queuing theory body of knowledge about waiting lines helps managers to better understand systems in...
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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 equip- ment.
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 “.
Queuing Theory Queuing Theory VariablesVariables
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 “.
Queuing Theory Queuing Theory VariablesVariables
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 “.
Queuing Theory VariablesQueuing Theory Variables
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.
Queuing Theory VariablesQueuing Theory Variables
• 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 !
Queuing Theory VariablesQueuing Theory Variables
• 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
Queuing Theory VariablesQueuing Theory Variables
“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.
QUEUING THEORYQUEUING THEORY
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.
QUEUING THEORYQUEUING THEORY
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
Single-Channel / Single-Single-Channel / Single-Phase ModelPhase Model
The Average Number Just Waiting in Line
Lq =λ
μ ( μ - λ )
2
Single-Channel / Single-Single-Channel / Single-Phase ModelPhase Model
Average Customer Time Spent in the System
W = ( μ - λ )
1
Single-Channel / Single-Single-Channel / Single-Phase ModelPhase Model
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
Single-Channel / Single-Phase Single-Channel / Single-Phase ModelModelAPPLICATIONAPPLICATION
The Average Number Just Waiting in Line
Lq = ( 20 )
30 ( 30 - 20 )= 1.33
2
Single-Channel / Single-Phase Single-Channel / Single-Phase ModelModelAPPLICATIONAPPLICATION
The Average Customer Time Spent in the System
W = 1
( 30 - 20 )= .10 hrs
( 6 minutes )
Single-Channel / Single-Phase Single-Channel / Single-Phase ModelModelAPPLICATIONAPPLICATION
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
Single-ChannelSingle-Phase
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
Multi-Channel Single-Phase Multi-Channel Single-Phase SystemsSystems
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
Multi-Channel Single-Phase Multi-Channel Single-Phase SystemsSystems
The average number of customers in the system:
λμ ( λ / μ ) L = . Po + λ / μ (M-1)! (Mμ-λ)
M
2
Multi-Channel Single-Phase Multi-Channel Single-Phase SystemsSystems
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
Multi-Channel Single-Phase Multi-Channel Single-Phase SystemsSystems
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
QM for WindowsQM for WindowsQUEUING APPLICATIONSQUEUING APPLICATIONS
Multi-ChannelSingle-Phase
Model
Queuing Theory ModelingQueuing Theory Modeling withwith
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
QM for WindowsQM for WindowsQUEUING APPLICATIONSQUEUING APPLICATIONS
Single-ChannelSingle-Phase
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.
Queuing Theory ModelingQueuing Theory Modeling withwith
Single-ChannelSingle-Phase
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
DistributionDistribution
NegativeNegativeExponentialExponentialService TimeService TimeDistributionDistribution
MULTI-CHANNELMULTI-CHANNELMULTI-SERVERMULTI-SERVER
MODELMODEL
2nd Example2nd Example
M M s
PoissonPoissonArrivalArrival
DistributionDistribution
NegativeNegativeExponentialExponentialService TimeService TimeDistributionDistribution
MULTI-CHANNELMULTI-CHANNELMULTI-SERVERMULTI-SERVER
MODELMODEL
3rd Example3rd Example
M M 1
PoissonPoissonArrivalArrival
DistributionDistribution
NegativeNegativeExponentialExponentialService TimeService TimeDistributionDistribution
SINGLE-CHANNELSINGLE-CHANNELSINGLE-SERVERSINGLE-SERVER
MODELMODEL
WITH FINITE POPULATIONWITH FINITE POPULATION
WITH FINITECALLING
POPULATION
ImposedImposedLegendLegend
4th Example4th Example
M D 3
Poisson Poisson ArrivalArrival
DistributionDistribution
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
PoissonPoissonArrivalArrival
DistributionDistribution
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
QUEUING THEORYQUEUING THEORY
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
Applied Management Science for Decision Making, 1e Applied Management Science for Decision Making, 1e © 2010 Pearson Prentice-Hall, Inc. Philip A. Vaccaro , PhD© 2010 Pearson Prentice-Hall, Inc. Philip A. Vaccaro , PhD
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 ).
The Post OfficeThe Post OfficeQueuing TheoryQueuing Theory
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
The Post OfficeThe Post OfficeQueuing TheoryQueuing Theory
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
The Post OfficeThe Post OfficeQueuing TheoryQueuing Theory
Problem 1
Given: λ = 20 and μ = 30 ( per hour rates ) and M = 2
The average length of the line ( Lq ) :
Lq = L – ( λ / μ )
Lq = .7541 – ( 20 / 30 ) = .0841
The Post OfficeThe Post OfficeQueuing TheoryQueuing Theory
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
The Post OfficeThe Post OfficeQueuing TheoryQueuing Theory
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
The Post OfficeThe Post OfficeQueuing TheoryQueuing Theory
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
The Post OfficeThe Post OfficeQueuing TheoryQueuing Theory
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 dis- tribution, 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 ).
The Post Office RevisitedThe Post Office RevisitedQueuing TheoryQueuing Theory
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 all three clerks idle?Po = .5122 BUT NOW SHOW ALL SUPPORTING CALCULATIONS
The Post Office RevistedThe Post Office RevistedQueuing TheoryQueuing Theory
Problem 2
Given: λ = 20 and μ = 30 ( per hour rates ) and M = 3
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
The Post Office RevisitedThe Post Office RevisitedQueuing TheoryQueuing Theory
Problem 2
Given: λ = 20 and μ = 30 ( per hour rates ) and M = 3
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:
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 all three clerks idle?Po = .5122 BUT NOW SHOW ALL SUPPORTING CALCULATIONS
The Post Office RevisitedThe Post Office RevisitedQueuing TheoryQueuing Theory
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
The Post Office RevisitedThe Post Office RevisitedQueuing TheoryQueuing Theory
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 all three clerks idle?Po = .5122 BUT NOW SHOW ALL SUPPORTING CALCULATIONS
The Post Office RevisitedThe Post Office RevisitedQueuing TheoryQueuing Theory
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.
The Computer TechnicianThe Computer TechnicianQueuing TheoryQueuing Theory
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! )
The Computer TechnicianThe Computer TechnicianQueuing TheoryQueuing Theory
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
The Computer TechnicianThe Computer TechnicianQueuing TheoryQueuing Theory
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! )
The Computer TechnicianThe Computer TechnicianQueuing TheoryQueuing Theory
Problem 3
The average number of computers not in working order:
L = Lq + ( 1 – Po )
L = .576 + ( 1 - .34 ) = 1.24
The Computer TechnicianThe Computer TechnicianQueuing TheoryQueuing Theory
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! )
The Computer TechnicianThe Computer TechnicianQueuing TheoryQueuing Theory
Problem 3
The probability that the system is empty :
Po = 0.344 ( as given )
The Computer TechnicianThe Computer TechnicianQueuing TheoryQueuing Theory
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! )
The Computer TechnicianThe Computer TechnicianQueuing TheoryQueuing Theory
Problem 3
The average time in the queue :
Wq = Lq
( N – L ) λ
Wq = .576
( 5 – 1.24 )( .706 )
= .217 of an hour
The Computer TechnicianThe Computer TechnicianQueuing TheoryQueuing Theory
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! )Note: Po = .344 or 34.4% ( no need to manually compute! )
The Computer TechnicianThe Computer TechnicianQueuing TheoryQueuing Theory
Problem 3
The average time in the system :
W = W q + ( 1 / μ )
W = .217 + ( 1 / 4 ) = .217 + .25 = .467 of an hour
Solved ProblemsSolved Problems
Queuing TheoryQueuing TheoryComputer-BasedComputer-Based
ManualManual
Applied Management Science for Decision Making, 1e Applied Management Science for Decision Making, 1e © 2010 Pearson Prentice-Hall, Inc. Philip A. Vaccaro , PhD© 2010 Pearson Prentice-Hall, Inc. Philip A. Vaccaro , PhD