heizer om10 mod_d-waiting-line models

10/16/2010

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DD Waiting-Line ModelsWaiting-Line Models

PowerPoint presentation to accompany PowerPoint presentation to accompany

D - 1© 2011 Pearson Education, Inc. publishing as Prentice Hall

p p yp p yHeizer and Render Heizer and Render Operations Management, 10e Operations Management, 10e Principles of Operations Management, 8ePrinciples of Operations Management, 8e

PowerPoint slides by Jeff Heyl

OutlineOutlineQueuing TheoryCharacteristics of a Waiting-Line System

Arrival Characteristics


Arrival CharacteristicsWaiting-Line CharacteristicsService CharacteristicsMeasuring a Queue’s Performance

Queuing Costs

Outline Outline –– ContinuedContinuedThe Variety of Queuing Models

Model A(M/M/1): Single-Channel Queuing Model with Poisson Arrivals and Exponential Service TimesM d l B(M/M/S) M lti l Ch l


Model B(M/M/S): Multiple-Channel Queuing ModelModel C(M/D/1): Constant-Service-Time ModelLittle’s LawModel D: Limited-Population Model

Outline Outline –– ContinuedContinued

Other Queuing Approaches


Learning ObjectivesLearning ObjectivesWhen you complete this module you When you complete this module you should be able to:should be able to:

1. Describe the characteristics of arrivals waiting lines and service


arrivals, waiting lines, and service systems

2. Apply the single-channel queuing model equations

3. Conduct a cost analysis for a waiting line

Learning ObjectivesLearning ObjectivesWhen you complete this module you When you complete this module you should be able to:should be able to:

4. Apply the multiple-channel queuing model formulas


queuing model formulas5. Apply the constant-service-time

model equations6. Perform a limited-population

model analysis

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Queuing TheoryQueuing Theory

The study of waiting linesWaiting lines are common situations


Useful in both manufacturing and service areas

Common Queuing Common Queuing SituationsSituations

Situation Arrivals in Queue Service ProcessSupermarket Grocery shoppers Checkout clerks at cash

registerHighway toll booth Automobiles Collection of tolls at boothDoctor’s office Patients Treatment by doctors and


nursesComputer system Programs to be run Computer processes jobs

Telephone company Callers Switching equipment to forward calls

Bank Customer Transactions handled by tellerMachine

maintenanceBroken machines Repair people fix machines

Harbor Ships and barges Dock workers load and unload

Table D.1

Characteristics of WaitingCharacteristics of Waiting--Line SystemsLine Systems

1. Arrivals or inputs to the systemPopulation size, behavior, statistical distribution

2 Queue discipline or the waiting line


2. Queue discipline, or the waiting line itself

Limited or unlimited in length, discipline of people or items in it

3. The service facilityDesign, statistical distribution of service times

Parts of a Waiting LineParts of a Waiting Line

Dave’s Car Wash

Population ofdirty cars

Arrivalsfrom thegeneral

population …

Queue(waiting line)

Servicefacility

Exit the system


Figure D.1

Enter Exit

Arrivals to the system Exit the systemIn the system

Arrival CharacteristicsSize of the populationBehavior of arrivalsStatistical distribution of arrivals

Waiting Line CharacteristicsLimited vs. unlimitedQueue discipline

Service CharacteristicsService designStatistical distribution of service

Arrival CharacteristicsArrival Characteristics1. Size of the population

Unlimited (infinite) or limited (finite)2. Pattern of arrivals

Scheduled or random, often a Poisson


distribution3. Behavior of arrivals

Wait in the queue and do not switch linesNo balking or reneging

Poisson DistributionPoisson Distribution

P(x) = for x = 0, 1, 2, 3, 4, …e-λλx

x!


where P(x) = probability of x arrivalsx = number of arrivals per unit of timeλ = average arrival ratee = 2.7183 (which is the base of the

natural logarithms)

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Poisson DistributionPoisson DistributionProbability = P(x) = e-λλx

x!

0.25 –

0.02 –

lity

0.25 –

0.02 –lit

y


0.15 –

0.10 –

0.05 –

–

Prob

abil

0 1 2 3 4 5 6 7 8 9Distribution for λ = 2

x

0.15 –

0.10 –

0.05 –

–

Prob

abil

0 1 2 3 4 5 6 7 8 9Distribution for λ = 4

x10 11

Figure D.2

WaitingWaiting--Line CharacteristicsLine Characteristics

Limited or unlimited queue lengthQueue discipline - first-in, first-out (FIFO) is most common


(FIFO) is most commonOther priority rules may be used in special circumstances

Service CharacteristicsService CharacteristicsQueuing system designs

Single-channel system, multiple-channel systemSingle-phase system, multiphase


Single phase system, multiphase system

Service time distributionConstant service timeRandom service times, usually a negative exponential distribution

Queuing System DesignsQueuing System Designs

Departuresafter service

Single channel single phase system

Queue

Arrivals Service facility

A family dentist’s office


Figure D.3

Single-channel, single-phase system

Single-channel, multiphase system

Arrivals Departuresafter service

Phase 1 service facility


Queue

A McDonald’s dual window drive-through


Queue

Most bank and post office service windows

Service facility

Channel 1


Figure D.3

Multi-channel, single-phase system

Arrivals

QueueDeparturesafter service

Service facility

Channel 2

Service facility

Channel 3


Queue

Some college registrations




Figure D.3

Multi-channel, multiphase system

Arrivals Departuresafter service

Channel 1


Channel 2

Channel 1


Channel 2

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Negative Exponential Negative Exponential DistributionDistribution

1.0 –0.9 –0.8 –0 7 –ce

tim

e ≥

1

Probability that service time is greater than t = e-µt for t ≥ 1µ = Average service ratee = 2.7183

Average service rate (µ) = 3 customers per hour⇒ Average service time = 20 minutes per customer


Figure D.4

0.7 –0.6 –0.5 –0.4 –0.3 –0.2 –0.1 –0.0 –

Prob

abili

ty th

at s

ervi

c

| | | | | | | | | | | | |

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00Time t (hours)

Average service rate (µ) = 1 customer per hour

Measuring Queue Measuring Queue PerformancePerformance

1. Average time that each customer or object spends in the queue

2. Average queue length3. Average time each customer spends in the


3. Average time each customer spends in the system

4. Average number of customers in the system5. Probability that the service facility will be idle6. Utilization factor for the system7. Probability of a specific number of customers

in the system

Queuing CostsQueuing CostsCost


Figure D.5

Total expected cost

Cost of providing service

Low levelof service

High levelof service

Cost of waiting time

MinimumTotalcost

Optimalservice level

Queuing ModelsQueuing Models

The four queuing models here all assume:

Poisson distribution arrivals


FIFO disciplineA single-service phase


Model Name ExampleA Single-channel Information counter

system at department store(M/M/1)


Table D.2

Number Number Arrival Serviceof of Rate Time Population Queue

Channels Phases Pattern Pattern Size DisciplineSingle Single Poisson Exponential Unlimited FIFO


Model Name ExampleB Multichannel Airline ticket

(M/M/S) counter


Table D.2


Channels Phases Pattern Pattern Size DisciplineMulti- Single Poisson Exponential Unlimited FIFOchannel

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Model Name ExampleC Constant- Automated car

service wash(M/D/1)


Table D.2


Channels Phases Pattern Pattern Size DisciplineSingle Single Poisson Constant Unlimited FIFO


Model Name ExampleD Limited Shop with only a

population dozen machines(finite population) that might break


Table D.2


Channels Phases Pattern Pattern Size DisciplineSingle Single Poisson Exponential Limited FIFO

Model A Model A –– SingleSingle--ChannelChannel

1. Arrivals are served on a FIFO basis and every arrival waits to be served regardless of the length of the queue

2 Arrivals are independent of preceding


2. Arrivals are independent of preceding arrivals but the average number of arrivals does not change over time

3. Arrivals are described by a Poisson probability distribution and come from an infinite population


4. Service times vary from one customer to the next and are independent of one another, but their average rate is known


5. Service times occur according to the negative exponential distribution

6. The service rate is faster than the arrival rate

Model A Model A –– SingleSingle--ChannelChannelλ = Mean number of arrivals per time periodµ = Mean number of units served per time period

Ls = Average number of units (customers) in the system (waiting and being served)


Table D.3

=

Ws = Average time a unit spends in the system (waiting time plus service time)

=

λµ – λ

1µ – λ


Lq = Average number of units waiting in the queue

= λ2

µ(µ – λ)


Table D.3

Wq = Average time a unit spends waiting in the queue

=

ρ = Utilization factor for the system

=

λµ(µ – λ)

λµ

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P0 = Probability of 0 units in the system (that is, the service unit is idle)

= 1 – λµ


Table D.3

Pn > k = Probability of more than k units in the system, where n is the number of units in the system

= λµ

k + 1

SingleSingle--Channel ExampleChannel Example

λ = 2 cars arriving/hour µ = 3 cars serviced/hour

Ls = = = 2 cars in the system on averageλµ – λ

23 - 2


Ws = = = 1 hour average waiting time in the system

Lq = = = 1.33 cars waiting in lineλ2

µ(µ – λ)

1µ – λ

13 - 2

22

3(3 - 2)

SingleSingle--Channel ExampleChannel Example

Wq = = = 2/3 hour = 40 minute average waiting time

λµ(µ – λ)

23(3 - 2)

λ = 2 cars arriving/hour µ = 3 cars serviced/hour


average waiting time

ρ = λ/µ = 2/3 = 66.6% of time mechanic is busy

λµP0 = 1 - = .33 probability there are 0 cars in the

system

SingleSingle--Channel ExampleChannel ExampleProbability of more than k Cars in the System

k Pn > k = (2/3)k + 1

0 .667.667 ← Note that this is equal to 1 - P0 = 1 - .331 .444


2 .2963 .198.198 ← Implies that there is a 19.8% chance that

more than 3 cars are in the system4 .1325 .0886 .0587 .039

SingleSingle--Channel EconomicsChannel EconomicsCustomer dissatisfaction

and lost goodwill = $10 per hourWq = 2/3 hour

Total arrivals = 16 per dayMechanic’s salary = $56 per day


Total hours customers spend waiting per day

= (16) = 10 hours23

23

Customer waiting-time cost = $10 10 = $106.6723

Total expected costs = $106.67 + $56 = $162.67

MultiMulti--Channel ModelChannel ModelM = number of channels openλ = average arrival rateµ = average service rate at each channel

P0 = for Mµ > λ1


Table D.4

P0 = for Mµ > λ1

M!1n!

MµMµ - λ

M – 1

n = 0

λµ

nλµ

M

+∑

Ls = P0 +λµ(λ/µ)M

(M - 1)!(Mµ - λ)2λµ

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MultiMulti--Channel ModelChannel Model

Ws = P0 + =λµ(λ/µ)M

(M - 1)!(Mµ - λ)21µ

Ls

λ


Table D.4

Lq = Ls – λµ

Wq = Ws – =1µ

Lq

λ

MultiMulti--Channel ExampleChannel Exampleλ = 2 µ = 3 M = 2

P0 = = 1

12!

1n!

2(3)2(3) - 2

123

n23

2

+∑12


2!n! 2(3) 2n = 0

3 3∑

Ls = + =(2)(3(2/3)2 2

31! 2(3) - 2 2

12

34

Wq = = .0415.083

2Ws = =3/42

38

Lq = – =23

34

112

MultiMulti--Channel ExampleChannel Example

Single Channel Two ChannelsP0 .33 .5L 2 cars 75 cars


Ls 2 cars .75 carsWs 60 minutes 22.5 minutesLq 1.33 cars .083 carsWq 40 minutes 2.5 minutes

Waiting Line TablesWaiting Line TablesPoisson Arrivals, Exponential Service Times

Number of Service Channels, M

ρ 1 2 3 4 5

.10 .0111

.25 .0833 .0039

.50 .5000 .0333 .003075 2 2500 1227 0147


Table D.5

.75 2.2500 .1227 .0147

.90 3.1000 .2285 .0300 .00411.0 .3333 .0454 .00671.6 2.8444 .3128 .0604 .01212.0 .8888 .1739 .03982.6 4.9322 .6581 .16093.0 1.5282 .35414.0 2.2164

Waiting Line Table ExampleWaiting Line Table ExampleBank tellers and customersλ = 18, µ = 20

From Table D 5

Utilization factor ρ = λ/µ = .90 Wq =Lq

λ


From Table D.5

Number of service windows M

Number in queue Time in queue

1 window 1 8.1 .45 hrs, 27 minutes

2 windows 2 .2285 .0127 hrs, ¾ minute3 windows 3 .03 .0017 hrs, 6 seconds4 windows 4 .0041 .0003 hrs, 1 second

ConstantConstant--Service ModelService Model

Lq = λ2

2µ(µ – λ)Average lengthof queue

Wq = λ2µ(µ – λ)

Average waiting timein queue


Table D.6

2µ(µ – λ)in queue

λµ

Ls = Lq + Average number ofcustomers in system

Ws = Wq + 1µ

Average time in the system

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ConstantConstant--Service ExampleService ExampleTrucks currently wait 15 minutes on averageTruck and driver cost $60 per hourAutomated compactor service rate (µ) = 12 trucks per hourArrival rate (λ) = 8 per hourCompactor costs $3 per truck

Current waiting cost per trip = (1/4 hr)($60) = $15 /trip


Net savings = $ 7 /trip

g p p ( )( ) p

Wq = = hour82(12)(12 – 8)

112

Waiting cost/tripwith compactor = (1/12 hr wait)($60/hr cost) = $ 5 /tripSavings withnew equipment = $15 (current) – $5(new) = $10 /trip

Cost of new equipment amortized = $ 3 /trip

Little’s LawLittle’s LawA queuing system in steady state

L = λW (which is the same as W = L/λLq = λWq (which is the same as Wq = Lq/λ


Once one of these parameters is known, the other can be easily foundIt makes no assumptions about the probability distribution of arrival and service timesApplies to all queuing models except the limited population model

LimitedLimited--Population ModelPopulation Model

Service factor: X =

Average number running: J = NF(1 - X)

Average number waiting: L = N(1 - F)

TT + U


Table D.7

Average number waiting: L N(1 F)

Average number being serviced: H = FNX

Average waiting time: W =

Number of population: N = J + L + H

T(1 - F)XF

LimitedLimited--Population ModelPopulation ModelD = Probability that a unit

will have to wait in queue

N = Number of potential customers

F = Efficiency factor T = Average service time

H = Average number of units b i d

U = Average time between it i


being served unit service requirements

J = Average number of units not in queue or in service bay

W = Average time a unit waits in line

L = Average number of units waiting for service

X = Service factor

M = Number of service channels

Finite Queuing TableFinite Queuing TableX M D F

.012 1 .048 .999

.025 1 .100 .997

.050 1 .198 .989

.060 2 .020 .9991 .237 .983


Table D.8

.070 2 .027 .9991 .275 .977

.080 2 .035 .9981 .313 .969

.090 2 .044 .9981 .350 .960

.100 2 .054 .9971 .386 .950

LimitedLimited--Population ExamplePopulation Example

Service factor: X = = 091 (close to 090)2

Each of 5 printers requires repair after 20 hours (U) of useOne technician can service a printer in 2 hours (T)Printer downtime costs $120/hourTechnician costs $25/hour


Service factor: X = = .091 (close to .090)

For M = 1, D = .350 and F = .960For M = 2, D = .044 and F = .998Average number of printers working:For M = 1, J = (5)(.960)(1 - .091) = 4.36For M = 2, J = (5)(.998)(1 - .091) = 4.54

2 + 20

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LimitedLimited--Population ExamplePopulation Example

Service factor: X = = 091 (close to 090)2

Each of 5 printers require repair after 20 hours (U) of useOne technician can service a printer in 2 hours (T)Printer downtime costs $120/hourTechnician costs $25/hour

Number of Technicians

Average Number Printers

Down (N - J)

Average Cost/Hr for Downtime(N - J)$120

Cost/Hr for Technicians

($25/hr)Total

Cost/Hr

1 .64 $76.80 $25.00 $101.80

2 46 $55 20 $50 00 $105 20


Service factor: X = = .091 (close to .090)

For M = 1, D = .350 and F = .960For M = 2, D = .044 and F = .998Average number of printers working:For M = 1, J = (5)(.960)(1 - .091) = 4.36For M = 2, J = (5)(.998)(1 - .091) = 4.54

2 + 202 .46 $55.20 $50.00 $105.20

Other Queuing ApproachesOther Queuing Approaches

The single-phase models cover many queuing situationsVariations of the four single-phase systems are possible


systems are possibleMultiphase models exist for more complex situations


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heizer om10 mod_d-waiting-line models

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