10 - queuing system - industrial engineering 2011 · pdf file10/02/2013 · 9.2...

Queuing ModelsQueuing ModelsQueuing ModelsQueuing Models

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9.1 Introduction9.1 Introduction

• Queuing is the study of waiting lines, or queues.

• The objective of queuing analysis is to design systems that enable organizations to perform

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systems that enable organizations to perform optimally according to some criterion.

• Possible Criteria

– Maximum Profits.

– Desired Service Level.

9.1 Introduction9.1 Introduction

• Analyzing queuing systems requires a clear understanding of the appropriate service measurement.

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measurement.

• Possible service measurements

– Average time a customer spends in line.

– Average length of the waiting line.

– The probability that an arriving customer must wait for service.

9.2 Elements of the Queuing Process9.2 Elements of the Queuing Process

• A queuing system consists of three basic components:– Arrivals: Customers arrive according to some arrival

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– Arrivals: Customers arrive according to some arrival pattern.

– Waiting in a queue: Arriving customers may have to wait in one or more queues for service.

– Service: Customers receive service and leave the system.

The Arrival ProcessThe Arrival Process

• There are two possible types of arrival processes

– Deterministic arrival process.

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– Deterministic arrival process.

– Random arrival process.

• The random process is more common in businesses.

• Under three conditions the arrivals can be modeled as a Poisson process

– Orderliness : one customer, at most, will arrive during any

The Arrival ProcessThe Arrival Process

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time interval.

– Stationarity : for a given time frame, the probability of arrivals

within a certain time interval is the same for all time intervals of

equal length.

– Independence : the arrival of one customer has no influence

on the arrival of another.

P(X = k) = (λt)ke- λt

k!

The Poisson Arrival ProcessThe Poisson Arrival Process

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Whereλ = mean arrival rate per time unit.

t = the length of the interval.

e = 2.7182818 (the base of the natural logarithm).k! = k (k -1) (k -2) (k -3) … (3) (2) (1).

HANK’s HARDWARE HANK’s HARDWARE –– Arrival ProcessArrival Process

• Customers arrive at Hank’s Hardware according to a Poisson distribution.

• Between 8:00 and 9:00 . an average of 6

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• Between 8:00 and 9:00 A.M. an average of 6

customers arrive at the store.

• What is the probability that k customers will arrive

between 8:00 and 8:30 in the morning (k = 0, 1, 2,…)?

• Input to the Poisson distribution

λ = 6 customers per hour.

HANK’s HARDWARE HANK’s HARDWARE ––An illustration of the Poisson distribution. An illustration of the Poisson distribution.

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k

λ = 6 customers per hour.t = 0.5 hour.λt = (6)(0.5) = 3.

(λt) e- λt

k !=

0

0!0.049787

0

1!

1

0.149361

2

2!0.224042

3

3!0.224042

1 2 3 4 5 6 7

P(X = k )=

8

0123

HANK’s HARDWARE HANK’s HARDWARE ––Using Excel for the Poisson probabilitiesUsing Excel for the Poisson probabilities

• Solution

– We can use the POISSON function in Excel to determine Poisson probabilities.

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determine Poisson probabilities.

– Point probability: P(X = k) = ?

• Use Poisson(k, λt, FALSE)

• Example: P(X = 0; λt = 3) = POISSON(0, 1.5, FALSE)

– Cumulative probability: P(X≤k) = ?

• Example: P(X≤3; λt = 3) = Poisson(3, 1.5, TRUE)

HANK’s HARDWARE HANK’s HARDWARE ––Excel PoissonExcel Poisson

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• Factors that influence the modeling of queues

– Line configuration

The Waiting Line CharacteristicsThe Waiting Line Characteristics

– Priority

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– Line configuration

– Jockeying

– Balking

– Priority

– Tandem Queues

– Homogeneity

• A single service queue.

• Multiple service queue with single waiting line.

• Multiple service queue with multiple waiting

Line ConfigurationLine Configuration

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• Multiple service queue with multiple waiting lines.

• Tandem queue (multistage service system).

• Jockeying occurs when customers switch lines once they perceived that another line is moving faster.

Jockeying and BalkingJockeying and Balking

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• Balking occurs if customers avoid joining the line when they perceive the line to be too long.

• These rules select the next customer for service.

• There are several commonly used rules:

– First come first served (FCFS).

Priority RulesPriority Rules

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– First come first served (FCFS).

– Last come first served (LCFS).

– Estimated service time.

– Random selection of customers for service.

Tandem QueuesTandem Queues

• These are multi-server systems.

• A customer needs to visit several service stations (usually in a distinct order) to complete

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stations (usually in a distinct order) to complete the service process.

• Examples

– Patients in an emergency room.

– Passengers prepare for the next flight.

• A homogeneous customer population is one in which customers require essentially the same type of service.

HomogeneityHomogeneity

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• A non-homogeneous customer population is one in which customers can be categorized according to: – Different arrival patterns

– Different service treatments.

• In most business situations, service time varies widely among customers.

• When service time varies, it is treated as a

The Service ProcessThe Service Process

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• When service time varies, it is treated as a random variable.

• The exponential probability distribution is used sometimes to model customer service time.

f(t) = µe-µt

µ = the average number of customers who can be served per time period.

The Exponential Service Time DistributionThe Exponential Service Time Distribution

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who can be served per time period.Therefore, 1/µµµµ = the mean service time.

The probability that the service time X is less than some “t.”

P(X ≤ t) = 1 - e-µt

Schematic illustration of the exponential Schematic illustration of the exponential distributiondistribution

The probability that service is completed within t time units

P(X ≤ t) = 1 - e-µt

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P(X ≤ t) = 1 - e-µt

X = t

HANK’s HARDWARE HANK’s HARDWARE –– Service timeService time

• Hank’s estimates the average service time to be 1/µ = 4 minutes per customer.

• Service time follows an exponential distribution.

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• Service time follows an exponential distribution.

• What is the probability that it will take less than 3 minutes to serve the next customer?

• We can use the EXPDIST function in Excel to determine exponential probabilities.

• Probability density: f(t) = ?

Using Excel for the Exponential ProbabilitiesUsing Excel for the Exponential Probabilities

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• Probability density: f(t) = ?

– Use EXPONDIST(t, µ, FALSE)

• Cumulative probability: P(X≤k) = ?

– Use EXPONDIST(t, µ, TRUE)

• The mean number of customers served per minute is ¼ = ¼(60) = 15 customers per hour.

• P(X < .05 hours) = 1 – e-(15)(.05) = ?

HANK’s HARDWARE HANK’s HARDWARE ––Using Excel for the Exponential ProbabilitiesUsing Excel for the Exponential Probabilities

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• P(X < .05 hours) = 1 – e-(15)(.05) = ?

• From Excel we have:

– EXPONDIST(.05,15,TRUE) = .5276

3 minutes = .05 hours

HANK’s HARDWARE HANK’s HARDWARE ––Using Excel for the Exponential ProbabilitiesUsing Excel for the Exponential Probabilities

=EXPONDIST(B4,B3,TRUE)

Exponential Distribution for Mu = 15

16.000

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0.0002.0004.0006.0008.000

10.00012.00014.00016.000

0.000 0.075 0.150 0.225 0.300 0.375

t

f(t)

=EXPONDIST(A10,$B$3,FALSE)Drag to B11:B26

• The memoryless property.– No additional information about the time left for the completion of a

service, is gained by recording the time elapsed since the service started.

The Exponential Distribution The Exponential Distribution --CharacteristicsCharacteristics

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started.

– For Hank’s, the probability of completing a service within the next 3 minutes is (0.52763) independent of how long the customer has been served already.

• The Exponential and the Poisson distributions are related to one another.

– If customer arrivals follow a Poisson distribution with mean rate λλλλ,their interarrival times are exponentially distributed with mean time 1////λ.λ.λ.λ.

9.3 Performance Measures of 9.3 Performance Measures of Queuing System Queuing System

• Performance can be measured by focusing on:

– Customers in queue.

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– Customers in the system.

• Performance is measured for a system in steady state.

Roughly, this is a transient

n


• The transient period occurs at the initial time of operation.

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is a transient period…

Time

time of operation.

• Initial transient behavior is not indicative of long run performance.

This is a steady state

n


• The steady state period follows the transient period. Roughly, this

is a transient

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steady state period………..

Time

transient period.

• Meaningful long run performance measures can be calculated for the system when in steady state.

is a transient period…

In order to achieve steady state, theeffective arrival rate must be less than


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λ< λ< λ< λ< kµµµµEach with

service rate of µ

λ< µλ< µλ< µλ< µ1 1 1 1 +µ+µ+µ+µ2222++++…+µ+µ+µ+µκκκκFor k servers

with service rates µι

λ< µλ< µλ< µλ< µFor one server

effective arrival rate must be less than the sum of the effective service rates .

k servers

P0 = Probability that there are no customers in the system.

Pn = Probability that there are “n” customers in the system.

L = Average number of customers in the system.

L = Average number of customers in the queue.

Steady State Performance MeasuresSteady State Performance Measures

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Lq = Average number of customers in the queue.

W = Average time a customer spends in the system.

Wq = Average time a customer spends in the queue.

Pw = Probability that an arriving customer must wait for service.

ρ = Utilization rate for each server (the percentage of time that each server is busy).

• Little’s Formulas represent important relationships between L, Lq, W, and Wq.

• These formulas apply to systems that meet the following conditions:

Little’s FormulasLittle’s Formulas

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– Single queue systems,

– Customers arrive at a finite arrival rate λ, and

– The system operates under a steady state condition.

L = λ W Lq = λ Wq L = Lq + λ////µ

For the case of an infinite population

• Queuing system can be classified by:

– Arrival process.

– Service process.

– Number of servers.

Example:Example:

Classification of QueuesClassification of Queues

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– Number of servers.

– System size (infinite/finite waiting line).

– Population size.

• Notation

– M (Markovian) = Poisson arrivals or exponential service time.

– D (Deterministic) = Constant arrival rate or service time.

– G (General) = General probability for arrivals or service time.

M / M / 6 / 10 / 20M / M / 6 / 10 / 20

9.4 M9.4 M////////MM////////1 Queuing System 1 Queuing System --AssumptionsAssumptions

– Poisson arrival process.

– Exponential service time distribution.

– A single server.

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– A single server.

– Potentially infinite queue.

– An infinite population.

The probability that

P0 = 1 – (λ////µ)

Pn = [1 – (λ////µ)](λ////µ)n

L = λ ////(µ – λ)

M / M /1 Queue M / M /1 Queue -- Performance MeasuresPerformance Measures

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the system more than

The probability thata customer waits in the system more than “t” is P(X>t) = e-(µµµµ - λλλλ)t

Lq = λ2 ////[µ(µ – λ)]

W = 1 ////(µ – λ)

Wq = λ ////[µ(µ – λ)]

Pw = λ //// µρ = λ //// µ

MARY’s SHOESMARY’s SHOES

• Customers arrive at Mary’s Shoes every 12 minutes on the average, according to a Poisson process.

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• Service time is exponentially distributed with an average of 8 minutes per customer.

• Management is interested in determining the performance measures for this service system.

MARY’s SHOESMARY’s SHOES -- SolutionSolution

– Input

λ = 1/12 customers per minute = 60/12 = 5 per hour.

µ = 1/ 8 customers per minute = 60/ 8 = 7.5 per hour.

– Performance Calculations

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– Performance Calculations

P0 = 1 - (λ////µ) = 1 - (5////7.5) = 0.3333

Pn = [1 - (λ////µ)](λ////µ)n = (0.3333)(0.6667)n

L = λ////(µ - λ) = 2

Lq = λ2////[µ(µ - λ)] = 1.3333

W = 1////(µ - λ) = 0.4 hours = 24 minutes

Wq = λ/[/[/[/[µ(µ - λ)] = 0.26667 hours = 16 minutes

P0 = 1 - (λ////µ) = 1 - (5////7.5) = 0.3333

Pn = [1 - (λ////µ)](λ////µ)n = (0.3333)(0.6667)n

L = λ////(µ - λ) = 2

Lq = λ2////[µ(µ - λ)] = 1.3333

W = 1////(µ - λ) = 0.4 hours = 24 minutes

Wq = λ/[/[/[/[µ(µ - λ)] = 0.26667 hours = 16 minutes

Pw = λ////µ = 0.6667

ρ = λ////µ = 0.6667

P(X<10min) = 1 – e-2.5(10/60)

= .565

µ –λ = 7.5 – 5 = 2.5 per hr.

=1-=B4/B5

MARY’s SHOESMARY’s SHOES --

Spreadsheet solutionSpreadsheet solution

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=A11-B4/B5

=1-B4/B5

=A11/B4

=C11-1/B5

=B4/B5

=H11*($B$4/$B$5)Drag to Cell AL11

=1-E11

=B4/(B5-B4)

12.5 M12.5 M////////MM////////k Queuing Systemsk Queuing Systems

• Characteristics

– Customers arrive according to a Poisson process at a mean rate λ.

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mean rate λ.

– Service times follow an exponential distribution.

– There are k servers, each of who works at a rate of µ customers (with kµ> λ).

– Infinite population, and possibly infinite line.

P

n k

k

k

n k

n

k0

0

1

1

1 1=

+

−

∑

=

−

! !λ

µλ

µµ

µ λ

M / M /k Queue M / M /k Queue -- Performance MeasuresPerformance Measures

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Pn

P

k kP

n

n

n

n k

=

≤

=

−

λµ

λµ

!

!

0

0

for n k.

P for n > k.n

( ) ( )W

k kP

k

=

− −+

λµ µ

µ λ µ1

12 0

!

M / M /k Queue M / M /k Queue -- Performance MeasuresPerformance Measures

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The performance measurements L, Lq, Wq,, can be obtained from Little’s formulas.

Pk

k

kP

w

k

=

−

10

!λ

µµ

µ λ

ρ λµ

=k

LITTLE TOWN POST OFFICELITTLE TOWN POST OFFICE

• Little Town post office is open on Saturdays

between 9:00 a.m. and 1:00 p.m.

• Data

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– On the average 100 customers per hour visit the officeduring that period. Three clerks are on duty.

– Each service takes 1.5 minutes on the average.

– Poisson and Exponential distributions describe the arrival and the service processes respectively.

LITTLE TOWN POST OFFICELITTLE TOWN POST OFFICE

• The Postmaster needs to know the relevant service measures in order to:

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– Evaluate the current service level.

– Study the effects of reducing the staff by one clerk.

• This is an M / M / 3 queuing system.

– Input


LITTLE TOWN POST OFFICE LITTLE TOWN POST OFFICE -- SolutionSolution

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µ = 40 customers per hour (60////1.5).

Does steady state exist (λ < kµ )?

λ = 100 < kµ = 3(40) = 120.

LITTLE TOWN POST OFFICE LITTLE TOWN POST OFFICE –– solution solution continued continued

• First P0 is found by

)40(310011001

1P

2 3n0

=

∑

44

045.

625.152

25.65.21

1

100)40(3

)40(3

40

100

!3

1

40

100

!n

12

0n

3n

=+++

=

−

+

∑

=

• P0 is used now to determine all the other performance measures.

LITTLE TOWN POST OFFICE LITTLE TOWN POST OFFICE ––Spreadsheet SolutionSpreadsheet Solution

M/M/k Queuing Model

INPUTS Value INPUTS ValueLambda = 100Mu = 40

Probabiility of n Customers

Server Cost =Goodwill Cost When Waiting =Goodwill Cost While Being Served =

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OUTPUTS# Servers L Lq W Wq Pw Rho Cost 0 1 2

123 6.011236 3.511236 0.060112 0.035112 0.702247 0.833333 0 0.044944 0.11236 0.1404494 3.033095 0.533095 0.030331 0.005331 0.319857 0.625 0 0.073695 0.184237 0.2302975 2.630371 0.130371 0.026304 0.001304 0.130371 0.5 0 0.0801 0.20025 0.2503136 2.533889 0.033889 0.025339 0.000339 0.047445 0.416667 0 0.08162 0.204051 0.2550637 2.50858 0.00858 0.025086 8.58E-05 0.015443 0.357143 0 0.08198 0.204951 0.2561898 2.502053 0.002053 0.025021 2.05E-05 0.004517 0.3125 0 0.082063 0.205157 0.2564469 2.50046 0.00046 0.025005 4.6E-06 0.001195 0.277778 0 0.082081 0.205201 0.25650210 2.500096 9.59E-05 0.025001 9.59E-07 0.000288 0.25 0 0.082084 0.20521 0.25651311 2.500019 1.87E-05 0.025 1.87E-07 6.34E-05 0.227273 0 0.082085 0.205212 0.25651512 2.500003 3.4E-06 0.025 3.4E-08 1.29E-05 0.208333 0 0.082085 0.205212 0.25651613 2.500001 5.79E-07 0.025 5.79E-09 2.43E-06 0.192308 0 0.082085 0.205212 0.25651614 2.5 9.28E-08 0.025 9.28E-10 4.27E-07 0.178571 0 0.082085 0.205212 0.25651615 2.5 1.4E-08 0.025 1.4E-10 7.02E-08 0.166667 0 0.082085 0.205212 0.256516

in the System where n =

9.6 M9.6 M////////GG////////1 Queuing System1 Queuing System

• Assumptions

– Customers arrive according to a Poisson process with a

mean rate λ.

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mean rate λ. – Service time has a general distribution with mean rate µ.

– One server.

– Infinite population, and possibly infinite line.

( )L =

+

− +

22

2 1

λ σ λµ

λλµ

9.6 M9.6 M////////GG////////1 Queuing System 1 Queuing System ––Pollaczek Pollaczek -- Khintchine Formula for LKhintchine Formula for L

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−

2 1 λ

µµ

Note: It is not necessary to know the particular service time distribution.Only the mean and standard deviation of the distribution are needed.

• Ted’s repairs television sets and VCRs.

• Data

– It takes an average of 2.25 hours to repair a set.

– Standard deviation of the repair time is 45 minutes.

TED’S TV REPAIR SHOPTED’S TV REPAIR SHOP

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– Standard deviation of the repair time is 45 minutes.

– Customers arrive at the shop once every 2.5 hours on the average, according to a Poisson process.

– Ted works 9 hours a day, and has no help.

– He considers purchasing a new piece of equipment.• New average repair time is expected to be 2 hours.

• New standard deviation is expected to be 40 minutes.

Ted wants to know the effects of using the new equipment on –

TED’S TV REPAIR SHOPTED’S TV REPAIR SHOP

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1. The average number of sets waiting for repair;

2. The average time a customer has to wait to get his repaired set.

• This is an M////G////1 system (service time is not exponential (note that σ σ σ σ ≠ 1////µ).

• Input

TED’S TV REPAIR SHOP TED’S TV REPAIR SHOP -- SolutionSolution

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– The current system (without the new equipment)λ = 1/ 2.5 = 0.4 customers per hour.

µ = 1/ 2.25 = 0.4444 costumers per hour.

σ = 45/ 60 = 0.75 hours.

– The new system (with the new equipment)µ = 1/2 = 0.5 customers per hour.

σ = 40/ 60 = 0.6667 hours.

9.7 M / M / k / F Queuing System9.7 M / M / k / F Queuing System

• Many times queuing systems have designs that limit their line size.

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• When the potential queue is large, an infinite queue model gives accurate results, even though the queue might be limited.

• When the potential queue is small, the limited line must be accounted for in the model.

• Poisson arrival process at mean rate λ.

• k servers, each having an exponential service time with mean rate µ.

Characteristics of MCharacteristics of M////////MM////////kk////////F Queuing SystemF Queuing System

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time with mean rate µ.

• Maximum number of customers that can be present in the system at any one time is “F”.

• Customers are blocked (and never return) if the system is full.

• A customer is blocked if the system is full.

• The probability that the system is full is PF (100PF% of

MM////////MM////////kk////////F Queuing System F Queuing System ––Effective Arrival RateEffective Arrival Rate

53

F F

the arriving customers do not enter the system).

• The effective arrival rate = the rate of arrivals that make it through into the system (λe).

λe = λ(1 - PF)

RYAN ROOFING COMPANYRYAN ROOFING COMPANY

• Ryan gets most of its business from customers who call and order service.

– When a telephone line is available but the secretary

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– When a telephone line is available but the secretary is busy serving a customer, a new calling customer is willing to wait until the secretary becomes available.

– When all the lines are busy, a new calling customer gets a busy signal and calls a competitor.

• Data– Arrival process is Poisson, and service process is

Exponential.


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– Each phone call takes 3 minutes on the average.

– 10 customers per hour call the company on the average.

– One appointment secretary takes phone calls from 3 telephone lines.

• Management would like to design the following system:

– The fewest lines necessary.

– At most 2% of all callers get a busy signal.


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• Management is interested in the following information:

– The percentage of time the secretary is busy.

– The average number of customers kept on hold.

– The average time a customer is kept on hold.

– The actual percentage of callers who encounter a busy signal.

• This is an M////M////1////3 system

• Inputλ = 10 per hour.

RYAN ROOFING COMPANY RYAN ROOFING COMPANY -- SolutionSolution

M/M/1/4 system M/M/1/5 system

See spreadsheet next

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λ = 10 per hour.

µ = 20 per hour (1////3 per minute).

• Excel spreadsheet gives: P0 = 0.533, P1 = 0.133, P3 = 0.06

6.7% of the customers get a busy signal.

This is above the goal of 2%.

P0 = 0.516, P1 = 0.258, P2 = 0.129, P3 = 0.065, P4 = 0.032

3.2% of the customers get the busy signalStill above the goal of 2%

P0 = 0.508, P1 = 0.254, P2 = 0.127, P3 = 0.063, P4 = 0.032P5 = 0.016

1.6% of the customers get the busy signalThe goal of 2% has been achieved.

RYAN ROOFING COMPANYRYAN ROOFING COMPANY --

SpreadsheetSpreadsheet SolutionSolution

58

12.8 M / M / 1 / / m Queuing Systems12.8 M / M / 1 / / m Queuing Systems

• In this system the number of potential customers is finite and relatively small.

• As a result, the number of customers already in the system affects the rate of arrivals of the remaining

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system affects the rate of arrivals of the remaining customers.

• Characteristics– A single server.

– Exponential service time, Poisson arrival process.

– A population size of a (finite) m customers.

PACESETTER HOMESPACESETTER HOMES

• Pacesetter Homes runs four different development projects.

• Data– At each site running a project is interrupted once every 20 working

days on the average.

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– At each site running a project is interrupted once every 20 working days on the average.

– The V.P. for construction handles each stoppage.

• How long on the average a site is non-operational?– If it takes 2 days on the average to restart a project’s progress (the

V.P. is using the current car).

– If it takes 1.875 days on the average to restart a project’s progress (the V.P. is using a new car)

PACESETTER HOMES PACESETTER HOMES –– SolutionSolution

• This is an M////M////1////////4 system, where:

– The four sites are the four customers.

– The V.P. for construction is the server.

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– The V.P. for construction is the server.

• Input

λ = 0.05 (1////20)

µ = 0.5 µ = 0.533

(1////2 days, using the current car) (1/1.875 days, using a new car).

Performance Current New

Measures Car Car

Average number of customers in the system 0.467 0.435

Summary of results

PACESETTER HOMES PACESETTER HOMES –– Solution Solution continuedcontinued

62

Average number of customers in the system L 0.467 0.435

Average number of customers in the queue Lq 0.113 0.100

Average number of days a customer is in the system W 2.641 2.437

Average number of days a customer is in the queue Wq 0.641 0.562

The probability that an arriving customer will wait Pw 0.353 0.334

Oveall system effective utilization-factor ρρρρ 0.353 0.334

The probability that all servers are idle Po 0.647 0.666

PACESETTER HOMES PACESETTER HOMES ––Spreadsheet SolutionSpreadsheet Solution

M/M/1//m Queuing Model

INPUTS Value0.05

Mu = 0.533334

Lambda =

m =

63

Probabiility of n Customers OUTPUTS in the System where n =# Servers L Lq W Wq Pw Rho 0 1 2 3 4

1 0.43451 0.10025 2.43734 0.56233 0.33427 0.33427 0.66573 0.24965 0.07021 0.01317 0.00123

9.9 Economic Analysis of 9.9 Economic Analysis of Queuing SystemsQueuing Systems

• The performance measures previously developed are used next to determine a minimal cost queuing system.

64

system.

• The procedure requires estimated costs such as:

– Hourly cost per server .

– Customer goodwill cost while waiting in line.

– Customer goodwill cost while being served.

WILSON FOODS WILSON FOODS TALKING TURKEY HOT LINETALKING TURKEY HOT LINE

• Wilson Foods has an 800 number to answer customers’ questions.

• If all the customer representatives are busy when a

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• If all the customer representatives are busy when a new customer calls, he/she is asked to stay on the line.

• A customer stays on the line if the waiting time is not longer than 3 minutes.

• Data

– On the average 225 calls per hour are received.

– An average phone call takes 1.5 minutes.How many customer

WILSON FOODS WILSON FOODS TALKING TURKEY HOT LINETALKING TURKEY HOT LINE

66

– An average phone call takes 1.5 minutes.

– A customer will stay on the line waiting at most 3 minutes.

– A customer service representative is paid $16 per hour.

– Wilson pays the telephone company $0.18 per minute when the customer is on hold or when being served.

– Customer goodwill cost is $20 per minute while on hold.

– Customer goodwill cost while in service is $0.05.

How many customer service representatives should be usedto minimize the hourly cost of operation?

WILSON FOODS WILSON FOODS –– SolutionSolution

• The total hourly cost model

Total hourly wages Average hourly goodwill cost for customers on hold

67TC(K) = Cwk + (Ct + gs)Lq + (Ct + gs)(L – Lq)

TC(K) = Cwk + CtL + gwLq + gs(L - Lq)

Total average hourly Telephone charge

cost for customers on hold

Average hourly goodwillcost for customers in service

• Input

Cw= $16

Ct = $10.80 per hour [0.18(60)]

g = $12 per hour [0.20(60)]

WILSON FOODS WILSON FOODS –– Solution continued Solution continued

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gw= $12 per hour [0.20(60)]

gs = $3 per hour [0.05(60)]

– The Total Average Hourly Cost =

TC(K) = 16K + (10.8+3)L + (12 - 3)Lq = 16K + 13.8L + 9Lq

• Assuming a Poisson arrival process and an Exponential service time, we have an M/M/K system.

λ = 225 calls per hour.

µ = 40 per hour (60////1.5).

WILSON FOODS WILSON FOODS –– Solution continuedSolution continued

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µ ////

– The minimal possible value for K is 6 to ensure that steady state exists (λ<Kµ).

– Excel MMk worksheet was used to generate results for L, Lq, and Wq.

• Summary of results of the runs for k=6,7,8,9,10

K L Lq Wq TC(K)6 18.1255 12.5 0.05556 458.64

WILSON FOODS WILSON FOODS –– Solution continuedSolution continued

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6 18.1255 12.5 0.05556 458.647 7.6437 2.0187 0.00897 235.658 6.2777 0.6527 0.0029 220.519 5.8661 0.2411 0.00107 227.1210 5.7166 0.916 0.00041 239.70

Conclusion: employ 8 customer service representatives.Conclusion: employ 8 customer service representatives.

WILSON FOODS WILSON FOODS –– Spreadsheet SolutionSpreadsheet Solution

M/M/k Queuing Model

INPUTS Value INPUTS ValueLambda = 225 16Mu = 40 22.8

13.8

Probabiility of n Customers

Server Cost =Goodwill Cost When Waiting =Goodwill Cost While Being Served =

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OUTPUTS# Servers L Lq W Wq Pw Rho Cost 0 1 2

123456 18.1255 12.5005 0.080558 0.055558 0.833367 0.9375 458.6364 0.001184 0.006659 0.018737 7.643727 2.018727 0.033972 0.008972 0.493467 0.803571 235.652 0.002742 0.015423 0.0433768 6.277703 0.652703 0.027901 0.002901 0.275586 0.703125 220.5066 0.003291 0.018514 0.052079 5.866105 0.241105 0.026072 0.001072 0.144663 0.625 227.1222 0.003492 0.019641 0.05524110 5.716569 0.091569 0.025407 0.000407 0.07122 0.5625 239.7128 0.003565 0.020056 0.05640711 5.659381 0.034381 0.025153 0.000153 0.032853 0.511364 254.4089 0.003592 0.020206 0.05683112 5.637531 0.012531 0.025056 5.57E-05 0.014202 0.46875 269.9107 0.003602 0.02026 0.05698113 5.629393 0.004393 0.02502 1.95E-05 0.00576 0.432692 285.7252 0.003605 0.020278 0.057032

in the System where n =Probabiility of n Customers

HARGROVE HOSPITAL MATERNITY HARGROVE HOSPITAL MATERNITY WARDWARD

• Hargrove Hospital is experiencing cutbacks, and is trying to reorganize operations to reduce operating costs.

• There is a trade off between

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• There is a trade off between – the costs of operating more birthing stations and

– the costs of rescheduling surgeries in the surgery room when women give birth there, if all the birthing stations are occupied.

• The hospital wants to determine the optimal number of birthing stations that will minimize operating costs.

HARGROVE HOSPITAL MATERNITY HARGROVE HOSPITAL MATERNITY WARDWARD

• Data

– Cutting one birthing station saves $25,000 per year.

– Building a birthing station costs $30,000.

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– Building a birthing station costs $30,000.

– Maternity in the surgery room costs $400 per hour.

– Six women on the average need a birthing station a day. The arrival process is Poisson.

– Every birthing process occupies a birthing station for two hours on the average.

• Solution• Analysis of the current situation

– Currently there are two birthing stations

//// //// ////

HARGROVE HOSPITALHARGROVE HOSPITAL

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– The current problem can be modeled as a M////G////2////2 queuing system.

– Using the MGkk Excel worksheet with λλλλ = 6 and µµµµ = 12/day we have:

• ρρρρ = .23077

• W = .0833 days

• Pw = .076923

• L = .46154

• P0 = .6154

7.7% of the arriving women are sent tothe surgery room to give birth.

• Solution – continued • The birthing stations problem can be modeled as a

M////G////k////k queuing model.

• Input


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• Input– λ = 6 women per day;

µ = 12 women per day (24/2);

– k = the number of birthing stations used

• The total cost for the hospital is

TC(k) = Cost of using the surgery room for maternity

+ Additional cost of operating k stations

• Solution – continued

• Average daily cost of using the surgery room for maternity:

Pk(λλλλ)(average time in the surgery room)(hourly cost)


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k

• Average additional daily cost of operating k stations

25,000/365 = $68.49 per day.

• Average daily total cost

TC(k) = Pk(λλλλ)(24/µµµµ)(Hourly cost) + 68.49k

= Pk(6)(24/12)(400) + 68.49k

= 4800Pk + 68.49k

Additional Total net

• From repeated runs of the MGkk worksheet to determine Pk we got the following results:

HARGROVE HOSPITAL HARGROVE HOSPITAL -- SolutionSolution

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k Pk $4800Pk

Additional

Cost of stations

Total net

average

Daily cost

1

2

3

4

.333333

.076923

.012658

.001580

$1,600

369.23

60.76

7.58

$ – 68.49

0

82.19

163.38

$1,531.51

364.23

142.95

171.96

Optimal

Current

HARGROVE HOSPITAL HARGROVE HOSPITAL ––Spreadsheet SolutionSpreadsheet Solution

M/G/k/k Queuing Model

INPUTS Value6

122

Lambda =Mu =k =

78

OUTPUTS# Servers L Lq W Wq Pw Rho 0 1 2 3

2 0.461538 0 0.083333 0 0.076923 0.230769 0.615385 0.307692 0.076923

in the System where n =Probabiility of n Customers

9.10 Tandem Queuing Systems9.10 Tandem Queuing Systems

• In a Tandem Queuing System a customer must visit several different servers before service is completed.

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BeverageMeats

• Examples– All-You-Can-Eat restaurant


9.10 Tandem Queuing Systems

80

BeverageMeats


9.10 Tandem Queuing Systems


81

BeverageMeats

– A drive-in restaurant, where first you place your order, then pay and receive it in the next window.

– A multiple stage assembly line.


9.10 Tandem Queuing Systems9.10 Tandem Queuing Systems

• For cases in which customers arrive according to a Poisson process and service time in each station is exponential, ….

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Total Average Time in the System = Sum of all average times at the individual stations

Total Average Time in the System = Sum of all average times at the individual stations

BIG BOYS SOUND, INC.BIG BOYS SOUND, INC.

• Big Boys sells audio merchandise.

• The sale process is as follows:

– A customer places an order with a sales person.

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– A customer places an order with a sales person.

– The customer goes to the cashier station to pay for

the order.

– After paying, the customer is sent to the pickup desk

to obtain the good.

• Data for a regular Saturday

– Personnel.

• 8 sales persons are on the job.

• 3 cashiers.

• 2 workers in the merchandise pickup area.


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• 2 workers in the merchandise pickup area.

– Average service times.

• Average time a sales person waits on a customer is 10 minutes.

• Average time required for the payment process is 3 minutes.

• Average time in the pickup area is 2 minutes.

– Distributions.

• Exponential service time at all the service stations.

• Poisson arrival with a rate of 40 customers an hour.

What is the average amount of time, a

Only 75% of the arriving customers make a purchase!


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What is the average amount of time, a customer who makes a purchase spends in the store?

BIG BOYS SOUND, INC. BIG BOYS SOUND, INC. –– SolutionSolution

• This is a Three Station Tandem Queuing System

Pickup deskM / M / 2(.75)(40)=30

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Sales ClerksM / M / 8

CashiersM / M / 3

W1 = 14 minutesW2 = 3.47 minutes

W3 = 2.67 minutes

Total = 20.14 minutes.

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