CH_13_QUEUING_ANALYSIS

Queuing Analysis

Chapter 13 13-1

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Chapter Topics 13-2

■ 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
■ Additional Types of Queuing Systems

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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.

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Elements of Waiting Line Analysis (1 of 2)

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.

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Elements of Waiting Line Analysis (2 of 2)

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.

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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.

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The Single-Server Waiting Line System (2 of 2)

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13-7

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).

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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

Symbols:
λ = the arrival rate (average number of arrivals/time period)
µ = the service rate (average number served/time period)
Customers must be served faster than they arrive (λ < µ) or an
infinitely large queue will build up.

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Single-Server Waiting Line System
Basic Single-Server Queuing Formulas (1 of 2)

Probability that no customers are in the queuing system:

P = 1− λµ 
0 
 


Probability that n customers are in the system:

n n 

 

1− 


λ λ λP = ⋅ P =  
µ µ µn 0  
  
 

Average number of customers in system: L = λ λ
µ−

Average number of customer in the waiting line: L = λ2 λ
q µ  µ − 
 

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Single-Server Waiting Line System
Basic Single-Server Queuing Formulas (2 of 2)

Average time customer spends waiting and being served:

W = µ 1 = L

−λ λ

Average time customer spends waiting in the queue:

W = µ  λ λ 
q  µ− 

Probability that server is busy (utilization factor): U = λµ

Probability that server is idle: I =1−U =1− λµ

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Single-Server Waiting Line System
Operating Characteristics: Fast Shop Market (1 of 2)

λ = 24 customers per hour arrive at checkout counter

µ = 30 customers per hour can be checked out

P =  1− λµ  = (1 - 24/30)
0  
 
 

= .20 probability of no customers in the system

L= λ = 24/(30 - 24) = 4 customers on the avg in the system
µ−λ

L = µ  λ 2 λ 
q  µ − 

= (24)2/[30(30 -24)] = 3.2 customers on the avg in the waiting line

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Single-Server Waiting Line System
Operating Characteristics for Fast Shop Market (2 of 2)

W = µ 1 = L = 1/[30 -24]

−λ λ

= 0.167 hour (10 min) avg time in the system per customer

W = µ  λ λ  = 24/[30(30 -24)]
q  µ− 

= 0.133 hour (8 min) avg time in the waiting line

U = λµ = 24/30

= .80 probability server busy, .20 probability server will be idle

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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 λ / µ < 1 and λ < µ.

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.

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Single-Server Waiting Line System
Effect of Operating Characteristics (1 of 6)

Manager wishes to test several alternatives for reducing customer
waiting time:

1. Addition of another employee to pack up purchases

2. Addition of another checkout counter.

Alternative 1: Addition of an employee

(raises service rate from µ = 30 to µ = 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 13-15

L = 1.5 customers on the average in the queuing
system

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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.

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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).

λ = 24/2 = 12 customers per hour per checkout counter
µ = 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

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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.

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Single-Server Waiting Line System
Effect of Operating Characteristics (5 of 6)

Table 13.1 13-19
Operating Characteristics for Each Alternative System

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Single-Server Waiting Line System
Effect of Operating Characteristics (6 of 6)

Figure 13.2 Cost Trade-Offs for Service Levels 13-20

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Single-Server Waiting Line System
Solution with Excel and Excel QM (1 of 2)

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Single-Server Waiting Line System
Solution with Excel and Excel QM (2 of 2)

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Single-Server Waiting Line System
Solution with QM for Windows

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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 single-server

model with undefined service times.

Queuing formulas:

P = 1− λµ W = L
0 q q

λ

L = λ 2σ 2 +  λ /µ 2 W =Wq + µ1
q  µ
2 
 
1−λ /

L = L + λµ U = λµ
q

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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:

P = 1− λµ = 1− 20 = .33 probability that machine not in use
0 30

L = λ 2σ 2 +  λ /µ 2 =  20 2 1/15 2 +  20 / 30 2
q  µ  
2    
 
1−λ / 2 1− 20 / 30 
 

= 3.33 employees waiting in line

L = L + λµ = 3.33 + (20 / 30)
q

= 4.0 employees in line and using the machine

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Single-Server Waiting Line System
Undefined Service Times Example (2 of 2)

Operating characteristics (continued):

W = L = 3.33 = 0.1665 hour = 10 minutes waiting time
q q 20

λ

W =Wq + µ1 = 0.1665+ 1 = 0.1998 hour
30
= 12 minutes in the system

U = λµ = 20 = 67% machine utilization
30

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Single-Server Waiting Line System
Constant Service Times Formulas

In the constant service time model there is no variability in
service times; σ = 0.

Substituting σ = 0 into equations:

L = λ 2σ 2 +  λ /µ 2 = λ 202 +  λ /µ 2 =  λ / µ 2 = λ2
q  µ  µ  λ
2   
  
1−λ / 2 1− λ / 21− /µ  2µ  µ − λ 
   

All remaining formulas are the same as the single-server
formulas.

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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.

λ = 10 cars per hour, µ = 60/4.5 = 13.3 cars per hour

L = λ2 λ ) = (10)2 −10) = 1.14 cars waiting
q 2µ(µ − 2(13.3)(13.3

W = L = 1.14 = 0.114 hour or 6.84 minutes waiting time
q q 10

λ

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Undefined and Constant Service Times
Solution with Excel

Exhibit 13.4

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Undefined and Constant Service Times
Solution with QM for Windows

Exhibit 13.5

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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:

P = 1− 1−λ /µ +1 P = ( P  λµ n for n ≤ M
0 (λ / µ n 0
)M ) 
 
 

L= λ/µ − (M +1)(λ / µ)M+1 L = L− λ (1− P )
1−λ / µ 1−(λ / µ)M+1 q M
µ

W = λ L W =W − µ1
(1− q
P )
M

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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 λ = 20, µ = 30, M = 4:

P = 1− 1−λ / µ +1 = 1−20/ 30 = .38 probability that system is empty
0 (λ / µ)M 1−(20/ 30)5

= P ) λµ n= M = (.38) 20 4 =
 30
P (    .076 probability that system is full
M   
0   

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Finite Queue Length Example (2 of 2)

Average queue lengths and waiting times:

L= λ/µ − (M +1)(λ / µ)M+1
1−λ / µ 1−(λ / µ)M+1

L= 20/ 30 − (5)(20/ 30)5 = 1.24 cars in the system
1−20/ 30 1−(20/ 30)5

L = L − λ (1− P ) = 1.24− 20(1− .076) = 0.62 cars waiting
q M 30
µ

W = L ) = 1.24 = 0.067 hours waiting in the system
20(1− .076)
λ(1− P M

W =W − µ1 = 0.067 − 1 = 0.033 hour waiting in line
q 30

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Finite Queue Model Example
Solution with Excel

Exhibit 13.6

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Finite Queue Model Example
Solution with QM for Windows

Exhibit 13.7

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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:

P = 1 n
0
N∑ N!  λµ 
n=0  
N − n)! 
(

where N = population size, and n = 1, 2,...N

P = N!  λµ n P L = N −  λ − µ  P)
n  0 q  0
N − n)!   λ (1−
( 


L = L + (1− P ) W = ( N L W =Wq + µ1
q 0 q q

− L)λ

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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?
λ = 1/200 hour = .005 per hour
µ = 1/3.6 hour = .2778 per hour
N = 20 machines

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Finite Calling Population Example (2 of 2)

P = 1 n = .652
0
2∑0 20!  .005 
n=0  .2778 
(20− n)! 

L = 20− .005+ .2778 1− .652  = .169 machines waiting
q .005 

L= .169+(1−.652)= .520 machines in the system

W = (20− .169 =1.74 hours waiting for repair
q .520)(.005)

W =1.74+ 1 = 5.33 hours in the system
.2778

…System seems woefully inadequate.

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Finite Calling Population Example
Solution with Excel and Excel QM (1 of 2)

Exhibit 13.8

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Finite Calling Population Example
Solution with Excel and Excel QM (2 of 2)

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Finite Calling Population Example
Solution with QM for Windows

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13-41

Multiple-Server Waiting Line (1 of 3)

Figure 13.3

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Multiple-Server Waiting Line (2 of 3)

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.

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Multiple-Server Waiting Line (3 of 3)

Customer Service 13-44
System at Biggs
Department Store

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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:

λ = arrival rate (average number of arrivals per time period)
µ = the service rate (average number served per time period)
per server (channel)
c = number of servers
c µ = mean effective service rate for the system (must exceed
arrival rate)

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Multiple-Server Waiting Line
Queuing Formulas (2 of 3)

P = nn=∑=c−01n1! n  1 c cµ = probability no customers in system
0  + cµ −
λµ   1 λµ   
     λ 
    
c!

P = c!c1n−c  λµ n P for n>c
n  0
 
 


P = 1  λµ n P for n< c = probability of n customers in system
n n  0
 
 


L= λµ(λ / µ)c P + λµ µ= average customers in the system
(c −1)!(cµ − λ )2 0

W = λL=average time customer spends in the system

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Multiple-Server Waiting Line
Queuing Formulas (3 of 3)

L = L− λµ = average number of customers in the queue
q

W =W − µ1 = L = average time customer is in the queue
q q

λ

P = 1 λµ c cµ λ P = probability customer must wait for service
w  cµ − 0

c! 


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Multiple-Server Waiting Line
Biggs Department Store Example (1 of 2)

λ = 10, µ = 4, c = 3

P =  0 1 1 2  3
0  
   10   
 1 10  + 1 10  + 1 4   + 1 10  3(4)
  4   4      4  3(4)−10

0! 1! 2! 3!
 

= .045 probability of no customers

L = (10)(4)(10 / 4)3 (.045)+ 10
(3−1)![3(4)−10]2 4

= 6 customers on average in service department

W = 6 = 0.60 hour average customer time in the service department
10

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Multiple-Server Waiting Line
Biggs Department Store Example (2 of 2)

L = 6 − 10
q 4

= 3.5 customers on the average waiting to be served

W = 3.5
q 10

= 0.35 hour average waiting time in line per customer

P = 1 10 3 3(4) (.045)
w  4 3(4)−10

3! 


= .703 probability customer must wait for service

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Multiple-Server Waiting Line
Solution with Excel

Exhibit 13.11

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