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Queuing Models

Chapter 9

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9.1 Introduction

Queuing is the study of waiting lines, or queues.

The objective of queuing analysis is to design

systems that enable organizations to performoptimally according to some criterion.

Possible Criteria

Maximum Profits.

Desired Service Level.

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9.1 Introduction

Analyzing queuing systems requires a clearunderstanding of the appropriate service

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.

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9.2 Elements of the Queuing Process

A queuing system consists of three basiccomponents: 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.

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The Arrival Process

There are two possible types of arrivalprocesses

Deterministic arrival process.

Random arrival process.

The random process is more common inbusinesses.

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Under three conditions the arrivals can be modeled as aPoisson process

Orderliness : one customer, at most, will arrive during anytime interval.

Stationarity: for a given time frame, the probability of arrivalswithin 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.

The Arrival Process

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P(X = k) =

Wherel = 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).

(lt)ke- lt

k!

The Poisson Arrival Process

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HANKs HARDWARE Arrival Process

Customers arrive at Hanks Hardware according to a

Poisson distribution.

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

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k

Input to the Poissondistribution

l = 6 customers per hour.t = 0.5 hour.lt = (6)(0.5) = 3.

(lt) e- lt

k !

=

0

0!

0.049787

0

1!

1

0.149361

2

2!

0.2240423

3!

0.224042

1 2 3 4 5 6 7

P(X = k )=

8

0123

HANKs HARDWARE

An illustration of the Poisson distribution.

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HANKs HARDWARE

Using Excel for the Poisson probabilities Solution

We can use the POISSON function in Excel to

determine Poisson probabilities. Point probability: P(X = k) = ?

Use Poisson(k, lt, FALSE)

Example: P(X = 0; lt = 3) = POISSON(0, 3, FALSE) Cumulative probability: P(Xk) = ?

Example: P(X3; lt = 3) = Poisson(3, 3, TRUE)

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HANKs HARDWARE

Excel Poisson

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Factors that influence the modeling of queues

Line configuration

Jockeying

Balking

The Waiting Line Characteristics

Priority

Tandem Queues

Homogeneity

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A single service queue.

Multiple service queue with single waiting line.

Multiple service queue with multiple waitinglines.

Tandem queue (multistage service system).

Line Configuration

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Jockeying occurs when customers switch linesonce they perceived that another line is moving

faster. Balking occurs if customers avoid joining the line

when they perceive the line to be too long.

Jockeying and Balking

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These rules select the next customer for service.

There are several commonly used rules:

First come first served (FCFS).

Last come first served (LCFS).

Estimated service time.

Random selection of customers for service.

Priority Rules

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Tandem Queues

These are multi-server systems.

A customer needs to visit several service

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

Examples

Patients in an emergency room. Passengers prepare for the next flight.

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A homogeneous customer population is one inwhich customers require essentially the same

type of service. A non-homogeneous customer population is one

in which customers can be categorized accordingto: Different arrival patterns

Different service treatments.

Homogeneity

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In most business situations, service time varieswidely among customers.

When service time varies, it is treated as arandom variable.

The exponential probability distribution is used

sometimes to model customer service time.

The Service Process

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f(t) = me-mt

m = the average number of customerswho can be served per time period.Therefore, 1/m = the mean service time.

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

P(X t) = 1 - e-mt

The Exponential Service Time Distribution

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Schematic illustration of the exponential

distribution

The probability that service is completed

within t time unitsP(X t) = 1 - e-mt

X = t

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HANKs HARDWARE Service time

Hanks estimates the average service time to be

1/m = 4 minutes per customer.

Service time follows an exponential distribution.

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

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We can use the EXPDIST function in Excel todetermine exponential probabilities.

Probability density: f(t) = ? Use EXPONDIST(t, m, FALSE)

Cumulative probability: P(Xk) = ?

Use EXPONDIST(t, m, TRUE)

Using Excel for the Exponential Probabilities

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The mean number of customers served perminute is = (60) = 15 customers per hour.

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

From Excel we have:

EXPONDIST(.05,15,TRUE) = .5276

HANKs HARDWARE

Using Excel for the Exponential Probabilities

3 minutes = .05 hours

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HANKs HARDWARE

Using Excel for the Exponential Probabilities=EXPONDIST(B4,B3,TRUE)

Exponential Distribution for Mu = 15

0.000

2.0004.0006.0008.000

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

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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 servicestarted.

For Hanks, the probability of completing a service within the next 3minutes is (0.52763) independent of how long the customer has beenserved already.

The Exponential and the Poisson distributions are related toone another. If customer arrivals follow a Poisson distribution with mean rate l,

their interarrival times are exponentially distributed with mean time

1/l.

The Exponential Distribution -

Characteristics

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9.3 Performance Measures of

Queuing Systems Performance can be measured by focusing on:

Customers in queue. Customers in the system.

Performance is measured for a system in steadystate.

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Roughly, thisis a transientperiod

n

Time


Queuing Systems

The transient periodoccurs at the initialtime of operation.

Initial transientbehavior is not

indicative of long runperformance.

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This is asteady stateperiod..

n

Time


Queuing Systems The steady state

period follows the

transient period. Meaningful long run

performancemeasures can be

calculated for thesystem when insteady state.

Roughly, thisis a transientperiod

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l< kmEach with

service rate ofml< m1 +m2++mk

For k serverswith service rates mi

l< mFor one server

In order to achieve steady state, the

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


Queuing Systems

k servers

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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.

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 waitfor service.

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

Steady State Performance Measures

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Littles Formulas represent important relationshipsbetween L, Lq, W, and Wq.

These formulas apply to systems that meet the

following conditions: Single queue systems,

Customers arrive at a finite arrival rate l, and

The system operates under a steady state condition.

L = l W Lq = l Wq L = Lq + l/m

Littles Formulas

For the case of an infinite population

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Queuing system can be classified by:

Arrival process.

Service process.

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.

Example:

M / M / 6 / 10 / 20

Classification of Queues

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9.4 M/M/1 Queuing System -Assumptions

Poisson arrival process.

Exponential service time distribution.

A single server.

Potentially infinite queue.

An infinite population.

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P0 = 1 (l/m)Pn = [1 (l/m)](l/m)nL = l /(ml)Lq = l

2 /[m(ml)]W = 1 /(ml)Wq = l /[m(ml)]Pw = l / mr = l / m

M / M /1 Queue - Performance Measures

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MARYs SHOES

Customers arrive at Marys Shoes every 12

minutes on the average, according to a Poissonprocess.

Service time is exponentially distributed with anaverage of 8 minutes per customer.

Management is interested in determining theperformance measures for this service system.

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MARYs SHOES - Solution

Input

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

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

Performance Calculations

P0 = 1 - (l/m) = 1 - (5/7.5) = 0.3333Pn = [1 - (l/m)](l/m)n = (0.3333)(0.6667)nL = l/(m - l) = 2Lq = l

2/[m(m - l)] = 1.3333W = 1/(m - l) = 0.4 hours = 24 minutesWq = l/[m(m - l)] = 0.26667 hours = 16 minutes

Pw = l/m =0.6667r = l/m =0.6667

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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

MARYs SHOES -

Spreadsheet solution

=B4/(B5-B4)

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12.5 M/M/k Queuing Systems Characteristics

Customers arrive according to a Poisson process at a

mean rate l.

Service times follow an exponential distribution.

There are k servers, each of whom works at a rate ofmcustomers (with km> l).

Infinite population, and possibly infinite line.

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P

n k

k

k

n k

n

k0

0

1

1

1 1=

+

-

=

-

! !

l

m

l

m

m

m l

Pn

P

k kP

n

n

n

n k

=

=

-

lm

lm

!

!

0

0

for n k.

P for n > k.n

M / M /k Queue - Performance Measures

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( ) ( )

W

k k

P

k

=

- -

+

lm m

m lm

1

12 0

!

The performance measurements L, Lq, Wq,, can be obtainedfrom Littles formulas.

Pk

k

kP

w

k

= -

10

!

lm

mm l

r

l

m= k

M / M /k Queue - Performance Measures

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LITTLE TOWN POST OFFICE

Little Town post office is open on Saturdays

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

Data

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 thearrival and the service processes respectively.

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LITTLE TOWN POST OFFICE

The Postmaster needs to know the relevantservice measures in order to:

Evaluate the current service level.

Study the effects of reducing the staff by one clerk.

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This is an M / M / 3 queuing system.

Input

l = 100 customers per hour.m = 40 customers per hour (60/1.5).

Does steady state exist (l < km )?

l = 100 < km = 3(40) = 120.

LITTLE TOWN POST OFFICE - Solution

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LITTLE TOWN POST OFFICE solution

continued First P0 is found by

045.

625.15225.65.21

1

100)40(3

)40(3

40

100

!3

1

40

100

!n

1

1

P 2

0n

3n0

=

+++

=

-

+

= =

P0 is used now to determine all the otherperformance measures.

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LITTLE TOWN POST OFFICESpreadsheet Solution

M/M/k Queuing Model

INPUTS Value INPUTS Value

Lambda = 100

Mu = 40

OUTPUTS

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

1

2

3 6.011236 3.511236 0.060112 0.035112 0.702247 0.833333 0 0.044944 0.11236 0.140449

4 3.033095 0.533095 0.030331 0.005331 0.319857 0.625 0 0.073695 0.184237 0.230297

5 2.630371 0.130371 0.026304 0.001304 0.130371 0.5 0 0.0801 0.20025 0.250313

6 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 0.256189

8 2.502053 0.002053 0.025021 2.05E-05 0.004517 0.3125 0 0.082063 0.205157 0.256446

9 2.50046 0.00046 0.025005 4.6E-06 0.001195 0.277778 0 0.082081 0.205201 0.256502

10 2.500096 9.59E-05 0.025001 9.59E-07 0.000288 0.25 0 0.082084 0.20521 0.256513

11 2.500019 1.87E-05 0.025 1.87E-07 6.34E-05 0.227273 0 0.082085 0.205212 0.256515

12 2.500003 3.4E-06 0.025 3.4E-08 1.29E-05 0.208333 0 0.082085 0.205212 0.256516

13 2.500001 5.79E-07 0.025 5.79E-09 2.43E-06 0.192308 0 0.082085 0.205212 0.256516

14 2.5 9.28E-08 0.025 9.28E-10 4.27E-07 0.178571 0 0.082085 0.205212 0.256516

15 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 =

Probabiility of n Customers

Server Cost =

Goodwill Cost When Waiting =

Goodwill Cost While Being Served =

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9.6 M/G/1 Queuing System Assumptions

Customers arrive according to a Poisson process with a

mean rate l.

Service time has a general distribution with mean rate m.

One server.

Infinite population, and possibly infinite line.

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( )L

=

+

-

+

2

2

2 1

l lm

lm

l

m

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

9.6 M/G/1 Queuing SystemPollaczek - Khintchine Formula for L

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Teds 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. 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.

TEDS TV REPAIR SHOP

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Ted wants to know the effects of using the newequipment on

1. The average number of sets waiting for repair;

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

TEDS TV REPAIR SHOP

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This is an M/G/1 system (service time is notexponential (note that 1/m).

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

m = 1/2.25 = 0.4444 costumers per hour.

= 45/60 = 0.75 hours. The new system (with the new equipment)

m = 1/2 = 0.5 customers per hour.

= 40/60 = 0.6667 hours.

TEDS TV REPAIR SHOP - Solution

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9.7 M / M / k / F Queuing System

Many times queuing systems have designs thatlimit their line size.

When the potential queue is large, an infinitequeue model gives accurate results, eventhough the queue might be limited.

When the potential queue is small, the limitedline must be accounted for in the model.

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Poisson arrival process at mean rate l.

k servers, each having an exponential servicetime with mean rate m.

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.

Characteristics of M/M/k/F Queuing System

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A customer is blocked if the system is full.

The probability that the system is full is PF (100PF% ofthe arriving customers do not enter the system).

The effective arrival rate = the rate of arrivals thatmake it through into the system (le).

le = l(1 - PF)

M/M/k/F Queuing SystemEffective Arrival Rate

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RYAN ROOFING COMPANY

Ryan gets most of its business from customerswho call and order service.

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

When all the lines are busy, a new calling customergets a busy signal and calls a competitor.

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Data Arrival process is Poisson, and service process is

Exponential.

Each phone call takes 3 minutes on the average.

10 customers per hour call the company on theaverage.

One appointment secretary takes phone calls from3 telephone lines.


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Management would like to design the following system:

The fewest lines necessary.

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

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.


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This is an M/M/1/3 system Input

l = 10 per hour.m = 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%

RYAN ROOFING COMPANY - Solution

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

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

P5 = 0.0161.6% of the customers get the busy signalThe goal of 2% has been achieved.

See spreadsheet next

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RYAN ROOFING COMPANY-SpreadsheetSolution

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12.8 M / M / 1 / / m Queuing Systems

In this system the number of potential customers isfinite and relatively small.

As a result, the number of customers already in the

system affects the rate of arrivals of the remainingcustomers.

Characteristics A single server.

Exponential service time, Poisson arrival process.

A population size of a (finite) m customers.

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PACESETTER HOMES

Pacesetter Homes runs four different development projects.

Data

At each site running a project is interrupted once every 20 workingdays 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 projects progress (theV.P. is using the current car).

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

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PACESETTER HOMES Solution

This is an M/M/1//4 system, where: The four sites are the four customers.

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

l = 0.05 (1/20)m = 0.5 m = 0.533

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

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Performance Current New

Measures Car Car

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 0.562The probability that an arriving customer will wait Pw 0.353 0.334

Oveall system effective utilization-factor r 0.353 0.334The probability that all servers are idle Po 0.647 0.666

Summary of results

PACESETTER HOMES Solutioncontinued

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PACESETTER HOMESSpreadsheet Solution

M/M/1//m Queuing Model

INPUTS Value

0.05

Mu = 0.53333

4


OUTPUTS in the System where n =

# Server 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

Lambda =

m =

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9.9 Economic Analysis ofQueuing Systems

The performance measures previously developedare used next to determine a minimal cost queuing

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.

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WILSON FOODSTALKING TURKEY HOT LINE

Wilson Foods has an 800 number to answercustomers questions.

If all the customer representatives are busy when anew customer calls, he/she is asked to stay on theline.

A customer stays on the line if the waiting time isnot longer than 3 minutes.

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Data

On the average 225 calls per hour are received.

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 customerservice representativesshould be used

to minimize thehourly cost of operation?

WILSON FOODSTALKING TURKEY HOT LINE

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

WILSON FOODS Solution

The total hourly cost model

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

Total hourly wages

Total averagehourly Telephone charge

Average hourly goodwill

cost for customers on hold

Average hourly goodwillcost for customers in service

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Input

Cw= $16

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

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

WILSON FOODS Solution continued

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

l = 225 calls per hour.

m = 40 per hour (60/1.5). The minimal possible value for K is 6 to ensure that

steady state exists (l

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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

7 7.6437 2.0187 0.00897 235.65

8 6.2777 0.6527 0.0029 220.51

9 5.8661 0.2411 0.00107 227.1210 5.7166 0.916 0.00041 239.70

Conclusion: employ 8 customer service representatives.

WILSON FOODS Solution continued

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WILSON FOODS Spreadsheet Solution

M/M/k Queuing Model

INPUTS Value INPUTS Value

Lambda = 225 16

Mu = 40 22.8

13.8

OUTPUTS

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

1

2

3

4

5

6 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 0.043376

8 6.277703 0.652703 0.027901 0.002901 0.275586 0.703125 220.5066 0.003291 0.018514 0.05207

9 5.866105 0.241105 0.026072 0.001072 0.144663 0.625 227.1222 0.003492 0.019641 0.055241

10 5.716569 0.091569 0.025407 0.000407 0.07122 0.5625 239.7128 0.003565 0.020056 0.056407

11 5.659381 0.034381 0.025153 0.000153 0.032853 0.511364 254.4089 0.003592 0.020206 0.056831

12 5.637531 0.012531 0.025056 5.57E-05 0.014202 0.46875 269.9107 0.003602 0.02026 0.056981

13 5.629393 0.004393 0.02502 1.95E-05 0.00576 0.432692 285.7252 0.003605 0.020278 0.057032



Server Cost =

Goodwill Cost When Waiting =

Goodwill Cost While Being Served =

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HARGROVE HOSPITAL MATERNITYWARD

Hargrove Hospital is experiencing cutbacks, and istrying to reorganize operations to reduce operatingcosts.

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

the costs of rescheduling surgeries in the surgery room whenwomen 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.

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HARGROVE HOSPITAL MATERNITYWARD

Data

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

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.

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Solution Analysis of the current situation

Currently there are two birthing stations The current problem can be modeled as a M/G/2/2 queuingsystem.

Using the MGkk Excel worksheet with l = 6 and m = 12/day wehave:

r = .23077 W = .0833 days

Pw = .076923

L = .46154

P0 = .6154

HARGROVE HOSPITAL

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

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Solution continued The birthing stations problem can be modeled as a

M/G/k/k queuing model. Input l = 6 women per day;

m = 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

HARGROVE HOSPITAL

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Solution continued

Average daily cost of using the surgery room for maternity:

Pk(l)(average time in the surgery room)(hourly cost) Average additional daily cost of operating k stations

25,000/365 = $68.49 per day.

Average daily total costTC(k) = Pk(l)(24/m)(Hourly cost) + 68.49k

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

= 4800Pk + 68.49k

HARGROVE HOSPITAL

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k Pk $4800PkAdditional

Cost ofstations

Total net

average

Daily cost

12

3

4

.333333

.076923

.012658

.001580

$1,600369.23

60.76

7.58

$ 68.490

82.19

163.38

$1,531.51364.23

142.95

171.96

Optimal

From repeated runs of the MGkk worksheet todetermine Pk we got the following results:

HARGROVE HOSPITAL - Solution

Current

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HARGROVE HOSPITALSpreadsheet Solution

M/G/k/k Queuing Model

INPUTS Value

6

12

2

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



Lambda =

Mu =

k =

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9.10 Tandem Queuing Systems

In a Tandem Queuing System a customer must visitseveral different servers before service is completed.

BeverageMeats

Examples

All-You-Can-Eat restaurant

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BeverageMeats



Examples


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BeverageMeats


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

A multiple stage assembly line.

Examples



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For cases in which customers arrive according toa Poisson process and service time in each

station is exponential, .Total Average Time in the System =

Sum of all average times at the individual stations

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BIG BOYS SOUND, INC.

Big Boys sells audio merchandise.

The sale process is as follows:

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.

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Data for a regular Saturday

Personnel.

8 sales persons are on the job.

3 cashiers.

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.


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What is the average amount of time, acustomer who makes a purchase spendsin the store?

Only 75% of the arriving customers make a purchase!


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BIG BOYS SOUND, INC. Solution

This is a Three Station Tandem Queuing System

Sales ClerksM / M / 8

CashiersM / M / 3

Pickup deskM / M / 2

W1 = 14 minutesW2 = 3.47 minutes

W3 = 2.67 minutes

Total = 20.14 minutes.

(.75)(40)=30

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