计算机网络实验3TCP实验Word文档格式.docx

1、 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 the TCP segment containing the HTTP POSTcommand Note that in order to find the POST command, youll need to dig intothe packet co

2、ntent field at the bottom of the Wireshark window, looking for asegment with a “POST” within its DATA field.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 se

3、gments in theTCP connection （including the segment containing the HTTP POST）At what time was each segment sent When was the ACK for each segment receivedGiven the difference between when each TCP segment was sent, and when itsacknowledgement was received, what is the RTT value for each of the sixseg

4、ments 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 measured RTT for the first segment, and then is computed using theEstimatedRTT equation on page 249 for all subsequent : Wireshark has a nice feature that

5、 allows you to plot the RTT foreach of the TCP segments sent. Select a TCP segment in the “listing ofcaptured packets” window that is being sent from the client to server. Then select: Statistics-TCP Stream Graph-Round Trip Time Graph.Segment 1Segment 2Segment 3Segment 4Segment 5Segment 6前6个报文段为,5,7

6、,8,10,11. 对应的ACK分别为 ,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:1Segment 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 tim

7、eRTT secondsEstimatedRTT=* EstimatedRTT+*SampleRTT接受到报文段1之后的EstimatedRTT为：EstimatedRTT=RTT for segment 1= second接受到报文段2之后的EstimatedRTT为：EstimatedRTT=*+*= sencond接受到报文段3之后的EstimatedRTT为：EstimatedRTT=*+*= second接受到报文段4之后的EstimatedRTT为：接受到报文段5之后的EstimatedRTT为：接受到报文段6之后的EstimatedRTT为：8. What is the leng

8、th of each of the first six TCP segments前6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 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

9、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. Are there any retransmitted segments in the trace file What did you check for （inthe trace） in order to answer this question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

10、. How much data does the receiver typically acknowledge in an ACK Can youidentify cases where the receiver is ACKing every other received segment 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 1ACK 21460ACK 3

11、ACK 4ACK 5ACK 67866ACK 790131147ACK 810473ACK 911933ACK 1013393ACK 1114853 报文段确认数据为2920bytes=1460*2 bytes，即=2920.12. What is the throughput （bytes transferred per unit time） for the TCP connectionExplain how you calculated this value.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，

12、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 秒）的时间和最后的 ACK（即第202 段秒） 时间的差值。因此，总传输时间是 = 秒。因此，TCP 连接的吞吐量为164090/ = KByte/sec13. Use the Time-Sequence-Graph（Stevens） plotting tool to view the sequence

13、 number versus time plot of segments being sent from the client to the 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 TCP that weve studied in the text.答

14、: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

15、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 dataData456552025ACK672920843809101158401213552714409649171560063007161718192021227300231631381922467322552722638122723522889229172053031186653220125332158534230453524505363738394041253974243268574428317452977746312374732697484950511752523358953

16、543504955365095637969573942958408895960616241781636443241654470166461616747621684908169707149973727351433745289375543537655813775727378795816581TCP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

17、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 。=9015（因为in flight data 从未超过9015）。823824228322843743374452035204666366648123812490157555609690161047611936133961485616316172081866820128215882304824508