计算机网络实验3TCP实验

－1－计算机网络实验报告三TCP实验1.WhatistheIPaddressandTCPportnumberusedbytheclientcomputer(source)thatistransferringthefiletogaia.cs.umass.edu?Toanswerthisquestion,it’sprobablyeasiesttoselectanHTTPmessageandexplorethedetailsoftheTCPpacketusedtocarrythisHTTPmessage,usingthe“detailsoftheselectedpacketheaderwindow”(refertoFigure2inthe“GettingStartedwithWireshark”Labifyou’reuncertainabouttheWiresharkwindows).答：clientcomputer(source)：IPaddress：192.168.1.102TCPportnumber：11612.WhatistheIPaddressofgaia.cs.umass.edu?OnwhatportnumberisitsendingandreceivingTCPsegmentsforthisconnection?答：theIPaddressofgaia.cs.umass.edu：IPaddress：128.119.245.12portnumber：803.Ifyouhavebeenabletocreateyourowntrace,answerthefollowingquestion:WhatistheIPaddressandTCPportnumberusedbyyourclientcomputer(source)totransferthefiletogaia.cs.umass.edu?答：Myclientcomputer:IPaddress:10.2.136.30－2－4.WhatisthesequencenumberoftheTCPSYNsegmentthatisusedtoinitiatetheTCPconnectionbetweentheclientcomputerandgaia.cs.umass.edu?WhatisitinthesegmentthatidentifiesthesegmentasaSYNsegment?答：sequencenumber:0；syn被设置为1说明是syn段。5.WhatisthesequencenumberoftheSYNACKsegmentsentbygaia.cs.umass.edutotheclientcomputerinreplytotheSYN?WhatisthevalueoftheACKnowledgementfieldintheSYNACKsegment?Howdidgaia.cs.umass.edudeterminethatvalue?WhatisitinthesegmentthatidentifiesthesegmentasaSYNACKsegment?答：ThesequencenumberoftheSYNACKsegmentsentbygaia.cs.umass.eduis:0；SYNACKsegment中ACKnowledgement的值为1；－3－ACKnowledgementnumber的值为SYN消息中sequencenumber加上1所得；SYN和Acknowledgementf都置为1说明这是一个SYNACKsegment.6.WhatisthesequencenumberoftheTCPsegmentcontainingtheHTTPPOSTcommand?NotethatinordertofindthePOSTcommand,you’llneedtodigintothepacketcontentfieldatthebottomoftheWiresharkwindow,lookingforasegmentwitha“POST”withinitsDATAfield.答：第四号报文段是包含HTTPPOST命令的TCPsegment.且报文段的序列号为1.－4－7.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.Segment1－5－Segment2Segment3Segment4Segment5－6－Segment6答：前6个报文段为No.4,5,7,8,10,11.对应的ACK分别为No.6,9,12,14,15,16.前6个报文段截图如下：报文段的序列号为每个报文段的首字节加1，所以序列号为：Segment1sequencenumber:1Segment2sequencenumber:566Segment3sequencenumber:2026Segment4sequencenumber:3486Segment5sequencenumber:4946Segment6sequencenumber:6406报文段的发送时间和相应ACK的到达时间如下表：:SendtimeACKreceivedtimeRTTsecondsSegment10.0264770.0539370.02746－7－Segment20.0417370.0772940.035557Segment30.0540260.1240850.070059Segment40.0546900.1691180.11443Segment50.0774050.2172990.13989Segment60.0781570.2678020.18964EstimatedRTT=0.875*EstimatedRTT+0.125*SampleRTT接受到报文段1之后的EstimatedRTT为：EstimatedRTT=RTTforsegment1=0.02746second接受到报文段2之后的EstimatedRTT为：EstimatedRTT=0.875*0.02764+0.125*0.035557=0.0285sencond接受到报文段3之后的EstimatedRTT为：EstimatedRTT=0.875*0.0285+0.125*0.070059=0.0337second接受到报文段4之后的EstimatedRTT为：EstimatedRTT=0.875*0.0337+0.125*0.11443=0.0438second接受到报文段5之后的EstimatedRTT为：EstimatedRTT=0.875*0.0438+0.125*0.13989=0.0558second接受到报文段6之后的EstimatedRTT为：EstimatedRTT=0.875*0.0558+0.125*0.18964=0.0725second8.WhatisthelengthofeachofthefirstsixTCPsegments?答：前6个段的长度分别为：565、1460、1460、1460、1460、1460字节。9.Whatistheminimumamountofavailablebufferspaceadvertisedatthereceivedfortheentiretrace?Doesthelackofreceiverbufferspaceeverthrottlethesender?－8－答：接收方通知给发送方的最低窗口大小为5840字节，即在服务器端传回的第一个ACK中的窗口大小。接收方的窗口大小没有抑制发送方的传输速率，因为窗口大小从5840逐步增加到62780，窗口大小始终大于发送方发送的分组的容量。10.Arethereanyretransmittedsegmentsinthetracefile?Whatdidyoucheckfor(inthetrace)inordertoanswerthisquestion？答：没有，从TCP报文段的序列号中可以得出以上结论。从上图中的时间—序号图可以看出，从源端发往目的端的序号逐渐递增，如果这其中有重传的报文段，则其序号中应该有小于其临近的分组序号的分组，在图中未看到这样的分组，所以没有被重传的分组。－9－11.HowmuchdatadoesthereceivertypicallyacknowledgeinanACK?CanyouidentifycaseswherethereceiverisACKingeveryotherreceivedsegment？答：右下图得，接收方在一个ACK确认的数据大小一般为1460字节。TheAcknowledgedsequencenumberandtheAcknowledgeddata:AcknowledgedsequencenumberAcknowledgeddataACK1566566ACK220261460ACK334861460ACK449461460ACK564061460ACK678661460ACK790131147ACK8104731460ACK9119331460ACK10133931460ACK11148531460报文段确认数据为2920bytes=1460*2bytes，即129541-12621=2920.12.Whatisthethroughput(bytestransferredperunittime)fortheTCPconnection?Explainhowyoucalculatedthisvalue.－10－答：TCP吞吐量计算很大程度上取决于所选内容的平均时间。作为一个普通的吞吐量计算，在这问题上，选择整个连接的时间作为平均时间段。然后，此TCP连接的平均吞吐量为总的传输数据与总传输时间的比值。传输的数据总量为TCP段第一个序列号（即第4段的1字节）和最后的序列号的ACK（第202段的164091个字节）之间的差值。因此，总数据是164091-1=1