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第1页上海应用技术学院2009—2010学年第1学期《网络原理ComputerNetworks》期末（A）试卷课程代码:学分:考试时间:分钟课程序号:6668,6780,6666班级：学号：姓名:我已阅读了有关的考试规定和纪律要求，愿意在考试中遵守《考场规则》，如有违反将愿接受相应的处理。试卷共8页，请先查看试卷有无缺页，然后答题。I.Abbreviations:(10pointsintotal)The(everydayEnglish)expansionsofthefollowingabbreviationsarewanted:A1-10.(1pointeach,10pointsintotal)A1.WAN=WideAreaNetworkA2.TCP=TransmissionControlProtocolA3.DNS=DomainNameSystem/ServiceA4.OSPF=OpenShortestPathFirstA5.SMTP=SimpleMailTransferProtocolA6.PPP=Point-to-PointProtocolA7.GPS=GlobalPositioningSystemA8.CSMA/CD=CarrierSenseMultipleAccesswithCollisionDetectionA9.IP=InternetProtocolA10.VLAN=VirtualLAN=VirtualLocalAreaNetwork题号一二三四五六七八九十总分得分第2页II.Concepts:(30pointsintotal)Givethedefinitionordescriptionofthefollowingconceptsornames:B1-5(6pointseach,30pointsintotal)B1.Whatisthemaintaskofthenetworklayer(intheOSIreferencemodel)?Solution(c5-P00)Thenetworklayerisconcernedwithgettingpacketsfromthesourcemachineallthewaytothedestinationmachine.B3.Wehavestudiedmanychannelallocationmethods.Pleasenamethreeofthem.B4.Whatapproachesdowetaketodoerrorcontrol?(orwhatmechanismsdoweusetodoerrorcontrol?)第3页B5.PleasedrawafiguretoshowtheOSIReferenceModel.Solution:（c1）P37B6.InFigII-1,oneofthe802.11MACsublayerprotocolisillustratedandfourstations,A,B,C,andD,areshown.WhichofthelasttwostationsdoyouthinkisclosesttoAandwhy?第4页III.Computation:(40pointsintotal)Theprocess/procedureorexplanationofcomputationandthecorrectresultsareallwanted.C1-5.(8pointseach,40pointsintotal).C1.Thefollowingcharacterencodingisusedinadatalinkprotocol:A:01000111;B:11100011;FLAG:01111110;ESC:11100000Showthebitsequencetransmitted(inbinary)forthefour-characterframe:AESCFLAGBAwheneachofthefollowingframingmethodsisused:(a)Charactercount.0000110…(b)Flagbyteswithbytestuffing.(c)Startingandendingflagbytes,withbitstuffing.第5页C2.Abitstream10011101istransmittedusingthestandardCRCmethoddescribedinthetext.Thegeneratorpolynomialisx4+x2+1.Showwhatcodethereceiver'sendreceivesifthereceiver'sendwillreceivecorrectcode.Solution:100111010000/10101theremainderis1000Sothereceiverwillreceive100111011000C3.Arouterhasthefollowing(CIDR)entriesinitsroutingtable:Address/maskNexthop195.46.56.0/22Interface0195.46.60.0/22Interface1defaultRouter1ForeachofthefollowingIPaddresses,whatdoestherouterdoifapacketwiththataddressarrives?a.195.46.53.20b.195.46.62.100c.192.53.56.78SOLUTION:a.53=00110101BBecausemaskis252(=1111100B)solutionis00110100=52SotheroutergoestoRouter1ifapacketwithaddress195.46.53.20arrives.b.62=00111110BBecausemaskis252(=11111100B)solutionis00111100SotheroutergoestoInterface1ifapacketwithaddress145.46.62.100arrives.c.router1b/c192.53.第6页C4.AlargenumberofconsecutiveIPaddressareavailablestartingat198.16.0.0.Supposethatfourorganizations,A,B,C,andD,request4000,2000,4000,and8000addresses,respectively,andinthatorder.Foreachofthese,givethefirstIPaddressassigned,thelastIPaddressassigned,andthemaskinthew.x.y.z/snotation.Solution(P478-40)Tostartwith,alltherequestsareroundeduptoapoweroftwo.Thestartingaddress,endingaddress,andmaskareasfollows:A:198.16.0.0–198.16.15.255writtenas198.16.0.0/20B:198.16.16.0–198.16.23.255writtenas198.16.16.0/21C:198.16.32.0–198.16.47.255writtenas198.16.32.0/20D:198.16.64.0–198.16.95.255writtenas198.16.64.0/19C5.ConsiderbuildingaCSMA/CDnetworkrunningat1Gbpsovera1-kmcablewithnorepeaters.Thesignalspeedinthecableis200,000km/sec.Whatistheminimumframesize?Solution(P340-21)对于1km电缆，单程传播时间为1/200000=5×10-6s，即5，来回路程传播时间为2t=10。为了能够按照CSMA/CD工作，最小帧的发射时间不能小于10。以1Gb/s速率工作，10可以发送的比特数等于：因此，最小帧是10000bit或1250字节长。第7页IV.Description:(20pointsintotal)abriefdescriptionwithallprincipalpointsiswanted.D1-2.(10pointseach,20pointsintotal)D1.Twotypesoftransmissiontechnologiesareinwidespreaduse.Theyarebroadcastlinksandpoint-to-pointlinks.a)Whatdowemeanby'broadcastnetworks'?Givethedefinitionorabriefdescription.[p15]Broadcastnetworkshaveasinglecommunicationchannelthatissharedbyallthemachinesonthenetwork.Shortmessages,calledpacketsincertaincontexts,sentbyanymachinearereceivedbyalltheothers.b)Whatdowemeanby'point-to-pointnetworks'?Givethedefinitionorabriefdescription.[p15]point-to-pointnetworksconsistofmanyconnectionsbetweenindividualpairsofmachines.Togofromthesourcetothedestination,apacketonthistypeofnetworkmayhavetofirstvisitoneormoreintermediatemachines.Oftenmultipleroutes,ofdifferentlengths,arepossible,sofindinggoodonesisimportantinpoint-to-pointnetworks.第8页D2.Theideabehindlinkstateroutingissimpleandcanbestatedasfiveparts.Whatarethefivepartsorstepsthateachroutermustdo?Solution:1).Discoveritsneighborsandlearntheirnetworkaddresses.2).Measurethedelayorcosttoeachofitsneighbors.3).Constructapackettellingallithasjustlearned.4).Sendthispackettoallotherrouters.5).Computetheshortestpathtoeveryotherrouter.