计算机网络实验3TCP实验(共15页).doc

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1、精选优质文档-倾情为你奉上计算机网络实验报告三TCP实验1. 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 to carry this HTTP me

2、ssage, 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：192.168.1.102 TCP port number：11612. What is the IP address of gaia.cs.umass.edu? On what

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

4、ter(source) to transfer the file to gaia.cs.umass.edu?答：My client computer:IP address:10.2.136.304. 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 it in the segment that identifies the segment a

5、s a SYN segment?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 ACKnowledgement field in the SYNACK segment? How did gaia.cs.umass.edu determine that value? What i

6、s it in the segment that 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 sequ

7、ence number of 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.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.

8、 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 theTCP connection (including the segment containing the HTTP POST)?At what time was each segment sent? When was the ACK for each segment receive

9、d?Given the difference between when each TCP segment was sent, and when itsacknowledgement was received, what is the RTT value for each of the sixsegments? What is the EstimatedRTT value (see page 249 in text) after thereceipt of each ACK? Assume that the value of the EstimatedRTT is equal tothe mea

10、sured RTT for the first segment, and then is computed using 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 TCP segments sent. Select a TCP segment in the “listing ofcaptured packets” window that is be

11、ing sent from the client to thegaia.cs.umass.edu server. Then select: Statistics-TCP Stream Graph-Round Trip Time Graph.Segment 1Segment 2Segment 3Segment 4Segment 5Segment 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：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:1Segm

12、ent 2 sequence number:566Segment 3 sequence number:2026Segment 4 sequence number:3486Segment 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 10.0.0.02746Segment 20.0.0.Segment 30.0.0.Segment 40.0.0.11443Segment 50.0.0.13989Seg

13、ment 60.0.0.18964EstimatedRTT=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.=0.0285 sencond接受到报文段3之后的EstimatedRTT为：EstimatedRTT=0.875*0.0285+0.125*0.=0.0337 second接受到报文段4之后的EstimatedR

14、TT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：EstimatedRTT=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个段的长度分别为：565、1460、1460、

15、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? 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.

16、Are there any retransmitted segments in the trace file? What did you check for (inthe trace) in order to answer this question？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。11. How much data does the receiver typically acknowledge i

17、n 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 dataACK 1566566ACK 220261460ACK 334861460ACK 449461460ACK 564061460ACK 6786

18、61460ACK 790131147ACK 8104731460ACK 9119331460ACK 10133931460ACK 11148531460 报文段确认数据为2920bytes=1460*2 bytes，即-12621=2920.12. What is the throughput (bytes transferred per unit time) for the TCP connection?Explain how you calculated 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

19、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 段的个字节）之间的差值。因此，总数据是 -1 = 字节。整个传输时间是第一个 TCP 段（即4号段0. 秒）的时间和最后的 ACK（即第202 段5.秒) 时间的差值。因此，总传输时间是5.-0. = 5.4294 秒。因此，TCP 连接的吞吐量为/5.4294 = 30.222 KByte/sec13. Use the Time-Sequence-Graph(Stevens) plotting tool to vi

20、ew the sequence number versus time plot of segments being sent from the client to the gaia.cs.umass.edu server. Can you identify where TCPs slow 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 T

21、CP 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-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

22、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 dataData41565Data55662025ACK65661460Data720262920Data834864380ACK920262920Data1049464380Data1164065840ACK1234864380Data1378665527ACK1440964917ACK1560063007ACK1678661147ACK1790130Data1890131460Data19104732920Data

23、20119334380Data21133935840Data22148537300Data23163138192ACK24104736732ACK25119335272ACK26133933812ACK27148532352ACK2816313892ACK29172050Data30172051460Data31186652920Data32201254380Data33215855840Data34230457300Data35245058192ACK36186656732ACK37201255272ACK38215853812ACK39230452352ACK4024505892ACK41

24、253970Data42253971460Data43268572920Data44283174380Data45297775840Data46312377300Data47326978192ACK48268576732ACK49283175272ACK50297773812ACK51312371752ACK52335890Data53335896732Data54350495272Data55365093812Data56379692352Data5739429892Data58408890ACK59350496732ACK60379693812ACK6140889892ACK6241781

25、0Data63417811460Data64432412920Data65447014380Data66461615840Data67476217300Data68490818192ACK69447015272ACK70476212352ACK71499730Data72499731460Data73514332920Data74528934380Data75543535840Data76558137300Data77572738192ACK78528935272ACK79558132352ACK80581650Data81581651460TCP的发送方会试探性的发送数据（即慢启动阶段），如

26、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，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.uma

27、ss.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。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

28、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 dataData161823Data178242283ACK198241460Data2022843743Data2137445203ACK2222841460Data2352046663Data2466648123ACK3337446663Data3481249015ACK3552047555ACK3666646096ACK3781241147ACK3890160Data3990161460Data40104762920Data41119364380Data42133

29、965840Data43148567300Data44163168192ACK49104766732ACK50119365272ACK51133963812ACK52148562352ACK5316316892ACK54172080Data55172081460Data56186682920Data57201284380Data58215885840Data59230487300Data60245088192ACK62186686732ACK63201285272ACK64215883812ACK65230482352ACK6624508892ACK67254000Data6825400146

30、0Data69268602920Data70283204380Data71297805840Data72312407300Data73327008192ACK75268606732ACK76283205272ACK77327003812ACK78297801752ACK79335920Data80335921460Data81350522920DataTCP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，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。专心-专注-专业