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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
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
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 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..
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 waiting units and then serve the unit with the highest priority first.the highest priority first.
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.
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.
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:
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.
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:
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.
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
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 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
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)
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.
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 service time exactly equals any specific value is 0.is 0.
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
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:
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
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
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
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.
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.
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?
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:
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?
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.
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
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
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.
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.
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.).
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
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.
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.)