2008 Prentice Hall, Inc. D – 1 Operations Management Module D – Module D – Waiting-Line Models Waiting-Line Models PowerPoint presentation to accompany PowerPoint presentation to accompany Heizer/Render Heizer/Render Principles of Operations Management, 7e Principles of Operations Management, 7e Operations Management, 9e Operations Management, 9e
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Operations ManagementOperations ManagementModule D – Module D – Waiting-Line ModelsWaiting-Line Models
PowerPoint presentation to accompany PowerPoint presentation to accompany Heizer/Render Heizer/Render Principles of Operations Management, 7ePrinciples of Operations Management, 7eOperations Management, 9e Operations Management, 9e
When you complete this module you When you complete this module you should be able to:should be able to:
1.1. Describe the characteristics of Describe the characteristics of arrivals, waiting lines, and service arrivals, waiting lines, and service systemssystems
2.2. Apply the single-channel queuing Apply the single-channel queuing model equationsmodel equations
3.3. Conduct a cost analysis for a Conduct a cost analysis for a waiting linewaiting line
Characteristics of Waiting-Characteristics of Waiting-Line SystemsLine Systems
1.1. Arrivals or inputs to the systemArrivals or inputs to the system Population size, behavior, statistical Population size, behavior, statistical
distributiondistribution
2.2. Queue discipline, or the waiting line Queue discipline, or the waiting line itselfitself Limited or unlimited in length, discipline Limited or unlimited in length, discipline
of people or items in itof people or items in it
3.3. The service facilityThe service facility Design, statistical distribution of service Design, statistical distribution of service
Arrivals to the systemArrivals to the system Exit the systemExit the systemIn the systemIn the system
Arrival CharacteristicsArrival Characteristics Size of the populationSize of the population Behavior of arrivalsBehavior of arrivals Statistical distribution Statistical distribution
of arrivalsof arrivals
Waiting Line Waiting Line CharacteristicsCharacteristics
Limited vs. Limited vs. unlimitedunlimited
Queue disciplineQueue discipline
Service CharacteristicsService Characteristics Service designService design Statistical distribution Statistical distribution
Probability that service time is greater than t = eProbability that service time is greater than t = e-µ-µtt for t for t ≥ 1≥ 1
µ =µ = Average service rate Average service ratee e = 2.7183= 2.7183
Average service rate Average service rate (µ) = (µ) = 1 customer per hour1 customer per hour
Average service rate Average service rate (µ) = 3(µ) = 3 customers per hour customers per hour Average service time Average service time = 20= 20 minutes per customer minutes per customer
1.1. Arrivals are served on a FIFO basis and Arrivals are served on a FIFO basis and every arrival waits to be served every arrival waits to be served regardless of the length of the queueregardless of the length of the queue
2.2. Arrivals are independent of preceding Arrivals are independent of preceding arrivals but the average number of arrivals but the average number of arrivals does not change over timearrivals does not change over time
3.3. Arrivals are described by a Poisson Arrivals are described by a Poisson probability distribution and come from probability distribution and come from an infinite populationan infinite population
4.4. Service times vary from one customer Service times vary from one customer to the next and are independent of one to the next and are independent of one another, but their average rate is another, but their average rate is knownknown
5.5. Service times occur according to the Service times occur according to the negative exponential distributionnegative exponential distribution
6.6. The service rate is faster than the The service rate is faster than the arrival ratearrival rate
== Mean number of arrivals per time periodMean number of arrivals per time period
µµ == Mean number of units served per time Mean number of units served per time periodperiod
LLss == Average number of units (customers) in Average number of units (customers) in the system (waiting and being served)the system (waiting and being served)
==
WWss== Average time a unit spends in the Average time a unit spends in the system (waiting time plus service time)system (waiting time plus service time)
PP00 == Probability of Probability of 00 units in the system units in the system (that is, the service unit is idle)(that is, the service unit is idle)
== 1 –1 –
PPn > kn > k ==Probability of more than k units in the Probability of more than k units in the system, where n is the number of units in system, where n is the number of units in the systemthe system
Probability of more than k Cars in the SystemProbability of more than k Cars in the System
kk PPn > kn > k = (2/3)= (2/3)k k + 1+ 1
00 .667.667 Note that this is equal toNote that this is equal to 1 - 1 - PP00 = 1 - .33 = 1 - .33
11 .444.444
22 .296.296
33 .198.198 Implies that there is aImplies that there is a 19.8% 19.8% chance that more thanchance that more than 3 3 cars are in the cars are in the systemsystem
MM == number of channels opennumber of channels open == average arrival rateaverage arrival rateµµ == average service rate at each average service rate at each channelchannel
Constant-Service ExampleConstant-Service ExampleTrucks currently wait Trucks currently wait 1515 minutes on average minutes on averageTruck and driver cost Truck and driver cost $60$60 per hour per hourAutomated compactor service rate Automated compactor service rate (µ) (µ) = 12 trucks per hour= 12 trucks per hourArrival rate Arrival rate (()) = 8= 8 per hour per hourCompactor costs Compactor costs $3$3 per truck per truck
Current waiting cost per trip Current waiting cost per trip = (1/4= (1/4 hr hr)($60) = $15)($60) = $15 //triptrip
Limited-Population ExampleLimited-Population Example
Service factor: X = Service factor: X = = .091 (= .091 (close to close to .090).090)
For M For M = 1,= 1, D D = .350= .350 and F and F = .960= .960
For M For M = 2,= 2, D D = .044= .044 and F and F = .998= .998
Average number of printers working:Average number of printers working:
For M For M = 1,= 1, J J = (5)(.960)(1 - .091) = 4.36= (5)(.960)(1 - .091) = 4.36
For M For M = 2,= 2, J J = (5)(.998)(1 - .091) = 4.54= (5)(.998)(1 - .091) = 4.54
222 + 202 + 20
Each of Each of 55 printers requires repair after printers requires repair after 2020 hours hours ((UU)) of use of useOne technician can service a printer in One technician can service a printer in 22 hours hours ((TT))Printer downtime costs Printer downtime costs $120/$120/hourhourTechnician costs Technician costs $25/$25/hourhour
Limited-Population ExampleLimited-Population Example
Service factor: X = Service factor: X = = .091 (= .091 (close to close to .090).090)
For M For M = 1,= 1, D D = .350= .350 and F and F = .960= .960
For M For M = 2,= 2, D D = .044= .044 and F and F = .998= .998
Average number of printers working:Average number of printers working:
For M For M = 1,= 1, J J = (5)(.960)(1 - .091) = 4.36= (5)(.960)(1 - .091) = 4.36
For M For M = 2,= 2, J J = (5)(.998)(1 - .091) = 4.54= (5)(.998)(1 - .091) = 4.54
222 + 202 + 20
Each of Each of 55 printers require repair after printers require repair after 2020 hours hours ((UU)) of use of useOne technician can service a printer in One technician can service a printer in 22 hours hours ((TT))Printer downtime costs Printer downtime costs $120/$120/hourhourTechnician costs Technician costs $25/$25/hourhour