Ch17 Modified

Copyright 2006 John Wiley & Sons, Inc.Copyright 2006 John Wiley & Sons, Inc.Beni AsllaniBeni Asllani

University of Tennessee at ChattanoogaUniversity of Tennessee at Chattanooga

Waiting Line AnalysisWaiting Line Analysisfor Service Improvementfor Service Improvement

Operations Management - 5th EditionOperations Management - 5th Edition

Chapter 17Chapter 17

Roberta Russell & Bernard W. Taylor, IIIRoberta Russell & Bernard W. Taylor, III

Copyright 2006 John Wiley & Sons, Inc.Copyright 2006 John Wiley & Sons, Inc. 17-17-22

Lecture OutlineLecture Outline

Elements of Waiting Line Analysis Waiting Line Analysis and Quality Single Server Models Multiple Server Model

Servers

Components of the Queuing SystemComponents of the Queuing System

CustomerArrivals

Waiting Line

Servicing System

Exit

Queue or


Waiting Line AnalysisWaiting Line Analysis

Operating characteristicsOperating characteristics average values for characteristics that describe the average values for characteristics that describe the

performance of a waiting line systemperformance of a waiting line system QueueQueue

A single waiting lineA single waiting line Waiting line system consists ofWaiting line system consists of

ArrivalsArrivals ServersServers Waiting line structuresWaiting line structures


Elements of a Waiting LineElements of a Waiting Line

Calling populationCalling population Source of customersSource of customers Infinite - large enough that one more customer can Infinite - large enough that one more customer can

always arrive to be servedalways arrive to be served Finite - countable number of potential customersFinite - countable number of potential customers

Arrival rate (Arrival rate (λλ)) Frequency of customer arrivals at waiting line Frequency of customer arrivals at waiting line

system system Typically follows Poisson distributionTypically follows Poisson distribution

Customer Service Population SourcesCustomer Service Population Sources

Population Source

Finite Infinite

Example: Number of machines needing repair when a company only has three machines.

Example: Number of machines needing repair when a company only has three machines.

Example: The number of people who could wait in a line for gasoline.

Example: The number of people who could wait in a line for gasoline.


Elements of a Waiting Line (cont.)Elements of a Waiting Line (cont.)

Service timeService time Often follows negative exponential Often follows negative exponential

distributiondistribution Average service rate = Average service rate = μμ

Arrival rate (Arrival rate (λλ) must be less than service ) must be less than service rate (rate (μμ) or system never clears out) or system never clears out


Elements of a Waiting Line (cont.)Elements of a Waiting Line (cont.)

Queue disciplineQueue discipline Order in which customers are servedOrder in which customers are served First come, first served is most commonFirst come, first served is most common

Length can be infinite or finiteLength can be infinite or finite Infinite is most commonInfinite is most common Finite is limited by some physicalFinite is limited by some physical


Basic Waiting Line StructuresBasic Waiting Line Structures

Channels are the number of parallel serversChannels are the number of parallel servers Single channelSingle channel Multiple channelsMultiple channels

Phases denote number of sequential servers the Phases denote number of sequential servers the customer must go throughcustomer must go through Single phaseSingle phase Multiple phasesMultiple phases

Steady stateSteady state A constant, average value for performance characteristics that A constant, average value for performance characteristics that

system will reach after a long timesystem will reach after a long time


Operating CharacteristicsOperating Characteristics

NOTATIONNOTATION OPERATING OPERATING CHARACTERISTICCHARACTERISTIC

LL Average number of customers in the system (waiting and Average number of customers in the system (waiting and being served)being served)

LLqq Average number of customers in the waiting lineAverage number of customers in the waiting line

WW Average time a customer spends in the system Average time a customer spends in the system (waiting and being served)(waiting and being served)

WWqq Average time a customer spends waiting in lineAverage time a customer spends waiting in line

PP00 Probability of no (zero) customers in the systemProbability of no (zero) customers in the system

PPnn Probability of Probability of nn customers in the system customers in the system

ρρ Utilization rate; the proportion of time the system is Utilization rate; the proportion of time the system is in usein use Table 16.1Table 16.1


Cost Relationship in Waiting Cost Relationship in Waiting Line AnalysisLine Analysis

Exp

ecte

d c

ost

sE

xpec

ted

co

sts

Level of serviceLevel of service

Total costTotal cost

Service costService cost

Waiting CostsWaiting Costs


Waiting Line Costs and Quality Waiting Line Costs and Quality ServiceService

Traditional view is that the level of Traditional view is that the level of service should coincide with minimum service should coincide with minimum point on total cost curvepoint on total cost curve

TQM approach is that absolute quality TQM approach is that absolute quality service will be the most cost-effective in service will be the most cost-effective in the long runthe long run

The Poisson Random VariableThe Poisson Random Variable The Poisson random variable The Poisson random variable xx is a model for data is a model for data

that represent the number of occurrences of a that represent the number of occurrences of a specified event in a given unit of time or space.specified event in a given unit of time or space.

• Examples:Examples:

• The number of calls received by a switchboard during a given period of time.

• The number of machine breakdowns in a day

• The number of traffic accidents at a given intersection during a given time period.

The Poisson Probability The Poisson Probability DistributionDistribution xx is the number of events that occur in a period is the number of events that occur in a period

of time or space during which an average of of time or space during which an average of such events can be expected to occur. The such events can be expected to occur. The probability of probability of kk occurrences of this event is occurrences of this event is

For values of k = 0, 1, 2, … The mean and standard deviation of the Poisson random variable are

Mean:

Standard deviation:

For values of k = 0, 1, 2, … The mean and standard deviation of the Poisson random variable are

Mean:

Standard deviation:

!)(

k

ekxP

k

ExampleExample

The average number of traffic accidents on a certain section of highway is two per week. Find the probability of exactly one accident during a one-week period.

!)1(

k

exP

k

2707.2!1

2 221

ee

The Exponential DistributionThe Exponential Distribution

Used to describe the amount of time between Used to describe the amount of time between occurrences of random eventsoccurrences of random events

Probability Distribution, for an Exponential Random Variable Probability Distribution, for an Exponential Random Variable

xx Probability Density function: Probability Density function:

Mean:Mean:

Standard Deviation:Standard Deviation:

)0(, xexf x

1

1


Shape of the distributionShape of the distributionis determined by the is determined by the value of value of

Mean is equal to Mean is equal to Standard deviationStandard deviation


To find the area A to the right of a,To find the area A to the right of a,

A can be calculatedA can be calculatedusing a calculator orusing a calculator orwith tableswith tables

aeaxPA


If If = .5, what is the p(a>5)? = .5, what is the p(a>5)?

From tables, A = .082085From tables, A = .082085 Probability that A > 5 isProbability that A > 5 is

.082085.082085

5.255. eeeA a


a)a) If If = .16, what are the = .16, what are the and and ??

b)b) What is the p(0<a<5)?What is the p(0<a<5)?

c)c) What is the p(0<a< What is the p(0<a< +2+2))

a)a) = = = 1/ = 1/ = 1/.16 = 6.25 = 1/.16 = 6.25


b) P(x>a) = eb) P(x>a) = e--aa

P(x>5) = eP(x>5) = e-(.16)5 -(.16)5 = e= e-.8-.8

= = .449329.449329 P(x<5) = 1-P(x>5)P(x<5) = 1-P(x>5) = 1-.449329 = .550671= 1-.449329 = .550671


c) What is the p(0 < a < c) What is the p(0 < a < +2+2)?)? Find the complement of the Find the complement of the

area above area above +2+2 P = 1-P(x>18.75) P = 1-P(x>18.75)

= 1- e = 1- e--(18.75) (18.75) = 1- e= 1- e-.16(18.75)-.16(18.75)

= 1- e= 1- e-3 -3 = 1- .049787= 1- .049787 = .950213= .950213


Single-server Models

All assume Poisson arrival rateAll assume Poisson arrival rate VariationsVariations

Exponential service timesExponential service times General (or unknown) distribution of service timesGeneral (or unknown) distribution of service times Constant service timesConstant service times Exponential service times with finite queue lengthExponential service times with finite queue length Exponential service times with finite calling Exponential service times with finite calling

populationpopulation


Basic Single-Server Basic Single-Server Model: AssumptionsModel: Assumptions

Poisson arrival ratePoisson arrival rate Exponential service timesExponential service times First-come, first-served queue disciplineFirst-come, first-served queue discipline Infinite queue lengthInfinite queue length Infinite calling populationInfinite calling population = mean arrival rate= mean arrival rate = mean service rate= mean service rate


Formulas for Single-Formulas for Single-Server ModelServer Model

LL = =

- -

Average number of Average number of customers in the systemcustomers in the system

Probability that no customers Probability that no customers are in the system (either in the are in the system (either in the queue or being served)queue or being served)

PP00 = 1 - = 1 -

Probability of exactly Probability of exactly nn customers in the systemcustomers in the system

PPnn = • = • PP00

nn

= 1 -= 1 -

nn

Average number of Average number of customers in the waiting linecustomers in the waiting line

LLqq = =

(( - - ))


Formulas for Single-Formulas for Single-Server Model (cont.)Server Model (cont.)

==

Probability that the server Probability that the server is busy and the customer is busy and the customer has to waithas to wait

Average time a customer Average time a customer spends in the queuing systemspends in the queuing system WW = = = =

11--

LL

Probability that the server Probability that the server is idle and a customer can is idle and a customer can be servedbe served

II = 1 - = 1 -

= 1 - == 1 - = P P00

Average time a customer Average time a customer spends waiting in line to spends waiting in line to be servedbe served

WWqq = = (( - - ))


A Single-Server Model EX. 17.1A Single-Server Model EX. 17.1

Given Given = 24 per hour, = 24 per hour, = 30 customers per hour = 30 customers per hour

Probability of no Probability of no customers in the customers in the systemsystem

PP00 = 1 - = 1 - = = 1 - = 1 - =

0.200.20

24243030

LL = = = 4 = = = 4Average number Average number of customers in of customers in the systemthe system

--

242430 - 2430 - 24

Average number Average number of customers of customers waiting in linewaiting in line

LLqq = = = 3.2 = = = 3.2(24)(24)22

30(30 - 24)30(30 - 24)22

(( - - ))


A Single-Server ModelA Single-Server Model

Average time in the Average time in the system per customer system per customer WW = = = 0.167 hour = = = 0.167 hour

11--

1130 - 2430 - 24

Average time waiting Average time waiting in line per customer in line per customer

WWqq = = = 0.133 = = = 0.133((--))

242430(30 - 24)30(30 - 24)

Probability that the Probability that the server will be busy and server will be busy and the customer must waitthe customer must wait

= = = 0.80= = = 0.80

24243030

Probability the Probability the server will be idleserver will be idle II = 1 - = 1 - = 1 - 0.80 = 0.20 = 1 - 0.80 = 0.20


Service Improvement AnalysisService Improvement Analysis

Possible AlternativesPossible Alternatives Another employee to pack up purchasesAnother employee to pack up purchases

service rate will increase from 30 customers to 40 service rate will increase from 30 customers to 40 customers per hourcustomers per hour

waiting time will reduce to only 2.25 minuteswaiting time will reduce to only 2.25 minutes Another checkout counterAnother checkout counter

arrival rate at each register will decrease from 24 to 12 per arrival rate at each register will decrease from 24 to 12 per hourhour

customer waiting time will be 1.33 minutescustomer waiting time will be 1.33 minutes Determining whether these improvements are worth the Determining whether these improvements are worth the

cost to achieve them is the crux of waiting line analysiscost to achieve them is the crux of waiting line analysis


Constant Service TimesConstant Service Times

Constant service times Constant service times occur with machinery occur with machinery and automated and automated equipmentequipment

Constant service times are a special case of the single-server model with undefined service times


Operating Characteristics for Operating Characteristics for Constant Service TimesConstant Service Times

PP00 = 1 - = 1 -Probability that no customersProbability that no customersare in systemare in system

Average number of Average number of customers in systemcustomers in system

LL = = LLqq + +

Average number of Average number of customers in queuecustomers in queue

LLqq = = 22

22(( - - ))


Operating Characteristics for Operating Characteristics for Constant Service Times (cont.)Constant Service Times (cont.)

= = Probability that the Probability that the server is busyserver is busy

Average time customer Average time customer spends in the systemspends in the system WW = = WWqq + +

11

Average time customer Average time customer spends in queuespends in queue WWqq = =

LLqq


Constant Service Times: Constant Service Times: Example 17.2Example 17.2

Automated car wash with service time = 4.5 minAutomated car wash with service time = 4.5 min

Cars arrive at rate Cars arrive at rate = 10/hour (Poisson) = 10/hour (Poisson)

= 60/4.5 = 13.3/hour= 60/4.5 = 13.3/hour

WWqq = = 1.14/10 = .114 hour or 6.84 minutes = = 1.14/10 = .114 hour or 6.84 minutesLLqq

(10)(10)22

2(13.3)(13.3 - 10)2(13.3)(13.3 - 10)LLqq = = = 1.14 cars waiting = = = 1.14 cars waiting

22

22(( - - ))


Finite Queue LengthFinite Queue Length A physical limit exists on length of queue M = maximum number in queue Service rate does not have to exceed arrival rate ()

to obtain steady-state conditions

PP00 = =Probability that no Probability that no customers are in systemcustomers are in system

1 - 1 - //

1 - (1 - (//))MM + 1 + 1

Probability of exactly Probability of exactly nn customers in systemcustomers in system

PPnn = ( = (PP00) for ) for nn ≤ ≤ MM

nn

LL = - = -Average number of Average number of customers in systemcustomers in system

//1 - 1 - //

((MM + 1)( + 1)(//))MM + 1 + 1

1 - (1 - (//))MM + 1 + 1


Finite Queue Length (cont.)Finite Queue Length (cont.)

Let Let PPMM = probability a customer will not join system = probability a customer will not join system

Average time customerAverage time customerspends in systemspends in system WW = =

LL

(1 - (1 -

PPMM))

LLqq = = L L - - (1- (1- PPMM))


Average time customer Average time customer spends in queuespends in queue WWqq = = W W --

11


Finite Queue: Example 17.3Finite Queue: Example 17.3First National Bank has waiting space for only 3 drive in window cars. = 20, = 30, M = 4 cars (1 in service + 3 waiting)

Probability that Probability that no cars are in no cars are in the systemthe system PP00 = = = 0.38 = = = 0.38

1 - 20/301 - 20/30

1 - (20/30)1 - (20/30)55

1 - 1 - //

1 - (1 - (//))M + 1M + 1

PPnn = ( = (PP00) = (0.38) = 0.076) = (0.38) = 0.076

Probability of Probability of exactly 4 cars in exactly 4 cars in the systemthe system

2020

3030

44

nn==MM

LL = - = 1.24 = - = 1.24Average number Average number of cars in the of cars in the systemsystem

//1 - 1 - //

((MM + 1)( + 1)(//))MM + 1 + 1

1 - (1 - (//))MM + 1 + 1


Finite Queue: Example (cont.)Finite Queue: Example (cont.)

Average time a carAverage time a carspends in the systemspends in the system W W = = 0.067 hr = = 0.067 hr

LL

(1 - (1 -

PPMM))

LLqq = = L L - = 0.62- = 0.62(1- (1- PPMM))

Average number of Average number of cars in the queuecars in the queue

Average time a carAverage time a carspends in the queuespends in the queue WWqq = = W W - = 0.033 hr- = 0.033 hr

11


Finite Calling PopulationFinite Calling Population

Arrivals originate from a finite (countable) populationN = population size

Probability of exactly Probability of exactly n n customers in systemcustomers in system PPnn = = PP00 where where nn = 1, 2, ..., = 1, 2, ..., NN

nn

NN!!

((NN - - nn)!)!

Average number of Average number of customers in queuecustomers in queue LLqq = = NN - (1- - (1- PP00))

++

Probability that no Probability that no customers are in systemcustomers are in system

PP00 = =

nn = 0 = 0

NN!!

((NN - - nn)!)!NN nn

11


Finite Calling Population (cont.)Finite Calling Population (cont.)

WWqq = = LLqq

((NN - - LL) )

Average time customer Average time customer spends in queuespends in queue

LL = = LLqq + (1 - + (1 - PP00))Average number of Average number of customers in systemcustomers in system

WW = = WWqq + +Average time customerAverage time customerspends in systemspends in system

11


20 trucks which operate an average of 200 days before breaking down ( = 1/200 day = 0.005/day)Mean repair time = 3.6 days ( = 1/3.6 day = 0.2778/day)

Probability that no Probability that no trucks are in the systemtrucks are in the system PP00 = 0.652= 0.652

Average number of Average number of trucks in the queuetrucks in the queue LLqq = 0.169= 0.169

Average number of Average number of trucks in systemtrucks in system LL = 0.169 + (1 - 0.652) = .520 = 0.169 + (1 - 0.652) = .520

Finite Calling Population: Finite Calling Population: Example 17.4Example 17.4

Average time truckAverage time truckspends in queuespends in queue

WWqq = 1.74 days = 1.74 days

Average time truckAverage time truckspends in systemspends in system

WW = 5.33 days = 5.33 days


Two or more independent servers serve a single waiting line

Poisson arrivals, exponential service, infinite calling population

s>

PP00 = =11

11s!s!

ssss

ss - - nn==ss-1-1

nn=0=0

11nn!!

nn

++

Basic Multiple-server Model

Computing P0 can be time-consuming.

Tables can used to find P0 for selected values of and s.


Probability of exactly Probability of exactly nn customers in the customers in the systemsystem

PPnn = =

PP00, , for for n n > > ss11

ss! ! ssn-sn-s

nn

PP00, , for for n n > > ss11

nn!!

nn

Probability an arrivingProbability an arrivingcustomer must waitcustomer must wait PPww = = PP00

11

ss!!

ssss - -

ss

Average number of Average number of customers in systemcustomers in system LL = = P P00 + +

((//))ss

((ss - 1)!( - 1)!(ss - - ))22

Basic Multiple-server Model (cont.)


WW = = LL

Average time customerAverage time customerspends in systemspends in system

= =

//ssUtilization factorUtilization factor

Average time customer Average time customer spends in queuespends in queue WWqq = = WW - = - =

11

LLqq

LLqq = = L L --


Basic Multiple-server Model (cont.)


Multiple-Server System: Multiple-Server System: Example 17.5Example 17.5

Student Health Service Waiting Room = 10 students per hour = 4 students per hour per service representatives = 3 representativess = (3)(4) = 12

PP00 = 0.045 = 0.045Probability no students Probability no students are in the systemare in the system

Number of students in Number of students in the service areathe service area LL = 6 = 6


Multiple-Server System: Multiple-Server System: Example (cont.)Example (cont.)

LLqq = = L L - - // = 3.5 = 3.5Number of students Number of students waiting to be servedwaiting to be served

Average time students Average time students will wait in linewill wait in line WWqq = = L Lqq// = 0.35 hours = 0.35 hours

Probability that a Probability that a student must waitstudent must wait PPww = 0.703= 0.703

Waiting time in the Waiting time in the service areaservice area WW = = LL / / = 0.60 = 0.60


Add a 4th server to improve serviceAdd a 4th server to improve service Recompute operating characteristicsRecompute operating characteristics

PP00 = 0.073 prob of no students = 0.073 prob of no students

LL = 3.0 students = 3.0 students WW = 0.30 hour, 18 min in service = 0.30 hour, 18 min in service LLqq = 0.5 students waiting = 0.5 students waiting

WWqq = 0.05 hours, 3 min waiting, versus 21 earlier = 0.05 hours, 3 min waiting, versus 21 earlier

PPww = 0.31 prob that a student must wait = 0.31 prob that a student must wait

Multiple-Server System: Multiple-Server System: Example (cont.)Example (cont.)


Copyright 2006 John Wiley & Sons, Inc.Copyright 2006 John Wiley & Sons, Inc.All rights reserved. Reproduction or translation All rights reserved. Reproduction or translation of this work beyond that permitted in section 117 of this work beyond that permitted in section 117 of the 1976 United States Copyright Act without of the 1976 United States Copyright Act without express permission of the copyright owner is express permission of the copyright owner is unlawful. Request for further information should unlawful. Request for further information should be addressed to the Permission Department, be addressed to the Permission Department, John Wiley & Sons, Inc. The purchaser may John Wiley & Sons, Inc. The purchaser may make back-up copies for his/her own use only make back-up copies for his/her own use only and not for distribution or resale. The Publisher and not for distribution or resale. The Publisher assumes no responsibility for errors, omissions, assumes no responsibility for errors, omissions, or damages caused by the use of these or damages caused by the use of these programs or from the use of the information programs or from the use of the information herein. herein.

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Ch17 Modified