Chapter 13 Queuing Analysis

Chapter 15 - Queuing Analysis 1

Chapter 13

Queuing Analysis

Introduction to Management Science

8th Edition

by

Bernard W. Taylor III


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


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


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


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)


Figure 15.1The Fast Shop Market Queuing System

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


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


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


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)


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)


= 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)


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


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.


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)


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.



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



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.



Table 15.1Operating Characteristics for Each Alternative System



Figure 15.2Cost Trade-Offs for Service Levels



Exhibit 15.1

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


Exhibit 15.2

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


Exhibit 15.3

Single-Server Waiting Line SystemSolution with QM for Windows


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


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)


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)


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


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


Exhibit 15.4

Undefined and Constant Service TimesSolution with Excel


Exhibit 15.5

Undefined and Constant Service TimesSolution with QM for Windows


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


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)


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)


Exhibit 15.6

Finite Queue Model ExampleSolution with Excel


Exhibit 15.7

Finite Queue Model ExampleSolution with QM for Windows


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


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)


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)


Exhibit 15.8

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


Exhibit 15.9

Finite Calling Population ExampleSolution with Excel and Excel QM (2 of 2)


Exhibit 15.10

Finite Calling Population ExampleSolution with QM for Windows


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)


Figure 15.3Customer Service

Queuing System

Multiple-Server Waiting Line (2 of 2)


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)


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

)()!()/(

!

!!



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

!



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


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)


Exhibit 15.11

Multiple-Server Waiting LineSolution with Excel


Exhibit 15.12

Multiple-Server Waiting LineSolution with Excel QM


Exhibit 15.13

Multiple-Server Waiting LineSolution with QM for Windows


Figure 15.4Single Queues with Single and Multiple

Servers in Sequence

Additional Types of Queuing Systems (1 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)

Additional Types of Queuing Systems (2 of 2)


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)


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



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



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



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):



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Chapter 13 Queuing Analysis

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