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Chapter 12 Waiting Line Models. The Structure of a Waiting Line System Queuing Systems Queuing System Input Characteristics Queuing System Operating Characteristics Analytical Formulas Single-Channel Waiting Line Model with Poisson Arrivals and Exponential Service Times - PowerPoint PPT Presentation
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Chapter 12Chapter 12Waiting 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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Structure of a Waiting Line SystemStructure of a Waiting Line System
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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Structure of a Waiting Line SystemStructure of a Waiting Line System
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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Structure of a Waiting Line SystemStructure of a Waiting Line System
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 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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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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Example: SJJT, Inc. (A)Example: SJJT, Inc. (A)
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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Example: SJJT, Inc. (A)Example: SJJT, Inc. (A)
Arrival Rate DistributionArrival Rate Distribution
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
15 15 Slide Slide
Example: SJJT, Inc. (A)Example: SJJT, Inc. (A)
Arrival Rate DistributionArrival Rate Distribution
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 = = 1 - .762 = .238.238
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Example: SJJT, Inc. (A)Example: SJJT, Inc. (A)
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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Example: SJJT, Inc. (A)Example: SJJT, Inc. (A)
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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Example: SJJT, Inc. (A)Example: SJJT, Inc. (A)
Service Time DistributionService Time Distribution
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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Example: SJJT, Inc. (A)Example: SJJT, Inc. (A)
Service Time DistributionService Time Distribution
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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Example: SJJT, Inc. (A)Example: SJJT, Inc. (A)
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 = 1/10 hour or 6 minutes6 minutes
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Example: SJJT, Inc. (A)Example: SJJT, Inc. (A)
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
= = 4/34/3
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Example: SJJT, Inc. (A)Example: SJJT, Inc. (A)
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 = 2/3 or 66.67%66.67%