数据、模型与决策(运筹学)课后习题和案例答案1.pdf

CHAPTER 14QUEUEING MODELSReview Questions14.1-1 Customers might be vehicles,machines,or other items.14.1-2 It might be a crew of people working together,a machine,a vehicle,or an electronic device.14.1-3 Mean arrival rate=1 /(mean interarrival time).14.1-414.1-5 The mean equals the standard deviation of the exponential distribution.14.1-6 Having random arrivals means that arrival times are completely unpredictable in the sensethat the chance of an arrival in the next minute always is just the same as for any otherminute.The only distribution of interarrival times that fits having random arrivals is theexponential distribution.14.1-7 The number of customers in the queue is the number of customers waiting for service tobegin.The number of customers in the system is the number in the queue plus the numbercurrently being served.14.1-8 Queueing models conventionally assume that the queue is an infinite queue and that thequeue discipline is first come first served.14.1-9 Mean service time=1 /(mean service rate).14.1-10 For the exponential distribution,the standard deviation equals the mean.For the degeneratedistribution,the standard deviation equals zero.For the Erlang distribution,the standarddeviation=(mean).14.1-11 The three parts of a label for queueing models provide information on the distribution ofservice times,the number of servers,and the distribution of interarrival times.14.2-1 In commercial service systems,outside customers receive service from commercialorganizations.Many examples are possible.14-114.2-2 In internal service systems,the customers receiving service are internal to the organization.Many examples are possible.14.2-3 In transportation service systems,either the customers or the servers are vehicles.Manyexamples are possible.14.3-1 When the customers are internal to the organization providing the service,it is moreimportant how many customers typically are waiting in the queueing system.14.3-2 Commercial service systems tend to place a greater importance on how long customerstypically have to wait.14.3-3 L=expected number of customers in the systemLq=expected number of customers in the queueW=expected waiting time in the system%=expected waiting time in the queue14.3-4 A queueing system is in a steady state condition if it is in its normal condition after operatingfor some time.14.3-5 W=%+(1/R)14.3-6 L=2lVandLq=ZWq14.3-7 L=Lq+(2/)14.3-8 Steady-state probabilities can also be used as measures of performance.14.4-1 Each Tech Rep should be assigned enough machines so that the Tech Rep will be activerepairing machines approximately 75%of the time.14.4-2 The issue is the increased number of complaints about intolerable waits fbr repairs on thenew copier.14.4-3 The average waiting time of customers before the Tech Rep begins the trip to the customersite to repair the machine should not exceed two hours.14.4-4 Four alternative approaches have been suggested.14.4-5 A team of management scientist and John Phixitt will analyze these approaches.14.4-6 The machines needing repair are the customers and the Tech Reps are the servers.14.5-1 2=expected number of arrivals per unit time =expected number of service completions per unit time(1/2)=expected interarrival time(1/)=expected service timep-utilization factor14.5-2(1)Interarrival times have an exponential distribution with a mean of(IM);(2)service timeshave an exponential distribution with a mean of(I/);(3)the queueing system has 1 server.14.5-3 Formulas are available for L,W,%,Lq,Pn,P(Wt),and P(Wq=0).14.5-4 p 114-214.5-5 The average waiting time until service begins is 6 hours.14.5-6 It would cost approximately$300 million annually.14.5-7 The M/G/l model differs in the assumption about service time.In this model the servicetimes can have any probability distribution.It is not even necessary to determine the form ofthis distribution.It is only necessary to estimate the mean and standard deviation of thedistribution.14.5-8 The M/D/l model assumes a degenerate service-time distribution.The M/Ek/1 modelassumes an Erlang distribution with shape parameter k.14.5-9 Decreasing the standard deviation decreases Lq,L,W,and%14.5-10 The total additional cost is a one-time cost of approximately$500 million.14.6-1 p=2/(s)which is the average fraction of time that individual servers are being utilizedserving customers.14.6-2 p 114.6-3 Explicit formulas are available for all the measures of performance considered fbr theM/M/l model.14.6-4 Three territories need to be combined in order to satisfy the new service standard.14.6-5 The M/M/s model has a great amount of variability in service times.The M/D/s model hasno variability.The M/Ek/s model provides a middle ground between the other two withsome variability in service times.14.7-1 When using priorities,more important customers are served ahead of others who havewaited longer.14.7-2 With nonpreemptive priorities,once a server has begun serving a customer,the service mustbe completed without interruption even if a higher priority customer arrives while thisservice is in process.With preemptive priorities,the lowest priority customer being served isejected back into the queue whenever a higher priority customer enters the queueing system.14.7-3 Except for using preemptive priorities,the assumptions are the same as for the M/M/l model.14.7-4 Except fbr using nonpreemptive priorities,the assumptions are the same as for the M/M/smodel.14.7-5 pt)=4.54E-05Wq=0.0759when t=110P=0.7511Prob(Wq t)=3.405E-0512when t=1nPn1300.251410.18751520.140625c)L=A/(/-2)=30/(60-30)=1 customerW=1/(60-30)=0.033 hours%=W-孙=30/60(60-30)=0.017 hoursLq=30(0.017)=0.5 customersPo=l-p=1-0.75=0.5Pi=(l-p)p=(1-0.75)(0.75)=0.25P2=Q-M =(1-0.75)(0.75)2=0.125There is a X-PP-P=1-0.5-0.25-0.125=12.5%chance of having more than 2customers at the checkout stand.d)(M/M/s model)BCDEGH3DataResults4A.=30(mean arrival rate)L=15N=60(mean service rate)Lq=0.56s=1(#servers)7W=0.033338Pr(W t)=9.3576E-14wq=0.016679when t=110p=0.511Prob(Wqt)=4.6788E-1412when t=1nPn1300.51410.251520.125e)The manager should adopt the new approach of adding another person to bag the groceries.14-714.11 a)Pa=-p=1-0.5=0.5Pi=(l-p)p=(l-0.5)(0.5)=0.25P2=(l-p)/r=(l-0.5)(0.5)2=0.125P3=(1-9=(l-0.5)(0.5)3=0.0625P4=(l-p)/=(l-0.5)(0.5)4=0.03125PO+P1+P2+P3+P4=O.5+O.25+O.125+O.O625+O.O3125=O.96875 or 96.875%of the time.b)96.875%of the time there are fewer than 4 in the system.(M/M71 model)BCDEGHI3DataResults4X=2(mean arrival rate)L=15p=4(mean service rate)Lq=0.56s=1(#servers)7w=0.58Pr(W t)=0.13533528Wq=0.259when t=110P=0.511Prob(Wq t)=0.0676676412when t=1nPnCumulative1300.50.51410.250.751520.1250.8751630.06250.93751740.031250.9687514.12 a)Tractor-trailer train(M/M/l model):BCDEGHI3DataResults4X=4(mean arrival rate)L=455(mean service rate)Lq=3.26S=1(#servers)7w=18Pr(W t)=0.36787944Wq=0.89when t=110P=0.811Prob(Wq t)=0.2943035512when t=1nPnCumulative1300.20.21410.160.361520.1280.4881630.10240.59041740.081920.67232The train does not meet any of the criteria.The average time is more than half-an-hour(W=1 hour),it is no more than an hour less than 80%of the time(Pr(W 1)=36.8%),andthere are three loads or fewer less than 80%of the time(尸0+P1+P2+尸3+P4=67.2%).14-8b)Forklift truck(M/M/s model):BCDEGHI3DataResults4X=4(mean arrival rate)L=1.556.66666667(mean service rate)J =0.96S=1(#servers)7W=0.375008Pr(W t)=0.06948345Wq=0.225009when t=110P 二0.611Prob(Wq t)=0.0416900712when t=1nPnCumulative1300.40.41410.240.641520.1440.7841630.08640.87041740.051840.92224The forklift truck meets all the criteria.The average time is less than hairan-hour(W=0.375 hours),it is no more than an hour more than 80%of the time(Pr(W 1)=6.9%),and there are three loads or fewer more than 80%of the time(尸0+P1+P2+P3+&=92.2%).c)Tractor-trailer train:L($20)+$50=(4)($20)+$50=$130/hourForklift truck:L($20)+$150=(1.5)($20)+$150=$18()/hourd)While the forklift truck has higher overall costs,it does a better job of meeting theadditional criteria.14.13 2=Z7W=8/120=0.0667 per minute =1/W+4=1/120+0.0667=0.075 per minute(M/M/l model)BCDEGH3DataResults4X=0.06666667(mean arrival rate)L=850.075(mean service rate)q=7.1111111116s=1(#servers)7w=1208Pr(W t)=0.99170129Wq=106.66666679when t=110p二0.88888888911Prob(Wqt)=0.8815122612when t=1nPn1300.11111111114.14 a)The customers are trucks to be loaded or unloaded and the servers are the crews.Thesystem currently has 1 server.14-9b)4 member crew(M/M/l model):BCDEGH3DataResults4X=1(mean arrival rate)L=0.33333333354(mean service rate)q=0.0833333336s=1(#servers)7w=0.333338Pr(W t)=0.04978707wq=0.083339when t=110p=0.2511Prob(Wqt)=0.0124467712when t=1nPn1300.75c)3 member crew(M/M/l model):BCDEGH3DataResults4X=1(mean arrival rate)L=0.55N=3(mean service rate)L=0.1666666676S=1(#servers)7w=0.58Pr(W t)=0.13533528wq=0.1666666679when t=110p=0.33333333311Prob(Wq t)=0.0451117612when t=1nPn1300.666666667d)2 member crew(M/M/l model):BCDEGH3DataResults4九=1(mean arrival rate)L=152(mean service rate)q =0.56s=1(#servers)7w=18Pr(W t)=|0.36787944wq=0.59when t=110p=0.511Prob(Wq t)=|0.1839397212when t=1nPn13I00.5e)A one person team should not be considered since that would lead to a utilization factor ofp=1 which does not enable the qeueuing system to reach a steady-state condition with amanageable load for the team.14-10f&g)Total cost=($20)(#on crew)+($30)(Lq)TC(4 members)=($20)(4)+($30)(0.0833)=$82.50/hourTC(3 members)=($20)(3)+($30)(0.167)=$65/hourTC(2 members)=($20)(2)+($30)(0.5)=$55/hourA crew of 2 people will minimize the expected total cost per hour.14.15 a)1/R=1 minute(M/M/l model):BCDEGH3DataResults4X=0.5(mean arrival rate)L=15尸1(mean service rate)Lq=0.56s=1(#servers)7W=28Pr(W t)=0.082085Wq=19when t=510P=0.511Prob(Wq t)=0.041042512when t=5nPn1300.5For l/|i=1.5 minutes(M/M/l model):BCDEGH3DataResults4X=0.5(mean arrival rate)L=35厂0.66666667(mean service rate)Lq=2.256s=1(#servers)7W=68Pr(W t)=0.43459821wq=4.59when t=510p 二0.7511Prob(Wqt)=0.3259486612when t=5nPn1300.25(l/)=2 minutes is not a feasible alternative since p=1.b)TC(l/=1 minute)=$1.60+($0.80)(l)=$2.40TC(l/=1.5 minute)=$0.90+($0.80)(3)=$3.30The grinder should be set so that the mean service time is 1 minute.14-1114.16 a)(M/M/l model):BCDEGHI3DataResults4X=10(mean arrival rate)L=15p=20(mean service rate)Lq=0.56s=1(#servers)7w=0.18Pr(W t)=0.00673795Wq=0.059when t=0.510P=0.511Prob(Wq t)=0.0033689712when t=0.5nPncumulative1300.50.51410.250.751520.1250.8751630.06250.93751740.031250.968751850.0156250.984375All the criteria are currently being satisfied.The average number of planes waiting toreceive clearance is less than 1 (Lq=0.5),98.44%of the time there are 4 or fewer planeswaiting to land,and 0.34%of the planes must wait more than 30 minutes fbr clearance.b)(M/M/l model):BCDEGHI3DataResults4A 二15(mean arrival rate)L=35M -20(mean service rate)Lq=2.256S=1(#servers)7w=0.28Pr(W t)=0.082085Wq=0.159when t=0.510P=0.7511Prob(Wq t)=|0.0615637512when t=0.5nPncumulative1300.250.251410.18750.43751520.1406250.5781251630.105468750.683593751740.0791015630.7626953131850.0593261720.822021484None of the criteria are now satisfied.The average number of planes waiting to land isnow 2.25,82.2%of the time there are 4 or fewer planes waiting to land,and 6.16%of theplanes must wait more than 30 minutes fbr clearance.14-12c)(M/M/s model):BCDEGHI3DataResults4X=25(mean arrival rate)L=2.051282051520(mean service rate)0.8012820516s=2(#servers)7w=0.0820512828Pr(W t)=0.00102172Wq=0.0320512829when t=0.510P=0.62511Prob(Wq t)=0.0002659112when t=0.5nPncumulative1300.2307692310.2307692311410.2884615380.5192307691520.1802884620.6995192311630.1126802880.8121995191740.070425180.88262471850.0440157380.926640437In this case,the first and third criteria are satisfied but the second is not.The averagenumber of planes waiting to land is().8 and only 0.02%of the planes wait more than 3()minutes for clearance,but only 92.7%of the time are there 4 or fewer planes waiting toland.14.17 Lq is unchanged and Wq is reduced by half.14.18 a)M/G/l Model:BCDEFG3DataResults4九=0.2(mean arrival rate)L=2.551/p.=4(expected service time)q=1.76Q=1(standard deviation)7S=1(#servers)w =12.58g8.5910p 二0.81112Po=0.213141516oLLWWq172.51.712.58.518443.220161933.32.516.512.52022.8214102112.51.712.58.52202.41.6128b)Lq is half with t)=0.60653066wq=0.59when t=0.5Proposal(M/Ek/1):BCDEFG3DataResults4X=0.25(mean arrival rate)L=0.81255N=0.5(mean service rate)Lq=0.31256k=4(shape parameter)7s=1(#servers)W=3.258Wq=1.25a)Under the current policy,an airplane loses 1 day of flying time as opposed to 3.25 daysunder the proposed policy.b)Under the current policy 1 airplane is losing flying time per day as opposed to 0.8125airplanes.c)The comparison in part b is the appropriate one for making the decision since it takes intoaccount that airplanes will not have to come in for service as often.14-1914.24 a)M/M/s model:BCDEGHI3Data214Results4A.=(mean arrival rate)L=2.1739130430.1739130431.0869565220.0869565220.55(mean service rate)Lq=6s=(#servers)7W=8Pr(W t)=0.00790187Wq=9when t=510P=11Prob(Wq t)=|7.8956E-0612when t=5nPncumulative1300.1304347830.2608695650.2608695650.1739130430.0869565220.0434782610.021739130.0108695650.0054347830.0027173910.1304347831410.3913043481520.6521739131630.8260869571740.9130434781850.9565217391960.978260872070.989130435218飞0.9945652170.99728260922All the guidelines are currently being met.The average number in line is 0.17,99.7%ofthe time there are 5 or fewer in line,and 0.000789%of customers wait more than 5minutes.b)(M/M/s model):BCDEGHI3DataResults4X=3(mean arrival rate)L=4.5283018875L I1(mean service rate)1 =1.5283018876s=4(#servers)7W=1.5094339628Pr(W t)=0.02390064Wq=0.5094339629when t=510P=0.7511Prob(Wq t)=0.0034325412when t=5nPncumulative1300.0377358490.0377358491410.1132075470.1509433961520.1698113210.3207547171630.1698113210.4905660381740.1273584910.6179245281850.0955188680.7134433961960.0716391510.7850825472070.0537293630.838811912180.0402970220.8791089332290.0302227670.9093317The first two guidelines will not be satisfied in one year but the third will be.The averagenumber in line is 1.53,90.9%of the time there are 5 or fewer in line,and 0.34%ofcustomers wait more than 5 minutes.14-20c)Five tellers will be needed in a year to meet all of the requirements(M/M/s model):BCDEGHI3DataResults4X=3(mean arrival rate)L=3.35422740551(mean service rate)4 =0.3542274056S=5(#servers)7W=1.1180758028Pr(W t)=0.0083184Wq=0.1180758029when t=510P=0.611Prob(Wq t)=1.0721E-0512when t=5nPncumulative1300.046647230.046647231410.1399416910.1865889211520.2099125360.3965014581630.2099125360.6064139941740.1574344020.7638483971850.0944606410.8583090381960.0566763850.9149854232070.0340058310.9489912542180.0204034990.9693947522290.0122420990.98163685114-2114.25 a)Increasing the utilization factor increases the length and duration of the queue.Theincreases are huge when p gets very close to 1.(M/M/s Model):_BCDEGH3DataResults4九=0.5(mean arrival rate)L=151(mean service rate)q=0.56s=1(#servers)7w=28Pr(W t)=0.082085匚wq=19when t=510p=0.511rob(Wq t)=0.041042512when t=5nPn1300