1 1 Slide © 2005 Thomson/South-Western Chapter 12, Part A Waiting Line Models n Structure of a...

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© 2005 Thomson/South-Western© 2005 Thomson/South-Western

Chapter 12, Part AChapter 12, Part AWaiting Line ModelsWaiting Line Models

Structure of a Waiting Line SystemStructure 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 Single-Channel Waiting Line Model with

Poisson Arrivals and Exponential Service Poisson Arrivals and Exponential Service TimesTimes

Multiple-Channel Waiting Line Model with Multiple-Channel Waiting Line Model with Poisson Arrivals and Exponential Service Poisson Arrivals and Exponential Service TimesTimes

Economic Analysis of Waiting LinesEconomic Analysis of Waiting Lines

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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 the priority determining the order of serviceservice

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

Structure of a Waiting Line SystemStructure of a Waiting Line 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 Frequently the arrival pattern is modeled as

a a Poisson processPoisson process.. Distribution of Service TimesDistribution of Service Times

• Service time is also usually a random Service time is also usually a random variable. 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 waiting units and then serve the unit with the highest priority first.the 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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Queuing SystemsQueuing Systems

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

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

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Queuing SystemsQueuing Systems

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 - General distribution (with a known mean and variance). a known mean and variance).

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

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 Queuing System Operating CharacteristicsCharacteristics

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 service service

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

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 = average time a unit spends in the systemsystem

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

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

Joe Ferris is a stock trader onJoe Ferris is a stock trader on

the floor of the New York Stockthe floor of the New York Stock

Exchange for the firm of Smith,Exchange for the firm of Smith,

Jones, Johnson, and Thomas, Inc. Jones, Johnson, and Thomas, Inc.

Stock transactions arrive at a meanStock transactions arrive at a mean

rate of 20 per hour. Each order received by rate of 20 per hour. Each order received by JoeJoe

requires an average of two minutes to requires an average of two minutes to process.process.

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MM//MM/1 Queuing System/1 Queuing SystemOrders arrive at a mean rateOrders arrive at a mean rate

of 20 per hour or one order everyof 20 per hour or one order every3 minutes. Therefore, in a 153 minutes. Therefore, in a 15minute interval the averageminute interval the averagenumber of orders arriving will benumber of 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 orders are received within a 15-minute period?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 Since Joe Ferris can process an order in an average time of 2 minutes (= 2/60 hr.), an average time of 2 minutes (= 2/60 hr.), then the mean service rate, then the mean service rate, µµ, is , is µµ = 1/(mean = 1/(mean service time), or 60/2.service 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 - e) = 1 - e--

µtµt. .

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 service time exactly equals any specific value is 0.is 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 SystemQuestionQuestion

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

AnswerAnswerThis 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 QueueQuestionQuestion

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

AnswerAnswerAverage 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 . Thus, the percentage of time he is processing orders is:processing orders is:

// = 20/30 = 20/30

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

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Formula SpreadsheetFormula SpreadsheetA B C D E F G H

1 202 3034 Po =1-H1/H25 Lg =H1^2/(H2*(H2-H1))6 L =H5+H1/H27 Wq =H5/H18 W =H7+1/H29 Pw =H1/H2

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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Spreadsheet SolutionSpreadsheet SolutionA 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


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MM//MM//kk Queuing System Queuing System

Multiple channels (with one central waiting Multiple channels (with one central waiting line)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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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 has begun a major advertising campaign which it believes will increase its business which it believes will increase its business 50%. To handle the increased volume, the 50%. To handle the increased volume, the company has hired an additional floor trader, company has hired an additional floor trader, Fred Hanson, who works at the same speed as Fred Hanson, who works at the same speed as Joe Ferris.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). , is 50% higher than that of problem (A). Thus, 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) = 1, /µ) = 1, the probability that neither Joe nor Fred will be the probability that neither Joe nor Fred will be working is: 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

P

n kk

k

n k

n

k0

0

1

1

( / )!

( / )!

( )

P

n kk

k

n k

n

k0

0

1

1

( / )!

( / )!

( )

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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?

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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. .

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

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


2

02 2

( ) (30)(30)(30 30) 1( ) (1/ 3)

( 1)!( ) (1!)(2(30) 30) 3

k

qL Pk k

2

02 2

( ) (30)(30)(30 30) 1( ) (1/ 3)

( 1)!( ) (1!)(2(30) 30) 3

k

qL Pk k

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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 . This was calculated earlier as 1/3.1/3.

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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

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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

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Creating Special Excel Function to Compute PCreating Special Excel Function to Compute P00

Select the Select the ToolsTools pull-down menu pull-down menuSelect the Select the MacroMacro option optionChoose the Choose the Visual Basic EditorVisual Basic EditorWhen the Visual Basic Editor appearsWhen the Visual Basic Editor appears Select the Select the InsertInsert pull-down menu pull-down menu Choose the Choose the ModuleModule option option

When the Module sheet appearsWhen the Module sheet appears Enter Enter Function Po (k,lamda,mu)Function Po (k,lamda,mu) Enter Visual Basic program (on next slide)Enter Visual Basic program (on next slide)

Select the Select the FileFile pull-down menu pull-down menuChoose the Choose the Close and Return to MS ExcelClose and Return to MS Excel optionoption

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Visual Basic Module for PVisual Basic Module for P00 Function Function

Function Po(k, lamda, mu)Function Po(k, lamda, mu)Sum = 0Sum = 0For n = 0 to k - 1For n = 0 to k - 1 Sum = Sum + (lamda/mu) ^ n / Sum = Sum + (lamda/mu) ^ n / Application.Fact(n)Application.Fact(n)NextNextPo = 1/(Sum+(lamda/mu)^k/Application.Fact(k))*Po = 1/(Sum+(lamda/mu)^k/Application.Fact(k))*

(k*mu/(k*mu-(k*mu/(k*mu-lamda)))lamda)))End FunctionEnd Function

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

Economic Analysis of Queuing SystemsEconomic Analysis of Queuing Systems

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

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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 waiting cost per minute for an order to be $.50. Floor traders hired will earn $20 per $.50. Floor traders hired will earn $20 per hour in wages and benefits. Using this hour in wages and benefits. Using this information compare the total hourly cost of information compare the total hourly cost of hiring 2 traders with that of hiring 3 traders.hiring 2 traders with that of hiring 3 traders.

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

+ ($30 waiting cost per hour) x (Average + ($30 waiting cost per hour) x (Average number number of orders of orders in the 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 = 30/hr. (since the average service time is 2 minutes (1/30 hr.).minutes (1/30 hr.).

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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

P

n kk

k

n k

n

k0

0

1

1

( / )!

( / )!

( )

P

n kk

k

n k

n

k0

0

1

1

( / )!

( / )!

( )

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

Thus,Thus,

LL = = LLqq + ( + ( / /µµ) = 16/15 + 4/3 = ) = 16/15 + 4/3 = 2.402.40

Total Cost = (20)(2) + 30(2.40) = $112.00 Total Cost = (20)(2) + 30(2.40) = $112.00 per hourper hour

2

02 2

( ) (40)(30)(40 30) 16( ) (1/ 5)

( 1)!( ) (1!)(2(30) 40) 15

k

qL Pk k

2

02 2

( ) (40)(30)(40 30) 16( ) (1/ 5)

( 1)!( ) (1!)(2(30) 40) 15

k

qL Pk k

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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

( / )!

( / )!

( )

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

Thus, Thus, LL = .1446 + 40/30 = 1.4780 = .1446 + 40/30 = 1.4780

Total Cost = (20)(3) + 30(1.4780) = $104.35 Total Cost = (20)(3) + 30(1.4780) = $104.35 per hourper hour

3

02 2

( ) (30)(40)(40 30)( ) (15/ 59) .1446

( 1)!( ) (2!)(3(30) 40)

k

qL Pk k

3

02 2

( ) (30)(40)(40 30)( ) (15/ 59) .1446

( 1)!( ) (2!)(3(30) 40)

k

qL Pk k

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System Cost ComparisonSystem Cost Comparison

WageWage Waiting Waiting TotalTotal

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

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 2 traders.that of 2 traders.

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

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

• Single-booth automatic car washSingle-booth automatic car wash

• Coffee vending machineCoffee vending machine

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Example: Ride ‘Em Cowboy!Example: Ride ‘Em Cowboy!

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

The mechanical pony rideThe mechanical pony ridemachine at the entrance to amachine at the entrance to avery popular J-Mart store providesvery popular J-Mart store provides2 minutes of riding for $.50. Children2 minutes of riding for $.50. Childrenwanting to ride the pony arrivewanting to ride the pony arrive(accompanied of course) according to a(accompanied of course) according to aPoisson distribution with a mean rate of 15 per Poisson distribution with a mean rate of 15 per hour.hour.

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What fraction of the time is the pony idle?What fraction of the time is the pony idle?

= 15 per hour= 15 per hour

= 60/2 = 30 per hour= 60/2 = 30 per hour

Utilization = Utilization = // = 15/30 = .5 = 15/30 = .5

Idle fraction = 1 – Utilization = 1 - .5 Idle fraction = 1 – Utilization = 1 - .5 = .5= .5

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What is the average number of children What is the average number of children waiting to ride the pony?waiting to ride the pony?

What is the average time a child waits for a What is the average time a child waits for a ride?ride?


2 2(15) = = .25 children

2 ( - ) 2(30)(30 - 15)

qL2 2(15)

= = .25 children2 ( - ) 2(30)(30 - 15)

qL

15 = = .01667 hours

2 ( - ) 2(30)(30 - 15)

qW15

= = .01667 hours2 ( - ) 2(30)(30 - 15)

qW

(or 1 minute)(or 1 minute)

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MM//GG//kk Queuing System Queuing System

Multiple channelsMultiple channels Poisson arrival-rate distributionPoisson arrival-rate distribution Arbitrary service timesArbitrary service times No waiting lineNo waiting line Infinite calling populationInfinite calling population Example:Example:

• Telephone system with Telephone system with kk lines. (When all lines. (When all kk lines are being used, additional callers get a lines are being used, additional callers get a busy signal.)busy signal.)

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Example: Allen-BoothExample: Allen-Booth


Allen-Booth (A-B) is an OTC marketAllen-Booth (A-B) is an OTC market

maker. A broker wishing to trade amaker. A broker wishing to trade a

particular stock for a client willparticular stock for a client will

call on a firm like Allen-Booth to call on a firm like Allen-Booth to

execute the order. If the marketexecute the order. If the market

maker's phone line is busy, a maker's phone line is busy, a

broker will immediately try broker will immediately try

calling another market maker tocalling another market maker to

transact the order.transact the order.

53 53 Slide

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A-B estimates that on the average, aA-B estimates that on the average, a

broker will try to call to execute abroker will try to call to execute a

stock transaction every two minutes.stock transaction every two minutes.

The time required to complete theThe time required to complete the

transaction averages 75 seconds.transaction averages 75 seconds.

A-B has four traders staffing itsA-B has four traders staffing its

phones. Assume calls arrivephones. Assume calls arrive

according to a Poisson distribution.according to a Poisson distribution.

54 54 Slide

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This problem can be modeled as an This problem can be modeled as an MM//GG//kk

system with block customers cleared with:system with block customers cleared with:

1/1/ = 2 minutes = 2/60 hour = 2 minutes = 2/60 hour

= 60/2 = 30 per hour= 60/2 = 30 per hour

1/1/µ µ = 75 sec. = 75/60 min. = = 75 sec. = 75/60 min. = 75/3600 hr.75/3600 hr.

µµ = 3600/75 = 48 per hour = 3600/75 = 48 per hour

55 55 Slide

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% of A-B’s Customers Lost Due to Busy Line% of A-B’s Customers Lost Due to Busy Line

First, we must solve for First, we must solve for PP00

where where kk = 4 = 4

11

PP00 = = 11 ++ (30/48)(30/48) ++ (30/48)2/2!(30/48)2/2! ++ (30/48)3/3!(30/48)3/3! ++

(30/48)4/4!(30/48)4/4!

PP00 = .536 = .536

continuedcontinued

0

0

1

( ) !k

i

i

Pi

0

0

1

( ) !k

i

i

Pi

56 56 Slide

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% of A-B’s Customers Lost Due to Busy Line% of A-B’s Customers Lost Due to Busy Line

Now, Now,

Thus, with four traders 0.3% of the potential Thus, with four traders 0.3% of the potential customers are lost.customers are lost.


4

4 0

( ) (30 48)(.536) .003

! 4!

k

P Pk

4

4 0

( ) (30 48)(.536) .003

! 4!

k

P Pk

57 57 Slide

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End of Chapter 12, Part AEnd of Chapter 12, Part A

1 1 Slide © 2005 Thomson/South-Western Chapter 12, Part A Waiting Line Models n Structure of a...

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