.otequal to the useful life of the challenger.Consider the SK-30 and EL-40 word processors, but assume that the lives have been changed as follows: ' SK-30 EL-40 Usefullife Remaining life 3 years 5 years ~ L = $342 ---- -" I 1'1 f 1;1' .. ... .h , ___ _ U I 418 REPLACEMENTANALYSIS Looking at this second case from an opportunity cost perspective, we use an annual cost comparison to calculate capital recovery effect of the first cost as follows: = = $80 Annualized first costEL-40= $1050(AjP, 10%,5) = $277 Annualized first COStSK-30 $200(Aj P, 10%,3) The difference in annualized first cost between the SK-30 and EL-40 is: AFCEL-40 - AFCSK-30 = $277 - $80 = $197 Now using a cashjlow perspective to look at the first costs of the defender and challenger, we can calculate the difference due to first cost between the SK-30 and EL-40. Annualized first costSK-30 = $O(Aj P, 10%, 3) = = $0 Annualized first costEL-40 ($1050 - 200)(Aj P, 10%,5) AFCEL-40 - AFCSK-30 = $224 - $0 = = $224 $224 When the remaining life of the defender (3 years) differs from that of the useful life of the challenger (5 years), analyses of the annualized first costs yield different results. The correct differenceof$197 is shownby using the opportunitycostapproach,.and aninaccurate difference of $224 is obtained if the cashjlow perspectiveis used. In the opportunity cost case the $200 is spread out over 3 years as a cost to the defender, yet in the cash flow case the opportunity cost is spread out over 5 years as a benefit to the challenger. Spreading the $200 over 3 years in one case and 5 years in the othercase does not produceequivalent annualized amounts. Because of the differencein the livesof the assets,the annualized$200 opportunity cost for the defender cannot be called an equivalentbenefit to the challenger. In the case of unequal lives, the correct method is to assign the current market value of the defender as its time zero opportunity costs, rather than subtractingthis amount from the first cost of the challenger. Because the cash flow approach yields an incorrect value when challenger and defender have unequal lives, the opportunity cost approach for assigning a first cost to the challenger and defender assets should always be used. Repeatability Assumptions Not Acceptable Under certain circumstances,the repeatability assumptionsdescribed earlier may not apply for a replacement analysis. In these cases replacement analysis techniques 1,2, and 3 may not be validmethodsfor comparison.For instance,a decisionmaker may set the studyperiod instead of assuming that there is an indefinite need for the asset. For example, co~r the case of phasing out production after a certain number of years-perhaps a person who is about to retire is closing down a business and selling all the assets. Other examples include production equipment such'as molds and dies that are no longer needed when a new model with new shapes is introduced. The study period could potentially be set at any number of years relative to the lives of the defender and the challenger, such as equal to the life of the defender, equal to the life of the challenger, less than the life of the defender, greater than the life of the challenger, -----. ---.----.--.--- - - --. --- What Is the Basic Comparison? 419 or somewhere between the lives of the defender and challenger. The essential principle in this case is that in setting the study period, the decision maker must use an appropriate method as described in earlier chapters. The analyst must be explicit about the economic costs and benefits of the challenger that is assumed for replacement of the defender (when replacement is made), as well as residual or salvage values of the alternatives at the end of the study period. In this case the repeatabilityreplacement assumptions do not apply, and thus the replacement analysis techniques are not necessarily valid. The analysis techniques in the decision map also may not apply when future challengers are not assumed to be identical to the current best challenger.This concept is discussed in the next section. A Closer Look at Future Challengers We defined the challenger as the best available alternative to replace the defender. But in time, the best available.alternativecan change. And given the trend in our technological society, it seems likely that future challengerswill be better than the present challenger.If so, the prospect of improvedfuture challengersmay affectthe present decision between the defender and the challenger. Figure 13-6 illustrates two possible estimates of future challengers. In many technological areas it seems likely that the equivalentuniform annual costs associated with future challengers will decreaseby a constant amount each year.In other fields,however,a rapidly changing technology will produce a sudden and substantially improved challenger-with decreased costs or increased benefits. The uniform decline curve of Figure 13-6 reflects the assumption that each future challenger has a minimum EUAC that is a fixed amount less than the previous year's challenger.This assumption, of course, is only one of many possible assumptionsthat could be made regarding future challengers. If futurechallengerswillbe betterthan thepresent challenger,whatimpactwill thishave on an analysis now? The prospect of better future challengers may make it more desirable to retain the defender and to reject the present challenger. By keeping the defender for now, we may be able to replace it later by a better future challenger. Or, to state it another way, the present challenger may be made less desirable by the prospect of improved future challengers.As engineeringeconomicanalysts,we mustfamiliarizeourselveswith potential FIGURE 13-6 Two possible ways the EUAC of future challengers may decline. Present Challenger o - - 1 2 Year 3 l 420 REPLACEMENT ANALYSIS technological advancesin assets targeted for replacement. This part of the decision process is much like the search for all available alternatives,from which we select the best. Upon finding out more about what alternatives and technologies are emerging, we will be better able to understand the repercussions of investing in the current best available challeng~r. Selecting the current best challenger asset can be particularly risky when (l) the costs are very high and/or (2) the useful minimum cost life of that challenger is relatively long (5-10 years or more). When one or both of these conditions exist, it may be bette~to keep or even augment our defender asset until better future challengers emerge. There are, of course, many assumptions that could be made regarding future challengers. However, if the replacement repeatability assumptions do not hold, the analysis techniques described earlier may not be valid. AFTER-TAX REPLACEMENT ANALYSIS As describedin Chapter 12, an after-taxanalysis adds an expandedperspectiveto our problems because various very important effectscan be included.We saw earlier that eXamining' problems on an after-tax basis provides greater realism and insight. This advantage also exists when one is considering the replacement problems discussed in this chapter. Tax effects have the potential to alter recommendations made in a before-tax-only analysis. After-tax effects may influence calculations in the remaining economic life of defender, in the economic life of challenger, and in the defender-challenger comparisons discussed earlier. Consequently, one should always perform these analyses on an after-tax basis. In this sectionwe illustrate the complicatingcircumstancesthat are introduced by an after-tax analysis. Marginal Costs on an After-Tax Basis As in the before-tax case, the defender-challenger comparison is sometimes based on the marginal costs of the defender on a year-to-yearbasis. Marginal costs on an after-taxbasis represent the cost that would be incurred through ownership of the asset in each year. On an after-taxbasis we must consider the effects of ordinary taxes as well as gains and losses due to asset disposal in calculating the after-tax marginal costs. Consider Example 13-10. Refer to Example 13-2, where we calculated the before-tax marginai costs for a new piece II of production machinery. Before-:tax~ual ,cost infonp.ation was given, as well as year'"en.d " salvagevaluesfor theassetbverits usefullife.Calculatethe after-taxD;1argimilcosts of thisasset a consid~ringthe additionalinformationbelqw. II · ,. D~p.rechltion.isby ~e stfatght-line'method, with S ,$0 imdn ,5ye¥8. Ordinary"incoine:1s ta:xe1i~at""aTate~Q~~40%.:= ::~ i:.o =:~~ · Assume that all gains and IQssesare t(lxedat a tate of 28%. . ~ ,.,. " -- - &a= ... -. = r;;1:1:;;;':1 I!O:II::;; =.""" 110" The after.,tax.~MARR is 10%~ ..' - -< -= r.;,;;: - -.- ===_ -: ~_~ III. II == -. After-Tax Replacement Analysis J - . ..' 421 -.~-~-~ cS9~UTIQ~j The after-taxmarginalcost of ownershipwill involvethe followingelements:forgone gain or los§, forgone interest, annual loss in after-taxvalue, and annual operatinglmaintenanceanctinsurance. We calculate the marginal cost of each of these in Table 13.;.1. TABLE13-1 Marginal Costs of Ownership Forgone Capital Gains/Losses Gainor Loss Book Value If Asset Sold Year,n MarketValue in Yearn in Yearn in Yearn 1 2 3 4 5 6 7 $18,000 13,000 9,000 6,000 4,000 3,000 2,500 $20,000 15,000 10,000 5,000 0 0 0 -$2000 -2000 -1000 1000 4000 3000 2500 Forgone Gain or Loss If Decideto Keepin Yearn 0 560 -2000( -0.28)..:... - 2000(-0.28) = 560 280 -1000(-0.28) = 1000(-0.28) = -280 40oo(-"0.28)== -1120 30oo(-0.28) = -840 t 4 5 6 . I I "I I Marginal O&M, Insurance Costs 1 Total O&M and Insurance Cost in Year n Marginal Cost in Year n Due to O&M and Insurance $25,000 18,000 $25,000(0.10)(0.60) = $1500 18,000(0.10)(0.60)= 1080 "$7,000 7,500 7,000(0.60) = $4200 7,500(0.60):4500 J 780 540 360 240 180 8,000 10,000 12,000 14,000 16,000 8,000(0.60)= 4800 10,000(0.60) = 6000 12,000(0.60) .. 7200 , t II .7 13,000 9,000 6,000 4,000 13,000(0.lQ)(0.60) = 9,000(0.10)(0.60) = 6,000(0.10)(0.60) 4,000(0.1O)(0.6(}) 3,000(0.10)(0.60) 3,000 J Ii, I I Forgone After-Tax Interest in Year n 2 3 I 0 560 560 280 -280 -1120 -840 Market Value Beginning of Year n 1 I, Forgone Gain/Los $ Forgone Interest Year,n MarginalCostin Yearn Dueto I = == 14,000(0.60) = 16,000(0.60) - i Ii II I 8400 9600 It Lost After-Tax Value After-TaxValue at Year,n = 1 2 3 4 5 Beginning Yearn ---.- 18,'000 + $25?000 18,560 13,560 9,280 5720 !!I~' _':---11;; = ==::=; ~D! $6440 5000 4280 3560 560= $18,560 '13;000+ 560 == 9,OOQ+ 280..:... 6,000.,.- 280 == 13,560 9,28Q 5,72,0 =:=:4.0Qil""i112~ =2~~~ _ =:= 3,000 2,880 2,160 7 Loss in After-Tax Value in Year n After~Tax Value at End Year n - 2.5Qp-" 11III II:IJ!i&:I~'1:z . ::t!llll1II1ZI !!QiIl:.lil:lliiI:=: iI!!Ib 840= 2,160 700;:::: 1,800 = II: 1m~ , r " ' ! .I! !!::284Q =' 720 360 I I , J. II ~ - .- - -.- -- --- - 422 REPLACEMENTANALYSIS Total Marginal Costs Due to Ownership Year,n 1 2 3 4 5 6 7 Forgone Gain/Loss in Year n Forgone After-TaxInterest in Year n $ 0 560 560 280 280 --1120 .."..840 $1500 1080 780 540 360 240 180 ..., Marginal Cost in Year n Due to O&M Lossin After-Tax Market Value and Insurance in Yearn $4200 4500 $6440 5000 4280 3560 2840 720 360 ". 4800 6000 7200 8400. 9600 Total Marginal Cost in Yearn $12,140 11,140 10,420 10,380 10,120 8,240 9,300 f.ir l'lotice that the total marginal cost for each year is thu,chdiffereIltfrom th~ pJ.~ginalcost for the ~~eass~t on a before...t~basis. _ 1:1 '~ After-Tax Cash Flows for the Challenger Finding the after-tax cash flowsfor the challenger assetis straightforward.Here we use the standard after-tax cash flow table and method developedin Chapter 12 to incorporate all the relevant tax effects, which include before-tax cash flows,depreciation, taxes, and gains and losses at disposal. After obtaining the after-tax cash flowsfor challenger, it is a simple taskto calculatethe EUACoverthe challenger'slife. . After-Tax Cash Flows for the Defender Finding the after-tax cash flows for the defender is not straightforward.The main complicating factor is that the defender is an asset that has been previously placed in service. As such, depreciation has been taken on the cost basis (amount being depreciated) up to the present, which changes the book value of the asset. Thus, when looking at the first cost of the defender on an after-taxbasis (value at time zero), we must consider the current market (salvage) value that can be obtained for the asset, as well as any depreciation recapture or losses at disposal presently associated with the defender. We have already suggested the use of the opportunity cost perspective for assigning a first cost to both the defender and challenger. For the challenger, this first cost will be the after-tax cash flow at time zero. Assuming use of equity financing, the first cost of the challenger is the purchase price plus installation cost. However, for the defender the after-tax first cost will be made up of theforgone market value of the asset at the present time plus any forgone gains or losses associated with keeping the asset. To develop the after-tax cash flow of the defender at time zero, it is often convenient to use the d~fender sign change procedure, described next. Defender Sign Change Procedure Thedefendersignchangeprocedureallowsusto findtheafter-taxcashflow(ATCF)forthe defenderassetin after-taxdefender~hallengercomparisons.Tofindthe time zeroATCF - - - -- -- ----- r After-Tax Replacement Analysis I I 423 of the defender use the following steps: r- 1. Assume you are selling the defender now (time 0). 2. Find the ATCFfor selling the defender at time O. 3. Then, because you are actually keeping the defender, not selling it, change all the signs (plus to minus, minus to plus) in the tax table used to develop the ATCF at time O. 4. Thus the ATCFfor selling now becomes the after-tax opportunity cost for keeping the defender. Assign this cost to the defender at time O. I ~ Find the after-tax first cost (today's cost) that should be assigned to a defender asset as described by the following data: First cost when implemented 5 years ago Depreciation method = $12,500 Current market value Remaining useful life Ann1ialcosts Annual " benefits Market value after useful life Compined tax rate (for ordinary income and capital gainsllosses) = $8000 = Straight-line depreciation (with estimated __ S = $2500,n = 10 years) 10years - $3000 = $4500 . = $1500 =34% SOLUTION The defender ~ign change procedure has four steps. " Steps 1 and 2 A$sm;P.ethat we will sell the defender ndw (Time zero), then: Yearl~ depreciation - ($12,500 - $2500)/10 = $1000 year :aefore-tax cash flow = +8000 Gainsl:1osses= Current market value - Current book value Current market value - $8000 Currentbookvalue__ Cost:basis=-Sumof depreciationcb.argesto date ~I := 12,500 - 5(1000) 12,500 - 5000 - $75QO ;;;;: 8000."", '7:S00'=$500:()rdin~gain'(deprecia.tion Tax.eson gain = 500(-0.34)= -$170 After-fax ca§h flow = 8000 -- 170 __$7830 --- :'re-c8.ptlirer= -, 424 REPLACEMENTANALYSIS Thus if we sell at the outset (n Before-Tax Cash Flow +8000 Year, n o (Sell) = 0) we have: "Depreciation Taxable Income +500 Income Taxes -170 After-Tax Cash Flow +$7830 Step 3 and 4 Change signs. The after-tax cash flow figure is the time zero opportunity cost for keeping the defender. Year, n (Keep) o Before-TaX Cash Flow -8000 Depreciation TaxaJ)Je Inco~e -500 After-tax Time-O costfor the defender Income Taxes +170. After.;TaX Cash Flow -$7830 = --$7830 Determine whether the SK-3"0laptop of Example 13-9 should be replaced by the EL-40 model. In addition to the data given in Example 13-9, the following estimates have been made: ··· · The SK-30 maintenance and service contract costs $80 a year. : The EL-40 will require no maintenance. Either laptop is expected to be used for the next 5 years. At the end of that time, the SK-30 will have no value, but the EL-40 probably could be sold for $250. .. The EL-40 laptop is faster and easier to use than the SK-30 model.This benefit is ~xiJlll!tl to save about $120 a year by reducing the needfor part-time employees. I Solve this problem with a MARR equal to 8% after taxes. Both laptop computers will be depreciated by straight-line depreciation using a 4-yeafclepreciablelife. The SK-30 is already 2 years old, so only 2 years of depreciationremain. The analysis period remains at 5 years. Assume a 3.4%corporate income tax rate. I I I ". = · =... {jdfflatiVM 0 Ur first option is to keep the SK-30rat her th " . after-tax .. II . . .. cashfiow~.9ver . . . .. .. .. ."" = · " " ..-=. ~ an sell it. Use the following data to compute the the 5:'YGar studyperiod.~ III J f lIP tJ After-TaxReplacement Analysis Before-Tax Year CashFlow 0 0 1 2 3 4 5 5 $200 -200 -80 -80 80 -80 -80 o Salvage (Sell) (Keep) Taxable Income Straight-line Depreciation 34% Income Taxes -$600* -480 -480 -80 -80 -80 ot o o o After-Tax CashFlow $404 -404t +8} +83 -53 -'-53 -53 0 $204* - 204* +163 +163 +27 +27 +27 0 +600 $400 400 425 *If sold for $200, there would be a $600 loss on disposal. If $600 of gains were offset by the loss during the year, there would be no gain to tax, saving 34% x $600 ',= $204. If the SK-30 is not sold, this loss is not realized, and the resulting income taxes will be a $204 higher than if it had been sold. tThis .~sthe sum of the $200 selling"price forgone plus the $204 income tax saving forgone. tGainILoss = Market value - Book value = 0 - [1600 - (4)400] = O. EUAC = [404 - 83(P f A, 8%, 2) + 53(P fA, 8%, 3)(P f F, 8%, 2)](Af P, 8%,5) =' -I I I = $93 Alternative B Our second option is to purchase an EL-40. Use the following datato compute the after-taxcash flow. Cash Flow Straight-Line Depreciation Taxable Income 34% Income Taxes After-Tax Cash Flow 0 1 -$1050 +120 $200 ---$80 +$27 -'-$1050 +1'4-7 2 3 4 5 +120 +120 +120 +120 +250 .. Year Before-Tax ,is = Market *ss 200 200 200 0 -80 -80 -80 +120 0* value - Book value = 250 - +27 +27 +27 +147 +147 +147 -79 +250 ......41 [1050 - (4)200] 0 = O. II COqJ.put~.the EUAC: - - !"'!' -- gUAC - - - -- -r- -. -.. '. ' [1050 -147(1:'/ A, 8%,4) .,.,.(79 + 250)(P f F, 8%, 5)](Af P, 8%, 5) _- -- --.. - 08! =:; = '::;; =;:;:;: ==... = _ "" ~ ... -=,$8.J . '- . .. - ~ ..Based~4~s~~~1~~i~~..EL4Qj$=tJJ~f~qP:9~lnnati~e,,=== - - -- ==:= O!! : ~= -- - - --- -- - - -- I I il "I 426 REPLACEMENTANALYSIS Solve Example 13-12 by computing the rate of return on the difference between alternatives. In Example 13-12,the two alternativeswere "keep the SK-30" and "buy an EL-40." The difference between the alternatives would be: Rather than minus Buy an EL-40 Alternative B Keep the SK-30 Alternative A SOLUTION; The after-tax cash flow for the differencebetween the alternativesmay be computed as follows: Year A" B 0 1 2 3 4 5 -$404 +83 +83 -53 -53 -53 -$1050 +147 +147 +147 +147 +329 B-A -$646 +64 +64 +200 +200 +382 The rate of return on the differencebetween the alternativesis computed as follows: PW of cost = PW of benefit 646 = 64(P / A, i, 2) + 200(P / A, i, 2)(P / F, i, 2) + 382(P / F, i, 5) Try i = 9.%: 646 ~ 64(1.759) + 200(1.759)(0.8417) + 382(0.6499) ~ 656.9 Try i ,f = 10%: 646 ~'64(1.736) + 200(1.736)(0.8264) + 382(0.6209) ~ 635.2 656.9 - 646.0 . The rate of return IRR ( ) - 9% + (0.10- 0.09) 656.9_ 635.2 = 9.5% The rate ofreturn is greaterthan the 8% after-taxMARR.Theincrement of investmentis desirable. iBuythe EL-40 model. Example 13-13illustratestheuse ofthe incrementalIRR methodin defender-ehallenger comparisonson anafter-taxbasis.In theexamplethis analysismethodis appropriatebecause the life of the SK-30 (defender)is equivalentto the economic life of the EL-40 (challenger). - -- - - --- - - . -....-.......... .- -. . . .. ... -. T I I After-Tax Replacement Analysis 427 Since these lives are the same, a present or future worth analysis also could have been used. , However,far more commonly the lives of the defender and the challenger are different. In such cases, if we are willing to make the assumption of a continuing requirement for the asset, then the comparisonmethod to use is the annual cash flowmethod.As in our previous discussion, this method allows a direct comparisonof the annual cash flowof the defender asset over its life against the annual cash flowof the challenger over its minimum cost life. The annual cash flow method is also appropriate for comparison when the lives are equal (as in Example 13-12),but more importantly,it is appropriate for the case of lives that are different, under the replacementrepeatabilityassumptions. Minimum Cost Life Problems . In this section we illustrate the effect that tax considerations can have on the calculation of the minimum cost life of the defender and the challenger. The c~culation of minimum EUAC on an after-tax basis can be affected by both the depreciation method used and by changes in the asset's market value over time, for either defender or challenger.Using an accelerated depreciation method (like MACRS) tends to reduce the after-tax costs early in the life of an asset. This effect alters the shape of the total EUAC curve-the concave shape can be shifted and the minimum EUAC changed. Example 13-14 illustrates the effect that taxes can have when either the straight-line or MACRS depreciation method is used. Some.new production machinery has a first cost of $100,000 and a useful °lifeof 10 years. Its estimated operating and maintenance (O&M) costs are $10,000 the first year, which will increase annually by $4000. The asset's before-tax market value will be $50,000 at the end of the first year and then will-decreaseby $5000 annually.Calculate the after-tax cash flowsusing MACRS depreciation.This propertyis a 7-yearMACRSproperty.The companyuses a 6% after tax MARR and is.;'subjectto a combined federal/state tax ~ateof 40%. SOLUTION To fin~ the minimum cost life of this new production machinery we first.find the after-tax cash flow effect of the O&M costs and depreciation (Table 13-2). Then, we find the ATCFsof disposal if the equipment is sold in each of the 10 years (Table 13-3). Finally in the closing section od spreadsheets, we combine these two ATQFs (in FiguI"e1~-7) and choose.the minimum cost lif~. . I In,I'Ta1;l1~~3-2, th~ O&M expense simply starts at $10,00~ ~d increases .at $~OOOpet; year. Thed~preclatlOn entrIes equal the 7-year rt .MACRS depreclat.10n values gIVen ill Table 11-3 :::: I, ~lUl!i~ned b)' t.qe:$1OO,OPQ fir~ ~q~.~~t~~gIJl~~phts.sim~~..&Mcostsminus: the depreciation values, is then multiplied 1;Iymn;lUsthe tax rate to detenn,iqe the impact of this taX,ableincome on taxes. The O&M expense plus taxes is the. Table 13-2 portion of t!Ie '" tpta1,A'tCF. 'II ~F .- - -- " .. --- -. -- ------- II '= = II' - -- - -- - = ~ == - REPLACEMENT ANALYSIS 428 -TABLE'13-2 ATCF for' O&M and Depreciation for Example 13-14 Year, t 1 2 3 4 5 6 7 8 9 10 O&M Expense -$10,000 -14,000 -18,000 -22,000 -26,000 -30,000 -34,000 -38,000 -42,000 -46,000 MACRS Depreciation,dt Taxable Income Taxes (at 40%) $14,290 24,490 17,490 12,490 8,930 8,920' 8,930 4,460 0 0 -$24,290 -38,490 -35,490 ....34,490 -34,930 -38,920 -42,930 ....42,460 -42,000 -46,000 $ 9,716 15,396 14,196 13,196 13,972 15,568 17,172 16,984 16,800 18,400 .- O&M Depreciation ATCF -$284 1,396 . -3,804 -8,204 -12,028 -14,432 -16,828 -21,016 --25,200 -27,600 . Regarding the market value data in this problem, it should be pointed out that the initial decrease of $50,000in Year1is not Uncommon.This is especiallytrue for custOJJ,1;.built equipment for a particular and unique applicationat a specificplant. Suchequipmentwouldnot be as valuable to others in the marketplace as to the company for which it was built. Also, the $100,000 first cost (cost basis) could ha,ve-inclpde£lcosts dlle to iJ.1s1allatioJ.1, facility modifications,orreJIloval of old equipment. The $50,000is realjstic for the market value of one-year-oldequipment. The next step is to determine the ATCFsthat would occur in each possible year of disposal. (The ATCFfor year 0 is easy; it is -$100,000.) For example, as shown in Table 13-3, in Year 1 there is a $35,710 loss as the book value exceeds the market value. The t:axsavings from tlri.sloss ..areadded to the salvage (market) value to determine the ATCF (If the asset is disposed ditring Year 1). This table assumes that depreciationis taken during the year of disposal and then calculates the recaptured depreciation (gain) or loss on the book value at the end of the year. I I t i I J i I I ,TABLE13-3 ATCF in Year of Disposal for Example 13-14 Year, Market Value Gain or Loss t BooJ< Value. I I I -- 1 2 3 " 4 II 5 6 7 - ---8 II9 10 - ~ $50,000 45,000 40,000 35,000 30,000 25,000 20,000 ~:= i5,oITo,=;:I 10,OQO 5,000 $85,710 61,220 43,730 31,240 22,310 13,39Q ___4,~60 u_ -=0 o o II; -- -$35,710 -16,220 -3,730 3,760 7,690 11,610 15~540 "5;000 10,00Q 5,000' Gain/Loss Tax (at40%) $14,284 6,488 1,492 1,504 -3,076' ..".4,644 6,216 == ~ffooo'~;;;; o - .~ .o.,i,OOQ """"-"--"'~''''''''''''.,,,,,,._. _ - ".' -4A')PQ _,.'" .'-- ~ ATCFIf I Disppsedof $64,284 51,488 41,492 33,496 26,924 20;356. 13,784 '101='9;'000Q,ooO 3,000 a l Summary SPREADSHEETS AND 429 REPLACEMENT ANALYSIS Spreadsheets are obviously useful in nearly all after-tax calculations.However, they are absolutely required for optimal life calculations in after-tax situations.Because MACRS is the tax law, the after-tax cash flows are different in every year. Thus, the NPV function and the PMT function are both needed to find the minimum EUACaftertaxes. Figure 13-7 illustrates the calculation of the minimum cost life for Example 13-14. "' "'. . A--'~" ~ _ t. .2. j Year <4 0 5 1 6 2 '7.. i..8. . 9 3 4 5 .10 6 .11 ,)2" 13 '9 . ~t1 .<-~lr-'-"C . -<Jl Table 13-2 Table 13-3 O&M & Depr. if disposed of ATCF ATCF '. J'i- .;E--- ::".'~.:\~.S; JL .<: " 6% InterestRate NPV EAC -$100,000 -$284 . $64,284 1,396 -39,623 $42,000 =PMT($D$I,A5,D5) 51,488 -53,201 29,018 -3,804 -8,204 -12,028 41,492 33,496 26,924 -67,382 -82,186 -97,587 25,208 23,718 23,167 -14,432 20,356 -113,530 23,088 optimallife 7 -16,828 13,784 -129,904 23,270 8 -21,016 -25,200 9,000 6,000 -146,610 -163,621 23,610 24,056 10 -27,600 15 16 17 3,oo.g..~-180,909 ____ . 24,580 =PMT($D$I,A14,D14) :~ ,~I =NPV($D$I,$B$5:BI4)+$I $4+PV($D$I,AI4,O,-C14) =NPV(i, B column) + year ( + presentvalueofa futuresalvage FIGURE 13-7 Spreadsheet for life with minimum after-tax cost. I I if In Figure 13-7, the NPV finds the present worth of the irregular cash flows from Period 1 through Period t for t = 1 to life. The PV function is used to find the PW of the salvage value. Then PMT can be used to find the EUAC over each potential life. Before-taxreplacement analysis can also be done this way. The spreadsheetblock function NPV is used to find the PW of cash flows from Period 1 to Period t. Note that the cell for Period 1 is an absolute address and the cell for period t is a relative address. This allows the fonnula to be copied. ,. UMMARY !II In selecting equipment for a new plant the question is, which of the machines available on the market will be more economical? But when a given piece of equipment is now performing the desired task, the analysis is more complicated. The existing equipment . (called the defender) is already in place, so the question is, shall we replace it now, or shall we keep it for one or more years? When a replacement is indicated, it will be by the best available replacement equipment (called the challenger). When we already have equipment, there may be a tendency to use past costs in the replacementanalysis. But only present and future costs are relevant. -- --- - I 430 REPLACEMENT ANALYSIS I J I , This chapter has presented three distinctly different replacement analysis techniques which are all relevant and appropriate depending upon the conditions of the cash flows for the defenderand the challenger.In all cases of analysisthe simplifyingreplace. . ment repeatability assumptions are accepted. These state that the defender will ulti- matelybe replacedby the currentbest challenger(as will any challengersimplemented. . in the future), and that we have an indefinite need for the service of the asset in question. In the usual case, marginalcost data are both availableand increasingon a year-to-year basis, and thus replacement analysis technique 1 allows a comparison of the marginal cost data of the defender, against the minimum EUAC of the challenger.In this case we should keep the defender as long as its marginal cost is less than the minimum EUAC of the challenger. When marginal cost data are available for the defender but are not increasing on a year-to-year basis, 'replacement analysis technique 2 calls for a comparison of the lowest EUAC of the defender against the minimum EUAC of the challenger. If the defender's lowest EUAC is smaller,we do not replace it yet. If the challenger's EUAC is less, we would select this asset in place of the defender today. If the cost "datafor the challenger and the defender do not change, we will replace the defender after the life that minimizes its EUAC when its marginal cost data exceed the minimum EUAC for the challenger. In the case of no marginal cost data available for the defender, replacement analysis technique 3 prescribesa comparisonof the EUACof thedefender over itsstated life, against the minimum EUAC of the challenger. As in the case of replacementanalysis technique2, we would select the alternativethat has the smallestEUAC.An important conceptwhen calculatingthe EUACof both defender and challenger is the first cost to be assigned to each alternative for calculation purposes. When the lives of the two alternativesare equivalent,either an opportunity cost or a cash flow approach may be used. However,in the more common case of different useful lives, only the opportunity cost approach accurately assigns an investment cost to the defender and challenger assets. It is important when performing engineering economic analyses to include the effects of taxes. In minimum economic life and marginal cost calculations, and in finding cash flows over the life of a defender or challenger, the effects of taxes can be significant. The replacement analysis techniques described on a before-taxbasis are also used for the aftertax case-the differencebeing that after-tax cash flowsare used in place of before-tax cash flows.Effects on an after-taxbasis include opportunitygains and losses at time zero, income taxes and depreciationover the assets' lives, and gains and losses at disposition time. The sign-changeprocedure can be used to determine opportunitygains and losses when one is assigning a first cost to the defender in after-taxproblems. Replacement analyses are vastly important, yet often ignored by companies as they invest in equipment and facilities. Investmentsin business and personal assets s~ould not be forgotten once an initial economic evaluationhas produced a "buy" recommendation. It is important to continue to evaluate assets over their respective life cycles to ensure that invested monies are continuing to yield the greatest benefit for the investor. Replacement analyses help us to ensure this. --- -- --- ...- T II Problems ~ 431 t f PROBLEMS F .-, i l i- f It ' Typically there are two alternatives in a replacement analysis. One alternative is to replace the defender now. The other alternative is which one of the following? (a) Keep the defender for its remaining useful life. (b) Keep the defender for another year and then reexamine the situation. (c) Keep the defender until there is an improved challenger that is better than the present challenger. (d) The answer to this question depends on the data available for the defender and challenger as well as the assumptions made regarding the period of needed service and future challengers. 13-2 The economic life of the defender can be obtained if certain estimates about the defender can be made. Assuming those estimates prove to be exactly correct, one can accurately predict the year when the defender should be replaced, even if nothing is known about the challenger. Is this assumption true or false? Explain. 13-1 13-3 A proposal has been made to replace a large heat ex- changer (3 years ago, the initial cost was $85,000) with a new, more efficient unit at a cost of $120,000. The existing heat exchanger is being depreciated by the MACRS method. Its present book value is $20,400, but it has no current value, since its scrap value just equals the cost to remove it from the plant. In preparing the before-tax economic analysis to see whether the existing heat exchanger should be replaced, the proper treatment of the $20,400 book value becomes an issue. Three possibilities are that the $20,400 book value of the old heat exchanger is: (a) Added to the cost of the new exchanger in the economic analysis. (b) Subtracted from the cost of the new exchanger in the economic analysis. (c) Ignored in this before-tax economic analysis. Which of the three possibilities is correct? 13-4 A machine tool, which has been used in a plant for 10 years, is being considered for replacement. It cost $9500 and was depreciated by MACRS depreciation using a 5-year recovery period. An equipment dealer indicates that the machine has no resale value. Maintenance on the machin~ tool has been a problem, with an $800 cost this year. Future annual mamtenance costs are expected to be higher. What is the economic life of this machine tool if it is kept in service? 13-5 A new $40,000bottlingmachinehas just been installed in a plant. It will have no salvage value when it is removed. The plant manager has asked you to estimate the economic service life for the machine, ignoring income taxes. He estimates that the annuai maintenance cost will be constant at $2500 per year. What service life will result in the lowest equivalent uniform annual cost? 13-6 Which one of the following is the proper dollar value of defender equipment to use in replacement analysis? (a) Original cost. (b) Present market value. (c) Present trade-in value. (d) Present book value. (e) Present replacement cost, if different from original cost. 13-7 You have the following options for a major equipment unit: (a) Buy new. (b) Trade in and buy a similar, rebuilt equipment from the manufacturer. (c) Have the manufacturer rebuild your equipment with all new available options. (d) Have the manufacturer rebuild your equipment to the original specifications. (e) Buy used equipment. State advantages and disadvantages of each option with respect to after-tax benefits. 13-8 Considerfollowingdatafor a defenderasset.What is the appropriate replacement analysis technique to - compare this asset against a:competing challenger? How is this method used? That is, what comparison is made, and how do we choose? BTCF in Year n Year, n (marginal costs) 1 -$2000 2 -1750 3 -1500 4 -1250 5 -1000 6 -1000 7 -1000 8 -1500 9 -2000 10 -3000 .. ~I . ..~ ------ - - - -- I... . 432 ... ..- -- ... .--- I REPLACEMENTANALYSIS 13-9 The Clap Chemical Company needs a large insulated stainless s~eeltank.for the expansion of its plant. Clap has located such a tank. at a recently closed brewery. The brewery has offered to sell the tank. for $15,000 delivered to the chemical plant. The price is so low that Clap believes it can sell the tank. at any future time and recover its $15,000 investment. The outside of the tank.is covered with heavy insulation that requires considerable maintenance with estimated costs as follows: Year o 1 2 3 4 5' Insulation Maintenance Cost $2000 500 1000 1500 2000 2500 13-10 The plant manager has just purchased a piece of unusual machinery for $10,000. Its resale value at the end of 1 year is estimated to be $3000, because the device is sought by antique collectors, resale value is rising at the rate of $500 per year. The maintenance cost is expected to be $300 per year for each of the first 3 years, and then it is expected to double each year after that. Thus the fourth-year maintenance will be $600; the fifth-year maintenance, $1200, and so on. Based on a 15% before-tax MARR, what life of this machinery has the lowest EUAC? 13-11 In a replacement analysis problem, the following facts are known: $12,000 None for the first 3 years $2000 at the end of the fourth year $2000 at the end of the fifth year Increasing $2500 per year after the fifth year ($4500 at the end of the sixth year, $7000 at the end of the seventh year, etc.) - - -- 13-12 An injection molding machine has a first cost of $1,050,000 and a salvage value of $225,000 in any year the machine is sold. The maintenance and operating cost is $235,000 with an annual gradient of $75,000. The MARR is 10%. What is the most economic life? 13-13 Bill's father read that, at the end of each year, an aut~ mobile is worth 25% less than it was at the beginning of the year. After a car is three years old, the rate of decline reduces to 15%. Maintenance and operating' costs, on the other hand, increase as the age of the car increases. Because of the manufacturer's warranty, first-year maintenance is very low. (a) Based on a 15% before-tax MARR, what life of the insulated tank.has the lowest EUAC? (b) Is it likely that the insulated tank.will be replaced by another tank. at the end of the period with the lowest EUAC? Explain. Initial cost Annual Maintenance Actual salvage value in any year is zero. Assume a 10% interest rate and ignore income taxes. Compute the life for this challenger having the lowest EUAC. (Answer: 5 years) -------- Age of Car (years) 1 2 3 4 5 6 7 Maintenance Expense $ 50 150 180 200 300 390 500 Bill decided this is a good economic analysis problem. Bill's dad wants to keep his annual cost of automobile ownership low. The car Bill's dad prefers costs $11,200 new. Should he buy a new or a used car and, if used, when would you suggest he buy it, and how long should it be kept? Give a practical, rather than a theoretical, solution. (Answer: Buy a 3-year-old car and keep it 3 years.) . 13-14 A professor of engineering economics owns a 1996 automobile. In the past 12 months, he has paid $2000 to replace the transmission, bought two new tires for $160, and installed a new tape deck for $110. He wants to keep the car for 2 more years because he invested money 3 years ago in a 5-year certificate of deposit, which is earmarked to pay for his dream machine, a red European sports car. Today the old car's engine failed. The professor has two alternatives. He can have the engine overhauled at a cost of $1800 and then most likely have to pay another $800 per year for the next 2 years for maintenance. The car will have no salvage value at that time. Alternatively, a colleague offered to make the professor a $5000 loan to buy -- -- - ... ... ... - - .. - ...-........-.. Problems another used car. He must pay the loan back in two equal installments of $2500 due at the end of Year 1 and Year 2, and at the end of the second year he must give the colleague the car. The "new" used car has an expected annual maintenance cost of $300. If the professor selects this alternative, he can sell his current vehicle to a junkyard for $1500. Interest is 5%. Using present worth analysis, which alternative should he select and why? ~ 13-15 The Ajax Corporation purchased a railroad tank car 8 years ago for $60,000. It is being depreciated by SOYD depreciation, assuming a 10-year depreciable life and a $7000 salvage value. The tank car needs to be reconditioned now at a cost of $35,000. If this is done, it is estimated the equipment will last for 1° more years and have a $10,000 salvage value at the end of the 1° years. On the other hand, the existing tank car could be sold now for $10,000 and a new tank car purchased for $85,000. The new tank car would be depreciated by MACRS depreciation. Its estimated actual salvage value would be $15,000. In addition, the new tank . car would save $7000 per year in maintenance costs, i compared to the reconditioned tank car. Based on a 15% before-tax rate of return, determine whether the existing tank car should be reconditioned or a new one purchased. (Note: The problem t statement provides more data than are needed, which is typical of real situations.) (Answer: Recondition the old tank car.) 13-16 The Quick Manufacturing Company, a large profitable corporation, is considering the replacement of a production machine tool. A new machine would cost $3700, have a 4-year useful and depreciable life, and have no salvage value. For tax. purposes, sumof-years'-digits depreciation would be used. The existing machine tool was purchased 4 years ago at a cost of $4000 and has been depreciated by straightline depreciation assuming an 8-year life and no salvage value. The tool could be sold now to a used equipment dealer for $1000 or be kept in service for another 4 years. It would then have no salvage value. The new machine tool would save about $900 per year in operating costs compared to the existing machine. Assume a 40% combined state and federal tax rate. (a) Compute the before-tax rate of return on the replacement proposal of installing the new machine rather than keeping the existing machine. 433 (b) Compute the after-tax rate of return on this replacement proposal. (Answer: (a) 12.6%) 13-17 The Plant Department of the local telephone company purchased four special pole hole diggers 8 years ago for $14,000 each. They have been in constant use to the present. Owing to an increased workload, additional machines will soon be required. Recently it was announced that an improved model of the digger has been put on the market. The new machines have a higher production rate and lower maintenance expense than the old machines, but will cost $32,000 each. The service life of the new machines is estimated to be 8 years, with salvage value estimated at $750 each. The four original diggers have an immediate salvage of $2000 each and an estimated salvage value of $500 each 8 years hence. The estimated average annual maintenance expe~e associated with the old machines is approximately $1500 each, compared with $600 each for the new machines. A field study and trial indicate that the workload would require three additional new machines if the old machines are continued in service. However, if the old machines are all retired from service, the present workload plus the estimated increased load could be carried by six new machines with an annual savings of $12,000 in operation costs. A personnel training program to prepare employees to run the machines will be necessary at an estimated cost of $700 per new machine. If the MARR is 9% before taxes, what should the company do? 13-18 Five years ago, Thomas Martin purchased and imple- mented productionmachinerythat had a first cost of . $25,000. At the time of the initial purchase it was estimated that yearly costs would be $1250, increasing by $500 in each year that followed. It was also estimated that the market value of this machinery would be only 90% of the previous year's value. It is currently projected that this machine will be useful in operations for 5 more years. There is a new machine available now that has a first cost of $27,900 and no yearly costs over its 5 year-minimum cost life. If Thomas Martin uses an 8% before-tax MARR, when, if at all, should he replace the existing machinery with the n~w unit? 13-19 Consider Problem 13-18 involving Thomas Martin. Suggest when, if at all, the old should be replaced with the new, if the values for the old machine are as follows. The old machine retains only 70% of its value -- ------- 434 I REPLACEMENTANALYSIS I in the market from year-to-year. The yearly costs of the old machine were $3000 in Year 1 and increase at 10% thereafter. 13-20 Mary O'Leary's company ships fine wool garments from County Cork, Ireland. Five years ago she purchased some new automated packing equipment having a fir~t cost of $125,000 and a MACRS class life of 7 years. The annual costs for operating, maintenance, and insurance, as well as market value data for each year of the equipment's lO-year useful life are as follows: Market AnnualCostsin Yearn for Valuein Year, n Operating Maintenance Insurance Yearn 1 2 3 4 5 6 7 8 9 10 $16,000 20,000 24,000 28,000 32,000 36,000 40,000 44,000 48,000 52,000 $ 5,000 10,000 15,000 20,000 25,000 30,000 35,000 40,000 45,000 50,000 $17,000 16,000 15,000 14,000 12,000 11,000 10,000 10,000 10,000 10,000 $80,000 78,000 76,000 74,000 72,000 70,000 68,000 66,000 64,000 62,000 Now Mary is looking at the remaining 5 years of her investment in this equipment, which she had initially evaluated on the basis of an after-tax MARR of 25% and a tax rate of35% on ordinary equipment. Assuming that the replacement repeatability assumptions are valid, answer the following questions. (a) What is the before-tax marginal cost for the remaining 5 years? (b) When, if at all, should Mary replace this packing equipm~tifa~wchall~~~withaminimum EUAC of $110,000, has been identified: Use the data from the table and the decision map from Figure 13-1. 13-21 Big-J Construction Company, Inc. (Big-J CC) is conducting routine periodic review of existing field equipment. The owner of Big-J CC has asked for a replacement evaluation of a paving machine now in use. A newer, more efficient machine is being considered. The old machine was purchased 3 years ago for $200,000, and yearly operating and maintenance costs are as follows. Big-J CC Uses a MARR of 20%; the current market value of the paver is $120,000. Estimates of Operating & Maintenance Cost and Market Value fof Next 7 Years (old paver) Year, n Operating Cost in Year n Maintenance Cost in Year n Market Value If Sold in Yearn 1 2 3 4 5 6 7 $15,000 15,000 17,000 20,000 25,000 30,000 35,000 $ 9,000 10,000 12,000 18,000 20,000 25,000 30,000 $85,000 65,000 50,000 40,000 35,000 30,000 25,000 Data for the new paving machine have been analyzed. Its most economic life is at 8 years with a minimum EUAC of $62,000.Make a recommendation to Big-J CC regding the paving machine in question. 13-22 VMIC Corp. has asked you to look at the follow41g data. The interest rate is 10%.After consideringthe data, answer questions (a)-(c). Year, n 1 2 3 4 5 6 7 Marginal Cost Data Defender EUACIf Kept n Years Challenger $2500 2400 2300 2550 2900 3400 4000 $4500 3600 3000 2600 2700 3500 4000 (a) What is the lowest EUAC of the defender? (b) What is the minimum cost life of the challenger? (c) When, if at all, should we replace the defender with the challenger? 13-23 SHOJ Enterprises has asked you to look at the following data. The interest rate is 10%. After considering the data, answer questions (a)-(c). Problems EUACIf Kept n Years Year,n Marginal CostData Defender 1 2 3 4 5 6 $3000 3150 3400 3800 4250 4950 $4500 4000 3300 4100 4400 6000 Challenger (a) What is the lowest EUAC of the defender? (b) What is the economic life of the challenger? (c) When, if at all, should we replace the defender with the challenger? 13-24 As proprietor of your own business, you are considering the option of purchasing a new high-efficiency machine to replace older machines currently in use. You believe that the new technology can be used to replace four of the older machines, each with a current market value of $600. The new machine will cost $5000 and will save the equivalent of 10,000 kW-hr of electricity per year over the older machines. After a period of 10 years neither option (new or old) will have any market value. If you use a before tax MARR of 25% and pay $0.075 per kilowatt-hour, would you replace the old machines today with the new one? 13-25 Fifteen years ago the Acme Manufacturing Company bought a propane-powered forklift truck for $4800. The company depreciated the forklift using straightline depreciation, a 12-year life, and zero salvage value. Over the years, the forklift has been a good piece of equipment, but lately the maintenance cost has risen sharply. Estimated end-of-year maintenance costs for the next 10 years are as follows: Year 1 2 3 4 5-10 Maintenance Cost $400 600 800 1000 1400/year The old forklift has no present or future net salvage value, as its scrap metal value just equals the cost to haul it away. A replacement is now being considered for the old forklift. A modem unit can be purchased for $6500. It has an economic life equal to its 10year depreciable life. Straight-line depreciation will be employed, with zero salvage value at the end of the 435 10-year depreciable life. At any time the new forklift can be sold for its book value. Maintenance on the new forklift is estimated to be a constant $50 per year for the next 10 years, after which maintenance is expected to increase sharply. Should Acme Manu_facturing keep its old forklift truck for the present, or replace it now with a new one? The firm expects an. 8% after-tax rate of return on its investments. Assume a 40% combined state-and-federal tax rate. (Answer: Keep the old forklift truck.) 13-26 A firm is concerned about the condition of some of its plant machinery. Bill James, a newly hired engineer, was assigned the task of reviewing the situation and determining what alternatives are available. After a careful analysis, Bill reports that there are five feasible, mutually exclusive alternatives. Alternative A: Spend $44,000 now repairing various items. The $44,000 can be charged as a current operating expense (rather than capitalized) and deducted from other taxable income immediately. These repairs are anticipated to keep the plant functioning for the next 7 years with operating costs remaining at present levels. Alternative B: Spend $49,000 to buy generalpurpose equipment. Depreciation would be straight line, with the depreciable life equal to the 7-year useful life of the equipment. The equipment will have no end-of-useful-life salvage value. The new equipment will reduce operating costs $6000 per year below the present level. Alternative C: Spend $56,000 to buy new specialized equipment. This equipment would be depreciated by sum-of-years' -digits depreciation over its 7-year useful life. This equipment would reduce operating costs $12,000 per year below the present level. It will have no end-ofuseful-life salvage value. Dr. ii'l Alternative D: This alternative is the same as Alternative B, except that this particular equipment would reduce operating costs $7000 per year below the present level. Alternative E: This is the "do-nothing" alternative. If nothing is done, future annual operating costs are expected to be $8000 above the present level. - -.- - -- - -- -- w- 436 l REPLACEMENT ANALYSIS I This profitable firm pays 40% corporate income taxes. InJheir economic analysis, they require a 10% after-tax rate of return. Which of the five alternatives should the firm adopt? (d) Should Fred buy the new SHSS unit if his MARR is 20%? You do not have to calculate the incremental rate of return; just show how you reach your decision. 13-27 Machine A has been completely overhauled f()r $9000 and is expected to last another 12 years. The $9000 was treated as an expense for tax purposes last year. Machine A can be sold now for $30,000 net after selling expenses, but will have no salvage value 12 years hence. It was bought new 9 years ago for $54,000 and has been depreciated since then by straight-line depreciation using a 12-year depreciable life. Because less output is now required, Machine A can now be replaced with a smaller machine: Machine B costs $42,000, has an anticipated life of 12 years, and would reduce operating costs $2500 per year. It would be depreciated by straight-line depreciation with a 12-year depreciable life and no salvage value. Both the income tax and capital gains tax rates are 40%. Compare the after-tax annual cost of the two machines and decide whether Machine A should be retained or replaced by Machine B. Use a 10% after-tax rate of return in the calculations. 13-29 (a) A new employee at CLL Engineering ConSulting 13-28 Fred's Rodent Control Corporation has been using a low-frequency sonar device to locate subterranean pests. This device was purchased 5 years ago for $18,000. The device has been depreciated using SOYD depreciation with an 8-year depreciable life and a salvage value of $3600. Presently, it could be sold to the cat next door for $7000. If it is kept for the next 3 years, its market value is expected to drop to $1600. A new lightweight subsurface heat-sensing searcher (SHSS) is available for $10,000 that would improve the annual net income by $500 for each of the next 3 years. The SHSS would be depreciated as a 5-year class property, using MACRS. At the end of 3 years, the SHSS should have a market value of $4000. Fred's Rodent Control is a profitable enterprise subject to a 40% tax rate. (a) Construct the after-tax cash flow for the old sonar unit for the next 3 years. (b) Construct the after-tax cash flow for the SHSS unit for the next 3 years. (c) Construct the after-tax cash flow for the difference between the SHSS unit and the old sonar unit for the next 3 years. - - - , , Inc., you are asked tojoin a team performingan economic analysis for a client. Your te~ seeltlS stumped on how to assign an after-tax first cost to the defender and challenger assets under consideration. Your task is to take the following data and find the ATCF for each alternative. There is no need for a complete analysis-your colleagues can handle that responsibility-they need help only with the time 0 ATCFs. CLL Inc. has a combined federal/state tax rate of 45% on ordinary income, depreciation recapture, and losses. Defender: This asset was placed in service 7 years ago. At that time the $50,000 cost basis was set up on a straight-line depreciation ' schedule with an estimated salvage value of $15,000 over its lO-year ADR life. This asset has a present market value of $30,060. Challenger: The new asset being considered has a first cost of $85,000 and will be depreciated by MACRS depreciation over its lO-year class life. This asset qualifies for'a 10% investment tax credit. (b) How would your calculations change if the' , present market value of the defender is $25,5OO? (c) How would your calculations change if the present market value of the defender is $18,0001 13-30 Foghorn Leghorn is considering the replacement of an old egg-sorting machine used with his Foggy's Farm Fresh Eggs business. The old ,egg machine is not quite running eggs-actly the way it was originally designed and will require an additional investment now of $2500 (expensed at time 0) to get it back in working shape. This old machine was purchased 6 years ago for $5000 and has been depreciated by the straight-line method at $500 per year. Six years ago the estimated salvage value for tax purposes was $1000. Operating expenses for the old machine are projected at $600 this next year and are increasing by $150 per year each year thereafter. Foggy projects that with refurbishing, machine will last another 3 years. Foggy believes that he could sell the old machine as-is today for $1000 to his friend Fido to sort bones. --- - -- - .. .. - _. . h_ __ ------- Problems He also believes he could sell it 3 years from now at the barnyard flea market for $500. The new egg-sorting machine, a deluxe model, has a purchase price of $10,000 and will last 6 years, at which time it will have a salvage value of $1000. The new machine qualifies as a MACRS 7-year property and will have operating expenses of $100 the first year, increasing by $50 per year thereafter. Foghorn uses an after-tax MARR of 18% and a tax rate of 35% on original income. (a) What was the depreciation life used with the defender asset (the old egg sorter)? (b) Calculate the after-tax cash flows for both the defender and challenger assets. (c) Use the annual cash flow method to offer a recommendation to Foggy. What assumptions did you make in this analysis? 1. I BC Junction purchased some embroidering equip- property. Maintenance 437 costs are estimated to be $1000 this next year and will increase by $1000 per year thereafter. The market (salvage) value for the equipment is $10,000 at the end of this year and declines by $1000 per year in the future. If BC Junction has an after-tax MARR of 30%, a marginal tax rate of 45% on ordinary income, and depreciation recapture an4 losses, what after-tax life of this previously purchased equipment has the lowest EUAC? Use a spreadsheet to develop your solution. ' 13-32 Reconsider the acquisition of packing equipment JJL Problem for Mary13-20. O'Leary's business, as described in Given the data tabulated there, and again using an after-tax MARR of 25% and a tax rate of 35% on ordinary income to evaluate the investment,determine the after-tax lowest EUACof the equipment. Use a spreadsheet to develop y~ur solution. JIlLment for their Denver facility 3 years ago for $15,000. This equipment qualified as MACRS 5-year !'" ,I. , , j' -.- - - ------ -------- ';~~~~~;~~ After Completing This Chapter... The student should be able to: Describe inflation, explain how it happens, and list its effects on purchasingpower. Define real and actual dollars and interest rates. Conduct constant dollar and the current dollar analyses. Define and utilize composite and commodity-specificprice indexes. · Develop and use cash flowsthat inflate at different interest rates and cash flowssubject to different interest rates per period. Incorporate the effects of inflation in before-tax and after-tax calculations.. Developspreadsheetsto solveengineeringeconomyproblemsthatincorporatethe effects of inflation and price change. ·· ·· ·· QUESTIONS TO CONSIDER 1. Flared jeans are all the rage these days. The Gap currently sells them for about $49.50 a pair. Flares were also in fashion back in the 1960s. Based on the BLS formula, what would a similar pair have sold for in 1967? 2. Some things have come down in price since the mid-1960s.Televisionsets, for instance. What are some others? . 3. Many of the products we commonly use today, such as cell phones and personal computers, didn't even exist in the 1960sconsumer market. How can this be inc1!}dedwhen calculating inflation over time? 4 ; _'_. _ _.~ ..... - -- ---- --- -- - -- - - - - - ---- - ------ Inflation and Price Change Austin Powers Shops at Wal-Mart Even an internationalman of mysteryhas to buy toothpasteonce in a while. And, of course, nobody could benefit from good dental hygiene more than Austin Powers. But prices certainly have changed since he entered suspended animationin the 1960s"In 1967(whenAustinPowers was at his originalshagadelicbest) a first-classpostage stamp cost 5 cents. And according to the online inflation calculator of the Bureau of Labor Statistics (BLS), a dollar bill that Austin Powers's nemesis,Dr.Evil, foldedin the pocket of his Nehru jacket that year would now be able to purchase only about 18% of what it could buy back then. !" ~ . I I .~.J ~ ---- - - - - - - -- - - - - 440 l INFLATION AND PRICECHANGE Thusfarwe haveassumedthat dollars in our analyseswereunaffectedby inflationorprice change. However,this assumption is not always valid or realistic. If inflation occurs in the general economy, or if there are price changes in economic costs and benefits, the impact can be substantialon both before- and after-taxanalyses. In this chapter we develop several key concepts and illustrate how inflation and price changes may be incorporated into our problems. MEANING AND EFFECTOF INFLATION Inflationis an importantconcept becausethe purchasingpowerof moneyusedin mostworld economiesrarely staysconstant.Rather,over time the amountof goods and servicesthatcan be purchased with a fixedamount of money tends to change.Inflationcauses U.S. dollarsto losepurchasing power.Thatis, whenpricesinflatewecanbuylesswiththe sameamount. of money.Inftation inakes future dollars less valuable than present dollars. Think about examples in your own life, or for an even starker comparison, ask your grandparents how much a loaf of bread or a new car cost 50 years ago. Then compare these prices with what you would pay today for the same items. This exercise will reveal the effect of inflation:as time passes, goods and services cost more, and more of the same monetaryunits are needed to purchase the same goods and services. Because of inflation, dollars in one period of time are not equivalent to dollars in another. We know from our previous study that engineering economic analysis requires that comparisons be made on an equivalent basis. So, it is important for us to be able to incorporate the effects of inflation in our analysis of alternatives. When the purchasing power of a monetary unit increases rather than decreases as time passes, the result is deflation. Deflation, very rare in the modern world, nonetheless potentially exists. Deflation has the opposite effect of inflation-one can purchase more with money in future years than can be purchased today. As such, deflation makes future dollars more valuable than current dollars. How Does Inflation Happen? Economists do not agree on all the sources of inflation, but most believe that the following effects influenceinflation either in isolation or in combination. Money supply: The amount of money in our national economy is thought to have an effecton its purchasingpower.If thereis too muchmoneyin the system (theFederal Reserve controls the flow of money) versus goods and services to purchase with that money,the value of dollars tends to decrease. When there are fewer dollars in the system,they become more valuable.The FederalReserve,through its influence on the money supply, seeks to increase the volume of money in the system at the same rate that the economy is growing. . Exchange rates: The strength of the U.S. dollar in world markets affects the profitability of international companies in those markets. Price rates may be adjusted to compensate for the relative strength or weakness of the dollar in the world market. As corporations' profits are weakened or eliminated in some markets ------ .-------. ------- -- -- I Meaning and Effect of Inflation 441 owing to fluctuations in exchange rates, prices may be raised in other markets to compensate. Cost-push inflation: This type of inflation develops as producers of goods and services "push" their increasing operating costs alongto the customerthrough higher prices. These operating costs include fabrication/manufacturing,marketing, and sales. Demand-pull inflation: This effect is realized when consumersspend money freely on goods and services. Often "free spending" is at the expense of consumer saving. As more and more people demand certain goods and services, the prices of those goods and services will rise (demand exceeding supply). A further considerationin analyzinghowinflationworksistheusuallydifferentratesat which the prices of goods and servicesrise versusthe wagesof workers.Do workersbenefit if, as their wages increase, the prices of goods and servicesincrease, as well? To determine. the net effect of differing rates of inflation, we must be able to make comparisons and understand costs and benefits from an equivalent perspective.In this chapter's examples, we willlearnhowto makesuchcomparisons. Definitions for Considering Inflation in Engineering Economy The following definitions are used throughout this chapter to illustrate how inflation and price change affect two quantities: interest rates and cash flows. Inflation rate (I): As suggestedearlier, the inflationrate captures the effect of goods and services costing more-a decrease in the purchasingpower of dollars. More money is required to buy a good or service whoseprice has inflated.The inflation rate is measured as the annual rate of increase in the number of dollars needed to pay for the same amount of goods and services. Real interest rate (if): This interest rate measures the "real" growth of our money excluding the effect of inflation. Because it does not include inflation, it is sometimes called the inflation-free interest rate. Market interest rate (i): This is the rate of interest that one obtains in the general marketplace. For instance, the interest rates on passbook savings, checking plus, and certificates of deposit quoted at the bank are all market rates. The lending interest rate for autos and boats is also a market rate.This rate is sometimes called the combined interest rate because it incorporatesthe effect of both real money growth and inflation.We can view i as follows: ;'1 '! ,I - I Market interest rate has in it "Real" growth of money and Effect of inflation rII I ,. " ',. ! ; j" -- -- -- " 442 INFLATION AND PRICECHANGE The mathematicalrelationship between the inflation,real and market interest rates is given as: i = i' + J + (i')(J) (14-1) S1,1ppose Tiger WqodSWaIltSltoinvest soWerecent golf winniIJ.gsin his hometown bank for one year. Currently, th.ebank is paying a rate of 55% compounded annually. Assume inflation is anticipateq to be 2% per year or 8%,per year for the year of Tiger's investment. In each case identify i, J, and i'. SOLUTION , If Inflation.Is 2% per Year F:t;ptnthe precedingdefinitions the inteJ;'estratethat the bankis paying is the market rate (i). The inflation rate (f) is given in the problem statement. What is left, then, is to find the real interesi rate (i'). i=5.5% " f 2% i'=? Sqlving for i' in Equation 14-1,weh.ave " ii' II II i ~ Ji i ,--,. + J + (i')(J) + (l')(f) J. , i'(1 + J) i' I -t (i ---/)/(1 + J) . -- (O.05~'-i0.02)/(1+ 0.02) .'Q.034or I 3.4% per ye~ I . 1 'ffiismeans that Tiger Woods wi11Ji!lye3.4% lIlorepurchasingpowerwitb. tl,ledollaJ,"sinve$ted 1 iniltb.ataccQuntthan b,ehad ayear ago.At tl,leend of the year hy cWlPUfchas.e3.4%D:}oJ;'egoods aq,dservigesthan he could have atth.~beginmng oftheyear. A$~Il eXCll1lple ofthegroWth of his tnQney,assUmehe was purchasiIlg golf balls that cost .$5eachalld tU!wheh.adillve$1:ed$loQOi'P h;sh.ometown bank accoUnt. ~'~Mthe beginning oftheyearh~could pUl"Ch~se: ia -- ... , - ,.-- !!Ii ~ -"- = 1;1-':;;a;-"~ -;0;:. -, !='iif.Dollars }(Umbe~o1~m.purqfi~t~y - ---~-- - -- today;,availabJe t~b~ halls · I ! J ::::5~~::::::j~ ~c~~~- __mJ: 1 -<-------..---.- ..-- Meaning and Effect of Inflation 443 At the end of the year he could purchase: Number of balls purchased at end of year = Dollars available for purchase at end of year Cost per ball at end of year In this case: Dollars available at end of year = ($I000)(F /P, 5.5%, 1) = $1055 Ball cost at end of year (inflated at 2%) = ($5)(1 + 0.02)1 = $5.10 Thus: Number of balls purchased at end of year = $1055/5.10 = 207 golf balls Tiger Woodscan, after one year, purchase 3.4% more golf balls than he could before. In this case 207 balls is about 3.4% more than 200 balls. If Inflation Is 8 % As for the lower inflation rate, we would solve for i': i' = (i - f)/O + f) = (0.055- 0.08)/(1 + 0.08) = -0.023 or -2.3% per year In this case we can see that the real growth in money has decreased by 2.3%, so that Tiger can now purchase 2.3% fewer balls with the money he had invested in the bank. Even though he has . more money at the end of the year, it is worth less, so he can purchase less. Regm:dless of howinflationbehavesoverthe year,the bankwillpay Tiger$1055at the end of theyear. However,as wehave seen,inflationcan greatlyaffectthe "real" growthof dollars over time. In que of his State of the Union.addresses, ]>residentGeorge H. W. Bush called inflation "that thie:fij'because .itstealsreal purchasing power frOIi1our dollars. Wehave seen this effect in this examnte. Let us continue the discussion of the effectsof inflationby focusing now on cash flows in our problems. We define dollars of two "types" in our analysis: . Actual dollars (A$): These are the dollars that we ordinarily think of when we think of money. These dollars circulate in our economy and are used for investments and payments. We can touch these dollars and often keep them in our purses and wallets-they are "actual" and exist physically.Sometimesthey are called inflated dollars because they carry any inflationthat has reduced their worth. I Problems 465 purchasing power than those originallyborrowed.This condition is advantageousto lenders at the expense of borrowers.Inflation and deflationhave opposite effects on the purchasing power of a monetary unit over time. To distinguish and account for the effect of inflation in our engineering economic analysis, we define inflation,real, and market interest rates. These interest rates are related by the following expression: . i = if + f + iff Each rate applies in a different circumstance, and it is important to apply the correct rate to the correct circumstance. Cash flowsare expressedin term.sof either actual or real dollars. The market interestrate should be used with actual dollars and the real interestrate should be used with real dollars. The different cash flows in our analysis may inflate or change at different interest rates when we look over the life cycle of the investment. Also, a single cash flow may inflate or deflate at different rates over time. These two circumstancesare handled easily by applying the appropriate inflationrate to each cash flow over the study period to obtain the actual dollar amounts occurring in each year. After the actual dollar quantities have been calculated, the analysis proceeds, as in earlier chapters, utilizing the market interest rate to . calculate the measure of merit of interest. Historical price change for single commoditiesand bundlesof commoditiesare tracked with price indexes. The Consumer Price Index (CPI) is an example of a composite index formed by a bundle of consumer goods. The CPI serves as a surrogate for general inflation in our economy. Indexes can be used to calculate the averageannual increase(or decrease) of the costs and benefits in our analysis. The historical data provide valuable information about how economic quantities may behave in the future over the long run. The effect of inflationon the computedrate of return for an investmentdepends on how future benefits respond to the inflation. If benefits produce constant dollars, which are not increased by inflation, the effect of inflationis to reduce the before-tax rate of return on the investment. If, on the other hand, the dollar benefits increase to keep up with the inflation, the before-tax rate of return will not be adversely affectedby the inflation.This outcome is not found when an after-tax analysis is made. Even if the future benefits increase to match the inflation rate, the allowable depreciation schedule does not increase. The result will be increased taxable income and income tax payments, which reduce the available after-tax benefits and, therefore, the after-taxrate of return.The importantconclusionis thatestimates of future inflationor deflationmay be important in evaluatingcapital expenditureproposals. '11 II' ~,. , .~ . BLEMS J" fine inflation in tenns of the purchasing power of lars. line and describe the relationships between the owing: actual and real dollars; and inflation, real !market (combined) interest rates. 14-3 How does inflationhappen-describe a few circumstances that cause prices in an economyto increase. 14-4 Is it necessary for inflation to be accounted for in an engineering economy study? What are the two approachesfor handlinginflation.insuch analyses? --- - l. r 444 l - INFLATION AND PRICECHANGE fforO periods C i' base (t) $ at (t) ~ ::J ~ R base (t) $ at n) ~A$at(n) I n A$ at (t) ~ I t n f for the tpe'od n, Time FIGURE 14-1 Relationship between i, f, i', A$, and R$. Real dollars (R$): These dollars are a bit harder to define.They are always expressed' in terms of some constant purchasing power "base" year, for example, 2002-based dollars. Real dollars are sometimescalled cQnstantdollars or constantpurchasing power dollars, and because they do not carry the effects of inflation,they are also known as inflation-freedollars. . Having defined market, inflation, and real interest rates as well as actual and real dollars, let us describe how these quantities relate. Figure 14-1 illustrates the relationship between these quantities. Figure 14-1illustrates the following principles: When dealingwith actual dollars(A$), use a market interestrate (i), and when discountingA$ overtime, alsouse i. When dealing with real dollars (R$), use a real interest rate (i'), and when discounting R$ over time, also use i'. Figure 14-1 shows the relationships between A$ and R$ that occur at the same period of time. Actual and real dollars are related by the inflation rate, in this case, over the period of years defined by n - t. To translate between dollars of one type to dollars of the other (A$ to R$ or R$ to A$) use the inflation rate for the appropriate number of periods. The following example illustrates many of these relationships. '!J:!. ''51: - __-' A urrivers!tyis considering replacing its staqiUl]1with a new facjlity~,When the prese:ntbuil4ing w~s completed in 1950, the total co§t was $1.2 million. At that time a wealthy alumnus gifted the u~iversity with $1.2 rhinion to beuse{l for a funire replacement.. University administrators are ncr considering building the new facility in the year 2Q05. . AJ~sumethe following: · Inflation is 6:0% perye~ troIn.i;9~Pto 2005. II . In 19~0the universityiny~~l¥t4~gi{t ata:lp.ar,ketjritere~trateof 8.Q%per year. = - -c-;:;;===II:=:< '--:!J!:I--""",.q."'='~~;-=<i ~- ,'. " n/ ",.,..~=,~,~'~,:;"~'-"" bl ill dfr ()Dfi f$ II . a e ne l, l " , A ~an. I.J..,"..Qm,.e pr6 em, . ~I =' ~ ' ~ - ' - - --- - -~ (b) How Inany flctual dolr~ji;vth~ye .'~.~OQ5 will tl.1egift be Worth? (c) flow much would the~~~tYA~4!:iilai~il} 2005 be in t~nns, <:>t .1950 purchasing i'~ " -- ","~)~e,:,¥&::~;'::?~y--'<",-~-- '. . _ _' __.""- ~~ ------- --- --- - -- '-- power? ... Meaning and Effect of Inflation Since 6.0% is the inflation rate (I) and 8.0% is the market interest rate (i), we can write if = (0.08 - 0.06)/0 + 0.06)= 0.01887, or 1.887% Therefore, $1,200,000 was the cost of the building in 1950. These were the actual. dollars (A.$) spent in 1950. FromFigure 14-1wearegoing fromactualclollarsat t, in )950, to actualdollarsat n, in 2005. To do so, we use the market interest fate andcompQuridthis amountfo:rWardiIlnm~55years, as illustrated in Figure 14-2. Actual dpllars in 2005 = Actual doUars.in.1950 (FI P, i,55 years) = $1,200,000(FIP~8%,55) . $82,701,600 FIGtJ~14-2 in 2005. DiscountingA$ in 1950to A$ $82,701,600 I I $1,200,000 I t . 8% 2005 195,9 I Now we"want to detefITIinethe amount of realdollars'--.basedinl;95rJ.=that(Jccur in theyear 2005, wnich are equivalent to the$82.7nll1li~n ft(jrn the solutiQp. to partb; tetl.ls'S()rV~ .tJjis problern;hvo ways. this approachletusdirectly'tJ;anslatethe actualilollars in the yegr 2005 to re.al,)950" based dollars in the.year 2005..FrornJ3igutel4-1 w~can'hse.tlJ,~i1J,fl~ti()I1 1"9te to.sn-ip 1.. trt 55 years of inflation from the actual. d6I1ars.We .dotlJiS by usip.gtlJ,eP{f faC.!9t;fo( 55.years at tlJ,einflation rat~.We arep.()tPbysic~llYI~lpvingtlJ,e.dQll(lf~i1J, tim~in tljiscase; rlther we are siInplyremovinginflatioufromthese dollars one' !~(lfata tft!Je"'tl},ecf1lf ~1ctordoes thatfor us. Tltis.isillustratedip. tJ:iefOllowing eqpatiop,and,Ifigur~14-J; II ~ J~'..x~!<i.-5.P~1:t~~d:.4oUars~mc.2QQ.$-," ..~(~~tp.Al,,-dPU~~-'f~Q.Q.~}Q~Ul,ik~5~):;:;:: II . \I . ~ ($82, 701,600)(P iF, 6%, 5~) .. .. ,. " $3,357,000 ---.-- 445 446 -I INFLATION AND PRICECHANGE l RB= 1950$ in 2005 3,357,000 A$ in 2005 = 82,701,600 ~ t 2005 Out 55 yearsof . Slrippmg 6% Inflation Time FIGURE 14-3 Translation of A$ in 20.0.5to R 195o.-baseddollars in 2005. 2. In this methodwe startby recognizingthattheactualdollars in 1950are exactlyequivalent to real 1950-hased dollars fhatexistin 1950.By definition,dollars thafhave today as the purchasing power base are the same as actualdollars today. 'Thus,actual.flollm-~in 20.0.4 are the same as real 2o.04.,-based dollars that occur in 20.0.4.So in this example, the $1.2 million can also be said to be reaLJ950-baseddollars that occur in 1950..As such, letus translate thosereal'dollm-sfrom 1950to the year 20.0.5. 'Since they are real dollar$,we use the real interest rate. ! Real 195Q-baseddollars in 20.05= (Real 1950-based dollm-s iIl195o.)(P 1P,i', 55) .. · ($1;20.0.,0.0.0)(FjP, 1.887%, 55) f .~ $3,355,0.00. I t FIGUllE14.,4 Translation of R 1950-based dollars in 1950.to R 195o.-based dollars in 20.0.5. ~ $3.355,qoo $1,200,000 t " 1950 ') I 1:887% 2005 Time (No~e:The difference between the answets to parts 1 and 2 is due t9 rounding themarkeOnterest rate"offto 1.887% versus caI"fYingitoutto mote significant digits. The amount of difference due to this rounding is less than 1%.'If vJewete to ca:Tythe caIcula.tlonof i' to a sufficientnumlJer6f digffs,the answers to the twoparts would,be identical.) From Example 14-2we can see therelationshipbetween dollars of differentpurchasing power bases, the selectionof appropriateinterestrates to use for moving dollars in time, and stripping out or adding in inflation.In that example the $1.2 million initially investedgrew in the 55-year period to over $82.7 million, which becomes the amount available to pay for construction of the new complex. Does this mean that the new stadium will be 82.~/1.2 or about70.times "better" than the one built in 195o.?The answerto that question is no, because in the year 20.0.5the purchasingpower of a dollar is less than it was in the year 1950..Assuming that construction costs increased at the rate of 6% per year given in the problem, then the amount available for the project in terms of 1950-based dollars is almost $3.4 million. This means that the new stadium will be about 3.4/1.2 or approximately 2.8 times "better" than the original one using real dollars-not the 70.times ratio if actual dollars are used. ' Meaning and Effectof Inflation 447 In 1924 Mr. O'Leary buried $1000worthof quartersin his backyard.Over the yearshe had always thought that the money would be a nice nest egg to give to his first grandchild.His first granddaughter, Gabrielle, arrived in 1994. At the time of her birth Mr. O'Leary was taking an economic analysis course in his spare time. He had learned that over the years 1924 to 1994, inflation averaged4.5%, the stock market increased an averageof 15%per year, and investments in guaranteed government obligations averaged 6.5% return per year. What was the relative purchasing power of the jar of quarters that Mr. O'Leary gave to his granddaughterGabrielle at the time of her birth? What might have been a better choice for his "backyard investment?" SOLUTION ! Mr. O'Leary's $1000 dollars are actual dollars in both 1924and in 1994. To obtain the real 1924-baseddollar equivalent of the $1000 that Gabriellereceivedin 1994 we wouldstrip70yearsofinflationoutofthosedollars.Asit turnedout,Gabrielle'sgrandfather gave her $45.90 worth of 1924 purchasing power at the time of her birth. Because inflation has "stolen" purchasing power from his stash of quarters during the 70-yearperiod, Mr.O'Leary gave his grandaughter much less than the amounthe first spaded underground.This loss of purchasing power caused by inflationover time is calculated as follows: Real 1924-based dollars in 1994 = (Actual - 1924) dollars in 1994)(P 1F, t, 1994 = $1000(P/ F, 4.5%,70) $45.90 On the other hand, if Mr. O'Leary had put his $1000 in the stock market in 1924,he would have madebabyGabriellean instantmultimillionaireby givingher $17,735,000.We calculatethis as follows: Actual dollars in 1994 = (Actual dollars in 1924)(F 1P, i, 1994- 1924) = $1000(FI P, -15%,70) = $17,735,000 At the tiple of Gabrielle's birth that $17.7 million translates to $814,069 in 1924 purchasing power. This is quite a bit different from the $45.90 in 1924 purchasing power calculated for the unearthedjar of quarters. Real 1924-based dollars in 1994 l_ - ~- = $17,735,OOO(P 1F, 4.5%, 70) = $814,069 Mr. O'Leary was nevera risk taker, so it is doubtfulhe would have chosen the stockIDarketforhis future gJ:IDIdchild's nest egg. If he had chosen guaranteed government obligations instead of his backyard, by 1994 the investmentwould have grown to $59,076 (actual dollars)-the equivalent of $271~ in 1924 purchasing power. I I" I I " ~= ~= Actuahiollars'in 1994~ ""(Acttial dollars in 1924)(F 1P/i, j 1994~ 1924) r I I .. ~~ - ~I ~ $1OO0(FIP, 6%, 70) = $59,076 'Real1924-basetl'dollafsihr994'~~$59',076'(~:P llIp,4.5%,70) --- $2712' 1:1 c 'III II ! ""';J 1 .... --- 448 INFLATION AND PRICECHANGE Obviously, either option would have been better than the choice Mr. O'Leary made. This example illustrates the effects of inflation and purchasing power, as well as the power of compounded interest over time. However, in Mr. O'Leary's defense, if the United States had experienced 70 years of deflation instead of inflation, he might have had the last laugh! There are in general two ways to approach an economic analysis problem after the potential effects of inflation have been recognized. The first is to conduct the analysis by systematicallyincludingthe effectsof inflation;the secondis to ignore the effectsof inflation in conducting the analysis. Each case requires a different approach. Incorporating inflation in the analysis: Use a market interest rate and actual dollars that include inflation. Ignoring 'inflation in the analysis: Use real dollars and a real interest rate that does not reflect inflation. ANALYSIS IN CONSTANT DOLLARS VERSUS THEN-CURRENT DOLLARS Performing an analysis requires that we distinguish cash flows as being either constant dollars (real dollars, expressed in terms of some purchasing power base) or then-current dollars (actual dollarsthat are then-currentwhen they occur). As previouslystated,constant (real) dollars require the use of a real interest rate for discounting, then-current dollars require a market (or combined) interest rate. It is not appropriate to mix these two dollar types when performingan analysis.If both types are stated in the problem, one must convert either the constant dollars to then-current dollars or the then-current dollars to constant dollars, so that a consistent comparisoncan be made. '(heWayg!<1te Corporationis interestedin evaluatingamajor newvideod!splaytechnology (VDT). Two competing computer innovationc:o!llpanieshave approached Waygate to develop the technologyfor the new VDT.Waygate believesthatboth companies will.be able to deliver equivalent products at the end of a 5,.yearpeli,od..Eromthe yearly development costs of theVD'r foJ."each ofihe two companies as given, detenjUnewhich Waj'gate shotildcboose.if the c0IJ.?0rateM1\RR (in}{estmeJ1t.market rate) .is 25%apd geJ1era.lpqcejnfla.nolli.s~sumed tope 3.5% perj'ear over the'mext5 years. tl..:_" _ '''' ",'-', _ " __.,',." ',' _' .',_."'- _' ,,' __,,',' _ _ _.<'" II CQmpany Alpha costs: DeveloP.lJleJ1t costs Will he $15Q;OOOtlIe'firstyeaJ.". ap.d W.mincrease ~ ;!::.:~ at,arate ot.5% oYeJ;'tbe~Ye~I!eriog! ~.;;: ~~ ~ == ~ ~,= = ~ .;;:~ =:: = = ':I'\': ' 0_. :: r =E==,rZ " CQmpany Be-faco~ts:DeveI9~J:l1ehtJcosts will bea COll$t!!Il.t'$1'5Q,OOO'petj'eMipte11Il$of. ',' ~ loday's dollars over the5'C"Yea.r~, ~~ ~== ~-, l1' - " . - -=,: == III - ;:;f -- - . , ==-~-;:;;;;:; I ~ Analysis in Constant Dollars Versus Then-Current Dollars 449 SOLUTION . . The costs for each of the two alternativesare as follows: Then-Current Costs Stated by Alpha $150,000 x (1.05)° = $150,000 Year 1 2 3 4 5 150,000 x (1.05)1 = 157,500 150,000 x (1.05)2 - 165,375 150,000 x (1.05)3 - 173,644 150,000 x (1.05)4 = 182,326 Constant Dollar Costs Stated by Beta $150,000 150,000 150,000 150,000 150,000 We inflate (or escalate) the stated yearlYocost given by Company Alpha by 5% per year to obtain' the then-current (actual) dollars each year. Company Beta's costs are given in terms of today-based constant dollars. Using a Constant Dollar Analysis Here we must convertthe then-currentcosts given by Company Alpha to constant today-based dollars. We do this by stripping the appropriate number of years of general inflation from each year's cost using (P / F, f, n) or (l + f)-n. Constant Dollar Costs Stated by 1\1pha Year 1 2 3 4 5 $150,000 x (1.035)-1 = $144,928 157,500 x (1.035)-2 = 165,375 x (1.035)-3 = 173,644 x (1.035)-4 182,326 x (1.035)-5= 147,02? 149,159 151,321 153,514 Constant Dollar Costs Stated by Beta $150,000 150,000 150,000 150,000 150,000 We use the real interestrate (i') calculated from Equation 14-1tq calculate the presentwor$ of costs for each alternative: i' t I = (i - PW of cost (Alpha) I f)/(1 + f) = (0.25- 0.035)/(1+ 0.035) = 0.208 = $144,928(P / F, 20.8%, 1) + $147,028(P I F, 20.8%, 2) i + $149, r + $153,514(P /F, 20.8%,5) -- $436,000 I 159(P / F, 20.8%, 3) + $151,321(P / F, 20.8%, 4) I?Wof cost (Beta) = $150,000(P/ A,20.8%, 5) = $150,000(2.9387)-- $441,000 Using a'Then-Current Dollar Amilysis ;II _ a ... . 1fere w~ must convert the constant dollar costs of Company Beta to then-current dollars. We do this by using (FIP, f, n) or (1 + f)n to "add in" the appropriate number of years of gen,eral . fl<>1.'t 0 e~c h y~ilt~s ' GP§t..: =' lll~tlOq\ ~ ~I:' '=Ii;;;;;:=; e: == =='" ~ _ ==11I:II: = = ~== !II= 11_ _. :I . _. - - - - ---.------ t - 450 --.-.---- INFLATION AND PRICECHANGE Year 1 2 3 4 5 Then-Current Costs Stated by Beta Then-Current Costs Stated by Alpha $150,000 x (1.05)° = $150,000 150,000 x (1.05)1 - 157,500 150,000 x (1.05)2 165,375 150,000 x (1.05)3 = 173,644 150,000 x (1.05)4 = 182,326 $150,000 x (1.035)1 150,000 x 150,000 x 150,000 x 150,000 x = $155,250 (1.035)2 = (1.035)3 = (1.035)4 = (1.035)5 = 160,684 166,308 172,128 178,153 Calculate the present worth of costs for each alternativeusing the market interest rate (i). PW of cost (Alpha) = $150,000(P I F, 25%, 1) + $157,500(P I F, 25%, 2) + $165,375(P I F, 25%,03) +'$173,644(P j F, 25%, 4) + $182,326(P I F, 25%, 5) PW of cost (Beta) = $155,250(P IF, 25%,1) ". $436,000 + $160,684(P I F, 25%, 2) + $166,308(P I F, 25%, 3)+ $172, 128(P I F, 25%, 4) +$178,153(PjF, 25%, 5) = $441,000 USing either a constant dollar or then-current dollar analysis Waygate should chose Company Alpha's offer, which has the lower present worth of costs. There may, of course, be intangible elements in the decision making that also should be considered. PRICE CHANGE WITH INDEXES We have already described the effects that inflation can have on money over time. Also, severaldefinitionsand relationshipsregardingdollars and interestrates have been given.We have seen that it is not appropriateto compare the benefits of an investment in 2004-1;>ased dollars againstcosts in 2006-baseddollars. This is likecomparing apples and oranges.Such compariS'8f1g~fbenefits and costs can be meaningful only if a standard purchasing power base of money is used. An often asked question is, How do I know what inflationrate to use in my studies?This is a validquestion.What can we use to measureprice changesovertime? What Is a Price Index? Price indexes are used as a means to describe the relative price fluctuation of goods and services in our national economy. They provide a historical record of the behavior of these quantities over time. Price indexes are tracked for specific commodities as.well as for bundles (composites) of commodities. As such, price indexes can b~ used to measure historical price changes for individual cost items (like labor and material costs) as well as general costs (like consumer products). In understanding the past price fluctuations we have more information for predicting the future behavior of those cash flows. - - - - -- ----- I' PriceChangewith Indexes 451 TABLE 14-1 Historic Prices of First-Class Mail, 1970-2003, and Letter Cost Index Year,n ClassMail LCI Annual Increasefor n Year,n ClassMail LCI Annual Increasefor n 1970 1971 1972 1973 1974 1975 1976 1977 1978 1979 1980 1981 1982 1983 1984 1985 1986 $0.06 0.08 0.08 0.08 0.10 0.13 0.13 0.13 0.15 0.15 0.15 0.20 0.20 0.20 0.20 0.22 0.22 100 133 133 133 166 216 216 216 250 25Q 250 333 333 333 333 367 367 .0.00% 33.33 0.00 0.00 25.00 30.00 0.00 0.00 15.74 0.00 0.00 33.33 0.00 0.00 0.00 10.00 0.00 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 $0.22 0.25 0.25 0.25 0.29 0.29 0.29 0.29 0.32 0.32 0.32 0.33 0.33 0.33 0.34 0.37 0.37 367 417 417 417 483 483 483 483 533 533 533 550 550 550 567 617 617 0.00% 13.64 0.00 0.00 16.00 0.00 0.00 0.00 10.34 0.00 0.00 3.13 0.00' 0.00 3.03 8.82 0.00 Cost of First- Cost of First- Table 14-1lists the historic prices of sending a first-classletter in the United States via the Postal Service from 1970to 2003. The cost is givenboth in terms of dollars (cents) and as measured by a fictitiousprice index that we could call the letter cost index (LCI). Notice two important aspects of the LCI. First, as with all cost or price indexes; the numbers used to express the change in price over time are based on some base year. With price (or cost) indexes the base year is always assigned a value of 100. Our LCI has a base year of 1970-thus for 1970, LCI = 100. The letter cost index value given in subsequent years is stated in relation to 1970as the base year. A second aspect to notice is that the LCI changes only when the cost of first-classpostage changes. In years when this quantity does not change (in other words, there was no price increase), the LCI is not affected. These general observations apply to all price indexes. In general, engineering economists are the "users" of cost indexes such as our hypothetical LCI. That is, cost indexes are calculated or tabulated by some other party, and our interest is in assessing what the index tells us about the historical prices and how these may affect our estimate of future costs. However, since we are "users" of indexes," rather than "compilers," it may be of interest to illustrate how the LCI in Table 14-1 was calculated. In Table 14-1,the LCI is assigned a value of 100because 1970 serves as our base year. In the following years the LCI is calculated on a year-to-year basis based on the annual percentage increase in first-class mail. Equation 14-2illustrates the arithmetic used. LCIyear,n = ([(cost(n) - cost 1970)/cost 1970] x 100%)+ 100 --___ _.__n ------ (14-2) -, 452 INFLATION AND PRICECHANGE For example, consider the LCI for the year 1980. We calculate the LCI as follows. LCI year 1980 = ([(0.15 - 0.06)/0.06] x 100) + 100 = 250 As mentioned, engineering economists are often in the business of using cost indexes to ~ project future cash flows. As such, our first job is to use a cost index to calculate annual cost increases for the items tracked by the index. To calculate the year-to-year percentage increase (or inflation) of prices tracked by an index, one can use Equation 14-3. Annual percentage increase, n = ([Index(n) - Index(n-1) ]/Index(n -1» x 100% (14-3) To illustrate the use of this equation, let us look at the percent change from 1977 to 1978 for the LCI just given. Annual percentage increase (1978) = [250 - 216]/216 x 100% = 15.74% For 1978the price of mailing a first-classletter increased by 15.74%over the previousyear. This is the value tabulated in Table 14-1. An engineering economist often wants to know how a particular cost quantity changes over time.Oftenwe areinterestedin calculatingthe averagerate of priceincreaseor inflation in some quantity,such as the cost of postage, over a period of time. For instance, one might want to know the average yearly increase in postal prices from 1970 to 2003. How do we calculate this quantity? Can we use Equation 14-2? If we were to use Equation 14-2 to calculate the percent change from 1970to 2003, we would obtain the following: % Increase (1971 to 1991) = (617 - 100)/100 x 100% = 517% . But how do we use this calculation to obtain the average rate of increase over those years? Should we divide 517% by 33 years (517/33 = 15.67%)? Of course not! As was establishedin earlier chapters,theconceptof compoundingpreclud~ssucha simpledivision. To do so would be to treat the interestrate as simpleinterest-where compoundingis not in effect. So the question remains: Howdo we calculate an equivalentaveragerate of increase in postage rates over a period of time? Let us start by thinking about the LCI. We have a number (index value) of 100 in year 1970 and another number, 617, in year 2003, and we want to know the interest rate that relates these two numbers. If we think of the index numbers as cash flows,it is easy to seethat we have a simpleinternal rate of return problem. Given this approach, let us calculate the average rate of increase in postage rates for the years under consideration. P=100 F = 617 n i =? = 33 years UsingF = P(1 + i)n 617 = 100(1+ i)33 i = (617/100)1/33- 1 i = 0.0567'= 5.7% In the same way, we can use a cost index to calculate the averagerate of increase over any period of years. Understanding of how costs have behaved historically should provide insight into how they may behave in the future. - --- -- . I' Price Change with Indexes 453 Composite Versus Commodity ~ndexes 'Cost indexes, in general, come in two types: commodity-specificindexes and composite indexes.Each type of index is useful to theengineeringeconomist.Commodity-specificindexes measure the historical changein pricefor specificitems-such as green beans or iron ore. Common commodities that are trackedby price indexesand used in engineering economic analysis include utility commodities,labor costs, andpurchaseprices. Commodity indexes,like our letter cost index, areusefulwhenaneconomicanalysisincludes individual cost items that are tracked by such indexes.Forexample,if we need to estimate the directlabor cost portion of a constructionproject,wecoulduse an indexthattracks the inflation,or escalation, of this particular cost over time.TheU.S. Departmentsof Commerce and Labor track many cost quantities through the Departmentof Economic Analysis and Bureau of Labor Statistics.For our example constructionproject,we wouldreference the appropriate labor index and investigate how this cost item had behavedin the past. This should give us valuable information about how to estimatethis cost in the future. . Congratulations!Yoursisterjust had a baby girlnamedVeronica.Whenyouhear this happy news, you fondly remember the fine pony you had as a young child, and you decide you would like to give your new niece a pony .forher fifth birthday.Youwouldlike to know how much you must put into your passbook savings account (earning 4% interestper year) today to purchase a pony and saddlp for Veronica5 years from now. But you have no idea how much a pony and saddle might cost in 5 years. I~ SOlUTI9N I~ Divide the;problem into the following steps: Step 1 trse commodity indexes to measure past pqce changesfor both ponies and saddles. Step 2 dan several dealers to get prices on the currentco~tsof ponies and saddles. Step 3 Use the averageprice increasevalues for eachcommodity(ponyandsaddle) to estimate ~ttat the cost of each will be in 5 yearS. . . Step 4 C1uculatehow much you would have to put intoyourbankaccoUnttoday to cover those cc>stsin 5 years. .. Step 1 A trip to tl)elibrary.reveals that, indeed, there are price indexesfor thetwo items that you would like to buy,j.n5 years: a pony and a saddle. From the indexesyoufindthat in the last 10 years the little pony;price index (LPP.I)has gone from 213 to 541, and the leathersaddleindex (LSI) from 1046 to 1229. For each commodity you calculate the averagerate ofprice increasefor the past I I 10 yea:t;s <1$: = ::: -. --- .' -.. -.. --- I = ii .;::: = ==:: A.veragepricechang~ II --- ::;;!:i!!!:=-== iii= ~ ~ :; = = = ~ -== = (LPPI) is 541= 213(1+i)1O solviyg tor i for ppQj.~s= 9.8% per year ;;;;;jjjji iiI = '= -- - - --- 454 - INFLATION AND PRICECHANGE Average price change (LSI) is 1229 solving for i for saddles = 1046(1+ i)10 = 1.6% per year Step 2 Call a stable and a tack shop to obtain the current price of a registeredpony at $600 and leather pony saddle at $350. Step 3 Using the current prices for both the pony and saddle, inflate these costs at the respectiveprice change rates calculated in Step 1. The assumption you are making is that the prices for ponies and saddles will change in the next 5 years at the same rate as the average of the last 10 years. The cost for the two items, and the total, in 5 years will be: Pony cost in 5 years Saddle cost in 5 years Total cost in 5 years = 600(1 + 0.098)5 = $958 = 350(1 + 0.016)5 = $378 = $958 + $378 = $1336 Step 4 Your final step is to calculate the amount that must be set aside today at 4% interest for 5 years to accumulate the future cost. Amount invested noW = 1336(1 + 0.04)-5 = $1098 So it will require $1098 today to make Veronicaa very happy pony rider on her fifth birthday! Composite cost indexes do not track historical prices for individual items. Instead, they measure the historical prices of groups or bundles of assets. Thus a composite index measures the overallprice change that is a composite of severaleffects. Examples of composite indexes include the ConsumerPrice Index (CPI) and the Producer Price Index (PPI). The CPI measures the effect of prices as experienced by consumers in the U.S. marketplace, and the PPI measures prices as felt by producers of goods in the U.S. economy. The CPI, an index calculated by the Bureau of Labor Statistics, tracks the cost of a standardbundleof consumergoods from year to year.This "consumer bundle" or "basket of consumer goods" is made up of common consum~rexpenses including housing, clothing, food, transportation, and entertainment. Because of its focus on consumer goods, people often use the CPI as a substitute measure for general inflation in the economy. 'Dtere are several problems with the use of the CPI in this manner, one being the assumption that all consumers purchase the same "basket of consumer goods" year after year. However, even with its deficiencies,the CPI enjoys popular identificationas an "inflation" indicator. Table 14-2givesthe yearly index values and annual percent increase in the CPI for the past 30 years. J PriceChangewith Indexes 455 TABLE14-2 CPllndex Values and Yearly Percentage Increases, 1973-2002 Year CPIValue* CPllncrease Year CPIValue. CPllncrease 1973 1974 1975 1976 1977 1978 1979 1980 1981 1982 1983 1984 1985 1986 1987 44.4 49.3 53.8 56.9 60.6 65.2 72.6 82.4 90.9 96.5 99.6 103.9 107.6 109.6 113.6 6.2% 11.0 9.1 5.8 6.5 7.6 11.3 13.5 10.3 6.2 3.2 4.3 3.6 1.9 3.6 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 118.3 124.0 130.7 136.2 140.3 144.5 148.2 152.4 156.9 160.5 163.0 166.6 172.2 177.1 179.1 4.1% 4.8 5.4 4.2 3.0 3.0 2.6 2.8 2.9 2.3 1.6 2.2 3.4 2.8 1.6 *Referencebase: 1982-1984 = 100. 15 o 1973 1976 1979 1982 1985 1988 Year 1991 1994 1997 2000 2002 FIGURE 14-5 CPI Inflation Rate over 30 years (1973-2002). Composite indexes can be used in much the same way as commodity-specificindexes, described earlier. That is, we can pick a single value from the table if we are interested in measuring the historic price for a single year, or we can calculate an average inflationrate or average rate of price increase as measured by the index over a time period extending several years. - - - - --- -- 456 INFLATION AND PRICECHANGE _l How to Use Price Indexes in Engineering Economic Analysis One may question the usefulness of historical data (as provided by price indexes) when engineering economic analysisdeals with economic effectsprojectedto occur in thefuture. However,both commodity-specificand composite indexesindeed are useful in manyanalyses. Engineeringeconomic analysisis concernedwith makingestimatesof future events:the outcomes of yearly costs and benefits,interest rates, salvagevalues;and tax rates are all examplesof suchestimates.Associatedwith these estimatesare varyingdegreesof uncertainty. The challenge for the engineering economist is to reduce this uncertainty for each estimate. Historical data provide a snapshot of how the quantities of interest have behaved in the past. Knowingthis past (historical)behavior shouldprovideinsight on how to estimate . their behavior in the future, as well as to reduce the uncertainty of that estimate. This is where the data that price indexes provide come into play. Although it is very dangerous to extrapolate past d~ta into the future in the short run, price index data can be useful in making estimates (especially when considered from a long-term perspective). In this way the engineering economist can use average historicalpercentage increases (or decreases) from commodity-specificand compositeindexes,alongwith data from market analysesand other sources, to estimate how economic quantities may behave in the future. One may wonder how both commodity-specificand composite price indexes may be used in engineering economic analyses. The answer to that question is reasonably straightforward. As we have established, price indexes can be useful in making estimates of future outcomes. The followingprinciple applies to commodity-specificand composite price indexes and such estimates: When the estimated quantities are items that are tracked by commodity specific indexes, then those indexes should be used to calculate average historical percentage increases (or decreases). I ., .I j I If no commodity-specificindexes are kept, one shoulduse an appropriate composite index to make this calculation. For example, to estimate electricusage costs for a turret lathe over a 5-year period, one would first want to refer to a commodity-specificindex that tracks this quantity. If such an index does not exist, one might use a specificindex for a very closely related commodity-perhaps, in this case, an index of electric usage costs of screw lathes. In the absence of such substitute or related commodity indexes, one could use appropriate composite indexes: there may be a composite index that tracks electric usage costs for industrial metal-cutting machinery.Or, as before, a related composite index.could be used. The key point is that one shouldtry to identify and use a price indexthat most closely relates to the quantity being estimated in the analysis. CASH FLOWS THAT INFLATEAT DIFFERENT RATES J j .!! I I I J Engineering economic analysis requires the estimation of various parameters. It is not uncommon that over time, these parameters will inflate or increase (or even decrease) at different rates. For instance, one parameter might increase 5% per year, another 15% per year, and a third decrease 3.5% per year. This phenomenon is important because of the various items of different types that are sometimes included in engineering economic analyses. Since we are looking at the behavior of cash flowsover time, we must have a way of handling this effect. ~ .' ~ iJ ~ _.. ~.~. ...~.._~ ~..~. h..'~_~.'~.. ~~ ~._ .~~ ~.~_.~.~~..~.~~.._-._----- Cash Flows That Inflate at Different Rates 457 On your first assignment as an engineer, your boss asks you to develop the utility cost portion of an estimate for the cost of a new manufacturing facility. After some research you define the problem as finding the present worth of utility costs given the following data: · · ·· · Your company uses a minimum attractive rate of return (MARR) = 35%as i. The project has a useful life of 25 years. Utilities to be estimated include electricity, water, and natural gas. The 35-year historical data reveal: Electricity costs increase at 8.5% per year Water costs increase at 5.5% per year Natural gas costs increase at 6.5% per year First-year estimates of the utility costs (in today's dollars) are as follows: Electricity cost will be $55,000 Water cost will be $18,000 Natural gas will be $38,000 SOLUTION For this problem we will take each of the utilities used in our manufacturing facility and inflate them independently at their respective historical annual rates. Once we have these actual dollar amounts (A$), we can total them and then discount each year's total at 35% for the appropriate number of periods back to the present. Year Electricity 1 55,000(1.085)° = $ 2 55,000(1.085)1 = 3 55,000a.085)2 = 4 55,000(1.085)3 = 5 55,000(1.085)4 = 6 55,000(1.085)5 = 7 55,000(1.085)6 = 8 55,000(1.085)7 = Water 55,000 59,675 64,747 70,251 76,222 82,701 89,731 97,358 Natural 18,000(1.055)° = $18,000 38,000(1.065)° 18,000(1.055)1 = 18,990 38,000(1.065)1 18,000(1.055)2 = 20,034 38,000(1.065)2 18,000(1.055)3= 21,136 38,000(1.065)3 18,000(1.055)4 = 22,299 38,000(1.065)4 18,000(1.055)5 = 23,525 38,000(1.065)5 18,000(1.055)6 = 24,819 38,000(1.065)6 18,000(1.055)7 = 26,184 38,000(1.065)7 Gas Total = $ 38,000 $111,000 = 40,470 119,135 = 43,101 127,882 137,289 = 45,902 = 48,886 147,407 = 52,063 158,290 = 55,447 169,997 = 59,051 182,594 I I I I 24 55,000(1.085)23 = L.. I 359,126 18,000(1.055)23= 61,671 25 55,000(1.085)24= 389,652 18,OOO(1.055}24 = 65,063 38,000(1.065)23 = 38,000(1.065)24 = 161,743 172,256 582,539 626,970 The present worth of the total yearly utility costs is: _ _ _ ~_ _ """" ___: =-=::iiiii:i ~~ == ;:;:;:; = ~ '== = i:;;= .. I!: a 'PW = $Qf,OOO(P/F,35"%,1)+$119'; 135(P/F, 35%, 2) +... +$6+6,970(P/F, 35%, 25) I -I. ~, I~' , - $5,,540,000 J - -- --- -- 458 INFLATION AND PRICECHANGE In Example 14-6 several parameters changed at different rates over the period of the economic study.By using therespectiveindividualinflationrates, theactual dollaramounts for each parameter were obtained in each year. Then, we used a market interest rate to discount these actual dollar amounts. Problems of this type can be handled by inflatingthe various parameters at their respective estimated inflationrates, combining these cash flows if appropriate, and then treating them as actual dollars that occur in those years. DIFFERENT INFLATION RATES PER PERIOD In this section we address the situation of inflation rates for the various cash flows in an analysis that are changing over the life of the study period. Rather than different inflation rates for different cash flows, in Example 14-7 the interest rate for the same cash flow is changing over time. A method for handling this situation is much like that of the preceding section. We can simply apply the inflation rates in the years in which they are projected to occur. We would do this for each of the individual cash flows over the entirety of the study period. Once we have all these actual dollar amounts, we can use the market interest rate and decision criteria to apply any of the measures of merit developedearlier. While working as a clerk atPiggly Wiggly,Elvis has learned much about the cost of different vegetables. The kitchen manager atHeartbr.eakHotel called..receIit1y, reqllestingElvis to estimate the raw materialcostover the next5 yearsto introduce succotash(limabeans andcom) to thebuffet line..,To develophis estimate, Elvis has used his advanced knowledgeof soil growing conditions, wor14demand, and govel1)IIlentsubsidyprograI11s for these two crops. He has estimated the .... follqwing data: ~ · Costs for lima beans willinfiate at3%per year for the next 3years and tl1~Iiat4% for the following.2 years. '., Co~tsfor com will inflate at8%per yearfor the next 2 years and then will decrease 2% in :: I I I t the (ollowing :3years. The ''kitchenmanager.Wants to know the~ql!iva.lent~nual cost ofpr()vidiI1gsuccofush on the buffet line over.the 5-yearperiod.Hisbefor:e-tilX .,1'v1A.R.Ris20%.Tb,e frn.lllager. estiIIJatesthat be wilfpeed an average of 50 pounds each()f be~s and (O()fIl each qay. Jiliebotel kitchen operates 6 days perwee1c, 52 week~per year. GliITentcosts are $0.35111:> fPtIim& bean$ and$0.80Ilb for ~!1' conn -, '.~" il . ''''',-' . ~. ';~~' ..... ....-...- I SOLUTlON1 ;- .,. .~J..:i,._",~",,,, II ~~ ;rodf~'s;cosHor;;Qll~,,¥~at:§,.§1.i:QR.W U = ~ J~ ~~ ~ Lima beans .,= :y~l}l~ f:., =., 1111II 1:1 ~Q~g~.w.fil~~s~:' "~ II II! ~ ~ '" II =--;;;p;;;;0:;~ = =;: 0.3$$ti1?x..~pJ1Jlij~y")y 6;day/Wkx$fl. ~¥Jyr ;;=~Q,$lib~~~_xi6.qaJ/~~~,$2wP~~ Ii': --- = = $S4()Q/yr j~~89~ _II l I ! , I Different Inflation Ratesper Period Year 0 1 2 3 4 5 EUAC , Lima Beans $5,460 5,460(1.03) 5,624(1.03) 5,793(1.03) 5,967(1.04) 6,206(1.04) - = 5,624 = = = = 5,793 5,967 6,206 6,454 Corn $12,480 12,480(1.08) = 13,478 13,478(1.08) = J4,55614,556(1.02)-1 = 14,271 14,271(1.02)'-1 = 13,991 13,991(1.02)-1 = 13,717 459 Total $19,102 20,349 :\: 20,238 20,197 20,171 [19, 102(PI F, 20%, 1) + 20,349(PI F, 20%, 2) + 20,238(P I F, 70%, 3) + 20, 197(P IF, 20%, 4) + 20,171(P IF, 20%, 5)](Alf,20%, -- 1$19,900 5) per year In Example 14-7,both today's cost for each vegetable and the respectiveinflationrates were used to calculate the yearly costs of purchasing the desired quantitiesover the 5-year period. As in Example 14-6,we obtained a total marginal cost (in terms of actual dollars) by combining the two individual yearly costs. We then calculated the equivalentuniform annual cost (EUAC)using the given market interest rate. Example 14-8provides anotherexample of how the effect of changesin inflationrates over time can affect an analysis. r~" If general price inflation is estimated to be 5% for the next 5 years, 7.5% for the 3 years after that, and 3% the following 5 years, at what market interest rate (i) would you have to investyour money to maintain a real purchasing power growthrate (i') of 10%during those years? r1ImII ill Years1-5 Youmustinvestat 0.10+ 0.050+ (0.10)(0.050) = 0.1150= 11.50%per year. In Years 6-8 you must invest at 0.10 + 0.075 + (9.10)(0.075) = 0.1825 = 18.25%per year. In Years9-13 yoUmust invest at 0.10 + 0.030 + (0.10)(0.030)- 0.1330 == 13.30%per year. (Note: This,exarqple illustrates a common problem in timing: the market interest rate necessary to maintain the real purchasing power of the investment must be calculated at a time when the -inflation rate is Qnly an estimate. Most interest-bearing investments have fixed, up-front rates I!'il that the investorwell understandsgoingjnto the investment.Variable.rateinvestmentsat"ethe I, 'j I lexception. On the!other hand, inflation is not quantified, and its effect on our real return is n2t ~mea'sured;ffiitil1hwe1fd1>fthe fear5T~eI&e~10 acliiev~ilieconOitibns~reqiiire=a1n fue erampie, -- one would either ijave to anticipate inflation and adjust one's investments accordingly, or accept l the !<;lct~hat theJe~ investment return (i') mar not tupI;;out to be wh~t w~ origjnalty r~9.uiredJ I- I r <~ TI ~_tdJ 460 I - INFLATION AND PRICECHANGE INFLATION - EFFECT ON AFTER-TAX CALCULATIONS Earlier we noted the impact of inflationon before-tax calculations.Wefound that if the subsequent benefits brought constant quantities of dollars, then inflationwill diminish the true value of the future benefitsand, hence, the real rate of return. If, however,the future benefits keep up with the rate of inflation, the rate of return will not be adversely affected by the inflation. Unfortunately, we are not so lucky when we consider a situation with income taxes,as illustratedby Example14-9. . A $12,000 investment will refum'anI1ualbenefits for 6 years with np salvage value at the end of th~period. Assume straight-linedepreciationand a46% incometaX t;ate.Th~probleJ1}islo solve for both before- and after-taxrates of return, the latter fortwo sitUations: 1. No inflation: the annual b~nefitsare copstant at$:4918Reryear. 2. Inflation equal to 5%:. the benefits from theinve$tin~nt;increa&eatthiss~e rate, so that they continue to be the ~uivaIent of$2~:I8 in -y-ear,.O-})ase&'dbllars. The benefit schedule for the two,situationsis asfollows: ~ ' iI Annual Benefit for II I Year BotJt Situations (Year-O.based dollars), No Inflation, Actuaf Dollars Received 5% Inflation Factor* 5 %loflatioJ1, Actual DoUars Received 1 2 3 4 5 6 $291~ 2918 2918 2918 2918 ?918 $2918 29{8. 2918 2.918 2918 291.8 (t05)1 (1.05)2 (1.05)3 (1.05)4 (1.05)5 (1.05),6 $3064 3217 3378 3547 3724 39~0 !I t II t \1 I II II *May be read from the 5% compound interest tabfe as (F / P, 5%, n). " , I ~ ., II PW of cost \I J:>Wof benefit II ~.... IL iF ~ ,. - -- -- 12,OQO -.;.' __;iiO - , '. ' . '.' " 12,000., 2918(P /...4,i, 6}1'! ~'" (f[Adr62 = -<:'''"".,, ~'4~\J :.:... A; ,.I!I:._~ ;.~ .: 2918,:",-.~;. 21' .__ Blfti1M __!lb.' .' ___,___",_~"""~'C~..,,,,,,-<-,-:.~AtW-,__ II __ __~_~ '", I ) Fr¥.m compound interest tables: - --- -------- I Inflation Effect on After-Tax Calculations 461 After- Tax R~te of Return, No Inflation Year o 1-6 Before-Tax Cash Flow -$12,000 +2,918 Straight-Line Depreciation Taxable Income 46% Income Taxes $2000 $918 -$422 PW of cost Actual Dollars, and Year-O-BasedDollars, After-Tax CaSh Flow -$12,000 +2,496 = PW of benefit 12,000 = 2496(P / A, i, 6) (P / A, i, 6) = 12,000 2496 = 4.81 From compound interest tables: after-tax rate of return equals6.7%. I I After- Tax Rate of Return, 5 % Inflation Before- Tax Year 0 1 2 3 4 5 6 Cash Flow -$12,000 + 3,064 +3,217 +3,378 +3,547 +3,724 +3,910 Straight-Line Depreciation Taxable Income 46% Income Taxes $2000 2000 2000 2000 2000 2000 $1064 1217 1378 1547 1724 1910 -$489 -560 ,.-634 -712 -793 -879 Actual Dollars, After-Tax Cash Flow -$12'000 .' +2,575 + 2,657 +2,744 + 2,835 +2,931 +3,031 ,I I I Converting to Year-O-BasedDollars and Solving for Rate of Return Year o 1 2 3 4 5 6 Actual Dollars, After-Tax ,Cash Flow -$12,000 +2,575 Conversion Factor x(1.05)-1 = +2,657 x (1.05)-2 = +2,744 X (1.05)-3 = +2,835 X (1.05)-4 Year-O-Based Dollars, After-Tax Cash Flow -$12,000 +2,452 +2,410 +2,370 = +2,931 x(1.05)-5:;::: +3,031 X(1.05)-6 +2,332 +2,297 +2,262 = Present Wodhat5% -$12,000 +2,335 +2,186 +2,047 +1,919 + 1.800 + 1,688 -25 Present Wodhat.4% -$12,000 +2,358 +2,228 +2,107 + 1,993 +1,888 +1,788 +362 Linear interpolation between 4 and 5%: -- -- From Example 14-9, we see that the before-tax rate of return for both situations (no inflation and 5% inflation) is the same. Equal before-tax rates of return are expected because the - - - --- -- II:, 462 - INFLATION AND PRICECHANGE benefitsin the inflationsituationincreasedin proportionto the inflation.This exampleshows that when future benefitsfluctuatewith changesin inflationordeflation,the effectsdo notalter the year-O-baseddollar estimates. Thus, no special calculations are needed in before-tax calculations when future benefits are expected to respond to inflation or deflationrates. The after-tax calculations illustrate a different result. The two situations, with equal before-tax rates of return, do not produce equal after-taxrates of return: Situation No inflation 5% inflation Before-Tax Rate of Return 12% 12% Mter- Tax Rate of Return 6.7% 4.9% Thus, 5% inflation .results in a smaller after-tax rate of return, even though the benefits increase at the same rate as the inflation. A review of the cash flow table reveals that while benefits increase, the depreciation schedule does not. Thus, the inflation results in increased taxable income and, hence, larger income tax payments; but there are not sufficient increases in benefits to offset these additional disbursements. The result is that while the after-tax cash flow in actual dollars increases, the augmented amount is not high enough to offset both inflation and increased income taxes. This effect is readily apparent when the equivalent year-O-based-dollar after-tax cash flow is examined. With inflation, the year-O-based-dollar after-tax cash flow is smaller than the year-O-baseddollar after-tax cash flow without inflation. Of course, inflation might cause equipment to have a salvage value that was not forecast, or a larger one than had been projected. This effect would tend to reduce the unfavorable consequences of inflation on the after-tax rate of return. USING SPREADSHEETSFOR INFLATION CALCULATIONS Spreadsheets are the perfect tool for incorporating consideration of inflation into analyses of economic problems. For example, next year's labor costs are likely to be estimated as equal to this year's costs times (1 + f), where f is the inflation rate. Thus each year's value is different, so we can't use factors for uniform flows,A. Also the formulas that link different years are easy to write. The result is problems that are very tedious to do by hand, and easy by spreadsheet. Example 14-10 illustrates two different ways to write the equation for inflating costs~ Example 14-11illustrates that inflationreduces the after-taxrate of return because inflation makes the depreciation deduction less valuable. I . i . I G9stsf9r construction of a small, remotemi,ne are for labor apd tratispOJ.1ati9n.Jjabor costs illliI!' III I" ar~ e5}pectedt~be $350,000 t4e first year,\Vit?inflatiop. of6%annuany.Up.it~~W0rta~on aCQsts.~e.' e~pected"19 ipJiate at5% annually,bJit th~ gqluII).yllqr1Jiatyrial befug:n]J)Y~Cl~h;lJlges e~cJi)e.;lt.~ iiiiie:6 amriirs,:tke ffiin~<fuati($tl c~m~ ~tlIn1t1~:tO iJ~-:4b;<)QO~q~;000,~ $R(),Q20,ap.d $30,000 in Yews J thro~gh .4.Theinflatiop~a~e f9rthe val~e 9f~~dol~~ is 7 %" .»;.~all ~; .is th£ egu.ival~nt;lQJ:luaJc()~tfo:rthj$ 4-YeID"i?:roject? ,Q~tie~n iif t4e JIDJl J!.§e§.l!!l i I .Qt. '''.. - M< .. .,;j,":: 111I 1;.; ~= ~i lEt' !;; ~~ . f' ~.;~ -- I f i , I J ~...I Using Spreadsheets for Inflation Calculations 463 SOlUT:ION' The data for labor costs can be stated so that no inflationneeds to be applied in Year 1: the cost is $350,000. In contrast, the transportation costs for Year 1 are determined by multiplying $40,000 by 1.05 (= 1 + f). Also in later years the labor costt = labor coStt-l (1 + f), while each transportation cost must be computed as the time-Ovalue times (1 + f)t. In Figure 14-6, the numbers in the Year0 (or real) dollar column equal the values in the actual dollars column divided by (I,03.Y. .. A 1 2 - Inflation Rate '. 3 4 B Year . . 6% , ..Labor C ., Costs 1 120',000 6 7 8 9 2 3 4 127,200 134,832 11 ,142,922 / ."'--........-.-.... ..:.. ., "...;- _-;..,.'_. .-'J-' . E ,- .--.- F -. G .-..-... H --7% InflationFree Int&esr - 3% - ,_. ", ''''.'' Transportation Costs Year 0 $s Actual $8 Total Total Actual $8 Real $s 193,350 192,713 179,387 182,251 176,360 159,383 ... . "42:000---162,000."'157,282--= Es/(l+$F$2)"A5 -40,00060,000 50;000 30,000 / =B7*(1 +$B$2) - - 5% . 5 10 D 66,150 57,881 36,465 / I / t I .1. , $571,732= NPV(Fl,F5:F8) =B8+D8 $168,791 ==.",.PMT(Fl,4,F9) =C8*(1 +$D$2)"A8 FIGURE 14-6 Spreadsheetforinflation. The equivalent annual cost equals $168,791. For the data of Example 14-9, calculate the IRR with and without inflation with MACRS depreciation. How are the results affectedby inflation by comparison with the earlier results. II! SOlUTlQN .= L.... Most of the formulas for this spreadsheet are given in rows 11 and 12 for the data in Year6. The benefits received are computed from the base value in cell B5. The depreciation is the MACRS percentage times the $120,000 spent in Year O.This value is not influenced by inflation, so the depreciation deduction is less valuable as inflation increases. The tax paid equals the tax rate times the taxable income, which equals dollars received minus the depreciation charge. Then ATCF (after-tax cash flow) equals the before-tax cash flow.minusthe tax paid. In~Fig~ 1<t.-7~ ifbfi'Ce="thllfm Yeaf2"thcde-prd!iatidniicharge is~nu.ge~nobgfito cause'-='thfs project to pe.y"negative" tax. for a firm, this meaI)sthat the deductionon this project will be used to offset inc.omefrom other projects. -- - - - - - - ~r \ ~: 464 INFLATION AND PRICE CHANGE A 1 2 3 4 5 6 7 8 9 10 I 11 0% - 12 B C D == InflationRate E 46% F = Tax Rate Actual $$ MACRS Actual $8 Actual $$ Actual$$ Received Deprec.% peprec. ATCF Tax -12000 -12000 2918 20.00% 2400 238 2680 2918 32.00% 840 -424 3342 2918 19.20% 282 2304 2636 2918 11.52% 706 ta82 2212 706 2918 11.52% 1382 2212 2918 5.76% ....6Ql. ,)024 ,)894 Year 0 1 2 3 4 5 6 Formulas for Yr q ./ / 13 14 5% Real$$ ATCF j -'-12000 2680 =:;:.F5/(1+$B$l)AA5 3342 2636 2212 2212 '1894 ' = Inflationi4te ;:;;;IRR 7.29% (BlO-IJtO)*$E$t " H #BlO-E10 ==..;-"$B$4",CI0 . /' ;:=$B$5*(1-j-$B$1)AA10 G 46% . ;:;;;Tax Rate > ,- 15 16 Year I 17 0 .... 18 1 3064 2().OR% 32.00% 20 .... 3 3217 3378 21 4 3547 22 5 3724 3910 11.52% 5.76% 19 I , ! I I I ~ I JI I I A1'CF Real$s ATCF -.-12000 '"'-12000 Actllal $8 .Actlla.l$s Tax ,. .. -12000 6 23 ! 24 Actual$$ MACRS ActuaUs Received Depree. % Pelmc. 2759 2627 3504 3178 494 2884 . 2,491 996. 2551 ." .2099. J()n 1481 2647 '.' 2400 ......305 . " 3840 ..,287 19.20% 2304 11.52% ' 1382 1382 .691 . > :!;; FIct.URE=:14-7 , e'After-taxJRR8::.with . .2074 1813 243() . .' . 5.68% ..., '..I:@.. w - .. MAGRS:anddn:flation~;:; ' 1/ The IRRs arethigher in this eXaJ1.lple(1~29%without' inflation yi';6.7% with straight':lfne ;: ,._ ._ .', : __ " '. . ,,' ", ,-'"' '", :_ ,_, .:: .', ,_ . " .,_ . - "'. .. ',', :,{i-~:'"":,, . ,.-~'ji:; --_,:co' ,..,' _ _ " . ...:. _','.'_ . ., , , ., :.' . ,". . .. _ _ ,', . _"..,'. ..., ,', __ "':'" . . .. _ '",' ,,' .i .. ... , deyreciationin Example 14-9,and5.<58%with irt'flationvs 4.9%) b~causeMACR,Ssupportsfaster depreciation, so the depreciation deductioQsare II1orevalllable..Also because the depreciationis fa~!er,the results are affected ~6mewhatlessby inf\a.qQn~ Specifically;witb,'MACRS5% inflation lowers the IRR by 1.6% and Withstrai~ht-:lilledepredation, 5% infliltionlowersthe 1M by 1.8%. SUMMARY Inflation is characterizedby rising prices for goods and services,while deflationproducesa fall in prices. An inflationarytrend makes future dollars have less purchasing power than present dollars. Inflationbenefits a long-term borrowerof money because paymeritof debt in the future is made with dollars that have reduced purchasing power. This advantage to borrowers is at the expense of lenders. Deflation has the opposite effect from inflation. If money is borrowed over a period of time in which deflation is occurring, then debt will be repaid with dollars that have more -- -- ----- 466 INFLATION AND PRICECHANGE 14-5 What is the Consumer Price Index (CPI)? What is the difference between commodity specific and composite price indexes? Can each be used in engineering economic analysis? 14-6 In Chapters 5 (Present Worth Analysis) and 6 (Annual Cash Flow Analysis) it is assumed that prices are stable and a machine purchased today for $5000 can be replaced for the same amount many years hence. In fact, prices have generally been rising, so the stable price assumption tends to be incorrect. Under what circumstances is it appropriate to use the "stable price" assumption when prices actually are changing? 14-7 An economist has predicted that there will be a 7% per year inflation of prices during the next 10 years. If this proves to be correct, how much will an item that presently sells for $10 bring a decade hence? (Answer: $19.67) 14-8 A man bought a 5% tax-free municipal bond. It cost $1000 and will pay $50 interest each year for 20 years. The bond will mature at the end of the 20 years and return the original $1000. If there is 2% annual inflation during this period, what rate of return will the investor receive after the effect of inflation has been accounted for? 14-9 A man wishes to set aside some money for his daughter's college education. His goal is to have a bank savings account containing an amount equivalent to $20,000 with today's purchasing power of the dollar, at the girl's 18th birthday. The estimated inflation rate is 8%. If the bank pays 5% compounded annually, what lump sum of money should he deposit in the bank savings account on the child's 4th birthday? (Answer: $29,670) . . 14-10 An economist has predicted that for the next 5 years, the United States will have an 8% annual inflation rate, followed by 5 years at a 6% inflation rate. This is equivalent to what average price change per year for the entire 1O-year period? 14-11 A newspaper reports that in the last 5 years, prices have increased a total of 50%. This is equivalent to what annual inflation rate, compounded annually? (Answer: 8.45%) 14-12 A South American country has had a high rate of inflation. Recently, its exchange rate was 15 cruzados per dollar; that is, one dollar will buy 15 cruzados in the foreign exchange market. It is likely that the country will continue to experience a 25% inflation rate and that the United States will continue at a 7% inflation rate. Assume that the exchange rate will vary the same as the inflation. In this situation, one dollar will buy how many cruzados 5 years from now? (Answer: 32.6) 14-13 An automobile manufacturer has an automobile that gets 10 kilometers per liter of gasoline. It is estimated that gasoline prices will increase at a 12% per year rate, compounded annually, for the next 8 years. This manufacturer believes that the automobile fuel consumption for its new automobiles should decline as fuel prices increase, so that the fuel cost will remain constant. To achieve this~ what must be the fuel rating, in kilometers per liter, of the automobiles 8 years hence? 14-14 An economist has predicted that during the next 6 years, prices in the United States will increase 55%. He expects a further increase of25% in the subsequent 4 years, so that prices at the end of 10 years will have increased to 180% of the present level. Compute the inflation rate, f, for the entire 10-year period. 14-15 Sally Johnson loaned a friend $10,000 at 15% interest, compounded annually. She is to repay the loan .in five equal end-of-year payments. Sally estimates the inflation rate during this period is 12%. After taking inflation into account, what rate of return is Sally receiving on the loan? Compute your answer to the nearest 0.1%. (Answer: 2.7%) 14-16 Dale saw that the campus bookstore is having a special on pads of computation paper normally priced at $3 a pad, now on sale for $2.50 a pad. This sale is unusual and Dale assumes the paper will not be put on sale again. On the other hand, Dale expects that there will be no increase in the $3 regular price, even though the inflation rate is 2% every 3 months. Dale believes that competition in the paper industry will keep wholesale and retail prices constant. He uses a pad of computation paper every 3 months. Dale considers 19.25% a suitable minimum attractive rate of return. Dale will buy one pad of paper for his immediate needs. How many extra pads of computation paper should he buy? (Answer: 4) 14-17 An investor wants a real rate of return if (rate of return without inflation) of 10% per year on any projects in which he invests. If the expected annual inflation rate for the next several years is 6%, what interest rate i should be used in project analysis calculations? Problems 14-18 (a) Compute the equivalent annual inflation rate, based on the consumer price index, for the period from 1981 to' 1986. (b) Using the equivalent annual inflation rate computed in part (a), estimate the consumer price index in 1996, working from the 1987 Consumer Price Index. 14-19 How much will a $20,000 automobile cost 10 years from now if inflation continues at an annual rate of 4% for the next decade? 14-20 You are considering the purchase, for $15,000, of an annuity that pays $2500 per year for the next 10 years. You want to have a real rate of return of 5%, and you estimate inflation will average 6% per year over the next 10 years. Should you buy the annuity? 14-21 Inflation is a reality for the general economy of the United States for the foreseeable future. Given this assumption, calculate the number of years it will take for the purchasing power of today's dollars to equal one-fifth of their present value. Assume that inflation will average 6% per year. 14-22 A homebuilder's advertising has the caption, "Inflation to Continue for Many Years." The advertisement continues with the explanation that if one buys a home now for $97,000, and inflation continues at a 7% annual rate, the home will be worth $268,000 in 15 years. According to the advertisement, by purchasing a new home now, the buyer will realize a profit of $171,000 in 15 years. Do you find this logic persuasive? Explain. 14-23 Sally Seashell bought a lot at the Salty Sea for $18,000 cash. She does not plan to build on the lot, but instead will hold it as an investment for 10 years. She wants a 10% after-tax rate of return after taking the 6% annual inflation rate into account. If income taxes amount to 15% of the capital gain, at what price must she sell the lot at the end of the 10 years? (Answer: $95,188) 14-24 A group of students decided to lease and run a gasoline service station. The lease is for 10 years. Almost immediately the students were ~onfronted with the need to alter the gasoline pumps to read in liters. The Dayton Company has a conversion kit available for $900 that may be expected to last 10 years. The firm also sells a $500 conversion kit that has a 5-year useful life. The students believe that any money not invested in the conversion kits may be invested elsewhere at a 10% interest rate. Income tax consequences are to be ignored in this problem. 467 (a) Assuming that future replacement kits cost the same as today, which alternative should be selected? (b) If one assumes a 7% inflation rate, which alternative should be selected? 14-25 Pollution control equipment must be purchased to remove the suspended organic material from liquid. being discharged from a vegetable packing plant. Two alternative pieces of equipment are available that would accomplish the task. A Filterco unit presently costs $7000 and has a 5-year useful life. A Duro unit, on the other hand, now costs $10,000 but will have a lO-year useful life. With inflation, equipment costs are rising at 8% per year, compounded annually, so when the Filterco unit needed to be replaced, the cost would be much more than $7000. Based on a lO-year analysis period, and a 20% minimum attractive rate of retuni, before taxes, which piece of pollution control equipment should be purchased? 14-26 The City of Columbia is trying to attract a new manufacturing business to the area. It has offered to install and operate a water pumping plant to provide service to the proposed plant site. This would cost $50,000 now, plus $5000 per year in operating costs for the next 10 years, all measured in year-O dollars. To reimburse the city, the new business must pay a fixed uniform annual fee, A, at the end of each year for 10 years. In addition, it is to pay the city $50,000 at the end of 10 years. It has been agreed that the city should receive a 3% rate of return, after taking an inflation rate, f, of 7% into account. Determine the amount of the uniform annual fee. (Answer:$12,100) . 14-27 Sam purchased a home for $150,000 with some creative financing. The bank, which agreed to lend Sam $120,000 for 6 years at 15% interest, took a first mortgage on the house. The Joneses, who sold Sam the house, agreed to lend Sam the remaining $30,000 for 6 years at 12% interest. They received a second mortgage on the house. Thus Sam became the owner without putting up any cash. Sam pays $1500 a month on the first mortgage and $300 a month on the second mortgage. In both cases these are "interes~only". loans, and the principal is due at the end of the loan. Sam rented the house, but after paying the taxes, insurance, and so on, he had only $800 left. so he was forced to put up $1000 a month of his own money to make the monthly payments on the mortgages. At I 468 INFLATION AND PRICECHANGE the end of 3 years, Sam sold the house for $205,000. After paying off the two loans and the real estate broker, he had $40,365 left. After taking an 8% inflation rate into account, what was his before-tax rate of return? 14-28 General price inflation is estimated to be 3% for the next 5 years, 5% the 5 years after that, and 8% the following 5 years. If you invest $10,000 at 10% for those 15 years, what is the future worth of your investment in terms of actual dollars at that time and in terms of real base-zero dollars at that time? 14-29 Ima Luckygirl recently found out that her grandfather has passed away and left her his Rocky Mountain Gold savings account. The account was originally opened 50 years ago when Ima's grandfather deposited $2500. He had not added to or subtracted from the account since then. If the account has Company C: A constant $65,000 per year in terms of Year-5-based purchasing power. 14-32 Calculate the future equivalent in Year 15 of: (a) Dollars having today's purchasing power. (b) Then-current purchasing power dollars, of $10,000 today. Use a market interest rate of 15% and an inflation rate of 8%. 14-33 A firm is having a large piece of equipment overhauled. It anticipates that the machine will be needed for the next 12 years. The firm has an 8% minimum attractive rate of return. The contractor has suggested three alternatives: (a) A complete overhaul for $6000 that should permit 12 years of operation. (b) A major overhaul for $4500 that can be expected to provide 8 years of service. At the end of 8 years, a minor overhaul would be needed. earned an average rate of 10% per year and inflation has been 4% per year, answer the following: (a) How much money "iscurrently in the account in actual dollars? (b) Express the answer to part (a) in terms of the purchasing power of dollars from 50 years ago. 14-30 Auntie Frannie wants to provide tuition for her twin nephews to attend a private school. She-anticipates sending a check for $2000 at the end of each of the next 8 years to apply to the cost of schooling. (a) If general price inflation, as well as tuition price inflation, is expected to average 5% per year for those 8 years, calculate the present worth of the gifts. Assume that the real interest rate will be 3% per year. (b) If Auntie Frannie wants her gifts to keep pace with inflation, what would be the present worth of her gifts? Again assume inflation is 5% and the real interest rate is 3%. (c) A minor overhaul now. At the end of 4 arid 8 years, additional minor overhauls would be needed. If minor overhauls cost $2500, which alternative should the firm select? If minor overhauls, which now cost $2500, increase in cost at +5% per year, but other costs remain unchanged, which alte~ative should the firm select? (Answers: Alt. (c); Alt. (a» 14-34 A couple in Ruston, Louisiana, must decide whether it is more economical to buy a home or to continue to rent during an inflationary period. Presently the couple rents a one-bedroom duplex for $450 a month plus $139 a month in basic utilities (heating and cooling). These costs tend to increase with inflation, and with the projected inflation rate of 5%, the couple's monthly costs per year over a 1O-yearplanning horizon are as follows. n 14-31 As a recent graduate, you are considering employment offers from three different companies. However, in an effort to confuse you and perhaps make their offers seem better, each company has used a different purchasing power base for expressing your annual salary over the next 5 years. If you anticipate inflation to be 6% for the next 5 years and your personal (real) MARR is 8%, which plan would you choose? Company A: A constant $50,000 per yefU"in terms of today's purchasing power. Company B: $45,000 the first year, with increases of $2500 per year thereafter. --- = Rent Utilities 1. 2 3 4 5 6 7 8 9 10 450 473 496 521 547 574.~ 603 633 665 698 139 146 153 161 169 177 186 196 205 216 The couple would like to live on the north side of the town, where an average home of 150 m2 of heating area costs $75,000. A local mortgage company will provide a loan for the property provided the couple makes a down payment of 5% plus estimated closing costs of 1% cash for the home. The couple prefers a 30-year fixed-rate mortgage with an 8% interest rate. Based on the couple's gross annual income, the couple falls in the 30% marginal income tax rate (federal plus state), and as such, buying a home will provide them some tax write-off. It is also estimated that the basic utilities for the home inflating at 5% --- -- Problems will cost $160 per month; insurance and maintenance also inflating at 5% ,will cost $50 per month. The home will appreciate in value about 6% per year. Assuming a nominal interest rate of 15.5%, which alternative will be more attractive to the couple on the basis of the present worth analysis? (Note: Realtor's sales commission here is 5%.) 35 Given the following data, calculate the present worth of the investment. First cost = $60,000 Salvage value = Project life = 10 years $15,000 MARR = 25% General price inflation = 4% per year Annualcost 1 = $4500in Year1and . inflating at 2.5% per year Annual cost 2 = $7000 in Year 1 and inflating at 10.0% per year Annual cost 3 = $10,000 in Year 1 and inflating at 6.5% per year Annual cost 4 = $8500 in Year 1 and inflating at -2.5% per year J6 Here is some information about a professor salary index (PSI). PSI Year Change in PSI 1991 1992 1993 1994 1995 1996 1997 1998 82 89 100 b 107 116 e 132 3.22% 8.50 a 4.00 c d 5.17 7.58 (a\ Calculate the unknown quantities a, b, c, d, e in me table. Review Equation 14..:3. (b) What is the base year of the PSI? How did you determine it? (c) Given the data for the PSI, calculate the average annual price increase in salaries paid to professors for between 1991 and 1995 and between 1992 and 1998. 7 From the data in Table 14-1 in the text calculate the average annual inflation rate of first class postage as measured by the LCI for the following years: (a) End of 1970 to end of 1979 (b) End of 1980 to end of 1989 (c) End of 1990 to end of 1999 8 From the data in Table 14-1 in the text, calculate the overall rate change of first-class postage as measured 469 by the LCI for the following decades: (a) The 1970s (1970-1979) (b) The 1980s (1980-1989) (c) The 1990s (1990-1999) 14-39 From the data in Table 14-2 in the text calculate the average annual inflation rate as measured by the CPI for the following years: (a) End of 1973 to end of 1982 (b) End of 1980 to end of 1989 (c) End of 1985 to end of 2002 14-40 Homeowner Henry is building a fireplace for the house he is constructing. He estimates that his fireplace will require 800 bricks. Answer the following: (a) If the cost of a chimney brick in 1978 was $2.10, calculate the material cost of Henry's project in 1998. The chimney brick index (CBI) was 442 in 1970 and is expected to be 618 in 1998. (b) Estimate the material cost of a similar fireplace to be built in the year 2008. What assumption did you make? 14-41 If a composite price index for the cost of vegetarian foods called eggs, artichokes and tofu (EAT) was at a value of 330 ten years ago, and has averaged an increase of 12% per year after that, calculate the current value of the index. 14-42 As the owner of Beanie Bob's Basement Brewery, you are interested in a construction project to increase production to offset competition from your crosstown rival, Bad Brad's Brewery and Poolhall. Construction cost percentage increases, as well as current cost estimates, for required construction costs are given in the table below over a three year period. Use a market interest rate of 25%, and assume that general price inflation is 5% over the 3-year period. Cost If Incurred Item Year 1 Year 2 Year 3 4.3% 3.2% 6.6% 14,000 35,000 2.0 1.6 2.5 2.1 3.0 3.6 9,000 85,000 5.8 5.0 6.0 4.5 7.5 4.5 Today Structural metal/concrete Roofing materials Heating/plumbing equipment/fixtures Insulation material Labor Cost Percentage Increase $120,000 (a) What would the costs be for labor in Years 1,2, and 3? ! ~I . i. " .. ,-fJ LJ.. -------- . --! 470 (b) What is the average percentage increase of labor cost .over the 3-year period? . (c) What is the present worth of the insulation cost of this project? (d) Calculate the future worth of the labor and insulation material cost portion of the project. (e) Calculate the present worth of the total construction project for Beanie Bob. 14-43 Philippe Marie wants to race in the Tour de France ten a . - INFLATION AND PRICECHANGE years from now. He wants to know what the cost of a custom-builtracing bicycle will be ten years from today. Calculate the cost given the following data. Item Frame Wheels Gearing system Braking system Saddle Finishes Current Cost $800 350 200 150 70 125 Cost Will Inflate x% per Year 2 % 10 5 3 2.5 8 14-44 Due to cost structures, trade policies and corporate changes the costs for three big automakers are estimated to vary over the next three-year period. These changes will be reflected in the purchase prices of their vehicles. Mary Clare will graduate in three years and will be buying a new vehicle-she is considering one model from each company. Which car should Mary Clare purchase three years from now, assuming everything but purchase price is equivalent? Price Will Inflate Automaker Current Price x% per Year X 4 % $27,500 1.5 Y 30,000 8 Z 25,000 14-45 Granny Viola has been saving money in the Bread & Butter mutual fund for 15 years. She has been a steady contributor to this fund over those years and has a pattern of putting $100 into the account every 3 months. If her original investment 15 years ago was $500 and interest in the account has varied as shown, what is the current value of her savings? Years Interest Earned in the Account 1-5 6-10 10-15 12% compounded quarterly 16 compounded quarterly 8 compounded quarterly 14-46 Andrew just purchased a new boat for $15,000 to use on the river near his home. He has received delivery of the boat, and agreed to the terms of the following loan: all principal and interest is due in three years (balloon loan), first year annual interest (on the purchase price) is set at 5%, this is to be adjusted up 1.5% per year for each of the following years of the loan. How much does Andrew owe to payoff the loan in three years? 14-47 You were recently looking at the historical pri~es paid for homes in a neighborhood that you are interested in. The data that you found, average price paid, is given below. Calculate on a year-to-year basis how home prices in this neighborhood have inflated (a-e in the table below). Year 5 years ago 4 years ago 3 years ago 2 years ago last year this year Average Home Price Inflation Rate for That Year $165,000 167,000 172,000 180,000 183,000 190,000 (a) (b) .(c) (d)" (e) (f, see below) (f) What is your estimate of the inflation rate for this year? 14-48 The U.S. tax laws provide for the depreciation of equipment based on original cost. Yet owing to substantial inflation, the replacement cost of equipment is often much greater than the original cost. What effect, if any, does this have on a firm's ability to buy new equipment to replace old equipment? 14-49 Sam Johnson inherited $85,000 from his father. Sam is considering investing the money in a house, which he will then rent to tenants. The $85,000 cost of the propertY,consists of $17,500 for the land, and $67,500 for the house. Sam believes he can rent the house and have $8000 a year net income left after paying the property taxes and other expenses. The house will be depreciated by straight-line depreciation using a 45-year depreciable life. (a) If the property is sold at the end of 5 years for its book value at that time, what after-tax rate of return will Sam receive? Assume that his incremental personal income tax rate is 34% for federal and state taxes. (b) Now assume there is 7% per year inflation, compounded annually. Sam will increase the rent 7% per year to match the inflation rate, so that after considering increased taxes and other expenses, the annual net income will go up 7% per year. -- Problems Assume Sam's incremental income tax rate remains at 34:.%for all ordinary taxable income related to the property. The value of the property is now projected to increase from its present $85,000 at a rate of 10% per year, compounded annually. If the property is sold at the end of 5 years, compute the rate of return on the after-tax cash flow in actual dollars. Also compute the rate of return on the after-tax cash flow in year-O dollars. 14-50 Tom Ward put $10,000 in a 5-year certificate of deposit that pays 12% interest per year. At the end of the 5 years the certificate will mature and he will receive his $10,000 back. Tom has substantial income from other sources and estimates that his incremental income tax rate is 42%. If the inflation rate is 7% per year, find his (a) before-tax rate of return, ignoring inflation (b) after-tax rate of return, ignoring inflation (c) after-tax rate of return, after taking inflation into account 14-51 Dick DeWolf and his wife have a total taxable income of $60,000 this year and file a joint federal income tax return. If inflation continues for the next 20 years at a 7% rate, compounded annually, Dick wonders what their taxable income must be in the future to provide them the same purchasing power, after taxes, as their present taxable income. Assuming the federal income tax rate table is unchanged, what must their taxable income be 20 years from now? 14-52 A small research device is purchased for $10,000 and depreciated by MACRS depreciation. The net benefits from the device, before deducting depreciation, a,re $2000 at the end of the first year and increasing $1000 per year after that (second year equals $3000, third year equals $4000, etc.), until the device is hauled to the junkyard at the end of 7 years. During the 7-year period there is an inflation rate f of 7%. This profitable corporation has a 50% combined federal and state income tax rate. If it requires a 12% after-tax rate of return on its investment, after taking inflation into account, should the device have been purchased? 14-53 When there is little or no inflation, a homeowner can expect to rent an unfurnished home for 12% of the market value of the property (home and land) per year. About 18 of the rental income is paid out for property taxes, insurance, and other operating expenses. Thus the net annual income to the owner is 10.5% 471 of the market value of the property. Since prices are relatively stable, the future selling price of the property often equals the original price paid by the owner. For a $150,000 property (where the land is estimated at $46,500 of the $150,000), compute the after-tax rate of return, assuming the selling price 59 months later (in December) equals the original pur-, chase price. Use modified accelerated cost recovery system depreciation beginning January 1. Also, assume a 35% income tax rate. (Answer: 6.84%) 14-54 (This is a continuation of Problem 14-53.) As inflation has increased throughout the world, the rental income of homes has decreased and a net annual rental income of 8% of the market value of the property is common. On the other hand, the market value of homes tends to rise about 2% per year more than the inflation rate. As a result, both annual net rental income, and the resale value of the property rise faster than the inflation rate. Consider the following situation. A $150,000 property (with the house valued at $103,500 and the land at $46,500) is purchased for cash in Year O. The market value of the property increases at a 12% annual rate. The annual rental income is 8% of the beginning-of-year market value of the property. Thus the rental income also increases each year. The general inflation rate f is 10%. The individual who purchased the property has an average income tax rate of 35%. (a) Use MACRS depreciation, beginning January 1, to compute the actual dollar after-tax rate of return for the owner, assuming he sells the property 59 months later (in December). (b) Similarly, compute the after-tax rate of return for the owner, after taking the general inflation rate into account, assuming he sells the property 59 months later. 14-55 Consider two mutually exclusive alternatives stated in year-O dollars. Both alternatives have a 3-year life with no salvage value. Assume the annual inflation rate is 5%, an income tax rate of 25%, and straight line depreciation. The minimum attractive rate of return (MARR) is 7%. Use rate ofreturn analysis to determine which alternative is preferable. A -$420 200 200 200 Year 0 1 2 3 - -- - -- B -$300 150 150 150 I ~;~~": <"". -'-:-'_.~'>< ';--~ :::-:-g -~...~ - -~ t'~~Ji~~~~~~~f~~~~~~. After Completing This Chapter... The student should be able to: · ·· ·· Define various sources of capital and the costs of those funds to the firm. Selecta firm's MARRbased on the opportunitycost approachfor analyzinginvestments. Adjust the firm's MARR to account for risk and uncertainty. Discuss the impact of inflation and the cost of borrowed money. Use spreadsheetsto developcumulativeinvestmentsand the opportunitycost of capital. QUESTIONS TO CONSIDER ~ 1. Many of BP's older sources of oil are now reaching the end of their useful life. How might this have affected the firm's decision to invest in Tyumen Oil? 2. Outside of Tyumen,there are few other oil companies left in Russia for competitors to buy.How might this have affected BP's decision? 3. Where else are the world's largest availableoil reserves located? How does the business climate in these areas compare to Russia's? 4. Most of the largest oil-producing nations are members of OPEC (the Organization of Petroleum Exporting Countries), but Russia is not. Is this good orbad forBP? - -- Selection of a MinimumAttractive Rate of Return BP Goes to Russia II II " " In the early 1990s,western investors flockedto Russia, hoping to reap a fortune as market~ opened up followingthefall of the SovietUnion.Manydidn't stay long once they discovered what it was like to do businessin a country where contracts wereoften impossibleto enforce and bribery was the norm. In 1998, when Russia devalued its currency and defaulted on debt obligations, most of the remaining investors fled in panic. Despite this dismalbusinessoutlook,British Petroleum (BP) announced in early 2003 that it was planning to pay $6.75 billion for a 50% interest in Tyumen Oil Company, Russia's fourth largest producer of oil. BP's decision is particularly striking in view of the company's own past history in Russia: in 1997 it bought a share in a small Russian oil company, only to lose part of its investmenta few years later after a bitter court battle. Moreover, Russia's state-owned pipeline infrastructure is outdated and inadequate-and the government has been slow to allow private companies to build and operate their own pipelines. Given all these drawbacks, BP would seem to be taking a big gambleby investingin Tyumen. . ..,oJ !~,~~~ . - - - --- - - - - - 474 I - SELECTION OF A MINIMUM ATTRACTIVERATEOF RETURN The preceding chapters have said very little about what interest rate or minimum attractive rate of return is suitablefor use in a particular situation.Since this problem is quite complex, there is no single answer that is always appropriate. A discussion of a suitable interestrate to use must inevitably begin with an examination of the sources of capital, followedby a look at the prospective investment opportunities and risk. Only in this way can an interest rate or minimum attractive rate of return be chosen intelligently. SOURCES OF CAPITAL In broad terms there are four sources of capital availableto a firm: money generated from the operation of the firm, borrowed money, sale of mortgage bonds, and sale of capital stock. Money Generated from the Operation of the Firm A major source of capital investment money is through the retention of profits resul.ting from the operation of the firm. Since only about half of the profits of industrial firins are paid out to stockholders, the half that is retained is an important source of funds for all purposes, including capital investments. In addition to profit, there is money generated in the business equal to the annual depreciation charges on existing capital assets if the firm is profitable. In other words, a profitable firm will generate money equal to its deprecia~on chargesplus its retained profits. Even a firm that earns zero profit will still generate money from operations equal to its depreciation charges. (A firm with a loss, of course, will have still less funds.) External Sources of Money When a firm requires money for a few weeks or months, it typically borrows from banks. Longer-term unsecured loans (of, say, 1-4 years) may also be arranged through banks. While banks undoubtedly finance a lot of capital expenditures,regular bank loans cannot be considered a source of permanent financing. Longer-term secured loans may be obtained from banks,insurance companies,pension funds, or even the public. The security for the loan is frequently a mortgage on specific property of the firm. When sold to the public, this financingis by mortgage bonds. The sale of stock in the firm is still another source of money.While bank loans and bonds represent debt that has a maturity date, stock is considered a permanent addition to the ownershipof the firm. Choice of Source of Funds Choosing the source of funds for capital expenditures is a decision for the firm's top executives,and it may require approval of the board of directors. When internal operations generate adequate funds for the desired capital expenditures,external sources of money are not likely to be used. But when the internal sources are inadequate, external sources must be employed or the capital expenditures will have to be deferred or canceled. ---- - --- - ! . u . .- - Cost of Funds 475 COST OF FUNDS 'Cost of Borrowed Money A first step in deciding on a minimum attractive rate of return might be to determine the interest rate at which money can be borrowed. Longer-term secured loans may be obtained from banks, insurance companies, or the variety of places in which substantial amounts of money accumulates (for example, the oil-producing nations). A large, profitable corporation might be able to borrow money at the prime rate, that is, the interest rate that banks charge their best and most sought-after customers. All other firms are charged an interest rate that is higher by one-half to several percentage points. In addition to the financial strength of the borrower and his ability to repay the loan, the interest rate will vary depending on the duration of the loan. Cost of Capital Another relevant interest rate is the cost of capital. The general assumption concerning the cost of capital is that all the money the firm uses for investmentsis drawn from all the components of the overall capitalization of the firm. The mechanics of the computation is given in Example 15-1. For a particular firm, the purchasers of common stock require an 11% rate of return, mortgage bonds are sold at a 7% interest rate, and bank loans are available at 9%. Compute the cost of capital for the following capital structure: . Rate of Annual Return Amount 9% $ 20 million Bank loan $1.8 million 1.4 7 20 Mortgage bonds 11 6.6 60 Common stock and retained earnings $100 million $9.8 million , SOLUTION r Interest payments on debt, like bank loans and mortgage bonds, are tax deductible business .expenses. Thus: After-tax interest cost = (Before-tax interest cost) x (1 - Tax rate) If,weass4me thaMl).e..firm,pays 40% income::taxes,,;the computat:j.onSbecome: ': -------- Bank loan After-tax interest cost .9%(1- 0.40) MOJ:tgageponds AfteJi:-taxinteJ;est cost ~~ 7~(1- - -- 0.40) ~ ---.. .: - -- 5.4% 4.2% --~ --- -- - - 476 - SELECTIONOF A MINIMUM ATTRACTIVERATEOF RETURN Dividends paid on the ownership in the finn (common stock + retained earnings) are not taX deductiple. Combining the three components, the after-tax interest cost for the $\00 million of capital is: $20 million (5.4%) + $20 millioll (4.2%) + $60p:)jllionOJ %) i z:... .. $8.52rp.illion .. $8.52II1illion Cost of.capItal ..>';';,.;..~ """ .. . 8.52% . . In an actual situation, the cost of capital is quite difficult to compute. The fluctuation in the price of common stock, for example, makes it difficultto pick a cost, and because of. the fluctuatingprospects of the finn, it is even more difficultto estimate the future benefits purchasers of the stock might expect to receive. Given the fluctuating costs and prospects of future benefits, what rate of return do stockholders require? There is no precise answer, but we can obtain an approximate answer.Similar assumptionsmust be made for the other. components of a finn's capitalization. . INVESTMENT OPPORTUNITIES An industrial finn can invest its money in many more places than are available to an individual. A finn has larger amounts of money, and this alone makes certain kinds of investment possible that are unavailable to individual investors, with their more limited investment funds. The U.S. government, for example, borrows money for short tenns of 90 or 180 days by issuing certificates called Treasury bills that frequently yield a .greater interest rate than savings accounts. The customary minimum purchase is $25,000. More important, however,is the fact that a finn conducts a business, which itself offers . manyinvestmentopportunities.Whileexceptionscan be found,a goodgeneraliza~onis that the opportunities for investment of money within the finn are superior to the investment opportunities outside the finn. Consider the available investment opportunities for a particular finn as outlined in Table 15-1. Figure 15-1 plots these projects by rate ofretum versus investment. The cumulative investmentrequired for all projects at or above a given rate of return is given in Figure 15-2. Figures 15-1 and 15-2 illustrate that a finn may have a broad range of investment opportunities availableat varying rates of return and with varying lives and uncertainties.It may take some study and searching to identify the better investmentprojects available to a finn. If this is done, the available projects will almost certainly exceed the money the firm budgets for capital investmentprojects. Opportunity Cost We see that there are two aspects of investing that are basically independent. One factor is the source and quantity of money availablefor capital investmentprojects. The other aspect is the investmentopportunities themselves that are available to the firm. These two situations are typically out of balance, with investment opportunities exceeding the available money supply. Thus some investment opportunities can be selected - - - -- - . --, . ___... . __.... _. ... ..d. Investment Opportunities 477 TABLE15-1 A Firm1sAvailable Investment Opportunities , Estimated Rate of Return Project Number Project Cost (x103) InvestmentRelated to Current Operations New equipmentto reduce labor costs $150 Other new equipmentto reduce labor costs 50 Overhaulparticular machine to reduce 50 material costs New test equipmentto reduce defective 100 products produced 1 2 3 4 New Operations Manufacture parts that previously had been purchased Further processing of products previously sold in semifinished form Further processing of other products 5 6 7 30% 45 38 40 200 35 100 28 200 18 250 25 New Production Facilities 8 Relocate production to new plant External Investments 9 10 11 12 Investment in a different industry Other investment in a different industry Overseas investment Purchase of Treasury bills 300 300 400 Unlimited 20 10 15 8 Unlimited $500 ,.... ...., 0 X '-' $400 .... c: Q) Q) $300 ;> c: .... .... 0 .21 0 $200 $100 o 45 40 38 35 30 28 25 20 18 Rate of Return (%) FIGURE 15-1 Rate of return versus project investment. 15 10 8 iir" .... ] 478 SELECTION OF A MINIMUM ATTRACTIVERATEOF RETURN 45 40 35 30 e B <U 25 'S 20 <U 15 Project Number t 1: 5 [ 6 o 8 I 9 I 7 I 11 10 I 12 200 400 600 800 1000 1200 1400 1600 1800 2000 2200 2400 CumulativeInvestment($ X 103) FIGURE 15-2 Cumulative investment required for all projects at or above a given rate of return. and many must be rejected. Obviously, we want to ensure that all the selected projects are better than the best rejected project. To do this, we must know something about the rate of return on the best rejected project. The best rejected project is the best opportunity forgone, and this in turn is called the opportunity cost. Opportunity cost = Cost of the best opportunity forgone = Rate of return on the best rejected project If one could predict the opportunity cost for some future period (like the next 12 months), this rate of return could be one way tojudge whetherto accept or rejectany proposed capital expenditure. Consider the situation represented by Figures 15-1 and 15-2. For a capital expenditurebudget of $1.2 million ($1.2 x 106),what is the opportunity cost? .SOLUTION\. 1-' '. ' FroqlFigtiry 15-2 we see tbat the eight projects with a rate of return of 20% Orqlore require a cUqlp.lativewvestqlent6f$1.2 (x 106).We would ta,keOPthese projects andreject the other four (7, 'trI, 10, and 12)witli rates Qfr;etuI'J:l of 18%~o{less.]he bestreject~d projectis 7, and irhas-ah 18%rate ofTettitn.1'huStheopPorWnity co,stis 18%, C""~'~":_;:;;:,'..": I l = ,_,,.,,,,:,,:",,,._:,4>:,,:,,, , "'." _' '/~:;.-;"-_.~ ... -- --,,-------.- --- "'_'_ =." _.~:.' -i:i..::. ,," _,,;',cM' _,__ Adjusting MARR to Account for Risk and Uncertainty 479 SELECTING A MINIMUM ATTRACTIVE RATE OF RETURN Focusing on the three concepts on the cost of money (the cost of borrowed money, the cost of capital, and opportunity cost), which, if any, of these values should be used as the minimum attractive rate of return (MARR) in economic analyses? Fundamentally,we know that unless the benefits of a project exceed its cost, we cannot add to the profitability of the firm. A lower boundary for the minimum attractive rate of. return mustbe the cost of the moneyinvestedin the project.It wouldbe unwise, for example, to borrow money at 8% and invest it in a project yielding a 6% rate of return. Further, we know that no firm has an unlimited ability to borrow money. Bankersand others who evaluate the limits of a firm's ability to borrow money-look at both the profitability of the firm and the relationship between the components in the firm's capital structure. This means that continuedborrowing of money will require that additional stock must be sold to maintain an acceptableratio between ownership and debt. In other words, borrowing for a particular investment project is only a block of money from the overall capital structure of the firm. This suggests that the MARR should not be less than the cost of capital. Finally, we know that the MARR should not be less than the rate of return on the best opportunity forgone. Stated simply, Minimum attractive rate of return should be equal to the largest one of the following: cost of borrowed money, cost of capital, or opportunity cost. ADJUSTING MARR TO ACCOUNT FOR RISK AND UNCERTAINTY We know from our study of estimatingthe future that what actually occurs is often different from the estimate. When we arefortunate enough to be able to assignprobabilitiesto a set of possible future outcomes,we call this a risk situation.Wesawin Chapter 10that techniques like expected value and simulation may be used when the probabilities are known. Uncertainty is the term used to describe the condition when the probabilities are not known. Thus, if the probabilities of future outcomes are known we have risk, and if they are unknown we have uncertainty. One. way to reduce the likelihood of undertaking projects that do not produce satis. . factory results is to pass up marginal projects. In other words, no matter what projects are undertaken, some will turn out better than anticipated and some worse. Some undesirable results can be preventedby selecting only the best projects and avoiding those whose expected results are closer to a minimum standard: then (in theory, at least) the selected projects will provide results above the minimum standard even if they do considerably worse than anticipated. In projects accompanied by normal business risk and uncertainty, the MARR is used without adjustment. For projects with greater than average risk or uncertainty, some firms increase the MARR. A preferable way deals explicitly with the probabilities using the techniques from Chapter 10.This may be more acceptableas an adjustmentfor uncertainty. When the interest rate (MARR) used in economic analysis calculations is raised to adjust for risk or uncertainty, greater emphasis is placed on immediate or short-term results and less emphasis on longer-term results. - - - -- 480 - SELECTIONOF A MINIMUM ATTRACTIVERATEOF RETURN Consider the two following alternatives:the MARR of Alt. B has been raised from 10% to 15% to take into account the greater risk and uncertainty that Alt. B's results may not be as favorable as indicated. What is the impact of this change of MARR on the decision? Year Alt. A -$80 10 20 o 1-10 11-20 Alt. B -$80 13.86 10 NPW Year Alt. A At 14.05% At 10% At 15% 0 1-10 11-20 -$80 10 20 -$80.00 52.05 27.95- -$80.00 61.45 47.38 -$80.00 50.19 24.81 0 --5.00 +28.83 . NPW Year Alt. 8 0 1-10 11....20 At 15.48% At 10% At 15 % -$80.00 68.31 11.99 - -$80.00 85.14 23.69 -$80.00 69.56 12.41 -$80 3.86 0 0 +28.83 +1.97 . Computations at MARR of 10% Ignoring R.isk~d Uncertainty . Both alternatives have the same positive NPW (+$28.83) at,a MARR o~10%. Also, the differences in the benefits schedules (A - B) produce a...10% increrp.ental rate of.rettlrn. (The calculations are qot shown here.) This must be tr'ueifNPW for the two alternatives is totemaJ,n constant at a MARR of 10%. Considering Risk.and Uncertainty with MARR of 10% I are equally desirable. Since Alt. B is believed to pave greater. risk arid Ai 10% ,both alternatives ~ uncertainty, a logical conclusion is to select Alt. A rather tpan B. t .. 'n::Cre:a~iMA.RR I L~ I -- A t.a :1\1 ARR of 15 . . . . p~eferred.qn<ler .-1000... ... . a ta15% "'::;: lt . A . = "" 1 ~ '"' ...~::: :::::: = ~~ e :=-='-~ has a ne. gative NPWand Alt. B has a.positive NPW .A71tematiyeJ3is .%,A . . .. . . tl1ese..cir~mrp.stance?:: . =!: . -- . . . .. .!ii!.'! 11\II,\:>::1 _ .. ~ .. ~ . '" >; ~ __ t ~::: . . ==='1:1=_11 . III oil ~ == = :;;:;;~ _ =..., j ..., J , A ] ~ ~ u._ .-. Adjusting MARR to Account for Risk and Uncertainty 481 Conclusion Based on a business-risk MARR of 10%, the two alternatives are equivalent. Recognizing some greater risk of failure for Alt. B makes A the preferred alternative. If the MARR is increased to 15%, to add a margin of safety against risk and uncertainty, the computed decision is to select B. Since Alt. B has been shown to be less desirable than A, the decision, based on a MARR of 15%, way be an unfortunate one. The difficulty is that the same risk adjustment (increase the MARR by 5%) is applied to both alternatives even though they have different amounts of risk. The conclusionto be drawn fromExample 15-3is that increasing the MARRto compensate for risk and uncertainty is only an approximate technique and may not always achieve the desired result. Nevertheless,it is common practice in industry to adjust the MARR upward to compensate for increased risk and uncertainty. Inflation and the Cost of Borrowed Money As inflationvaries,whatis its effecton the cost of borrowedmoney? A widelyheld viewhas been that interest rates on long-term borrowing, like 20-year Treasury bonds, will be about 3% more than the inflationrate. For borrowers this is the real-that is, after-inflation--cost of money,and for lenders the real return on loans. If inflationrates were to increase, it would follow that borrowing rates would also increase. All this suggests a rational and orderly situation, about as we might expect. Unfortunately, things have not worked out this way. Figure 15-3 shows that the real interest rate has not always been about 3% and, in fact, there have been long periods 10 8 6 '-' 4 CIJ 'i;j ... VJ 2 £= 0 ..... <;J -2 CIJ -4 -6 I I 1975 , , , I I 1980 1985 1990 1995 2000 Year FIGURE 15-3 The real interest rate. The interest rate on 20-year Treasury bonds minus the inflation rate, f, as measured by changes in the Consumer Price Index. ] 482 SELECTIONOF A MINIMUM ATTRACTIVERATEOF RETURN - during which the real interest rate was negative. Can this be possible? Would anyone invest money at an interest rate several percentage points below the inflation rate? Well consider this: when the U.S. inflation rate was 12%, savings banks were paying 51/2%o~ regular passbook deposits. And there was a lot of money in those accounts. While there must be a relationship between interest rates and inflation, Figure 15-3 suggests that it is complex. REPRESENTATIVE VALUES OF MARR USED IN INDUSTRY We argued that the minimum attractive rate of return should be established at the highest one of the following: cost of borrowed money, cost of capital, or the opportunity cost. The cost of borrowed money will vary from enterprise to enterprise, with the lowest, rate being the prime interest rate. The prime rate may change several times in a year; it is widely reported in newspapers and business publications. As we pointed out, the interest rate for firms that do not qualify for the prime interest rate may be 1/z% to several percentage points higher. . The cost of capital of a firm is an elusive value. There is no widely accepted way to compute it; we know that as a composite value for the capital structure of the finn, it conventionally is higher than the cost of borrowed money. The ,cost of capital must consider the market valuation of the shares (common stock, etc.) of the firm, which may fluctuate widely,depending on future earnings prospects of the'firm,We cannot generalize on representativecosts of capital. Somewhat related to cost of capital is the computation of the return on total capital (long-term debt, capital stock, and retained earnings) actually achieved by firms. Fortune magazine, among others, does an annual analysis of the rate of return on total capital. The after-tax rate of return on total capital for individual firms ranges from 0% to about 40% and averages 8%. Business Week does a periodic survey of corporate performance. This magazine reports an after-tax rate of return on common'stock and retaine~ earnings. We would expect the values to be higher than the rate of return on total capital, and this is the case. The after-taxreturn on common stock and retained earnings ranges from 0% to about 65% with an average of 14%. , . When discussing MARR, firms can usually be divided into two general groups. First, there are firms that are struggling along with an inadequate supply of investment capital or are in an unstable situation or unstable industry. These firms cannot or do not invest money in anything but the most critical projects with very high rates of return and a rapid return of the capital invested. Often these firms use a payback period of one year or less, before income taxes. For an investment project with a 5-year life, this correspop.ds to about a 60% after-tax rate of return. When these firms do rate of return analysis, they reduce the MARR to possibly 25 to 30% after income taxes. There is potentially a substantial difference between a one-year before-tax payback period and a 30% after-tax MARR, but this appare.ntly does not disturb firms that specify this type of dual criteria. The second group of firms represents the bulk of all enterprises. They are in a more stable situation and take a longer-range view of capital investments. Their greater money supply enables them to invest in capital investment projects that firms in the first group - ---- Representative Values of MARR Used in Industry 483 will reject. Like the first group, this group of firms also uses payback and rate of return ,analysis. When small capital investments (of about $500 or less) are considered, payback period is often the only analysis technique used. The criterion for accepting a proposal may be a before-tax payback period not exceeding 1or 2 years. Larger investmentprojects are analyzed by rate of return. Where there is a normal level of business risk, an after-tax MARR of 12 to 15% appears to be widely used. The MARR is increased when there is greater risk involved. In Chapter 9 we saw that payback period is not a proper method for the economic analysis of proposals. Thus, industrial use of payback criteria is not recommended.Fortunately, the trend in industry is toward greater use of accurate methods and less use of payback period. Note that the values ofMARR givenearlier are approximations.But the values quoted appear to be opportunity costs, rather than cost of borrowed money or cost of capital. This indicates that firms cannot or do not obtain money to fund projects whose anticipated rates of return are nearer to the cost of borrowed money or cost of capital. While one could make a case that good projects are needlesslybeing rejected, one reason that firmsoperate as th~y do is that they can focus limited resources of people, management, and time on a smaller number of good projects. One cannot leave this section without noting that the MARR used by enterprises is so much higher than can be obtained by individuals. (Where can you get a 30% after-tax rate of return without excessiverisk?) The reason appears to be that businesses are not obliged to compete with the thousands of individuals in any region seeking a place to invest $2000 with safety, whereas the number of people who could, or would want to invest $500,000 in a business is far smaller.This diminishedcompetition,combinedwith a higher risk, appears to explain at least some of the difference. Spreadsheets, Cumulative Investments, and the Opportunity Cost of Capital As shown in earlier chapters, spreadsheets make computing rates of return dramatically easier. In addition, spreadsheets can be used to sort the projects by rate of return and then calculate the cumulative first cost. This is accomplished through the following steps. . 1. Enter or calculate each project's rate of return. 2. Select the data to be sorted. Do not include headings, but do include all information on the row that goes with each project. 3. Select the SORT tool (found in the menu under DATA),identify the rate of return column as the first key, and a sort order of descending.Also ensure that row sorting is selected. Sort. 4. Add a column for the cumulative first cost. This column is compared with the capital limit to identify the opportunity cost of capital and which projects should be funded. ~ r f: Example 15-4 illustrates these steps. - - - 484 1 SELECTIONOF A MINIMUM ATTRACTIVERATEOF RETURN A finn has a budget of $800,000 for projects this year. Which of the following projects shouldbe accepted? What is the opportunity cost of capital? Project First Cost Annual Benefit Salvage Value Life (years) A $200,000 $25,000 15 $50,000 10 B 250,000 47,000 -25,000 C 150,000 17,500 15 20,000 10 D 100,000 20,000 15,000 20 E 200,000 24,000 25,000 15 F 300,000 35,000 15,000 o 10 G 100,000 18,000 H 200,000 22,500 20 15,000 o 25 I 350,000 50,000 SO~UTlor,,{ The first step is to use the RATEfunction to find the rate of return for each project. The results of this step are shown iI1the top portion of FigUre15-4. Next the projects are sorted in descending A 1 - -- D E F First Cost Salvage Value Life IRR 200,000 25,000 50,000 2 3 B 250,000 47,000 4 C 150,000 17,000 20,000 5 D 100,000 10 16.0% E 200,000 20,OqO 24,000 15,000' 6 25,000 20 10.6% 7 F.... 300,000 35,000 15,000 8 G' 100,000 18,000 9 H 200,000 22,500 10 I 350,000 50,000 15,000 0 20,000 50,000 II Projects Sortedby IRR D 100,000 I 350,000 I 14 B 15 G 16 E 17 A 200,000 25,000 18 H 200,000 22,500 19 ;"C 150,000 17,500 20 F 300,000 fIGIJRE '1 I C Annual Benefit Project A 11 12 13 l. B , 250,000 47,000 100,000 18,000 24,000 200,000 35,000 -25,000 12.8% 15 8.6% = RATE(E2,C2, -B2,D2) , I , 15 8.2% 12.4% 20 9.6% I- 25 13.7% I Cumulative First Cost 15,000 I 0 10 25 16.0% 13.7% 100,000 450,000. 10 12.8% 7oo,006 10 20 12.4% o ", -2,000 0 25,000 I 10.2% H 10 I. _ 15 10 G " .. 800,000 10.6% 1,000,000 50,000 15 10.2% I 1,200,000 15,0()() 20 9.6% 1,400,000 8;6% , 155();000 8.2% 1,850,000 20,000 15 ! 15 15,000 I -.,...,-:-,- ;. ~ i 15.4 Spreadshee~;.forfindingi opportunity cost of capital. "'"- --- - :i == I i I J. I 1 - --- -- ~ - Problems 485 order by thep-rates of return. Finally, the cumulativefirst cost is computed. Projects D, I, B, and G should be funded. The opportunity cost of capital is 12.4%if defined as the last project funded and 10.6%if defined as the first project rejected. " SUMMARY There are four general sources of capital available to an enterprise. The most important one is money generated from the operation of the firm. This has two components: there is the portion of profit that is retained in the business; in addition, a profitable firm generates funds equal to its depreciationcharges that are available for reinvestment. The three other sources of capital are from outside the operation of the enterprise: 1. Borrowed money from banks, insurance companies, and so forth. 2. Longer-term borrowing from a lending institution or from the public in the form of mortgage bonds. . 3. Sale of equity securities like common or preferred stock. Retainedprofits and cash equalto depreciationcharges aretheprimary sourcesof investment capital for most firms, and the only sources for many enterprises. In selecting a value of MARR, three values are frequently considered: 1. Cost of borrowed money. 2. Cost of capital. This is a composite cost of the components of the overall capitalization of the enterprise. 3. Opportunity cost. This refers to the cost of the opportunity forgone; stated more simply, opportunity cost is the rate of return on the best investmentproject that is rejected. The MARR should be equal to the highest one of these three values. When there is a risk aspect to the problem (probabilities are known or reasonably estimated), this can be handled by techniques like expected value and simulation. Where thereis uncertainty (probabilitiesofthe variousoutcomesarenotknown),thereare analytical techniques, but they are less satisfactory.A method commonly used to adjust for risk and uncertaintyis to increasethe MARR.This method has the effectof distortingthe time-valueof-moneyrelationship.The effectis to discountlonger-termconsequencesmoreheavilythan short-term consequences, which mayor may not be desirable. Other possibilities might be to adjust the discounted cash flows or the lives of the alternatives. PROBLEMS 15-1 Examine the financial pages of your newspaper (or the Wall St. Journal) and detenmne the current interest rate on the following securities, and explain why the interest rates are differentfor these different bonds. --- (a) U.S. Treasury bond due in 5 years. (b) General obligation bond of a municipal district, city, or a state due in 20 years. (c) Corporate debenture bond of a U.s. industrial firm due in 20 years. ---- ;~ --, 486 SELECTION OF A MINIMUM ATTRACTIVERATEOF RETURN 15.2 Considerfour mutually exclusivealternatives: A B C Initial cost $0 $50 $100 16.27 Uniform annual 0 9.96 benefit 0% 10% 15% Computedrate of return 15-6 A factory has a $100,000 capital budget. Determine D $25 5.96 JJL mty ",:hich project(s! should be funded and the opportu_ cost of capItal. ~. 20% Project A B C D Each alternative has a lO-year usefu11ife and no salvage value. Over what range of interest rates is C the preferred alternative? (Answer: 4.5% < i :S 9.6%) 15-3 Frequently we read in the newspaper that one should lease an automobile rather than buying it. For a typical 24-month lease on a car costing $9400, the monthly lease charge is about $267. At the end of the 24 months, the car is returned to the lease company (which owns the car). As an alternative, the same car could be bought with no down payment and 24 equal monthly payments, with interest at a 12% nominal annual percentage rate. At the end of 24 months the car is fully paid for. The car would then be worth about half its original cost. (a) Over what range of nominal before-tax interest rates is leasing the preferred alternative? (b) What are some of the reasons that would make leasing more desirable than is indicated in (a)? 15-4 Assume you have $2000 available for investment for a 5-year period. You wish to invest the moneynot just spend it on fun things. There are obviously many alternatives available. You should be willing to assume a modest amount of risk of loss of some or all of the money if this is necessary, but not a great amount of risk (no investments in poker games or at horse races). How would you invest the money? What is your minimum attractive rate of return? Explain. 15-5 There are many venture capital syndicates that consist of a few (say, eight or ten) wealthy people who combine to make investments in small and (hopefully) growing businesses. Typically, the investors hire a young investment manager (often an engineer with an MBA) who seeks and analyzes investment opportunities for the group. Would you estimate that the MARR sought by this group is more or less than 12%? Explain. -- _ First Cost $50,000 50,000 50,000 50,000 Annual Life Salvage Benefits (years) Value 5 $13,500 $5000 10 9,000 0 5 13,250 1000 8 9,575 6000 15-7 Chips USA is considering the following projects to improve their production Project 1 2 Y 4 5 6 7 _ process. Chips have a short life, so a 3-year horizon is used in evaluation. Which projects should be done if the budget is $70,000? What is the opportunity cost of capital? First Cost $20,000 30,000 10,000 5,000 25,000 15,000 40,000 Benefit $11,000 14,000 ~,000 2,400 13,000 7,000 21,000 15-8 National Motors's Rock Creek plant is considering the following projects to improve the company's production process. Which projects should be done if the budget is $500,000? What is the opportunity cost of capital ? Project 1 2 3 4 5 6 7 First Cost $200,000 300,000 100,000 50,000 250,000 150,000 400,000 Annual Benefit $50,000 70,000 40,000 12;500 75,000 32,000 125,000 Life (years) 15 10 5 10 5. 20 5 1\ Problems . 487 9 The WhatZit Company has decided to fund six of nine 15-10 Which projects should be done if the budget is I 8I:iiiiiiEiii Iiii. project proposals for ~e coming budget year. Deter- mine the next capital budget forWhatZit.What is the $100,000? What is the opportunity cost of capital? MARR? Project A B C D E F G H I First Cost $15,000 20,000 30,000 25,000 40,000 50,000 35,000 60,000 75,000 Annual Benefits $ 4,429 6,173 9,878 6,261 11,933 11,550 6,794 12,692 14,058 Life (years) 4 4 4 5 5 5 8 8 8 Project 1 2 3 4 5 6 7 Life Annual First Cost Benefit (years) 20 $4000 $20,000 20 3200 20,000 30 3300 20,000 15 4500 20,000 25 4500 20,000 10 5800 20,000 15 4000 20,000 Salvage Value $20,000 10,000 - 20,000 10,000 .,-;:' j II"! 1 1 i -- --~-- After Completing This Chapter... The student should be able to: · Distinguish the unique objective and viewpoint of public decisions. Explain methods for determining the interest rates for evaluatingpublic projects. · Us.ethe benefit--costratio to analysis projects. · Distinguish between the conventionaland modified versions of the benefit--costratio. Use an incremental benefit--costratio to evaluate a set of mutually exclusive projects. Discuss the impact of financing,duration, and politics in public investment analysis. · ·· QUESTIONS TO CONSIDER ~ 1. The situation described here is not unique to the Tar...,PamlicoBasin. In fact, it exists to some degree nationwide. So why has Congress"beenslow.to impose environmental requirements on the agricultural sector? 2. Manufacturersin the Tar-Pamlico Basin eventuallywere able to work out an agreement under which they paid local farmers to reduce pollutant runoff. What was the highest amount they were willing to pay per kilogram of pollutants avoided? 3. What was the lowest amount farmers were willing to accept per kilogram of pollutant to sign onto the agreement? " --- -- --- ,-~ Economic Analysis in the Public Sector Hogs and Heavy Industry Industrial facilities in North Carolina's Tar-Pamlico river basin knew they had a costly problem on their hands a few years ago. State environmentalofficials had discovered high levels of water pollutants flowing into a nearby estuary. Industrial plants would have to cut their pollution discharges significantly, the officials said. New regulations would be forthcoming. There wasjust one snag in this plan: most of the pollutants weren't coming from industrial plants. They were coming from local farms. The Tar-Pamlico Basin's economy had a large agricultural component, including many hog and chicken farms. These farms were contributing substantial quantities of water pollutants in the form of runoff from manure, fertilizer, and pesticides. So why didn't the environmental regulators go after the farmers? Simple: They weren't coveredby the Clean WaterAct, and the factories were. The region's manufacturershadbeen heavilyregulatedfor yearsand had alreadyreduced their pollutants well below required levels. Trying to extract the last trace of pollutants from factory effluents would be extremelycostly and wouldnot have donemuch to help the environment in any case. When regulators ran the required cost analysis on their proposed regulations, they found that manufacturerswould have to pay as much as $500 to eliminate one kilogram of targeted pollutants. By contrast, farmers could cutpollutants significantly,at aboutonetenth the cost to manufacturers, simply by instituting practices such as constructing retention ponds, which would help keep pollutants from running off their property into waterways. - - - - 490 ECONOMIC ANALYSISIN THE PUBLIC SECTOR So far we have consideredeconomic analysis for companiesin the private sector,where the main objectives are to generateprofits for growth and to reward current stockholders. Investment decisions for private-sectorcompaniesinvolve evaluating the costs at:J.d benefits associatedwith prospectiveprojectsin termsof life-cyclecashflowstreams.In earlier chapters we developed several methods for calculating measures-of-meritand making decisions. Public organizations, such as federal, state, and local governments, port authorities, school districts, and government agencies also make investment decisions. Fo! these decision-making bodies, economic analysis is complicated by several factors that do not affect companies in the private sector. These factors include the overall purpose of investment, project financing sources, expected project duration, effects of politics, beneficiaries of investment, and the multipurpose nature of investments. The overall mission in the public sector is the same as that in the private sector-to make prudent investment decisions that promote the overall objectives of the organization. The primary economic decision measure-of-merit used in the public sector is the benefit-cost (B/C) ratio. This measure is calculated as a ratio of the equivalent worth of the benefits of investmentin a project to the equivalent worth of costs. If the BIC ratio is greater than 1.0, the project under evaluation is accepted; if not, it is rejected. The B/C ratio is used to evaluate both single investments and sets of mutually exclusive projects (where the incremental B/C ratio is used). The uncertainties of quantifying cash flows, long project lives, and low interest rates all tend to lessen the degree of reliability of the B/C ratio. There are two versions of the B/C ratio: conventional and modified. Both provide consistent recommendations to decision makers for single investment decisions and for decisions involving sets of mutually exclusive alternatives.Nevertheless, the BIC ratio is a widely used and accepted measure in government economic analysis and decision making. INVESTMENT OBJECTIVE Organizationsexist to promote the overall goals of those they serve. In private-sectorcompanies, investment decisions are based on increasing the wealth and economic stability of the organization.Beneficiariesof investmentsgenerally are clearly identifiedas the owners and/or stockholders of the company.. . In the public sector,the purpose of investmentdecisions is sometimes ambiguous. For people in the United States of America, the Preamble to the Constitution establishes the overall theme or objective for why the public body exists: We the People of the United States, in order to form a more perfect Union, establish justice, insure domestic tranquillity, provide for the common defense, promote the general welfare [italics supplied], and secure the blessings of liberty to ourselves and our posterity, do ordain and establish this Constitution for the United States of America. The catch phrase promote the general welfare serves as a guideline for public decision making. But what does this phrase mean? At best it is a general guideline, at worst it is a vague slogan that can be used to justify any action. Projects one segment of the citizenry deems to be useful and necessary may not be so viewed by other groups of citizens. ---- Investment Objective 491 In government economic analysis, it is not always easy to distinguish which investments promQtethe "general welfare" and which do not. Consider the case of a dam construction project to provide water, electricity, flood control, and recreational facilities. Such a project might seem to be advantageous for the entire population of the region. But on closer inspection, decision makers must consider that the dam will require the loss of land upstream due to backed up water. Farmers will lose pasture or crop land, and nature lovers will lose canyon lands. Or perhaps the land to be lost is a pivotal breeding ground for protected species, and environmentalists will oppose the project. The project may also have a negative impact on towns, cities, and states downstream. How will it affect their water supply? Thus, a project initially deemed to have many benefits, on closer inspection, reveals many conflicting aspects. Such conflictingaspects are characteristic of investment and decision making in the public sector. Public investment decisions are more difficult than those in the private sector owing to the many people, organizations, and political units potentially affected by investments. Opposition to a proposal is more likely in public investment decisions than in those made by private-sectorcompanies because for everygroup that benefitsfrom a particular project, there is usually an opposinggroup.Many conflictsin opinionarise whenthe project involves the use of public lands, including industrial parks, housing projects, business districts, roadways, sewageand power facilities, and landfills.Opposition may be based on the belief that development of any kind is bad or that the proposed development should not be near "our" homes, schools, or businesses. Consider the decision that a small town might face when considering whether to establish a municipal rose garden, seemingly a beneficial public investmentwith no adverse consequences. However, an economic analysis of the project must consider all effects of the project, including potential unforeseen outcomes. Where will visitors park their vehicles? Will increased auto travel around the park necessitate new trafficlights and signage? Will traffic and visitors to the park increase noise levels to adjacent homes? Will the garden's special varieties of roses create a disease hazard for local gardens? Will the garden require concentrated levels of fertilizers and insecticides, and where will these substances wind up after they have been applied? Clearly, many issues must be addressed. What appeared to be a simple proposal for a city rose garden, in fact, brings up many aspects to be considered. Our simplerose garden illustrateshow effectson all parties involvedmustbe identified, even for projects that seem very useful. Public decision makers must reach a compromise between the positive effects enjoyed by some groups and the negative effects on other groups. The overall objectiveis to make prudent decisions thatpromote the general welfare, but in the public sector the decision process is not so straightforward as in the private sector. The Flood Act of 1936 specifiedthat waterway improvementsfor flood control could be made as long as "the benefits to whomsoever they accrue [italics supplied] are in excess of the estimatedcosts." Perhaps the overall generalobjectiveof investmentdecision analysis in government should be a dual one: to promote the general welfare and to ensure that the value to those who can potentially benefit exceeds the overall costs to those who do not benefit. .. - - - - - - - - i 492 ECONOMIC ANALYSIS IN THE PUBLIC SECTOR VIEWPOINT FOR ANALYSIS When governmentalbodies do economic analysis, an important concern is the proper viewpoint of the analysis. A look at industry will help to explain how the viewpoint,or perspective, from which an analysis is conducted influences the final recommendation.Economic analysis, both governmental and industrial, must be based on a viewpoint. In the case of industry the viewpointis obvious-a companyin theprivate sectorpays the costs and counts its benefits.Thus, both the costs and benefits are measuredfrom the perspectiveof the firm. Costs and benefits that occur outside the firm are referred to as external consequences (Figure 16-1).In years past, private-sector companiesgenerally ignored the externalconsequences of their actions. Ask anyone who has lived near a cement plant, a slaughterhouse, or a steel mill about external consequences! More recently, government has forced industry to reduce pollution and other undesirable external consequences, with the "resultthat today many companies are evaluating the consequences of their action from a broader, or community-oriented, viewpoint. D Smoke Stack Internal Consequences External Consequences FIGURE 16-1 Internal and external consequences for an industrial plant. The council members of a small town that levies taxes can be expected to take the "viewpoint of the town" in making decisions: unless it can be shown "that the money from taxes can be used effectively, the town council is unlikely to spend it. But what happens when the money is contributed to the town by the federal government, as in "revenue sharing" "or by means of some other federal grant? Often the federal government pays a share"of project costs varying from 10 to 90%. Example 16-1illustrates the viewpointproblem that is created. A municipal project will cost $1 million. The federal government will pay 50% of the cost if the project is undertaken. Although the original economic analysis showed that the PW of benefits was $1:5 million, a subsequent detailed analysis by~the town engineer indicates a more'realistic estimate of the PW of benefits is $750,000. The town council must decide whether to proceed with the project. What would you advise? --..- .. - -- ------ - --- ; -- " ~ _. .__ u _. ... . --. -. ..- ~,....-_._..- Selecting an Interest Rate 493 From the viewpoint of the town, the project is still a good one. If the town puts up half the cost ($500,000) it will receive all the benefits ($750,000). On the other hand, from an overall viewpoint, the revised estimate of $750,000 of benefits does not justify the $1 million expenditure. This illustrates the dilemma caused by varying viewpoints. For economic efficiency, one does not want to encourage the expenditure of money, regardless of the source, unless the benefits at least equal the costs. Possible viewpoints that may be taken include those of an individual, a business firm or corporation, a town or city district, a city, a state, a nation, or a group of nations. To avoid suboptimizing, the proper approach is to take a viewpoint at least as broad as those who pay the costs and those who receive the benefits. When the costs and benefits are totally' confined to a town, for example, the town's viewpoint seems to be an appropriate basis for the analysis. But when the costs or the benefits are spread beyond the proposed viewpoint, then the viewpoint should be enlarged to this broader population. Other than investmentsin defense and social programs, most of the benefits provided by government projects are realized at a regional or local level. Projects, such as dams for electricity, flood control and recreation, and transportation facilities, such as roads, bridges, and harbors, all benefit most those in the region in which they are constructed. Even smaller-scaleprojects, such as the municipal rose garden, although funded by public monies at a local or state level,provide mostbenefitto those nearby.As in the case of private decision making, it is important to adopt an appropriate and consistent viewpoint and to designate all the costs and benefits that arise from the prospectiveinvestmentbased on that perspective.To shift perspectivewhen quantifyingcosts and benefitscould greatly skew the results of the analysis and subsequent decision, and thus such changes in perspective are inappropriate. ELECTINGAN INTERESTRATE Several factors, not present for nongovernment firms, influence the seJection of an appropriate interest rate for use in economic analysis.in the government sector. Recall that for private-sector companies the overall objective is wealth maximization. An appropriate interest rate for use in evaluating projects is selected consistent with this goal. Many nongovernment firms use cost of capital or opportunity cost concepts when setting an interest rate. The objectiveof public investment,on the other hand, involvesthe use of public resources to promote the general welfare and to secure the benefits of a given project to whomsoevermay accruebenefits,as long as thosebenefitsoutweighthe costs. The settingof an interest rate for use in investmentanalysis is less clear-cutin this case. Severalalternative concepts have been suggested for how governmentshould set this rate; these are discussed next. . -, 494 ECONOMIC ANALYSISIN THE PUBLIC SECTOR No Time-Value-of-Money Concept In government, monies are obtained through taxation and spent about as quickly as they are obtained. Often, there is little time delay between collecting money from taxpayers and spending it. (Remember that the federal government and many states collect taxes every paycheck in the form of withholding tax.) The collection of taxes, like their disbursement, although based on an annual budget, is actually a continuous process. Using this line of reasoning, some would argue that there is little or no time lag between collecting and spending tax dollars. As such, they would advocate the use of a 0% interest rate for economic analysis of public projects. Cost of Capital Concept Another consideration in the determination of interest rates in public investments is that most levels of government(federal, state, and local) borrow money for capital expenditures in addition to collecting taxes. Where money is borrowed for a specificproject, one ~ne of reasoning is to use an interestrate equal to the cost of borrowedmoney.This argumentis less valid for state and local governmentsthan for private firmsbecause the federal government, through the income tax laws, subsidizes state and local bonded debt. If a state or one of its political subdivisions (like a county, city, or special assessment district) raises money through the sale of bonds, the interest paid on these bonds is exempt from federal taxes. In this way the federal government is subsidizing the debt, thereby encouraging investors to purchase such bonds. These bonds, called municipal bonds, can be either general obligation or revenue bonds. General obligationmunicipal bonds pay interest and are retired (paid oft) through taxes raised by the issuing government unit. A school district may use property taxes it receives to finance bond debt for construction of new language labs.-Revenue bonds, on the other hand, are not supported by the taxing authority of the government-unit; rather, they are supported by revenues earned by the project being funded. As an example, the city of Athens, Ohio, would use toll revenues from a new bridge over the Hocking River to retire debt on revenue bonds sold for the bridge's construction. For those who purchase municipal bonds, the instruments' tax-free status means that the expected return on this investment is somewhat less than that required of fully taXedbond investments (of similar risk). As a rough estimate, when fully taxed bonds yield an 8% interest rate, municipal bonds might make interest payments at a rate of 6%. The difference of 2% represents the effect of the preferred treatment for federal taxation, and hence a form of hidden federal subsidy on tax-free bonds. This means that the cost of capital approach is sometimes skewed for debt incurred through the use of long-term bonds. Opportunity Cost Concept Opportunitycost, which relates to the interest rate on the best opportunityforgone, may take two forms in governmental economic analysis: government opportunity cost and taxpayer opportunity cost. Inpublic decision making, if the interest rate is based on the opportunity - - - - - -- -- ------- _..---- -. . -....-. 1\ Selecting an Interest Rate 495 cost to a government agency or other governing body, this interest rate is known as government opportunity cost. In this case the interest rate is set at that of the best prospective project for which fundingis not available.One disadvantageof the government opportunity cost concept is that different agencies and subdivisions of government will have different opportunities. Therefore, political units could potentially set different interest rates for use in economic analysis, and a project that may be rejected in one branch owing to an inadequate rate of return may be accepted in another.Differing interest rates lead to inconsistent evaluation and decision making across government. Dollars used for public investments are generally gathered through taxation of the citizenry. The concept of taxpayer opportunity cost suggests that a correct interest rate to use in evaluating public investments is that which the taxpayer could have received if the government had not collected those dollars through taxation. This philosophy holds that through taxation the government is taking awaythe taxpayers'opportunityto use the same dollars for investment."The interest rate that the government requires should not be less than what the taxpayer would have received. This compelling argument is supported in general by the Office of Management and Budget (OMB) A94 directive that stipulates a 7% interest rate be used in economic analysis for a wide range of federal projects. It is not economically desirable to take money from a taxpayer with a 12%opportunity cost, for example, and invest it in a governmentproject yielding 4%. Recommended Concept The general rule of thumb in setting an interest rate for government investments has been to select the largest of the cost of capital, the government opportunitycost, or the taxpayer opportunity cost interest rates. However, as is the case in the private sector, there is no hard and fast rule universally applied in all decision circumstances.Setting an interest rate for use in economic analysis is at the discretion of the government entity performing the analysis. Consider the sevengovernmententitiesgivenin Table 16-1.The interestrates used by these decision-making bodies for evaluatinginvestments could all potentially be set at different levels. Setting these interest rates would involvea management decision based on both objective (cost of capital, etc.) and subjective (risk attitudes, etc.) factors considered by each unit. TABLE16-1 Examples of Interest Rates Used in Government Economic Analysis Interest Rate Used (%) Government Entity 4 6 3 5 6.5 5 8 U.S. armed services State agency Federal agency for highway transportation safety City port authority City school board State waterway commission City of Any town - - --- -- - -- 496 ECONOMIC ANALYSISIN THE PUBLIC SECTOR THE BENEFIT-COST RATIO The benefit-cost ratio was described briefly in Chapter 9 as one of the alternative economic analysis methods for evaluating prospective projects. This method is used almost exclusively in public investment analysis, and because of the magnitude of public dOllars committedeach year through such analysis,the benefit-cost ratio deservesour attentionand understanding. One of the primary reasons for the use of the benefit-cost ratio (B/C ratio) in public decision making is its simplicity. The ratio is formed by calculating the equivalentworth of the benefits accrued through investmentin a project divided by the equivalentworth of the costs of the project. The benefit-cost ratio can be shown as follows: . B/C ratio = Equivalent worth of net benefits Equivalent worth of costs - PW benefits FW benefits PW costs - FW costs AW benefits AW costs Notice that any of the equivalent worth methods (present, future, and annual) can be used to calculate this ratio. Each formulation of the ratio will produce an identical result, as . illustrated in Example 16-2. Demonstrate that for a highway expansion project with data given as follows,the same BIC ratio is obtained using the present, future, and annual worth formulations. $1,500,000 Initial costs of expansion 65,000 Annual costs for operating/maintenance 225,000 Annual savings and benefits to travelers 300,000 Scrap value after useful life Useful life of investment 30 years 8% Interest rate UsingPresentWorth PW benefits - 225,000(P I A, 8%, 30) + 300,000(P IF, 8%, 30) =$2,563,000 PW costs - 1,500,QOO+ 65,000(P I A, 8%, 30) - $2,232,000 U$in~FUtureWorth FW benefits FWcosts= = 225,;00(F I A,=8%,30)+ 1,500,000(F - 300,000 - $2{79'0,'Q"bd"= - ;:; I P, 8%,30) + 65,000(FI A, 8%,30) - $22,460,000 .. - - - - --- .----------- --- -. ----- The Benefit-Cost Ratio 497 Using Annu~ Worth AW benefits = 225,000+ 300,000(AIF, 8%,30) = $227,600 AW costs = 1,500,000(AI P, 8%,30) + 65,000 = $198,200 BIC ratio = 2,563,000 = 25,790,000 = 227,600 = 1.15 . .. ... ....... . . One can see that the ratio provided by each of these methods produces the sameresult: 1.15. - -. . -~~- .~. - ;.."JI An economic analy~is is performed to assist in the objective of making a decision. When one is using the BIC ratio, the decision rule is in two parts: If the BIC ratio is ; 1.0,thenthe decisionshouldbe to "invest." If the BIC ratiois < 100,then the decisionshouldbe "do not invest." Cases of a BIC ratiojust equal to 1.0 are analogous to the case of a calculated n.etpresent worth of $0 or an IRR analysis that yields i = MARR%. In other words, the decision measure is just at the breakeven criteria. In such cases a detailed analysis of the input variables and their estimates is necessary, and one should consider the merlts of other available opportunities for the targeted funds. But, if the BIC ratio is greater than or less than 1.0, the recommendationis clear. The B/C ratio is a numerator/denominatorrelationship between the equivalentworths (EW) of benefits and costs: . . B/C ratio = EW of net benefits to whomsoeverthey may accrue EW of costs to the sponsors of the project The numerator and denominator aspects of the ratio are sometimes interpreted and used in different fashions. For instance, the conventional B/C ratio defines the numerator and denominator as follows: EW of net benefits Conventional BIC ratio = EW of initial costs + EW of operating and maintenance costs This is the formulation of the B/C ratio used in Example 16-2. However,there is another version of the ratio, called the modified BI C ratio. Using a modifiedversion, the numerator and denominator are defined in a different manner: the annual operating and maintenanc~ costs to the users are subtracted in the numerator, whereas they are added as a cost in the denominator.In this case the ratio becomes: M Ole d fi d B/C ratio O 0 = EW of net benefits - EW of operating and maintenance costs EW of initial costs -- 498 -, -- -- _.---- ----- - -- ----------------------------------------------------------------------------------------- ECONOMIC ANALYSIS IN THE PUBLIC SECTOR For decision making, the two versions of the benefit-cost ratio will produce the same recommendation on whether to invest or not invest in the project being considered. The numericBIe ratiofor the twoversionswill not alwaysbe the same,but the recommendation will be. This fact is illustrated in Example 16-3. Consider the highway expansion project from Example 16-2. Let us use the present worth formulation of conventionaland modified versions to calculate the BIC ratio. , \SOlUTION Using the Conventional B / C Ratio . 225,000(P I A, 8%,30) + 300,000(PI F, 8%,30) 1 BI C ratio = = .15 1,500,000 + 65,000(P IA, 8%,30) Using the Modified BIC Ratio . B/C raho = 225,000(P I A, 8%, 30) + 300,000(P I F, 8%,30) - 65,000(P / A, 8%, 30) 1,500,000 = 1.22 Whether the conventional or the modified ratio is used, the recommendation is to invest in the highway expansion project. The ratios are not identical in magnitude (1.15 vs 1.22), but the decision is the same. It is important when one is using the conventional and modified BIC ratios not to directly compare the magnitudes of the two versions. Evaluating a project with one version may produce a higher ratio than i&produced with the other version, but this does not imply that the project is somehow better. The net benefitsto the users of governmentprojects are the differencebetween the expected benefitsfrom investmentminus the expecteddisbenefits.Disbenefitsare the negative effectsof governmentprojects felt by some individualsor groups. For example,considerthe National Park System in the United States. Development projects by the skiing or lumber industries might provide enormous benefits to the recreation or construction sectors while creating simultaneous disbenefits for environmental groups. Table 16-2illustrates some of the primary benefits and disbenefits of several example public investments. . INCREMENTAL BENEFIT-COST ANALYSIS In Chapter 9 we discussed using the incremental benefit-cost ratio in economic decision analysis.As is thecase with theinternalrate ofreturn (IRR) decisionmethod, theincremental B/C ratio shouldbe used in comparing sets of mutually exclusivealternatives.This method produces a result that is consistent with the result produced by optimizing the present worth --- u' . u. Incremental Benefit-Cost Analysis TABLE 16-2 499 Example Benefits and Disbenefits for Public Investments Public Project Primary Benefits Primary Disbenefits New city airport on the exterior of town More flights, new businesses Increased travel time to airport, more traffic on outer belt Interstate bypass around town Quicker commute times, reduced congestion on surface roads Lost sales to businesses on surface roads, lost agricultural lands New metro subway system Faster commute times, less pollution Lost jobs due to bus line closing, less access to service (fewer stops) Creation of a city disposal facility versus sending waste out of state Less costly, faster and more responsive to customers Objectionable sight and smells, lost market value to homeowners, lost pristine forest land Construction of a nuclear Cheaper energy costs, new industry in area Perceived environmental risks energy generation facility of the decision alternatives over their respective life cycles. As with the incremental IRR method, it is not proper to simply calculate the BIC ratio for each alternativeand choose the one with the highest value. Rather, an incrementalapproach is called for. The incremental approach used with the IRR method can be slightly augmentedfor use with the BIC ratio. Elements of the Incremental Benefit-Cost Ratio Method 1. Identify all relevant alternatives. Decision makers identify the set of alternatives from which a choice is ultimatelymade. In this contextit is importantto identify all relevant and competitive alternatives. Decision rules or models can recommend a best course of action only from the set of identified alternatives. If a better alternative than those in the considered set exists, it will never be selected, and the solution will be suboptimal. For benefit-cost ratio problems, the "do-nothing" option is always the "base case" from which the incremental methodology proceeds. . 2. (Optional) Calculate the BI C ratio of each competing alternative in the set. In this optional step one calculates the BIC ratio of each of the competing alternatives based on the total cash flows for each alternative by itself. Once the individual B/C ratios have been calculated, the alternatives with a ratio less than 1.0 are eliminated from further consideration.The alternativeswith a BIC ratio greaterthan 1.0 remain in the set of feasible alternatives. This step gets the "poor performers" out of the way before the incremental procedure is initiated. This step may be omitted,however,because the incremental analysis method will eliminate the subpar alternativesin due time. 3. Rank order the projects. The incremental IRR method is used to analyze each of the mutually exclusive alternativesof the set, arraying the alternativesin order of increasing first cost. However,when one is using the incremental B/C ratio, the alternatives must be ordered according to increasing size of the denominator of the BIe ratio. (The rank order will be the same regardless of whether one uses the present worth, annual worth, or future -- ""'- 500 - ECONOMIC ANALYSIS IN THE PUBLIC SECTOR worth of costs to form the denominator.) To form the list the denominator of the B IC ratio (cost portion of the ratio) is first calculated for each of the feasible alternatives and then placed into an ascending rank order from low to high cost. The "do-nothing" alternative always becomes the first on the ordered list. 4. Identify the increment under consideration. In this step the increment under considerationis identified.The firstincrement taken under considerationis alwaysthat of going from the do-nothingoption to the first of the feasible alternatives.As the analysisproceeds, any identified increment is always in reference to some previouslyjustified alternative. S. Calculate the B I C ratio on the considered incremental cash flows. Upon identify_ ing the increment under consideration, it is necessary to calculate the incrementalbenefits as well as the incrementalcosts.This stepis accomplishedby findingthe cash flowsthat represent the difference (.6.) between the two alternatives under consideration. For two alternatives X and Y, the incremental benefits (.6.B)and incremental costs (.6.C)of going from Alternative X to Alternative Y must be determined. The increment can b.ewritten as (X --+ Y) to signify going from X to Y or as (Y - X) to signify the cash flows of Y minus cash flows of X. Both modes identify the incremental costs and benefits of investing in alternative Y, where X is a previouslyjustified (or base) alternative:The .6.B and .6.Cvalues are used to calculate the overall incremental BIC ratio (.6.B/.6.C)of the increment. 6. Use the incremental BIC ratio to make a decision. The incremental B/C ratio (.6.BI .6.C) calculated in Step 5 is evaluated as follows: if the ratio.is greater than 1.0, then the increment is desirable or justified; if the ratio is less than 1.0, it is not desirable, or is not justified. If an increment is accepted, the alternative associated with that additional increment of investment becomes the base from which the next increment is formed. In the case of an increment that is not justified, the alternative associated with the additional increment is rejected and the previously justified alternative is maintained as the base for formation of the next increment. 7. Iterate to Step 4 until all increments (projects) have been considered. The incremental method requires that the entire list of ranked feasible alternativesbe evaluated.All pairwise comparisons are made such that the additional increment being considered is examined with respect to a previouslyjustified alternative.The incrementalmethod continues until all alternatives have been evaluated. 8. Select the appropriaie alternative from the sei of mutually exclusive competing projects. Mter all alternatives (and associated increments) have been considered, the incremental BIC ratio method calls for selection of the alternativethat is associated with the lastjustified increment. In this way, it is assured that a maximum investmentis made such that each ratio of equivalentworth of incremental benefitsto equivalentworth of incremental costs is greater than 1.0. (A common error in applying the incremental B/C method is selecting the alternativewith the largest incremental BIC ratio, which is inconsistent with the objective of maximizing investment size with incremental BIC ratios above 1.0.) Both the conventional and modified versions of the B/C ratio can be used with the incrementalBIC ratio methodologyjust described,but the two versionsshouldnot bemixed in the same problem. Such an approach could affect the rank ordei and cause confusion and errors. Instead, one of the two versions should be consistently used throughout the analysis. --- -- -- ~----- Ill- - ---....... I Incremental Benefit-Cost Analysis 501 Examples 16-4 and 16-5illustrate the use of the incremental B/C ratio and showhow the conventionaland modified versions can be used with this procedure to evaluate sets of mutually exclusive alternatives. A midwesternindustrial state is consideringthe construction and operation of facilities to provide electricity to several state-owned properties. Electricity will be provided via two coal-burnirig power plants and a distribution network wired to the properties targeted for conversion.A group studying the proposal has identifiedgeneral cost and benefit categories for the project as follows. Primary costs: Construction of the power plantfacilities; cost of installing the power distribution network; life-cycle maintenance and operating costs. Primary benefits: Eliminationofpayments to thecurrent electricityprovider;creationofjobs for construction, operation, and maintenance of the facilities and distribution network; revenue from selling excess power to utility companies; increased employment at coal mines in the state. 4 Assume that there have been four competing designs identified for the power plants. Each design affects costs and benefits in a unique way. Given the following data for the four mutually exclusive design alternatives, use the conventional BIC ratio method to recommend a course of action. I ;~ "" ~. '.. :' I " . Values (x $104) for Competing Design Alternatives III II , ~.. . .. IV ~ .,., Project costs Plant constructioncost ! Annual operatiIig and maintenance cost Project benefits Anoual savings from utility payments 1 .mue from over;capacity Annual effect of jobs created $12,500 120 $11,000 480 $12,500 450 $16,800 145 580 700' 400 700 550 750 950 200 150 1,300 250 500 Other data Project life, in years Discounting rate (MARR) 45 8% 45 8% 45 8% 45 8% II Alternatives I through IV constitute a set of mutually exclusive choices because we will select one and oWy one, of tp~ d~sign QpqoY§,fot;.tl:1~PoWef~g~~Qts,J)lerefore, : an~incremental B I C ratio methodis usedto obtainthe solution.Let us use the incrementalmethodas describedabove: - Step 1 The alternatives are do nothin~, and designs I, II,m, and IV. - ~ j:i : I "~f~ ~' .::;:!, " " " -.- , - , , - ---- 502 - ECONOMIC ANALYSISIN THE PUBLIC SECTOR Step 2 In this optional step we calculate the conventionalB/C ratio for each alternativebased on individual cash flows.We will use the ratio of the PW of benefitsto costs. B/C ratio (I) = (580 + 700 + 400)(P / A, 8%,45)/[12,500 B/C ratio (III) = (200+ 950+ 150)(P/ A, 8%,45)/[12,500 + 325(P/ A, 8%,45)] = 0.96 . B/C ratio (IV) = (1300 + 250 + 500)(P / A, 8%,45)/[16,800 + 120(P/ A, 8%,45)] = 1.46 B/C ratio (II) = (700 + 550 + 750)(P / A, 8%,45)/[11,000 + 480(P / A, 8%, 45)] = 1.44 + 145(P/ A, 8%,45)] = 1.34 Alternative I, n, and IV all have B/C ratios greater than 1.0 and thus are included in the feasible set. Alternativeill does not meet the acceptabilitycriterion and should be eliminated from further consideration. However,to illustrate that Step 2 is optional, we will continue with.all four design alternativesin the original feasible set. Step 3 Here we calculate the PW of costs for each alternative in the feasible set (remember we are keeping alternativeill along for the ride). The denominator of the conventional B/C r~tio includes first cost and annual O&M costs. We calculate the PW of costs for each alternative as follows. PW costs (I) = PW costs (II) = PW costs (III) = PW costs (IV) = 12,500+ 120(P/A, 8%,45)= $13,953 11,000 + 480(P / A, 8%, 45) = $16,812 12,500 + 325(P / A, 8%, 45) = $16,435 16,800 + 145(P/ A, 8%,45) = $18,556 . . The rank order from low to high value of the B/C ratio denominator is as follows: do nothing, I, ill, n, IV. Step 4 From the ranking list, the first increment considered is that of going from "do nothing" to Alternative I (do nothing -+ Alternative I). The analysis proceeds from this point. Steps 5 and 6 We proceed through the analysis, designating the incremental cash flows and calculating Ll(B/C) until all feasible alternatives have b~en considered. Each additional increment taken under considerationmust be based on the lastjustified increment. - - -- -------- ] ---- - .- I.\) I Incremental Benefit-Cost Analysis 503 :z <10 As .anexample of these calculations, consider the third increment (I ~ II). = 11,000- 12,500= -$1500 APlant construction cost = 480 - 120= $360 = -1500 + 360(P/ A, 8%,45) = $2859 AAnnualO&Mcost PW of ACosts or AAnnual utility payment savings AAnnual overcapacityrevenue dAnnual benefits of newjobs PW of dBenefits = 16,812 - = (120 - = $3875 AB/C ratio (PW dB)/(PW dC) 13,953 = $2859 = 700 - 580 = $120 = 550 -700 = -$150 = 750 - 400 = $350 150 + 350)(P / A, 8%,45) = 3875/2850 = 1.36 Step 8 The analysis in the table proceeded as follows:do nothing to Alternative I was justified (AB / C ratio = 1.46), Alternative I became the new base; Alternative I to Alternative ill was not justified (AB/C ratio = -'-1.15), Alternative I remained base; Alternative I td Alternative II was justified (dB/C ratio = 1.36), Alternative II became the base; = Alternative II to Alternative IV was not justified (.AB/C ratio 0.35), and Aiternativell became the recommended power plant design alternative because it is the one associated with the last justified increment. All alternatives in the feasible set were considered before the recommendation was produced. Notice that Alternative ill, even though included in the feasible set, did not affect the recommendation and was eliminated through the incremental method. Notice also that the first increment considered (do nothing ~ I) was not selected even though it had the largest AB/C ratio (1.45). Selection is based on the alternative associated with the last justified increment (in this case, Alt. II). Let us reconsider Example 16-4, this time using the modified B/C ratio to analyze the set of cr 'peting design alternatives.Again we will use the present worth method. .SOLUTION, Here we use the modified B/C ratio. Step 1 The alternatives are still do nothing and designs I, II, ill, and IV. Step 2 We calculate the modified B/C ratio using the PW of benefits and costs~ B/C ratio (I) = (580 + 700 + 400 - 120)(P/ A, 8%, 45)/(12,500) = 1.51 B/C ratio (II) = (700+ 550+ 750- 480)(P/ A,=-:8%,45)/(11,000) 1.67 ..:;;;:r;~~=:tiC I L r . B/C ratio (III) = (200 + 950 + 150 - 325)(P / A, 8%,45)/(12,500) = 0.95 I B/C ratio (IV) = (1300+ 250+ 500- 145)(P/ A, 8%=45)/(16,&00)-).37 -.J_;":':: - ----- - -.. 1 504 ECONOMIC ANALYSISIN THE PUBLIC SECTOR , Step 3 - Again Alternative ill would be eliminated from further considerationbecause its B/C ratio is less than 1.0. In this case we will eliminate it from the feasible set, which now becomes do nothing and alternativedesigns I, II, and IV. The PW of costs for each alternative in the feasible set: = $12,500 PW Costs (I) PW Costs (II) = $11,000 PW Costs (IV) = $16,800 The appropriate rank order is now do nothing, II, I, IV.Notice that the modified B/C ratio produces a rank order different from that yielded by the conventional version in Example 16-4. Step 4 The firstincrementis now(donothing~ Alternative11).The methodproceedsfrom this point. Steps 5 and 6 These steps are accomplished as follows. Increment Incremental Effects Plant construction cost PW of Costs Annual utility payment savings Annua1 overcapacity revenue Annua1 benefits of new jobs Annual O&M disbenefit PW of Benefits B/C ratio (pW B)/(PW C) Is increment justified? (Do Nothing -+ II) (II -+ I) $11,000 11,000 700 550 750 480 18,405 1.67 Yes $ 1500 1500 -120 150 -350 -360 484 0.32 No (II -+ IV) $ 5800 5800 600 -300 -250 -335 4662 0.80 No As an example of the calculations in the forgoing table, consider the third increment (II ~ IV). .6.Plant construction cost PW of !!:.Costs !!:.Annualutility payment savings !!:.Annualovercapacity revenue .6.Annual benefits of new jobs .6.Annual O&M disbenefit PW of !!:.Benefits f = .6.B/C ratio, (PW~~B:)/(P~ ~C) " = 16,800 ~ 11,000 = $5800 = $5800 = = = 1300"7700 = $600 250 - 550= -$300 500- 750= -$250 145- 480 = -$335 = (600 - 300 - 250 + 335)(P I A, 8%,45) 0;; 4fi62/5800 = 0~8~Q ~~ =~ =.. = $4662 ,y-.. When the modified version of the B IC ratio is used, Alt. II emerges as the recommended power plant design-just as it did when we used the conventional B/C ratio. .~ .. iI I J J -- -- Other Effectsof Public Projects 505 OTHER EFFECTS OF PUBLIC PROJECTS , Three areas remain that merit discussion in describing the differencesbetween government and nongovernment economic analysis: (1) financing government versus nongovernment projects, (2) the typical length of government versus nongovernment project lives, and (3) the general effects of politics on economic analysis. Project Financing Governmentalandnongovernmentorganizationsdifferin the wayinvestmentsin equipment, facilities, and other projects are financed.In general, nongovernmentfirms rely on monies from individual investors (through stock and bond issuance), private lenders, and retained earningsfrom operations. These sources serveas the pool from which investmentdollarsfor projects come. Management'sjob in the nongovernmentfirmis to match financialresources with projects in a way that keeps the firm growing, produces an efficient and productive environment,and continues to attract investors and future lenders of capital. On the other hand, the government sector often uses taxation and municipal bond issuance as the source of investment capital. In government, taxation and revenue from operations is adequate to financeonly modest projects. However,public projects tend to be large in scale (roadways, bridges, etc.), which means that for many public projects 100% of the investment costs must be borrowed-unlike those in the private sector. To prevent excessive public borrowing and to assure timely debt repayment, the U.S. government, through constitutional and legislativechannels,has placed restrictions on governmentdebt. These restrictions include the following: 1. Local government bodies are limited in their borrowing to a specified percentage of the assessed value of the property in their taxation district. This means that governmententities in areas with high property valuescan borrowmore total monies than those in areas where assessed values are lower. 2. For new construction,borrowedfunds attained through the sale of bonds require the approval of local voters (sometimes by a two-thirds majority) through the election process. An example might be a $2 million levy to fund a new municipaljail. If approved by voters, a $2 million increase wouldraise the property taxes of landowners in the city's tax district by $2 for every $1000 of assessed property value. These additional tax revenues would then be used to retire the debt on the bonds sold to finance the cityjail project. 3. Repaymentof public debt must be made in a preset period of time and in accordance with a specificplan. This is often the case with monies borrowedby issuing bonds, since bond interest payments and maturity dates are set at the time of issuance. Limitations on the use and sources of borrowed monies make funding projects in the public sector much different from this process in the private sector.Private-sectorfirms are seldom able to borrow 100% of required funds for projects, as can be done in the public sector, but at the same time, private entities do not face restrictions on debt retirement or the uncertainty of voter approval. I ~ l -. ....... 506 - ECONOMIC ANALYSIS IN THE PUBLIC SECTOR -, Project Duration Another aspect distinguishing government projects from those in the private sector is the typical duration or project life of the investment. In the private sector, projects most often have a projected or intended life ranging between 5 and 15 years. On some occasionsthe project life is shorter and in others longer,but a majorityof projects fall in this interval.Complex advanced manufacturingtechnologies, like computer-aidedmanufacturing or flexible automated manufacturingcells, typically have projects lives at the longer end of this range. In the 1980ssomecriticized U.S. manufacturingmanagersfor short-sightedviewsof capital investments in such technologies. At that time short-sightedness and lack of investments were blamed for the overall loss of competitivenessin such key U.S. industries as textiles, steel, electronics, automotive, and machine tools. Government projects typically have lives in the range of 20 to 50 years (or longer). Typical projects include federal highways, city water/sewer infrastructure, county dumps, and state libraries and museums. These projects, by nature, have a longer useful life than a typical project in the private sector. There are exceptions to this rule because private finns invest in facilities and other long-range projects, and government entities also invest in projects with shorter-term lives. But, in general, investmentduration in the government sector is longer than in the private sector. Governmentprojects,because they tend to be long range and large scale,usuallyrequire substantial funding in the early stages. The highway, water/sewer,and library projectsjust mentioned could each require millions of dollars in design, surveying, and construction costs. Because of this requirement, it is in the best interest of decision makers who are advocates of such projects to spread that first cost over as many years as possible to reduce the annual cost of capital recovery.This tendency to use longer project lives to downplay the effects of a large first cost can affect the desirability of the project, as measured by the BIC ratio. Another aspect closely associated with managing the size of the capitalrecovery cost in a BIC ratio analysis is the interest rate used for discounting. Lower interest rates reduce the size of the capital recovery cost by reducing the penalty of having money tied up in a project. Example 16-6 illustrates the effects that project life and interest rate can have on the analysis and acceptability of a project. Consider aproject that has been approvedby local votersto build a newjunior high school,needed because of increased (and projected)population growth. Information for the project is as follows: $10,000,000 5,500,000 500,000 350,000 500,000 1,500,000 Building first costs (design, planning, and construction) Initial cost for roadway and parking facilities Firsti!I cost to equip and furnish facility I L Annual operating and maintenance costs Annual savings from rented space Annual beqefits to community - - ----- -- - -- -- - -- --- -- .-. -. .. .. .- '..- .... Other Effectsof Public Projects 507 With this project we examine the effect that varying project lives and interest rates have on the conventional BjC ratio. Project lives at 15,30, and 60 years and interest rates at: 3, 10, and 15% are used to calculate the ratio for the investment. The ratio for each combination of project life and interest rate is tabulated as follows: Conventional Benefit-Cost Ratio for Various Combinations of Project Lifeand Interest Rate Interest Project Life(years) 3% 10% 15% 15 30 60 1.24 1.79 2.24 0.86 1.03 1.08 0.69 0.76 0.77 As an example of how the ratios are calculated, suppose that life = 30 years and interest rate = 10%. 1,500,000+ 500,000 ConventionalBjC ratio = (10,000+ 5,500,000+ 500,000)(Aj P, 10%,30)+ 350,000 = 1.03 From these numbers one can see the effect of project life and interest on the analysis and recommendation. At the lower interest rate, the project has BjC ratios above 1.0 in all cases of project,life, while at the higher rate the ratios are all less than 1.0. At an interest rate of 10% the recommendation to invest changes from no at a life of 15 years to yes at 30 and 60 years. By manipulating these two parameters (project life and interest rate), it is possible to reach entirely different conclusions regarding the desirability of the project. The key point is that those advocating '~nvestment" are well advised to use lower interest rates and longer project life in their example calculations. Project Politics .. . To some degree political influences are felt in nearly every decision made in any organization. Predictably, some individual or group will support its own particular interestsover competing views. This actuality exists in both government and nongovernmentorganizations. In government the effects of politics are continuouslyfelt at a111evelsbecauseof the large-scale and multipurpose nature of projects and because government decision making involves the use of the citizenry's common pool of money. To illustrate on a small scale situations faced by government, compare the decisionmaking process a family may face when planning an evening out. As most familiescan attest, this decision is not always easy--even when Dad is footing the bill! Imagine the increased level of conflict that would arise if every member of the family were to contribute to the tab. Perhaps the choices would then be dine out, go to a ball game, have a shopping spree at the mall, loan the money to Mom, or put it into the family bank account. This scenario characterizes governmentinvestmentdecisions-individnals and-"~ groupswith - -- .~ - --- -- --- - - - - -- -- - I --.---. --. ... -. --- --- . - . - - -. - --. .- I 508 - ECONOMIC ANALYSIS IN THE PUBLIC SECTOR different values and views spending a common pool of money. As with the family decision , the parties involved often have squabbles, form alliances, and maneuver politically. The guideline for public decision making, as set forward in the Preamble to the United States Constitution, is to promote the general welfare of citizens. However, it is impossible to please everyone all the time. The term "general welfare" implies that the architects of this document understood that the political process would produce opposition, but at the same time they empowered decision makers to act in a representative way. As mentioned previously, government projects tend to be large in scale. Therefore, the time required to plan, design, fund, and construct such projects is usually several years. However, the political process tends to produce government leaders who support short-tenn decision making (because many government terms of office, either elected or appointed, are relatively short). Therein lies another difference between government and nongovernment decision making-short-term decision making, long-term projects. Because goveriunent decision makers are in the public eye more than those in the private sector,governmentaldecisionsare generallymore affectedby "politics."As such,the decisions that public officialsmake may not always be the best from an overallperspective. If a particular situation exposes a public official to ridicule, he may choose an expedient action to eliminate negative exposure (whereas a more careful analysis might have been better). Or, such a decision maker may placate a small, but vocal, political group over the interest of the majority of citizens by committing funds to a favored project (at the expenseof otherbetterprojects).Or, a publicdecisionmakermay avoidcontroversyby declining to make a decision on an important, but politically charged,issue (whereasit would be in the overall interest of the citizenry if action were taken). Indeed, the role of politics in government decision making is more complex and far ranging than in the private sector. Consider again Example 16-4, where we evaluated power plants designs. Remember that governmentprojects are often opposed and supported by differing groups in the populace. As such, decision makers become very aware of potential political aspects when they are considering such projects. For the electric power plant decision, several political considerations may affect any evaluation of funding this project. . · The governor has been a strong advocate of workers' rights and has received abundant campaign support from organizedlabor (which is especially important in an industrialized state). By championing this project, the governor should be seen as pro-labor, thereby benefiting his bid for reelection, even if the project is not funded. · The regulated electric utility provideJ;s in the state are strongly against this project, claiming that it would directly compete with rp.emand take away some of their biggestcustomers~ The proviClersllave a stroT1globby and key contacts with ttie stiie'siItilities corniniSsion. A senior state senator has already protested that this project is the first step toward "rampant socialism in this great state." -- --- ----- ---------- -- ... ou . .- ., ..---- Summary 509 · Business leaders in the municipalitieswhere the two facilities would be constructed are in · · · · favor of the project because it would create more jobs and increase the tax base. These leaders promote the project as a win-win opportunity for government and industry, where the state can benefit by reducing costs, and the electric utilities can improve their service by focusing more effectivelyon residential customers and their needs. The lieutenantgovernoris promotingthisproject,proclaimingthatit is anexcellentexample of "initiating proactive and creative solutions to the problems that this state faces." Federal and state regulatory agencies are closely watching this project with respect to the Clean Air Act. Speculation is that the state plans to use a high-sulfur grade of local coal exclusively.Thus "stack scrubbers" would be required, or the high-sulfur coal would have to be mixed with lower-sulfurcoal imported from other states to bring the overall air emissions in line with federal standards. The governor is using this opportunity to make the point that "the people of this state don't need regulators to tell us if we can use our own coal!" The state's coal operators and mining unions are very much in favor of this project. They see the increased demands for coal and the governor's pro-labor advocacyas very positive. They plan to lobby the legislature strongly in favor of the project. Land preservation and environmentalgroups are strongly opposing the proposed project. They have studied the potential negative impacts of this project on theland and on water and air quality,as well as on the ecosystemand wildlife,in the areas where thetwo facilities wouldbe constructed. Environmentalistshave started a public awareness campaign urging the governor to act as the "chief steward" of the natural beauty and resources of the state. Will the project be funded? We can only guess. Clearly, however, we can see the competing influences that can be, and often are, part of decision making in the public sector. SUMMARY Economic analysis and decision making in government is notably different from these processes in the private sectorbecause the basic objectivesof the public and private sectors are fundamentally different.Governmentinvestmentsin projects seek to maximize benefits to the greatest number of citizens, while minimizing the disbenefits to citizens and costs to the government. Private firms, on the other hand, are focused primarily on maximizing stockholder wealth. Several factors, not affecting private firms, enter into the decision-making process in government. The source of capital for public projects is limited primarily to taxes and bonds. Government bonds issued for project construction are subject to legislativerestric.tions on debt not required for private firms. Also, raising tax and bond monies involves sometimes long and politically charged processes not present in the private sector. In addition, government projects tend to be larger in scale than those of competitive firms and to affect many more people and groups in the population. All these factors slow down the process and make investmentdecision analysis more difficultfor government decision -- -- -- I¥; 510 1 ECONOMIC ANALYSISIN THE PUBLIC SECTOR makers than for those in the private sector. Another difference between the public and private sectors lies in how the interest rate (MARR) is set for economic studies. In the private sector, considerations for setting the rate include the cost of capital and opportunity costs. In government, establishing the interest rate is complicated by uncertainty in speci_ fying the cost of capital and the issue of assigning opportunity costs to taxpayers or to the government. Thebenefit-costratio is widelyusedto evaluateandjustify government-fundedprojects. This measure of merit is the ratio of the equivalentworth of benefitsto the equivalentworth of costs. This ratio can be calculated using PW, AW,or FW methods. A B/C ratio greater than 1.0 indicates that a project should be invested in if funding sources are available. For considering mutually exclusive alternatives, an incremental method should be used to evaluate the merits of additional cost. This method results in the recommendation of the project with the highest investment cost that can be incrementallyjustified. Two versions of the BIC ratio, the conventional and modified BIC ratios, produce identical recommendations when single projects or sets of competing alternatives are being considered. The differencebetween the two ratios is in the way that annual operating and maintenancecosts are handled-as an added cost in the denominator in the former, or as a subtracted benefit in the numerator in the latter. PROBLEMS 16-1 Economic analysis and decision making in the public sector is often called "a multi-actor or multistakeholder decision problem." Explain what this phrase means. . 16-2 Compare the general underlying objective of public decision making versus private decision making. 16-3 In the text the authors describe a general recommendation regarding the scope of viewpoint that is appropriate in public decision making. What do they suggest? What example is given to higWight the dilemma of viewpoint in public decision making? 16-4 In government projects, what is meant by the phrase "most of the benefits are local." What conflict does this create for the federal government in terms of funding projects from public monies. , 16-5 Discuss the alternative concepts that can be employed when setting the discounting rate for economic analysis in the public sector. What is the authors' final recommendation for setting this rate? 16-6 What is the essential difference between the conventional and modified versions of the benefit-cost ratio? Is it possible for these two measures to provide conflicting recommendations regarding invest/do-notinvest decisions? 16-7 List the potential costs, benefits, and disbenefits that should be considered when one is evaluating a nuclear power plant construction. 16-8 Describe how a decision maker can use each of the following to "skew" the results of a BjC ratio analysis in favor of his or her own position on funding projects: (a) (b) (c) (d) . Conventional versus modified ratios. Interest rates. Project duration. Benefits, costs, and disbenefits. 16-9 Think about a major government construction project under way in your state, city, or region. Are the decision makers who originally analyzed and initiated the project currently in office? How can politicians use "political posturing" with respect to government projects? 16-10 Consider the following investment opportunity: Initial cost Additional cost at end of Year 1 Benefit at end of Year 1 Annual benefit per year at end of Years 2-10 -- -- - - - - -- $100,000 150,000 o 20,000 1\ y. Problems With interest at 7%, what is the benefit-cost ratio for this project? (Ans~er: 0.51) -11 A government agency has estimated that a flood control project has costs and benefits that are parabolic, according to the equation Required first costs Annual benefits to users Annual disbenefits to users Annual cost to government Project life Interest rate 511 $1,200,000 $500,000 $25,000 $125,000 35 years 10% (Present worth of benefits)2 - 22(Present worth of cost) + 44 = 0 where both benefits and costs are stated in millions of dollars. What is the present worth of cost for the optimal size project? .12 The Highridge Water District needs an additional supply of water from Steep Creek. The engineer has selected two plans for comparison: Gravity plan: Divert water at a point 10 miles up Steep Creek and carry it through a pipeline by gravity to the district. Pumping plan: Divert water at a point near the district and pump it through 2 miles of pipeline to the district. The pumping plant can be built in two stages, with half-capacity installed initially and the other half 10 years later. Use a 4O-year analysis period and 8% interest. Salvage values can be ignored. During the first 10 years, the average use of water will be less than during the remaining 30 years. Use the conventional benefit-cost ratio method to select the more economical plan. Gravity Initial investment $2,800,000 Additional investment 0 in tenth year 10,000 Operation, maintenance, replacements, per year 0 Average power cost per year (first 10 years) 0 Average power cost per year (next 30 years) Pumping $1,400,000 200,000 25,000 50,000 100,000 (Answer: Pumping plan) 3 Calculate the conventional and modified benefit-cost ratio for the investment represented by the following data. 16-14 For the data benefits and BjC ratio is ing methods: worth. given in Problem 16-13, for handling costs, demonstrate that the calculated the same using the each of the followpresent worth, annual worth, and future 16-15 Big City Carl, a local politician, is advancing a project for the construction of a new dock and pier system at the river to attract new commerce to the city. A committee appointed by the mayor (an opponent of Carl's) has developed the following estimates for the effects of the project. Cost to wreck and remove current facilities Material, labor, and overhead for new construction Annual operating and maintenance expenses Annual benefits from new commerce Annual disbenefits to sportsmen in area Project life Interest rate 2,750,000 185,000 550,000 35,000 20 years 8% (a) Using the conventional BjC ratio, determine whether the project should be funded. (b) After studying the numbers given by the committee, Big City Carl argued that the project life should be at least 25 years and more likely closer to 30 years. How did he arrive at this estimate, and why is he making this statement? 16-16 Two different routes, which entail driving across a mountainous section, are being considered for a highway construction project. The first route (the high road) will require building several bridges and navigates around the highest mountain points, thus requiring more roadway. The second alternative (the low road) will require the construction of several tunnels, but takes a more direct approach through the mountainous area. Projected travel volume for this new section of road is 2500 cars per day. Given the following data, use the modified BjC ratio to determine --..-------. -- $ 750,000 - ----- ': 512 - ECONOMIC ANALYSIS IN THE PUBLIC SECTOR which alternative should be recommended. Assume that projec~ life is 45 years and i Averageconstruction cost per mile Number of miles required Annual benefit per car per mile Annual O&M costs per mile end of 30 years. Maintenance will be $75,000 per year for the first 15 years, $125,000 per year for the next 15 years, and $250,000 per year for the final 10 years. Plan B is a two-stage program: $450,000 is required immediately (including money for special equipment), followed by $50,000 at the end of 15 years. Maintenance will be $100,000 per year for the first 15 years and $125,000 for each of the subsequent years. At the end of 40 years, it is believed that the equipment can be sold for $150,000. (a) Use a conventional benefit-cost ratio analysis to determine which plan should be chosen. = 6%. The High The Low Road Road $200,000 $450,000 35 $0.015 10 $0.045 $2000 $10,000 16-17 The federal government proposes to construct a multipurpose water project to provide water for irrigation and municipal use. In addition, flood control and recreation benefits will be realized. The estimated benefits of the project computed for 10 year periods for the next 50 years are given in Table P16-17. The annual benefits may be assumed to be one-tenth of the decade benefits. The operation and maintenance cost of the project is estimated to be $15,000 per year. Assume a 50-year analysis period with no net project salvage value. (a) If an interest rate of 5% is used, and a benefitcost ratio of unity, what capital expenditure can be justified to build the water project now? (b) If the interest rate is changed to 8%, how does this change the justified capital expenditure? 16-18 The city engineer has prepared two plans for the construction and maintenance of roads in the city park. Both plans are designed to provide the anticipated road and road maintenance requirements for the next 40 years. The minimum attractive rate of returri used by the city is 7%. Plan A is a three-stage development program: $300,000 is to be spent immediately, followed by $250,000 at the end of 15 years and $300,000 at the (b) If you favored Plan B, what value of MARR . would you use in the computations? Explain. 16-19 The state is considering eliminating a railroad grade crossing by building an overpass. The new Structure, together with the needed land, would cost $1.8 million. The analysis period is assumed to be 30 years based on the projection that either the railroad or the highway above it will be relocated by then. Salvage value of the bridge (actually, the net value of the land on either side of the railroad tracks) 30 years hence is estimated to be $100,000. A 6% interest rate is to be used. At present, about 1000 vehicles per day are delayed by trains at the grade crossing. Trucks represent 40%, and 60% are other vehicles. Time for truck drivers is valued at $18 per hour and for other drivers at $5 per hour. Average time saving per vehicle will be 2 minutes if the overpass is built. No time saving occurs for the railroad. The installation will save the railroad an annual expehse of $48,000 now spent f-orcrossing guards. During the preceding 10-year period, the rli}.lroadhas paid out $600,000 in settling lawsuits and awident cases related to the grade crossing. The proposed TABLE P16-17 Data Decades Purpose Municipal Irrigation Flood control Recreation Totals First Second Third Fourth Fifth $ 40,000 350,000 150,000 60,000 - $ 50,000 370,000 150,000 70,000 $ 60,000 370,000 150,000 80,000 $ 70,000 360,000 150,000 80,000 $110,000 350,000 150,000 - 90,000 $600,000 $640,000 $660,000 $660,000 $700,000 - - - - -- - -- - I - -- -- -- -- - .-. ..-.- . - -- Problems project will entirely eliminate both these expenses. The state estimates th~t the new overpass will save it about $6000 per year in expenses directly due to the accidents. The overpass, if built, will belong to the state. Should the overpass be built? If the overpass is built, how much should the railroad be asked to contribute to the state as its share of the $1,800,000 construction cost? :0 An existing two-lane highway between two cities, 10 miles apart, is to be converted to a four-lane divided freeway. The average daily traffic (ADT) on the new freeway is forecast to average 20,000 vehicles per day over the next 20 years. Trucks represent 5% of the total traffic. Annual maintenance on the existirig highway is $1500 per lane-mile. The existing acci- 6 rt per mile 18 rt per mile Time saving Autos Trucks 3 rt per minute 15 rt per minute Averageaccident cost $1200 I" '! If a 5% interest rate is used, which of the three proposed plans should be adopted? (Answer: Plan C) 16-21 The local highway department is preparing an economic analysis to see whether reconstruction of the pavement on a mountain road is justified. The number of vehicles traveling on the road increases each year, hence the benefits to the motoring public of the pavement reconstruction also increase. Based on a traffic count, the benefits are projected as follows: d, + rate is 4.58 per million vehicle miles (MVM). Tlu..:e alternate plans of improvement are now under consideration. Year 2001 2002 2003 2004 2005 2006 Plan A: Improve along the existing development by adding two lanes adjacent to the existing lanes at a cost of $450,000 per mile. It is estimated that this plan will reduce auto travel time by 2 minutes and truck travel time by 1 minute. The Plan A estimated accident rate is 2.50 per MVM. Annual maintenance is estimated to be $1250 per lanemile. Plan B: Improve along the existing alignment with grade improvementsat a cost of $650,000 per mile. Plan B would add two additional lanes, and it is estimated that this plan would :educe auto and truck travel time by 3 minutes each. The accident rate on this improved road is estimated to be 2.40 per MVM. Annual maintenanceis estimated to be $1000 per lanemile. Plan C: Constructa new freewayon new alignment at a cost of $800,000 per mile. It is estimated that this plan would reduce auto travel time by 5 minutes and truck travel time by 4 minutes.Plan C is 0.3 mile longer than A or B. The estimated accident rate for C is 2.30 per MVM. Annual maintenance is estimated to be $1000per lane-mile. Plan C includes abandonment of the existing highway with no salvage value. Incrementaloperating cost Autos Trucks 513 End-of-Year Benefit $10,000 12,000 14,000 16,000 18,000 20,000 and so on, increasing $2000 per year The reconstructed pavement will cost $275,000 when it is installed and will have a 15-year useful life. The construction period is short, hence a beginning-ofyear reconstruction will result in the end-of.:year benefits listed in the table. Assume a 6% interest rate. The reconstruction, if done at all, must be done not later than 2006. Should it be done, and if so, in what . year? 16-22 A section of road in the state highway system needs repair at a cost of $150,000. At present, the volume of traffic on the road is so low that few motorists would benefit from the work. However, future traffic is expected to increase, with resulting increased motorist benefits. The repair work will produce benefits for 10 years after it is completed. The highway'planning department is examining five mutually exclusive alternatives concerning the road repair. Should the road be repaired and, if so, when should the work be done? Use a 15% MARR. ----- - ------ --- -------- II 16-23 A 50-meter tunnel must be constructed as part of a new aqueduct system for a city. Two alternatives are being considered. One is to build a full-capacity tunnel now for $500,000. The other alternative is to build a half-capacity tunnel now for $300,000 ana then to build a second parallel half-capacity tunnel 20 years hence for $400,000. The cost to repair the tunnellining every 10 years is estimated to be $20,000 for the full-capacity tunnel and $16,000 for each halfcapacity tunnel. Determine whether the full-capacity tunnel or the half-capacity tunnel should be constructed now. Solve the problem by the conventional benefit-cost ratio analysis, using a 5% interest rate and a 50-year analysis period. There will be no tunnel lining repair at the end of the 50 years. 16-24 The Fishery and Wildlife Agency of Ireland is considering four mutually exclusive design alternatives (Table P16-24) for a major salmon hatchery.. This agency of the Irish government uses the following B/C ratio for decision making: B IC ratio = EW(Net benefits) EW(Capital recovery cost) + EW(O&M cost) Using an interest rate of 8% and a project life of 30 years, recommend which of the designs is best. 16-25 Six mutually exclusive investments have been identified for evaluation by means of the benefit-cost ratio method. Assume a MARR of 10%, an equal project life of 25 years for all alternatives, and the data in Table P16-25. TABLEP16-24 Data Irish Fishery Design Alternatives First cost Annual benefits Annual O&Mcosts Annual disbenefits Salvage value A B C D $9,500,000 2,200,000 550,000 350,000 1,000,000 $12,500,000 1,500,000 175,000 150,000 6,000,000 $14,000,000 1,000,000 325,000 75,000 3,500,000 $15,750,000 2,500,000 145,000 700,000 7,500,000 ---.-..-.----- Problems TABLEP16-25 Data Mutually Exclusive Alternatives 1 Annualized net costs to sponsor ($M) Annualized 2 3 4 5 6 15.5 13.7 16.8 10.2 17.0 23.3 515 A B C $9500 $18,500 $22,000 3200 5,000 9,800 1000 2,750 6,400 6000 4,200 14,000 15 15 15 Project life, in years MARR 12% 12% 12% Initial investment Annual savings Annual costs Salvagevalue 20.0 16.0 15.0 13.7 22.0 25.0 net benefits to Answer the following questions. users ($M) (a) Use the conventional B/C ratio to evaluate the alternatives and make a recommendation. (b) Use the modified B/C ratio to evaluate the alternatives and make a recommendation. (c) Use a present worth analysis to evaluate the alternatives and make a recommendation. (a) Use annual worth and the B/C ratio to identify the better alternative. (b) If this were a set of independent alternatives, how would you conduct a comparison? Mr. O'Gratias, a top manager in his company, has been asked to consider the following mutually exclusive investment alternatives. --~ - (d) Use an internal rate ofretum analysis to evaluate the alternatives and make a recommendation. (e) Use the simple payback period to evaluate the alternatives and make a recommendation. r'r:~;L. .~ ,~~~~j'.I. ''''-. - ~':. -'...... After Completing This Chapter... The student should be able to: · Identify capital expenditureproject proposals of severaltypes, including mutually. exclusive alternatives and single project proposals. Identify and reject unattractive alternatives,and select the best alternativefrom each project proposal. Use the rate of return and present worth methods to ration capital among projects. Rank project proposals by MARR. · ·· QUESTIONS TO CONSIDER :;. 1. Cancelling projects, particularly ones in which large investments have already been made, is a difficult process in any business. At the Wne that Donald Rumsfeld was preparing to cancel the Crusader project, he would also have had to consider the case against cancellation. Some pressures to keep the project would have come, for example, from Congressionalrepresentativesfrom districts where the Crusader would be built, or who sawmeritin the originalArmy argumentsin favorof the system,or from someArmy representanves.What factors would a decision-makerlike Rumsfeld have to consider to decide among competing interests? 2. In the past, the militaryhas generally soughtto order large quantities of specificweapons in order to keep the unit cost down. In the more uncertain military future, the Pentagon may have to order smaller quantities, but within a wider range of components, in order to keep up with swiftly changing battlefield conditions. How could this affect defense department budgeting decisions? 3. From thepoint of viewof a purchasingfirm(in this case, the US Government),you would need to consideryour firm's budgets, needs, and politicalpressures within the firm when allocating capital. What other concerns do you need to allow for when rationing capital? 4. From the point of view of an engineering firm (in this case, Crusader's manufacturer), you would need to consider the long-term prospects of your customer contracts. How would you prepare your firm to recover from the cancellationof an important contract? What effect would this have on your capital allocation decisions? - '--'.~- -. - - - -- - - - - - - -- - j - - - -- - - - --- - - - - - ,L," @OO~P11~ffi W Rationing Capital Among Competing Projects The Caissons Went Rolling Along Army commanders like field artillery, and who can blame them? When you're engaged in ground operations, there's nothing like a 155 mm howitzer to get your point across in a hurry. The U.S. Army was particularly pleased with the Crusad~r Advanced Field Artillery System, its newest artillery piece, an automatic-loading,rapid-fire, self-propelled "big gun" that used modem robotics and digital technology a generation ahead of other available military systems. It was called the Crusader Advanced Field Artillery System, and it wasjust whatthe Armyneeded. ..... . . Because of its range advantage, the Crusader was able to effectively cover as much territory as four older-style artillery pieces. It could also hit multiple targets in only one-third as much time, and with less collateral damage. Not only that, it offered its crew protection against unconventional attack from nuclear, biological, and chemical weapons-a crucial consideration in modem warfare. It was even being produced on time and under budgeteven while changingthe design in response to changes in Army requirements. Naturally, Donald Rumsfeld killed it. But the Secretary of Defense had his reasons. Despite its advantages, the Crusader also had some significant drawbacks. For one thing, it was a heavy land vehicle, which limited the ways it could be transported to the battlefield.And while everyone agreed that it would have been perfect for repelling a Soviet invasion of West Germany,it was likely to be less suitablefor a style of warfare that deemphasized land forces. n In announcing his plan to terminate the Crusader program over II the objectionsof army generals, Rumsfeld stated, ''This decision is not about anyone weapon system, but ... about a strategy of warfare-a strategy that drives the choices that we must make abouthow best to prepare our total forces for the future." The secretary's stated goal was to modernize the U.S. military with leaner, more futuristic weapons that could fight "asymmetrical" wars against terrorist organizationsand "rogue" states. While the argumentson how best to achievethat goal continued, Rumsfeld's statements were taken to favor air-mobile forces. - -- - --- -, .--..- -- -- ... 518 . RATIONING CAPITAL AMONG COMPETING PROJECTS - Sofrom this cabinetofficer'spointof view,the Crusaderhad to go. The Anny had spent a long timejustifying the Crusader project to Congress under one set of requirements but change in Defense Policy meant that money needed to be freed up for differentprojects: We have until now dealt with situations where, at some interest rate, we choose eachproject's best mutually exclusivealternative.Thus, we were assuming that there is an ample amountof moneyto make all desiredcapitalinvestmentsin these projects. But the conceptof scarcity of resources is fundamentalto a free market economy.It is throughthis mechanism that more economically attractiveactivities are encouraged at the expense of less desirable activities. For industrial firms, there are often more ways of spending money than there is money that is available. The result is that we must select from available alternatives the more attractive projects and reject-or, at least, delay-the less attractiveprojects. This problem of rationing capital among competing projects is one part of a two-pan problem called capital budgeting. In planning its capital expenditures, an industrial firm is faced with two questions: "Where will money for capital expenditures-come from?" and, "How shall we allocate available money among the various competing projects?" In Chapter 15 we discussed the sources of money for capital expenditures as one aspect in deciding on an appropriate interest rate for economic analysis calculations. Thus, the first problem has been treated. Throughout this book, we have examined for any given project two or more feasible alternatives. We have, therefore, sought to identify in each project the most attractive alternative. For the sake of simplicity, we have looked at these projects in an isolated setting-almost as if a firm had just one project it was considering. In the business world, we know that this is rarely the case. A firm will find that there are a great many projects.that are economically attractive. This situation raises two problems not previously considered: 1. Howdo yourankprojectsto showtheirorderof economicattractiveness?. 2. What do you do if there is not enough money to pay the costs of all economically attractive projects? In this chapter we will look at the typical situation faced by a firm: multiple attractive projects, with an inadequate money supply to fund them all. To do this, we review our concepts of capital expenditure situations and available alternatives. Then we su~arize the various techniques that have been presented for determining whether an alternative. is economically attractive. First we screen all alternatives to find those that merit further consideration. Then we will select the best alternative from each~ject, as~umingthat there is no shortage of money.The next step will be the addition of a budget constraint. If we find that there is not enough money to fund the best alternativefrom each project, we will have to do what we can with the limited amount of money available.It will become important that we have a technique for accurately ranking the various competing projects in order of economic attractiveness.All this is designed to answer the question, How shall we allocate available money among the various competingprojects-?--CAPITAL EXPENDITURE PROJECT PROPOSALS At the beginning of the book, we described decision making as the process of selecting the best alternative to achieve the desired objective in a given situation or problem. By carefully defining our objective and the model, the given situation is reduced to one of selecting the best from the feasible alternatives. In this chapter we call the engineering decision-making ---- -, Capital Expenditure Project Proposals 519 process for a givensituationor problem a project proposal. Associatedwith variousproject propqsals are their particular available alternatives.For a firmwith manyprojectproposals, the following situation may result: Capital Expenditure Proposals Project Alternatives 1. Acquire additional manufacturing facility A B C A B 2. Replace old grinding machine 3. Produce parts for the assembly line Lease an existing building Construct a new building Contract for the manufacturing to be doneoverseas Purchase semiautomatic machine Purchase automatic machine A Make the parts in the plant B Buy the parts from a subcontractor Our task is to apply economic analysis techniques to this more complex problem. Mutually Exclusive Alternatives and Single Project Proposals Until nowwe havedealtwith mutuallyexclusivealternatives,thatis, selectingonealternative results in rejecting the other alternatives being considered.Even in the simplestproblems encountered, the question was one of selection between alternatives.Should, for example, Machine A or Machine B be purchased to perform the necessarytask? Clearly,thepurchase of oneof the machinesmeantthatthe other one wouldnotbepurchased.Sinceeithermachine would perform the task, the selection of one precludes the possibilityof selectingthe other one as well. Even in the case of multiple alternatives,we have been consideringmutuallyexclusive alternatives.A typical example was: What size pipeline shouldbe installedto supplywater to a remote constructionsite? Only one alternativeis to be selected.This is differentfrom the situation for single project proposals, where only one course of actionis outlined.Consider Example 17-1. The general manager of a manufacturingplant has received the followingproject proposalsfrom the various operating departments: 1. The foundry wishes to purchase a new ladle to speed up the casting operation. 2. The machine shop has asked for some new inspection equipment. 3. The painting department reports that improvements must be made to the spray booth to conform to new air pollution standards. 4. 'Theoffice.::manager wants to buy:a larger".more modern safe. For each project there is a single course of action proposed. Note that the singleproject proposals are also independent,for there is no interrelationshipor interdependenceamongthem.The general ~ger~.a~de.~ideJo=;allQcat~0ney;;:for.none;=some;::o~l:of the~various.'Pr{)ject proposals~ -. - L__ ________ --- --- --- .J -'-- 520 --- _..----- - RATIONING CAPITAL AMONG COMPETING PROJECTS ] SOlUTIQN Each of the four project proposals has a single course of action. The general manager could, for example, buy the office manager a new safe and buy the inspection equipment for the machine shop. But he could also decide to not buy the office manager a safe or the equipment for the machine shop. There is, then, an alternativeto buying the safe for the office manager: not to buy . him the safe-to do nothing. Similarly,he could decide to do nothing about the request for the machine shop inspection equipment. Naturally, there are do-nothing alternatives for each of the four single project proposals: IA. lB. 2A. 2B. 3A. 3B. 4A. 4B. Purchase the foundry a new ladle. Do nothing. (Do not purchase a new ladle.) Obtain the inspection equipment for the machine shop. Do nothing. (Do not obtain the inspection equipment.) Make improvements to the spray booth in the painting department. Do nothing. (Make no improvements.) Buy a new safe for the officemanager. Do nothing. (Let him use the old safe!) One can adopt Alt. 1A (buy the ladle) or 1B (do not buy the ladle), but not both. We find that what we considered to be a single course of action is really a pair of mutually exclusive alternatives. Even Alt. 3 is in this category.The originally stated single proposal was: The painting departmentreports that improvementsmust be made to the spray booth to conformto newair pollutionstandards. Since the painting department must make the improvements, is there actually another alternative? Although at first glance we might not think so, it may be possible to change the paint, or the spray equipment, and thereby solve the air pollution problem without any improvements to the spray booth. In this situation, there does not seem to be a practical do-nothing alternative, for failure to comply with the air pollution standards might result in large fines or even shutting down the plant. But if there is not a practical,do-nothing alternative,there might be a number of do-something-else alternatives. . We conclude that all project proposals may be considered to have mutually exclusive alter-: natives. Identifying and Rejecting Unattractive Alternatives It is clear that no matter what the circumstancesmay be, we want to eliminate from further considerationany alternativethatfailsto meettheminimumlevelof economicattractiveness, provided one of the other alternatives does meet the criterion. Table 17-1 summarizes five techniques that may be used. At first glance it appears that many calculations are required, but this appearance is misleading.Any of the fivetechniqueslisted in Table 17-1may be used to determinewhether to reject an alternative.Each will produce the same decision regarding RejectIDon't reje..o. ---- 1;: Rationing Capital by Rateof Return 521 TABLE 17-1 Criteria for Rejecting Unattractive Alternatives For'Each Alternative Compute Reject Alternative When Do Not Reject Alternative When Rate of return, i Present worth, PW Annual cost. EUAC Annual benefit, EUAB Benefit-cost ratio, BjC Net present worth, NPW i < MARR PW of benefits < PW of costs EUAC > EUAB i :::MARR PW of benefits :::PW of costs EUAC:::EUAB B/C < 1 NPW < 0 B/C::: 1 NPW :::0 TABLE17-2 Criteria for Choosing the BestAlternative from Among Mutually Exclusive Alternatives Situations Neither Input Nor Output Fixed (Neither the costs nor the benefits for the alternatives Fixed Input (The cost of each alternative is the same.) FixedOutput (The benefits from each alternative are the same.) given are the same.) Present worth Maximize present worth of benefits Minimize present worth of cost Maximize net present worth Annual cash flow Maximize equivalent unifonn annual benefits Minimize Maximize EAUB Maximize benefit-cost ratio Maximize benefit-costratio Analysis Method Benefit-cost ratio Rate of return - EAUC equivalent unifonn annual cost Incrementalbenefit-cost ratio analysisis required Incrementalrate of return analysisis required Selecting the Best Alternative from Each Project Proposal The task of selecting the best alternative from among two or more mutually exclusive alternatives has been a primary subject of this book. Since a project proposal is the same form of problem, we may use any of the severalmethodsdiscussedin Chapters5 through 9. The criteria are summarized in Table 17-2. . TIONING CAPITAL BY RATE OF RETURN One way of looking at the capitalrationingproblem is through the use of rate of return. The technique for selecting from among independent projects is illustrated by Example 17-2. These projects are independent, so their only interdependenceis the budget constraint on the total accepted set of projects. - - - -- -- 1 522 RATIONING CAPITAL AMONG COMPETING PROJECTS Nine independentprojects are being considered. Figure 17-1may be prepared from the following data. Project 1 2 3 4 5 6 7 8 9 Uniform Annual Benefit (thousands) $23.85 39.85 34.72 20.00 20.00 18.00 94.64 47.40 7.00 Cost (thousands) $100 200 50 100 100 100 300 300 50 Useful Life (years) 10 10 2 6 10 10 4 10 10 Salvage Value (thousands) $ 0 0 0 100 100 100 0 100 50 Computed Rate of Return 20% 15 25 20 20 18 10 12 14 If a capital budget of $650,000 is available,which projects should be selected? SOLUTION Looking at the nine projects, we see that some are expectedto produce a larger rate of return than others. It is natural that if we are to select from among them, we will pick those with a higher rate of return. When the projects are arrayed by rate of return, as in Figure 17-1, the correct choice (of Projects 3, 1,4,5,6, and 2) is readily apparent. 30 - '- f3 o f 1 50 4 5 6 2 9 8 7 - - 150 250 350 450 650 700 1000 CumulativeCost of Projects(thousandsof dollars) FIGURE 1.!7-'lGumulative cost of projects versus rate ofretum., ... iii' ..r __ - - - - - - 1300 Rationing Capital by Rateof Return 523 .... pl I.. ~ .,. I.. I.. """" c::: .... ::I """" .... V ~ .... o .';'" I.. <,W ~I I I.. "'i-, II) 'i;j ~ Cutoff """""""" 11-.... ~ Rate of '8' ~ Return -I.. """"II- .~. -rTl o Capital Budget CumulativeCostof Projects($) FIGURE 17-2 Location of the cutoff rate of return. In Example 17-2,the rate of returnwas computedfor each project and then the projects were arranged in order of decreasing rate of return. For a fixed amount of money in the capital budget, the projects are selectedby going down the list until the money is exhausted. Thus when we use this procedure, we cut off approving projects at the point where the money runs out. This point is called the cutoff rate of return. Figure 17-2 illustrates the general situation. For any set of ranked projects and any capital budget, the rate of return at which the budget is exhausted is the cutoff rate of return. In Figure 17-2 the cost of each individual project is smallin comparisonto the capitalbudget.The cumulativecost curveis a relatively smooth curveproducing a specificcutoffrate of return.Lookingback at Figure 17-1,we see the curve is actually a step function. For Example 17-2,the cutoff rate of return is between 14 and 15% for a capital budget of $650,000. Significance of the Cutoff Rate of Return Cutoff rate of return is determined by comparisonof an established capital budget and the available projects. One must examine all the projects and all the money for some period of time (like one year) to compute the cutoff rate of return. It is a computation relating known projects to a known money supply.For this period of time, the cutoff rate of return is the opportunity cost (rate of return on the opportunity or project forgone) and also the minimum attractiverate of return. In other words, the minimum attractive rate of return to get a project accomplishedis the cutoff rate of return. MARR = Cutoff rate of return = Opportunity cost We generally use the minimum attractive rate of return to decide whether to approve an individual project even though we do not know exactly what other projects will be proposed during the year. In this situation, we cannot know whether the MARR is equal to the cutoff - -- --.-- ----- - .-- --". . -.. . -- -" . .. . ....., 524 - RATIONING CAPITAL AMONG COMPETING PROJECTS rate of return. When the MARR is different from the cutoff rate of return, incorrect decisions may occur. This will be illustrated in the next section. RATIONING CAPITAL BY PRESENT WORTH METHODS Throughout this book we have chosen from among project alternatives to maximize net present worth. If we can do the same thing for a group of projects, and we do no~exceed the availablemoney supply,then the capital budgeting problem is solved. But more frequently in capital budgeting problem we will be unable to accept all desirable projects. We, therefore, have a task not previouslyencountered. We must choose the best from the larger group of acceptableprojects. Lorie and Savagel showed that a proper technique is to use a multiplier, p, to decrease the attractivenessof an alternative in proportion to its use of the scarce supply of mon~y. The revised criterion is NPW - p(PW of cost) (17-1) where p is a multiplier. If a value of p were selected (say, 0.1), then some alternatives with a positive NPW will have a negative [NPW - p(PW of cost)]. This new criterion will reduce the number of favorablealternatives and thereby reduce the combined cost of the projects meeting this more severe criterion. By trial and error, the multiplier p is adjusted until the total cost of the projects meeting the [NPW - p(PW of cost)] criterion equals the available money supply-the capital budget. . Use the present worth method to determine which of the nine independent projects of Example 17-2 should be included in a capital budget of $650,000. The minimum attractive rate of return has been set at 8%. . Uniform Useful Computed Salvage Cost Annual Benefit Life Value NPW (thousands) (thousands) (years) Jthousands) (thous~ds) Project 10 1 $23.85 $ 0 $60.04 $100 10 67.40 2 200 39.85 o 34.72 2 11.91 3 50 o 6 55.48 4 100 20.00 100 10 5 20.00 80.52 100 100 10 18.00 67.1.0 100 6 100 4 13.46 7 94.64 o 300 10 64.38 47.40 100 8 300 10 20.-13 7.00 50 9 50 1 Lorie, J. and L. Savage, ''Three Problems in Rationing Capital," Journal of Business, October 1955, pp.229-239. ------- \1,"':' If' I Rationing Capital by Present Worth Methods '.' .__ ,_"'_" j.c. 525 ,_,_,."" SQ! lJ.JjION' Locating a value of pin [NPW - p(PW of cost)] by trial and error, we use the following table, where all amounts are in thousands of dollars. Cost Computed NPW $ 100 200 50 100 100 100 300 300 50 $60.04 67.40 11.91 55.48 80.52 67.10 13.46 64.38 20.13 'oject 1 2 3 4 5 6 7 8 9 - Trialp = 0.20 [NPW Trial p - p(PW of cost)) Cost $40.04 27.40 1.91 35.48 60.52 47.10 -46.54 4.38 10.13 [NPW - p(PW $1300 Cost of cost)] $35.04 17.40 -0.59 30.48 55.52 42.10 -61.54 10.62 7.63 $ 100 200 50 100 100 100 - = 0.25 300 50 $100 200 100 100 100 :-;; ..:'1:' ~t 50 $1000 $650 For a value of p equal to 0.25, the best selection order is computed to be Projects 1,2,4,5,6, and 9. (Note: the smallest value for p is 0.239.) Alternate Formation of Example 17-3 The preceding answer does not agree with the solution obtained in Example 17-2.The difficultyis that the interest rate used in the present worth calculationsis not equal to the computedcutoffrate of return. In Example 17-2the cutoff rate of return was between 14and 15%,say 14.5%. Wewi)l recompute the present worth solution using MARR = 14.5%, again with amounts in thousands of dollars. I Project 1 2 3 4 5 6 7 89 --'10- _ ____ _ - -- Computed NPW at 14.5% $22.01 3.87 6.81 21.10 28.14 17.91 -27.05 --,::31.69 -1.28 Cost $100 200 50 100 100 100 300 300 50 '::!!:: = ====:;;::;= ---- , Cost of Projects with Positive NPW $100 200 50 100 100 100 '" =11I $650 :;.n.. -- I I ::; = + --.. -- .-.-- 526 1 RATIONING CAPITAL AMONG COMPETING PROJECTS . , SQ1VJ!9N At a MARR of 14.5%the best set of projects is the same as computed in Example 17-2, namely, Projects 1,2,3,4,5, and 6, and their cost equals the capital budget. One can see that only projects with a rate of return greater than MARR can have a positive NPW at this interestrate. With MARR equal to the cutoff rate of return, we must obtain the same solution by either the rate of return or present worth method. c .,_~ ~ ~ Example 17-4outlines the present worth method for the more elaboratecase of independent projects with mutually exclusive alternatives. A company is preparing its capital budget for next year according to the steps in the flowchart shown in Figure 17-3.The amounthas been set at $250,000by the Board of Directors.The MARR of 8% is believedto be close to the cutoffrate of return. The followingproject proposals are being considered. Project Proposals Proposal1 Alt. A Alt. B Alt. C Alt. D Proposal 2 Alt.' A Alt.B Proposal 3 Alt.A Alt. B Alt. C Cost (thousands) Uniform Annual Benefit (thousands) $100 150 200 0 $23.85 32.20 39.85 0 $0 0 0 10 10 10 $60.04 66.06 67.40 0 50 0 14.92 0 0 5 9.57 0 100 150 0 18.69 19.42 0 25 125 10 10 36.99 38.21 0 Salvage Value (thousands) Useful Life (years) Computed NPW (thousands) Which project alternatives should be selected, based on present worth methods? SbLUTION I . . . , Tl1~following tabulation shows that to maximize NPW, we would choose Alternatives 1C, 2A, anQ.3.1I. The to.tal cost of these three projects is $400,000. Since the capital budget is only $250,000, we cannot fund ffiese projects. 1'5 pen1ilize a1fpr6jec1s 1)1propodion~tb tlfeif cost, "we will use Equation 17-1 with its multiplier, p. As a first trial, a value of p -:..0.10 is selected and thelalternatiyes with the large~t .[NPW, - p(PW of cost)] selected. --... II I I J \ \, I I Rationing Capital by PresentWorth Methods I Project Proposals . 2 3 ComputeNPW for each alternative. From each project proposal, select alternative with the largest positive NPW. Reject any alternative for which NPW is negative. 5 4 Compute for each alternative with positive NPW: NPW - p(pW of cost) 527 From each project proposal select, alternative with the largest positive NPW - p(pW of cost) Omit any project for which all values are negative. , ,. Proposal I 1A IB IC IV Donothing Proposal2 2A 2B Do nothing Proposal3 3A 3B 3C Donothing This is the desiredset of projects. == For Fixed-Sum-of-Money Constraint Compute Total Required Money for This Set of Projects Compute Total Required Money for this set of projects Fixed-Sum-of- Money Constraint Fixed sum is equal to or greater than amount required for this set of projects. This is the desired set of projects. Fixed sum is less than amount required for this set of projects. Select an initial value of p and continue with calculations in columns 4 and 5. FIGURE,17...3Steps in computing-acapital budget = - -.. - -...- Fixed sum is less than amount required for this set of projects. Fixed sum is Fixed sum is equal to greater than amount amount required required for this for this set of projects. set of projects. Decrease p Increasep and This is the desired set of and recompute recompute columns columns projects. 4 and 5. 4 and 5. - ill :Ii~j --~ --- --~ --. :! - ------ II !II .L. -,~ 528 RATIONING CAPITAL AMONG COMPETING PROJECTS p=O.10 Project Proposals - Cost (thousands) NPW (thousands) $100 150 200 0 $60.04 66.06 67.40 0 Alternative with Largest Positive NPW (thousands) Alt. [NPW Alternative with Largest Positive [NPW p(PW of cost)] (thousands) - - p(PW of cost)] (thousands) Cost Alt. Cost IB $150 4.57 0 2A 50 26.99 23.21 0 3A 100. Proposal 1 Alt. A Alt. B Alt.C Alt. D Proposal 2 Alt.A 9.57 50 0 Alt. B Proposal 3 Alt.A Alt.B Alt.C . 1C $200 2A 50 3B 150 $50.04 51.06 47.40 0 0 36.99 38.21 0 100 150 0 $400 The first trial with p $300 = 0.10 selects Alternatives 1B, 2A, and 3A with a total cost of $300,000. This still is greater than the $250,000 capital budget. Another trial is needed with of p. Select p = 0.15 and recompute. alarger value p=O.1S' Alternative with Largest Positive [NPW Project Proposals - Proposal 1 Alt.A Alt. B Alt. C Alt. D Proposal 2 Alt. A Alt. B Proposal3 Alt. A Alt. B Alt.C -- - Cost (thousands) NPW (thousands) $100 150 200 0 $60.04 66.06 67.40 $45.04 43.56 37.40 o o [NPW - p(PW of cost)] (thousands) 50 0 9.57 2.07 o o 100 150 0 36.99 38.21 0 The second trial, witllp : - p(PW of cost» (thousands) 21.99 15.71 0 -. - :; i8i::i:::: :; :: i!!: Alt. Cost lA $100 2A 50 3A 100 ;:;; =iii== ;;;:. $250 = 0.15, points to AlternatjveslA,2A, and3A for a total cost Qf$250,000. -nus ~qmi1s~the "dipital"budget,:::hence'=is.the;desired 'wt!()f""Prejeets~ == ~ -~ .. ~ -J!!! ~ \~ __.,_~_ "" 4" J2' _. J - -- I.\} ---~-- III Rationing Capital by Present Worth Methods 529 Solve Example 17-4 by the rate of return method. For project proposals with two or more alternatives, incremental rate of return analysis is required. The data from Example 17-4 and the computed rate of return for each alternative and each increment of investment are as follows: Incremental Analysis Uniform Annual Salvage Computed ,mbination Cost Benefit Value Rate of Return Alternatives (thousands) (thousands) (thousands) )POScu1 A B-A B C-B C-A C b $100 $23.85 150 32.20 $ 0 20.0% 0 17.0 200 0 39.85 0 0 0 50 0 14.92 0 0 0 15.0 0 100 18.69 25 15.0 150 0 19.42 0 125 0 12.0 0 Uniform Annual Computed Salvage Cost Benefit Value Rate of Return (thousands) (thousands) (thousands) 10.6% $ 50 $ 8.35 $ 0 50 100 7.65 16.00 0 0 8.6 9.6 50 0.73 100 8.3 .15.0 0 )posal 2 B iposal 3 4- 9-A r1 The various separableincrements of investmentmay be ranked by rate of return. They areplotted in a graph of cumulativecost versus rate of return in Figure 17-4.The ranking of projects by rate of return gives the following: Project lA 2A 3A --------- :;; IB in place of lA lC in place of IB 3B in place of 3A ."" ~-~ ~ ~~::::iii:iii :: .- I 1 I 530 RATIONING CAPITAL AMONG COMPETING PROJECTS 20 .-.. 15.0% 15 e ! 10.6% .... 0 10 8.6% 8.3% - - "0 B =' U 5 lA o 2A 3A IB - 1A lC IB 3B 100 150 250 300 350 CumulativeCost of InvestmentIncrements(thousandsof dollars) 3A 400 FIGURE 17-4 Cumulative cost versus incremental rate of return. For a budget of $250,000, the selected projects are lA, 2A, and 3A. Note that if a budget of $300,000 were available, IB would replace lA, making the proper set of projects IB,2A, and 3A. At a budget of $400,000, lC would replace IB; and 3B would replace 3A, making the selected projects lC, 2A, and 3B. These answers agree with the computations in Example 17-4. RANKINGPROJECTPROPOSALS Closely relatedto theproblem of capitalbudgetingis the matterof rankingprojectproposals. We will first examine a method of ranking by present worth methods and then show that project rate of return is not a suitable method of ranking projects. Anyone who has ever bought firecrackersprobably used the practical ranking criterion of "biggest bang for the buck" in making a selection. This same criterion-stated more elegantly-may be used to correctly rank independent projects. Rank independent projects according to their value of net present worth divided by the present worth of cost. The appropriate interest rate is MARR (as a reasonable estimate of the cutoff rate ofreturn). Example 17-6illustrates the method of computation. --- - Ranking Project Proposals . 531 I ~ Rank the following nine independent projects in their order of desirability, based on a 14.5% minimum attractiverate of return. (Tofacilitate matters, the necessarycomputationsare included in the tabulation.) Cost "Oject (thousands) 1 $100 2 200 3 50 4 100 5 100 6 100 7 300 8 300 9 50 Uniform Annual Benefit (thousands) $23.85 39.85 34.72 20.00 20.00 18.00 94.64 47.40 7.00 Useful Life (years) 10 10 2 6 10 10 4 10 10 Salvage Value (thousands) $ 0 0 0 100 100 100 0 100 50 Computed Rate of Return 20% 15 25 20 20 18 10 12 14 Computed NPWat 14.5% (thousands) $22.01 3.87 6.81 21.10 28.14 17.91 - 27.05 -31.69 -1.28 Computed NPW/Cost (thousands) 0.2201 0.0194 0.1362 0.2110 0.2814 0.1791 -0.0902 -0.1056 -0.0256 SOLUTION I Ranked by NPw IPW of cost, the projects are listed as follows: Project 5 1 4 6 3 2 9 7 8 NPWIPW of Cost 0.2814 0.2201 0.2110 0.1791 0.1362 0.0194 -0.0256 -0.0902 -0.1056 Rate of Return 20% 20 20 18 25 15 14 10 12 The rate of return tabulation illustrates that it is not a satisfactory ranking criterion and would have given a different ranking from the present worth criterion. In Example 17-6, the projects are ranked according to the ratio NPWIPW of cost. In Figure 17-3,the criterion used is [NPW - p(pW of cost)]. If one were to compute the value of p at which [NPW - p(PW of cost)] = 0, we would obtain p = (NPWIPWof cost). Thus the multiplier p is the ranking criterion at the point at which [NPW - p(PW of cost)] = 0."." If independent projects can be ranked in their order of desirability,then the selection of projects to be.include~ in a cap.italbudget is a simple task. One may proc~eddown ~e li~t of ranked projects until the capItal budget has been exhausted.The only difficultyWIththis ~~ ::'.11, ~. ~~, I, '. i][.£- _.. u _ __ .- - - 532 -- --- RATIONING CAPITAL AMONG COMPETING PROJECTS scheme occurs, occasionally,when the capital budget is more than enough for n projects but too little for n + 1 projects. In Example 17-6, a capital budget of $300,000 is just right to fund the top three projects. But a capital budget of $550,000 is more than enough for the top five projects (sum = $450,000) but not enough for the top six projects (sum = $650,000). When we have this lumpiness problem, it may not be possible to say with certainty that the best use of a capital budget of $550,000is to fund the top five projects. There may be some other set of projects that makes better use of the available $550,000. While some trial-and-error computationsmay indicatethe proper set of projects, more elaboratetechniques are needed to prove optimality. As a practical matter,a capitalbudget probably has some flexibility.If in Example 17-6 the tentativecapital budgetis $550,000, then a careful examinationof Project 2 will dictate whether to expand the capitalbudget to $650,000 (to be able to includeProject 2) or to drop back to $450,000 (and leaveProject 2 out of the capital budget). . SUMMARY Prior to this chapter we had assumed that all worthwhile projects are approved and implemented. But industrial firms, like individuals and governments,are typically faced with more good projects than can be funded with the money available.The task is to select the best projects and reject, or at least delay, the rest. Alternatives are mutually exclusive when the acceptance of one "effectivelyprevents the adoption of the other.This could be"becausethe alternativesperform the same function (like Pump A vs Pump B) or would occupy the same physical location (like a gas station vs a hamburger stand). If a project has a single alternativeof doing something, "weknow there is likely to be a mutually exclusive alternative of doing nothfug-or possibly doing something else. A project proposal may be thought of as having two or more mutually exclusive alternatives.Projects are assumed in this chapter to be independent. Capital may be rationed among competing investmentopportunities by either rate of return or present worth methods. The results may not always be the same for these two methods in many practical situations~ If projects are ranked by rate of return, a proper procedure is to go down the list until the capital budget has been exhausted. The rate of return at this point is the cutoff rate of return. This procedure gives the best group of projects, but does not necessarily have them in the proper priority order. MaximizingNPWis an appropriatepresent worthselectioncriterionwhen the available projects do not exhaust the money supply. But if the amount of money required for the best alternative from each project exceeds the available money, a more severe criterion is imposed: adopt only alternatives and projects that have a positive NPW - p(PW of cost). The value of the multiplier p is chosen by trial and error until the alternatives and projects meeting the criterionjust equal the available capital budget money. It has been shown in earlier chapters that the usual business objective is to maximize NPW, and this is not necessarily the same as maximizing rate of return. One suitable procedure is to use the ratio (NPW/PW of cost) to rank the projects. This present worth ranking method will order the projects so that, for a limited capital budget, NPW will be .". " --- - -- -- Problems 533 maximized. Weknow that MARR mustbe adjusted from time to time to reasonablybalance the cO,stof the projects that meet the MARR criterion and the available supply of money. This adjustment of the MARR to equal the cutoff rate of return is essential for the rate of return and present worth methods to yield compatible results. Another way of ranking is by incremental rate of return analysis. Once a ranking has been made, we can go down the list and accept the projects until the money runs out. There is a theoretical difficulty if the capital budget contains more money than is required for n projects, but not enough for one more (n + 1projects). As a practical matter,capital budgets are seldom inflexible, with the result that some additional money may be allocated if the (n + l)st project looks like it should be included. ~OBLEMS Ear 1.. of the following 10 independent projects has a 10.,) ~ar life and no salvage value. Project 1 2 3 4 5 6 7 8 9 10 Cost (thousands) $ 5 15 10 30 5 20 5 20 5 10 Uniform Annual Benefits (thousands) $1.03 3.22 1.77 4.88 1.19 3.83 1.00 3.69 1.15 2.23 Cost Project Proposal (thousands) Project 1 Alt.A $25 Alt. B 50 Alt. C 10 Computed Rate of Return 16% 17 12 10 20 14 15 13 19 18 Project 2 Alt. A Alt. B Project 3 Alt. A Alt. B Project 4 Computed Rate of Return $4.61 9.96 2.39 13% 15 20 20 35 4.14 6.71 16 14 25 10 10 5.56 2.15 1.70 18 17 11 (a) Use rate of return methods to determine which set of projects should be uQdertaken if the MARR is 10%. (b) Use rate of return methods to determine which set of projects should,be undertaken if the capital budget is limited to $100,000. (c) For a budget of $100,000, what interest rate should be used in rationing capital by present worth methods? (Limit your answer to a value for which there is a compound interest table available in the appendix). (d) Using the interest rate determined in part (c), rank order the eight different investment opportunities by means of the present worth method. (e) For a budget of $100,000 and the ranking in part (d), which of the investment opportunities should be selected? The projects have been proposed by the staff of the Ace Card Company. The MARR of Ace has been 12% for several years. (a) If there is ample money available, what projects should Ace approve? (b) Rank order all the acceptable projects in their order of desirability. (c) If only $55,000 is available, which projects should be approved? At Miami Products, four project proposals (three with mutually exclusive alternatives) are being considered. All the alternatives have a lO-year useful life and no salvage value. -- -- Uniform Annual Benefits (thousands) - - -- - - _.'_ .u _ .. ..- 534 1 - RATIONINGCAPITALAMONG COMPETINGPROJECTS TABLE P17-3 Data "Oh" Rating of Gift If Given to Various FamilyMembers Prospective Gift - 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. $20 box of candy $12 box of cigars $16 necktie $20 shirt or blouse $24 sweater $30 camera $ 6 calendar $16 magazinesubscription $18 book $16 game Father Mother Sister Brother Aunt Uncle Cousin 4 3 2 5 3 1 0 4 3 2 4 0 0 3 4 5 0 3 4 2 2 0 0 4 5 2 1 4 2 3 1 1 3 4 4 5 0 4 3 2 5 0 0 4 3 1 1 3 4 2 2 1 3 1 4 2 0 1 0 1 3 2 2 4 2 0 1 3 3 2 17-3 AI Dale is planning his Christmas shopping for seven people. To quantify how much his various relatives would enjoy receiving items from a list of prospective gifts, AI has assigned appropriateness units (called "ohs") for each gift if given to each of the seven people. A rating of 5 ohs. represents a gift that the recipient would really like. A rating of 4 ohs indicates the recipient would like it four-fifths as much; 3 ohs, three-fifths as much, and so forth. A zero rating indicates an inappropriate gift that cannot be given to that person. The objective is to select the most appropriate set of gifts for the seven people (that is, maximize total ohs) that can be obtained with the selected budget. (a) How much will it cost to buy the seven gifts the people would like best, if there is ample money for Christmas shopping? (b) If the Christmas shopping budget is set at $112, which gifts should be purchased, and what is their total appropriateness rating in ohs? (c) If the Christmas shopping budget must be cut to $90, which gifts should be purchased, and what is their total appropriateness rating in ohs? (Answer: (a), $168) The foUowing facts are to be used in solving Problems 17-4 through 17.7 In assembling data for the Peabody Company annual capital budget, five independent projects are being considered. Detailed examinationby the staff has resulted in the identificationof three to six mutually exclusive dosomething alternativesfor each project. In addition, each project has a do-nothing alternative. The projects and their alternatives are listed at the top of next page. Each project concerns operations at Peabody's St. Louis brewery. The plant was leased from another firm many years ago, and the lease expires 16 years. from now. For this reason, the analysis period for all projectS is 16 years. Peabody considers 12% to be the minimum attractive rate of return. . In solving the Peabody problems, an important assumption concerns the situation at the end of the useful life of an alternative when the alternative has a useful life less than the 16-year analysis period. Two replacement possibilities are listed. Assumptio.n 1: When an alternative has a useful life less than 16 years, it will be replaced by a new alternative with the same useful life as the original. This may need to occur more than once. The new alternative will have a 12% computed rate of return and, hence, a NPW =0 at 12%. Assumption 2: When an alternative has a useful life less than 16 years, it will be replaced at the end of its useful life by an identical alternative (one with the same cost, uniform annual benefit, useful life, and salvage value as the original alternative). 17-4 For an unlimited supply of money, and replacement Assumption 1, which project alternatives should Peabody select? Solve the problem by present worth methods. (Answer: Alternatives 1B, 2A, 3F, 4A, and 5A) ------ I 1\\.''''' Problems Cost (thousands) Unifonn Annual Benefit (thousands) $40 10 55 30 15 535 Useful Life (years) Computed Rate of Return $13.52 1.87 18.11 6.69 3.75 2 16 4 8 2 $20 5 0 0 15 10% 18 12 15 25 10 5 5 15 1.91 1.30 0.97 5.58 16 8 8 4 2 0 2 0 18 20 15 18 20 5 10 15 10 15 2.63 0.84 1.28 2.52 3.50 2.25 16 16 16 16 4 16 10 0 0 0 0 15 12 15 10 15 15 15 .D :t 5 .A B 10 5 5 15 2.61 0.97 0.90 3.34 8 16 16 8 0 0 5 0 20 18 18 15 5 10 0.75 3.50 e 15 2.61 8 4 8 5 0 5 15 15 12 IOSal :ct 1 t.A t.B t.e t.D t.E ct 2 t.A .B :.e :.D ct 3 :.A :.B .e .D .E .F ;t4 .A .B .e For an unlimited supply of money, and replacementAssumption2, whichproject alternativesshould Pea' ly select? Solvethe problem by present worth methods. Periodically they present their findings concerning business opportunities. On a particular occasion, they presented the following investment opportunities: Project A: This is a project for the utilization of the commercial land the financier already owns. There are three mutually exclusive alternatives. For an unlimited supply of money, and replacementAssumption2, whichprojectalternativesshould Peabody select? Solve the problem by rate of re:urn methods. (Hint: By careful inspection of the l.lternatives,you should be able to reject about half )f them. Even then the problem requires lengthy ;alculations.) · · "or a capital budget of $55,000, and replacenent Assumption 2, which project alternatives :hould Peabody select? (Answer: Alternatives .E, 2A, 3F, 4A, and 5A) \. financier has a staff of three people whose job it s to examine possible business ventures for him. --- · Ai. Sell the land for $500,000. A2. Lease the property for a car-washing business. An annual income, after all costs (property taxes, etc.) of $98,700 would be received at the end of each year for 20 years. At the end of the 20 years, it is believed that the property could be sold for $750,000. A3. Construct an office building on the land. The building will cost $4.5 million to construct and will not produce any net income for the first 2 years. The probabilities of various levels of rental income, --- ------ ----- f' ' ., 1 " t End-of-Useful-Life Salvage Value (thousands) ect It. : ,: - --- I 536 RATIONING CAPITAL AMONG COMPETING PROJECTS after all expenses, for the subsequent 18 years are as follows: Annual Rental Income $1,000,000 1,100,000 1,200,000 1,900,000 Probability 0.1 0.3 0.4 0.2 The property (building and land) probably can be sold for $3 million at the end of 20 years. Project Eo'An insurance company is seeking to borrow money for 90 days at 13 3/4%per annum, compounded continuously. Project C: A financier owns a manufacturing company. The firm desires additional working capital to allow it to increase its inventories of raw materials and finished products. An investment of $2 million will allow the company to obtain sales that in the past the company had to forgo. The additional capital will increase company profits by $500,000 a year. The financier can recover this additional investment by ordering the company to reduce its inventories and to return the $2 million. For planning purposes, assume the additional investment will be returned at the end of 10 years. Project D: The owners of Sunrise magazine are seeking a loan of $500,000 for 10 years at a 16% interest rate. Project E: The Galveston Bank has indicated a willingness to accept a deposit of any sum of money over $100,000, for any desired duration, at a 14.06% interest rate, compounded monthly. It seems likely that this interest rate will be available from Galveston, or some other bank, for the next several years. Project F: A car rental company is seeking a loan of $2 million to expand its fleet of automobiles. The Company offers to repay the loan by paying $1 million at the end of Year 1 and $1,604,800 · · at theendof Year2. . If there is $4 million available for investment now (or $4.5 million if the Project A land is sold), which projects should be selected? What is the MARR in this situation? If there is $9 million available for investment now (or $9.5 million if the Project A land is sold), which projects should be selected? -- -- --- 17-9 The Raleigh Soap Company has been offered a 5-year contract to manufacture and package a leading brand of soap for Taker Bros. It is understood that the contract will not be extended past the 5 years because Taker Bros. plans to build its own plant nearby. The contract calls for 10,000 metric tons (one metric ton equals 1000 kg) of soap a year. Raleigh normally produces 12,000 metric tons of soap a year, so production for the 5-year period would be increased to 22,000 metric tons. Raleigh must decide what changes, if any, to make to accommodate this increased production. Five projects are under consideration. Project 1: Increase liquid storage capacity. Raleigh has been forced to buy caustic soda in . tank truck quantities owing to inadequate storage capacity. If another liquid caustic soda tank is installed to hold 1000 cubic meters, the caustic soda may be purchased in railroad tank car quantities at a more favorable price. The result would be a saving of 0.1 cent per kilogram of soap. The tank, which would cost $83,400, has no net salvage value. Project 2: Acquire another sulfonation unit. The present capacity of the plant is limited by the sulfonation unit. The additional 12,000 metric tons of soap cannot be produced without an additional sulfon~tion unit. Another unit can be installed for $320,000. Project 3: Expand the packaging department. With the new contract, the packaging department must either work two 8-hour shifts or have another packaging line installed. If the two-shift operation is used, a 20% wage premium must be paid for the second shift. This premium would amount to $35,000 a year. The second packaging line could be installed for $150,000. It would have a $42,000 salvage value at the end of 5 years. Project 4: Build a new warehouse. The existing warehouse will be inadequate for the greater production. It is estimated that 400 square meters of additional warehouse is needed. A new warehouse can be built on a lot beside the existing warehouse for $225,000, including-the land. The annual taxes, insurance, and other ownership costs would be $5000 a year. It is believed the warehouse could be sold at the end of 5 years for $200,000. -- -- Problems Project 5: Lease a warehouse. An alternative to building an adqitional warehouse would be to lease warehouse space. A suitable warehouse one mile away could be leased for $15,000 per year. The $15,000 includes taxes, insurance, and so forth. The annual cost of moving materials to this more remote warehouse would be $34,000 a year. 17-11 Mike Moore's microbrewery is considering production of a new ale called Mike's Honey Harvest Brew. To produce this new offering he is considering two independent projects. Each of these projects has two mutually exclusive alternatives and each alternative has a useful life of 10 years and no salvage value. Mike's MARR is 8%. Information regarding the projects and alternatives are given in the following table: The contract offered by Taker Bros. is a favorable one, which Raleigh Soap plans to accept. Raleigh management has set a 15% before-tax minimum attractive rate of return as the criterion for any of the projects. Which projects should be undertaken? Project A B C D E F G H I J Initial Cost (thousands) $10 15 5 20 15 30 25 10 5 10 Cost Project! Alternative Project 1. Purchase new fermenting tanks Alt. A: 5000-gallon tank Alt. B: 15,000-gallon tank Project 2. Purchase bottle filler and capper Alt. A: 2500-bottle/hour machine Alt. B: 5000-bottle/hour machine Ten capital spending proposals have been made to the budget committee as the members prepare the annual budget for their firm. Each independent project has a 5-: .life and no salvage value. Uniform Annual Benefit (thousands) $2.98 5.58 1.53 5.55 4.37 9.81 7.81 3.49 1.67 3.20 537 Computed Rate of Return 15% 25 16 12 14 19 17 22 20 18 $ 5000 Annual Benefit 10,000 $1192 1992 15,000 3337 25,000 4425 Use incremental rate of return analysis to complete the following worksheet. Proj./Alt. lA IB-IA 2A 2B-2A Annual Cost, P Benefit, A AlP, i, 10 IRR 0.2385 20% $1192 $ 5,000 800 0.1601 5,000 15,000 10,000 . 3337 Use this information to determine: (a) Which projects should be funded if only $15,000 is available. ~ ...~ ~ (b) The cutoff rate of return if only $15,000 is available. (c) Which projects should be funded if $25,000 is available. (a) sed on a MARR of 14%, which projects should be approved? (b) Rank order allthe projectsin orderof desirability. (c) If only $85,000 is available, which projects shouldbe approved? -- - -- - -- .. After Completing This Chapter... The students should be able to: Describe the links between engineering economy and accounting. Describe the objectives of general accounting, explain what financial transactions are, and show how they are important. Use a firm's balance sheet and associated financialratios to evaluate the firm's health. .. Use a firm's income statement and associated financial ratios to evaluate the firm's performance. Use traditional absorption costing to calculate product costs. Understand the greater accuracy in product costs available with activity-based costing (ABC). ·· · ·· QUESTIONS TO CONSIDER ~ 1. ABB's adoption of activity-based costing received widespread publicity and boosted the company's reputation for innovative inanagement. How might this have affected. investors' assumptions about the company's other accounting practices? 2. Outside analystsestimated that ABB lost over $690 million in 2001.Yetmany investors were still stunned by its poor showing in late 2002. What does this say about the relationship between financial accounting and investorconfidence? 3. ABB was not alone in its financial misery, of course. How did ABB's accounting problems compare with those of well-known American companies such as Eoron and WorldCom? .,.. ---- ------- Accou nting and Engineering Economy The Two Faces of ABB In the late 1990s,ABB Ltd was flyinghigh.The European conglomeratewas an engineering giant with a global network of operations and forward-lookingmanagement. Unlike many of its competitors, ABB was also committed to modernizing its accounting system. ABB had previously followed the traditional practice of assigning overhead costs to its divisions on a roughly equal basis. But this practice tended to obscure the fact that some activities incurred far more costs than others. So the company spent substantialtime and resources switching over to an activity-based costing (ABC) system, which assigns costs to the activities that actually produce them. The results were quite positive, allowingABB to zeroin onareas whereit couldcutcosts mosteffectively. But in other respects, ABB's accounting practices were less than proactive. One of its most seriousproblems arose at CombustionEngineering, an.American subsidiarythat was exposed to numerous asbestos liability claims. For years, ABB had downplayed the extent of this potential liability, despite warnings from outside analysts. Finally, in late 2002, ABB admitted that its asbestos liability exceeded the subsidiary's total asset value. But there was more bad news still. The company also issued a very poor third-quarter earnings report, after earlier assuring investors that ABB was on target to improve its earnings and decrease its debt. When incredulous investors asked how this could have happened, ABB management blamed "poor internal reporting." By this point, ABB's stock price had nose-dived, and credit rating agencies viewed the company's bonds as little better than junk. Today, ABB is struggling to survive. It has sent Combustion Engineering into bankruptcy in an attempt to limit asbestos claims, and has sold off other subsidiaries. It has also slashed thousands of jobs and embarked on a wide array of cost-cutting projects. ABB now claims to have put the worst behind it. Investors must certainly be hoping this is true. --.----- --_.- 540 l ACCOUNTING AND ENGINEERING ECONOMY Engineering economyfocuseson the financialaspects of projects, while accountingfocuses on the financialaspectsof firms.Thus the applicationof engineeringeconomyis mucheasier if one has some understanding of accounting principles. In fact, one important accounting topic, depreciation, was the subject of an earlier chapter. THE ROLE OF ACCOUNTING Accounting data are used to value capital equipment, to decide whether to make or buy a part, to determine costs and set prices, to set indirect cost rates, and to make product mix decisions. Accounting is used in private-sector firms and public-sector agencies, but for simplicity this chapter uses "the firm" to designate both. Accountants track the costs of projects and products, which are the basis for estimating future costs and revenues. The engineeringeconomy,accounting,and managerialfunctionsare interdependent.. As shown in Figure' 18-1, data and communications flow between them. Whether carried out by a single person in a small firm or by distinct divisions in a large firm, all are needed. · · · performance,allocatesresources,and selectsand directspersonnel. Engineering economy analyzes the economic impact of design alternativ.esand projects over their life cycles. Accounting determinesthe dollar impact of past decisions, reports on the economic viability of a unit or firm, and evaluatespotential funding sources. Management allocatesavailableinvestmentfunds to projects, evaluatesunit and firm Accountil1g . MlDagemel1t About past Analyzing Summarizing Reporting Financial indicators Economic trends About past and future Capital budgeting Decision making Setting goals Assessing impacts Analyzing risk Planning Controlling Record keeping Cost acquisitions . . ., . Engifi;;ringEcort6my About future Feasibility of alternatives Collect/analyze data Estimate Evaluate projects Recommend Audit Identify needs Trade-offslconstraints I Data and Communication Data and Communication Data and Communication Budgeting Estimating I FIGURE 18-1 The accounting, managerial, and engineering economy functions. Accounting for Business Transactions A business transaction involvestwo parties and the exchange of dollars (or the promise of dollars) for a product or service. Each day, millions of transactions occur between firms and their customers, suppliers, vendors, and employees. Transactions are the lifeblood of the business world and are most often stated in monetary terms. The accountiQ.gfunction records, analyzes, and reports these exchanges. Transactions can be as simple as payment for a water bill, or as complex as the international transfer of millions of dollars of buildings, land, equipment, inventory,and othe1" -- -- - - - 0..-_ f j + - ... -.. - .-.- . ........-.- The Balance Sheet 541 assets. Also, with transactions, one business event may lead to another-all of which need to be.accounted for. Consider,for example,the process of selling a robot or bulldozer.This simple act involves several related transactions: (1) equipment released from inventory, (2) equipment shipped to the purchaser, (3) invoicing the purchaser, and finally (4) collecting from the purchaser. Transactionaccountinginvolvesmore thanjustreporting: it includes finding,synthesizing, summarizing,and analyzingdata.For the engineering economist,historicaldata housed in the accounting function are the foundation for estimates of future costs and revenues. Most accounting is done in nominal or stable dollars. Higher market values and costs due to inflationare less objectivethan cost data, and with a goingconcern, accountantshave decided that objectivity should be maintained. Similarly, most assets are valued at their acquisition cost adjusted for depreciation and improvements. To be conservative, when market value is lower than this adjusted cost, the lower value is used. This restrains the interests of managementin maXimizinga firm's apparent value. If a firmmust be liquidated, then current market value must be estimated. The accounting function provides data for general accounting and cost accounting. This chapter's presentation begins with the balance sheet and income statement, which are the two key summaries of financial transactions for general accounting. This discussion includes some of the basic financial ratios used for short- and long-term evaluations. The chapter concludes with a key topic in cost accounting-allocating indirect expenses. ::BALANCESHEET The primary accounting statements are the balance sheet and income statement. The balance sheet describes the firm's financial condition at a specifictime, while the income statement describes the firm's performance over a period of time-usually a year. The balance sheet lists the firm's assets, liabilities, and equityon a specifieddate. This is a picture of the organization's financial health or a snap shot in time. Usually, balance sheets are taken at the end of the quarter and fiscal year. The balance sheet is based on the fundamental accounting equation: Assets = Liabilities + Equity (18-1) Figure 18-2illustrates thebasic format of the balance sheet.Noticein thebalance sheet, as in Equation 18-1, that assets are listed on the left-hand side and liabilities and equity are on the right-hand side. The fact that the firm's resources are balanced by the sources of funds is the basis for the name of the balance sheet. Assets In Equation 18-1 and Figure 18-2 assets are owned by the firm and have monetary value. Liabilities are the dollar claims against the firm. Equity represents funding from the firm and its owners (the shareholders). In Equation 18-1 assets are always balanced by the sum of the liabilities and the equity.Retained earnings are set so that equity equals assets minus liabilities. On a balance sheet, assets are listed in order of decreasing liquidity,that is, according to how quickly each one can be convertedto cash. Thus, current assets are listed first, and I 542 ACCOUNTING AND ENGINEERINGECONOMY BalanceSheet for EngineeredIndustries,December31, 2004 (all amountsin $1OOOs) - Assets liabilities Cun-ent assets Cash Accounts receivable Securities Inventories (minus)Bad debt provision 1940 950 4100 1860 --80 8770 Total cun-ent assets Cun-entliabilities Accounts payable Notes payable Accrued expense Totalcun-entliabilities Long-termliabilities Fixed assets Land Plant and equipment (minus)Accumulateddepr. Total fixedassets - 950 2i8O 1200 335 6500 -- 2350 4485 Other assets 140 420 560 Prepays/deferredcharges Intangibles Total other assets Total assets 1150 80 13,815 Equity Preferred stock Common stock Capital surplus Retained earnings Total equity 110 650 930 8745 10,435 Total liabilities and equity 13,815 - FIGURE 18-2 Sample balance sheet. within that category in order of decreasing liquidity are listed cash, receivables,securities, and inventories. Fixed assets, or property, plant, and equipment, are used to produce and deliver goods and/or services, and they are not intended for sale. Lastly, items su~h as . . prepayments and intangibles such as patents are listed. The term "receivables" comes from the manner of handling billing and payment for most business sales. Rather than requesting immediate payment for every transaction by check or credit card, most businesses record each transaction and then once a month a bill for all transactions. The total that has been billed less payments already received is called accounts receivable, or receivables. Liabilities On the balance sheet, liabilities are divided into two major classifications-short term and long term. The short-term or current liabilities are expenses, notes, and other payable accounts that are due within one year from the balance sheet date.Long-termliabilitiesinclude mortgages, bonds, and loans with later due dates.For Engineered Industriesin Figure 18-2, total current and long-term liabilities are $2,180,000and $1,200,000,respectively.Oftenin performing engineeringeconomic analyses, the working capital for a project must be estimated. The total amountof working capital.availablemay be calculatedusing Equation 18-2 as the difference between current assets and current liabilities. Working capital = Current assets - Current liabilities (18-2) ~ -. The Balance Sheet 543 For Engineered Industries, there would be $4,670,000 - $2,180,000= $2,490,000 available in working capital. Equity Equity is also called owner's equity or net worth. It includes the par value of the owners' stockholdingsand the capital surplus,which are the excess dollars brought in overpar value when the stock was issued. Retainedearnings are dollars a firmchooses to retain rather than paying out as dividends to stockholders.Equity is the dollar quantity that alwaysbrings the balance sheet, and thus the fundamentalaccounting equation, into balance. For Engineered Industries, total equity value is listed at $10,435,000. From Equation 18-1 and the assets, liabilities, and equity values in Figure 18-2, we can write the balance as follows: Assets = Liabilities + Equity Assets(current, fixed,other) = Liabilities(currentand long-term) + Equity (4,670,000 + 4,485,000 + 560,000) = (2,180,000 + 1,200,000) + 10,435,000 $13,815,000 = $13,815,000 An example of owner's equity is ownership of a home. Most homes are purchased by means of a mortgage loan that is paid off at a certain interest rate over 15 to 30 years. At any point in time, the difference between what is owed to the bank (the remaining balance on the mortgage) and what the house is worth (its appraised market value) is the owner's equity. In this case, the loan balance is the liability,and the home's value is the asset-with equity being the difference. Over time, as the house loan is paid off, the owner's equity increases. The balance sheet is a very useful tool that shows one view of the firm's financial condition at a particular point in time. Financial Ratios Derived from Balance Sheet Data . One common way to evaluate the firm's health is through ratios of quantities on the balance sheet. Firms in a particular industry will typically have similar values, and exceptions will often indicate firms with better or worse performance.Twocommon ratios used to analyze the firm's current position are the current ratio and the acid-test ratio. A firm's current ratio is the ratio of current assets to current liabilities, as in Equation 18-3. Current Ratio = Current assets/Current liabilities (18-3) This ratio provides insight into the firm's solvency over the short term by indicating its ability to cover current liabilities. Historically, firms aim to be at or above a ratio of 2.0; however, this depends heavily on the industry as well as the individual firm's management practices and philosophies. The current ratio for Engineered Industries in Figure 18-2 is above 2 (4,670,000/2,180,000 = 2.14). Both working capital and the current ratio indicate the firm's ability to meet currently maturing obligations. However, neither describes the type of assets owned. The acid-test --- u 544 .... ... .--- ACCOUNTING AND ENGINEERINGECONOMY ratio or quick ratio becomes important when one wishes to consider the finn's ability to pay debt "instantly."The acid-test ratio is computedby dividing a finn's quick assets (cash, receivables, and market securities) by total current liabilities, as in Equation 18-4. Acid-test ratio = Quick assets/Total current liabilities (18-4) Current inventoriesare excluded from quick assetsbecause of the time required to sell these inventories, collect the receivables, and subsequentlyhave the cash on hand to reduce debt. For Engineered Industries in Figure 18-2 the calculated acid-test ratio is well below the current ratio [(1,940,000 + 950,000)/2,180,000 = 1.33]. Workingcapital, current ratio, and acid-test ratio all provide an indication of the finn's financialhealth (status). A thorough financial evaluationwouldconsiderall three, including comparisons with values from previous periods and with broad-based industry standards. THE INCOME STATEMENT The income statement or profit and loss statement summarizes the finn's revenues and expenses over a month, quarter,or year. Rather than being a snapshotlike the balance sheet, the income statement encompasses a period of business activity. The income statement is used to evaluate revenue and expenses that occur in the interval between consecutive balance sheet statements. The income statement reports the finn's net income (profit) or loss by subtracting expenses from revenues. If revenues minus expenses is positive in Equation 18-5 there has been a profit, if negative a loss has occurred. Revenues - Expenses = Net profit (Loss) (18-5) Revenue, as in Equation 18-5, serves to increase ownershipin a finn, while expenses serve to decrease ownership. Figure 18-3 is an example income statement. To aid in analyzing perfonnance, the income statement in Figure 18-3 separates operating and nonoperating activities and shows revenues and expenses for each. Operating revenues are made up of sales revenues(minus returns and allowances),whilenonopex:ating revenues come from rents and interest receipts. . . Operating expenses produce the products and services that generate the finn's revenue stream of cash flows. Typical operating expenses include cost of goods sold, selling and promotion costs, depreciation, general and administrativecosts, and lease payments. Cost of goods sold (COGS) includes the labor, materials, and indirect costs of production. Engineers designproduction systems, and theyareinvolvedin labor loading, specifying materials, and make/buy decisions. All these items affect a finn's cost of goods sold. Good engineeringdesign focuses not only on technical functionalitybut also on cost-effectiveness as the design integrates the entire production system. Also of interest to the engin~ering economist is depreciation(see Chapter 11)-which is the systematic"writing off" of a capital expense over a period of years. This noncash expenseis importantbecause it represents a decrease in value in the finn's capital assets. The operating revenues and expenses are typically shown first, so that the finn's operating income from its products and services can be calculated. Also shown on the income The Income Statement 545 Income Statement for Engineered Industries for End of Year 2005 (all amounts in $1000) Operating revenues and expenses OperatingRevenues Sales (minus)Returns and allowances Total operating revenues 28,900 -870 28,030 Operatingexpenses Cost of goods and services sold Labor Materials Indirect cost Selling and promotion 6140 4640 2280 930 1850 900 510 Depreciation . General and administrative Lease payments Total operating expense 17,250 10,780 Total operating income Nonoperating revenues and expenses Rents Interest receipts Interest payments Total Nonoperating income 400 180 -120 460 Net income before taxes Income taxes 11,240 3930 7310 Net profit (loss) for Year 2005 FIGURE 18-3 Sample income statement. statement are non-operatingexpenses such as interest payments on debt in the form of loans or bonds. . . From the data in Figure 18-3, Engineered Industries has total expenses (operating = $17,250,000 and nonoperating= $120,000) of $17,370,000. Total revenues are $28,610,000 (= $28,030,000 + $400,000 + $180,000). The net after-tax profit for year 2005 shown in Figure 18-3as $7,310,000,but it can also be calculated using Equation 18-5 as: Net profits (Loss) = Revenues - Expenses [before taxes] $11,240,000 = 28,610,000 - 17,370,000 [before taxes] and with $3,930,000taxespaid thus $7,310,000 = 11,240,000 - 3,930,000 [after taxes] - - -- - -- -- - - --- --- -- "1 546 ACCOUNTING AND ENGINEERINGECONOMY Financial Ratios Derived from Income Statement Data Interest coverage, as given in Equation 18-6, is calculated as the ratio of total income to interest payments-where total income is total revenuesminus all expenses except interest payments. Interest coverage = Total income/Interest payments (18-6) The interest coverageratio (which for industrial firms should be at least 3.0) indicates how much revenue must drop to affect the firm's ability to financeits debt. With an interest coverage ratio of 3.0, a firm's revenue would have to decrease by two-thirds (unlikely) before it became impossible to pay the interest on the debt. The larger the interest coverage ratio the better. Engineered Industries in Figure 18-3 has an interest coverage ratio of (28,610,000 - 17,250,000)/120,000 = 94.7 Another importantfinancialratio based on the income statementis the net profit ratio. This ratio (Equation 18-7)equals net profits divided by net sales revenue.Net salesrevenue equals sales minus returns and allowances. Net profit ratio = Net profit/Net sales revenue (18-7) This ratio provides insight into the cost efficiency of operations as well as a firm's ability to convert salesinto profits. For Engineered Industries in Figure 18-3, the net profit ratio is 7,310,000/28,030,000 = 0.261 = 26.1%. As with other financial measures, the net profit ratio is best evaluated by comparisons with other time periods and industry benchmarks. Linking the Balance Sheet, Income Statement, and Capital Transactions The balance sheet and the income statement are separate, but linked documents. Understanding how the two are linked together helps clarify each. Accounting describes these links as the articulation between these reports. The balance sheet shows a firm's assets, liabilities, and equity at a particular point in time, whereas the income statement summarizes revenues and expenses over a time interval. These tabulationscanbe visualizedas a snapshotat the period's beginning (a balance sheet), a video summary overthe period (the income statement),and a snapshot at the period's end (another balance sheet).The income statementand changesin the balance sheets summarize the business transactionsthat have occurred during that period. There aremany linksbetween these statementsand the cash flowsthatmake up business transactions, but for engineeringeconomic analysis the followIngare the most important. 1. Overallprofitorloss (incomestatement) and the starting and ending equity (balance sheets). 2. Acquisition of capital assets. 3. Depreciation of capital assets. The overall profit or loss during the year (shownon the income statement) is reflected in the change in retained earnings between the balance sheets at the beginning and end of - . - Traditional Cost Accounting -......-... 547 the year.To find the change in retained earnings (RE), one must also subtract any dividends distribp.tedto the owners and add the value of any new capital stock sold: REbeg+ Net income/Loss + New stock - Dividends = REend When capital equipment is purchased, the balance sheet changes,but the income statement does not. If cash is paid, then the cash asset account decrease equals the capital equipment account increase-no change in total assets. If a loan is used, then the capital equipment account increases, and so does the liability item for loans. In both cases the equity accounts and the income statement are unchanged. The depreciation of capital equipment is shown as a line on the income statement. The depreciation for that year equals the change in accumulated depreciation between the beginning and the end of the year-after subtraction of the accumulated depreciation for any asset that is sold or disposed of during that year. Example 18-1 applies these relationships to with the data in Figures 18-1 and 18-2. For simplicity,assume that Engineered Industries will not pay dividendsin 2005 and did not sell any capital equipment. It did purchase $4 million in capital equipment. What can be said about the values on the balance sheet at the end of 2005, using the linkagesjust described? SOLUTION; First, the net profit of $7,310,000 will be added to the retained earnings from the end of 2004 to find the new retained earnings at the end of 2005: RE12l3112005 = $7,310,000+ $8,745,000== $16,055,000 Second, the fixed assets shown at the end of 2005 would increase from $6,500,000 to I $10,500,000. (Note: This is a major investment of the retained earnings in the finn's physical t assets.) _ I fhird,the accumulated depreciation would increase by the $1,850,000"in depreciation shown in the 2005 income statement from the $2,350,000 shown in the 2004 balance sheet. The new accumulated depreciation on the 2005 balance sheet would be $4,200,000. Combined with the change in the amount of capital equipment, the new fixed asset total for 2005 would equal: $335,000 +$10,500,000 - $4,200,000 = $6,635,000 TIONAL COST ACCOUNTING A finn's cost-accounting system collects, analyzes, and reports operational perfonnance data (costs, utilization rates, etc.). Cost-accounting data are used to developproduct costs, to determine the mix of labor, materials, and other costs in a production setting, and to evaluate instances of outsourcing or subcontracting. .-. ___A _.+__.._ 548 -.----.- ACCOUNTING AND ENGINEERING ECONOMY Direct and Indirect Costs Costs incurred to produce a product or service are traditionally classified as either direct or indirect (overhead). Direct costs come from activities directly associated with the final product or serviceproduced.Examplesincludematerialcostsandlabor costsfor engineering design, component assembly,painting, and drilling. Some organizational activities are difficult to link to specific projects, products, or services. For example, the receiving and shipping areas of a manufacturingplant are"used by all incoming materials and all outgoing products. Materials and products differ in their weight, size, fragility,value, number of units, packaging, and so on, and the receiving and shipping costs depend on all these factors. Also, different materials arrive and different products are shipped together, so these costs are intermingled and often cannot be tied directly to each product or material. Other costs, such as the organization'smanagement,sales,and administrativeexpenses, are difficultto link directly to individualproducts or services.Theseindirect or overheadexpenses also includemachine depreciation,engineering and technical support, and customer warranties. " Indirect Cost Allocation To allocate indirect costs to different departments, products, and services, accountantsuse quantitiessuchas direct-laborhours,direct-laborcosts,materialcosts,andtotaldirectcost.~ One of these is chosen to be the burden vehicle. The total of all indirect or overheadcosts is divided by the total for the burden vehicle. For example, if direct-laborhours is the burden vehicle, then overhead will be allocated based on overhead dollar per direct-labor hour. Then each product, project, or department will absorb (or be allocated) overhead costs, based on the number of direct-labor hours each has. This is the basis for calling traditional costing systems absorption costing. For decision making, the problem is that the absorbed costs represent average, not incremental, performance. Four common ways of allocating overhead are direct-labor hours, direct-labor cost, direct-materials cost, and total direct cost. The first two differ significantly only if the cost .per hour of labor differs for different products. Example 18-2 uses direct-labor and direct-materials cost to illustrate the difference choices of burde~ vehicle. Industrial Robots does not manufacture its own motors or computer chips. Its premium product differs from its standard product in having heavier-duty motors and more computer chips for greater flexibility. - As a result, Industrial Robots manufactures a higher fractidh 6f the standarilprotluet's v"IDlTe- i itself, and it purchases a higher fraction of the premium product's value. Use the following data to allocate $850,000 ~,-. ~ 0" in overhead . ~on the basis of labor cost and materials cost. .a a ~ It 15 " . --- Traditional Cost Accounting Number of units per year Labor cost (each) Materials cost (each) Standard 750 $400 $550 549 Premium 400 $500 $900 SOLUTION First, the labor and material costs for the standard product, the premium product, and in total are calculated. Number of units per year Labor cost (each) Materials cost (each) Labor cost Materials cost Standard 750 $ 400 550 300,000 412,500 Premium 400 $ 500 900 200,000 360,000 Total $500,000 772,500 Then the allocated cost per labor dollar, $1.70, is found by dividing the $850,000 in overhead by the $500,000 in total labor cost. The allocated cost per material dollar, $1.100324, is found by dividing the $850,000 in overhead by the $772,500 in materials cost. Now, the $850,000 in allocated overhead is split between the two products using labor costs and material costs. Labor cost Overheadllabor Allocation by labor Standard $300,000 1.70 510,000 Premium $200,000 1.70 340,000 Total $500,000 Material cost Overhead/material Allocation by material 412,500 1.100324 453,884 360,000 1.100324 396,117 0 850,000 850,000 abor cost is the burden vehicle, then 60% of the $850,000 in overhead is allocated to the standard product. If material cost is the burden vehicle, then 53.4% is allocated to the standard product. In both cases, the $850,000 has been split between the two products. Using total direct costs wouldproduce another overheadallocationbetween these two values.However,for decision making about product mix and product prices, incremental overhead costs must be analyzed. All the allocation or burden vehicles are based on an average cost of overhead per unit of burden vehicle. ~ Problems with Traditional Cost Accounting ~! Allocation of indirect costs can distort product costs and the decisions based on those costs. To be accurate, the analyst must determine which indirect or overhead expenses will be changed because of an engineering project. In other words, what are the incremental cash -- -- --- ii, !!ii 11:([' ,.J " ' ';.f; I, I ;,~ 1 - , U~:~<~;~ j 550 ACCOUNTINGAND ENGINEERINGECONOMY Direct Material Overhead FIGURE 18-4 Activity-based costing versus traditional overhead allocation. (Based on an example by Kim LaScola Needy.) flows?For example, vacation and sick leave accrual may be part of overhead, but will they change if the labor content is changed? The changes in costs incurred must be estimated. Loadings, or allocations, of overheadexpenses cannot be used. . This issue has become very important because in some firms, automation has reduced direct-labor content to less than 5% of the product's cost. Yet in some of these firms, ~e basis for allocating overhead is still direct-labor hours or cost. Other firmsare shifting to activity-basedcosting (ABC),l where each activity"is linked to specific cost drivers, and the number of dollars allocated as overhead is minimized. Figure 18-4illustratesthe differencebetween activity-basedcosting andtraditionaloverhead allocations.2 " Other Problems to Watch For Centralized accounting systems have ofteribeen accused by project managers of being too slow or being "untimely."Because engineering economyis not concerned with the problem of daily project control, this is a less critical issue. However,if an organization establishes multiple files and systems so that project managers (and others) have the timely data they need, then the level of accuracy in one or all systems may be low. As a result, analy!)ts makingcostestimateswillhaveto considerotherinternaldata sources. " There are several cases in which data on equipment or inventory values may be questionable. When inventory is valued on a "last in, first out" basis, the remaining inventory may be valued too low. Similarly,land valued at its acquisition cost is likely to be significantly undervalued.Finally,capitalequipmentmay be valued at either a low or a high value, depending on allowable depreciationtechniques and companypolicy. 1Liggett, Hampton R., Jaime Trevino, and Jerome P. Lavelle, "Activity-Based Cost Management Systems, in an Advanced Manufacturing Environment." In Economic and Financial Justificq.tion of Advanced Manufacturing Technologies, 1992, H. R. Parsaei (editor), Elsevier Science. 2Tippet, Donald D., and Peter Hoekstra, "Activity-Based Costing: a Manufacturing Management Decision-Making Aid," Engineering Management Journal, Vol. 5, No.2, June 1993, American Society for Engineering Management, pp. 37-42. - - - -- __u - ._.--. _.- --. -- \,," f Problems 551 ROBLEMS .1 Why is it important for engineers and managers to understand accounting principles? Name a few ways that they can do so. Cash Net accounts and notes receivable Retailers' inventories 2 Explain the accounting function within a firm. What does this function do, and why is it important? What types of data does it provide? Prepaid expenses Accounts and notes payable (short term) Accumulated liabilities 3 Develop short definitions for the following terms: balance sheet, income statement, and fundamental accounting equation. 4 Explain the difference between short-term and longterm liabilities. :; List the two primary general accounting statements. What is each used for and how do they differ? Which ;ost important? ) acid-test ratio less than 1.0, will the company eventually go bankrupt and out of business? Explain your answer. Calculatethe equityof the GravelConstructionCompany if it has $1 million worth of assets? Gravel has $127,000 in current liabilities and $210,000in longterm liabilities. Scarmack's Paint Company has annual sales of $500,000 per year. If there is a profit of $1000 per day, 6 days per week operation, what is the total yearly b'- .,ess expense? All calculations are on a before-tax b!1.~..s. (Answer: $188,000) o Mama L's Baby Monitor Company has current assets of $5 million and current liabilities of $2 million. Give the company's working capital and current ratio. (Answer: $3 million, 2.50) 1 Laila's Surveying Inc. had revenues of $100,000 in 2004. Expenses totaled $60,000. What was her net profit (or loss)? ~ From the following data, taken from the balance sheet of Petey's Widget Factory, determine the working capital, current ratio, and acid-test ratio. 87,000 18-13 The general ledger of the Fly-Buy-Nite (FBN) Engineering Company contained the following account balances (partial listing) at the end of June, 2004. Construct an income statement. What is the net income before taxes and the net profit (or loss) after taxes? FBN has a tax rate of 27%. What is the advantage of comparing financial statements across periods or against industry benchmarks rather than in looking at statements associated with a single date or period? If a firm has a current ratio less than 2.0 and an $ 90,000 175,000 210,000 6,000 322,000 Administrative expenses Subcontracted services Development expenses Interest expense Sales revenue Selling expenses Amount (thousands) $ 2,750 18,000 900 200 30,000 4,500 18-14 For Gee-Whiz Devices calculate the following: working capital, current ratio, and acid-test ratio. Gee-Whiz Devices Balance Sheet Data Cash Market securities Net accounts and notes receivable Retailers' inventories Prepaid expenses Accounts and notes payable (short term) Accumulated liabilities to date $100,000 45,000 150,000 200,000 8,000 315,000 90,000 (Answer: $90,000; 1.22; 0.73) 18-15 For Evergreen Environmental Engineering (EEE), determine the working capital, current ratio, and acidtest ratio. Evaluate the company's economic situation with respect to its ability to payoff debt. . ~ 552 ACCOUNTING AND ENGINEERINGECONOMY EEEBalance Sheet Data (thousands) Cash Securities Accounts receivable Retailers' inventories Prepaid expenses Accounts payable Accumulatedliabilities 18-19 For J &W Graphics Supply, compute the current ratio. Is this a financially healthy company? Explain. $110,000 40,000 160,000 250,000 3,000 351,000 89,000 J&W Graphics Supply Balance Sheet Data (thousands) Assets Cash Inventories Accountsreceivable Bad debt provision Liabilities Notes payable Accrued expenses Accounts payable 18-16 For Magdalen Industries, compute the net income before taxes and net profit (or loss). Taxes for the year were $1 million. Calculate the interest coverage and net profit ratio. Is the interest coverage acceptable? Explain why or why not. Magdalen Industries Balance Sheet Data (millions) Revenues Total operating revenue (including sales of $48 million) Total nonoperating revenue Expenses. Total operating expenses Total nonoperating expenses (interest payments) $51 Sutton Manufacturing (thousands) 70 7 Assets Current assets Cash $870 Accounts 450 receivable 1200 Inventory Prepaid expenses 50 10 Other expenses Turbo Start Balance Sheet Data (thousands) Accounts payable Selling expense Sales revenue Owner's equity Income taxes Cost of goods sold Accounts receivable $ 1,000 5,000 50,000 4,500 2,000 30,000 15,000 (Answer: $13,000,000) 18-18 Thrbo Start (Problem 18-17) has current assets totaling $1.5 million (this includes $500,000 in current inventory) and current liabilities totaling $50,000. Find the current ratio and acid-test ratio. Are the ratios at 500 125 1050 18-20 For Sutton Manufacturing, determine the current ratio and the acid-test ratio. Are these values acceptable? Why or why not? 35 18-17 Find the net income ofThrbo Start given the following data. $1740 900 2500 -75 Balance Sheet Data Total current assets 2670 Net fixed assets Land Plant and equipment Notes receivable Intangibles Other assets Total assets 1200 2800 200 20 - 100 6900 Liabilities Current liabilities $500 Notes payable Accountspayable 600 Accruals Taxespayable Currentportion long-termdebt Total current liabilities Long-termdebt Officerdebt (subordinated) Total liabilities 200 30 100 1430 2000 200 - 3630 Common stock Capital surplus Retained earnings Total net worth 1670 400 1200 3270 Totalliabilities 6900 - and net Worth J desirable levels? Explain. . - - ... - --- -- Problems J-21 For Andrew's Electronic Instruments, calculate the interest coverage and net profit ratio. Is Andrew's business healthy? ' Income Statement for Andrew's Electronics for End of Year2004 (thousands) Revenues Operaing revenues Sales (minus) Returns Total operating revenues Nonoperating revenues Interest receipts Stock revenues Total nonoperating revenues Total revenues, R Expenses Operating expenses Cost of goods and services sold Labor Materials Indirect cost Selling and promotion Depreciation General and administrative Lease payments Total operating expenses Nonoperating expenses Interest payments Total nonoperating expenses Total expenses, E Net income before Taxes, R - E Incomes taXes Net profit (Loss) for the year 2004 553 Item Model S Model M Model G Direct-materialcosts $3,800,000 $1,530,000 $2,105,000 Direct-laborcosts 600,000 380,000 420,000 Direct-laborhours 64,000 20,000 32,000 (Answer: $132; $93; $84) $395 -15 380 50. 25 75 455 200 34 68 20 30 10 10 372 22 22 394 61 30 31 22 LeGaroutte Industries makes industrial pipe manufacturing equipment. Use direct-labor hours as the burden vehicle, and compute the total cost per unit for eachmodelgiveninthe table.Totalmanufacturing indirect costs are $15,892,000,and there are 100,000 units manufacturedper year for Model S, 50,000 for Model M, and 82,250for Model G. 18-23 Par Golf Equipment Company produces two types of golf bag: the standard and deluxe models. The total indirect cost to be allocated to the two bags is $35,000. Determine the net revenue that Par Golf can expect from the sale of each bag. (a) Use direct-labor cost to allocate indirect costs. (b) Use direct-materials cost to allocate indirect costs. Data Item Direct-labor cost Direct-material cost Selling price Units produced Standard $50,000 35,000 60 1800 Deluxe $65,000 47,500 95 1400 18-24 RLW-II Enterprises estimated that indirect manufacturing costs for the year would be $60 million and that 12,000 machine-hours would be used. (a) Compute the predetermined indirect cost application rate using machine hours as the burden vehicle. (b) Determine the total cost of production for a product with direct material costs of $1 million, directlabor costs of $600,000, and 200 machine-hours. (Answers: $5000, $2.6 million) 18-25 Categorize each of the following costs as direct or indirect. Assume that a traditional costing system is in place. Machine run costs Machine depreciation Material handling costs Cost of materials Overtime expenses Machine operator wages Utility costs Support (administrative) staff salaries Cost to market the product Cost of storage Insurance costs Cost of product sales force Engineering drawings Machine labor Cost of tooling and fixtures - - --- - Introduction to Spreadsheets Computerizedspreadsheets are availablenearly everywhere,and they can be easily applied to economic analysis. In fact, spreadsheets were originally developed to analyze financial data, and they are often credited with initiating the explosivegrowth in demand for desktop computing. A spreadsheet is a two-dimensional table, whose cells can contain numerical values, labels, or formulas. The software automatically updates the table when an entry is changed, and there are powerfultools for copyingformulas,creatinggraphs, and formatting results. THE ELEMENTSOF A SPREADSHEET A spreadsheet is a two-dimensional table that labels the columns in alphabetical order A to Z, AA to AZ, BA to BZ, up to IV (256 columns). The rows are numbered from 1 to 65,536. Thus a cell of the spreadsheet is specifiedby its column letter and row number.For example, A3 is the third row in column A and AA6 is the sixth row in the twenty-seventh . column. Each cell can contain a label, a numerical value, or a formula. A label is any cell where the contents should be treate~ as text. Arithmetic cannot be performed on labels. Labels are used for variable names, row and column headings, and explanatory notes. In Excel any cell which contains more than a simple number, such as 3.14159, is treated as a label, unless it begins with an =, which is the signal for a formula. Thus 2*3 and Bl+B2 are labels. Meaningful labels can be wider than a normal column. One solution is to allowthose cells to "wrap" text, which is one of the "alignment" options. The table heading row (row 8) in Example A-I has turned this on by selecting row 8, right clicking on the row, and selecting wrap text under the alignment tab. A numerical value is any number.Acceptable formats for entry or display include percentages, currency,accounting, scientific,fractions, date, and time. In addition the number of decimal digits, the display of $ symbols, and commas for "thousands" separators can be adjusted. The format for cells can be changed by selecting a cell, a block of cells, a row, a column, or the entire spreadsheet. Then right click on the selected area, and a menu that includes "format cells" will appear. Then number formats, alignment, borders, fonts, and patterns can be selected. 554 ... - . .... Copy Command 555 Formulasmust begin with an =, such as =3*4A2or =BI+B2. They can include many funptions-financial, statistical,trigonometric, etc. (and others can be defined by the user). The formula for the "current" cell is displayedin theformula bar at the top of the spreadsheet. The value resulting from the formula is displayed in the cell in the spreadsheet. Often the printed out spreadsheet will be part of a report or a homework assignment and the formulas must be explained. Here is an easy way to place a copy of the formula in an adjacent or nearby cell. (1) Convert the cell with the formula to a label by inserting a space before the = sign. (2) Copy that label to an adjacent cell by using cut and paste. Do not drag the cell to copy it, as any formula ending with a number (even an address like B4) will have the number automaticallyincremented.(3) Convertthe original formula back into a formula by deleting the space. EFINING VARIABLESIN A DATA BLOCK The cell AI, top left comer, is the HOME cell for a spreadsheet. Thus, the top left area is where the data block should be placed. This data block should have every variable in .the spreadsheet with an adjacent label for each. This data block supports a basic principle of good spreadsheet modeling, which is to use variables in your models. The data block in Example A-I contains entered data-the loan amount (AI), the number of payments (A2), and the interest rate (A3), and computed data-the payment (A4). Then instead of using the loan amount of $5000 in a formula, the cell reference Al is used. Even if a value is only referenced once, it is better to include it in the data block By using one location to define each variable,you can change any value at one place in the spreadsheet and have the entire spreadsheetinstantly recomputed. Even for simple homeworkproblems you should use a data block. 1. 2. 3. 4. Youmay be able to use it for another problem. Solutions to simple problems may grow into solutions for complex problems. Good habits, like using data blocks, are easy t~ maintain once they are established. It makes the assumptions clear if you've estimated a value or for grading. In the real world, data blocks are even more important. Most problems are solved more than once, as more and more accurate values are estimated. Often the spreadsheetis revised to add other variables,time periods, locations,etc. Without data blocks, it is hard to change a spreadsheet and the likelihood of missing a required change skyrockets. If you want your formulas to be easier to read, you can name your variables. Note: In current versions of Excel, the cell's location or name is displayed at the left of the formula bar. Variablenames can be entered here. They will then automatically be applied if cell addresses are entered by the point and click method. If cell addresses are entered as A2, then A2 is what is displayed. To change a displayed A2, to the name of the cell (LoanAmount), the process is to click on insert, click on name, click on apply, and then select the names to be applied. tPYCOMMAND The copy commandandrelative/absoluteaddressingmake spreadsheetmodels easy to build. If the range of cells to be copied contains only labels, numbers, and functions, then the copy - -- - - 556 INTRODUCTION TO SPREADSHEETS command is easy to use and understand. For example, the formula =EXP(1.9) would be copied unchanged to a new location. However,cell addresses are usually part of the range being copied, and their absolute and relative addresses are treated differently. An absolute address is denoted by adding $ signs before the column and/or row. . For examplein FigureA-la, $A$4 is the absoluteaddressfor the interestrate.Whenan absolute address is copied, the column and/or row that is fixed is copied unchanged.Thus $A$4 is completely fixed, $A4 fixes the column, and A$4 fixes the row. One common Use for absolute addresses is any data block entry, such as the interest rate. When entering or editing a formula,changingbetween A4, $A$4, A$4, $A4, and A4 is most easily done using the F4 key, which scrolls an address through the choices. In contrast a relative addressis best interpreted as directions from one cell to another. For example in Figure A-I a, the balance due in year t equals the balance due in year t - 1 minus the principal payment in year t. Specifically for the balance due in year 1, DlO contains =D9-ClO. From cell DlO, cell D9 is one row up and ClO is one column to the left, so the formula is really (contents of 1 up) minus (contents of 1 to the left). When a cell containing a relative address is copied to a new location, it is these directions that are copied to determine any new relative addresses. So if cell DlO is copied to cell F14, the formula is =F13-EI4. Thus to calculate a loan repayment schedule, as in Figure A-I, the row of formulas is created and then copied for the remaining years. Table 3-1 shows four repayment schedules for a loan of $5000 to be repaid over 5 years at an interest rate of 8%. Use a spreadsheet to calculate the amortization schedule for the constant principal payment option. SOLUTION The first step is to enter the loan amount, number of periods, and interest rate into a data block in the top left part of the spreadsheet. The next step is to calculate the constant principal payment amount, which was given as $1252.28 in Table 3-1. The factor approach to finding this value is given in Chapter 3 and the spreadsheet function is explained in Chapter 4; The next step is to identify the columns for the amortizatiqu schedule. These are the year, interest owed, principal payment, and balance due. Because some of these labels arewider than a normal column, the cells are formatted so that the textwraps (rowheight increases automatically). The initial balance is shown in the year 0 row. Next, the formulas for the first year are written, as shown m.figure A-I a. The interest owed (cellBlO) equals the interest rate ($A$4) times the balance due for year 0 (D9). The principal payment (cell ClO) equals the annual payment ($A$6) minus the interest owed and paid (B1O). Finally, the balance due (cell DlO) equals the balance due for the previous year (D9) minus the ., ~ .. )?rin<:ipalpaym€?nt.(C~O):T~e re~ults are shown in Figure A-la. - '= Now ~ells AlO toMDlOare selecred for yefu-1. Brdra~ginglloWn ... .. . d"nthe right comer'ofD10, the entire row can be copied for years 2 through 5. Note if cut andpaste is used,thenitis necessary toc()I11pletethe year column separately (dragging increme,ntsthe year, b1Jtcuttinga,ndpasting does not). The results are shown in Figure A-Ib. . ---- --j. - - . . .u . ...... Copy Command , 1 2 3 4 5 6 7 8 9 10 11 A Entered Data 5000 5 8% B C D 557 E LoanAmount Numberof Payments InterestRate ComputerData $1,252.28 Loan Payment Year 0 1 Interest Owed 400.00 Principal Payment Balance Due 852.28 5000.00 4147.72 =D9-ClO / 12 $A$4*D9 =$A$6-BlO (a) '~":::: A B C D 8 Year Interest Owed Principal Payment Balance Due 9 0 10 1 400.00 852.28 4147.72 11 2 331.82 920.46 3227.25 12 3 258.18 994.10 2233.15 13 4 178.65 1073.63 1159.52 , 14 15 5 92.76 1159.52 0.00 , 16 E 5000.00 =Dl3-CI4 / =$A$6-BI4 = $A$4*D 13 (b) FIGURE A-I (a) Year 1 amortization schedule. (b) Completed amortization schedule. This appendix has introduced the basics of spreadsheets.Chapter 2 uses spreadsheets and simple bar charts to draw cash flow diagrams. Chapters 4-15 each have spreadsheet sections. These are designed to develop spreadsheet modeling skills and to reinforce your understanding of engineering economy. As current spreadsheetpackages are built around using mice to click on cells and items in charts, there is usually an intuitive connection between what you would like to do and how to do it. It seems the best way to learn how to use the spreadsheetpackage is to simply play around with it. In addition, as you look at the menu choices, you will find new commands that you hadn't thought of but find useful. t I I Compound Interest Tables Values of Interest Factors When N Equals Infinity Single Payment: (F/P, i, 00) = 00 (P/F, i, 00) = 0 Arithmetic Gradient Series: (A/G, i, 00) = l/i (PIG, i, 00) = l/P I ----- Uniform Payment Series: (A/F, i, 00) = 0 (A/P, i, 00) = i (F/A, i, 00) = 00 (PIA, i, 00) = l/i :1 ;, ._---.._----jI : I COMPOUND INTERESTTABLES Compound Interest Factors 0/0 Single Payment 561 1/20/0 Arithmetic Gradient Uniform Payment Series Compound Amount Factor Find F Given P FIP Present Worth Factor Find P Given F PIF Sinking Fund Factor Find A Given F AIF Capital Recovery Factor Find A Given P AlP Compound Amount Factor Find F Given A FIA Present Worth Factor Find P Given A PIA Gradient Uniform Series Find A Given G AIG Gradient Present Worth Find P Given G PIG n 1.005 1.010 1.015 1.020 1.025 1.030 1.036 1.041 1.046 1.051 1.056 1.062 1.067 1.072 1.078 1.083 1.088 1.094 1.099 1.105 1.110 1.116 1.122 1.127 1.133 1.138 1.144 1.150 1.156 ,.161 1.197 1.221 1.270 1.283 1.296 1.349 1.418 1.432 1.490 1.520 1.567 1.614 1.647 1.680 1.819 3.310 6.023 10.957 .9950 .9901 .9851 .9802 .9754 .9705 .9657 .9609 .9561 .9513 .9466 .9419 .9372 .9326 .9279 .9233 .9187 .9141 .9096 .9051 .9006 .8961 .8916 .8872 .8828 .8784 .8740 .8697 .8653 .8610 .8356 .8191 .7871 .7793 .7716 .7414 .7053 .6983 .6710 .6577 .6383 .6195 .6073 .5953 .5496 .3021 .1660 .0913 1.0000 .4988 .3317 .2481 .1980 .1646 .1407 .1228 .1089 .0978 .0887 .0811 .0746 .0691 .0644 .0602 .0565 .0532 .0503 .0477 .0453 .0431 .0411 .0393 .0377 .0361 .0347 .0334 .0321 .0310 .0254 .0226 .0185 .0177 .0169 .0143 .0120 .0116 .0102 .00961 .00883 .00814 .00773 .00735 .00610 .00216 .00100 .00050 1.0050 .5038 .3367 .2531 .2030 1.000 2.005 3.015 4.030 5.050 6.076 7.106 8.141 9.182 10.228 11.279 12.336 13.397 14.464 15.537 16.614 17.697 18.786 19.880 20.979 22.084 23.194 24.310 25.432 26.559 27.692 28.830 29.975 31.124 32.280 39.336 44.159 54.098 56.645 59.218 69.770 83.566 86.409 98.068 104.074 113.311 122.829 129.334 135.970 163.880 462.041 1004.5 1 991.5 0.995 1.985 2.970 3.951 4.926 5.896 6.862 7.823 8.779 9.730 10.677 11.619 12.556 13.489 14.417 15.340 16.259 17.173 18.082 18.987 19.888 20.784 21.676 22.563 23.446 24.324 25.198 26.068 26.933 27.794 32.871 36.172 42.580 44.143 45.690 51.726 58.939 60.340 65.802 68.453 72.331 76.095 78.543 80.942 90.074 139.581 166.792 181.748 0 0.499 0.996 1.494 1.990 2.486 2.980 3.474 3.967 4.459 4.950 5.441 5.931 6.419 6.907 7.394 7.880 8.366 8.850 9.334 9.817 10.300 10.781 11.261 11.741 12.220 12.698 13.175 13.651 0 0.991 2.959 5.903 9.803 14.660 20.448 27.178 34.825 43.389 52.855 63.218 74.465 86.590 99.574 113.427 128.125 143.668 160.037 177.237 195.245 214.070 233.680 254.088 275.273 297.233 319.955 343.439 367.672 392.640 557.564 681.341 959.928 1 035.70 1 113.82 1 448.65 1 913.65 2012.35 2424.65 2 640.67 2 976.08 3324.19 3 562.80 3 806.29 4 823.52 13415.56 21 403.32 27 588.37 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 60 70 72 80 84 90 96 100 104 120 .240 360 480 .1696 . .1457 .1278 .1139 .1028 .0937 .0861 .0796 .0741 .0694 .0652 .0615 .0582 .0553 .0527 .0503 .0481 .0461 .0443 .0427 .0411 .0397 .0384 .0371 .0360 .0304 .0276 .0235 .0227 .0219 .0193 .0170 .0166 .0152 .0146 .0138 .0131 .0127 .0124 .0111 .00716 .00600 .00550 --- 14.127. 16.962 18.836 22.544 23.463 24.378 28.007 32.468 33.351 36.848 38.576 41.145 43.685 45.361 47.025 53.551 96.113 128.324 151.795 . '. ..........- 562 COMPOUND INTERESTTABLES Compound Interest Factors 3/4% n Present Worth Factor Find P Given F PIF Sinking Fund Factor Find A Given F AIF Capital Recovery Factor Find A Given P AlP Compound Amount Factor Find F Given A FIA Present Worth Factor Find P Given A PIA Gradient Uniform Series Find A Given G AIG 1.0000 .4981 .3308 .2472 .1970 .1636 .1397 .1218 .1078 .0967 .0876 .0800 .0735 .0680 .0632 .0591 .0554 .0521 .0492 .0465 .0441 .0420 .0400 .0382 .0365 .0350 .0336 .0322 .0310 .0298 .0243 .0215 .0174 .0166 .0158 .0133 .0109 .0105 .00917 .00859 .00782 .00715 .00675 .00638 1.0075 .5056 .3383 .2547 .2045 .1711 .1472 .1293 .1153 .1042 .0951 .0875 .0810 .0755 .0707 .0666 .0629 .0596 .0567 .0540 .0516 .0495 .0475 .0457 .0440 .0425 .0411 .0397 .0385 .0373 .0318 .0290 .0249 .0241 .0233 .0208 .0184 .0180 .0167 .0161 .0153 .0147 .0143 .0139 1.000 2.008 3.023 4.045 5.076 6.114 7.160 8.213 9.275 10.344 11.422 12.508 13.602 14.704 15.814 16.932 18.059 19.195 20.339 21.491 22.653 23.823 25.001 26.189 27.385 28.591 29.805 31.029 32.261 33.503 41.153 46.447 57.521 60.395 63.312 75.425 91.621 95.008 109.074 116.428 127.881 139.858 148.147 156.687 0.993 1.978 2.956 3.926 4.889 5.846 6.795 7.737 8.672 9.600 10.521 11.435 12.342 13.243 14.137 15.024 15.905 16.779 17.647 18.508 19.363 20.211 21.053 21.889 22.719 23.542 24.360 25.171 25.976 26.775 31.447 34.447 40.185 41.567 42.928 48.174 54.305 55.477 59.995 62.154 65.275 68.259 70.175 72.035 0 0.499 0.996 1.492 1.986 2.479 2.971 3.462 3.951 4.440 4.927 5.412 5.897 6.380 6.862 7.343 7.822 8.300 8.777 9.253 9.727 10.201 10.673 11.143 11.613 12.081 12.548 13.014 13.479 13.942 16.696 18.507 22.070 22.949 23.822 27.268 31.465 32.289 35.540 37.137 39.496 41.812 43.332 44.834 0 0.987 2.943 5.857 9.712 14.494 20.187 26.785 34.265 42.619 51.831 61.889 72.779 84.491 97.005 110.318 124.410 139.273 154.891 171.254 188.352 206.170 224.695 243.924 263.834 284.421 305.672 327.576 350.122 373.302 525.038 37.519 886.899 953.911 1 022.64 1313.59 1 708.68 1 791.33 2 132.23 2 308.22 2 578.09 2 854.04 3 040.85 3229.60 193.517 78.942 50.653 3 998.68 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 60 70 72 80 84 90 96 100 104 120 667.901 1 830.8 4681.5 111.145 124.282 129.641 85.422 107.115 119.662 9 494.26 13 312.50 15513.16 240 360 480 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 60 70 72 80 84 90 96 tOO t04 1.309 1.348 1.431 1.453 1.475 1.566 1.687 1.713 1.818 1.873 1.959 2.049 2.111 2.175 .9926 .9852 .9778 .9706 .9633 .9562 .9490 .9420 .9350 .9280 .9211 .9142 .9074 .9007 .8940 .8873 .8807 .8742 .8676 .8612 .8548 .8484 .8421 .8358 .8296 .8234 .8173 .8112 .8052 .7992 .7641 .7416 .6986 .6882 .6780 .6387 .5927 .5839 .5500 .5338 .5104 .4881 .4737 .4597 l20 2.451 .4079 .00517 .0127 O 160 180 6.009 14.731 36.111 .1664 .0679 .0277 .00150 .00055 .00021 .00900 .00805 .00771 1.008 1.015 1.023 1.030 1.038 1.046 1.054 1.062 1.070 1.078 1.086 1.094 1.102 1.110 1.119 1.127 1.135 1.144 1.153 1.161 1.170 1.179 1.188 1.196 1.205 1.214 1.224 1.233 1.242 1.251 . Arithmetic Gradient Uniform Payment Series Single Payment Compound Amount Factor Find F Given P FIP 3/4% - - - Gradient Present Worth Find P Given G PIG n = ,c':<f. ... COMPOUND INTEREST TABLES I 0 1% Compound Interest Factors Single - Payment Present Compound Amount Worth Factor Factor Find F Find P Given P Given F FIP PIF Compound Amount Factor Find F Given A FIA Present Worth Factor Find P Given A PIA Gradient Uniform Series Find A Given G AIG Gradient Present Worth Find P Given G PIG n 1.000 2.010 3.030 4.060 5.101 6.152 7.214 8.286 9.369 10.462 11.567 12.682 13.809 14.947 16.097 17.258 18.430 19.615 20.811 22.019 23.239 24.472 25.716 26.973 28.243 29.526 30.821 32.129 33.450 34.785 0.990 1.970 2.941 3.902 4.853 5.795 6.728 7.652 8.566 9.471 10.368 11.255 12.134 13.004 13.865 14.718 15.562 16.398 17.226 18.046 18.857 19.660 20.456. 21.243 22.023 22.795 23.560 24.316 25.066 25.808 0 0.498 0.993 1.488 1.980 2.471 2.960 3.448 3.934 4.418 4.900 5.381 5.861 6.338 6.814 7.289 7.761 8.232 8.702 9.169 9.635 10.100 10.563 11.024 11.483 11.941 12.397 12.852 13.304 13.756 . 0 0.980 2.921 5.804 9.610 14.320 19.917 26.381 33.695 41.843 50.806 60.568 71.112 82.422 94.481 107.273 120.783 134.995 149.895 165.465 181.694 198.565 216.065 234.179 252.892 272.195 292.069 312.504 333.486 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 29 355.001 30 43.077 48.886 61.223 64.463 67.769 30.107 32.835 37.974 39.196 40.394 16.428 18.178 21.598 22.436 23.269 494.620 596.854 820.144 879.417 939.916 36 40 48 50 52 .0222 .0199 .0196 .0182 .0177 81.670 100.676 104.710 121.671 130.672 44.955 50.168 26.533 30.470 31.239 34.249 35.717 1 192.80 1 528.64 1 597.86 1879.87 2023.31 .00690 .00625 .00587 .00551 .00435 .0169 .0163 .0159 .0155 .0143 144.863 159.927 170.481 181.464 230.039 59.161 61.528 63.029 64.471 69.701 2 240.56 2459.42 2605.77 2752.17 3334.11 .00101 .00029 .00008 .0110 .0103 .0101 989.254 3495.0 11 764.8 90.819 97.218 99.157 37.872 39.973 41.343 42.688 47.835 75.739 89.699 95.920 60 70 72 80 84 90 96 100 104 120 6 878.59 8 720.43 9511.15 240 360 480 Capital Recovery Factor Find A Given P AlP .9901 .9803 .9706 .9610 .9515 .9420 .9327 .9235 .9143 .9053 .8963 .8874 .8787 .8700 .8613 .8528 .8444 .8360 .8277 .8195 .8114 .8034 .7954 .7876 .7798 .7720 .7644 .7568 .7493 .7419 1.0000 .4975 .3300 .2463 .1960 .1625 .1386 .1207 .1067 .0956 .0865 .0788 .0724 .0669 .0621 .0579 .0543 .0510 .0481 .0454 .0430 .0409 .0389 .0371 .0354 .0339 .0324 .0311 .0299 .0287 1.0100 .5075 .3400 .2563 .2060 .1725 .1486 .1307 .1167 .1056 .0965 .0888 .0824 .0769 .0721 .0679 .0643 .0610 .0581 .0554 .0530 .0509 .0489 .0471 .0454 .0439 .0424 .0411 .0399 .0387 1.431 1.489 1.612 1.645 1.678 .6989 .6717 .6203 .6080 .5961 .0332 .0305 .0263 .0255 .0248 1.817 2.007 2.047 2.217 2.307 .5504 .4983 .4885 .4511 .4335 .0232 .0205 .0163 .0155 .0148 .0122 .00993 .00955 .00822 .00765 2.449 2.599 2.705 2.815 3.300 .4084 .3847 .3697 .3553 .3030 .0918 .0278 .00843 10.893 35.950 118.648 --- Arithmetic Gradient Uniform Payment Series Sinking Fund Factor Find A Given F AIF 1.010 1.020 1.030 1.041 1.051 1.062 1.072 1.083 1.094 1.105 1.116 1.127 1.138 1.149 1.161 1.173 1.184 1.196 1.208 1.220 1.232 1.245 1.257 1.270 1.282 1.295 1.308 1.321 1.335 1.348 563 _ _ 51.150 54.888 56.648 . - .4 COMPOUND INTERESTTABLES 1/4% CompoundInterestFactors Present Worth Factor Find P Given F PIF Sinking Fund Factor Find A Given F AIF Capital Recovery Factor Find A Given P AlP Compound Amount Factor Find F Given A FIA Present Worth Factor Find P Given A PIA Gradient Uniform Series Find A Given G AIG 1.013 1.025 1.038 1.051 1.064 1.077 1.091 1.104 1.118 1.132 1.146 1.161 1.175 1.190 .9877 .9755 .9634 .9515 .9398 .9282 .9167 .9054 .8942 .8832 .8723 .8615 .8509 .8404 1.0000 .4969 .3292 .2454 .1951 .1615 .1376 .1196 .1057 .0945 .0854 .0778 .0713 .0658 1.0125 .5094 .3417 .2579 .2076 .1740 .1501 .1321 .1182 .1070 .0979 .0903 .0838 .0783 1.000 2.013 3.038 4.076 5.127 6.191 7.268 8.359 9.463 10.582 11.714 12.860 14.021 15.196 0.988 1.963 2.927 3.878 4.818 5.746 6.663 7.568 8.462 9.346 10.218 11.079 11.930 12.771 0 0.497 0.992 1.485 1.976 2.464 2.951 3.435 3.918 4.398 4.876 5.352 5.827 6.299 0 0.976 2.904 5.759 9.518 14.160 19.660 25.998 33.152 41.101 49.825 59.302 69.513 80.438 1.205 .8300 .0610 .0735 16.386 13.601 6.769 92.058 1.220 1.235 1.251 1.266 1.282 15 .8197 .8096 .7996 .7898 .7800 .0568 .0532 .0499 .0470 .0443 .0693 .0657 .0624 .0595 .0568 17.591 18.811 20.046 21.297 22.563 14.420 15.230 16.030 16.849 17.599 7.237 7.702 8.166 8.628 9.088 104.355 117.309 130.903 145.119 159.940 16 17 18 19 20 I 1.298 1.314 1.331 1.347 1.364 .7704 .7609 .7515 .7422 .7330 .0419 .0398 .0378 .0360 .0343 .0544 .0523 .0503 .0485 .0468 23.845 25.143 26.458 27.788 29.136 18.370 19.131 19.882 20.624 21.357 9.545 10.001 10.455 10.906 11.355 175.348 191.327 207.859 224.930 242.523 21 22 23 24 25 ) 7 i ) ) ) ) J ) t 1.381 1.399 1.416 1.434 1.452 .7240 .7150 .7062 .6975 .6889 .0453 .0439 .0425 .0413 .0402 11.803 12.248 12.691 13.133 13.572 260.623 279.215 298.284 317.814 337.792 26 27 28 29 30 .6394 .6084 .5509 .5373 .5242 .0347 .0319 .0278 .0270 .0263 30.500 31.881 32.280 34.696 36.129 45.116 51.490 65.229 68.882 72.628 22.081 22.796 23.503 24.200 24.889 1.564 1.644 1.845 1.861 1.908 .0328 .0314 .0300 .0288 .0277 .0222 .0194 .0153 .0145 .0138 28.847 31.327 35.932 37.013 38.068 16.164 17.852 21.130 21.930 22.722 466.297 559.247 759.248 811.692 864.960 36 40 48 50 52 2.107 2.386 2.446 2.701 2.839 .4746 .4191 .4088 .3702 .3522 .0113 .00902 .00864 .0238 .0215 .0211 88.575 110.873 115.675 136.120 42.035 46.470 47.293 50.387 25.809 29.492 30.205 32.983 1 084.86 1 370.47 1 428.48 1 661.89 60 70 72 80 3.059 3.296 .3269 .3034 .00680 .00607 .00545 .0193 .0186 .0179 147.130 164.706 183.643 51.822 53.846 55.725 34.326 36.286 38.180 1778.86 1953.85 2 127.55 84 90 96 3.463 3.640 4.440 19.716 87.543 388.713 .2887 .2747 .2252 .0507 .0114 .00257 .00507 .00474 .00363 .00067 .00014 .00003 56.901 58.021 61.983 75.942 79.086 79.794 39.406 40.604 45.119 67.177 75.840 78.762 2 242.26 2355.90 2 796.59 5 101.55 5997.91 6284.74 100 104 120 240 360 480 Compound Amount Factor Find F Given P FIP a ) 7 ) ) L ! a ) 7 ) ) Arithmetic Gradient Uniform Payment Series Single Payment L ! 11/4% .00735 .---...& .. .0176 .0172 .0161 .0132 .0126 .0125 197.074 211.190 275.220 1497.3 6 923.4 31017.1 Gradient Present Worth Find P Given G PIG n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 , I , i , I I I 1 --, J J ...... . - COMPOUND INTERESTTABLES ,2 0;;0 Compound Interest Factors 11/20/0 Arithmetic Gradient Uniform Payment Series Single Payment Compound Amount Factor Find F Given P FIP Present Worth Factor Find P Given F PIF Sinking Fund Factor Find A Given F AIF Capital Recovery Factor Find A Given P AlP Compound Amount Factor Find F Given A FIA Present Worth Factor Find P Given A PIA Gradient Uniform Series Find A Given G AIG Gradient Present Worth Find P Given G PIG 1.015 1.030 1.046 1.061 1.077 1.093 1.110 1.126 1.143 1.161 1.178 1.196 1.214 1.232 1.250 1.269 1.288 1.307 1.327 1.347 1.367 1.388 1.408 1.430 1.451 1.473 1.495 1.517 1.540 1.563 1.709 1.814 2.043 2.105 2.169 2.443 2.835 2.921 3.291 3.493 3.819 4.176 4.432 4.704 5.969 35.632 212.700 1 269.7 .9852 .9707 .9563 .9422 .9283 .9145 .9010 .8877 .8746 .8617 .8489 .8364 .8240 .8118 .7999 .7880 .7764 .7649 .7536 .7425 .7315 .7207 .7100 .6995 .6892 .6790 .6690 .6591 .6494 .6398 .5851 .5513 .4894 .4750 .4611 .4093 .3527 .3423 .3039 .2863 .2619 .2395 .2256 .2126 .1675 .0281 .00470 .00079 1.0000 .4963 .3284 .2444 .1941 .1605 .1366 .1l86 .1046 .0934 .0843 .0767 .0702 .0647 .0599 .0558 .0521 .0488 .0459 .0432 .0409 .0387 .0367 .0349 .0333 .0317 .0303 .0290 .0278 .0266 .0212 .0184 .0144 .0136 .0128 .0104 .00817 .00781 .00655 .00602 .00532 .00472 .00437 .00405 .00302 .00043 .00007 .00001 1.0150 .5113 .3434 .2594 .2091 .1755. .1516 .1336 .1l96 .1084 .0993 .0917 .0852 .0797 .0749 .0708 .0671 .0638 .0609 .0582 .0559 .0537 .0517 .0499 .0483 .0467 .0453 .0440 .0428 .0416 .0362 .0334 .0294 .0286 .0278 .0254 .0232 .0228 .0215 .0210 .0203 .0197 .0194 .0190 .0180 .0154 .0151 .0150 1.000 2.015 3.045 4.091 5.152 6.230 7.323 8.433 9.559 10.703 11.863 13.041 14.237 15.450 16.682 17.932 19.201 20.489 21.797 23.124 24.470 25.837 27.225 28.633 30.063 31.514 32.987 34.481 35.999 37.539 47.276 54.268 69.565 73.682 77.925 96.214 122.363 128.076 152.710 166.172 187.929 211.719 228.802 246.932 331.286 2 308.8 14 113.3 84 577.8 0.985 1.956 2.912 3.854 4.783 5.697 6.598 7.486 8.360 9.222 10.071 10.907 11.731 12.543 13.343 14.131 14.908 15.673 16.426 17.169 17.900 18.621 19.331 20.030 20.720 21.399 22.068 22.727 23.376 24.016 27.661 29.916 34.042 35.000 35.929 39.380 43.155 43.845 46.407 47.579 49.210 50.702 51.625 52.494 55.498 64.796 66.353 66.614 0 0.496 0.990 1.481 1.970 2.456 2.940 3.422 3.901 4.377 4.851 5.322 5.791 6.258 6.722 7.184 7.643 8.100 8.554 9.005 9.455 9.902 10.346 10.788 11.227 11.664 12.099 12.531 12.961 13.388 15.901 17.528 20.666 21.428 22.179 25.093 28.529 29.189 31.742 32.967 34.740 36.438 37.529 38.589 42.518 59.737 64.966 66.288 0 0.970 2.883 5.709 9.422 13.994 19.400 25.614 32.610 40.365 48.855 58.054 67.943 78.496 89.694 101.514 113.937 126.940 140.505 154.611 169.241 184.375 199.996 216.085 232.626 249.601 266.995 284.790 302.972 321.525 439.823 524.349 703.537 749.955 796.868 988.157 1 231.15 1 279.78 1 473.06 1 568.50 1 709.53 1 847.46 1 937.43 2 025.69 2359.69 3 870.68 4310.71 4415.74 . 565 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 36 40 48 50 52 60 70 72 80 84 90 96 100 104 120 240 360 480 - -- - - - - - - - COMPOUND INTERESTTABLES % 1 2 3 4 5 6 7 8 9 .0 1 .2 3 4 5 6 7 8 9 0 1 2 3 :I 5 7 ) ) 2% Compound Interest Factors Single Payment 567 Uniform Payment Series - Arithmetic Gradient Compound Amount Factor Find F Given P FIP Present Worth Factor Find P Given F PIF Sinking Fund Factor Find A Given F AIF Capital Recovery Factor Find A Given P AlP Compound Amount Factor Find F Given A FIA Present Worth Factor Find P Given A PIA Gradient Uniform Series Find A Given G AIG Gradient Present Worth Find P Given G PIG n 1.020 1.040 1.061 1.082 1.104 1.126 1.149 1.172 1.195 1.219 1.243 1.268 1.294 1.319 1.346 1.373 1.400 1.428 1.457 1.486 1.516 1.546 1.577 1.608 1.641 1.673 1.707 1.741 1.776 1.811 2.040 2.208 2.587 2.692 2.800 3.281 4.000 4.161 4.875 5.277 5.943 6.693 7.245 7.842 10.765 115.887 1 247.5 13 429.8 .9804 .9612 .9423 .9238 .9057 .8880 .8706 .8535 .8368 .8203 .8043 .7885 .7730 .7579 .7430 .7284 .7142 .7002 .6864 .6730 .6598 .6468 .6342 .6217 .6095 .5976 .5859 .5744 .5631 .5521 .4902 .4529 .3865 .3715 .3571 .3048 .2500 .2403 .2051 .1895 .1683 .1494 .1380 .1275 .0929 .00863 .00080 .00007 1.0000 .4951 .3268 .2426 .1922 .1585 .1345 .1165 .1025 .0913 .0822 .0746 .0681 .0626 .0578 .0537 .0500 .0467 .0438 .0412 .0388 .0366 .0347 .0329 .0312 .0297 .0283 .0270 .0258 1.0200 .5151 .3468 .2626 .2122 .1785 .1545 .1365 .1225 .1113 .1022 .0946 .0881 .0826 .0778 .0737 .0700 .0667 .0638 .0612 .0588 .0566 .0547 .0529 .0512 .0497 .0483 .0470 .0458 .0447 .0392 .0366 .0326 .0318 .0311 .0288 .0267 .0263 .0252 .0247 .0240 .0235 .0232 .0229 .0220 .0202 .0200 .0200 1.000 2.020 3.060 4.122 5.204 6.308 7.434 8.583 9.755 10.950 12.169 13.412 14.680 15.974 17.293 18.639 20.012 21.412 22.840 24.297 25.783 27.299 28.845 30.422 32.030 33.671 35.344 37.051 38.792 40.568 51.994 60.402 79.353 84.579 90.016 114.051 149.977 158.056 193.771 213.865 247.155 284.645 312.230 342.090 488.255 5 744.4 62 326.8 671 442.0 0.980 1.942 2.884 3.808 4.713 5.601 6.472 7.325 8.162 8.983 9.787 10.575 11.348 12.106 12.849 13.578 14.292 14.992 15.678 16.351 17.011 17.658 18.292 18.914 19.523 20.121 20.707 21.281 21.844 22.396 25.489 27.355 30.673 31.424 32.145 34.761 37.499 37.984 39.744 40.525 41.587 42.529 43.098 43.624 45.355 49.569 49.960 49.996 0 0.495 0.987 1.475 1.960 2.442 2.921 3.396 3.868 4.337 4.802 5.264 5.723 6.178 6.631 7.080 7.526 7.968 8.407 8.843 9.276 9.705 10.132 10.555 10.974 11.391 11.804 12.214 12.621 13.025 15.381 16.888 19.755 20.442 21.116 23.696 26.663 27.223 29.357 30.361 31.793 33.137 33.986 34.799 37.711 47.911 49.711 49.964 0 0.961 2.846 5.617 9.240 13.679 18.903 24.877 31.571 38.954 46.996 55.669 64.946 74.798 85.200 .96.127 107.553 119.456 131.812 144.598 157.793 171.377 185.328 199.628 214.256 229.196 244.428 259.936 275.703 291.713 392.036 461.989 605.961 642.355 678.779 823.692 999.829 1 034.050 1 166.781 1 230.413 1 322.164 1409.291 1464.747 1518.082 1 710.411 2 374.878 2 483.567 2 498.027 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 60 70 72 80 84 90 96 100 104 120 240 360 480 . .0247 .0192 .0166 .0126 .0118 .0111 .00877 .00667 .00633 .00516 .00468 .00405 .00351 .00320 .00292 .00205 .00017 .00002 .._4'____ - - __ ____ _'_0" INTERESTTABLES COMPOUND 569 CompoundInterestFactors Arithmetic Gradient UniformPaymentSeries Single Payment Compound Amount Factor FindF Given P FIP Present Worth Factor FindP Given F PIF Sinking Fund Factor Find A Given F AIF 1.030 1.061 1.093 1.126 1.159 1.194 1.230 1.267 1.305 1.344 1.384 1.426 1.469 1.513 1.558 1.605 1.653 1.702 1.754 1.806 1.860 1.916 1.974 2.033 2.094 2.157 2.221 2.288 2.357 2.427 2.500 2.575 2.652 2.732 2.814 3.262 3.782 4.384 5.082 5.892 6.830 7.918 9.179 10.641 12.336 14.300 16.578 19.219 .9709 .9426 .9151 .8885 .8626 .8375 .8131 .7894 .7664 .7441 .7224 .7014 .6810 .6611 .6419 .6232 .6050 .5874 .5703 .5537 .5375 .5219 .5067 .4919 .4776 .4637 .4502 .4371 .4243 .4120 .4000 .3883 .3770 .3660 .3554 .3066 .2644 .2281 .1968 .1697 .1464 .1263 .1089 .0940 .0811 .0699 .0603 .0520 1.0000 .4926 .3235 .2390 .1884 .1546 .1305 .1125 .0984 .0872 .0781 .0705 .0640 .0585 .0538 .0496 .0460 .0427 .0398 .0372 .0349 .0327 .0308 .0290 .0274 .0259 .0246 .0233 .0221 .0210 .0200 .0190 .0182 .0173 .0165 .0133 .0108 .00887 .00735 .00613 .00515 .00434 .00367 .00311 .00265 .00226 .00193 .00165 Capital Recovery Factor FindA Given P AlP 1.0300 .5226 .3535 .2690 .2184 .1846 .1605 .1425 .1284 .1172 .1081 .1005 .0940 .0885 .0838 .0796 .0760 .0727 .0698 .0672 .0649 .0627 .0608 .0590 .0574 .0559 .0546 .0533 .0521 "';-.0510 .0500 .0490 .0482 .0473 .0465 .0433 .0408 .0389 .0373 .0361 .0351 .0343 .0337 .0331 .0326 .0323 .0319 .0316 - --- Compound Amount Factor FindF Given A FIA Present Worth Factor FindP Given A PIA 1.000 2.030 3.091 4.184 5.309 6.468 7.662 8.892 10.159 11.464 12.808 14.192 15.618 17.086 18.599 20.157 21.762 23.414 25.117 26.870 28.676 30.537 32.453 34.426 36.459 38.553 40.710 42.931 45.219 47.575 50.003 52.503 55.078 57.730 60.462 75.401 92.720 112.797 136.072 163.053 194.333 0.971 1.913 2.829 3.717 4.580 5.417 6.230 7.020 7.786 8.530 9.253 9.954 10.635 11.296 11.938 12.561 13.166 13.754 14.324 14.877 15.415 15.937 16.444 16.936 17.413 17.877 18.327 18.764 19.188 19.600 20.000 20.389 20.766 21.132 21.487 23.115 24.519 25.730 26.774 27.676 28.453 29.123 29.702 30.201 30.631 31.002 31.323 31.599 .230.594 . 272.631 321.363 377.857 443.349 519.272 607.287 - Gradient Uniform Series Find A Given G AIG o 0.493 0.980 1.463 1.941 2.414 2.882 3.345 3.803 4.256 4.705 5.148 5.587 6.021 6.450 6.874 7.294 7.708 8.118 8.523 8.923 9.319 9.709 10.095 10.477 10.853 11.226 11.593 11.956 12.314 12.668 13.017 13.362 13.702 14.037 15.650 17.156 18.558 19.860 21.067 22.184 23.215 24.163 25.035 25.835 26.567 27.235 27.844 Gradient Present Worth FindP Given G PIG o 0.943 2.773 5.438 8.889 13.076 17.955 23.481 29.612 36.309 43.533 51.248 59.419 68.014 77.000 86.348 96.028 106.014 116.279 126.799 137.549 148.509 . 159.656 170.971 -182.433 194.026 205.731 217.532 229.413 241.361 253.361 265.399 277.464 289.544 301.627 361.750 420.632 477.480 531.741 583.052 631.201 676.087 717.698 756.086 791.353 823.630 853.074 879.854 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 90 95 100 -- --- -- -_.._.__._--_._- - --_.._-- - ----- r-t::: .:::: - COMPOUND INTEREST TABLES 571 .I I I Yo Amount Factor Find F Given P FIP Fund Factor Find A Given F AIF Recovery Factor Find A Given P AlP 1.040 1.082 1.125 1.170 1.217 1.265 1.316 1.369 1.423 1.480 1.539 1.601 1.665 1.732 1.801 1.873 1.948 2.026 2.107 2.191 2.279 2.370 2.465 2.563 2.666 2.772 2.883 2.999 3.119 3.243 3.373 3.508 3.648 3.794 3.946 4.801 5.841 7.107 8.646 10.520 12.799 15.572 18.945 23.050 28.044 34.119 41.511 50.505 .9615 .9246 .8890 .8548 .8219 .7903 .7599 .7307 .7026 .6756 .6496 .6246 .6006 .5775 .5553 .5339 .5134 .4936 .4746 .4564 .4388 .4220 .4057 .3901 .3751 .3607 .3468 .3335 .3207 .3083 .2965 .2851 .2741 .2636 .2534 .2083 .1712 .1407 .1157 .0951 .0781 .0642 .0528 .0434 .0357 .0293 .0241 .0198 1.0000 .4902 .3203 .2355 .1846 .1508 .1266 .1085 .0945 .0833 .0741 .0666 .0601 .0547 .0499 .0458 .0422 .0390 .0361 .0336 .0313 .0292 .0273 .0256 .0240 .0226 .0212 .0200 .0189 .0178 .0169 .0159 .0151 .0143 .0136 .0105 .00826 .00655 .00523 .00420 .00339 .00275 .00223 .00181 .00148 .00121 .00099 .00081 1.0400 .5302 .3603 .2755 .2246 .1908 .1666 .1485 .1345 .1233 .1141 .1066 .1001 .0947 .0899 .0858 .0822 .0790 .0761 .0736 .0713 .0692 .0673 .0656 .0640 .0626 .0612 .0600 .0589 .0578 .0569 .0559 .0551 .0543 .0536 .0505 .0483 .0466 .0452 .0442 .0434 .0427 .0422 .0418 .0415 .0412 .0410 .0408 Sinking Capital ' . ( ,ru:: Arithmetic Gradient Uniform Payment Series Present Worth Factor Find P Given F PIF Compound I ".51 Gradient Gradient Amount Factor Find F Given A FIA Present Worth Factor Find P Given A PIA Uniform Series Find A Given G AIG Present Worth Find P Given G PIG n 1.000 2.040 3.122 4.246 5.416 6.633 7.898 9.214 10.583 12.006 13.486 15.026 16.627 18.292 20.024 21.825 23.697 25.645 27.671 29.778 31.969 34.248 36.618 39.083 41.646 44.312 47.084 49.968 52.966 56.085 59.328 62.701 66.209 69.858 73.652 95.025 121.029 152.667 191.159 237.990 294.968 364.290 448.630 551.244 676.089 827.981 1 012.8 1 237.6 0.962 1.886 2.775 3.630 4.452 5.242 6.002 6.733 7.435 8.111 8.760 9.385 9.986 10.563 11.118 11.652 12.166 12.659 13.134 13.590 14.029 14.451 14.857 15.247 15.622 15.983 16.330 16.663 16.984 17.292 17.588 17.874 18.148 18.411 18.665 19.793 20.720 21.482 22.109 22.623 23.047 23.395 23.680 23.915 24.109 24.267 24.398 24.505 0 0.490 0.974 1.451 1.922 2.386 2.843 3.294 3.739 4.177 4.609 5.034 5.453 5.866 6.272 6.672 7.066 7.453 7.834 8.209 8.578 8.941 9.297 9.648 9.993 10.331 10.664 10.991 11.312 11.627 11.937 12.241 12.540 12.832 13.120 14.476 15.705 16.812 17.807 18.697 19.491 20.196 20.821 21.372 21.857 22.283 22.655 22.980 0 0.925 2.702 5.267 8.555 12.506 17.066 22.180 27.801 33.881 40.377 47.248 54.454 61.962 69.735 77.744 85.958 94.350 102.893 111.564 120.341 129.202 138.128 147.101 156.104 165.121 174.138 183.142 192.120 201.062 209.955 218.792 227.563 236.260 244.876 286.530 325.402 361.163 393.689 422.996 449.201 472.479 493.041 511.116 526.938 540.737 552.730 563.125 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 95 100 Compound - ",\ ';" 'i; 4% CompoundInterestFactors Single Payment f' . -- - - - - 1j ' ,;!"" ;.,ill" ;' ' ,#..: ' :} :; r . ,I I . :..7,j . II'" I I - - -- - -- -"- - - . -----_.- - -. -_.- -- --_._-_.._.._---. --- - COMPOUNDINTEREST TABLES 573 5% CompoundInterestFactors Present Worth Factor Find P Given A PIA Gradient Uniform Series Find A Given G AIG Gradient Present Worth Find P Given G PIG n 1.000 2.050 3.152 4.310 5.526 6.802 8.142 9.549 11.027 12.578 14.207 15.917 17.713 19.599 21.579 23.657 25.840 28.132 30.539 33.066 35.719 38.505 41.430 44.502 47.727 51.113 54.669 58.402 62.323 66.439 70.761 75.299 80.063 85.067 0.952 1.859 2.723 3.546 4.329 5.076 5.786 6.463 7.108 7.722 8.306 8.863 9.394 9.899 10.380 10.838 11.274 11.690 12.085 12.462 12.821 13.163 13.489 13.799 14.094 14.375 14.643 14.898 15.141 15.372 15.593 15.803 16.003 16.193 0 0.488 0.967 1.439 1.902 2.358 2.805 3.244 3.676 4.099 4.514 4.922 5.321 5.713 6.097 6.474 6.842 7.203 7.557 7.903 8.242 8.573 8.897 9.214 9.524 9.827 10.122 10.411 10.694 10.969 11.238 11.501 11.757 12.006 0 0.907 2.635 5.103 8.237 11.968 16.232 20.970 26.127 31.652 37.499 43.624 49.988 56.553 63.288 70.159 77.140 84.204 91.327 98.488 105.667 112.846 120.008 127.140 134.227 141.258 148.222 155.110 161.912 168.622 175.233 181.739 188.135 194.416 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 .0611 90.320 16.374 12.250 200.580 35 .00828 .00626 .00478 .00367 .0583 .0563 .0548 .0537 120.799 159.699 209.347 272.711 17.159 17.774 18.256 18.633 13.377 14.364 15.223 15.966 229.545 255.314 277.914 297.510 40 45 50 55 .00283 .00219 .00170 .00132 .00103 .00080 .00063 .00049 .00038 .0528 .0522 .0517 .0513 .0510 .0508 .0506 .0505 .0504 353.582 456.795 588.525 756.649 971.222 1 245.1 1 594.6 2040.7 2610.0 18.929 19.161 19.343 19.485 19.596 19.684 19.752 19.806 19.848 16.606 17.154 17.621 18.018 18.353 18.635 18.871 19.069 19.234 314.343 328.691 340.841 351.072 359.646 366.800 372.749 377.677 381.749 60 65 70 75 80 85 90 95 100 Present Worth Factor Find P Given F PIF Sinking Fund Factor Find A Given F AIF Capital Recovery Factor Find A Given P AlP 1.050 .9524 1.102 1.158 1.216 1.276 1.340 1.407 1.477 1.551 .9070 .8638 .8227 .7835 .7462 .7107 .6768 .6446 .629 .6139 1.710 1.796 1.886 1.980 2.079 2.183 2.292 2.407 2.527 2.653 2.786 2.925 3.072 3.225 3.386 3.556 3.733 3.920 4.116 322 4.538 4.765 5.003 5.253 .5847 .5568 .5303 .5051 .4810 .4581 .4363 .4155 .3957 .3769 .3589 .3419 .3256 .3101 .2953 .2812 .2678 .2551 .2429 .2314 .2204 .2099 .1999 .1904 1.0000 .4878 .3172 .2320 .1810 .1470 .1228 .1047 .0907 .0795 .0704 .0628 .0565 .0510 .0463 .0423 .0387 .0355 .0327 .0302 .0280 .0260 .0241 .0225 .0210 .0196 .0183 .0171 .0160 .0151 .0141 .0133 .0125 .0118 1.0500 .5378 .3672 .2820 .2310 .1970 .1728 .1547 .1407 .1295 .1204 .1128 .1065 .1010 .0963 .0923 .0887 .0855 .0827 .0802 .0780 .0760 .0741 .0725 .0710 .0696 .0683 .0671 .0660 .0651 .0641 .0633 .0625 .0618 5.516 .1813 .0111 7.040 .1420 .1113 .0872 .0683 .0535 .0419 .0329 .0258 .0202 .0158 .0124 .00971 .00760 Compound Amount Factor Find F Given P FIP 8.985 11.467 14.636 18.679 23.840 30.426 38.832 49.561 63.254 80.730 103.034 131.500 -.------- Arithmetic Gradient Uniform Payment Series Single Payment - Compound Amount Factor Find F Given A FIA --- - --- - , I I,I!. . Ie'" lirl L -rl' II: I ;IIi: r1: ."'i, I B! (I:H - - - - ..---- .______....__n_ -- - -, .,' "'-'j .',; ,-, - COMPOUNDINTEREST TABLES 575 Arithmetic Gradient Uniform Payment Series Compound Amount Factor Find F Given P FIP Present Worth Factor Find P Given F PIF Sinking Fund Factor Find A Given F AIF Capital Recovery Factor Find A Given P AlP Compound Amount Factor Find F Given A FIA Present Worth Factor Find P Given A PIA Gradient Uniform Series Find A Given G AIG Gradient Present Worth Find P Given G PIG n 1.070 1.145 1.225 1.311 1.403 1.501 1.606 1.718 1.838 )67 2.105 2.252 2.410 2.579 2.759 2.952 3.159 3.380 3.617 3.870 4.141 4.430 4.741 5.072 5.427 5.807 6.214 6.649 7.114 2 8.145 8.715 9.325 9.978 10.677 14.974 21.002 29.457 41.315 57.947 81.273 113.990 159.877 224.235 314.502 441.105 618.673 867.720 .9346 .8734 .8163 .7629 .7130 .6663 .6227 .5820 .5439 .5083 .4751 .4440 .4150 .3878 .3624 .3387 .3166 .2959 .2765 .2584 .2415 .2257 .2109 .1971 .1842 .1722 .1609 .1504 .1406 .1314 .1228 .1147 .1072 .1002 .0937 .0668 .0476 .0339 .0242 .0173 .0123 .00877 .00625 .00446 .00318 .00227 .00162 .00115 1.0000 .4831 .3111 .2252 .1739 .1398 .1156 .0975 .0835 .0724 .0634 .0559 .0497 .0443 .0398 .0359 .0324 .0294 .0268 .0244 .0223 .0204 .0187 .0172 .0158 .0146 .0134 .0124 .0114 .0106 .00980 .00907 .00841 .00780 .00723 .00501 .00350 .00246 .00174 .00123 .00087 .00062 .00044 .00031 .00022 .00016 .00011 .00008 1.0700 .5531 .3811 .2952 .2439 .2098 .1856 .1675 .1535 .1424 .1334 .1259 .1197 .1143 .1098 .1059 .1024 .0994 .0968 .0944 .0923 .0904 .0887 .0872 .0858 .0846 .0834 .0824 .0814 1.000 2.070 3.215 4.440 5.751 7.153 8.654 10.260 11.978 13.816 15.784 17.888 20.141 22.551 25.129 27.888 30.840 33.999 37.379 40.996 44.865 49.006 53.436 58.177 63.249 68.677 74.484 80.698 87.347 94.461 102.073 110.218 118.934 128.259 138.237 199.636 285.750 406.530 575.930 813.523 1 146.8 1 614.1 2269.7 3 189.1 4478.6 6287.2 8 823.9 12381.7 0.935 1.808 2.624 3.387 4.100 4.767 5.389 5.971 6.515 7.024 7.499 7.943 8.358 8.745 9.108 9.447 9.763 10.059 10.336 10.594 10.836 11.061 11.272 11.469 11.654 11.826 11.987 12.137 12.278 0 0.483 0.955 1.416 1.865 2.303 2.730 3.147 3.552 3.946 4.330 4.703 5.065 5.417 5.758 6.090 6.411 6.722 7.024 7.316 7.599 7.872 8.137 8.392 8.639 8.877 9.107 9.329 9.543 9.749 9.947 10.138 10.322 10.499 10.669 11.423 12.036 12.529 12.921 13.232 13.476 13.666 13.814 13.927 14.015 14.081 14.132 14.170 0 0.873 2.506 4.795 7.647 10.978 14.715 18.789 23.140 27.716 32.467 37.351 42.330 47.372 52.446 57.527 62.592 67.622 72.599 77.509 82.339 87.079 91.720 96.255 100.677 104.981 109.166 113.227 117.162 120.972 124.655 128.212 131.644 134.951 138.135 152.293 163.756 172.905 180.124 185.768 190.145 193.519 196.104 198.075 199.572 200.704 201.558 202.200 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 95 100 .0806 . .0798 .0791 .0784 .0778 .0772 .0750 .0735 .0725 .0717 .0712 .0709 .0706 .0704 .0703 .0702 .0702 .0701 .0701 il1 7% Compound Interest Factors Single Payment I 12.409 12.532 12.647 12.754 12.854 12.948 13.332 13.606 13.801 13.940 14.039 14.110 14.160 14.196 14.222 14.240 14.253 14.263 14.269 . 1!§!';;';i!!,.'S' I ---, 576 8% 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 95 100 8% CompoundInterestFactors UniformPaymentSeries SinglePayment n - COMPOUND INTEREST TABLES Compound Amount Factor Find F Given P FIP Present Worth Factor Find P Given F PIF 1.080 1.166 1.260 1.360 1.469 1.587 1.714 1.851 1.999 2.159 2.332 2.518 2.720 2.937 3.172 3.426 3.700 3.996 4.316 4.661 5.034 5.437 5.871 6.341 6.848 7.396 7.988 8.627 9.317 10.063 10.868 11.737 12.676 13.690 14.785 21.725 31.920 46.902 68.914 101.257 148.780 218.607 321.205 471.956 693.458 1 018.9 1497.1 2 199.8 .9259 .8573 .7938 .7350 .6806 .6302 .5835 .5403 .5002 .4632 .4289 .3971 .3677 .3405 .3152 .2919 .2703 .2502 .2317 .2145 .1987 .1839 .1703 .1577 .1460 .1352 .1252 .1159 .1073 .0994 .0920 .0852 .0789 .0730 .0676 .0460 .0313 .0213 .0145 .00988 .00672 .00457 .00311 .00212 .00144 .00098 .00067 .00045 Sinking Fund Factor Find A Given F AIF 1.0000 .4808 .3080 .2219 .1705 .1363. .1121 .0940 .0801 .0690 .0601 .0527 .0465 .0413 .0368 .0330 .0296 .0267 .0241 .0219 .0198 .0180 .0164 .0150 .0137 ..0125 .0114 .0105 .00962 .00883 .00811 .00745 .00685 .00630 .00580 .00386 .00259 .00174 .00118 .00080 .00054 .00037 .00025 .00017 .00012 .00008 .00005 .00004 Capital Recovery Factor Find A Given P AlP Compound Amount Factor Find F Given A FIA 1.000 2.080 3.246 4.506 5.867 7.336 8.923 10.637 12.488 14.487 16.645 18.977 21.495 24.215 27.152 30.324 33.750 37.450 41.446 45.762 50.423 55.457 60.893 66.765. 73.106 79.954 87.351 95.339 103.966 113.283 123.346 134.214 145.951 158.627 172.317 259.057 386.506 573.771 848.925 1 253.2 1 847.3 2720.1 4 002.6 5887.0 8655.7 12 724.0 18 701.6 27 484.6 1.0800 .5608 .3880 .3019 .2505 .2163 .1921 .1740 .1601 .1490 .1401 .1327 .1265 .1213 .1168 .1130 .1096 .1067 .1041 .1019 .0998 .0980 .0964 .0950 .0937 .0925 .0914 .0905 .0896 .0888 .0881 .0875 .0869 .0863 .0858 .0839 .0826 .0817 .0812 .0808 .0805 .0804 .0802 .0802 .0801 .0801 .0801 .0800 -- - ArithmeticGradient Present Worth Factor Find P Given A PIA Gradient Uniform Series Find A Given G AIG Gradient Present Worth Find P Given G PIG 0.926 1.783 2.577 3.312 3.993 4.623 5.206 5.747 6.247 6.710 7.139 7.536 7.904 8.244 8.559 8.851 9.122 9.372 9.604 9.818 10.017 10.201 10.371 10.529 10.675 10.810 10.935 11.051 11.158 11.258 11.350 11.435 11.514 11.587 11.655 11.925 12.108 12.233 12.319 12.377 12.416 12.443 12.461 12.474 12.482 12.488 12.492 12.494 0 0.481 0.949 1.404 1.846 2.276 2.694 3.099 3.491 3.871 4.240 4.596 4.940 5.273 5.594 5.905 6.204 6.492 6.770 7.037 7.294 7.541 7.779 8.007 8.225 8.435 8.636 8.829 9.013 9.190 9.358 9.520 9.674 9.821 9.961 10.570 11.045 11.411 11.690 11.902 12.060 12.178 12.266 12.330 12.377 12.412 12.437 12.455 0 0.857 2.445 4.650 7.372 10.523 14.024 17.806 21.808 25.977 . 30.266 34.634 39.046 43.472 47.886 52.264 56.588 60.843 65.013 69.090 73.063 76.926 80.673 84.300 87.804 91.184 94.439 97.569 100.574 103.456 106.216 108.858 111.382 113.792 116.092 126.042 133.733 139.593 144.006 147.300 149.739 151.533 152.845 153.800 154.492 154.993 155.352 155.611 - - -- - 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 90 IS 1" -_.-------- r- - ----- -- !j I ffi: ..' COMPOUNDINTEREST TABLES 577 I I ";ir; . I "I 9% Compound Interest Factors ) Single Payment I JIC n' , I Arithmetic Gradient Uniform Payment Series I Compound Amount Factor Find F Given P Present Worth Factor Find P Given F Sinking Fund Factor Find A Given F Capital Recovery Factor Find A Given P Compound Amount Factor Find F Given A Present Worth Factor Find P Given A Gradient Uniform Series Find A Given G Gradient Present Worth Find P Given G FIP PIF AIF AlP FIA PIA AIG 1.000 0.917 0 PIG 0 2.090 3.278 1.759 2.531 0.478 0.943 0.842 2.386 4.573 5.985 7.523 9.200 11.028 13.021 15.193 17.560 20.141 22.953 26.019 29.361 33.003 36.974 41.301 46.019 51.160 56.765 62.873 69.532 76.790 84.701 93.324 102.723 112.968 124.136 136.308 149.575 164.037 179.801 196.983 215.711 337.883 525.860 815.085 1260.1 1 944.8 2 998.3 4619.2 7 113.3 10950.6 3.240 3.890 4.486 5.033 5.535 5.995 6.418 6.805 7.161 7.487 7.786 8.061 8.313 8.544 8.756 8.950 9.129 9.292 9.442 9.580 9.707 9.823 9.929 10.027 10.116 10.198 10.274 10.343 10.406 10.464 10.518 10.567 10.757 10.881 10.962 11.014 11.048 11.070 11.084 11.094 11.100 1.393 1.828 2.250 2.657 3.051 3.431 3.798 4.151 4.491 4.818 5.133 5.435 5.724 6.002 6.269 6.524 6.767 7.001 7.223 7.436 7.638 7.832 8.016 8.191 8.357 8.515 8.666 8.808 8.944 9.072 9.193 9.308 9.796 10.160 10.430 10.626 10.768 10.870 10.943 10.994 11.030 4.511 7.111 10.092 13.375 16.888 20.571 24.373 28.248 32.159 36.073 39.963 43.807 47.585 51.282 54.886 58.387 61.777 65.051 68.205 71.236 74.143 76.927 79.586 82.124 84.542 86.842 g9.028 91.102 93.069 94.931 96.693 98.359 105.376 110.556 114.325 117.036 118.968 120.334 121.294 121.965 122.431 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 1.090 .9174 1.0000 1.0900 1.188 1.295 1.412 .8417 .7722 .4785 .3051 .5685 .3951 .7084 .6499 .5963 .5470 .5019 .4604 .4224 .3875 .3555 .3262 .2992 .2745 .2519 .2311 .2120 .1945 .1784 .1637 .1502 .1378 .1264 .1160 .1064 .0976 .0895 .0822 .0754 .0691 .0634 .0582 .0534 .0490 .0318 .0207 .0134 .00874 .00568 .00369 .00240 .00156 .00101 .2187 .1671 .1329 .1087 .0907 .0768 .0658 .0569 .0497 .0436 .0384 .0341 .0303 .0270 .0242 .0217 .0195 .0176 .0159 .0144 .0130 .0118 .0107 .00973 .00885 .00806 .00734 .00669 .00610 .00556 .00508 .00464 .00296 .00190 .00123 .00079 .00051 .00033 .00022 .00014 .00009 .3087 .2571 .2229. .1987 .1807 .1668 .1558 .1469 .1397 .1336 .1284 .1241 .1203 .1170 .1142 .1117 .1095 .1076 .1059 .1044 .1030 .1018 .1007 .0997 .0989 .0981 .0973 .0967 .0961 .0956 .0951 .0946 .0930 .0919 .0912 .0908 .0905 .0903 .0902 .0901 .0901 1.539 1.677 1.828 1.993 2.172 2.367 2.580 2.813 3.066 3.342 3.642 3.970 4.328 4.717 5.142 5.604 6.109 6.659 7.258 7.911 8.623 9.399 10.245 11.167 12.172 13.268 14.462 15.763 17.182 18.728 20.414 31.409 48.327 74.358 114.409 176.032 270.847 416.731 641.193 986.555 I I n Ii 1 2 3 1517.9 .00066 .00006 .0901 16854.9 11.104 11.055 122.753 85 2 335.5 .00043 .00028 .00018 .00004 .00003 .00002 .0900 .0900 .0900 25939.3 39916.8 61422.9 11.106 11.108 11.109 11.073 11.085 11.093 122.976 123.129 123.233 90 95 100 3 593.5 5529.1 --- -- - -- r", -::?:?{ :.,':/......:ff. :,..;:0;-:;...-:__ I >III ;'..:1 I ,I, r - 578 COMPOUNDINTEREST TABLES 10% 1 0% CompoundInterestFactors Single Payment n Compound Amount Factor Find F Given P FIP Present Worth Factor Find P Given F PIF 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 95 100 1.100 1.210 1.331 1.464 1.611 1.772 1.949 2.144 2.358 2.594 2.853 3.138 3.452 3.797 4.177 4.595 5.054 5.560 6.116 6.728 7.400 8.140 8.954 9.850 10.835 11.918 13.110 14.421 15.863 17.449 19.194 21.114 23.225 25.548 28.102 45.259 72.891 117.391 189.059 304.482 490.371 789.748 1 271.9 2 048.4 3 299.0 5313.0 8556.7 13 780.6 .9091 .8264 .7513 .6830 .6209 .5645 .5132 .4665 .4241 .3855 .3505 .3186 .2897 .2633 .2394 .2176 .1978 .1799 .1635 .1486 .1351 .1228 .1117 .1015 .0923 .0839 .0763 .0693 .0630 .0573 .0521 .0474 .0431 .0391 .0356 .0221 .0137 .00852 .00529 .00328 .00204 .00127 .00079 .00049 .00030 .00019 .00012 .00007 Arithmetic Gradient Uniform Payment Series Sinking Fund Factor Find A Given F AIF 1.0000 .4762 .3021 .2155 .1638 .1296 .1054 .0874 .0736 .0627 .0540 .0468 .0408 .0357 .0315 .0278 .0247 .0219 .0195 .0175 .0156 .0140 .0126 .0113 .0102 .00916 .00826 .00745 .00673 .00608 .00550 .00497 .00450 .00407 .00369 .00226 .00139 .00086 .00053 .00033 .00020 .00013 .00008 .00005 .00003 .00002 .00001 .00001 Capital Reovery Factor Find A Given P AlP Compound Amount Factor Find F Given A FIA Present Worth Factor Find P Given A PIA Gradient Uniform Series Find A Given G AIG Gradient Present Worth Find P Given G PIG 1.1000 .5762 .4021 .3155 .2638 .2296 .2054 .1874 .1736 .1627 .1540 .1468 .1408 .1357 .1315 .1278 .1247 .1219 .1195 .1175 .1156 .1140 .1126 .1113 .1102 .1092 .1083 .1075 .1067 .1061 .1055 .1050 .1045 .1041 .1037 .1023 .1014 .1009 .1005 .1003 .1002 .1001 .1001 .1000 .1000 .1000 .1000 .1000 1.000 2.100 3.310 4.641 6.105 7.716 9.487 11.436 13.579 15.937 18.531 21.384 24.523 27.975 31.772 35.950 40.545 45.599 51.159 57.275 64.003 71.403 79.543 88.497 98.347 109.182 121.100 134.210 148.631 164.494 181.944 201.138 222.252 245.477 271.025 442.593 718.905 1 163.9 1 880.6 3 034.8 4893.7 7 887.5 12709.0 20474.0 32979.7 53 120.3 85 556.9 137796.3 0.909 1.736 2.487 3.170 3.791 4.355 4.868 5.335 5.759 6.145 6.495 6.814 7.103 7.367 7.606 7.824 8.022 8.201 8.365 8.514 8.649 8.772 8.883 8.985 9.077 9.161 9.237 9.307 9.370 9.427 9.479 9.526 9.569 9.609 9.644 9.779 9.863 9.915 9.947 9.967 9.980 9.987 9.992 9.995 9.997 9.998 9.999 9.999 0 0.476 0.937 1.381 1.810 2.224 2.622 3.004 3.372 3.725 4.064 4.388 4.699 4.996 5.279 5.549 5.807 6.053 6.286 6.508 6.719 6.919 7.108 7.288 7.458 7.619 7.770 7.914 8.049 8.176 8.296 8.409 8.515 8.615 8.709 9.096 9.374 9.570 9.708 9.802 9.867 9.911 9.941 9.961 9.974 9.983 9.989 9.993 0 0.826 2.329 4.378 6.862 9.684 12.763 16.029 19.421 22.891 . 26.396 29.901 33.377 36.801 40.152 43.416 46.582 49.640 52.583 55.407 58.110 60.689 63.146 65.481 67.696 69.794 71.777 73.650 75.415 77.077 78.640 80.108 81.486 82.777 83.987 88.953 92.454 94.889 96.562 97.701 98.471 98.987 99.332 99.561 99.712 99.812 99.877 99.920 n 1 2 3 4 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 95 100 - I ...J -- - ------ -.----.-..--...- -) , COMPOUND INTEREST TABLES 579 2% 12% CompoundInterestFactors Arithmetic Gradient Uniform Payment Series Single Payment Present Worth Factor Find P Given A PIA Gradient Uniform Series Find A Given G AIG Gradient Present Worth Find P Given G PIG 1.000 2.120 3.374 4.779 6.353 8.115 10.089 12.300 14.776 17.549 20.655 24.133 28.029 32.393 37.280 42.753 48.884 55.750 63.440 72.052 81.699 92.503 104.603 118.155 133.334 150.334 169.374 190.699 214.583- 0.893 1.690 2.402 3.037 3.605 4.111 4.564 4.968 5.328 5.650 5.938 6.194 6.424 6.628 6.811 6.974 7.120 7.250 7.366 7.469 7.562 7.645 7.718 7.784 7.843 7.896 7.943 7.984 8.022 0 0.472 0.925 1.359 1.775 2.172 2.551 2.913 3.257 3.585 3.895 4.190 4.468 4.732 4.980 5.215 5.435 5.643 5.838 6.020 6.191 6.351 6.501 6.641 6.771 6.892 7.005 7.110 0 0.797 2.221 4.127 6.397 8.930 11.644 14.471 17.356 20.254 23.129 25.952 28.702 31.362 33.920 36.367 38.697 40.908 42.998 44.968 46.819 48.554 50.178 51.693 53.105 54.418 55.637 56.767 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 .1241 241.333 8.055 57.814 58.782 29 30 .00369 .00328 .00292 .00260 .00232 .1237 .1233 .1229 .1226 .1223 271.293 304.848 342.429 384.521 431.663 8.085 8.112 8.135 8.157 8.176 7.207 7.297 7.381 . 7.459 7.530 7.596 7.658 59.676 60.501 61.261 61.961 62.605 31 32 33 34 35 .0107 .00610 .00346 .00196 .00111 .00130 .00074 .00042 .00024 .00013 767.091 1 358.2 2 400.0 4 236.0 7471.6 8.244 8.283 8.304 8.317 8.324 7.899 8.057 8.160 8.225 8.266 65.116 1 581.9 2 787.8 4913.1 8 658.5 15 259.2 .00063 .00036 .00020 .00012 .00007 .00008 .00004 .00002 .00001 .00001 .1213 .1207 .1204 .1202 .1201 .1201 .1200 .1200 .1200 .1200 13 173.9 23223.3 40 933.8 72 145.7 127 151.7 8.328 8.330 8.332 8.332 8.333 8.292 8.308 8.318 8.324 8.328 69.058 69.210 69.303 69.359 69.393 40 45 50 55 60 65 70 75 80 85 26891.9 47 392.8 83 522.3 .00004 .00002 .00001 .1200 .1200 .1200 224 091.1 394 931.4 696010.5 8.333 8.333 8.333 8.330 8.331 8.332 69.414 69.426 69.434 90 95 100 Present Worth Factor Find P Given F PIF Sinking Fund Factor Find A Given F AIF Capital Recovery Factor Find A Given P AlP 1.120 1.254 1.405 1.574 1.762 1.974 2.211 2.476 2.773 3.106 3.479 3.896 4.363 4.887 5.474 6.130 6.866 7.690 8.613 9.646 10.804 12.100 13.552 15.179 17.000 19.040 21.325 23.884 26.750 .8929 .7972 .7118 .6355 .5674 .5066 .4523 .4039 .3606 .3220 .2875 .2567 .2292 .2046 .1827 .1631 .1456 .1300 .1161 .1037 .0926 .0826 .0738 .0659 .0588 .0525 .0469 .0419 .0374 1.0000 .4717 .2963 .2092 .1574 .1232 .0991 .0813 .0677 .0570 .0484 .0414 .0357 .0309 .0268 .0234 .0205 .0179 .0158 .0139 .0122 .0108 .00956 .00846 .00750 .00665 .00590 .00524 .00466 1.1200 .5917 .4163 .3292 .2774 .2432 .2191 .2013 .1877 .1770 .1684 .1614 .1557 .1509 .1468 .1434 .1405 .1379 .1358 .1339 .1322 .1308 .1296 .1285 .1275 .1267 .1259 .1252 .1247 29.960 .0334 .00414 33.555 37.582 42.092 47.143 52.800 .0298 .0266 .0238 .0212 .0189 93.051 163.988 289.002 509.321 897.597 Compound Amount Factor Find F Given P FIP Compound Amount Factor Find F Given A FIA ---- . . 66.734 67.762 68.408 68.810 L n 1 2 3 4 5 6 7 8 9 10 n I -580 COMPOUND INTEREST TABLES 15% Single Payment n 1 2 3 4 5 6 7 8 1u 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 ..:JV 31 32 33 34 35 40 45 50 55 60 65 70 75 80 85 15% CompoundInterestFactors Compound Amount Factor Find F Given P FIP 1.150 1.322 1.521 1.749 2.011 2.313 2.660 3.059 3.518 4.046 4.652 5.350 6.153 7.076 8.137 9.358 10.761 12.375 14.232 16.367 18.822 21.645 24.891 28.625 32.919 37.857 43.535 50.066 57.575 66.212 76.144 87.565 100.700 115.805 133.176 267.864 538.769 1 083.7 2179.6 4 384.0 8817.8 17735.7 35 672.9 71 750.9 144316.7 Arithmetic Gradient Uniform Payment Series Present Worth Factor Find P Given F PIF Sinking Fund Factor Find A Given F AIF Capital Recovery Factor Find A Given P AlP Compound Amount Factor Find F Given A FIA Present Worth Factor Find P Given A PIA Gradient Uniform Series Find A Given G AIG Gradient Present Worth Find P Given G PIG n .8696 .7561 .6575 .5718 .4972 .4323 .3759 .3269 .2843 .2472 .2149 .1869 .1625 .1413 .1229 .1069 .0929 .0808 .0703 .0611 .0531 .0462 .0402 .0349 .0304 .0264 .0230 .0200 .0174 .0151 .0131 .0114 .00993 .00864 .00751 .00373 .00186 .00092 .00046 .00023 .00011 .00006 .00003 .00001 .00001 1.0000 .4651 .2880 .2003 .1483 .1142 .0904 .0729 .0596 .0493 .0411 .0345 .0291 .0247 .0210 .0179 .0154 .0132 .0113 .00976 .00842 .00727 .00628 .00543 .00470 .00407 .00353 .00306 .00265 .00230 .00200 .00173 .00150 .00131 .00113 .00056 .00028 .00014 .00007 .00003 .00002 .00001 1.1500 .6151 .4380 .3503 .2983 .2642 .2404 .2229 .2096 .1993 .1911 .1845 .1791 .1747 .1710 .1679 .1654 .1632 .1613 .1598 .1584 .1573 .1563 .1554 .1547 .1541 .1535 .1531 .1527 .1523 .1520 .1517 .1515 .1513 .1511 .1506 .1503 .1501 .1501 .1500 .1500 .1500 .1500 .1500 .1500 1.000 2.150 3.472 4.993 6.742 8.754 11.067 13.727 16.786 20.304 24.349 29.002 34.352 40.505 47.580 55.717 65.075 75.836 88.212 102.444 118.810 137.632 159.276 184.168 212.793 245.712 283.569 327.104 377.170 434.745 500.957 577.100 664.666 765.365 881.170 1 779.1 3585.1 7217.7 14524.1 29 220.0 58 778.6 118231.5 237812.5 478 332.6 962 104.4 0.870 1.626 2.283 2.855 3.352 3.784 4.160 4.487 4.772 5.019 5.234 5.421 5.583 5.724 5.847 5.954 6.047 6.128 6.198 6.259 6.312 6.359 6.399 6.434 6.464 6.491 6.514 6.534 6.551 6.566 6.579 6.591 6.600 6.609 6.617 6.642 6.654 6.661 6.664 6.665 6.666 6.666 6.666 6.667 6.667 0 0.465 0.907 1.326 1.723 2.097 2.450 2.781 3.092 3.383 3.655 3.908 4.144 4.362 4.565 4.752 4.925 5.084 5.231 5.365 5.488 5.601 5.704 , 5.798 5.883 5.961 6.032 6.096 6.154 6.207 6.254 6.297 6.336 6.371 . 6.402 6.517 6.583 6.620 6.641 6.653 6.659 6.663 6.665 6.666 6.666 0 0.756 2.071 3.786 5.775 7.937 10.192 12.481 14.755 16.979 19.129 21.185 23.135 24.972 26.693 28.296 29.783 31.156 32.421 33.582 34.645 35.615 36.499 37.302 38.031 38.692 39.289 39.828 40.315 40.753 41.147 41.501 41.818 42.103 42.359 43.283 43.805 44.096 44.256 44.343 44.390 44.416 44.429 44.436 44.440 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 - --- COMPOUNDINTEREST TABLES 581 8% Compound Amount Factor Find F Given P FIP Present Worth Factor Find P Given F PIF Sinking Fund Factor Find A Given F AIF Capital Recovery Factor Find A Given P AlP Compound Amount Factor Find F Given A FIA Present Worth Factor Find P Given A PIA Gradient Uniform Series Find A Given G AIG Gradient Present Worth Find P Given G PIG ;:: n .8475 1.0000 1.1800 1.000 0.847 0 0 1 .7182 .6086 .5158 .4371 .4587 .2799 .1917 .1398 .6387 .4599 .3717 .3198 2.180 3.572 5.215 7.154 1.566 2.174 2.690 3.127 0.459 0.890 1.295 1.673 0.718 1.935 3.483 5.231 2 3 4 5 2.700 5.234 6.176 7.288 8.599 10.147 .3704 .3139 .2660 .2255 .1911 .1059 .0824 .0652 .0524 .0425 .2859 .2624 .2452 .2324 .2225 9.442 12.142 15.327 19.086 23.521 3.498 3.812 4.078 4.303 4.494 2.025 2.353 2.656 2.936 3.194 7.083 8.967 10.829 12.633 14.352 6 7 '8 9 10 .1619 .1372 .1163 .0985 .0348 .0286 .0237 .0197 .2148 .2086 .2037 .1997 28.755 34.931 42.219 50.818 4.656 4.793 4.910 5.008 3.430 3.647 3.845 4.025 15.972 17.481 18.877 20.158 11 12 13 14 11.974 .0835 .0164 .1964 60.965 5.092 4.189 21.327 15 14.129 16.672 19.673 23.214 27.393 .0708 .0600 .0508 .0431 .0365 .0137 .0115 .00964 .00810 .00682 .1937 .1915 .1896 .1881 .1868 72.939 87.068 103.740 123.413 146.628 5.162 5.222 5.273 5.316 5.353 4.337 4.471 4.592 4.700 4.798 22.389 23.348 24.212 24.988 25.681 16 17 18 19 20 32.324 38.142 45.008 53.109 62.669 73.949 87.260 102.966 .121.500 143.370 .0309 .0262 .0222 .0188 .0160 .0135 .0115 .00971 .00823 .00697 .00575 .00485 .00409 .00345 .00292 .00247 .00209 .00177 .00149 .00126 .1857 .1848 .1841 .1835 .1829 .1825 .1821 .1818 .1815 .1813 174.021 206.345 244.487 289.494 342.603 405.272 479.221 566.480 669.447 790.947 5.384 5.410 5.432 5.451 5.467 5.480 5.492 5.502 5.510 4.885 4.963 5.033 5.095 5.150 5.199 5.243 5.281 5.315 26.300 26.851 27.339 27.772 28.155 28.49"'4 28.791 29.054 29.284 21 22 23 24 25 26 27 28 29 5.517 . 5.345 29.486 169.177 199.629 .00591 .00501 .00107 .00091 .1811 .1809 5.523 5.528 29.664 29.819 30 31 32 .00425 .00360 .00305 .00133 .00058 .00025 .00011 .00005 .00002 .00001 .00077 .00065 .00055 .00024 .00010 .00005 .00002 .00001 .1808 .1806 .1806 .1802 .1801 .1800 .1800 .1800 .1800 .1800 5.371 5.394 5.415 29.955 30.074 30.177 30.527 30.701 30.786 30.827 30.846 30.856 30.860 33 34 j5 40 45 50 55 60 65 70 235.562 277.963 327.997 750.377 1 716.7 3 927.3 8 984.8 20555.1 47025.1 107581.9 1303.1 1538.7 1 816.6 4 163.2 9 531.6 21813.0 49910.1 114 189.4 261 244.7 597 671.7 L' IT 5.536 5.539 5.548 5.552 5.554: 5.555 5.555 5.555 5.556 5.433 5.449 5.502 5.529 5.543 5.549 5.553 5.554 5.555 ".- .... .f,: 8::". 1.180 934.317 1 103.5 1 I 1.392 1.643 1.939 2.288 3.185 3.759 4.435 :i.Ii Arithmetic Gradient Uniform Payment Series Single Payment If 18% CompoundInterestFactors I . Ii ,.L'lI, I' 'II . " 11111 I ]11,1, t ;Iilll 'II, .. °1'11 , . j , .d! ;:;i . "ii'I 582 COMPOUNDINTEREST TABLES 200/0 200/0 CompoundInterestFactors Single Payment Arithmetic Gradient Uniform Payment Series Present Worth Factor Find P Given F PIF Sinking Fund Factor Find A Given F AIF Capital Recovery Factor Find A Given P AlP Present Worth Factor Find P Given A PIA Gradient Uniform Series Find A Given G AIG Gradient Present Worth Find P Given G PIG n 0.833 1.528 2.106 2.589 2.991 3.326 3.605 3.837 4.031 4.192 4.327 4.439 4.533 4.611 4.675 4.730 4.775 4.812 4.843 4.870 4.891 4.909 4.925 4.937 4.948 4.956 0 0.455 0.879 1.274 1.641 1.979 2.290 2.576 2.836 3.074 3.289 3.484 3.660 3.817 3.959 4.085 4.198 4.Z98 4.386 4.464 4.533 4.594 4.647 4.694 4.735 4.771 0 0.694 1.852 3.299 4.906 6.581 8.255 9.883 11.434 12.887. 14.233 15.467 16.588 17.601 18.509 19.321 20.042 20.680 21.244 21.739 22.174 22.555 22.887 23.176 23.428 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 23.646 23.835 23.999 24.141 24.263 26 27 28 29 30 24.368 24.459 24.537 24.604 24.661 24.847 24.932 24.970 24.987 24.994 31 32 33 34 35 n Compound Amount Factor Find F Given P FIP 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 1.200 1.440 1.728 2.074 2.488 2.986 3.583 4.300 5.160 6.192 7.430 8.916 10.699 12.839 15.407 18.488 22.186 26.623 31.948 38.338 46.005 55.206 66.247 79.497 95.396 114.475 .8333 .6944 .5787 .4823 .4019 .3349 .2791 .2326 .1938 .1615 .1346 .1122 .0935 .0779 .0649 .0541 .0451 .0376 .0313 .0261 .0217 .0181 .0151 .0126 .0105 .00874 1.0000 .4545 .2747 .1863 .1344 .1007 .0774 .0606 .0481 .0385 .0311 .0253 .0206 .0169 .0139 .0114 .00944 .00781 .00646 .00536 .00444 .00369 .00307 .00255 .00212 .00176 1.2000 .6545 .4747 .3863 .3344 .3007 .2774 .2606 .2481 .2385 .2311 .2253 .2206 .2169 .2139 .2114 .2094 .2078 .2065 .2054 .2044 .2037 .2031 .2025 .2021 .2018 27 28 29 30 137.371 164.845 197.814 237.376 .00728 .00607 .00506 .00421 .2015 .2012 .2010 .2008 681.853 819.223 984.068 1 181.9 4.964 4.970 4.975 4.979. 31 32 33 34 35 284.852 341.822 410.186 492.224 590.668 .00351 .00293 .00244 .00203 .00169 .00068 .00027 .00011 .00004 .00002 .00147 .00122 .00102 .00085 .00070 .00059 .00049 .00041 .00034 .2007 .2006 .2005 .2004 .2003 1 419.3 1 704.1 2045.9 2 456.1 2 948.3 4.982 4.985 4.988 4.990 4.992 4.802 4.829 4.853 4.873 4.891 4.906 4.919 4.931 4.941 .2001 .2001 .2000 .2000 .2000 7343.9 18281.3 45 497.2 113219.0 281 732.6 4.997 4.999 4.999 5.000 5.000 4.973 4.988 4.995 4.998 4.999 40 45 50 55 60 1 469.8 3 657.3 9 100.4 22 644.8 56347.5 .00014 .00005 .00002 .00001 Compound Amount Factor Find F Given A FIA 1.000 2.200 3.640 5.368 7.442 9.930 12.916 16.499 20.799 25.959 32.150 39.581 48.497 59.196 72.035 87.442 105.931 128.117 154.740 186.688 225.026 271.031 326.237 392.484 471.981 567.377 -- - - . 40 45 50 55 60 'J '.," " ,', " ',f '"!. "/ COMPOUND INTERESTTABLES 583 I I 5% 25 Compound Interest Factors Single Payment Compound Amount Factor Find F Given P FIP 1.250 1.563 1.953 2.441 3.052 Sinking Fund Factor Find A Given F AIF Capital Recovery Factor Find A Given P AlP .8000 .6400 .5120 .4096 .3277 1.0000 .4444 .2623 .1734 .1218 1.2500 .6944 .5123 .4234 .3718 .2621 .2097 .1678 .1342 .1074 .3388 .3163 .3004 .2888 .2801 % Present Worth Factor Find P Given A PIA Gradient Uniform Series Find A Given G AIG 0.800 1.440 1.952 2.362 2.689 0 0.444 0.852 1.225 1.563 0 0.640 1.664 2.893 4.204 1 2 3 4 5 11.259 15.073 19.842 25.802 33.253 2.951 3.161 3.329 3.463 3.571 1.868 2.142 2.387 2.605 2.797 5.514 6.773 7.947 9.021 9.987 6 7 8 9 10 42.566 54.208 68.760 86.949 109.687 3.656 3.725 3.780 3.824 3.859 2.966 3.115 3.244 3.356 3.453 10.846 11.602 12.262 12.833 13.326 11 12 13 14 15 138.109 173.636 218.045 273.556 342.945 3.887 3.910 3.928 3.942 3.954 13.748 14.108 14.415 14.674 14.893 3.963 3.970 3.976 3.981 3.985 3.537 3.608 3.670 3.722 3.767 3.805 3.836 3.863 3.886 3.905 .15.078 15.233 15.362 15.471 15.562 16 17 18 19 20. 21 22 23 24 25 3.992 3.921 3.935 3.946 15.637 15.700 15.752 26 27 28 Compound Amount Factor Find F Given A FIA 1.000 2.250 3.813 5.766 8.207 Gradient Present Worth Find P Given G PIG 1,1 n 3.815 4.768 5.960 7.451 9.313 11.642 14.552 18.190 22.737 28.422 .0859 .0687 .0550 .0440 .0352 .0888 .0663 .0504 .0388 .0301 .0235 .0184 .0145 .0115 .00912 35.527 44.409 55.511 69.389 86.736 .0281 .0225 .0180 .0144 .0115 .00724 .00576 .00459 .00366 .00292 .2735 .2684 .2645 .2615 .2591 .2572 .2558 .2546 .2537 .2529 108.420 135.525 169.407 211.758 264.698 .00922 .00738 .00590 .00472 .00378 .00233 .00186 .00148 .00119 .00095 .2523 .2519 .2515 .2512 .2509 429.681 538.101 673.626 843.033 1 054.8 330.872 413.590 516.988 .00302 .00242 .00193 .00076 .00061 .00048 .2508 .2506 .2505 1 319.5 1 650.4 2 064.0 646.235 807.794. .00155 .00124 .00039 .00031 .2504 .2503 2 580.9 3 227.2 3.994 3.995 3.955 3.963 15.796 15.832 29 .30 .00099 .00025 .2502 4035.0 3.996 3.969 15.861 31 1 009.7 t: Arithmetic Gradient Uniform Payment Series Present Worth Factor Find P Given F PIF .i :l' 3.988 . 3.990 1 262.2 .00079 .00020 .2502 5044.7 3.997 3.975 15.886 32 1577.7 1 972.2 2 465.2 .00063 .00051 .00041 .00016 .00013 .00010 .2502 .2501 .2501 6 306.9 7 884.6 9 856.8 3.997 3.998 3.998 3.979 3.983 3.986 15.906 15.923 15.937 33 34 35 7 523.2 22 958.9 70 064.9 213 821.2 .00013 .00004 .00001 .00003 .00001 .2500 .2500 .2500 .2500 30088.7 91 831.5 280255.7 855 280.7 3.999 4.000 4.000 4.000 3.995 3.998 3.999 4.000 15.977 15.991 15.997 15.999 40 45 50 55 ,'.11 J'I ' ,II Iii I ;, i r 'I j :i r " "p I , ., I ,. ',.. Ii " , I I I I ).1 .... 584 COMPOUND INTEREST TABLES 300/0 Single Payment n 1 2 3 4 5 6 300/0 CompoundInterest Factors Compound Amount Factor Find F Given P FIP 1.300 1.690 2.197 2.856 3.713 4.827 Arithmetic Gradient Uniform Payment Series Present Worth Factor Find P Given F PIF Sinking Fund Factor Find A Given F AIF Capital Recovery Factor Find A Given P AlP .7692 .5917 .4552 .3501 .2693 .2072 1.0000 .4348 .2506 .1616 .1106 .0784 1.3000 .7348 .5506 .4616 .4106 Compound Amount Factor Find F Given A FIA 1.000 2.300 3.990 6.187 9.043 Present Worth Factor Find P Given A PIA Gradient Uniform Series Find A Given G AIG Gradient Present Worth Find P Given G PIG n 0.769 1.361 1.816 2.166 2.436 0 0.435 0.827 1.178 1.490 0 0.592 1.502 2.552 3.630 1 2 3 4 5 7 8 9 10 6.275 8.157 10.604 13.786 .1594 .1226 .0943 .0725 .0569 .0419 .0312 .0235 .3784 .3569 .3419 .3312 .3235 12.756 17.583 23.858 32.015 42.619 2.643 2.802 2.925 3.019 3.092 1.765 2.006 2.216 2.396 2.551 4.666 5.622 6.480 .7.234 7.887 11 12 13 14 15 17.922 23.298 30.287 39.374 51.186 66.542 86.504 112.455 146.192 190.049 .0558 .0429 .0330 .0254 .0195 .0177 .0135 .0102 .00782 .00598 .3177 .3135 .3102 .3078 .3060 56.405 74.327 97.625 127.912 167.286 3.147 3.190 3.223 3.249 3.268 2.683 2.795 2.889 2.969 3.034 8.445 8.917 9.314 9.644 9.917 6 7 8 9 -10 11 12 13 14 15 .0150 .0116 .00889 .00684 .00526 .00458 .00351 .00269 .00207 .00159 218.472 285.014 371.518 483.973 630.165 3.283 3.295 3.304 3.311 3.316 3.089 3.135 3.172 3.202 3.228 10.143 10.328 10.479 10.602 10.702 16 17 18 19 20 820.214 1 067.3 1 388.5 1 806.0 2 348.8 3.320 3.323 3.325 3.327 3.329 3.330 3.331 3.331 3.332 3.332 3.248 3.265 3.278 3.289 3.298 10.783 10.848 . 10.901 10.943 10.977 21 22 23 24 25 3.305 3.311 3.315 3.319 3.322 11.005 11.026 11.044 11.058 11.069 26 27 28 29 30 3.324 3.326 3.328 3.329 3.330 11.078 11.085 11.090 11.094 11.098 31 32 33 34 35 3.332 3.333 11.107 11.110 40 45 16 17 18 19 20 . 21 22 23 24 25 26 27 28 2Y 30 247.064 321.184 417.539 542.800 705.640 .00405 .00311 .00239 .00184 .00142 .00122 .00094 .00072 .00055 .00043 .3046 .3035 .3027 .3021 .3016 .3012 .3009 .3007 .3006 .3004 917.332 1 192.5 1 550.3 2 015.4 2 620.0 .00033 .00025 .00019 .00015 .00011 .3003 .3003 .3002 .3001 .3001 5 164.3 6714.6 8 730.0 31 32 33 34 35 3 406.0 4427.8 5756.1 7 483.0 9 727.8 .00109 .00084 .00065 .00050 . .00038 .00029 .00023 .00017 .00013 .00010 .00009 .00007 .00005 .00004 .00003 .300 1 .3001 .3001 .3000 .3000 11 350.0 14756.0 19 183.7 24939.9 32 422.8 40 45 36 118.8 134 106.5 .00003 .00001 .00001 .3000 .3000 3 054.4 3 971.8 . 120392.6 447018.3 3.332 3.333 3.333 3.333 3.333 3.333 3.333 '". - -- - ._---------- '. 1! ifJ I COMPOUND INTERESTTABLES 5% 35 % CompoundInterestFactors Compound Amount Factor Find F Given P FIP 1.350 1.822 2.460 3.322 4.484 6.053 8.172 11.032 14.894 20.107 27.144 36.644 49.470 66.784 90.158 121.714 164.314 221.824 299.462 404.274 545.769 736.789 994.665 1 342.8 1 812.8 2447.2 3 303.8 4460.1 021.1 " 128.5 10 973.5 14 814.3 19 999.3 26 999.0 36448.7 Arithmetic Gradient Uniform Payment Series Single Payment 585 Present Worth Factor Find P Given F PIF Sinking Fund Factor Find A Given F AIF Capital Recovery Factor Find A Given P AlP Compound Amount Factor Find F Given A FIA Present Worth Factor Find P Given A PIA Gradient Uniform Series Find A Given G AIG Gradient Present Worth Find P Given G PIG n .7407 .5487 .4064 .3011 .2230 .1652 .1224 .0906 .0671 .0497 .0368 .0273 .0202 .0150 .0111 .00822 .00609 .00451 .00334 .00247 .00183 .00136 .00101 .00074 .00055 .00041 .00030 .00022 .00017 1.0000 .4255 .2397 .1508 .1005 .0693 .0488 .0349 .0252 .0183 .0134 .00982 .00722 .00532 .00393 .00290 .00214 .00158 .00117 .00087 .00064 .00048 .00035 .00026 .00019 .00014 .00011 .00008 .00006 1.3500 .7755 .5897 .5008 .4505 .4193.3988 .3849 .3752 .3683 .3634 .3598 .3572 .3553 .3539 .3529 .3521 .3516 .3512 .3509 .3506 .3505 .3504 .3503 .3502 .3501 .3501 .3501 .3501 1.000 2.350 4.173 6.633 9.954 14.438 20.492 28.664 39.696 54.590 74.697 101.841 138.485 187.954 254.739 344.897 466.611 630.925 852.748 1 152.2 1 556.5 2 102.3 2839.0 3 833.7 5176.5 6989.3 9 436.5 12740.3 17200.4 0.741 1.289 1.696 1.997 2.220 2.385 2.508 2.598 2.665 2.715 2.752 2.779 2.799 2.814 2.825 2.834 2.840 2.844 2.848 2.850 2.852 2.853 2.854 2.855 2.856 2.856 2.856 2.857 2.857 0 0.426 0.803 1.134 1.422 1.670 1.881 2.060 2.209 2.334 2.436 2.520 2.589 2.644 2.689 2.725 2.753 2.776 2.793 2.808 2.819 2.827 2.834 2.839 2.843 2.847 2.849 2.851 2.852 0 0.549 1.362 2.265 3.157 3.983 4.717 5.352 5.889 6.336 6.705 7.005 7.247 7.442 7.597 7.721 7.818 7.895 7.955 8.002 8.038 8.067 8.089 8.106 8.119 8.130 8.137 8.143 8.148 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 .00012 .00004 2.857 2.857 2.857 2.857 2.857 2.857 2.853 8.152 30. .00003 .00002 .00002 .00001 .00001 .3500 .3500 .3500 .3500 .3500 .3500 23 221.6 .00009 .00007 .00005 .00004 .00003 2.854 2.855 2.855 2.856 2.856 8.154 8.157 8.158 8.159 8.160 31 32 33 34 35 31 350.1 42323.7 57137.9 77 137.2 104 136.3 - --- -- - I I>li jiii' - - --- COMPOUNDINTEREST TABLES 587 450/0 SinglePayment n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ZO U !2 !3 !4 6 7 !8 !9 10 11 12 i3 4 .5 450/0 CompoundInterestFactors UniformPaymentSeries - ArithmeticGradient Compound Amount Factor Find F Given P FIP Present Worth Factor Find P Given F PIF Sinking Fund Factor Find A Given F AIF Capital Recovery Factor Find A Given P AlP Compound Amount Factor Find F Given A FIA Present Worth Factor Find P Given A PIA Gradient Uniform Series Find A Given G AIG Gradient Present Worth Find P Given G PIG 1.450 2.103 3.049 4.421 6.410 9.294 13.476 19.541 28.334 41.085 59.573 86.381 125.252 181.615 263.342 381.846 553.677 802.831 1164.1 1 688.0 2447.5 3 548.9 5 145.9 7461.6 10 819.3 15 688.0 22747.7 32 984.1 47 826.9 69349.1 100556.1 145806.4 211 419.3 306 558.0 444 509.2 .6897 .4756 .3280 .2262 .1560 .1076 .0742 .0512 .0353 .0243 .0168 .0116 .00798 .00551 .00380 .00262 .00181 .00125 .00086 .00059 .00041 .00028 .00019 .00013 .00009 .00006 .00004 .00003 .00002 .00001 .00001 .00001 1.0000 .4082 .2197 .1316 .0832 .0543 .0361 .0243 .0165 .0112 .00768 .00527 .00362 .00249 .00172 .00118 .00081 .00056 .00039 .00027 .00018 .00013 .00009 .00006 .00004 .00003 .00002 .00001 .00001 .00001 1.4500 .8582 .6697 .5816 .5332 :5043 .4861 .4743 .4665 .4612 .4577 .4553 .4536 .4525 .4517 .4512 .4508 .4506 .4504 .4503 .4502 .4501 .4501 .4501 .4500 .4500 .4500 .4500 .45QO .4500 .4500 .4500 .4500 .4500 .4500 1.000 2.450 4.553 7.601 12.022 18.431 27.725 41.202 60.743 89.077 130.162 189.735 276.115 401.367 582.982 846.325 1 28.2 1 781.8 2584.7 3748.8 5 436;7 7 884.3 11433.2 16579.1 24040.7 34 860.1 50548.1 73 295.8 106279.9 154 106.8 223 455.9 324 012.0 469818.5 681237.8 987 795.9 0.690 1.165 1.493 1.720 1.876 1.983 2.057 2.109 2.144 2.168 2.185 2.196 2.204 2.210 2.214 2.216 2.218 2.219 2.220 2.221 2.221 2.222 2.222 2.222 2.222 2.222 2.222 2.222 2.222 2.222 2.222 2.222 2.222 2.222 2.222 0 0.408 0.758 1.053 1.298 1.499 1.661 1.791 1.893 1.973 2.034 2.082 2.118 2.145 2.165 2.180 2.191 2.200 2.206 2.210 2.214 2.216 2.218 2.219 2.220 2.221 2.221 2.221 2.222 2.222 2.222 2.222 2.222 2.222 2.222 0 0.476 1.132 1.810 2.434 2.972 3.418 3.776 4.058 4.277 4.445 4.572 4.668 4.740 4.793 4.832 4.861 4.882 4.898 4.909 4.917 4.923 4.927 4.930 4.933 4.934 4.935 4.936 4.937 4.937 4.938 4.938 4.938 4.938 4.938 n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 . t1 ".f;' .:, 15 16 17 18 19 20 I ,'I III 24 25 26 27 28 29 30 31 32 33 34 35 .I i: I' I I ' I ' , 'I I ;I' " '" -.--- -- ---- 588 COMPOUND INTEREST TABLES 50% Single Payment Compound Amount Factor Find F Given P FIP n 1 2 3 4 5 6 7 8 lu 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 3u 31 32 50% CompoundInterestFactors 1.500 2.250 3.375 5.063 7.594 11.391 17.086 25.629 38.443 57.665 86.498 129.746 194.620 291.929 437.894 656.814 985.261 1 477.9 2216.8 3 325.3 4987.9 7481.8 11 222.7 16834.1 25 251.2 . 37876.8 56815.1 85 222.7 127834.0 191751.1 287 626.6 431439.9 Arithmetic Gradient Uniform Payment Series Present Worth Factor Find P Given F PIF Sinking Fund Factor Find A Given F AIF Capital Recovery Factor Find A Given P AlP Compound Amount Factor Find F Given A FIA Present Worth Factor Find P Given A PIA Gradient Uniform Series Find A Given G AIG Gradient Present Worth Find P Given G PIG n .6667 .4444 .2963 .1975 .1317 .0878 .0585 .0390 .0260 .0173 .0116 .00771 .00514 .00343 .00228 .00152 .00101 .00068 .00045 .00030 .00020 .00013 .00009 .00006 .00004 .00003 .00002 .00001 .00001 .00001 1.0000 .4000 .2105 .1231 .0758 .0481 .0311 .0203 .0134 .00882 .00585 .00388 .00258 .00172 .00114 .00076 .00051 .00034 .00023 .00015 .00010 .00007 .00004 .00003 .00002 .00001 .00001 .00001 1.5000 .9000 .7105 .6231 .5758 .5481 .5311 .5203 .5134 .5088 .5058 .5039 .5026 .5017 .5011 .5008 .5005 .5003 .5002 .5002 .5001 .5001 .5000 .5000 .5000 .5000 .5000 .5000 .5000 .5000 .5000 .5000 1.000 2.500 4.750 8.125 13.188 20.781 32.172 49.258 74.887 113.330 170.995 257.493 387.239 581.859 873.788 1 311.7 1968.5 2 953.8 4431.7 6648.5 9 973.8 14961.7 22 443.5 33 666.2 50 500.3 75751.5 113628.3 170443.4 255666.1 383500.1 575251.2 862877.8 0.667 1.111 1.407 1.605 1.737 1.824 1.883 1.922 1.948 1.965 1.977 1.985 1.990 1.993 1.995 1.997 1.998 1.999 1.999 1.999 2.000 2.000 2.000 2.000 2.000 2.000 2.000 2.000 2.000 2.000 2.000 2.000 0 0.400 0.737 1.015 1.242 1.423 1.565 1.675 1.760 1.824 1.871 1.907 1.933 1.952 1.966 1.976 1.983 1.988 1.991 1.994 1.996 1.997 1.998 1.999 1.999 1.999 2.000 2.000 2.000 2.000 2.000 2.000 0 0.444 1.037 1.630 2.156 2.595 2.947 3.220 3.428 3.584 3.699 3.784 3.846 3.890 3.922 3.945 3.961 3.973 3.981 3.987 3.991 3.994 3.996 3.997 3.998 3.999 3.999 3.999 4.000 4.000 4.000 4.000 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 - --- - COMPOUNDINTEREST TABLES 589 0% 60% CompoundInterestFactors Single Payment Arithmetic Gradient Uniform Payment Series Compound Amount Factor Find F Given P FIP Present Worth Factor Find P Given F PIF Sinking Fund Factor Find A Given F AIF Capital Recovery Factor Find A Given P AlP Compound Amount Factor Find F Given A FIA Present Worth Factor Find P Given A PIA Gradient Uniform Series Find A Given G AIG Gradient Present Worth Find P Given G PIG 1.600 2.560 4.096 6.554 10.486 16.777 26.844 42.950 68.719 109.951 175.922 281.475 450.360 720.576 1152.9 1 844.7 2951.5 4 722.4 7 555.8 12089.3 19342.8 30 948.5 49517.6 79 228.1 126765.0 202 824.0 324 518.4 519229.5 .6250 .3906 .2441 .1526 .0954 .0596 .0373 .0233 .0146 .00909 .00568 .00355 .00222 .00139 .00087 .00054 .00034 .00021 .00013 .00008 .00005 .00003 .00002 .00001 .00001 1.0000 .3846 .1938 .1080 .0633 .0380 .0232 .0143 .00886 .00551 .00343 .00214 .00134 .00083 .00052 .00033 .00020 .00013 .00008 .00005 .00003 .00002 .00001 .00001 1.6000 .9846 .7938 .7080 .6633 .6380 .6232 .6143 .6089 .6055 .6034 .6021 .6013 .6008 .6005 .6003 .6002 .6001 .6011 .6000 .6000 .6000 .6000 .6000 .6000 .6000 .6000 .6000 1.000 2.600 5.160 9.256 15.810 26.295 43.073 69.916 112.866 181.585 291.536 467.458 748.933 1 199.3 1 919.9 3 072.8 4917.5 7 868.9 12591.3 20 147.1 32 236.3 51 579.2 82 527.6 132045.2 211 273.4 338 038.4 540 862.4 865 380.9 0.625 1.016 1.260 1.412 1.508 1.567 1.605 1.628 1.642 1.652 1.657 1.661 1.663 1.664 1.665 1.666 1.666 1.666 1.666 1.667 1.667 1.667 1.667 1.667 1.667 1.667 1.667 1.667 0 0.385 0.698 0.946 1.140 1.286 1.396 1.476 1.534 1.575 1.604 1.624 1.638 1.647 1.654 1.658 1.661 1.663 1.664 1.665 1.666 1.666 1.666 1.666 1.666 1.667 1.667 1.667 0 0.391 0.879 1.337 1.718 2.016 2.240 2.403 2.519 2.601 2.658 2.697 2.724 2.742 2.754 2.762 2.767 2.771 2.773 2.775 2.776 2.777 2.777 2.777 2.777 2.778 2.778 2.778 -- -_._- - - - n 1 2 3 4 5 6 "1 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 590 COMPOUND INTERESTTABLES Continuous Compounding-Single Payment Factors "",. ' -,\- 1 rn Compound Amount Factor ern Find F Given P FfP Present Worth Factor e-rn Find P Given F PfF .01 .02 .03 .04 .05 .06 .07 .08 .09 .10 11 .12 .13 .14 .15 .16 .17 .18 .19 .20 .21 .22 3 .24 .25 .26 .27 .28 .29 .30 .31 32 .33 1.0101 1.0202 1.0305 1.0408 1.0513 1.0618 1.0725 1.0833 1.0942 1.1052 1.1163 1.1275 1.1388 1.1503 1.1618 1.1735 1.1853 1.1972 1.2092 1.2214 1.2337 1.2461 1.2586 1.2712 1.2840 1.2969 1.3100 1.3231 1.3364 1.3499 1.3634 1.3771 1.3910 .9900 .9802 .9704 .9608 .9512 .9418 .9324 .34 1.4049 .35 .36 .37 .38 .39 .40 .41 .42 .43 .44 .45 .46 .47 .48 .49 .50 1.4191 1.4333 1.4477 1.4623 1.4770 1.4918 1.5068 1.5220 1.5373 1.5527 1.5683 1.5841 1.6000 1.6161 1.6323 1.6487 Present Worth Factor e-rn Find P Given F PfF Compound Amount Factor ern rn Find F Given P FfP .51 .52 .53 .54 .55 .56 .57 .58 .59 .60 .61 .62 .63 .64 .65 .66 .67 .68 .69 .70 .71 .72 .73 .74 .75 .76 .77 .78 .79 .80 .81 .82 .83 1.6653 1.6820 1.6989 1.7160 1.7333 1.7507 1.7683 1.7860 1.8040 1.8221 1.8404 1.8589 1.8776 1.8965 1.9155 1.9348 1.9542 1.9739 1.9937 2.0138 2.0340 2.0544 2.0751 2.0959 2.1170 2.1383 2.1598 2.1815 2.2034 2.2255 2.2479 2.2705 2.2933 .7118 .84 2.3164 .4317 .7047 .6977 .6907 .6839 .6771 .6703 .6637 .6570 .6505 .6440 .6376 .6313 .6250 .6188 .6126 .6065 .85 .86 .87 .88 .89 .90 .91 .92 .93 .94 .95 .96 .97 .98 .99 1.00 2.3396 2.3632 2.3869 2.4109 2.4351 2.4596 2.4843 2.5093 2.5345 2.5600 2.5857 2.6117 2.6379 2.6645 2.6912 2.7183 .4274 .4232 .4190 .4148 .4107 .4066 .4025 .3985 .3946 .3906 . .3867 .3829 .3791 .3753 .3716 .3679 .9231. .9139 .9048 .8958 .8869 .8781 .8694 .8607 .8521 .8437 .8353 .8270 .8187 .8106 .8025 .7945 .7866 .7788 .7711 .7634 .7558 .7483 .7408 .7334 .7261 .7189 -- ---- .6005 .5945 .5886 .5827 .5769 .5712 .5655 .5599 .5543 .5488 .5434 '.5379 .5326 .5273 .5220 .5169 .5117 .5066 .5016 .4966 .4916 .4868 . ,4819 .4771 .4724 .4677 .4630 .4584 .4538 .4493 .4449 .4404 .4360 - - ----------------.-- AASHTO. 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John Wiley & Sons, 1887. - ---- A Absorption costing, 548 .ccounting: absorption costing, 548 acid-test or quick ratio, 543~ assets, 541-42 balance sheet, 541~ cost, 541 cost of goods sold, 544 current ratio, 543 direct costs, 548 equity, 543 financial ratios from balance sheet, 543~ financial ratios from income statement, 546 fundamental accounting equation, 541 general, 541 income statement, 544-47 indkect cost allocation, 548-49 indirect costs, 548 interest coverage, 546 liabilities, 542 linking balance sheet and income statement, 546-47 net profit ratio, 546 problems with traditional cost accounting, 549-50 role of, 540-41 traditional cost accounting, 547-50 working capital, 542 After-tax: cash flows and spreadsheets, 385-86 economic analysis, 380-83 inflation, 460-62 ininimum cost life problems, 427-28 rate of return estimating, 385 replacement analysis, 420-27 Amortization schedule, 190 Analysis period: common multiple of lives, 186 for continuing requkement, 186 different from lives, 151-54 infinite, 154-57, 187-88 lives equal, 147-51, 186 other, 188-89 rate ofreturn, 219 Annual cash flow calculations, 178-82 Arithmetic gradient, 98-100 present worth factor, 100 uniform series factor, 100 Asset depreciation range (ADR), 347 Assets, 541-42 Audit results, 15 Average cost, 28 B Balance sheet, 541~ links with income statement, 546-47 Benefi~costratio,490,496-98 analysis, 274-80 graph, 247-49, 251, 252,257, 260-61, 278,279 incremental analysis, 498-504 Benefits estimating, 50 Beta approximation of mean, 308 Book cost, 36 Book value, 341 Brainstorming, 11 Breakeven analysis,285-89 .. chart, 31-32, 285 spreadsheets, 289-92 Business expenses, 374 c Capital budgeting, 518 Capital expenditures, 374 Capital gains, 356-57 and losses, 384 Capital rationing, 518-32 by present worth method~, 524-30 593 594 INDEX Capital rationing-Cont. ranking project proposals, 530-32 by rate of return, 521-24 Capitalized cost, 154-57 Cash cost, 35 Cash flow: categories, 51 computing, 62-63 diagrams, 50-52, 75-76 Challenger, 400 after-tax cash flows, 422 future, 419-20 minimum cost life, 402-4 Chance node, 316 Compound interest, 65-66 Compounding subperiods; 109 Constant dollars, 444 versus then-current dollars, 448-50 Consumer price index, 454-55 -:ontinuous alternatives benefit-cost ratio graph, 279 Continuous compounding, 115-21 capital recovery factor, 118 compound amount factor, 117 effective interest rate, 110 present worth factor, 117 series compound amount factor, 119 series present worth factor, 119 sinking fund factor, 118 Continuous uniform cash flow with continuous compounding 120-21 Corporate income taxes: after-tax analysis, 380-83 after-tax cash flows and spreadsheets, 385-86 business expenses, 374 capital expenditures, 374 capital gains and losses, 384 estimating after-tax rate of return, 385 investment tax credit, 384-85 rates, 377-78 taxable income, 374-75 20staccounting, 547-50 problems with traditional, 549-50 Cost: average, 28 book,36 of borrowed money, 475 of capital, 475-76 cash,35 depletion, 360-61 estimating, 38-49 fixed, 28 of goods sold, 544 incremental, 34-35 indexes, 44-45 life-cycle, 36-38 life-cycle committed, 37 life-cycle design change, 37 marginal, 28-29 nonrecurring, 34 opportunity, 32-33 recurring, 34 sunk, 32 variable, 28 Cost-push inflation, 441 Cutoff rate of return, 523 D Data block in spreadsheets, 555 Decision node, 316 Decision-making process, 6-15 Declining balance depreciation, 346-47 Defender, 400 after-tax cash flows, 422-27 lowest EUAC, 411-13 marginal cost data, 404-7 marginal cost data not available, 415-18 marginal costs are increasing, 407-13 marginal costs not increasing, 41;3-15 Deflation definition, 440 Demand-pull inflation, 441 Depletion, 360-62, 400 Depreciable life, 339 Depreciated versus expensed, 339 Depreciation, 546 asset disposal, 356-59 basics,338-41 calculation fundamentals, 341-42 capital gains, 356-57 declining balance, 346-47 historical methods, 342-47 historical methods compared with MACRS, 355-56 losses, 356-57 MACRS (see MACRS) recapture, 356-57 spreadsheets, 362-64 straight-line, 342-44 sum-of-years' -digits, 344-45 unit -of-production, 359-60 Descartes' rule of signs, 231-32 Deterioration due to aging, 400 Direct costs, 548 E Economic criteria fixed input, 12, 146, 182 benefi~costratio,274 fixed output, 12, 146, 182 benefi~costratio,274 .. INDEX neitherinput nor output fixed, 12-13, 146, 182 fixed benefit-cost ratio, 274 Economic decision trees', 316-22 Economic decision-making for current costs, 15-18 Effective interest rate, 109 End-of-year convention, 144 Engineering economic analysis: assumptions, 144-46 borrowed money viewpoint, 145 choosing a method, 261-62 identifying and rejecting unattractive alternatives, 520-21 ignoring inflation, 448 incorporating inflation, 448 price indexes, 456 role of, 5-6 viewpoint, 145 Equity, 543 Equivalence, 68-73 c:I'~nds on interest rate, 71-73 Estk...dted data, 304-6 range,306-8 Estimating: accuracy, 39 by analogy, 40 benefits, 50 cost indexes, 44-45 costs, 3~9 detailed, 39 difficulties, 39-40 learning curve, 47-49 per-unit model, 41 power-sizing model, 45-46 rough, 38 segmenting model, 43 semidetailed, 38-39 triangulation, 47 3xchange rates, 440-41 3xpected value, 313 3xpl'.J'I!;ed versus depreciated, 339 3xtl aarket consequences, 10, 14 ~actor: arithmetic gradient present worth, 100 arithmetic gradient uniform series, 100 geometric series present worth, 108 notation, 74 relationships between, 97-98 single payment compound amount, 74 single payment present worth, 76 uniform series capital recovery, 89 uniform series compound amount, 87 uniform series present worth, 90 uniform series sinking fund, 88 595 Financial ratios from: balance sheet, 543-44 income statement, 546 Fixed cost, 28 Functional notation, 74 Future worth analysis, 272-73 G Geometric gradient, 105-8 Geometric series present worth factor, 108 I Income statement, 544-47 links with balance sheet, 546-47 Incremental analysis, 246-63 benefi~costratio,498-504 graphical, 246-52 multiple alternatives, 250 spreadsheets, 262-63 unlimited alternatives, 258-59 Incremental cost, 34-35 Incremental rate of return, 212-18, 252-58 Indirect cost, 8, 548 Induectcostallocation,548-49 Individual income tax: capital gains and losses, 384 itemized deductions, 373 personal exemption, 373 rates, 375-76 standard deduction, 373 taxable income, 372-73 Infinite analysis period, 154-57 Inflation: actual dollars, 443 after-tax calculations, 460-62 cash flows at different rates, 456-58 causes, 440-41 composite versus commodity indexes, 453-55 constant dollars versus then-current dollars, 448-50 consumer price index, 454-55 cost of borrowed money, 481-82 definition, 440 deflation, 145-46 different rates per period, 458-59 engineering economic analysis ignoring, 448 engineering economic analysis including, 448 indexes, 450-56 producer price index, 454 rate, 441-42 real and market interest rates, 441-42 real dollars, 444 spreadsheets, 462-64 -- -- - - - - -- --- r 596 INDEX Insurance, 320-22, 323 Intangible consequ.ences, 10, 14 Intangible property, 340 Interest: compound, 65-66 formula notation, 73 on loan, 191 Interest rate: adjusting MARR for risk, 479-81 cost of capital, 475-76 difficulties in solving for, 229-40 effective, 109 equivalence, 71-73 inflation and cost of borrowed money, 481-82 MARR values used in industry, 482-83 nominal, 109 opportunity cost government, 495 opportunity cost of capital, 476-78 opportunity cost taxpayer, 495 plot versus PW, 209-10 real and market, 441-42 selecting, 474-85 selecting in public sector, 493-95 spreadsheets and opportunity cost of capital, 483-85 simple, 64-65 Internal rate of return, 204-5 modified, 238-40 Interpolation, 92, 208 Investment tax credit, 384-85 Itemized deductions, 373 J N Nominal interest rate, 109 Nonrecurring cost, 34 o Joint probabilitydistributions,311-13 Obsolescence, 338, 400 Opportunity cost, 32-33 of capital, 476-78, 495, 523 perspective for defender's first cost, 417-18 Optimistic estimate, 306-8 Outcome node, 316 L Learning curve, 47-49 Liabilities, 542 examples, 353-54 percentage derivation, 351-53 percentage tables, 349-51 property class, 348-50 recovery period, 348-50 spreadsheet using VDB, 363-64 Marginal cost, 28-29 Marginal cost data defender, 404-7 Market consequences, 10, 14 Market interest rate, 441-42 Minimum attractive rate of return: selecting, 474-85, 479 spreadsheets, 483-85 Minimum cost life: after-tax, 427-28 challenger, 402-4 Modified accelerated cost recovery system (see MACRS) Modified internal rate of return, 238-40 Money supply, 440 Most likely estimate, 306-8 Multiple alternatives, 158 Multiple rates of return, 231-32 how many, 234-38 Multiple sign changes projects with, 232-34 Mutually exclusive alternatives, 519 . Life-cycle cost, 36-38 committed, 37 design change, 37 Liquidity or speed of return versus profitability, 282, 284 Lives: different from analysis period, 151-54 equal analysis period, 147-51 Loan: balance due, 191-92 paying off early, 192 repayment plans, 67, 69-71 Lorie and Savage, 524 M MACRS, 347-53 compared to historical methods, 355-56 ------ . -.---.-.. p Payback period, 280-84 limitations, 282 Percentage depletion, 361-62 Personal exemption, 373 Personal property, 340 Pessimistic estimate, 306-8 Power-sizing model, 45-46 Present worth: analysis with benefit-cost graphs, 260-261 techniques, 147-62 Price indexes, 450-56 Probability, 308-9 INDEX joint distributions, 31.1-13 ofloss, 313 Profit and loss statement, 544 Project duration public sector, 506-07 Project financing public sector, 505 Project politics public sector, 507-09 Project proposal, 518 Property types, 340 Pruned branch, 316 Public sector: analysis viewpoint, 492-93 investment objective, 490-91 project duration, 506-07 project financing, 505 project politics, 507-09 selecting interest rate, 493-95 PW plot versus interest rate, 209-10 R Rate of return: analysis, 212 calculating, 205-9 difficulties in solving for, 229-40 Real interest rate, 441-42 Real property, 340 Recovery period, 339 Recurring cost, 34 Repeatability assumptions not acceptable, 418-19 Replacement analysis: after-tax, 420-27 decision map, 401 defender and challenger first costs, 415-18 future challengers, 419-20 reasons, 400 repeatability assumptions, 408-9 repeatability assumptions not acceptable, 418-19 spreadsheets, 429 tp"J,nique 1: increasing defender's marginal costs, 407-13 Sensitivity analysis, 285-89 spreadsheets, 289-92 Shadow prices, 10 Simple interest, 64-65 Simulation,326-29 Single payment: compound amount factor, 74 formulas, 73-81 present worth factor, 76 Sources of capital, 474 Spreadsheets, 122-26 absolute address, 556 after-tax cash flows, 385-86 to analyze loans, 190-93 amortization schedule, 190 annuity functions, 122-23 block functions, 123-24 breakeven analysis, 289-92 copycommand,555-56 cumulative investments and opportunity cost of capital, 483-85 data block, 555 depreciation, 362-64 drawing cash flow diagrams, 52 evaluating how many roots, 234-38 formulas, 555 GOAL SEEK, 262-63 graphing, 124-26 incremental analysis, 262-63 inflation, 462-64 inkoduction, 554-57 present worth, 162-64 rate of return analysis, 220-21 relative address, 556 replacement analysis, 429 sensitivity analysis, 289-92 Standard deduction, 373 Standard deviation, 322 Statistically independent, 311 Skaight-line depreciation, 342-44 Sum-of-years' -digits depreciation, 344-45 Sunk cost, 32, 145 . technique 2: defender's marginal costs not increasing, 413-15 technique 3: defender's marginal cost data not available, 415-18 Risk,320-24 standard deviation, 322 of unplanned replacement, 400 versus return, 324-26 Rule of 78's, 137 s 597 T Tax rates: combined federal and state, 379 corporate, 377-78 individual, 375-76 selecting for economy studies, 380 Taxable income: firms, 374-75 individual,372-73 Time value of money, 64 u Uniform series: capital recovery factor, 89 compound amount factor, 87 formulas, 86-90 -; ~.. ""r. I : ~ l 598 INDEX Crusader advanced field artillery system, 516-17 green design, 270-71 inkjet cartridges, 176-77 RFID chips, 202-3 Superfund, 302-3 Tappan Zee Bridge, 244-45 tobacco bonds, 60-61 Webvan, 26-27 wind turbines, 370-71 World Trade Center, 2-3 Uniform series: capital recovery factor-Cont. present worth factor, 90 sinking fund factor, 88 v Variable cost, 28 Vignette: ABB and activity-based costing, 538-39 agricultural pollution, 488-89 Anne Scheiber's bonanza, 84-85 Austin Powers baseball going broke, 336-37 Boeing versus Airbus, 142-43 BP and Russian oil, 472-73 chip making fabrication plant, 398-99 w Work breakdown structure, 44 Working capital, 542 -- Economic Criteria Method of Analysis ,,-- PRESENT WORTH ANNUAL CASH FLOW FUTURE WORTH BENEFIT-COST RATIO RATE OF RETURN Neither Input Nor Output Fixed Fixed Input Fixed Output Maximize PW of Benefits Minimize PW of Costs Maximize Equivalent Uniform Annual Benefits (EUAB) Minimize Equivalent Uniform Annual Cost (EUAC) Maximize FW of Benefits Minimize FW of Costs Maximize (FW of Benefits-FW of Costs), or Maximize Net Future Worth Maximize Benefit-Cost Ratio Two Alternatives: Compute the incremental Maximize Benefit-Cost Ratio Maximize (PW of Benefits-PW of Costs), or Maximize Net Present Worth Maximize (EUAB-EUAC) Benefit-Cost ratio (tlB / tlC) on the increment of investment between the alternatives. If tlB / tlC :::: 1, choose higher-cost alternative; if not, choose lower-cost alternative. Three or more Alternatives: Iricremental analysis is required (see Ch. 9).. Two Alternatives: Compute the incremental rate of return (tlROR) on the increment of investment between the alternatives. If tlROR ::::minimum attractive rate of return, choose the higher-cost alternative; if not, choose lower-cost alternative. Three or more Alternatives: Incremental analysis is required (see Ch. 8). -- I. ~.~ . ..1 ! It. T ow in a ninth edition, Engineering Economic Analysis offers comprehensive coverage of financial and economic deci~ sion making for eng~neering projects, with an emphasis on problem solving, life.:cycle costs, and the time value of ley. The spreadsheet material from the previous edition has been expanded, allowing students to create and analyze 'e realistic cash-flow models. The authors' concise, accessible writing style and practical emphasis make this text ideal mdergraduate engineering economy courses. ruRESOFTHENINTHEDmON I. ~wpedagogical features include: Chapter-opening vignettes Chapter objectives Ie new two-color interior design will improve readability, generate student interest, and facilitate comprehension of ~material. new chapter has been added: Chapter 18, Accounting and Engineering Economy. lapter 10, Probability and Uncertainty, has been completely rewritten to emphasize how to make good choices by nsidering the uncertainty that is part of every engineering economy application. Iapt- 13, Replacement Analysis, has been rewritten to clarify the comparison of existing assetswith newer alternatives. 'peh..uX7A, Difficulties Solving for an Interest Rate, has been thoroughly revised to use the power of spreadsheets' to ve problems. .d-of-chapter problems are reorganized and updated throughout. :ompanion website is available: www.oup.com/us/engineeringeconomy. ANCED STUDENT SUPPORT :aged with every copy of the text: Idy Guide by Donald G. Newnan and Edward Wheeler ) featuring interactive Excel@spreadsheets and tutorials ANCEDINSTRUcroRSUPPORT ) containing exam files by Meenakshi Sundaram, Excel@spreadsheets, and PowerPoin~ slides (ISBN 0-19-517423-2) lutions Manual (ISBN 0-19-517424-0) ISBN 0-19-516807-0 -- - --