计算机网络实验3：Wireshark-Lab-TCP

Lab3WiresharkLab:TCPSTEPS•Startupyourwebbrowser.Gothe•Nextgoto•UsetheBrowsebuttoninthisformtoenterthenameofthefile(fullpathname)onyourcomputercontainingAliceinWonderland(ordosomanually).Don’tyetpressthe“Uploadalice.txtfile”button.•NowstartupWiresharkandbeginpacketcapture(Capture-Options)andthenpressOKontheWiresharkPacketCaptureOptionsscreen(we’llnotneedtoselectanyoptionshere).•Returningtoyourbrowser,pressthe“Uploadalice.txtfile”buttontouploadthefiletothegaia.cs.umass.eduserver.Oncethefilehasbeenuploaded,ashortcongratulationsmessagewillbedisplayedinyourbrowserwindow.•StopWiresharkpacketcapture.•First,filterthepacketsdisplayedintheWiresharkwindowbyentering“tcp”(lowercase,noquotes,anddon’tforgettopressreturnafterentering!)intothedisplayfilterspecificationwindowtowardsthetopoftheWiresharkwindow.QUESTIONS1.WhatistheIPaddressandTCPportnumberusedbytheclientcomputer(source)thatistransferringthefiletogaia.cs.umass.edu?Toanswerthisquestion,it’sprobablyeasiesttoselectanHTTPmessageandexplorethedetailsoftheTCPpacketusedtocarrythisHTTPmessage,usingthe“detailsoftheselectedpacketheaderwindow”.TheIPaddressis192.168.1.102.TheTCPportnumberis1161.2.WhatistheIPaddressofgaia.cs.umass.edu?OnwhatportnumberisitsendingandreceivingTCPsegmentsforthisconnection?TheIPaddressis128.119.245.12.TheTCPportnumberis80.3.WhatistheIPaddressandTCPportnumberusedbyyourclientcomputer(source)totransferthefiletogaia.cs.umass.edu?TheIPaddressis172.18.40.131.TheTCPportnumberis51458.4.WhatisthesequencenumberoftheTCPSYNsegmentthatisusedtoinitiatetheTCPconnectionbetweentheclientcomputerandgaia.cs.umass.edu?WhatisitinthesegmentthatidentifiesthesegmentasaSYNsegment?ThesequencenumberoftheTCPSYNsegmentis0.TheSYNflagissetto1identifiesthesegmentasaSYNsegment.5.WhatisthesequencenumberoftheSYNACKsegmentsentbygaia.cs.umass.edutotheclientcomputerinreplytotheSYN?WhatisthevalueoftheACKnowledgementfieldintheSYNACKsegment?Howdidgaia.cs.umass.edudeterminethatvalue?WhatisitinthesegmentthatidentifiesthesegmentasaSYNACKsegment?ThesequencenumberoftheSYNACKsegmentsentbygaia.cs.umass.eduis0.ThevalueoftheAcknowledgementfieldintheSYNACKsegmentis1.Thevalueisadding1totheinitialsequencenumberofSYNsegment.TheSYNflagandAcknowledgementflaginthesegmentaresetto1identifiesthesegmentasaSYNACKsegment.6.WhatisthesequencenumberoftheTCPsegmentcontainingtheHTTPPOSTcommand?NotethatinordertofindthePOSTcommand,you’llneedtodigintothepacketcontentfieldatthebottomoftheWiresharkwindow,lookingforasegmentwitha“POST”withinitsDATAfield.ThesequencenumberoftheTCPsegmentcontainingtheHTTPPOSTcommandis1.7.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.Wecangettable1:Table1NumberTime(s)SeqNoACKNo40.026477150.04173756660.05393756670.054026202680.05469348690.0772942026100.0774054946110.0781576406120.1240853486130.1241857866140.1691184946150.2172996406160.2678027866Then,wecangettable2formtable1.Table2NOSendTime(s)AckTime(s)RTT(s)10.0264770.0539370.0274620.0417370.0772940.03555730.0540260.1240850.07005940.054690.1691180.11442850.0774050.2172990.13989460.0781570.2678020.189645EstimatedRTTafterthereceiptoftheACKofsegment1:EstimatedRTT=0.02746sEstimatedRTTafterthereceiptoftheACKofsegment2:EstimatedRTT=0.875*0.02746+0.125*0.035557=0.0285sEstimatedRTTafterthereceiptoftheACKofsegment3:EstimatedRTT=0.875*0.0285+0.125*0.070059=0.0337sEstimatedRTTafterthereceiptoftheACKofsegment4:EstimatedRTT=0.875*0.0337+0.125*0.114428=0.0438sEstimatedRTTafterthereceiptoftheACKofsegment5:EstimatedRTT=0.875*0.0438+0.125*0.139894=0.0558sEstimatedRTTafterthereceiptoftheACKofsegment6:EstimatedRTT=0.875*0.0558+0.125*0.189645=0.0725s8.WhatisthelengthofeachofthefirstsixTCPsegments?AccordingtoTable1,wecangetthatthefirstTCPsegment’slengthis565bytes.Theother5is1460bytes.9.Whatistheminimumamountofavailablebufferspaceadvertisedatthereceivedfortheentiretrace?Doesthelackofreceiverbufferspaceeverthrottlethesender?Theminimumamountofavailablebufferspaceadvertisedatthereceivedfortheentiretraceis5840bytes.Wecanseethatthesenderisneverthrottlebecauseofthelackofreceiverbufferspa