QUEUING THEORY Body of knowledge about waiting lines Helps managers to better understand systems in...

QUEUING THEORYQUEUING THEORY

Body of knowledge about waiting lines

Helps managers to better understand systems in manufacturing, service, and maintenance

Provides competitive

advantage and cost saving

A QUEUE REPRESENTS ITEMS

OR PEOPLE AWAITING SERVICE

Applied Management Science for Decision Making, 1e Applied Management Science for Decision Making, 1e © 2012 Pearson Prentice-Hall, Inc. Philip A. Vaccaro , PhD© 2012 Pearson Prentice-Hall, Inc. Philip A. Vaccaro , PhD

Queue CharacteristicsQueue Characteristics

Average number of Average number of customers in a linecustomers in a line

Average number of Average number of customers in a service customers in a service facilityfacility

Probability a customer Probability a customer must waitmust wait

Average time a Average time a customer spends in a customer spends in a waiting line.waiting line.

Average time a Average time a customer spends in a customer spends in a service facilityservice facility

Percentage of time a Percentage of time a service facility is busyservice facility is busy

Queuing System ExamplesQueuing System Examples

SystemSystem CustomersCustomers ServersServersGrocery StoreGrocery Store ShoppersShoppers Checkout ClerksCheckout Clerks

Phone SystemPhone System Phone CallsPhone Calls Switching EquipmentSwitching Equipment

Toll HighwayToll Highway VehiclesVehicles TollgateTollgate

RestaurantRestaurant Parties of DinersParties of Diners Tables & WaitstaffTables & Waitstaff

FactoryFactory ProductsProducts WorkersWorkers

The Father of Queuing The Father of Queuing TheoryTheory

Danish engineer, who, in 1909 experimented with fluctuating demand in telephone traffic in Copenhagen.

In 1917, he published a report addressing the delays in auto- matic telephone dialing equipment.

At the end of World War II, his work was extended to more general problems, including waiting lines in business.

AGNER K. ERLANGAGNER K. ERLANG

Lack of Managerial Intuition Lack of Managerial Intuition Surrounding Waiting LinesSurrounding Waiting Lines

Queuing theory is not a matter of common sense. It is one of those applications where diligent, intelligent managers will arrive at drastically wrong solutions if they fail to thoroughly appreciate and understand the mathematics involved.

CostCost

MinimumMinimumTotalTotalCostCost

Low LevelLow LevelOf ServiceOf Service

High LevelHigh LevelOf ServiceOf Service

Optimal ServiceOptimal ServiceLevelLevel

Cost of Waiting Time

( time x value of time )

Cost of ProvidingService

( salaries + benefits )

Total CostTotal Cost

THE QUEUING COSTTHE QUEUING COSTTRADE-OFFTRADE-OFF

Aspects of a Queuing Aspects of a Queuing ProcessProcess

SYSTEM ARRIVALSSYSTEM ARRIVALS

THE QUEUE ITSELFTHE QUEUE ITSELF

THE SERVICE FACILITYTHE SERVICE FACILITY

The Calling PopulationThe Calling Population

The source of all systemThe source of all system arrivalsarrivals

It is usually of infinite size It is usually of infinite size

Theoretically, any personTheoretically, any person or object can enter the or object can enter the service facility duringservice facility during operating hoursoperating hours

PoissonPoisson Arrival Arrival DistributionDistribution

.25

.20

.15

.10

.05

.00

0 1 2 3 4 5 6 7 8 9 10

PoissonProbabilityDistribution

for λ = 2

(estimated mean arrival rate)

X ( the number of arrivals )

P( probability )

PoissonPoisson Arrival Arrival DistributionDistribution

.25

.20

.15

.10

.05

.00

0 1 2 3 4 5 6 7 8 9 10

PoissonProbabilityDistribution

for λ = 4

(estimated mean arrival rate)

X ( the number of arrivals )

P( probability )

EstablishEstablishing A Discrete ing A Discrete Poisson Poisson Arrival DistributionArrival Distribution

Given any averageGiven any average arrival rate ( arrival rate ( λλ ) in seconds, minutes, hours, days: ) in seconds, minutes, hours, days:

P ( X ) = ε λ X!

( FOR X = 0,1,2,3,4,5, etc. )

Where : P ( X ) = probability of X arrivals X = number of arrivals per time unit λ = the average arrival rate ε = 2.7183 ( base of the natural logarithm )

- λ x

EXAMPLE

If the average arrival rateper hour is two people

( λ = 2 ) , what is theprobability of three ( 3 )

arrivals per hour?

SolutionSolution

Given Given λλ = 2 : = 2 :

P ( 3 ) = 2.7183 2 3 !

3

= [ 1 / 7.389 ] x 8 (3)(2)(1)

= .1353 x 8 = .1804 ≈

- 2

618%

P ( X ) =

- λ ε λ

X

X !

The Remaining The Remaining ProbabilitiesProbabilitiesGIVEN THAT GIVEN THAT λλ = 2 = 2

P ( 0 arrivals ) = 14%P ( 1 arrival ) = 28%P ( 2 arrivals ) = 28%P ( 3 arrivals ) = 18%P ( 4 arrivals ) = 9%P ( 5 arrivals ) = 4%P ( 6 arrivals ) = 2%P ( 7 arrivals ) = 1%P ( 8 arrivals ) = .8%P ( 9 arrivals ) = .6%P ( => 10 “ ) = 0%

Poisson Probability TableFor a given value of λ , entry indicates the probability of obtaining a specified value of ‘X’

0 .1653 .1496 .1353

1 .2975 .2842 .2707

2 .2678 .2700 .2707

33 .1607.1607 .1710.1710 .1804

4 .0723 .0812 .0902

5 .0260 .0309 .0361

6 .0078 .0098 .0120

7 .0020 .0027 .0034

8 .0005 .0006 .0009

9 .0001 .0001 .0002

X λ = 1.8 λ = 1.9 λ = 2.0

EXAMPLE

Precise Precise TerminologyTerminology

THIS DISTRIBUTION MAYTHIS DISTRIBUTION MAYOR MAY NOT BE POISSONOR MAY NOT BE POISSON

DISTRIBUTED.DISTRIBUTED.

The discrete arrivalprobability distribution,based on the average arrival rate ( λ ) which was computed from

the actual systemobservations

Theoretical Distribution

The actual discrete arrivalprobability distributionthat was constructed

from the actual system observations

Observed Distribution

THIS CAN BE ESTABLISHED BY A THIS CAN BE ESTABLISHED BY A GOODNESS – OF - FITGOODNESS – OF - FIT HYPOTHESIS TEST HYPOTHESIS TEST

The theoretical poisson arrival probability distribution must be

statistically identical to the observed arrival probability distribution

If the two probability distributions are notfound to be statistically identical, we areforced to study and solve the problem

via simulation modeling

Service TimesService Times

.25

.20

.15

.10

.05

.00

PROBABILITY

0 30 60 90 120 150 180 210seconds

Service timesnormally

follow a negative exponentialprobabilitydistribution

THE PROBABILITY A CUSTOMER

WILL REQUIRE THAT SERVICE

TIME

Queue DisciplineQueue Discipline

BalkingBalking, , renegingreneging, and, and jockeying jockeying are not are not permitted in the service system.permitted in the service system.

JockeyingJockeying is the switching from one waiting is the switching from one waiting line to another.line to another.

JOCKEYING CAN BE DISCOURAGED BY PLACINGBARRICADES SUCH AS MAGAZINE RACKS AND

IMPULSE ITEM DISPLAYS BETWEEN WAITING LINES

Queuing Theory Queuing Theory VariablesVariables

Lambda ( λ ) is the average arrival rate

of people or items into the service system.

It can be expressed in seconds, minutes, hours, or days.

From the Greek small letter “ L “.


Mu ( μ ) is the average service rate of the service system.

It can be expressed as the number of people or items processed per second, minute, hour, or day.

From the Greek small letter “ M “.


Rho ( ρ ) is the % of time that the service facility is busy on the average.

It is also known as the

utilization rate. From the Greek small

letter “ R “.

“BUSY” IS DEFINED AS AT LEAST ONE PERSON OR ITEM IN THE SYSTEM

Queuing Theory VariablesQueuing Theory Variables

Mu ( M ) is a channel or service point in the ser-vice system.

Examples are gasoline pumps, checkout coun-ters, vending machines, bank teller windows.

From the Greek large letter “ M “.


Phases are the number of service points that must be negotiated by

a customer or item before leaving the service system.

They have no symbol. A CARWASH TAKES A VEHICLETHROUGH SEVERAL PHASES:PRE-WASH, WASH, WAX, ANDDRY BEFORE IT IS ALLOWED

TO LEAVE THE FACILITY.


• Po or ( 1 – ρ ) is the percentage of time that the service facility is idle.

• L is the average number of people or items in the service system both waiting to be served and currently being served.

• Lq is the average number of people or items in the waiting line ( queue ) only !


• W is the average time a customer or item spends in the service system, both waiting and receiving service.

• Wq is the average time a customer or item spends in the waiting line ( queue ) only.

• Pw is the probability that a customer or item must wait to be served.

The average number of customers or items processed by the entire

service system


“Mμ” is the effective service rate.*

* [ NUMBER OF SERVERS ] x [ AVERAGE SERVICE RATE PER SERVER ]

It can be expressed in seconds, minutes, hours, or days

IMPORTANT IMPORTANT CONSIDERATIONCONSIDERATION

The average service rate must always exceed the average arrival rate.

Otherwise, the queue will grow to infinity.

μ > λ

THERE WOULD BE NO SOLUTION !

Single-Channel / Single-Single-Channel / Single-Phase SystemPhase System

ONE WAITING LINE or QUEUEONE WAITING LINE or QUEUE ONE SERVICE POINT or CHANNELONE SERVICE POINT or CHANNEL

EXIT

Dual-Channel / Single-Phase Dual-Channel / Single-Phase SystemSystem

ONE OR TWO WAITING LINESONE OR TWO WAITING LINES TWO DUPLICATE SERVICE POINTSTWO DUPLICATE SERVICE POINTS

EXITEXIT

No JockeyingPermitted

Between Lines

Dual-Channel / Triple-PhaseDual-Channel / Triple-PhaseSystemSystem

TWO IDENTICAL SERVICE CHANNELS. EACH CHANNEL HAS 3 DISTINCT SERVICE POINTS ( A-B-C )

EXITEXIT

A A

BB

C C

ENTER ENTER JockeyingIs Permitted

Between Lines !

The Service SystemThe Service SystemCOMPRISED OF TWO GROUPSCOMPRISED OF TWO GROUPS

SUPERMARKET SHOPPERS ARE NOT IN THE SUPERMARKET SHOPPERS ARE NOT IN THE SERVICE SYSTEM UNTIL THEY MOVE SERVICE SYSTEM UNTIL THEY MOVE

TO THE CHECKOUT AREATO THE CHECKOUT AREA

RESTAURANT PATRONS ENTER THE SERVICERESTAURANT PATRONS ENTER THE SERVICE SYSTEM AS SOON AS THEY ARRIVESYSTEM AS SOON AS THEY ARRIVE

Customers or itemswaiting to be served or

processedCustomers or items

currently being servedor processed

Queuing Theory Queuing Theory LimitationsLimitations

BJ’s WHOLESALE CLUB HAS FOURTEEN (14) CHECKOUTS.

HOWEVER,THEY COULD BE DIVIDED INTO CONTRACTOR, EXPRESS, CASH-ONLY,

AND CREDIT-CARD-ONLY SUBSYSTEMS.

Formulae only accommodate eight ( 8 )channels and / or eight ( 8 ) phases

If service systems exceed the above,it may be possible to divide them intosub-systems for separate analyses.

Stealth Queuing SystemsStealth Queuing SystemsNORMAL CHARACTERISTICS MISSINGNORMAL CHARACTERISTICS MISSING

VISITING NURSES, PLUMBERS,

ELECTRICIANS

BROKEN MACHINES WAITING FOR A MECHANIC, OR SEATED PATIENTS

IN A DENTIST’S OFFICE, ORWORK-IN-PROCESS INVENTORY

WAITING FOR PROCESSING.

Fixed channels may be replaced by

mobile servers who carry portable

equipment and make housecalls.

Moving waiting linesmay be replaced bysitting customers

or stockpiled items.

Behavioral ConsiderationsBehavioral Considerations

Customer willingness to wait depends on what is perceived as reasonable.

Waiting lines that are

always moving are perceived as less

painful.

Customer willingness to wait is higher if they know that others are also waiting their

turn.

Customers should be permitted to perform the services that they can easily provide for themselves.


Behavioral ConsiderationsBehavioral Considerations

Well projected waiting

times allow customers to adjust their expectations

and therefore their aggravation.

If customers are kept If customers are kept

busy, their waiting time busy, their waiting time

may not be construed as may not be construed as wasted time.wasted time.

Customers should be rewarded with price discounts or gifts if

they must wait beyond a certain period of time.


FILLING OUT SURVEYS AND FORMS, BEING ENTERTAINED

IT SHOWS THAT THE FIRM VALUESTHEIR TIME AND IS WILLING TO PAY

THEM FOR IT IF THE WAIT IS TOO LONG

Single-Channel / Single-Single-Channel / Single-Phase ModelPhase Model

The Average Number of Customers in the System

L =λ

μ - λ


The Average Number Just Waiting in Line

Lq =λ

μ ( μ - λ )

2


Average Customer Time Spent in the System

W = ( μ - λ )

1


Percentage of Time the System is Busy

ρ = μ

λ

Therefore:

μ = 30

λ = 20

M = 1

Single-Channel / Single-Phase Single-Channel / Single-Phase ModelModel

A clerk can serve thirty customers per hour on

average.

Twenty customers arrive each hour on average.

APPLICATIONAPPLICATION

Single-Channel / Single-Phase Single-Channel / Single-Phase ModelModelAPPLICATIONAPPLICATION

The Average Number of Customers in the System

L = 20

( 30 - 20 )= 2


The Average Number Just Waiting in Line

Lq = ( 20 )

30 ( 30 - 20 )= 1.33

2


The Average Customer Time Spent in the System

W = 1

( 30 - 20 )= .10 hrs

( 6 minutes )


The Percentage of Time the System is Busy

ρ = 20

30 = 67%

QM for WindowsQM for WindowsQUEUING APPLICATIONSQUEUING APPLICATIONS

Single-ChannelSingle-Phase

Model

WE SCROLL TO“WAITING LINES”

One ( 1 ) Clerk On-Duty

Twenty ( 20 ) Arrivals per Hour

Thirty ( 30 ) Customers Can BeServed per Hour

The probability of exactly seven (7)persons in the system is 2%

( = k )

The probability of seven or fewerpersons in the system is 96%

( <= k )

The probability of more than seven (7)persons in the system is 4%

( > k )

Queuing Theory ModelingQueuing Theory Modeling withwith


Model

Templateand

Sample Data

Multi-Channel Single-Phase Multi-Channel Single-Phase SystemsSystems

This system is a singlewaiting line serviced bymore than one server.It assumes:

an infinite calling population a first-come, first-served queue discipline a poisson arrival rate negative exponential service times

Additional Parameters

M = number of servers or channels

Mμ = mean effective service rate for

the facility


The probability that the service facility is idle:

Po = 1

n = M-1 n M Σ 1/n! (λ / μ) + 1 (λ / μ) . Mμ n=0 M! Mμ-λ


The average number of customers in the system:

λμ ( λ / μ ) L = . Po + λ / μ (M-1)! (Mμ-λ)

M

2


The average number of customers in the queue:

Lq = L – ( λ / μ )

The average time a customer spends in the system:

W = L / λ

The average waiting time in the queue:

Wq = W – ( 1 / μ ) or Lq / λ

The probability that all the system’s servers are currently busy:

1 λ . Mμ . PoPw = M! μ Mμ-λ

M

Multi-Channel Single-Phase Multi-Channel Single-Phase SystemsSystems Application ExampleApplication Example

A bank has three loan officers on duty, each of whom can servefour customers per hour. Every hour, ten loan applicants arriveat the loan department and join a common queue. What are thesystem’s operating characteristics?

0 1 2 3

1 (10 / 4 ) + 1 (10 / 4 ) + 1 ( 10 / 4 ) + 1 . ( 10 / 4 ) . 3(4) 0! 1! 2! 3! 3(4) - 10

1 Po =

= .045 = 4.5%

Multi-Channel Single-Phase Multi-Channel Single-Phase SystemsSystemsApplication Example ContinuedApplication Example Continued

L = ( 10 )( 4 )( 10 / 4 ) . ( .045 ) + [ 10 / 4 ] = ( 3 – 1 ) ! [ 3 ( 4 ) – 10 ]

3

2

L = ( 40 )( 2.5 )3

2 ! [ 12 - 10 ] 2. ( .045 ) + 2.5 =

L = ( 40 )( 15.625 )

( 2 )( 1 ) [ 2 ] 2x .045 + 2.5 =

L = [ ( 625 / 8 ) x .045 ] + 2.5 =

L = 3.515625 + 2.5 ≈ 6.0


Application Example ContinuedApplication Example Continued

Lq = 6 – [10/4] = 3.5

W = 6/10 = .60 hours ( 36 minutes )

Wq = 3.5 / 10 = .35 hours ( 21 minutes )

1 10 . 3(4) . (.045) = .703 = 70.3% 3! 4 3(4)-10

3

Pw =

L = 6


Multi-ChannelSingle-Phase

Model

Templateand

Sample Data

Finite Calling Population Finite Calling Population ModelModel ApplicationApplication

A shop has fifteen (15) machines which are repaired in thesame order in which they fail. The machines fail accordingto a poisson distribution, and the service times are expo-nentially distributed.One (1) mechanic is on-duty. A machine fails on average,every forty (40) hours. The average repair takes 3.6 hours.

N = 15 machines

λ = 1/40th of a machine per hour = .0250 machine per hour

μ = 1/3.6th of a machine per hour = .2778 machine per hour

Finite Calling Population Finite Calling Population ModelModel

N = size of the finite calling population

Probability that the system is empty:

N! λ (N – n)! μ

1

Σn = 0

N nPo =

Finite Calling Population Finite Calling Population ModelModelAverage length of the queue

Lq = N – λ + μ ( 1 – Po ) λ

Finite Calling Population Finite Calling Population ModelModel

Average number of customers (items) in the system:

L = Lq + ( 1 – Po )

Average waiting time in the queue:

( N – L ) λ

Lq

Average time in the system:

W = Wq + ( 1 / μ )

Wq =

Finite Calling Population Finite Calling Population ModelModel ApplicationApplication

15! .0250 ( 15 – n )! .2778

15

Σn = 0

1

n= .0616 = 6.16%Po =

Lq = 15 – .0250 + .2778 ( 1 - .0616 ) = 3.63 machines .0250

Finite Calling Population Finite Calling Population ModelModelAPPLICATIONAPPLICATION

L = 3.63 + ( 1 - .0616 ) = 4.57 machines

Wq = = 13.94 hours ( 15 – 4.57 ) ( .0250 )

W = 13.94 + ( 1 / .2778 ) = 17.54 hours

3.63

Finite Calling Population Finite Calling Population ModelModelPerformance Summary

Number

of

Mechanics

Utilization

Rate

( ρ )

Average

Wait Line

(Lq)

Average

Number In

System (L)

Average

Wait

(Wq)

Average

Time In

System (W)

Probability

of Waiting

(Pw)

M = 1 .9384

3.63

Machines

4.57

Machines

13.94

Hours

17.54

Hours 91.14%

M = 2 .6008

.4464

Machines

1.648

Machines

1.337

Hours

4.9372

Hours 39.49%

M = 3 .4109

.0678

Machines

1.300

Machines

.198

Hours

3.7977

Hours 11.47%

M = 4 .3094

.0099

Machines

1.247

Machines

.0287

Hours

3.6284

Hours .0251

Finite Calling Population Finite Calling Population ModelModelCost Summary

Number

of

Mechanics

Total

Hourly

Wage

Average

Number In

System (L)

Total Hourly

Opportunity

Cost

Total

Cost

M = 1 $24.00 4.57

Machines

$13,710.00 $13,734.00

M = 2 $48.00 1.648

Machines

4,944.00 $4,992.00

M = 3 $72.00 1.300

Machines

$3,900.00 $3,972.00

M = 4 $96.00 1.247

Machines

$3,741.00 $3,837.00

Assume Machine HourlyOpportunity Cost of $3,000.00



Finite Calling PopulationModel

The Mechanic Problem

operational analysis and

cost analysis

We assume that a mechanic earns$24.00 per hour on average.

We also assume that theopportunity cost of an

out-of-service machine is $3,000.00 per hour.



Finite Calling PopulationModel

The Mechanic Problem

( operational analysis only )

Templateand

Sample Data

Kendall-Lee Kendall-Lee ConventionConvention

Widely accepted classification system for queuing

models.

Indicates the pattern of arrivals, the service time

distribution, and the number of channels in a model.

Often encountered in queuing software.

Known also as the Kendall Notation.

Basic Three - Symbol Basic Three - Symbol NotationNotation

Arrival Distribution

Service Time

Distribution

Service Channels

Open

11stst 2 2ndnd 3 3rdrd

Where:Where: M = poisson distributionM = poisson distribution D = constant (deterministic) rateD = constant (deterministic) rate G = general distribution with mean and variance knownG = general distribution with mean and variance known m / s = number of channels or serversm / s = number of channels or servers

The TemplateThe Template

1st Example1st Example

M M 1

PoissonPoissonArrivalArrival

DistributionDistribution

Negative Negative ExponentialExponentialService TimeService TimeDistributionDistribution

SINGLE-CHANNELSINGLE-CHANNELSINGLE-SERVERSINGLE-SERVER

MODELMODEL

2nd Example2nd Example

M M m

Poisson Poisson ArrivalArrival


NegativeNegativeExponentialExponentialService TimeService TimeDistributionDistribution

MULTI-CHANNELMULTI-CHANNELMULTI-SERVERMULTI-SERVER

MODELMODEL

2nd Example2nd Example

M M s




MULTI-CHANNELMULTI-CHANNELMULTI-SERVERMULTI-SERVER

MODELMODEL

3rd Example3rd Example

M M 1




SINGLE-CHANNELSINGLE-CHANNELSINGLE-SERVERSINGLE-SERVER

MODELMODEL

WITH FINITE POPULATIONWITH FINITE POPULATION

WITH FINITECALLING

POPULATION

ImposedImposedLegendLegend

4th Example4th Example

M D 3

Poisson Poisson ArrivalArrival


ConstantConstantService TimeService TimeDistributionDistribution

TRIPLE-CHANNELTRIPLE-CHANNELTRIPLE-SERVERTRIPLE-SERVER

MODELMODEL

MD3 ExampleMD3 Example

3 stamping machines in a work center 5 second fixed stamp time per inserted disc blank steel discs follow a poisson arrival pattern into the center

5th Example5th Example

M G 2



The Normal CurveThe Normal CurveService TimeService TimeDistributionDistribution

DUAL-CHANNELDUAL-CHANNELDUAL-SERVERDUAL-SERVER

MODELMODEL

MG2 ExampleMG2 Example

An office copy room has 2 copiers Copy job time is normally distributed The mean copy job time is 2 minutes The standard deviation is 30 seconds Employees follow a poisson arrival pattern into the copy room

μ = 2.0 min

σ = 30 secs

M = 2

What We Have SeenWhat We Have Seen

The Convenience Store Clerk

The Loan Officers

The Mechanic

M / M / 1

M / M / m or M / M / s

M / M / 1 with finite calling population

PROBLEMPROBLEM MODELMODEL


Applied Management Science for Decision Making, 1e Applied Management Science for Decision Making, 1e © 2012 © 2012 Pearson Prentice-Hall, Inc. Philip A. Vaccaro , PhDPearson Prentice-Hall, Inc. Philip A. Vaccaro , PhD

Solved ProblemsSolved Problems

Queuing TheoryQueuing TheoryComputer-BasedComputer-Based

ManualManual


The Post OfficeThe Post OfficeQueuing TheoryQueuing Theory

Problem 1

A post office has a single line for customers to use while waiting forthe next available postal clerk. There are two postal clerks who workat the same rate. The arrival rate of customers follows a poisson dis-tribution, while the service time follows an exponential distribution.The average arrival rate is one customer every three ( 3 ) minutes andthe average service rate is one customer every two ( 2 ) minutes foreach of the two clerks. The facility is idle 50% of the time ( Po = .50 ).


REQUIREMENT:

1. What is the average length of the line?

2. How long does the average person spend waiting for a clerk to become available?

3. What proportion of the time are both clerks idle?Po = .50 BUT NOW SHOW ALL SUPPORTING CALCULATIONS


Problem 1

Given: λ = 20 and μ = 30 ( per hour rates ) and M = 2

The average number of customers in the system ( L ) * :

L = ( 20 x 30 ) ( .67 )

( 2 - 1 )! (2 x 30 – 20)

x ( .50 ) + .67

2

2

L =600 ( .4489 )

( 60 – 20 )2

X ( .50 ) + .67 = .754165

* * L needs to be calculated before Lq can be found.L needs to be calculated before Lq can be found.

Po = .50


Problem 1


The average length of the line ( Lq ) :

Lq = L – ( λ / μ )

Lq = .7541 – ( 20 / 30 ) = .0841


REQUIREMENT:





W = L / λ = ( .7541 / 20 ) = .0377 of an hour or 2.25 minutes

Problem 1

The average time in the system ( W ) :

The average time in the queue ( Wq ) :

Wq = Lq / λ = ( .0841 / 20 ) = .0042 of an hour or .25 minutes


REQUIREMENT:





1 20 1 20 1 x 20 x 2(30)

0! 30 1! 30 (2)(1) 30 2(30) - 20

1Po =

00 1122

++

Po =

[ 1 + .67 ] + .50 ( .4489 ) ( 1.5 )

1

Po = .50 - THE PROBABILITY THAT THE POST OFFICE IS IDLE

++

The Post Office RevisitedThe Post Office RevisitedQueuing TheoryQueuing Theory

Problem 2

A post office has a single line for customers to use while waiting for the next available postal clerk. There are three postal clerks who work at the same rate. The arrival rate of customers follows a poisson distribution, while the service time follows an exponential distribution. The average arrival rate is one customer every three ( 3 ) minutes and the average service rate is one customer every two minutes for each of the three clerks. The facility is idle 51.22% of the time ( Po = .5122 ).


REQUIREMENT:



3. What proportion of the time are all three clerks idle?Po = .5122 BUT NOW SHOW ALL SUPPORTING CALCULATIONS

The Post Office RevistedThe Post Office RevistedQueuing TheoryQueuing Theory

Problem 2


The average number of customers in the system ( L ) * :

L = ( 20 x 30 ) ( .67 )

( 3 - 1 )! (3 x 30 – 20)

x ( .5122 ) + .67

3

2

L =600 ( .300763 )

2! ( 90 – 20 )2

X ( .5122 ) + .67 = .6794

* * L needs to be calculated before Lq can be found.L needs to be calculated before Lq can be found.

Po = .5122Given


Problem 2


The average length of the line ( Lq ) :

Lq = L – ( λ / μ )

Lq = .6794 – ( 20 / 30 ) = .0094

The Post Office RevistedThe Post Office RevistedQueuing TheoryQueuing Theory

REQUIREMENT:





W = L / λ = ( .6794 / 20 ) = .0339 of an hour or 2.028 minutes

Problem 2

The average time in the system ( W ) :

The average time in the queue ( Wq ) :

Wq = Lq / λ = ( .0094 / 20 ) = .00047 of an hour or .028 minutes


REQUIREMENT:





1 20 1 20 1 20 1 x 20 x 3(30)

0! 30 1! 30 2! 30 (3)(2)(1) 30 3(30) - 20

1

Po = 0 1 3

++

Po =

[ 1 + .67 + .2244 ] + .166 ( .3007 ) ( 1.285 )

1

Po ≈ .5122 THE PROBABILITY THAT THE POST OFFICE IS IDLE

++

2

++

The Computer TechnicianThe Computer TechnicianQueuing TheoryQueuing Theory

Problem 3

A technician monitors a group of five (5) computers that run anautomated manufacturing facility. It takes an average of fifteen(15) minutes ( exponentially distributed ) to adjust a computerthat developes a problem. The computers run for an average ofeighty-five (85) minutes ( poisson distributed ) without requiringadjustments.


Problem 3

REQUIREMENT :

1. What is the average number of computers waiting for adjustment?2. What is the average number of computers not in working order?3. What is the probability that the system is empty?4. What is the average time in the queue?5. What is the average time in the system?

Note: Po = .344 or 34.4% ( no need to manually compute! )


Problem 3

Given: λ = 60/85 = .706 computers ; μ = 4 computers ; N = 5 M = 1 ( technician ) ; Po = .344

Average number of computers waiting for adjustment :

Lq = N - λ + μ (1 – Po )

λ

Lq = 5 – 4.706 ( .66 ) = 5 – 4.4 = .576

.706


Problem 3

REQUIREMENT :




Problem 3

The average number of computers not in working order:

L = Lq + ( 1 – Po )

L = .576 + ( 1 - .34 ) = 1.24


Problem 3

REQUIREMENT :




Problem 3

The probability that the system is empty :

Po = 0.344 ( as given )


Problem 3

REQUIREMENT :




Problem 3

The average time in the queue :

Wq = Lq

( N – L ) λ

Wq = .576

( 5 – 1.24 )( .706 )

= .217 of an hour


Problem 3

REQUIREMENT :


Note: Po = .344 or 34.4% ( no need to manually compute! )Note: Po = .344 or 34.4% ( no need to manually compute! )


Problem 3

The average time in the system :

W = W q + ( 1 / μ )

W = .217 + ( 1 / 4 ) = .217 + .25 = .467 of an hour

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