Queueing theory

ADVANCED
OPERATIONS
RESEARCH
By: -
Hakeem–Ur–Rehman
IQTM–PU 1
RA O
QUEUEING THEORY

QUEUEING THEORY: AN INTRODUCTION
 Queuing theory is one of the most widely used quantitative analysis
techniques. The three basic components are: Arrivals, Service facilities &
Actual waiting line
7:00 7:10 7:20 7:30 7:40 7:50 8:00
Patient
Arrival
Time
Service
Time
1
2
3
4
5
6
7
8
9
10
11
12
0
5
10
15
20
25
30
35
40
45
50
55
4
4
4
4
4
4
4
4
4
4
4
4
A SOMEWHAT ODD SERVICE PROCESS

Patient
Arrival
Time
Service
Time
1
2
3
4
5
6
7
8
9
10
11
12
0
7
9
12
18
22
25
30
36
45
51
55
5
6
7
6
5
2
4
3
4
2
2
3
Time
7:10 7:20 7:30 7:40 7:50 8:007:00
Patient 1 Patient 3 Patient 5 Patient 7 Patient 9 Patient 11
Patient 2 Patient 4 Patient 6 Patient 8 Patient 10 Patient 12
0
1
2
3
2 min. 3 min. 4 min. 5 min. 6 min. 7 min.
Service times
Numberofcases
A MORE REALISTIC SERVICE
PROCESS

7:00 7:10 7:20 7:30 7:40 7:50
Inventory
(Patients at
lab)
5
4
3
2
1
0
8:00
7:00 7:10 7:20 7:30 7:40 7:50 8:00
Wait time
Service time
Patient
Arrival
Time
Service
Time
1
2
3
4
5
6
7
8
9
10
11
12
0
7
9
12
18
22
25
30
36
45
51
55
5
6
7
6
5
2
4
3
4
2
2
3
VARIABILITY LEADS TO WAITING TIME

Input:
• Unpredicted
Volume swings
• Random arrivals
(randomness is the rule,
not the exception)
• Incoming quality
• Product Mix
Resources:
• Breakdowns / Maintenance
• Operator absence
• Set-up times
Tasks:
• Inherent variation
• Lack of SOPs
• Quality (scrap / rework)
Routes:
• Variable routing
• Dedicated machines
Buffer
Processing
Especially relevant in service operations
(what is different in service industries?):
• emergency room
• air-line check in
• call center
• check-outs at cashier
VARIABILITY: WHERE DOES IT
COME FROM?

TEST FOR STATIONARY ARRIVAL
Call
Arrival
Time, ATi
1
2
3
4
5
6
7
6:00:29
6:00:52
6:02:16
6:02:50
6:05:14
6:05:50
6:06:28
Time6:01 6:02 6:03 6:04 6:05 6:066:00
Call 1
The concept of inter-arrival times
Call 2 Call 3 Call 4 Call 5 Call 6 Call 7
IA1 IA2 IA3 IA4 IA5 IA6
Inter-Arrival
Time, IAi=ATi+1 -ATi
00:23
01:24
00:34
02:24
00:36
00:38
0
20
40
60
80
100
120
140
160
0:15
2:00
3:45
5:30
7:15
9:00
10:45
12:30
14:15
16:00
17:45
19:30
21:15
23:00
Seasonality over the course of a day
Time
Number of customers
Per 15 minutes
0
20
40
60
80
100
120
140
160
0:15
2:00
3:45
5:30
7:15
9:00
10:45
12:30
14:15
16:00
17:45
19:30
21:15
23:00
Time
Number of customers
Per 15 minutes

TEST FOR STATIONARY ARRIVAL (Cont…)
Test for stationary arrivals
0
6:00:00 7:00:00 8:00:00 9:00:00 10:00:00
0
100
200
300
400
500
600
700
Cumulative
Customers
6:00:00 7:00:00 8:00:00 9:00:00 10:00:006:00:00 7:00:00 8:00:00 9:00:00 10:00:00
Time
Expected arrivals
if stationary
Actual, cumulative
arrivals
Test for stationary arrivals
0
7:15:00 7:18:00 7:21:00 7:24:00 7:27:00 7:30:00
0
10
20
30
40
50
60
70
7:15:00 7:18:00 7:21:00 7:24:00 7:27:00 7:30:00
Time
Cumulative
Customers
Stationary
Arrivals?
How to analyze a demand / arrival process?
YES NO
Break arrival process up into
smaller time intervals
YES
Exponentially distributed inter-
arrival times?
NO
 Compute a: average interarrival time
 CVa=1
 Compute a: average interarrival time
 CVa= St.dev. of interarrival times / a

Blocked calls
(busy signal)
Abandoned calls
(tired of waiting)
Calls
on Hold
Sales reps
processing
calls
Answered
Calls
Incoming
calls
Call center
Financial consequences
Lost throughput Holding cost
Lost goodwill
Lost throughput (abandoned)
$$$ Revenue $$$Cost of capacity
Cost per customer
 At peak, 80% of calls
dialed received a busy
signal.
 Customers getting
through had to wait on
average 10 minutes
 Extra telephone
expense per day for
waiting was $25,000.
AN EXAMPLE OF A SIMPLE QUEUING
SYSTEM

ARRIVAL CHARACTERISTICS:
POISSON DISTRIBUTION
!
P(X)
X
e x


.00
.05
.10
.15
.20
.25
.30
.35
0 1 2 3 4 5 6 7 8 9 10 1
1X
P(X)P(X),  = 2
.00
.05
.10
.15
.20
.25
.30
0 1 2 3 4 5 6 7 8 9 10 1
1X
P(X)
P(X),  = 4

SERVICE TIME CHARACTERISTICS:
EXPONENTIAL DISTRIBUTION
Probability
(forIntervalsof1Minute)
30 60 90 120 150 180
Average Service Time of 1 Hour
Average Service Time of 20 Minutes
X
MinutePerServedNumberAverageμ
0μ0,for xμef(x) μx

 

KENDALL NOTATION FOR
QUEUING MODELS
Kendall notation consists of a basic three-
symbol form.
Arrival Service Time Number of Service
Distribution Distribution Channels Open
Where,
M = Poisson distribution for the number of occurrences
(or exponential times)
D = Constant (deterministic rate)
G = General distribution with mean and variance known
M/M/2
Single channel with Poisson arrivals and
exponential service times and two channels

Average flow
time T
Theoretical Flow Time
Utilization 100%
Increasing
Variability
Waiting Time Formula
Service time factor
Utilization factor
Variability factor







 








21
22
pa CVCV
nutilizatio
nutilizatio
TimeActivityqueueinTime
Inflow Outflow
Inventory
waiting Iq
Entry to system DepartureBegin Service
Time in queue Tq Service Time p
Flow Time T=Tq+p
Flow rate
THE WAITING TIME FORMULA

THE CONCEPT OF POOLING
Independent Resources
2x(m=1)
Pooled Resources
(m=2)
The concept of pooling
How pooling can reduce waiting time
0.00
10.00
20.00
30.00
40.00
50.00
60.00
70.00
60% 65%
m=1
m=2
m=5
m=10
70% 75% 80% 85% 90% 95%
Waiting
Time Tq
[sec]
Utilization u








 


















21
221)1(2
pa
m
CVCV
nutilizatio
nutilizatio
m
timeActivity
queueinTime
Waiting Time Formula for Multiple (m) Servers
Inflow Outflow
Inventory
waiting Iq
Entry to system DepartureBegin Service
Time in queue Tq Service Time p
Flow Time T=Tq+p
Inventory
in service Ip
Flow rate
WAITING TIME FORMULA FOR MULTIPLE,
PARALLEL RESOURCES

1. Collect the following data:
- number of servers, m
- activity time, p
- interarrival time, a
- coefficient of variation for interarrival (CVa ) and processing time (CVp )
2. Compute utilization: u= ma
p

3. Compute expected waiting time
Tq







 


















21
221)1(2
pa
m
CVCV
nutilizatio
nutilizatio
m
timeActivity
4. Based on Tq , we can compute the remaining performance measures as
Flow time T=Tq+p
Inventory in service Ip=m*u
Inventory in the queue=Iq=Tq/a
Inventory in the system I=Ip+Iq
QUEUEING THEORY
FORMULAS

qp
p
qq
q
pa
m
q
III
muI
T
a
I
pTT
CVCV
u
u
m
p
T
am
p
p
m
au











 



















*
*
1
21
1*
1
221)1(2
Server
Inflow Outflow
Inventory
waiting Iq
Entry to
system
DepartureBegin
Service
Waiting Time Tq Service Time p
Flow Time T=Tq+p
Flow unit
Inventory
in service Ip
Utilization (Note: make sure <1)
Time related measures
Inventory related measures (Flow rate=1/a)
SUMMARY OF QUEUING ANALYSIS

Customers send e-mails to a help desk of an online
retailer every 2 minutes, on average, and the standard
deviation of the inter-arrival time is also 2 minutes.
The online retailer has three employees answering e-
mails. It takes on average 4 minutes to write a
response e-mail. The standard deviation of the service
times is 2 minutes.
a. Estimate the average customer wait before being
served.
b. How many e-mails would there be, on average,
that have been submitted to the online retailer but
not yet answered?
QUESTION: ONLINE
RETAILER

My-law.com is a recent start-up trying to cater to customers in search of legal services who are
intimidated by the idea of talking to a lawyer or simply too lazy to enter a law office. Unlike
traditional law firms, My-law.com allows for extensive interaction between lawyers and their
customers via telephone and the Internet. This process is used in the upfront part of the
customer interaction, largely consisting of answering some basic customer questions prior to
entering a formal relationship.
In order to allow customers to interact with the firm's lawyers, customers are encouraged to send
e-mails to my-lawyer@My-law.com. From there, the incoming e-mails are distributed to the
lawyer who is currently “on call.” Given the broad skills of the lawyers, each lawyer can respond
to each incoming request. E-mails arrive from 8 a.m. to 6 p.m. at a rate of 10 e-mails per hour
(coefficient of variation for the arrivals is 1). At each moment in time, there is exactly one lawyer
“on call,” that is, sitting at his or her desk waiting for incoming e-mails. It takes the lawyer, on
average, 5 minutes to write the response e-mail. The standard deviation of this is 4 minutes.
λ= 10 emails per hour  Inter-arrival time (a): = 10 emails / 60 min  a = 6 min,
CVa = 1 ;
processing time: p = 5 min, CVp = 4 min / 5 min = 0.8
a. What is the average time a customer has to wait for the response to his/her e-mail, ignoring
any transmission times? Note: This includes the time it takes the lawyer to start writing the e-
mail and the actual writing time.
TS = ?
(Tq) waiting time = 5 min * [5/6 / (1-5/6)] * [(12 + 0.82)/2] = 20.5 min
Total Response Time (Ts) = waiting time + processing time = 25.5 min
b. How many e-mails will a lawyer have received at the end of a 10-hour day?
emails per day = 10 emails per hour * 10 hours = 100 emails per day
QUESTION: (My-Law.Com)

c. When not responding to e-mails, the lawyer on call is encouraged to actively
pursue cases that potentially could lead to large settlements. How much time on a
10-hour day can a My-law.com lawyer dedicate to this activity (assume the lawyer
can instantly switch between e-mails and work on a settlement)?
Idle Time = Probability that there in no customer in the system=?
idle time = (1- utilization) * 10 hours = 1.66 hours
To increase the responsiveness of the firm, the board of My-law.com proposes a new
operating policy. Under the new policy, the response would be highly standardized,
reducing the standard deviation for writing the response e-mail to 0.5 minute. The
average writing time would remain unchanged.
d. How would the amount of time a lawyer can dedicate to the search for large
settlement cases change with this new operating policy?
The average amount of time would not change, since utilization is not dependent
on the variance or standard deviation, but on the average processing and inter-
arrival time.
e. How would the average time a customer has to wait for the response to his/her e-
mail change? Note: This includes the time until the lawyer starts writing the e-mail
and the actual writing time.
processing time: p = 5 min, CVp = 0.5 min / 5 min = 0.1
waiting time = 5 min * [5/6 / (1-5/6)] * [(12 + 0.12)/2] = 12.63 min
total response time = waiting time + processing time = 17.63 min
QUESTION: (My-Law.Com) Cont…

The airport branch of a car rental company maintains a fleet of 50 SUVs.
The interarrival time between requests for an SUV is 2.4 hours, on
average, with a standard deviation of 2.4 hours. There is no indication of
a systematic arrival pattern over the course of a day. Assume that, if all
SUVs are rented, customers are willing to wait until there is an SUV
available. An SUV is rented, on average, for 3 days, with a standard
deviation of 1 day.
We approach this problem as though the rental car is the “server”.
a = 2.4 hours, p = 72 hours, CVa = (2.4/2.4) = 1,
CVp = (24/72) = 0.33, and m = 50 cars.
a. What is the average number of SUVs parked in the company's lot?
To determine the number of cars on the lot, we can look at the
utilization rate of our “servers” = (1/a) / (m/p) = (1/2.4) / (50/72) =
60%.
Therefore, on average 60% of the cars are in use or 30 cars, so on
average 20 cars are in the lot.
QUESTION: (Car Rental Company)

b. Through a marketing survey, the company has discovered that if it
reduces its daily rental price of $80 by $25, the average demand would
increase to 12 rental requests per day and the average rental duration will
become 4 days. Is this price decrease warranted? Provide an analysis!
We assume that the standard deviations DO NOT change. If the average demand
is increased to 12 rentals per day, then a = 2 hours. If the average rental
duration increases to 4 days, then p = 96 hours. These values raise the
utilization rate to (1/2) / (50/96) = 96%. This means that 48 cars are rented on
average. With the initial rate average revenue per day = $80*30 cars = $2400.
With the proposed rate average revenue per day = $55*48 = $2640. Therefore,
the company should make the proposed changes.
c. What is the average time a customer has to wait to rent an SUV? Please
use the initial parameters rather than the information in (b).
the average wait time = 0.019 hours or 1.15 minutes.
d. How would the waiting time change if the company decides to limit all
SUV rentals to exactly 4 days? Assume that if such a restriction is
imposed, the average inter-arrival time will increase to 3 hours, with the
standard deviation changing to 3 hours.
the average wait time is computed as 0.046 hours.
QUESTION: (Car Rental Company)

The following situation refers to Tom Opim, a first-year MBA student. In order to pay the
rent, Tom decides to take a job in the computer department of a local department store.
His only responsibility is to answer telephone calls to the department, most of which are
inquiries about store hours and product availability. As Tom is the only person answering
calls, the manager of the store is concerned about queuing problems.
Currently, the computer department receives an average of one call every 3 minutes,
with a standard deviation in this inter-arrival time of 3 minutes.
Tom requires an average of 2 minutes to handle a call. The standard deviation in this
activity time is 1 minute.
The telephone company charges $5.00 per hour for the telephone lines whenever they
are in use (either while a customer is in conversation with Tom or while waiting to be
helped).
Assume that there are no limits on the number of customers that can be on hold and
that customers do not hang up even if forced to wait a long time.
inter-arrival time (λ): 1 call / 3 min  a = 3 min, CVa = 3 min / 3 min = 1
processing time: p = 2 min, CVp = 1 min / 2 min = 0.5
a. For one of his courses, Tom has to read a book (The Pole, by E. Silvermouse). He can
read 1 page per minute. Tom's boss has agreed that Tom could use his idle time for
studying, as long as he drops the book as soon as a call comes in. How many pages
can Tom read during an 8-hour shift?
idle time = (1 – U) = (1 – 2/3) =(1 / 3) * 8 hours = 160 min  pages read = 160 min
* 1 page/min = 160 pages
QUESTION: (Tom Opim)

b. How long does a customer have to wait, on average, before talking to Tom?
waiting time = 2 min * [2/3 / (1-2/3)] * [(12 + 0.52)/2] = 2.5 min
c. What is the average total cost of telephone lines over an 8-hour shift? Note that
the department store is billed whenever a line is in use, including when a line is
used to put customers on hold.
The average total time a line is used per customer = average wait time + average processing
time.
 In this case, the average total time per customer = 2.5 + 2 = 4.5 minutes per customer.
 There are an average of 20 customers per hour, so the average number of minutes per
hour = 20*4.5 = 90 minutes.
 Thus, the total per hour charge = (90/60) * $5 = $7.50 per hour or $60 for 8 hours.
Another way to approach the same problem:
 Average number of callers at any given time (Is) = average number of callers on hold at any
given time (Iq) + average number of callers talking to Tom at any given time (IP) .
 We can calculate Iq = R * Tq where the flow rate = 1 call / 3 minutes and Tq = 2.5
minutes.
 Thus, Iq = 0.83 calls. The average number of callers talking to Tom must be a number
between 0 and 1, and is equal to Tom’s utilization = 0.67.
 So, the average number of callers at any given time = 0.83 + 0.67 = 1.5 callers. The
line charge for 8 hours = $5*8 = $40 per line. Therefore the total cost over an 8-hour
shift = 1.5*40 = $60.
QUESTION: (Tom Opim)

Atlantic Video, a small video rental store in Philadelphia, is open 24 hours a
day, and—due to its proximity to a major business school—experiences
customers arriving around the clock. A recent analysis done by the store
manager indicates that there are 30 customers arriving every hour, with a
standard deviation of interarrival times of 2 minutes. This arrival pattern is
consistent and is independent of the time of day. The checkout is currently
operated by one employee, who needs on average 1.7 minutes to check out a
customer. The standard deviation of this check-out time is 3 minutes, primarily
as a result of customers taking home different numbers of videos.
inter-arrival time: 30 customers per hour = 1 customer / 2 min
 a = 2 min, CVa = 1
process time: p = 1.7 min, CVp = 3 min / 1.7 min = 1.765
a. If you assume that every customer rents at least one video (i.e., has to go
to the checkout), what is the average time a customer has to wait in line
before getting served by the checkout employee, not including the actual
checkout time (within 1 minute)?
waiting time = 1.7 min * [(1.7/2)/(1-1.7/2)] * [(12 + 1.7652)/2] = 19.82 min
QUESTION: (Atlantic Video)

b. If there are no customers requiring checkout, the employee is sorting returned videos, of
which there are always plenty waiting to be sorted. How many videos can the employee sort
over an 8-hour shift (assume no breaks) if it takes exactly 1.5 minutes to sort a single video?
 utilization = 1.7 min / 2 min = 0.85
 idle time = 0.15 * 8 hrs = 72 min / 1.5 min per sort = 48 sorts
c. What is the average number of customers who are at the checkout desk, either waiting or
currently being served (within 1 customer)?
 Using R = minimum of 1/a and 1/p, we calculate R = 0.5.
 Thus, the average number of customers in line waiting I = R*Tq = 0.5*19.8 = 9.9
customers.
 In addition, an average of 0.85 customers (equal to the utilization u) are being served
at any given time. So the average number of customers at the check-out desk = 9.9 +
.85 = 10.75.
d. Now assume for this question only that 10 percent of the customers do not rent a video at all
and therefore do not have to go through checkout. What is the average time a customer has
to wait in line before getting served by the checkout employee, not including the actual
checkout time (within 1 minute)? Assume that the coefficient of variation for the arrival
process remains the same as before.
 Because only 90% of customers go through checkout, the inter-arrival time of paying
customers changes: 27 customers per hour = 1 customer / 2.22 min
 waiting time = 1.7 min * [(1.7/2.22)/(1-1.7/2.22)] * [(12 + 1.7652)/2] = 11.38 min

e. As a special service, the store offers free popcorn and sodas for customers
waiting in line at the checkout desk. (Note: The person who is currently
being served is too busy with paying to eat or drink.) The store owner
estimates that every minute of customer waiting time costs the store 75
cents because of the consumed food. What is the optimal number of
employees at checkout? Assume an hourly wage rate of $10 per hour.
 The average person waits 19.81 minutes, and 30 customers arrive in
one hour, so there are approximately 19.82*30=594.6 minutes of wait
time per hour.
 This costs the store .75*594.6 = $445.95 for wait time.
 If 2 servers are used, we can apply the formula, and using the same
methodology, calculate a cost of .75*(0.88*30) = $19.79.
 Finally, if 3 servers are used, the cost is .75*(0.162*30) = $3.65.
 Adding any more employees would not be cost effective. The most cost
effective number of employees is 3.

RentAPhone is a new service company that provides European mobile phones to
American visitors to Europe. The company currently has 80 phones available at Charles
de Gaulle Airport in Paris. There are, on average, 25 customers per day requesting a
phone. These requests arrive uniformly throughout the 24 hours the store is open.
(Note: This means customers arrive at a faster rate than 1 customer per hour.) The
corresponding coefficient of variation is 1.
Customers keep their phones on average 72 hours. The standard deviation of this time
is 100 hours.
Given that RentAPhone currently does not have a competitor in France providing
equally good service, customers are willing to wait for the telephones. Yet, during the
waiting period, customers are provided a free calling card. Based on prior experience,
RentA-Phone found that the company incurred a cost of $1 per hour per waiting
customer, independent of day or night.
To answer this question, we must first set up the problem where we
consider the phones as “servers”.
a. What is the average number of telephones the company has in its store?
 m = 80, a = 24 hours/25 customers = 0.96 hours average inter-arrival time
 p = 72 hours. So, utilization = p / (m*a) = 72/(80*.96) = 94%.
 This means that .94*80 phones = 75 phones are in use, and 5 phones are available
on average.
QUESTION: (RentAPhone)

b. How long does a customer, on average, have to wait for the phone?
 We are given that CVa = 1 and CVp = 100/72 = 1.39. Using the wait time formula, we
calculate an average wait time of 9.89 hours.
c. What are the total monthly (30 days) expenses for telephone cards?
 We first need to calculate the average number of people in the queue.
 We know that the flow rate R = demand rate = 1/.96 = 1.04 customers/hour.
 Time in the queue Tq = 9.89 hours, so Iq = Tq * R = 9.89 * 1.04 = 10.31 people in
the queue on average.
 So we can multiply $1 * 24 hours/day * 30 days * 10.31 people in the queue =
$7419.87.
d. Assume RentAPhone could buy additional phones at $1,000 per unit. Is it worth it to buy
one additional phone? Why?
 We can repeat the same calculations for m=81 and obtain Iq=7.379. The new
expenses would be 5312, thus it would pay to buy at least one additional phone.
e. How would waiting time change if the company decides to limit all rentals to exactly 72
hours? Assume that if such a restriction is imposed, the number of customers requesting a
phone would be reduced to 20 customers per day.
QUESTION: (RentAPhone) (Cont…)

Queueing theory

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Queueing theory