Introduction to Management Science
8th Edition
by
Bernard W. Taylor III
Chapter 13
Queuing Analysis
Chapter 15 - Queuing Analysis
1
Chapter Topics
Elements of Waiting Line Analysis
The Single-Server Waiting Line System
Undefined and Constant Service Times
Finite Queue Length
Finite Calling Problem
The Multiple-Server Waiting Line
Addition Types of Queuing Systems
Chapter 15 - Queuing Analysis
2
Overview
Significant amount of time spent in waiting lines by people,
products, etc.
Providing quick service is an important aspect of quality
customer service.
The basis of waiting line analysis is the trade-off between
the cost of improving service and the costs associated with
making customers wait.
Queuing analysis is a probabilistic form of analysis.
The results are referred to as operating characteristics.
Results are used by managers of queuing operations to
make decisions.
Chapter 15 - Queuing Analysis
3
Elements of Waiting Line Analysis
Waiting lines form because people or things arrive at a service
faster than they can be served.
Most operations have sufficient server capacity to handle
customers in the long run.
Customers however, do not arrive at a constant rate nor are they
served in an equal amount of time.
Waiting lines are continually increasing and decreasing in
length.and approach an average rate of customer arrivals and an
average service time, in the long run.
Decisions concerning the management of waiting lines are based
on these averages for customer arrivals and service times.
They are used in formulas to compute operating characteristics
of the system which in turn form the basis of decision making.
Chapter 15 - Queuing Analysis
4
The Single-Server Waiting Line System (1 of 2)
Components of a waiting line system include arrivals
(customers), servers, (cash register/operator), customers in
line form a waiting line.
Factors to consider in analysis:
The queue discipline.
The nature of the calling population
The arrival rate
The service rate.
Chapter 15 - Queuing Analysis
5
The Single-Server Waiting Line System (2 of 2)
Figure 15.1
The Fast Shop Market Queuing System
Chapter 15 - Queuing Analysis
6
Single-Server Waiting Line System
Component Definitions
Queue Discipline: The order in which waiting customers
are served.
Calling Population: The source of customers (infinite or
finite).
Arrival Rate: The frequency at which customers arrive at a
waiting line according to a probability distribution (frequently
described by a Poisson distribution).
Service Rate: The average number of customers that can
be served during a time period (often described by the
negative exponential distribution).
Chapter 15 - Queuing Analysis
7
Single-Server Waiting Line System
Single-Server Model
Assumptions of the basic single-server model:
An infinite calling population
A first-come, first-served queue discipline
Poisson arrival rate
Exponential service times
Symbology:
 = the arrival rate (average number of arrivals/time period)
 = the service rate (average number served/time period)
Customers must be served faster than they arrive ( < ) or
an infinitely large queue will build up.
Chapter 15 - Queuing Analysis
8
Single-Server Waiting Line System
Basic Single-Server Queuing Formulas (1 of 2)
Probability that no customers are in the queuing system:


Po  1  






Probability that n customers are in the system:
n
n


 




  
Pn     Po     1  


 






Average number of customers in system:
waiting line:
2

Lq  
     
Chapter 15 - Queuing Analysis
L 
 
and
9
Single-Server Waiting Line System
Basic Single-Server Queuing Formulas (2 of 2)
Average time customer spends waiting and being served:
W  1 L
  
Average time customer spends waiting in the queue:
Wq 

     
Probability that server is busy (utilization factor):
Probability that server is idle:
Chapter 15 - Queuing Analysis
U 

I 1U 1 

10
Single-Server Waiting Line System
Characteristics for Fast Shop Market (1 of 2)
 = 24 customers per hour arrive at checkout counter
 = 30 customers per hour can be checked out


Po  1    (1 - 24/30)






 .20 probabilit y of no customers in the system
L    24/(30 - 24)  4 customers on the avg in the system
 
2

Lq  
     
 (24)2/[30(30 - 24)]  3.2 customers on the avg in the waiting line
Chapter 15 - Queuing Analysis
11
Single-Server Waiting Line System
Characteristics for Fast Shop Market (2 of 2)
W  1  L  1/[30 - 24]
  
 0.167 hour (10 min) avg time in the system per customer
Wq 
  24/[30(30 - 24)]
     
 0.133 hour (8 min) avg time in the waiting line
U 
  24/30
 .80 probabilit y server busy, .20 probabilit y server will be idle
Chapter 15 - Queuing Analysis
12
Single-Server Waiting Line System
Steady-State Operating Characteristics
Because of steady-state nature of operating characteristics:
Utilization factor, U, must be less than one: U < 1,or
 /  < 1 and  < .
The ratio of the arrival rate to the service rate must be
less than one or, the service rate must be greater than
the arrival rate.
The server must be able to serve customers faster than
the arrival rate in the long run, or waiting line will grow
to infinite size.
Chapter 15 - Queuing Analysis
13
Single-Server Waiting Line System
Effect of Operating Characteristics (1 of 6)
Manager wishes to test several alternatives for reducing
customer waiting time:
Addition of another employee to pack up purchases
Addition of another checkout counter.
Alternative 1: Addition of an employee (raises service rate
from  = 30 to  = 40 customers per hour).
Cost $150 per week, avoids loss of $75 per week for
each minute of reduced customer waiting time.
System operating characteristics with new parameters:
Po = .40 probability of no customers in the system
L = 1.5 customers on the average in the queuing system
Chapter 15 - Queuing Analysis
14
Single-Server Waiting Line System
Effect of Operating Characteristics (2 of 6)
System operating characteristics with new parameters
(continued):
Lq = 0.90 customer on the average in the waiting line
W = 0.063 hour average time in the system per customer
Wq = 0.038 hour average time in the waiting line per customer
U = .60 probability that server is busy and customer must wait
I = .40 probability that server is available
Average customer waiting time reduced from 8 to 2.25
minutes worth $431.25 per week.
Chapter 15 - Queuing Analysis
15
Single-Server Waiting Line System
Effect of Operating Characteristics (3 of 6)
Alternative 2: Addition of a new checkout counter ($6,000
plus $200 per week for additional cashier).
 = 24/2 = 12 customers per hour per checkout counter
 = 30 customers per hour at each counter
System operating characteristics with new parameters:
Po = .60 probability of no customers in the system
L = 0.67 customer in the queuing system
Lq = 0.27 customer in the waiting line
W = 0.055 hour per customer in the system
Wq = 0.022 hour per customer in the waiting line
U = .40 probability that a customer must wait
I = .60 probability that server is idle
Chapter 15 - Queuing Analysis
16
Single-Server Waiting Line System
Effect of Operating Characteristics (4 of 6)
Savings from reduced waiting time worth $500 per week $200 = $300 net savings per week.
After $6,000 recovered, alternative 2 would provide $300 281.25 = $18.75 more savings per week.
Chapter 15 - Queuing Analysis
17
Single-Server Waiting Line System
Effect of Operating Characteristics (5 of 6)
Table 15.1
Operating Characteristics for Each Alternative System
Chapter 15 - Queuing Analysis
18
Single-Server Waiting Line System
Effect of Operating Characteristics (6 of 6)
Figure 15.2
Cost Trade-Offs for Service Levels
Chapter 15 - Queuing Analysis
19
Single-Server Waiting Line System
Solution with Excel and Excel QM (1 of 2)
Exhibit 15.1
Chapter 15 - Queuing Analysis
20
Single-Server Waiting Line System
Solution with Excel and Excel QM (2 of 2)
Exhibit 15.2
Chapter 15 - Queuing Analysis
21
Single-Server Waiting Line System
Solution with QM for Windows
Exhibit 15.3
Chapter 15 - Queuing Analysis
22
Single-Server Waiting Line System
Undefined and Constant Service Times
Constant, rather than exponentially distributed service
times, occur with machinery and automated equipment.
Constant service times are a special case of the singleserver model with undefined service times.
Queuing formulas:

Po 1 
Lq 
Wq  Lq

2


2
2
     /  

21  /  


L  Lq  
Chapter 15 - Queuing Analysis

1
W Wq  


U 
23
Single-Server Waiting Line System
Undefined Service Times Example (1 of 2)
Data: Single fax machine; arrival rate of 20 users per hour,
Poisson distributed; undefined service time with mean of 2
minutes, standard deviation of 4 minutes.
Operating characteristics:
 1 20  .33 probabilit y that machine not in use
Po 1 
30
Lq 
2


2
2
     /  

21  /  







2
2 
2


20 1/15   20 / 30






21 20 / 30

 3.33 employees waiting in line


  3.33  (20 / 30)
L  Lq  
 4.0 employees in line and using the machine
Chapter 15 - Queuing Analysis
24
Single-Server Waiting Line System
Undefined Service Times Example (2 of 2)
Operating characteristics (continued):
Wq  Lq  3.33  0.1665 hour  10 minutes waiting time
 20
1  0.1665  1  0.1998 hour
W Wq  
30
 12 minutes in the system
  20  67% machine utilizatio n
U 
30
Chapter 15 - Queuing Analysis
25
Single-Server Waiting Line System
Constant Service Times Formulas
In the constant service time model there is no variability in
service times;  = 0.
Substituting  = 0 into equations:
2 2 2 
2 
2




2
2
  /  
     /    0    /  
2



Lq 

 
 





21  /  
21  /  
21  /   2     






All remaining formulas are the same as the single-server
formulas.
Chapter 15 - Queuing Analysis
26
Single-Server Waiting Line System
Constant Service Times Example
Car wash servicing one car at a time; constant service time
of 4.5 minutes; arrival rate of customers of 10 per hour
(Poisson distributed).
Determine average length of waiting line and average
waiting time.
 = 10 cars per hour,  = 60/4.5 = 13.3 cars per hour
Lq 
(10)2
2 
 1.14 cars waiting
2 (   ) 2(13.3)(13.3 10)
Wq  Lq  1.14  0.114 hour or 6.84 minutes waiting time
 10
Chapter 15 - Queuing Analysis
27
Undefined and Constant Service Times
Solution with Excel
Exhibit 15.4
Chapter 15 - Queuing Analysis
28
Undefined and Constant Service Times
Solution with QM for Windows
Exhibit 15.5
Chapter 15 - Queuing Analysis
29
Finite Queue Length
In a finite queue, the length of the queue is limited.
Operating characteristics, where M is the maximum number
in the system:
n


1


/

  for n  M
Po 
Pn  (Po) 



1 ( /  )M 1
M 1

/

(
M

1
)(

/

)
L

1  / 
1 ( /  )M 1
W
L
 (1 PM )
Chapter 15 - Queuing Analysis
Lq  L   (1PM )
1
Wq W  
30
Finite Queue Length Example (1 of 2)
Metro Quick Lube single bay service; space for one vehicle
in service and three waiting for service; mean time between
arrivals of customers is 3 minutes; mean service time is 2
minutes; both inter-arrival times and service times are
exponentially distributed; maximum number of vehicles in
the system equals 4.
Operating characteristics for  = 20,  = 30, M = 4:
Po 
1  /   1 20 / 30  .38 probabilit y that system is empty
1 ( /  )M 1 1 (20 / 30)5

PM  (Po) 










nM










4
 (.38) 20  .076 probabilit y that system is full
30
Chapter 15 - Queuing Analysis
31
Finite Queue Length Example (2 of 2)
Average queue lengths and waiting times:
M 1

/

(
M

1
)(

/

)
L

1  / 
1 ( /  )M 1
5
(
5
)(
20
/
30
)
20
/
30
L

 1.24 cars in the system
1 20 / 30 1 (20 / 30)5
Lq  L   (1PM ) 1.24  20(1.076)  0.62 cars waiting
30
W
L
 1.24
 0.067 hours waiting in the system
 (1 PM ) 20(1.076)
1  0.067  1  0.033 hour waiting in line
Wq W  
30
Chapter 15 - Queuing Analysis
32
Finite Queue Model Example
Solution with Excel
Exhibit 15.6
Chapter 15 - Queuing Analysis
33
Finite Queue Model Example
Solution with QM for Windows
Exhibit 15.7
Chapter 15 - Queuing Analysis
34
Finite Calling Population
In a finite calling population there is a limited number of
potential customers that can call on the system.
Operating characteristics for system with Poisson arrival
and exponential service times:
1
Po 
n
N
N!   
 ( N  n)!  


n0
where N  population size, and n  1, 2,...N
n
 Po
Pn  N! 
( N  n)!










L  Lq  (1 Po)
Chapter 15 - Queuing Analysis





Lq  N 
(1 Po)
 





Wq 
Lq
( N  L)
1
W Wq  
35
Finite Calling Population Example (1 of 2)
Wheelco Manufacturing Company; 20 machines; each
machine operates an average of 200 hours before breaking
down; average time to repair is 3.6 hours; breakdown rate
is Poisson distributed, service time is exponentially
distributed.
Is repair staff sufficient?
 = 1/200 hour = .005 per hour
 = 1/3.6 hour = .2778 per hour
N = 20 machines
Chapter 15 - Queuing Analysis
36
Finite Calling Population Example (2 of 2)
Po 
1
 .652
n

20 20!  .005 
 (20  n)!.2778 


n0
Lq  20  .005 .2778 1.652  .169 machines waiting


.005
L  .169  (1.652)  .520 machines in the system
Wq 
.169
1.74 hours waiting for repair
(20 .520)(.005)
W 1.74 
1  5.33 hours in the system
.2778
System seems inadequate.
Chapter 15 - Queuing Analysis
37
Finite Calling Population Example
Solution with Excel and Excel QM (1 of 2)
Exhibit 15.8
Chapter 15 - Queuing Analysis
38
Finite Calling Population Example
Solution with Excel and Excel QM (2 of 2)
Exhibit 15.9
Chapter 15 - Queuing Analysis
39
Finite Calling Population Example
Solution with QM for Windows
Exhibit 15.10
Chapter 15 - Queuing Analysis
40
Multiple-Server Waiting Line (1 of 2)
In multiple-server models, two or more independent servers
in parallel serve a single waiting line.
Biggs Department Store service department; first-come,
first-served basis.
Chapter 15 - Queuing Analysis
41
Multiple-Server Waiting Line (2 of 2)
Figure 15.3
Customer Service
Queuing System
Chapter 15 - Queuing Analysis
42
Multiple-Server Waiting Line
Queuing Formulas (1 of 3)
Assumptions:
First-come first-served queue discipline
Poisson arrivals, exponential service times
Infinite calling population.
Parameter definitions:
 = arrival rate (average number of arrivals per time
period)
 = the service rate (average number served per time
period) per server (channel)
c = number of servers
c  = mean effective service rate for the system (must
exceed arrival rate)
Chapter 15 - Queuing Analysis
43
Multiple-Server Waiting Line
Queuing Formulas (2 of 3)
Po  
1
 probabilit y no customers in system
c


 
nc11   

1
    c


  
  



 

  c   
n
!
c
!

 

 

 n0

Pn 
n
 
n
1  Po for n  c
c!cnc 










n
 Po for n  c  probabilit y of n customers in system
Pn  1
n










c

(

/

)
   average customers in the system
L
Po  
(c 1)!(c   )2
W  L  average time customer spends in the system

Chapter 15 - Queuing Analysis
44
Multiple-Server Waiting Line
Queuing Formulas (3 of 3)
  average number of customers in the queue
Lq  L  
1  Lq  average time customer is in the queue
Wq W  

c c


1
 
Pw    
Po  probabilit y customer must wait for service
c!  c  
Chapter 15 - Queuing Analysis
45
Multiple-Server Waiting Line
Biggs Department Store Example (1 of 2)
 = 10,  = 4, c = 3
1
Po  
0  1  2   3



3(4)
 1 10 
110   1 10    1 10 

  

 4 
 4  
 4  3(4) 10
0! 4 
1
!
2
!
3
!










 .045 probabilit y of no customers
3
(
10
)(
4
)(
10
/
4
)
L
(.045)  10
4
(3 1)![3(4) 10]2
 6 customers on average in service department
W 6
10
 0.60 hour average customer time in the service department
Chapter 15 - Queuing Analysis
46
Multiple-Server Waiting Line
Biggs Department Store Example (2 of 2)
Lq  6 10
4
 3.5 customers on the average waiting to be served
Wq  3.5
10
 0.35 hour average waiting time in line per customer





3





3(4) (.045)
Pw  1 10
3! 4 3(4) 10
 .703 probabilit y customer must wait for service
Chapter 15 - Queuing Analysis
47
Multiple-Server Waiting Line
Solution with Excel
Exhibit 15.11
Chapter 15 - Queuing Analysis
48
Multiple-Server Waiting Line
Solution with Excel QM
Exhibit 15.12
Chapter 15 - Queuing Analysis
49
Multiple-Server Waiting Line
Solution with QM for Windows
Exhibit 15.13
Chapter 15 - Queuing Analysis
50
Additional Types of Queuing Systems (1 of 2)
Figure 15.4
Single Queues with
Single and Multiple
Servers in Sequence
Chapter 15 - Queuing Analysis
51
Additional Types of Queuing Systems (2 of 2)
Other items contributing to queuing systems:
Systems in which customers balk from entering
system, or leave the line (renege).
Servers who provide service in other than first-come,
first-served manner
Service times that are not exponentially distributed or
are undefined or constant
Arrival rates that are not Poisson distributed
Jockeying (i.e., moving between queues)
Chapter 15 - Queuing Analysis
52
Example Problem Solution (1 of 5)
Problem Statement: Citizens Northern Savings Bank loan
officer customer interviews.
Customer arrival rate of four per hour, Poisson distributed;
officer interview service time of 12 minutes per customer.
Determine operating characteristics for this system.
Additional officer creating a multiple-server queuing
system with two channels. Determine operating
characteristics for this system.
Chapter 15 - Queuing Analysis
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Example Problem Solution (2 of 5)
Solution:
Step 1: Determine Operating Characteristics for the SingleServer System
 = 4 customers per hour arrive,  = 5 customers per
hour are served
Po = (1 -  / ) = ( 1 – 4 / 5) = .20 probability of no
customers in the system
L =  / ( - ) = 4 / (5 - 4) = 4 customers on average in
the queuing system
Lq = 2 / ( - ) = 42 / 5(5 - 4) = 3.2 customers on
average in the waiting line
Chapter 15 - Queuing Analysis
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Example Problem Solution (3 of 5)
Step 1 (continued):
W = 1 / ( - ) = 1 / (5 - 4) = 1 hour on average in the
system
Wq =  / (u - ) = 4 / 5(5 - 4) = 0.80 hour (48 minutes)
average time in the waiting line
Pw =  /  = 4 / 5 = .80 probability the new accounts
officer is busy and a customer must wait
Chapter 15 - Queuing Analysis
55
Example Problem Solution (4 of 5)
Step 2: Determine the Operating Characteristics for the
Multiple-Server System.
 = 4 customers per hour arrive;  = 5 customers
per hour served; c = 2 servers
1
Po  
n   c  c 
n  c 11    
1    


  









n!   c!   c   
 n 0


 .429 probabilit y no customers in system
c
)

/

(


Po  
L
(c 1)!(c   )2
 0.952 average number of customers in the system
Chapter 15 - Queuing Analysis
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Example Problem Solution (5 of 5)
Step 2 (continued):

Lq  L  
 0.152 average number of customers in the queue
1  Lq
Wq W  

 0.038 hour average time customer is in the queue

c
1
 
Pw     c Po
c!  c  
 .229 probabilit y customer must wait for service
Chapter 15 - Queuing Analysis
57
Chapter 15 - Queuing Analysis
58