计算机网络实验3Wireshark Lab TCPWord文档下载推荐.docx

计算机网络实验3Wireshark Lab TCPWord文档下载推荐.docx

《计算机网络实验3Wireshark Lab TCPWord文档下载推荐.docx》由会员分享，可在线阅读，更多相关《计算机网络实验3Wireshark Lab TCPWord文档下载推荐.docx（9页珍藏版）》请在冰豆网上搜索。

•First,filterthepacketsdisplayedintheWiresharkwindowbyentering“tcp”（lowercase,noquotes,anddon’tforgettopressreturnafterentering!

）intothedisplayfilterspecificationwindowtowardsthetopoftheWiresharkwindow.

QUESTIONS

1.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:

Table1

Number

Time（s）

SeqNo

ACKNo

4

0.026477

1

5

0.041737

566

6

0.053937

7

0.054026

2026

8

0.05469

3486

9

0.077294

10

0.077405

4946

11

0.078157

6406

12

0.124085

13

0.124185

7866

14

0.169118

15

0.217299

16

0.267802

Then,wecangettable2formtable1.

Table2

NO

SendTime（s）

AckTime（s）

RTT（s）

0.02746

2

0.035557

3

0.070059

0.114428

0.139894

0.189645

EstimatedRTTafterthereceiptoftheACKofsegment1:

EstimatedRTT=0.02746s

EstimatedRTTafterthereceiptoftheACKofsegment2:

EstimatedRTT=0.875*0.02746+0.125*0.035557=0.0285s

EstimatedRTTafterthereceiptoftheACKofsegment3:

EstimatedRTT=0.875*0.0285+0.125*0.070059=0.0337s

EstimatedRTTafterthereceiptoftheACKofsegment4:

EstimatedRTT=0.875*0.0337+0.125*0.114428=0.0438s

EstimatedRTTafterthereceiptoftheACKofsegment5:

EstimatedRTT=0.875*0.0438+0.125*0.139894=0.0558s

EstimatedRTTafterthereceiptoftheACKofsegment6:

EstimatedRTT=0.875*0.0558+0.125*0.189645=0.0725s

8.WhatisthelengthofeachofthefirstsixTCPsegments?

AccordingtoTable1,wecangetthatthefirstTCPsegment’slengthis565bytes.Theother5is1460bytes.

9.Whatistheminimumamountofavailablebufferspaceadvertisedatthereceivedfortheentiretrace?

Doesthelackofreceiverbufferspaceeverthrottlethesender?

Theminimumamountofavailablebufferspaceadvertisedatthereceivedfortheentiretraceis5840bytes.Wecanseethatthesenderisneverthrottlebecauseofthelackofreceiverbufferspace.

10.Arethereanyretransmittedsegmentsinthetracefile?

Whatdidyoucheckfor（inthetrace）inordertoanswerthisquestion?

Therearen’tanyretransmittedsegmentsinthetracefile.WecanchecktheTime-Sequence-Graph（Stevens）.Inthegraph,therearen’ttwodotsinthesamey-line.

11.HowmuchdatadoesthereceivertypicallyacknowledgeinanACK?

CanyouidentifycaseswherethereceiverisACKingeveryotherreceivedsegment（seeTable3.2onpage257inthetext）.

Thereceivertypicallyacknowledge1460bytesinanACK.

80thisACKingeveryotherreceivedsegment.Itacks76thand77thtwosegment.

12.Whatisthethroughput（bytestransferredperunittime）fortheTCPconnection?

Explainhowyoucalculatedthisvalue.

Thelastsegment’ssequencenumberis164091.Sothetotaldatais164091-1=164090bytes.Thewholetimeis5.455830–0.026477=5.429353s.Sothethroughputis164090/5.429353=30.222KB/sec.

•SelectaTCPsegmentintheWireshark’s“listingofcaptured-packets”window.Thenselectthemenu:

TCPStreamGraph->

Time-Sequence-Graph（Stevens）.Youshouldseeaplotthatlookssimilartothefollowingplot,whichwascreatedfromthecapturedpacketsinthepackettracetcp-etherealtrace-1inhttp:

//gaia.cs.umass.edu/wireshark-labs/wireshark-traces.zip

13.UsetheTime-Sequence-Graph（Stevens）plottingtooltoviewthesequencenumberversustimeplotofsegmentsbeingsentfromtheclienttothegaia.cs.umass.eduserver.CanyouidentifywhereTCP’sslowstartphasebeginsandends,andwherecongestionavoidancetakesover?

CommentonwaysinwhichthemeasureddatadiffersfromtheidealizedbehaviorofTCPthatwe’vestudiedinthetext.

TCPslowstartbeginsatthestartoftheconnection.SowhentheHTTPPOSTsegmentissentout,theslowstartbegins.ButifwewanttoidentifywhereTCP’sslowstartphaseendsandwherecongestionavoidancetakesover,wehavetoknowthevalueofcongestionwindowsizeofthissender.Sadly,wecan’tgetthevaluedirectly.Allwecandoistoestimateitbytheamountofdatawithoutacknowledgement.BecauseweknowthatLastByteSend–LastbyteAcked<

=min{CongWin,RcvWin}andtheRcvWinislargeenough.ButevenwegetthelowerboundoftheTCPwindowsize,itisstillhardtoidentifywhereTCP’sslowstartphaseendsandwherecongestionavoidancetakesover.Afterall,theamountofdatawithoutacknowledgementdoesn’tequaltothevalueofcongestionwindowsize.

TheidealizedbehaviorofTCPthatwe’vestudiedinthetext:

TCPsenderwilltrytosendmoredata.WhenitgetanACK,itwillsendmoredata.Butoncecongestionhappened,theCongwinwilldropdowntoahalf.Butinpractice,TCPbehavioralsodependsontheapplication.Forexample,insomewebapplications,thewebobject’ssizeisverysmall.Sobeforetheendofslowstart,thetransmissionisover.Thenwewillgetalongdelayandasmallthroughput.

14.Answereachoftwoquestionsaboveforthetracethatyouhavegatheredwhenyoutransferredafilefromyourcomputertogaia.cs.umass.edu.

=min{CongWin,RcvWin}andtheRcvWinislargeenough.ButevenwegetthelowerboundoftheTCPwindowsize,itisstillhardtoidentifywher