D – 1 Operations Management Module D – Waiting-Line Models © 2006 Prentice Hall, Inc. PowerPoint presentation to accompany Heizer/Render Principles of.

D – 1

Operations ManagementOperations ManagementModule D – Waiting-Line ModelsModule D – Waiting-Line Models

© 2006 Prentice Hall, Inc.

PowerPoint presentation to accompanyPowerPoint presentation to accompany Heizer/Render Heizer/Render Principles of Operations Management, 6ePrinciples of Operations Management, 6eOperations Management, 8e Operations Management, 8e

D – 2

Common Queuing Situations

Situation Arrivals in Queue Service Process

Supermarket Grocery shoppers Checkout clerks at cash register

Highway toll booth Automobiles Collection of tolls at booth

Doctor’s office Patients Treatment by doctors and nurses

Computer system Programs to be run Computer processes jobs

Telephone company Callers Switching equipment to forward calls

Bank Customer Transactions handled by teller

Machine maintenance Broken machines Repair people fix machines

Harbor Ships and barges Dock workers load and unload

Table D.1Table D.1

D – 3

Characteristics of Waiting-Line Systems

1. Arrivals or inputs to the system Population size, behavior, statistical

distribution

2. Queue discipline, or the waiting line itself Limited or unlimited in length, discipline of

people or items in it

3. The service facility Design, statistical distribution of service

times

D – 4

Parts of a Waiting Line

Figure D.1Figure D.1

Dave’s Dave’s Car WashCar Wash

enterenter exitexit

Population ofPopulation ofdirty carsdirty cars

ArrivalsArrivalsfrom thefrom thegeneralgeneral

population …population …

QueueQueue(waiting line)(waiting line)

ServiceServicefacilityfacility

Exit the systemExit the system

Arrivals to the systemArrivals to the system Exit the systemExit the systemIn the systemIn the system

Arrival Characteristics Size of the population Behavior of arrivals Statistical distribution

of arrivals

Waiting Line Characteristics

Limited vs. unlimited

Queue discipline

Service Characteristics Service design Statistical distribution

of service

D – 5

Arrival Characteristics

1. Size of the population Unlimited (infinite) or limited (finite)

2. Pattern of arrivals (statistical distribution) Scheduled or random, often a Poisson

distribution

3. Behavior of arrivals Wait in the queue and do not switch lines

Balking or reneging

D – 6

Poisson Distribution

PP((xx)) = for x = for x = 0, 1, 2, 3, 4, …= 0, 1, 2, 3, 4, …ee--xx

xx!!

wherewhere P(x)P(x) == probability of x arrivalsprobability of x arrivals

xx == number of arrivals per number of arrivals per unit of timeunit of time

== average arrival rateaverage arrival rate

ee == 2.71832.7183 (which is the (which is the base of the natural logarithms)base of the natural logarithms)

D – 7

Poisson DistributionProbability = PProbability = P((xx)) = =

ee--xx

x!x!

0.25 0.25 –

0.02 0.02 –

0.15 0.15 –

0.10 0.10 –

0.05 0.05 –

–

Pro

bab

ility

Pro

bab

ility

00 11 22 33 44 55 66 77 88 99

Distribution for Distribution for = 2 = 2

xx

0.25 0.25 –

0.02 0.02 –

0.15 0.15 –

0.10 0.10 –

0.05 0.05 –

–

Pro

bab

ility

Pro

bab

ility

00 11 22 33 44 55 66 77 88 99

Distribution for Distribution for = 4 = 4

xx1010 1111


D – 8

Waiting-Line Characteristics

Limited or unlimited queue length

Queue discipline: first-in, first-out is most common

Other priority rules may be used in special circumstances

D – 9

Service Characteristics

Queuing system designs Single-channel system, multiple-channel

system

Single-phase system, multiphase system

Service time distribution Constant service time

Random service times, usually a negative exponential distribution

D – 10

Queuing System Designs


DeparturesDeparturesafter serviceafter service

Single-channel, single-phase systemSingle-channel, single-phase system

Queue

ArrivalsArrivals

Single-channel, multiphase systemSingle-channel, multiphase system

ArrivalsArrivalsDeparturesDeparturesafter serviceafter service

Phase 1 service facility


Service facility

Queue

Your family dentist’s officeYour family dentist’s office

McDonald’s dual window drive-throughMcDonald’s dual window drive-through

D – 11


Figure D.3Figure D.3Multi-channel, single-phase systemMulti-channel, single-phase system

ArrivalsArrivals

Queue

Most bank and post office service windowsMost bank and post office service windows


Service facility

Channel 1

Service facility

Channel 2

Service facility

Channel 3

D – 12


Figure D.3Figure D.3Multi-channel, multiphase systemMulti-channel, multiphase system

ArrivalsArrivals

Queue

Some college registrationsSome college registrations



Channel 1


Channel 2


Channel 1


Channel 2

D – 13

Negative Exponential Distribution


1.0 1.0 –

0.9 0.9 –

0.8 0.8 –

0.7 0.7 –

0.6 0.6 –

0.5 0.5 –

0.4 0.4 –

0.3 0.3 –

0.2 0.2 –

0.1 0.1 –

0.0 0.0 –

Pro

bab

ility

th

at s

ervi

ce t

ime

Pro

bab

ility

th

at s

ervi

ce t

ime

≥ 1

≥ 1

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

0.000.00 0.250.25 0.500.50 0.750.75 1.001.00 1.251.25 1.501.50 1.751.75 2.002.00 2.252.25 2.502.50 2.752.75 3.003.00

Time t in hoursTime t in hours

Probability that service time is greater than t = eProbability that service time is greater than t = e-µ-µtt for t for t ≥ 1≥ 1

µ =µ = Average service rate Average service ratee e = 2.7183= 2.7183

Average service rate Average service rate (µ) = (µ) = 1 customer per hour1 customer per hour

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

D – 14

Measuring Queue Performance

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

2. Average queue length

3. Average time in the system

4. Average number of customers in the system

5. Probability the service facility will be idle

6. Utilization factor for the system

7. Probability of a specified number of customers in the system

D – 15

Queuing Costs


Total expected costTotal expected cost

Cost of providing serviceCost of providing service

CostCost

Low levelLow levelof serviceof service

High levelHigh levelof serviceof service

Cost of waiting timeCost of waiting time

MinimumMinimumTotalTotalcostcost

OptimalOptimalservice levelservice level

D – 16

Queuing Models

Table D.2Table D.2

ModelModel NameName ExampleExample

A Single channel Information counter system at department store(M/M/1)

NumberNumber NumberNumber ArrivalArrival ServiceServiceofof ofof RateRate TimeTime PopulationPopulation QueueQueue

ChannelsChannels PhasesPhases PatternPattern PatternPattern SizeSize DisciplineDiscipline

SingleSingle SingleSingle PoissonPoisson ExponentialExponential UnlimitedUnlimited FIFOFIFO

D – 17

Queuing Models

Table D.2Table D.2


BB Multichannel Multichannel Airline ticketAirline ticket (M/M/S) (M/M/S) counter counter



Multi-Multi- SingleSingle PoissonPoisson ExponentialExponential UnlimitedUnlimited FIFOFIFO channelchannel

D – 18

Queuing Models

Table D.2Table D.2


CC Constant Constant Automated car Automated car service service wash wash(M/D/1)(M/D/1)



SingleSingle SingleSingle PoissonPoisson ConstantConstant UnlimitedUnlimited FIFOFIFO

D – 19

Queuing Models

Table D.2Table D.2


DD Limited Limited Shop with only a Shop with only a population population dozen machines dozen machines

(finite)(finite) that might break that might break



SingleSingle SingleSingle PoissonPoisson ExponentialExponential LimitedLimited FIFOFIFO

D – 20

Model A - Single Channel

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

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

D – 21


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

D – 22


== Mean number of arrivals per time periodMean number of arrivals per time period

µµ == Mean number of units served per time Mean number of units served per time periodperiod

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

==

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

==

µ – µ –

11µ – µ –

Table D.3Table D.3

D – 23


LLqq== Average number of units waiting in the Average number of units waiting in the queuequeue

==

WWqq== Average time a unit spends waiting in Average time a unit spends waiting in the queuethe queue

==

pp == Utilization factor for the systemUtilization factor for the system

==

22

µ(µ – µ(µ – ))

µ(µ – µ(µ – ))

µµ

Table D.3Table D.3

D – 24


PP00 == Probability of 0 units in the system Probability of 0 units in the system (that is, the service unit is idle)(that is, the service unit is idle)

== 1 –1 –

PPn > kn > k ==Probability of more than Probability of more than kk units in the units in the system, where system, where nn is the number of units in is the number of units in the systemthe system

==

µµ

µµ

k k + 1+ 1

Table D.3Table D.3

D – 25

Single Channel Example

== 2 2 cars arriving/hourcars arriving/hourµµ= 3 = 3 cars serviced/hourcars serviced/hour

LLss= = = 2= = = 2 cars in cars in

the system on averagethe system on average

WWss = = = = 1= = 1 hour hour

average waiting time in the average waiting time in the systemsystem

LLqq== = = 1.33 = = 1.33

cars waiting in linecars waiting in line

22

µ(µ – µ(µ – ))

µ – µ –

11µ – µ –

22

3 - 23 - 2

11

3 - 23 - 2

2222

3(3 - 2)3(3 - 2)

D – 26


WWqq= = = =

= 40 = 40 minute average minute average waiting timewaiting time

pp= = /µ = 2/3 = 66.6% /µ = 2/3 = 66.6% of time mechanic is of time mechanic is busybusy

µ(µ – µ(µ – ))

223(3 - 2)3(3 - 2)

µµ

PP00= 1 - = .33= 1 - = .33 probability probability

there are there are 00 cars in the system cars in the system

== 2 2 cars arriving/hourcars arriving/hourµµ= 3 = 3 cars serviced/hourcars serviced/hour

D – 27


Probability of More Than k Cars in the SystemProbability of More Than k Cars in the System

kk PPn > kn > k = (2/3)= (2/3)k k + 1+ 1

00 .667.667 Note that this is equal toNote that this is equal to 1 - 1 - PP00 = 1 - .33 = 1 - .33

11 .444.444

22 .296.296

33 .198.198 Implies that there is aImplies that there is a 19.8% 19.8% chance that more thanchance that more than 3 3 cars are in the cars are in the systemsystem

44 .132.132

55 .088.088

66 .058.058

77 .039.039

D – 28

Single Channel EconomicsCustomer dissatisfactionCustomer dissatisfaction

and lost goodwill and lost goodwill = $10= $10 per hour per hour

WWqq = 2/3= 2/3 hour hour

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

Total hours Total hours customers spend customers spend waiting per daywaiting per day

= (16) = 10 = (16) = 10 hourshours2233

2233

Customer waiting-time cost Customer waiting-time cost = $10 10 = $106.67= $10 10 = $106.672233

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

D – 29

Measuring Queue Performance

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

2. Average queue length

3. Average time in the system

4. Average number of customers in the system

5. Probability the service facility will be idle

6. Utilization factor for the system

7. Probability of a specified number of customers in the system

D – 30

Queuing Models

Table D.2Table D.2


BB Multichannel Multichannel Airline ticketAirline ticket (M/M/S) (M/M/S) counter counter



Multi-Multi- SingleSingle PoissonPoisson ExponentialExponential UnlimitedUnlimited FIFOFIFO channelchannel

D – 31

PP00 = = Probability of 0 units in the system (that Probability of 0 units in the system (that is, the service unit is idle)is, the service unit is idle)

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

Lq = Lq = Average number of units waiting in the queueAverage number of units waiting in the queue

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

Wq Wq = Average time a unit spends waiting in the = Average time a unit spends waiting in the queuequeue

Multi-Channel Model

D – 32

Multi-Channel Model

MM == number of channels opennumber of channels open == average arrival rateaverage arrival rateµµ == average service rate at each average service rate at each channelchannel

PP00 = for M = for Mµ > µ > 11

11MM!!

11nn!!

MMµµMMµ - µ -

M M – 1– 1

n n = 0= 0

µµ

nnµµ

MM

++∑∑

LLss = P = P00 + +µ(µ(/µ)/µ)MM

((M M - 1)!(- 1)!(MMµ - µ - ))22

µµ

Table D.4Table D.4

D – 33

Multi-Channel Example

= 2 µ = 3 = 2 µ = 3 M M = 2= 2

PP00 = = = = 11

1122!!

11nn!!

2(3)2(3)

2(3) - 22(3) - 2

11

n n = 0= 0

2233

nn

2233

22

++∑∑

11

22

LLss = + = = + =(2)(3(2/3)(2)(3(2/3)22 22

331! 2(3) - 21! 2(3) - 2 22

11

22

33

44

WWqq = = .0415= = .0415.083.083

22WWss = = = =

3/43/4

22

33

88LLqq = – = = – =22

3333

44

11

1212

D – 34

Multi-Channel Example

Single Channel Two Channels

P0 .33 .5

Ls 2 cars .75 cars

Ws 60 minutes 22.5 minutes

Lq 1.33 cars .083 cars

Wq 40 minutes 2.5 minutes

D – 35

Queuing Models

Table D.2Table D.2


CC Constant Constant Automated car Automated car service service wash wash(M/D/1)(M/D/1)



SingleSingle SingleSingle PoissonPoisson ConstantConstant UnlimitedUnlimited FIFOFIFO

D – 36

Constant Service Model

Table D.5Table D.5

LLqq = = 22

2µ(µ – 2µ(µ – ))Average lengthAverage lengthof queueof queue

WWqq = = 2µ(µ – 2µ(µ – ))

Average waiting timeAverage waiting timein queuein queue

µµ

LLss = L = Lqq + + Average number ofAverage number ofcustomers in systemcustomers in system

WWss = W = Wqq + + 11µµ

Average waitingAverage waitingtime in systemtime in system

D – 37

Net savingsNet savings = $ 7 /= $ 7 /triptrip

Constant Service ExampleTrucks currently wait 15 minutes on averageTrucks currently wait 15 minutes on averageTruck and driver cost $60 per hourTruck and driver cost $60 per hourAutomated compactor service rate (µ) = 12 trucks per hourAutomated compactor service rate (µ) = 12 trucks per hourArrival rate (Arrival rate () = 8 per hour) = 8 per hourCompactor costs $3 per truckCompactor costs $3 per truck

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

WWqq = = = = hourhour882(12)(12 - 8)2(12)(12 - 8)

111212

Waiting cost/tripWaiting cost/tripwith compactorwith compactor = (1/12= (1/12 hr wait hr wait)($60/)($60/hr costhr cost)) = $ 5 /= $ 5 /triptrip

Savings withSavings withnew equipmentnew equipment

= $15 (= $15 (currentcurrent) - $ 5 () - $ 5 (newnew)) = $10 = $10 //triptrip

Cost of new equipment amortizedCost of new equipment amortized = = $ 3 /$ 3 /triptrip

D – 1 Operations Management Module D – Waiting-Line Models © 2006 Prentice Hall, Inc. PowerPoint presentation to accompany Heizer/Render Principles of.

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