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DD Waiting-Line ModelsWaiting-Line Models
PowerPoint presentation to accompany PowerPoint presentation to accompany
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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!
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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
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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
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(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
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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
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Figure D.3
Single-channel, single-phase system
Single-channel, multiphase system
Arrivals Departuresafter service
Phase 1 service facility
Phase 2 service facility
Queue
A McDonald’s dual window drive-through
Queuing System DesignsQueuing System Designs
Queue
Most bank and post office service windows
Service facility
Channel 1
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Figure D.3
Multi-channel, single-phase system
Arrivals
QueueDeparturesafter service
Service facility
Channel 2
Service facility
Channel 3
Queuing System DesignsQueuing System Designs
Queue
Some college registrations
Phase 2 service facility
Phase 1 service facility
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Figure D.3
Multi-channel, multiphase system
Arrivals Departuresafter service
Channel 1
Phase 2 service facility
Channel 2
Channel 1
Phase 1 service facility
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
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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
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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
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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
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FIFO disciplineA single-service phase
Queuing ModelsQueuing Models
Model Name ExampleA Single-channel Information counter
system at department store(M/M/1)
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Table D.2
Number Number Arrival Serviceof of Rate Time Population Queue
Channels Phases Pattern Pattern Size DisciplineSingle Single Poisson Exponential Unlimited FIFO
Queuing ModelsQueuing Models
Model Name ExampleB Multichannel Airline ticket
(M/M/S) counter
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Table D.2
Number Number Arrival Serviceof of Rate Time Population Queue
Channels Phases Pattern Pattern Size DisciplineMulti- Single Poisson Exponential Unlimited FIFOchannel
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Queuing ModelsQueuing Models
Model Name ExampleC Constant- Automated car
service wash(M/D/1)
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Table D.2
Number Number Arrival Serviceof of Rate Time Population Queue
Channels Phases Pattern Pattern Size DisciplineSingle Single Poisson Constant Unlimited FIFO
Queuing ModelsQueuing Models
Model Name ExampleD Limited Shop with only a
population dozen machines(finite population) that might break
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Table D.2
Number Number Arrival Serviceof of Rate Time Population Queue
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
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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
Model A Model A –– SingleSingle--ChannelChannel
4. Service times vary from one customer to the next and are independent of one another, but their average rate is known
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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)
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Table D.3
=
Ws = Average time a unit spends in the system (waiting time plus service time)
=
λµ – λ
1µ – λ
Model A Model A –– SingleSingle--ChannelChannel
Lq = Average number of units waiting in the queue
= λ2
µ(µ – λ)
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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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Model A Model A –– SingleSingle--ChannelChannel
P0 = Probability of 0 units in the system (that is, the service unit is idle)
= 1 – λµ
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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systems are possibleMultiphase models exist for more complex situations
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