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1、ANDREWS.TANENBAUMmedium-speedline,alow-speedline,ornoline.Ifittakes100msofcomputertime------COMPUTERNETWORKSFOURTHEDITIONPROBLEMSOLUTIONStogenerateandinspecteachtopology,howlongwillittaketoinspectallofthem?(E)CollectedandModifiedByYanZhenXing,MailTo:zxyan@ecnu.cn将路由器称为A,B,C,D和E.Classify:
2、EàEasy,MàMiddle,HàHard,DàDelete则有10条可能的线路;AB,AC,AD,AE,BC,BD,BE,CD,CE,和DEGreen:ImportantRed:MasterBlue:VIOthers:KnowGrey:Unnecessary10每条线路有4种可能性(3速度或者不是线路),拓扑的总数为4=1,048,576。Chapter1IntroductionProblems检查每个拓扑需要100ms,全部检查总共需要104,857.6秒,或者稍微超过292.AnalternativetoaLANissimplyabigtimesharingsy
3、stemwithterminalsfor个小时。allusers.Givetwoadvantagesofaclient-serversystemusingaLAN.(M)9.Agroupof2n-1routersareinterconnectedinacentralizedbinarytree,witha使用局域网模型可以容易地增加节点。routerateachtreenode.Routericommunicateswithrouterjbysendingamessage如果局域网只是一条长的电缆,且不会因个别的失效而崩溃(例如采用镜像服务totherootofthet
4、ree.Therootthensendsthemessagebackdowntoj.Derivean器)的情况下,使用局域网模型会更便宜。approximateexpressionforthemeannumberofhopspermessageforlargen,assumingthatallrouterpairsareequallylikely.(H)使用局域网可提供更多的计算能力和更好交互式接口。这意味着,从路由器到路由器的路径长度相当于路由器到根的两倍。若在树中,3.Theperformanceofaclient-serversystemisinfluencedb
5、ytwonetworkfactors:根深度为1,深度为n,从根到第n层需要n-1跳,在该层的路由器为0.50。thebandwidthofthenetwork(howmanybits/secitcantransport)andthelatency-从根到n-1层的路径有router的0.25和n-2跳步。因此,路径长度l为:(howmanysecondsittakesforthefirstbittogetfromtheclienttotheserver).Giveanexampleofanetworkthatexhibitshighbandwidthandhighl
6、atency.Thengivel=0.5*(n-1)+0.25*(n-2)+0.125*(n-3)……anexampleofonewithlowbandwidthandlowlatency.(E)结果化简为l=n-2,平均路由路径为2n-4。横贯大陆的光纤连接可以有很多千兆位/秒带宽,但是由于光速度传送要越过数10.Adisadvantageofabroadcastsubnetisthecapacitywastedwhenmultiple千公里,时延将也高。hostsattempttoaccessthechannelatthesametime.Asasimplistic
7、example,相反,使用56kbps调制解调器呼叫在同一大楼内的计算机则有低带宽和较低的supposethattimeisdividedintodiscreteslots,witheachofthenhostsattemptingto时延。usethechannelwithprobabilitypduringeachslot.Whatfractionoftheslotsare4.Besidesbandwidthandlatency,whatotherparameterisneededtogiveagoodwastedduetocolli