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Chapter 15 - Queuing Analysis 1
Chapter 13
Queuing Analysis
Introduction to Management Science
8th Edition
by
Bernard W. Taylor III
Chapter 15 - Queuing Analysis 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
Addition Types of Queuing Systems
Chapter Topics
Chapter 15 - Queuing Analysis 3
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.
Overview
Chapter 15 - Queuing Analysis 4
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.
Elements of Waiting Line Analysis
Chapter 15 - Queuing Analysis 5
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.
The Single-Server Waiting Line System (1 of 2)
Chapter 15 - Queuing Analysis 6
Figure 15.1The Fast Shop Market Queuing System
The Single-Server Waiting Line System (2 of 2)
Chapter 15 - Queuing Analysis 7
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).
Single-Server Waiting Line SystemComponent Definitions
Chapter 15 - Queuing Analysis 8
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.
Single-Server Waiting Line SystemSingle-Server Model
Chapter 15 - Queuing Analysis 9
Probability that no customers are in the queuing system:
Probability that n customers are in the system:
Average number of customers in system: and waiting line:
1Po
1
nPo
nPn
2
Lq
L
Single-Server Waiting Line SystemBasic Single-Server Queuing Formulas (1 of 2)
Chapter 15 - Queuing Analysis 10
Average time customer spends waiting and being served:
Average time customer spends waiting in the queue:
Probability that server is busy (utilization factor):
Probability that server is idle:
LW
1
Wq
U
11 UI
Single-Server Waiting Line SystemBasic Single-Server Queuing Formulas (2 of 2)
Chapter 15 - Queuing Analysis 11
= 24 customers per hour arrive at checkout counter
= 30 customers per hour can be checked out
system the in customers no ofy probabilit .20
24/30) - (1 1
Po
system the in avg the on customers 4 24) - 24/(30
L
line waitingthe in avg the on customers 3.2 24)]- 30(24)2/[30(
2
Lq
Single-Server Waiting Line SystemCharacteristics for Fast Shop Market (1 of 2)
Chapter 15 - Queuing Analysis 12
Single-Server Waiting Line SystemCharacteristics for Fast Shop Market (2 of 2)
customer per system the in time avg min) (10 hour 0.167
24]- 1/[30
LW 1
line waitingthe in time avg min) (8 hour 0.133
24)]- 24/[30(30
Wq
idle be willservery probabilit .20 busy, servery probabilit .80
24/30
U
Chapter 15 - Queuing Analysis 13
Single-Server Waiting Line SystemSteady-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 14
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
Single-Server Waiting Line SystemEffect of Operating Characteristics (1 of 6)
Chapter 15 - Queuing Analysis 15
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.
Single-Server Waiting Line SystemEffect of Operating Characteristics (2 of 6)
Chapter 15 - Queuing Analysis 16
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
Single-Server Waiting Line SystemEffect of Operating Characteristics (3 of 6)
Chapter 15 - Queuing Analysis 17
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.
Single-Server Waiting Line SystemEffect of Operating Characteristics (4 of 6)
Chapter 15 - Queuing Analysis 18
Table 15.1Operating Characteristics for Each Alternative System
Single-Server Waiting Line SystemEffect of Operating Characteristics (5 of 6)
Chapter 15 - Queuing Analysis 19
Figure 15.2Cost Trade-Offs for Service Levels
Single-Server Waiting Line SystemEffect of Operating Characteristics (6 of 6)
Chapter 15 - Queuing Analysis 20
Exhibit 15.1
Single-Server Waiting Line SystemSolution with Excel and Excel QM (1 of 2)
Chapter 15 - Queuing Analysis 21
Exhibit 15.2
Single-Server Waiting Line SystemSolution with Excel and Excel QM (2 of 2)
Chapter 15 - Queuing Analysis 22
Exhibit 15.3
Single-Server Waiting Line SystemSolution with QM for Windows
Chapter 15 - Queuing Analysis 23
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:
1Po
/
/
12
222Lq
LqL
LqWq
1WqW
U
Single-Server Waiting Line SystemUndefined and Constant Service Times
Chapter 15 - Queuing Analysis 24
machine the using and line in employees 4.0
3020333
line in waitingemployees 3.33
302012
23020
2151
220
12
222
use in not machine thaty probabilit .33 302011
)/(.
/
//
/
/
LqL
Lq
Po
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:
Single-Server Waiting Line SystemUndefined Service Times Example (1 of 2)
Chapter 15 - Queuing Analysis 25
nutilizatio machine 67% 3020
system the in minutes 12
hour 0.1998 301166501
time waitingminutes 10 hour 1665020333
U
WqW
LqWq
.
..
Operating characteristics (continued):
Single-Server Waiting Line SystemUndefined Service Times Example (2 of 2)
Chapter 15 - Queuing Analysis 26
In the constant service time model there is no variability in service times; = 0.
Substituting = 0 into equations:
All remaining formulas are the same as the single-server formulas.
22
12
2
12
2202
12
222
/
/
/
/
/
/Lq
Single-Server Waiting Line SystemConstant Service Times Formulas
Chapter 15 - Queuing Analysis 27
time waitingminutes 6.84 or hour 0.114 10141
waitingcars 1.14 103133132
2102
2
.
).)(.()(
)(
LqWq
Lq
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
Single-Server Waiting Line SystemConstant Service Times Example
Chapter 15 - Queuing Analysis 28
Exhibit 15.4
Undefined and Constant Service TimesSolution with Excel
Chapter 15 - Queuing Analysis 29
Exhibit 15.5
Undefined and Constant Service TimesSolution with QM for Windows
Chapter 15 - Queuing Analysis 30
1 1
1 11
111
M n for 11
1
WWqPMLW
PMLLqMMML
nPoPn
MPo
)(
)()/(
)/)((/
/
)()/(/
In a finite queue, the length of the queue is limited.
Operating characteristics, where M is the maximum number in the system:
Finite Queue Length
Chapter 15 - Queuing Analysis 31
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:
full is system thaty probabilit .076 4
302038
empty is system thaty probabilit .38 530201
3020111
1
)(.)(
)/(/
)/(/
MnPoPM
MPo
Finite Queue Length Example (1 of 2)
Chapter 15 - Queuing Analysis 32
Average queue lengths and waiting times:
line in waitinghour 0.03330106701
system the in waitinghours 0.067 076120
2411
waitingcars 0.62 30
0761202411
system the in cars 1.24 530201
53020530201
3020
11
111
.
).(.
)(
).(.)(
)/()/)((
//
)/()/)((
//
WWq
PMLW
PMLLq
L
MMML
Finite Queue Length Example (2 of 2)
Chapter 15 - Queuing Analysis 34
Exhibit 15.7
Finite Queue Model ExampleSolution with QM for Windows
Chapter 15 - Queuing Analysis 35
1 1
1
2,...N 1, n and size, population N where
0
1
WqWLNLqWqPoLqL
PoNLqPon
nNNPn
nN
n nNN
Po
)()(
)()!(
!
)!(!
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:
Finite Calling Population
Chapter 15 - Queuing Analysis 36
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
Finite Calling Population Example (1 of 2)
Chapter 15 - Queuing Analysis 37
System seems inadequate.
system the in hours 3352778
1741
repair for waitinghours 74100552020
169
system the in machines 5206521169
waitingmachines 1696521005
277800520
65220
0 2778005
2020
1
..
.
.))(..(
.
.).(.
...
..
.
..
)!(!
W
Wq
L
Lq
n
n
n
Po
Finite Calling Population Example (2 of 2)
Chapter 15 - Queuing Analysis 38
Exhibit 15.8
Finite Calling Population ExampleSolution with Excel and Excel QM (1 of 2)
Chapter 15 - Queuing Analysis 39
Exhibit 15.9
Finite Calling Population ExampleSolution with Excel and Excel QM (2 of 2)
Chapter 15 - Queuing Analysis 40
Exhibit 15.10
Finite Calling Population ExampleSolution with QM for Windows
Chapter 15 - Queuing Analysis 41
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.
Multiple-Server Waiting Line (1 of 2)
Chapter 15 - Queuing Analysis 42
Figure 15.3Customer Service
Queuing System
Multiple-Server Waiting Line (2 of 2)
Chapter 15 - Queuing Analysis 43
Multiple-Server Waiting LineQueuing 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 44
system the in spends customer time average
system the in customers average 21
system in customers n ofy probabilit c n for 1
cn for 1
system in customers noy probabilit 11
01
1
LW
Pocc
cL
Pon
nPn
Pon
cnccPn
ccc
ccn
n
n
n
Po
)()!()/(
!
!!
Multiple-Server Waiting LineQueuing Formulas (2 of 3)
Chapter 15 - Queuing Analysis 45
service for waitmust customery probabilit 1
queue the in is customer time average 1
queue the in customers of number average
Po
ccc
cPw
LqWWq
LLq
!
Multiple-Server Waiting LineQueuing Formulas (3 of 3)
Chapter 15 - Queuing Analysis 46
department service the in time customer average hour 600
106
department service in average on customers 6
410045
2104313
3410410
customers no ofy probabilit 045
1043433
410
31
2
410
21
1
410
11
0
410
01
1
.
)(.])([)!(
)/)()((
.
)()(
!!!!
W
L
Po
Multiple-Server Waiting LineBiggs Department Store Example (1 of 2)
= 10, = 4, c = 3
Chapter 15 - Queuing Analysis 47
service for waitmust customery probabilit .703
0451043
433
410
31
customer per line in time waitingaverage hour 350
1053
served be to waitingaverage the on customers 53
4106
)(.
)()(
!
.
.
.
Pw
Wq
Lq
Multiple-Server Waiting LineBiggs Department Store Example (2 of 2)
Chapter 15 - Queuing Analysis 50
Exhibit 15.13
Multiple-Server Waiting LineSolution with QM for Windows
Chapter 15 - Queuing Analysis 51
Figure 15.4Single Queues with Single and Multiple
Servers in Sequence
Additional Types of Queuing Systems (1 of 2)
Chapter 15 - Queuing Analysis 52
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)
Additional Types of Queuing Systems (2 of 2)
Chapter 15 - Queuing Analysis 53
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.
Example Problem Solution (1 of 5)
Chapter 15 - Queuing Analysis 54
Solution:
Step 1: Determine Operating Characteristics for the Single-Server 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
Example Problem Solution (2 of 5)
Chapter 15 - Queuing Analysis 55
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
Example Problem Solution (3 of 5)
Chapter 15 - Queuing Analysis 56
system the in customers of number average 9520
21
system in customers noy probabilit 429
11
01
1
.
)()!()/(
.
!!
Pocc
cL
ccc
ccn
n
n
n
Po
Step 2: Determine the Operating Characteristics for the Multiple-Server System.
= 4 customers per hour arrive; = 5 customers per hour served; c = 2 servers
Example Problem Solution (4 of 5)
Chapter 15 - Queuing Analysis 57
service for waitmust customery probabilit 229
1
queue the in is customer time average hour 0380
1
queue the in customers of number average 1520
.
!
.
.
Po
ccc
cPw
LqWWq
LLq
Step 2 (continued):
Example Problem Solution (5 of 5)
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