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ID:43076785
大小:512.03 KB
页数:9页
时间:2019-09-26
《数据、模型与决策(运筹学)课后习题和案例答案014s》由会员上传分享,免费在线阅读,更多相关内容在工程资料-天天文库。
1、CDSUPPLEMENTTOCHAPTER14THEFINITEQUEUEANDFINITECALLINGPOPULATIONVARIATIONSOFTHEM/M/SMODELReviewQuestions14s-1WhenthesystemisfullwithKcustomers,anynewarrivingcustomersleavewithoutenteringthesystem.14s-2Pkistheprobabilitythatthesystemisfullandthereforetheprobabilitythat
2、anamvingcustomeristurnedaway.14s-3Averagefractionoftimethatserversarebusy=—―—.14s-4Thecallingpopulationforaqueueingsystemisthepopulationofallpotentialcustomerswhomightneedtocometothesystemforservice.Afinitecallingpopulationisonethatissmallenoughthatthenumberofcustome
3、rsalreadyinthequeueingsystemaffectsthemeanarrivalrateofnewcustomers.14s-5Uponcompletingservice,thetimeuntilamemberofthecallingpopulationneedsserviceagainhasanexponentialdistributionwithameanofN/k.14s-6Nz14s-72Utilizationofservers=—.sp14s-8No,thisapproximationshouldno
4、tbeusedwhenpisnearly1unlessKorNishuge.Problems14s.1a)With0spaces,42.9%ofcustomerswillbelost.ABcDEFG1TemplateforM/M/sFiniteQueueModel23DataResults4X=0.25(meanarrivalrate)L=0.42857145P=0.333(meanservicerate)Lq=06s=1(#servers)7K=1(maxcustomers)w=38wq=0910p=0.751112nPn13
5、00.57142861410.4285714b)With2spaces,15.4%ofcustomerswillbelost.ABcDEFG1TemplateforM/M/sFiniteQueueModel23DataResults4X=0.25(meanarrivalrate)L=1.14857145=0.333(meanservicerate)Lq=0.51428576s=1(#servers)7K=3(maxcustomers)w=5.43243248wq=2.4324324910p=0.751112nPn1300.365
6、71431410.27428571520.20571431630.15428571c)With4spaces,7.2%ofcustomerswillbelost.ABCDEFG1TemplateforM/M/sFiniteQueueModel23DataResults4X=0.25(meanarrivalrate)L=1.70092075p=0.333(meanservicerate)u=1.0050496s=1(#servers)7K=5(maxcustomers)w=7.33290658wq=4.3329065910p=0.
7、751112nPn1300.30412831410.22809621520.17107221630.12830411740.09622811850.072171114s.2a&b)Withspacefor2cars,21.1%ofcustomerswillbelost,厶=0.74cars,andW=2.8minutes.ABcDEFGIH1TemplateforM/M/sFiniteQueueModel23DataResults4X=20(meanarrivalrate)L=0.7368421530(meanservicera
8、te)0.21052636S=1(#servers)rrdnutes7K=2(maxcustomers)w=0.04666672.88wq=0.01333330.8910p=0.66666671112nPn1300.4736842I1410.3157895152
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