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1 1 Slide Slide
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
2 2 Slide Slide
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.
3 3 Slide Slide
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..
4 4 Slide Slide
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.
5 5 Slide Slide
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
6 6 Slide Slide
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
7 7 Slide Slide
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
8 8 Slide Slide
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.
9 9 Slide Slide
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
10 10 Slide Slide
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
11 11 Slide Slide
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
12 12 Slide Slide
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.
13 13 Slide Slide
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
14 14 Slide Slide
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= =
15 15 Slide Slide
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 – ))
16 16 Slide Slide
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
17 17 Slide Slide
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.
18 18 Slide Slide
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?
19 19 Slide Slide
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.
20 20 Slide Slide
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
21 21 Slide Slide
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
22 22 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 = .238= 1 - .762 = .238
23 23 Slide Slide
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.
24 24 Slide Slide
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%
25 25 Slide Slide
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 .
26 26 Slide Slide
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%
27 27 Slide Slide
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 6 minutes = 1/10 hour or 6 minutes
28 28 Slide Slide
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
= 4/3= 4/3
29 29 Slide Slide
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 66.67%= 2/3 or 66.67%
30 30 Slide Slide
Example: SJJT, Inc. (A)Example: SJJT, Inc. (A)
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
31 31 Slide Slide
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
32 32 Slide Slide
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k
k
w
kkn
n
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n
kn
n
n
LWL
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kkk
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33 33 Slide Slide
General Operating CharacteristicsGeneral Operating Characteristics
1
WW
)λ
L(or W λWL
)λ
L (or W λW L
:Equations Flow sLittle'
q
qqqq
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?
35 35 Slide Slide
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.
36 36 Slide Slide
Example: SJJT, Inc. (B)Example: SJJT, Inc. (B)
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.
37 37 Slide Slide
Example: SJJT, Inc. (B)Example: SJJT, Inc. (B)
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?
38 38 Slide Slide
Example: SJJT, Inc. (B)Example: SJJT, Inc. (B)
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
P
n kk
k
n k
n
k0
0
1
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( / )!
( / )!
( )
P
n kk
k
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( )
39 39 Slide Slide
Example: SJJT, Inc. (B)Example: SJJT, Inc. (B)
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
Example: SJJT, Inc. (B)Example: SJJT, Inc. (B)
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.
41 41 Slide Slide
Example: SJJT, Inc. (B)Example: SJJT, Inc. (B)
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
Example: SJJT, Inc. (B)Example: SJJT, Inc. (B)
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 )
Probability an order must wait
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
Example: SJJT, Inc. (B)Example: SJJT, Inc. (B)
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 )
Probability an order must wait
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
Example: SJJT, Inc. (C)Example: SJJT, Inc. (C)
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
Example: SJJT, Inc. (C)Example: SJJT, Inc. (C)
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
Example: SJJT, Inc. (C)Example: SJJT, Inc. (C)
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
( / )!
( / )!
( )
48 48 Slide Slide
Example: SJJT, Inc. (C)Example: SJJT, Inc. (C)
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
Example: SJJT, Inc. (C)Example: SJJT, Inc. (C)
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
Example: SJJT, Inc. (C)Example: SJJT, Inc. (C)
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
Example: SJJT, Inc. (C)Example: SJJT, Inc. (C)
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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