计算机网络实验三参考答案

1.WhatistheIPaddressandTCPportnumberusedbytheclientcomputer(source)thatistransferringthefiletogaia.cs.umass.edu?Toanswerthisquestion,it’sprobablyeasiesttoselectanHTTPmessageandexplorethedetailsoftheTCPpacketusedtocarrythisHTTPmessage,usingthe“detailsoftheselectedpacketheaderwindow”(refertoFigure2inthe“GettingStartedwithWireshark”Labifyou’reuncertainabouttheWiresharkwindows).Ans:IPaddress:192.168.1.102TCPport:11612.WhatistheIPaddressofgaia.cs.umass.edu?OnwhatportnumberisitsendingandreceivingTCPsegmentsforthisconnection?Ans:IPaddress:128.119.245.12TCPport:80Ifyouhavebeenabletocreateyourowntrace,answerthefollowingquestion:3.WhatistheIPaddressandTCPportnumberusedbyyourclientcomputer(source)totransferthefiletogaia.cs.umass.edu?ANS:IPaddress:10.211.55.7TCPport:492654.WhatisthesequencenumberoftheTCPSYNsegmentthatisusedtoinitiatetheTCPconnectionbetweentheclientcomputerandgaia.cs.umass.edu?WhatisitinthesegmentthatidentifiesthesegmentasaSYNsegment?ANS:sequencenumber:0SynSet=1identifiesthesegmentasaSYNsegment5.WhatisthesequencenumberoftheSYNACKsegmentsentbygaia.cs.umass.edutotheclientcomputerinreplytotheSYN?WhatisthevalueoftheACKnowledgementfieldintheSYNACKsegment?Howdidgaia.cs.umass.edudeterminethatvalue?WhatisitinthesegmentthatidentifiesthesegmentasaSYNACKsegment?ANS:Thesequencenumber:0ACKnowledgementnumber:1whichissequencenumberplus1BoththesequenceflagandtheACKnowledgementflagbeensetas1,identifiesthesegmentasSYNACKsegment.6.WhatisthesequencenumberoftheTCPsegmentcontainingtheHTTPPOSTcommand?NotethatinordertofindthePOSTcommand,you’llneedtodigintothepacketcontentfieldatthebottomoftheWiresharkwindow,lookingforasegmentwitha“POST”withinitsDATAfield.Ans:Thesequencenumber:17.ConsidertheTCPsegmentcontainingtheHTTPPOSTasthefirstsegmentintheTCPconnection.WhatarethesequencenumbersofthefirstsixsegmentsintheTCPconnection(includingthesegmentcontainingtheHTTPPOST)?Atwhattimewaseachsegmentsent?WhenwastheACKforeachsegmentreceived?GiventhedifferencebetweenwheneachTCPsegmentwassent,andwhenitsacknowledgementwasreceived,whatistheRTTvalueforeachofthesixsegments?WhatistheEstimatedRTTvalue(seepage249intext)afterthereceiptofeachACK?AssumethatthevalueoftheEstimatedRTTisequaltothemeasuredRTTforthefirstsegment,andtheniscomputedusingtheEstimatedRTTequationonpage249forallsubsequentsegments.Note:WiresharkhasanicefeaturethatallowsyoutoplottheRTTforeachoftheTCPsegmentssent.SelectaTCPsegmentinthe“listingofcapturedpackets”windowthatisbeingsentfromtheclienttothegaia.cs.umass.eduserver.Thenselect:Statistics-TCPStreamGraph-RoundTripTimeGraph.Segment1Segment2Segment3Segment4Segment5Segment6ANS:SequencenumberSenttimeACKreceivedtimeRTTEstimatedRTTvalueLength/bytesSegment110.0264770.0539370.027460.02746565Segment25660.0417370.0772940.0355570.0284721460Segment320260.0540260.1240850.0700590.0336701460Segment434860.0546900.1691180.114430.0437651460Segment549460.0774050.2172990.139890.0557811460Segment664060.0781570.2678020.189640.0725131460EstimatedRTT=0.875*LastEstimatedRTT+0.125*sampleRTTAfterSegment1:EstimatedRTT=0.02746AfterSegment2:EstimatedRTT=0.875*0.02746+0.125*0.035557=0.028472AfterSegment3:EstimatedRTT=0.875*0.028472+0.125*0.070059=0.033670AfterSegment4:EstimatedRTT=0.875*0.033670+0.125*0.11443=0.043765AfterSegment5:EstimatedRTT=0.875*0.043765+0.125*0.13989=0.055781AfterSegment6:EstimatedRTT=0.875*0.055781+0.125*0.18964=0.0725138.WhatisthelengthofeachofthefirstsixTCPsegments?(seeQ7)9.Whatistheminimumamountofavailablebufferspaceadvertisedatthereceivedfortheentiretrace?Doesthelackofreceiverbufferspaceeverthrottlethesender?ANS:Theminimumamountofbufferspace(receiverwindow)advertisedatgaia.cs.umass.edufortheentiretraceis5840bytes;Thisreceiverwindowgrowssteadilyuntilamaximumreceiverbuffersizeof62780bytes.Thesenderisneverthrottledduetolackingofreceiverbufferspacebyinspectingthistrace.10.Arethereanyretransmittedsegmentsinthetracefile?Whatdidyoucheckfor(inthetrace)inordertoanswerthisquestion?ANS:Therearenoretransmittedsegmentsinthetracefile.WecanverifythisbycheckingthesequencenumbersoftheTCPsegmentsinthetracefile.Allsequencenumbersareincreasing.sothereisnoretramstmittedsegment.11.HowmuchdatadoesthereceivertypicallyacknowledgeinanACK?CanyouidentifycaseswherethereceiverisACKingeveryotherreceivedsegment(seeTable3.2onpage257inthetext).ANS:Accordingtothisscreenshot,thedatareceivedbytheserverbetweenthesetwoACKsis1460bytes.therearecaseswherethereceiverisACKingeveryothersegment2920bytes=1460*2bytes.Forexample64005-61085=292012.Whatisthethroughput(bytestransferredperunittime)fortheTCPconnection?Explainhowyoucalculatedthisvalue.ANS:totalamountdata=164091-1=164090bytes#164091bytesforNO.202segmentand1bytesforNO.4segmentTotaltransmissiontime=5.455830–0.026477=5.4294SothethroughputfortheTCPconnectioniscomputedas164090/5.4294=30.222KByte/sec.13.UsetheTime-Sequence-Graph(Stevens)plottingtooltoviewthesequencenumberversustimeplotofsegmentsbeingsentfromtheclienttothegaia.cs.umass.eduserver.CanyouidentifywhereTCP’sslowstartphasebeginsandends,andwherecongestionavoidancetakesove