计算机网络实验报告-实验五 探索TCP协议.doc

1、西范大学陕师西范大学陕师 算机网 计络 算机网 计络告实验报告实验报实验五 探索实验五 探索 TCPTCP 协议协议一、实验目的一、实验目的 1.熟悉并掌握 wireshark的操作。2.通过实验分析并掌握TCP 是如何进行数据传输的，及其如何实现拥塞控制、流量控制等二、实验器材二、实验器材 1.PC机电脑一台。2下载 wireshark软件并安装三、实验内容三、实验内容通过实验分析并掌握 TCP是如何进行数据传输的，及其如何实现拥塞控制、流量控制等四、问题回答四、问题回答1.Capturing a bulk TCP transfer from your computer to a remot

2、eServer1Go the http:/gaia.cs.umass.edu/wiresharklabs/alice.txt and retrieve an ASCII copy of Alice in Wonderland.Store this file.somewhere on my computer.Next go to http:/gaia.cs.umass.edu/wireshark-labs/TCP-wireshark-file1.html.Then I cansee a screen that looks like:,2start up Wireshark and begin p

3、acket capture.Returning to browser,press the“Upload alice.txt file”button to upload the file to the gaia.cs.umass.edu server.Stop Wireshark packet capture.The Wireshark window like this:Then opening the Wireshark captured packet file tcpethereal-trace-1 in http:/gaia.cs.umass.edu/wireshark-labs/wire

4、shark-traces.zip2.A first look at the captured trace1.What is the IP address and TCP port number used by the client computer(source)that is transferring the file to gaia.cs.umass.edu?To answer this question,itsprobably easiest to select an HTTP message and explore the details of the TCPpacket used t

5、o carry this HTTP message,using the“details of the selected packetheader window”(refer to Figure 2 in the“Getting Started with Wireshark”Lab ifyoure uncertain about the Wireshark windows).答：client computer(source)：IP address：192.168.1.1023ip addressTcp port numberTCP port number：11612.What is the IP

6、 address of gaia.cs.umass.edu?On what port number is it sendingand receiving TCP segments for this connection?答：the IP address of gaia.cs.umass.edu：IP address：128.119.245.12port number：803.If you have been able to create your own trace,answer the following question:Whatis the IP address and TCP port

7、 number used by your client computer(source)to transferthe file to gaia.cs.umass.edu?4ip address:port number答：My client computer:IP address:10.2.136.305ip address:Tcp port numberThen Wireshark do this,select Analyze-Enabled Protocols.Then uncheck the HTTP box andselect OK.now see an Wireshark window

8、 that looks like:3.TCP BasicsAnswer the following questions；4.What is the sequence number of the TCP SYN segment that is used to initiate theTCP connection between the client computer and gaia.cs.umass.edu?What is itin the segment that identifies the segment as a SYN segment?6Sequence numberSyn is 1

9、答：sequence number:0；syn 被设置为1说明是syn 段。5.What is the sequence number of the SYNACK segment sent by gaia.cs.umass.eduto the client computer in reply to the SYN?What is the value of the ACKnowledgementfield in the SYNACK segment?How did gaia.cs.umass.edu determine that value?Whatis it in the segment th

10、at identifies the segment as a SYNACK segment?答：The sequence number of the SYNACK segment sent by gaia.cs.umass.edu is:0；SYNACK segment 中 ACKnowledgement 的值为1；ACKnowledgement number 的值为 SYN 消息中 sequence number 加上 1所得；SYN 和Acknowledgement f 都置为 1说明这是一个 SYNACK segment.6.What is the sequence number of

11、the TCP segment containing the HTTP POSTcommand?Note that in order to find the POST command,youll need to dig intothe packet content field at the bottom of the Wireshark window,looking for asegment with a“POST”within its DATA field.7Sequence numberThe value of the ACKnowledgement field8Sequence numb

12、erPost commandNO.4 segment答：第四号报文段是包含 HTTP POST 命令的 TCP segment.且报文段的序列号为 1.7.Consider the TCP segment containing the HTTP POST as the first segment in theTCP connection.What are the sequence numbers of the first six segments in theTCPconnection(including the segment containing the HTTP POST)?At wha

13、t time was eachsegment sent?When was the ACK for each segment received?Given the differencebetween when each TCP segment was sent,and when itsacknowledgement wasreceived,what is the RTT value for each of the sixsegments?What is the EstimatedRTTvalue(see page 249 in text)after thereceipt of each ACK?

14、Assume that the value of theEstimatedRTT is equal tothe measured RTT for the first segment,and then is computedusing theEstimatedRTT equation on page 249 for all subsequent segments.Note:Wireshark has a nice feature that allows you to plot the RTT foreach of the TCPsegments sent.Select a TCP segment

15、 in the“listing ofcaptured packets”window that isbeing sent from the client to thegaia.cs.umass.edu server.Then select:Statistics-TCPStream Graph-Round Trip Time Graph.9Segment 1-6 areNo.4,5,7,8,10Segment 1Segment 2Segment 3Segment 410Sequence number=1Sequence number=566Sequence number=2026Segment 5

16、Segment 6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：11Sequence number=3486Sequence number=4946Sequence number=6406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：Segment 1 sequence number:1Segment 2 sequence number:566Segment 3 sequence number:2026Segment 4 sequence number:3486Segmen

17、t 5 sequence number:4946Segment 6 sequence number:6406报文段的发送时间和相应ACK 的到达时间如下表：:Send timeACK received timeRTT secondsSegment 10.0264770.0539370.02746Segment 20.0417370.0772940.035557Segment 30.0540260.1240850.070059Segment 40.0546900.1691180.11443Segment 50.0774050.2172990.13989Segment 60.0781570.267

18、8020.18964EstimatedRTT=0.875*EstimatedRTT+0.125*SampleRTT接受到报文段1 之后的 EstimatedRTT 为：EstimatedRTT=RTT for segment 1=0.02746 second接受到报文段2 之后的 EstimatedRTT 为：EstimatedRTT=0.875*0.02764+0.125*0.035557=0.0285 sencond接受到报文段3 之后的 EstimatedRTT 为：EstimatedRTT=0.875*0.0285+0.125*0.070059=0.0337 second接受到报文段4

19、 之后的 EstimatedRTT 为：EstimatedRTT=0.875*0.0337+0.125*0.11443=0.0438 second接受到报文段5 之后的 EstimatedRTT 为：EstimatedRTT=0.875*0.0438+0.125*0.13989=0.0558 second接受到报文段6 之后的 EstimatedRTT 为：12EstimatedRTT=0.875*0.0558+0.125*0.18964=0.0725 second8.What is the length of each of the first six TCP segments?答：前 6

20、个段的长度分别为：565、1460、1460、1460、1460、1460 字节。9.What is the minimum amount of available buffer space advertised at the receivedfor the entire trace?Does the lack of receiver buffer space ever throttle thesender?13接收方通知给发送方的 最低大小为 5840字节答：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

21、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.Are there any retransmitted segments in the trace file?What did you check for(inthe trace)in order to answer this question？14接收方窗口最大为62780 字节答：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

22、有被重传的分组。11.How much data does the receiver typically acknowledge in an ACK?Can youidentify cases where the receiver is ACKing every other received segment？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。The Acknowledged sequence number and the Acknowledged data:Acknowledged sequence numberAcknowledged data15ACK

23、1566566ACK 220261460ACK 334861460ACK 449461460ACK 564061460ACK 678661460ACK 790131147ACK 8104731460ACK 9119331460ACK 10133931460ACK 11148531460 报文段确认数据为2920bytes=1460*2 bytes，即 129541-12621=2920.12.What is the throughput(bytes transferred per unit time)for the TCP connection?Explain how you calculat

24、ed this value.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=164090 字节。整个传输时间是第一个 TCP 段（即4号段0.026477 秒）的时间和最后的 ACK（即第202 段5.455830秒)时间的差值。因此，总传输时间是5.455830-0.026477=5

25、.4294 秒。因此，TCP 连接的吞吐量为164090/5.4294=30.222 KByte/sec13.Use the Time-Sequence-Graph(Stevens)plotting tool to view the sequence number versus time16129541-12621=2920 bytesThe last ACKThe fist ACKplot of segments being sent from the client to the gaia.cs.umass.edu server.Can you identify where TCPs slo

26、w start phase begins and ends,and where congestion avoidance takes over?Comment on ways in which the measured data differs from the idealized behavior of TCP that weve studied in the text.答:慢启动阶段即从HTTP POST 报文段发出时开始，但是无法判断什么时候慢启动结束，拥塞避免阶段开始。慢启动阶段和拥塞避免阶段的鉴定取决于发送方拥塞窗口的大小。拥塞窗口的大小并不能从时间序号图（time-sequence

27、-graph）直接获得。然而在一个发送方中未被确认的数据量（即in flight 数据量）不会超过CongWin（拥塞窗口）和RcvWindow（接收窗口）中的最小值，即LastByteSend-LastByteAcked=8192(因为in flight data 从未超过8192)。但是，从第10题（即从时间序号图）得，没有分组丢失（不管是超时，还是三个冗余ACK），因此无法判断什么时候慢启动结束，拥塞避免阶段开始。17TypeNo.Seq.ACKed seq.in flight dataData41565Data55662025ACK65661460Data720262920Data83

28、4864380ACK920262920Data1049464380Data1164065840ACK1234864380Data1378665527ACK1440964917ACK1560063007ACK1678661147ACK1790130Data1890131460Data19104732920Data20119334380Data21133935840Data22148537300Data23163138192ACK24104736732ACK25119335272ACK26133933812ACK27148532352ACK2816313892ACK29172050Data3017

29、205146018发出但未被确认的数据。Data31186652920Data32201254380Data33215855840Data34230457300Data35245058192ACK36186656732ACK37201255272ACK38215853812ACK39230452352ACK4024505892ACK41253970Data42253971460Data43268572920Data44283174380Data45297775840Data46312377300Data47326978192ACK48268576732ACK49283175272ACK5029

30、7773812ACK51312371752ACK52335890Data53335896732Data54350495272Data55365093812Data56379692352Data5739429892Data58408890ACK59350496732ACK60379693812ACK6140889892ACK62417810Data63417811460Data64432412920Data65447014380Data66461615840Data67476217300Data68490818192ACK69447015272ACK70476212352ACK71499730D

31、ata7249973146019Data73514332920Data74528934380Data75543535840Data76558137300Data77572738192ACK78528935272ACK79558132352ACK80581650Data81581651460TCP的发送方会试探性的发送数据（即慢启动阶段），如果太多的数据使网络拥塞了，那么发送方会根据AIMD算法进行调整。但是在实际中，TCP的行为主要依赖于应用程序怎么设计。在这次抓包中，在发送方还可以发送数据的时候，已经没有数据可发了。在web应用中，有些web对象比较小，在慢启动还没有结束之前，传送就结束啦，

32、因此，传送小的web对象受到TCP慢启动阶段的影响，导致较长的延迟。14.Answer each of two questions above for the trace that you have gathered when you transferred a file from your computer to gaia.cs.umass.edu。答：慢启动阶段即从HTTP POST 报文段发出时开始，但是无法判断什么时候慢启动结束，拥塞避免阶段开始。慢启动阶段和拥塞避免阶段的鉴定取决于发送方拥塞窗口的大小。拥塞窗口的大小并不能从时间序号图（time-sequence-graph）直接获得

33、。然而在一个发送方中未被确认的数据量（即in flight 数据量）不会超过CongWin（拥塞窗口）和RcvWindow（接收窗口）中的最小值，即LastByteSend-LastByteAcked=9015(因为in flight data 从未超过9015)。但是，从第10题（即从时间序号图）得，没有分组丢失（不管是超时，还是三个冗余ACK），因此无法判断什么时候慢启动结束，拥塞避免阶段开始。TypeNo.Seq.ACKed seq.in flight dataData161823Data178242283ACK198241460Data2022843743Data2137445203A

34、CK2222841460Data2352046663Data2466648123ACK3337446663Data3481249015ACK3552047555ACK366664609620ACK3781241147ACK3890160Data3990161460Data40104762920Data41119364380Data42133965840Data43148567300Data44163168192ACK49104766732ACK50119365272ACK51133963812ACK52148562352ACK5316316892ACK54172080Data551720814

35、60Data56186682920Data57201284380Data58215885840Data59230487300Data60245088192ACK62186686732ACK63201285272ACK64215883812ACK65230482352ACK6624508892ACK67254000Data68254001460Data69268602920Data70283204380Data71297805840Data72312407300Data73327008192ACK75268606732ACK76283205272ACK77327003812ACK78297801752ACK79335920Data80335921460Data81350522920Data21TCP的发送方会试探性的发送数据（即慢启动阶段），如果太多的数据使网络拥塞了，那22么发送方会根据AIMD算法进行调整。但是在实际中，TCP的行为主要依赖于应用程序怎么设计。在这次抓包中，在发送方还可以发送数据的时候，已经没有数据可发了。在web应用中，有些web对象比较小，在慢启动还没有结束之前，传送就结束啦，因此，传送小的web对象受到TCP慢启动阶段的影响，导致较长的延迟。23