Queuing Theory

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Waiting Line ModelsWaiting Line Models

The Structure of a Waiting Line SystemThe Structure of a Waiting Line System Queuing SystemsQueuing Systems Queuing System Input CharacteristicsQueuing System Input Characteristics Queuing System Operating CharacteristicsQueuing System Operating Characteristics Analytical FormulasAnalytical Formulas Single-Channel Waiting Line Model with Poisson Single-Channel Waiting Line Model with Poisson

Arrivals and Exponential Service TimesArrivals and Exponential Service Times Multiple-Channel Waiting Line Model with Multiple-Channel Waiting Line Model with

Poisson Arrivals and Exponential Service TimesPoisson Arrivals and Exponential Service Times Economic Analysis of Waiting LinesEconomic Analysis of Waiting Lines

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Structure of a Waiting Line SystemStructure of a Waiting Line System

Queuing theoryQueuing theory is the study of waiting lines. is the study of waiting lines. Four characteristics of a queuing system are: Four characteristics of a queuing system are:

•the manner in which customers arrivethe manner in which customers arrive

•the time required for servicethe time required for service

•the priority determining the order of servicethe priority determining the order of service

•the number and configuration of servers in the number and configuration of servers in the system.the system.

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Distribution of ArrivalsDistribution of Arrivals•Generally, the arrival of customers into the Generally, the arrival of customers into the

system is a system is a random eventrandom event. .

•Frequently the arrival pattern is modeled as a Frequently the arrival pattern is modeled as a Poisson processPoisson process..

Distribution of Service TimesDistribution of Service Times•Service time is also usually a random variable. Service time is also usually a random variable.

•A distribution commonly used to describe A distribution commonly used to describe

service time is the service time is the exponential distributionexponential distribution..

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Queue DisciplineQueue Discipline

•Most common queue discipline is Most common queue discipline is first come, first come, first served (FCFS)first served (FCFS). .

•An elevator is an example of last come, first An elevator is an example of last come, first served (LCFS) queue discipline.served (LCFS) queue discipline.

•Other disciplines assign priorities to the Other disciplines assign priorities to the waiting units and then serve the unit with the waiting units and then serve the unit with the highest priority first.highest priority first.

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Single Service ChannelSingle Service Channel

Multiple Service ChannelsMultiple Service Channels

SS11SS11

SS11SS11

SS22SS22

SS33SS33

CustomerCustomerleavesleaves

CustomerCustomerleavesleaves

CustomerCustomerarrivesarrives

CustomerCustomerarrivesarrives

Waiting lineWaiting line

Waiting lineWaiting line

SystemSystem

SystemSystem

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Examples of Internal Service SystemsExamples of Internal Service SystemsThat Are Queueing SystemsThat Are Queueing Systems

Type of SystemType of System CustomersCustomers Server(s)Server(s)

Secretarial servicesSecretarial services EmployeesEmployees SecretarySecretary

Copying servicesCopying services EmployeesEmployees Copy machineCopy machine

Computer Computer programming servicesprogramming services

EmployeesEmployees ProgrammerProgrammer

Mainframe computerMainframe computer EmployeesEmployees ComputerComputer

First-aid centerFirst-aid center EmployeesEmployees NurseNurse

Faxing servicesFaxing services EmployeesEmployees Fax machineFax machine

Materials-handling Materials-handling systemsystem

LoadsLoads Materials-Materials-handling unithandling unit

Maintenance systemMaintenance system MachinesMachines Repair crewRepair crew

Inspection stationInspection station ItemsItems InspectorInspector

Production systemProduction system JobsJobs MachineMachine

Semiautomatic Semiautomatic machinesmachines

MachinesMachines OperatorOperator

Tool cribTool crib Machine Machine operatorsoperators

ClerkClerk

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Examples of Transportation Service Examples of Transportation Service SystemsSystems

That Are Queueing SystemsThat Are Queueing Systems

Type of SystemType of System CustomersCustomers Server(s)Server(s)

Highway tollboothHighway tollbooth CarsCars CashierCashier

Truck loading dockTruck loading dock TrucksTrucks Loading crewLoading crew

Port unloading areaPort unloading area ShipsShips Unloading crewUnloading crew

Airplanes waiting to Airplanes waiting to take offtake off

AirplanesAirplanes RunwayRunway

Airplanes waiting to Airplanes waiting to landland

AirplanesAirplanes RunwayRunway

Airline serviceAirline service PeoplePeople AirplaneAirplane

Taxicab serviceTaxicab service PeoplePeople TaxicabTaxicab

Elevator serviceElevator service PeoplePeople ElevatorElevator

Fire departmentFire department FiresFires Fire truckFire truck

Parking lotParking lot CarsCars Parking spaceParking space

Ambulance serviceAmbulance service PeoplePeople AmbulanceAmbulance

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Queuing SystemsQueuing Systems

A A three part codethree part code of the form of the form AA//BB//kk is used to is used to describe various queuing systems. describe various queuing systems.

AA identifies the arrival distribution, identifies the arrival distribution, BB the service the service (departure) distribution and (departure) distribution and kk the number of the number of channels for the system. channels for the system.

Symbols used for the arrival and service Symbols used for the arrival and service processes are: processes are: MM - Markov distributions - Markov distributions (Poisson/exponential), (Poisson/exponential), DD - Deterministic - Deterministic (constant) and (constant) and GG - General distribution (with a - General distribution (with a known mean and variance). known mean and variance).

For example, For example, MM//MM//kk refers to a system in which refers to a system in which arrivals occur according to a Poisson arrivals occur according to a Poisson distribution, service times follow an exponential distribution, service times follow an exponential distribution and there are distribution and there are kk servers working at servers working at identical service rates. identical service rates.

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Queuing System Input CharacteristicsQueuing System Input Characteristics

= the average arrival = the average arrival raterate

1/1/ = the average = the average timetime between arrivals between arrivals

µ µ = the average service = the average service raterate for each server for each server

1/1/µ µ = the average service = the average service timetime

= the standard deviation of the service = the standard deviation of the service timetime

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Queuing System Operating CharacteristicsQueuing System Operating Characteristics

PP0 0 = probability the service facility is idle = probability the service facility is idle

PPnn = probability of = probability of nn units in the system units in the system

PPww = probability an arriving unit must wait for = probability an arriving unit must wait for serviceservice

LLqq = average number of units in the queue = average number of units in the queue awaiting awaiting serviceservice

LL = average number of units in the system = average number of units in the system

WWqq = average time a unit spends in the queue = average time a unit spends in the queue awaiting serviceawaiting service

WW = average time a unit spends in the system = average time a unit spends in the system

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Analytical FormulasAnalytical Formulas

For nearly all queuing systems, there is a For nearly all queuing systems, there is a relationship between the average time a unit relationship between the average time a unit spends in the system or queue and the spends in the system or queue and the average number of units in the system or average number of units in the system or queue. queue.

These relationships, known as These relationships, known as Little's flow Little's flow equationsequations are: are:

LL = = WW and and LLqq = = WWqq

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Analytical FormulasAnalytical Formulas

When the queue discipline is FCFS, analytical When the queue discipline is FCFS, analytical formulas have been derived for several different formulas have been derived for several different queuing models including the following: queuing models including the following: •MM//MM/1/1

•MM//MM//kk

•MM//GG/1/1

•MM//GG//kk with blocked customers cleared with blocked customers cleared

•MM//MM/1 with a finite calling population/1 with a finite calling population Analytical formulas are not available for all Analytical formulas are not available for all

possible queuing systems. In this event, possible queuing systems. In this event, insights may be gained through a simulation of insights may be gained through a simulation of the system. the system.

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M/M/1 Queuing SystemM/M/1 Queuing System

Single channelSingle channel Poisson arrival-rate distributionPoisson arrival-rate distribution Exponential service-time distributionExponential service-time distribution Unlimited maximum queue lengthUnlimited maximum queue length Infinite calling populationInfinite calling population Examples:Examples:

•Single-window theatre ticket sales boothSingle-window theatre ticket sales booth

•Single-scanner airport security stationSingle-scanner airport security station

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Notation for Single-Server Queueing ModelsNotation for Single-Server Queueing Models

= = Mean arrival rateMean arrival rate for customers for customers= Expected number of arrivals per unit time= Expected number of arrivals per unit time

1/1/ = expected interarrival time= expected interarrival time

= = Mean service rateMean service rate (for a continuously busy (for a continuously busy server)server)

= Expected number of service completions per unit = Expected number of service completions per unit timetime

= expected service time= expected service time

= the = the utilizationutilization factorfactor= the average fraction of time that a server is busy = the average fraction of time that a server is busy

serving customersserving customers= =

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AssumptionsAssumptions

1.1.Interarrival timesInterarrival times have an exponential distribution with a mean of have an exponential distribution with a mean of 1/1/..

2.2.Service timesService times have an exponential distribution with a mean of have an exponential distribution with a mean of 1/1/..

3.3.The queueing system has one server.The queueing system has one server.

• The The expected number of customers in the systemexpected number of customers in the system is is

LL = = 1 –1 – = = – –

• The The expected waiting time in the systemexpected waiting time in the system is is

WW = (1 / = (1 / ))LL = 1 / ( = 1 / ( – – ))

• The The expected waiting time in the queueexpected waiting time in the queue is is

WWqq = = WW – 1/ – 1/ = = / [ / [(( – – )])]

• The The expected number of customers in the queueexpected number of customers in the queue is is

LLqq = = WWqq = = 22 / [ / [(( – – )] = )] = 22 / (1 – / (1 – ))

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TheThe probability of having exactly probability of having exactly nn customers in the system is customers in the system is

PPnn = (1 – = (1 – ))nn

Thus,Thus,PP00 = 1 – = 1 – PP11 = (1 – = (1 – ))PP22 = (1 – = (1 – ))22

::::

The probability that the The probability that the waiting time in the systemwaiting time in the system exceeds exceeds tt isis

PP((WW > > tt) = ) = ee––(1–(1–))tt for for tt ≥ 0 ≥ 0

The probability that the The probability that the waiting time in the queuewaiting time in the queue exceeds exceeds tt is is

PP((WWqq > > tt) = ) = ee––(1–(1–)t)t for for tt ≥ 0 ≥ 0

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Problem:Problem:

Consider the situation where the mean arrival Consider the situation where the mean arrival rate is one customer every 4 minutes and the rate is one customer every 4 minutes and the mean service time is 2.5 minutes. Calculate the mean service time is 2.5 minutes. Calculate the followingfollowing

•Average no. of customer in the systemAverage no. of customer in the system

•Average queue lengthAverage queue length

•Average time a customer spends in the Average time a customer spends in the systemsystem

•Average time a customer waits before being Average time a customer waits before being served.served.

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Problem:Problem:

Arrivals at a telephone booth are considered to Arrivals at a telephone booth are considered to be Poisson, with an average time of 10 minutes be Poisson, with an average time of 10 minutes between one arrival and the next. The length of between one arrival and the next. The length of a phone call is assumed to be exponentially a phone call is assumed to be exponentially distributed with mean 3 minutes.distributed with mean 3 minutes.

What is the probability that a person arriving at What is the probability that a person arriving at the booth will have to wait?the booth will have to wait?

The telephone department will install a second The telephone department will install a second booth when convinced that an arrival would booth when convinced that an arrival would expect to have to wait at least three minutes for expect to have to wait at least three minutes for the phone. By how much must the flow of the phone. By how much must the flow of arrivals be increased in order to justify a second arrivals be increased in order to justify a second booth?booth?

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Example: SJJT, Inc. (A)Example: SJJT, Inc. (A)

MM//MM/1 Queuing System/1 Queuing System

Joe Ferris is a stock trader on the floor of Joe Ferris is a stock trader on the floor of the New York Stock Exchange for the firm of the New York Stock Exchange for the firm of Smith, Jones, Johnson, and Thomas, Inc. Stock Smith, Jones, Johnson, and Thomas, Inc. Stock transactions arrive at a mean rate of 20 per transactions arrive at a mean rate of 20 per hour. Each order received by Joe requires an hour. Each order received by Joe requires an average of two minutes to process.average of two minutes to process.

Orders arrive at a mean rate of 20 per Orders arrive at a mean rate of 20 per hour or one order every 3 minutes. Therefore, hour or one order every 3 minutes. Therefore, in a 15 minute interval the average number of in a 15 minute interval the average number of orders arriving will be orders arriving will be = 15/3 = 5. = 15/3 = 5.

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Arrival Rate DistributionArrival Rate Distribution

QuestionQuestion

What is the probability that no orders What is the probability that no orders are received within a 15-minute period?are received within a 15-minute period?

AnswerAnswer

P P ((xx = 0) = (5 = 0) = (500e e -5-5)/0! = )/0! = e e -5-5 = .0067= .0067

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QuestionQuestion

What is the probability that exactly 3 What is the probability that exactly 3 orders are received within a 15-minute period?orders are received within a 15-minute period?

AnswerAnswer

P P ((xx = 3) = (5 = 3) = (533e e -5-5)/3! = 125(.0067)/6 = )/3! = 125(.0067)/6 = .1396.1396

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QuestionQuestion

What is the probability that more than 6 What is the probability that more than 6 orders arrive within a 15-minute period? orders arrive within a 15-minute period?

AnswerAnswer

P P ((xx > 6) = 1 - > 6) = 1 - P P ((xx = 0) - = 0) - P P ((xx = 1) - = 1) - P P ((xx = = 2) 2)

- - P P ((xx = 3) - = 3) - P P ((xx = 4) - = 4) - P P ((xx = = 5)5)

- - P P ((xx = 6) = 6)

= 1 - .762 = .238= 1 - .762 = .238

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Service Rate DistributionService Rate Distribution

QuestionQuestion

What is the mean service rate per hour?What is the mean service rate per hour?

AnswerAnswer

Since Joe Ferris can process an order in an Since Joe Ferris can process an order in an average time of 2 minutes (= 2/60 hr.), then the average time of 2 minutes (= 2/60 hr.), then the mean service rate, mean service rate, µµ, is , is µµ = 1/(mean service = 1/(mean service time), or 60/2.time), or 60/2.

= 30/hr.= 30/hr.

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Service Time DistributionService Time Distribution

QuestionQuestion

What percentage of the orders will take less What percentage of the orders will take less than one minute to process? than one minute to process?

AnswerAnswer

Since the units are expressed in hours, Since the units are expressed in hours,

P P ((TT << 1 minute) = 1 minute) = P P ((TT << 1/60 hour). 1/60 hour).

Using the exponential distribution, Using the exponential distribution, P P ((TT << t t ) = 1 - ) = 1 - ee-µt-µt. .

Hence, Hence, P P ((TT << 1/60) = 1 - e 1/60) = 1 - e-30(1/60)-30(1/60)

= 1 - .6065 = .3935 = = 1 - .6065 = .3935 = 39.35%39.35%

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QuestionQuestion

What percentage of the orders will be What percentage of the orders will be processed in exactly 3 minutes?processed in exactly 3 minutes?

AnswerAnswer

Since the exponential distribution is a Since the exponential distribution is a continuous distribution, the probability a continuous distribution, the probability a service time exactly equals any specific value is service time exactly equals any specific value is 0 .0 .

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QuestionQuestion

What percentage of the orders will require What percentage of the orders will require more than 3 minutes to process?more than 3 minutes to process?

AnswerAnswer

The percentage of orders requiring more The percentage of orders requiring more than 3 minutes to process is:than 3 minutes to process is:

P P ((TT > 3/60) = > 3/60) = ee-30(3/60)-30(3/60) = = ee -1.5-1.5 = .2231 = = .2231 = 22.31% 22.31%

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Average Time in the SystemAverage Time in the System

QuestionQuestion

What is the average time an order must What is the average time an order must wait from the time Joe receives the order until it wait from the time Joe receives the order until it is finished being processed (i.e. its turnaround is finished being processed (i.e. its turnaround time)?time)?

AnswerAnswer

This is an This is an MM//MM/1 queue with /1 queue with = 20 per hour = 20 per hour and and = 30 per hour. The average time an order = 30 per hour. The average time an order waits in the system is:waits in the system is: WW = 1/(µ - = 1/(µ - ) )

= 1/(30 - 20)= 1/(30 - 20)

= 1/10 hour or 6 minutes = 1/10 hour or 6 minutes

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Average Length of QueueAverage Length of Queue

QuestionQuestion

What is the average number of orders Joe What is the average number of orders Joe has waiting to be processed?has waiting to be processed?

AnswerAnswer

Average number of orders waiting in the Average number of orders waiting in the queue is:queue is:

LLqq = = 22/[µ(µ - /[µ(µ - )] )]

= (20)= (20)22/[(30)(30-20)]/[(30)(30-20)]

= 400/300 = 400/300

= 4/3= 4/3

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Utilization FactorUtilization Factor

QuestionQuestion

What percentage of the time is Joe What percentage of the time is Joe processing orders?processing orders?

AnswerAnswer

The percentage of time Joe is processing The percentage of time Joe is processing orders is equivalent to the utilization factor, orders is equivalent to the utilization factor, //. . Thus, the percentage of time he is processing Thus, the percentage of time he is processing orders is:orders is:

// = 20/30 = 20/30

= 2/3 or 66.67%= 2/3 or 66.67%

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SolutionSolutionA B C D E F G H

1 202 3034 Po 0.3335 Lg 1.3336 L 2.0007 Wq 0.0678 W 0.1009 Pw 0.667

Operating Characteristics Probability of no orders in system Average number of orders waiting Average number of orders in system Average time an order waits Average time an order is in system

Poisson Arrival Rate Exponential Service Rate

Probability an order must wait

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MM//MM//kk Queuing System Queuing System

Multiple channels (with one central waiting line)Multiple channels (with one central waiting line) Poisson arrival-rate distributionPoisson arrival-rate distribution Exponential service-time distributionExponential service-time distribution Unlimited maximum queue lengthUnlimited maximum queue length Infinite calling populationInfinite calling population Examples:Examples:

•Four-teller transaction counter in bankFour-teller transaction counter in bank

•Two-clerk returns counter in retail storeTwo-clerk returns counter in retail store

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33 33 Slide Slide

General Operating CharacteristicsGeneral Operating Characteristics

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34 34 Slide Slide

Problem:Problem:

A Tax consulting firm has four service stations (counters) A Tax consulting firm has four service stations (counters) in its office to receive people who have problems and in its office to receive people who have problems and complaints about their income, wealth and sales taxes. complaints about their income, wealth and sales taxes. Arrivals average 80 persons in an 8 hour service day. Each Arrivals average 80 persons in an 8 hour service day. Each tax advisor spends irregular amount of time servicing the tax advisor spends irregular amount of time servicing the arrivals which have been found to have an exponential arrivals which have been found to have an exponential distribution. The average service time is 20 minutes. distribution. The average service time is 20 minutes.

Calculate the average no. of customers in the system, Calculate the average no. of customers in the system, average no. of customers waiting to be serviced, average no. of customers waiting to be serviced, average time a customer spend in the system, average time a customer spend in the system, average waiting time for a customer in queue. average waiting time for a customer in queue. Calculate how many hours each week does a tax advisor Calculate how many hours each week does a tax advisor

spend performing his job? spend performing his job? What is the probability that a customer has to wait before What is the probability that a customer has to wait before

he gets service? he gets service? What is the expected no. of idle tax advisors at any What is the expected no. of idle tax advisors at any

specified time?specified time?

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Example: SJJT, Inc. (B)Example: SJJT, Inc. (B)

MM//MM/2 Queuing System/2 Queuing System

Smith, Jones, Johnson, and Thomas, Inc. Smith, Jones, Johnson, and Thomas, Inc. has begun a major advertising campaign which it has begun a major advertising campaign which it believes will increase its business 50%. To believes will increase its business 50%. To handle the increased volume, the company has handle the increased volume, the company has hired an additional floor trader, Fred Hanson, hired an additional floor trader, Fred Hanson, who works at the same speed as Joe Ferris.who works at the same speed as Joe Ferris.

Note that the new arrival rate of orders, Note that the new arrival rate of orders, , , is 50% higher than that of problem (A). Thus, is 50% higher than that of problem (A). Thus, = 1.5(20) = 30 per hour.= 1.5(20) = 30 per hour.

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Sufficient Service RateSufficient Service Rate

QuestionQuestion

Why will Joe Ferris alone not be able to Why will Joe Ferris alone not be able to handle the increase in orders?handle the increase in orders?

AnswerAnswer

Since Joe Ferris processes orders at a Since Joe Ferris processes orders at a mean rate of mean rate of µµ = 30 per hour, then = 30 per hour, then = = µµ = 30 = 30 and the utilization factor is 1. and the utilization factor is 1.

This implies the queue of orders will This implies the queue of orders will grow infinitely large. Hence, Joe alone cannot grow infinitely large. Hence, Joe alone cannot handle this increase in demand.handle this increase in demand.

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Probability of Probability of nn Units in System Units in System

QuestionQuestion

What is the probability that neither Joe nor What is the probability that neither Joe nor Fred will be working on an order at any point in Fred will be working on an order at any point in time?time?

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Probability of Probability of nn Units in System (continued) Units in System (continued)

AnswerAnswer

Given that Given that = 30, = 30, µµ = 30, = 30, kk = 2 and ( = 2 and ( /µ) /µ) = 1, the probability that neither Joe nor Fred will = 1, the probability that neither Joe nor Fred will be working is: be working is:

= 1/[(1 + (1/1!)(30/30)1] + = 1/[(1 + (1/1!)(30/30)1] + [(1/2!)(1)2][2(30)/(2(30)-30)][(1/2!)(1)2][2(30)/(2(30)-30)]

= 1/(1 + 1 + 1) = 1/3 = .333= 1/(1 + 1 + 1) = 1/3 = .333

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Average Time in SystemAverage Time in System

QuestionQuestion

What is the average turnaround time for What is the average turnaround time for an order with both Joe and Fred working?an order with both Joe and Fred working?

40 40 Slide Slide


Average Time in System (continued)Average Time in System (continued)

AnswerAnswer

The average turnaround time is the The average turnaround time is the

average waiting time in the system, average waiting time in the system, WW. .

µµ(( / /µµ))kk (30)(30)(30/30)(30)(30)(30/30)22 LLqq = = PP0 0 = =

(1/3) = 1/3(1/3) = 1/3 ((kk-1)!(-1)!(kµkµ - - ) )22 (1!)((2)(30)-30)) (1!)((2)(30)-30))22

LL = = LLqq + ( + ( / /µµ) = 1/3 + (30/30) = 4/3 ) = 1/3 + (30/30) = 4/3

WW = = LL//(4/3)/30 = 4/90 hr. = 2.67 (4/3)/30 = 4/90 hr. = 2.67 min.min.

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Average Length of QueueAverage Length of Queue

QuestionQuestion

What is the average number of orders What is the average number of orders waiting to be filled with both Joe and Fred waiting to be filled with both Joe and Fred working?working?

AnswerAnswer

The average number of orders waiting to The average number of orders waiting to be filled is be filled is LLqq. This was calculated earlier as 1/3 . This was calculated earlier as 1/3 . .

42 42 Slide Slide


Formula SpreadsheetFormula Spreadsheet

A B C D E F G H1 k 22 303 3045 Po =Po(H1,H2,H3)6 Lg # #7 L =H6+H2/H38 Wq =H6/H29 W =H8+1/H310 Pw =H2/H3

Mean Arrival Rate (Poisson) Mean Service Rate (Exponential )


Number of Channels

Average time (hrs) an order waits Average time (hrs) an order is in system

Operating Characteristics Probability of no orders in system Average number of orders waiting Average number of orders in system

43 43 Slide Slide


Spreadsheet SolutionSpreadsheet Solution

A B C D E F G H1 k 22 303 3045 Po 0.3336 Lg 0.3337 L 1.3338 Wq 0.0119 W 0.04410 Pw 1.000

Mean Arrival Rate (Poisson) Mean Service Rate (Exponential )


Number of Channels

Average time (hrs) an order waits Average time (hrs) an order is in system

Operating Characteristics Probability of no orders in system Average number of orders waiting Average number of orders in system

44 44 Slide Slide

Example: SJJT, Inc. (C)Example: SJJT, Inc. (C)

Economic Analysis of Queuing SystemsEconomic Analysis of Queuing Systems

The advertising campaign of Smith, Jones, The advertising campaign of Smith, Jones, Johnson and Thomas, Inc. (see problems (A) and Johnson and Thomas, Inc. (see problems (A) and (B)) was so successful that business actually (B)) was so successful that business actually doubled. The mean rate of stock orders arriving doubled. The mean rate of stock orders arriving at the exchange is now 40 per hour and the at the exchange is now 40 per hour and the company must decide how many floor traders to company must decide how many floor traders to employ. Each floor trader hired can process an employ. Each floor trader hired can process an order in an average time of 2 minutes.order in an average time of 2 minutes.

45 45 Slide Slide


Economic Analysis of Queuing SystemsEconomic Analysis of Queuing Systems

Based on a number of factors the Based on a number of factors the brokerage firm has determined the average brokerage firm has determined the average waiting cost per minute for an order to be $.50. waiting cost per minute for an order to be $.50. Floor traders hired will earn $20 per hour in Floor traders hired will earn $20 per hour in wages and benefits. Using this information wages and benefits. Using this information compare the total hourly cost of hiring 2 traders compare the total hourly cost of hiring 2 traders with that of hiring 3 traders.with that of hiring 3 traders.

46 46 Slide Slide


Economic Analysis of Waiting LinesEconomic Analysis of Waiting Lines

Total Hourly Cost Total Hourly Cost

= (Total salary cost per hour)= (Total salary cost per hour)

+ (Total hourly cost for orders in the system) + (Total hourly cost for orders in the system)

= ($20 per trader per hour) x (Number of traders) = ($20 per trader per hour) x (Number of traders)

+ ($30 waiting cost per hour) x (Average + ($30 waiting cost per hour) x (Average number ofnumber of orders in the orders in the system) system)

= 20= 20kk + 30 + 30LL..

Thus, Thus, LL must be determined for must be determined for kk = 2 = 2 traders and for traders and for kk = 3 traders with = 3 traders with = 40/hr. and = 40/hr. and = = 30/hr. (since the average service time is 2 minutes 30/hr. (since the average service time is 2 minutes (1/30 hr.).(1/30 hr.).

47 47 Slide Slide


Cost of Two ServersCost of Two Servers

PP00 = 1 / = 1 / [1+(1/1!)(40/30)]+[(1/2!)(40/30)2(60/(60-40))][1+(1/1!)(40/30)]+[(1/2!)(40/30)2(60/(60-40))]

= 1 / [1 + (4/3) + (8/3)] = 1 / [1 + (4/3) + (8/3)]

= 1/5= 1/5

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48 48 Slide Slide


Cost of Two Servers (continued)Cost of Two Servers (continued)

Thus,Thus,

µµ(( / /µµ))kk (40)(30)(40/30)2 (40)(30)(40/30)2 LLqq = = PP00 = =

(1/5) = 16/15 (1/5) = 16/15 ((kk-1)!(-1)!(kµkµ - - ) )22 1!(60-40)2 1!(60-40)2

LL = = LLqq + ( + ( / /µµ) = 16/15 + 4/3 = ) = 16/15 + 4/3 = 12/512/5

Total Cost = (20)(2) + 30(12/5) = $112.00 Total Cost = (20)(2) + 30(12/5) = $112.00 per hourper hour

49 49 Slide Slide


Cost of Three ServersCost of Three Servers

PP00 = 1/[[1+(1/1!)(40/30)+(1/2!)(40/30)2]+ = 1/[[1+(1/1!)(40/30)+(1/2!)(40/30)2]+

[(1/3!)(40/30)3(90/(90-[(1/3!)(40/30)3(90/(90-40))] ]40))] ]

= 1 / [1 + 4/3 + 8/9 + 32/45] = 1 / [1 + 4/3 + 8/9 + 32/45]

= 15/59= 15/59

P

n kk

k

n k

n

k0

0

1

1

( / )!

( / )!

( )

P

n kk

k

n k

n

k0

0

1

1

( / )!

( / )!

( )

50 50 Slide Slide


Cost of Three Servers (continued)Cost of Three Servers (continued)

(30)(40)(40/30)3 (30)(40)(40/30)3

Hence, Hence, LLqq = (15/59) = = (15/59) = 128/885 = .1446128/885 = .1446

(2!)(3(30)-40)2(2!)(3(30)-40)2

Thus, Thus, LL = 128/885 + 40/30 = 1308/885 (= = 128/885 + 40/30 = 1308/885 (= 1.4780)1.4780)

Total Cost = (20)(3) + 30(1308/885) = $104.35 Total Cost = (20)(3) + 30(1308/885) = $104.35 per hourper hour

51 51 Slide Slide


System Cost ComparisonSystem Cost Comparison

WageWage Waiting Waiting Total Total

Cost/HrCost/Hr Cost/HrCost/Hr Cost/HrCost/Hr

2 Traders2 Traders $40.00 $40.00 $82.00 $82.00 $112.00$112.00

3 Traders3 Traders 60.00 60.00 44.35 44.35 104.35104.35

Thus, the cost of having 3 traders is less than Thus, the cost of having 3 traders is less than that of that of

2 traders.2 traders.

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Queuing Theory