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Heizer om10 mod_d

1. 1. D - 1© 2011 Pearson Education, Inc. publishing as Prentice HallD Waiting-Line ModelsPowerPoint presentation to accompanyHeizer and RenderOperations Management, 10ePrinciples of Operations Management, 8ePowerPoint slides by Jeff Heyl
2. 2. D - 2© 2011 Pearson Education, Inc. publishing as Prentice HallOutline Queuing Theory Characteristics of a Waiting-LineSystem Arrival Characteristics Waiting-Line Characteristics Service Characteristics Measuring a Queue’s Performance Queuing Costs
3. 3. D - 3© 2011 Pearson Education, Inc. publishing as Prentice HallOutline – Continued The Variety of Queuing Models Model A(M/M/1): Single-ChannelQueuing Model with Poisson Arrivalsand Exponential Service Times Model B(M/M/S): Multiple-ChannelQueuing Model Model C(M/D/1): Constant-Service-Time Model Little’s Law Model D: Limited-Population Model
4. 4. D - 4© 2011 Pearson Education, Inc. publishing as Prentice HallOutline – Continued Other Queuing Approaches
5. 5. D - 5© 2011 Pearson Education, Inc. publishing as Prentice HallLearning ObjectivesWhen you complete this module youshould be able to:1. Describe the characteristics ofarrivals, waiting lines, and servicesystems2. Apply the single-channel queuingmodel equations3. Conduct a cost analysis for awaiting line
6. 6. D - 6© 2011 Pearson Education, Inc. publishing as Prentice HallLearning ObjectivesWhen you complete this module youshould be able to:4. Apply the multiple-channelqueuing model formulas5. Apply the constant-service-timemodel equations6. Perform a limited-populationmodel analysis
7. 7. D - 7© 2011 Pearson Education, Inc. publishing as Prentice HallQueuing Theory The study of waiting lines Waiting lines are commonsituations Useful in bothmanufacturingand serviceareas
8. 8. D - 8© 2011 Pearson Education, Inc. publishing as Prentice HallCommon QueuingSituationsSituation Arrivals in Queue Service ProcessSupermarket Grocery shoppers Checkout clerks at cashregisterHighway toll booth Automobiles Collection of tolls at boothDoctor’s office Patients Treatment by doctors andnursesComputer system Programs to be run Computer processes jobsTelephone company Callers Switching equipment toforward callsBank Customer Transactions handled by tellerMachinemaintenanceBroken machines Repair people fix machinesHarbor Ships and barges Dock workers load and unloadTable D.1
9. 9. D - 9© 2011 Pearson Education, Inc. publishing as Prentice HallCharacteristics of Waiting-Line Systems1. Arrivals or inputs to the system Population size, behavior, statisticaldistribution2. Queue discipline, or the waiting lineitself Limited or unlimited in length, disciplineof people or items in it3. The service facility Design, statistical distribution of servicetimes
10. 10. D - 10© 2011 Pearson Education, Inc. publishing as Prentice HallParts of a Waiting LineFigure D.1Dave’sCar WashEnter ExitPopulation ofdirty carsArrivalsfrom thegeneralpopulation …Queue(waiting line)ServicefacilityExit the systemArrivals to the system Exit the systemIn the systemArrival Characteristics Size of the population Behavior of arrivals Statistical distributionof arrivalsWaiting LineCharacteristics Limited vs.unlimited Queue disciplineService Characteristics Service design Statistical distributionof service
11. 11. D - 11© 2011 Pearson Education, Inc. publishing as Prentice HallArrival Characteristics1. Size of the population Unlimited (infinite) or limited (finite)2. Pattern of arrivals Scheduled or random, often a Poissondistribution3. Behavior of arrivals Wait in the queue and do not switch lines No balking or reneging
12. 12. D - 12© 2011 Pearson Education, Inc. publishing as Prentice HallPoisson DistributionP(x) = for x = 0, 1, 2, 3, 4, …e-xx!where P(x) = probability of x arrivalsx = number of arrivals per unit of time = average arrival ratee = 2.7183 (which is the base of thenatural logarithms)
13. 13. D - 13© 2011 Pearson Education, Inc. publishing as Prentice HallPoisson DistributionProbability = P(x) =e-xx!0.25 –0.02 –0.15 –0.10 –0.05 ––Probability0 1 2 3 4 5 6 7 8 9Distribution for  = 2x0.25 –0.02 –0.15 –0.10 –0.05 ––Probability0 1 2 3 4 5 6 7 8 9Distribution for  = 4x10 11Figure D.2
14. 14. D - 14© 2011 Pearson Education, Inc. publishing as Prentice HallWaiting-Line Characteristics Limited or unlimited queue length Queue discipline - first-in, first-out(FIFO) is most common Other priority rules may be used inspecial circumstances
15. 15. D - 15© 2011 Pearson Education, Inc. publishing as Prentice HallService Characteristics Queuing system designs Single-channel system, multiple-channel system Single-phase system, multiphasesystem Service time distribution Constant service time Random service times, usually anegative exponential distribution
16. 16. D - 16© 2011 Pearson Education, Inc. publishing as Prentice HallQueuing System DesignsFigure D.3Departuresafter serviceSingle-channel, single-phase systemQueueArrivalsSingle-channel, multiphase systemArrivalsDeparturesafter servicePhase 1servicefacilityPhase 2servicefacilityServicefacilityQueueA family dentist’s officeA McDonald’s dual window drive-through
17. 17. D - 17© 2011 Pearson Education, Inc. publishing as Prentice HallQueuing System DesignsFigure D.3Multi-channel, single-phase systemArrivalsQueueMost bank and post office service windowsDeparturesafter serviceServicefacilityChannel 1ServicefacilityChannel 2ServicefacilityChannel 3
18. 18. D - 18© 2011 Pearson Education, Inc. publishing as Prentice HallQueuing System DesignsFigure D.3Multi-channel, multiphase systemArrivalsQueueSome college registrationsDeparturesafter servicePhase 2servicefacilityChannel 1Phase 2servicefacilityChannel 2Phase 1servicefacilityChannel 1Phase 1servicefacilityChannel 2
19. 19. D - 19© 2011 Pearson Education, Inc. publishing as Prentice HallNegative ExponentialDistributionFigure D.41.0 –0.9 –0.8 –0.7 –0.6 –0.5 –0.4 –0.3 –0.2 –0.1 –0.0 –Probabilitythatservicetime≥1| | | | | | | | | | | | |0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00Time t (hours)Probability that service time is greater than t = e-µt for t ≥ 1µ = Average service ratee = 2.7183Average service rate (µ) =1 customer per hourAverage service rate (µ) = 3 customers per hour Average service time = 20 minutes per customer
20. 20. D - 20© 2011 Pearson Education, Inc. publishing as Prentice HallMeasuring QueuePerformance1. Average time that each customer or objectspends in the queue2. Average queue length3. Average time each customer spends in thesystem4. Average number of customers in the system5. Probability that the service facility will be idle6. Utilization factor for the system7. Probability of a specific number of customersin the system
21. 21. D - 21© 2011 Pearson Education, Inc. publishing as Prentice HallQueuing CostsFigure D.5Total expected costCost of providing serviceCostLow levelof serviceHigh levelof serviceCost of waiting timeMinimumTotalcostOptimalservice level
22. 22. D - 22© 2011 Pearson Education, Inc. publishing as Prentice HallQueuing ModelsThe four queuing models here all assume: Poisson distribution arrivals FIFO discipline A single-service phase
23. 23. D - 23© 2011 Pearson Education, Inc. publishing as Prentice HallQueuing ModelsTable D.2Model Name ExampleA Single-channel Information countersystem at department store(M/M/1)Number Number Arrival Serviceof of Rate Time Population QueueChannels Phases Pattern Pattern Size DisciplineSingle Single Poisson Exponential Unlimited FIFO
24. 24. D - 24© 2011 Pearson Education, Inc. publishing as Prentice HallQueuing ModelsTable D.2Model Name ExampleB Multichannel Airline ticket(M/M/S) counterNumber Number Arrival Serviceof of Rate Time Population QueueChannels Phases Pattern Pattern Size DisciplineMulti- Single Poisson Exponential Unlimited FIFOchannel
25. 25. D - 25© 2011 Pearson Education, Inc. publishing as Prentice HallQueuing ModelsTable D.2Model Name ExampleC Constant- Automated carservice wash(M/D/1)Number Number Arrival Serviceof of Rate Time Population QueueChannels Phases Pattern Pattern Size DisciplineSingle Single Poisson Constant Unlimited FIFO
26. 26. D - 26© 2011 Pearson Education, Inc. publishing as Prentice HallQueuing ModelsTable D.2Model Name ExampleD Limited Shop with only apopulation dozen machines(finite population) that might breakNumber Number Arrival Serviceof of Rate Time Population QueueChannels Phases Pattern Pattern Size DisciplineSingle Single Poisson Exponential Limited FIFO
27. 27. D - 27© 2011 Pearson Education, Inc. publishing as Prentice HallModel A – Single-Channel1. Arrivals are served on a FIFO basis andevery arrival waits to be servedregardless of the length of the queue2. Arrivals are independent of precedingarrivals but the average number ofarrivals does not change over time3. Arrivals are described by a Poissonprobability distribution and come froman infinite population
28. 28. D - 28© 2011 Pearson Education, Inc. publishing as Prentice HallModel A – Single-Channel4. Service times vary from one customerto the next and are independent of oneanother, but their average rate isknown5. Service times occur according to thenegative exponential distribution6. The service rate is faster than thearrival rate
29. 29. D - 29© 2011 Pearson Education, Inc. publishing as Prentice HallModel A – Single-ChannelTable D.3 = Mean number of arrivals per time periodµ = Mean number of units served per time periodLs = Average number of units (customers) in thesystem (waiting and being served)=Ws = Average time a unit spends in the system(waiting time plus service time)=µ – 1µ – 
30. 30. D - 30© 2011 Pearson Education, Inc. publishing as Prentice HallModel A – Single-ChannelTable D.3Lq = Average number of units waiting in thequeue=Wq = Average time a unit spends waiting in thequeue= = Utilization factor for the system=2µ(µ – )µ(µ – )µ
31. 31. D - 31© 2011 Pearson Education, Inc. publishing as Prentice HallModel A – Single-ChannelTable D.3P0 = Probability of 0 units in the system (that is,the service unit is idle)= 1 –Pn > k = Probability of more than k units in thesystem, where n is the number of units inthe system=µµk + 1
32. 32. D - 32© 2011 Pearson Education, Inc. publishing as Prentice HallSingle-Channel Example = 2 cars arriving/hour µ = 3 cars serviced/hourLs = = = 2 cars in the system on averageWs = = = 1 hour average waiting time inthe systemLq = = = 1.33 cars waiting in line2µ(µ – )µ – 1µ – 23 - 213 - 2223(3 - 2)
33. 33. D - 33© 2011 Pearson Education, Inc. publishing as Prentice HallSingle-Channel ExampleWq = = = 2/3 hour = 40 minuteaverage waiting time = /µ = 2/3 = 66.6% of time mechanic is busyµ(µ – )23(3 - 2)µP0 = 1 - = .33 probability there are 0 cars in thesystem = 2 cars arriving/hour µ = 3 cars serviced/hour
34. 34. D - 34© 2011 Pearson Education, Inc. publishing as Prentice HallSingle-Channel ExampleProbability of more than k Cars in the Systemk Pn > k = (2/3)k + 10 .667  Note that this is equal to 1 - P0 = 1 - .331 .4442 .2963 .198  Implies that there is a 19.8% chance thatmore than 3 cars are in the system4 .1325 .0886 .0587 .039
35. 35. D - 35© 2011 Pearson Education, Inc. publishing as Prentice HallSingle-Channel EconomicsCustomer dissatisfactionand lost goodwill = \$10 per hourWq = 2/3 hourTotal arrivals = 16 per dayMechanic’s salary = \$56 per dayTotal hourscustomers spendwaiting per day= (16) = 10 hours2323Customer waiting-time cost = \$10 10 = \$106.6723Total expected costs = \$106.67 + \$56 = \$162.67
36. 36. D - 36© 2011 Pearson Education, Inc. publishing as Prentice HallMulti-Channel ModelTable D.4M = number of channels open = average arrival rateµ = average service rate at each channelP0 = for Mµ > 11M!1n!MµMµ - M – 1n = 0µnµM+∑Ls = P0 +µ(/µ)M(M - 1)!(Mµ - )2µ
37. 37. D - 37© 2011 Pearson Education, Inc. publishing as Prentice HallMulti-Channel ModelTable D.4Ws = P0 + =µ(/µ)M(M - 1)!(Mµ - )21µLsLq = Ls –µWq = Ws – =1µLq
38. 38. D - 38© 2011 Pearson Education, Inc. publishing as Prentice HallMulti-Channel Example = 2 µ = 3 M = 2P0 = =112!1n!2(3)2(3) - 21n = 023n232+∑12Ls = + =(2)(3(2/3)2 231! 2(3) - 2 21234Wq = = .0415.0832Ws = =3/4238Lq = – =2334112
39. 39. D - 39© 2011 Pearson Education, Inc. publishing as Prentice HallMulti-Channel ExampleSingle Channel Two ChannelsP0 .33 .5Ls 2 cars .75 carsWs 60 minutes 22.5 minutesLq 1.33 cars .083 carsWq 40 minutes 2.5 minutes
40. 40. D - 40© 2011 Pearson Education, Inc. publishing as Prentice HallWaiting Line TablesTable D.5Poisson Arrivals, Exponential Service TimesNumber of Service Channels, Mρ 1 2 3 4 5.10 .0111.25 .0833 .0039.50 .5000 .0333 .0030.75 2.2500 .1227 .0147.90 3.1000 .2285 .0300 .00411.0 .3333 .0454 .00671.6 2.8444 .3128 .0604 .01212.0 .8888 .1739 .03982.6 4.9322 .6581 .16093.0 1.5282 .35414.0 2.2164
41. 41. D - 41© 2011 Pearson Education, Inc. publishing as Prentice HallWaiting Line Table ExampleBank tellers and customers = 18, µ = 20From Table D.5Utilization factor  = /µ = .90 Wq =LqNumber ofservice windows MNumberin queue Time in queue1 window 1 8.1 .45 hrs, 27 minutes2 windows 2 .2285 .0127 hrs, ¾ minute3 windows 3 .03 .0017 hrs, 6 seconds4 windows 4 .0041 .0003 hrs, 1 second
42. 42. D - 42© 2011 Pearson Education, Inc. publishing as Prentice HallConstant-Service ModelTable D.6Lq = 22µ(µ – )Average lengthof queueWq =2µ(µ – )Average waiting timein queueµLs = Lq +Average number ofcustomers in systemWs = Wq +1µAverage timein the system
43. 43. D - 43© 2011 Pearson Education, Inc. publishing as Prentice HallNet savings = \$ 7 /tripConstant-Service ExampleTrucks currently wait 15 minutes on averageTruck and driver cost \$60 per hourAutomated compactor service rate (µ) = 12 trucks per hourArrival rate () = 8 per hourCompactor costs \$3 per truckCurrent waiting cost per trip = (1/4 hr)(\$60) = \$15 /tripWq = = hour82(12)(12 – 8)112Waiting cost/tripwith compactor = (1/12 hr wait)(\$60/hr cost) = \$ 5 /tripSavings withnew equipment= \$15 (current) – \$5(new) = \$10 /tripCost of new equipment amortized = \$ 3 /trip
44. 44. D - 44© 2011 Pearson Education, Inc. publishing as Prentice HallLittle’s Law A queuing system in steady stateL = W (which is the same as W = L/Lq = Wq (which is the same as Wq = Lq/ Once one of these parameters is known, theother can be easily found It makes no assumptions about the probabilitydistribution of arrival and service times Applies to all queuing models except the limitedpopulation model
45. 45. D - 45© 2011 Pearson Education, Inc. publishing as Prentice HallLimited-Population ModelTable D.7Service factor: X =Average number running: J = NF(1 - X)Average number waiting: L = N(1 - F)Average number being serviced: H = FNXAverage waiting time: W =Number of population: N = J + L + HTT + UT(1 - F)XF
46. 46. D - 46© 2011 Pearson Education, Inc. publishing as Prentice HallLimited-Population ModelD = Probability that a unitwill have to wait inqueueN = Number of potentialcustomersF = Efficiency factor T = Average service timeH = Average number of unitsbeing servedU = Average time betweenunit servicerequirementsJ = Average number of unitsnot in queue or inservice bayW = Average time a unitwaits in lineL = Average number of unitswaiting for serviceX = Service factorM = Number of servicechannels
47. 47. D - 47© 2011 Pearson Education, Inc. publishing as Prentice HallFinite Queuing TableTable D.8X M D F.012 1 .048 .999.025 1 .100 .997.050 1 .198 .989.060 2 .020 .9991 .237 .983.070 2 .027 .9991 .275 .977.080 2 .035 .9981 .313 .969.090 2 .044 .9981 .350 .960.100 2 .054 .9971 .386 .950
48. 48. D - 48© 2011 Pearson Education, Inc. publishing as Prentice HallLimited-Population ExampleService factor: X = = .091 (close to .090)For M = 1, D = .350 and F = .960For M = 2, D = .044 and F = .998Average number of printers working:For M = 1, J = (5)(.960)(1 - .091) = 4.36For M = 2, J = (5)(.998)(1 - .091) = 4.5422 + 20Each of 5 printers requires repair after 20 hours (U) of useOne technician can service a printer in 2 hours (T)Printer downtime costs \$120/hourTechnician costs \$25/hour
49. 49. D - 49© 2011 Pearson Education, Inc. publishing as Prentice HallLimited-Population ExampleService factor: X = = .091 (close to .090)For M = 1, D = .350 and F = .960For M = 2, D = .044 and F = .998Average number of printers working:For M = 1, J = (5)(.960)(1 - .091) = 4.36For M = 2, J = (5)(.998)(1 - .091) = 4.5422 + 20Each of 5 printers require repair after 20 hours (U) of useOne technician can service a printer in 2 hours (T)Printer downtime costs \$120/hourTechnician costs \$25/hourNumber ofTechniciansAverageNumberPrintersDown (N - J)AverageCost/Hr forDowntime(N - J)\$120Cost/Hr forTechnicians(\$25/hr)TotalCost/Hr1 .64 \$76.80 \$25.00 \$101.802 .46 \$55.20 \$50.00 \$105.20
50. 50. D - 50© 2011 Pearson Education, Inc. publishing as Prentice HallOther Queuing Approaches The single-phase models cover manyqueuing situations Variations of the four single-phasesystems are possible Multiphase modelsexist for morecomplex situations
51. 51. D - 51© 2011 Pearson Education, Inc. publishing as Prentice HallAll rights reserved. No part of this publication may be reproduced, stored in a retrievalsystem, or transmitted, in any form or by any means, electronic, mechanical, photocopying,recording, or otherwise, without the prior written permission of the publisher.Printed in the United States of America.