计网-第三章作业

计网-第三章作业

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时间:2017-11-08

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1、Chapter3注:括弧中标题号为第四版教材中对应的习题号1.(R14)SupposeHostAsendstwoTCPsegmentsbacktobacktoHostBoveraTCPconnection.Thefirstsegmenthassequencenumber90;thesecondhassequencenumber110.a.Howmuchdataisinthefirstsegment?b.Supposethatthefirstsegmentislostbutthesecondsegm

2、entarrivesatB.IntheacknowledgmentthatHostBsendstoHostA,whatwillbetheacknowledgmentnumber?答:a.[90,109]=20bytesb.acknumber=90,对第一个报文段确认2.(R15)Trueorfalse?a.ThesizeoftheTCPRcvWindowneverchangesthroughoutthedurationoftheconnection.b.supposeHostAissendingH

3、ostBalargefileoveraTCPconnection.ThenumberofunacknowledgedbytesthatAsendscannotexceedthesizeofthereceivebuffer.c.HostAissendingHostBalargefileoveraTCPconnection.AssumeHostBhasnodatatosendHostA.HostBwillnotsendacknowledgmentstoHostAbecauseHostBcannotpi

4、ggybacktheacknowledgmentondata.d.TheTCPsegmenthasafieldinitsheaderforRcvWindow.e.SupposeHostAissendingalargefiletoHostBoveraTCPconnection.Ifthesequencenumberforasegmentofthisconnectionism,thenthesequencenumberforthesubsequentsegmentwillnecessarilybem+

5、1.f.SupposethatthelastSampleRTTinaTCPconnectionisequalto1sec.ThecurrentvalueofTimeoutIntervalfortheconnectionwillnecessarilybe>=1sec.g.SupposeHostAsendsonesegmentwithsequencenumber38and4bytesofdataoveraTCPconnectiontoHostB.Inthissamesegmenttheacknowle

6、dgmentnumberisnecessarily42.答:a.Fb.Tc.F:即使没有数据传送,也会进行单独确认d.Te.F:按字节编号,不按报文段编号f.Fg.F:B->A的确认号不一定为38+4=423.(R17)Trueorfalse?ConsidercongestioncontrolinTCP.Whenthetimerexpiresatthesender,thethresholdissettoonehalfofitspreviousvalue.答:F:应为当前拥塞窗口的一半,而不是阈值的

7、一半。4.(P3)UDPandTCPuse1scomplementfortheirchecksums.Supposeyouhavethefollowingthree8-bitbytes:01101010,01001111,01110011.Whatisthe1scomplementofthesumofthese8-bitbyte?(NotethatalthoughUDPandTCPuse16-bitwordsincomputingthechecksum,forthisproblemyouarebe

8、ingaskedtoconsider8-bitsums.)Showallwork..WhyisitthatUDPtakescomplementofthesum;thatis,whynotjustusethesum?Withthe1scomplementscheme,howdoesthereceiverdetecterrors?Isitpossiblethata1-biterrorwillgoundetected?Howabouta2-biterror?答:01101010+0100

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