2022年2022年计算机网络实验三参考答案 .pdf

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1、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, its probably easiest to select an HTTP message and explore the details of the TCP packet used to carry this HTTP message, using the “details

2、 of the selected packet header window” (refer to Figure 2 in the “Getting Started with Wireshark” Lab if youre uncertain about the Wireshark windows). Ans: IP address:192.168.1.102 TCP port:11612. What is the IP address of gaia.cs.umass.edu? On what port number is it sending and receiving TCP segmen

3、ts for this connection? Ans: IP address:128.119.245.12 TCP port:80 If you have been able to create your own trace, answer the following question: 3. What is the IP address and TCP port number used by your client computer (source) to transfer the file to gaia.cs.umass.edu? ANS: IP address :10.211.55.

4、7 TCP port:49265 4. What is the sequence number of the TCP SYN segment that is used to initiate the TCP connection between the client computer and gaia.cs.umass.edu? What is it in the segment that identifies the segment as a SYN segment? ANS: sequence number: 0 Syn Set = 1 identifies the segment as

5、a SYN segment 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 1 页，共 7 页 - - - - - - - - - 5. What is the sequence number of the SYNACK segment sent by gaia.cs.umass.edu to the client computer in reply to the SYN? What is the value of the ACKnowledgement field in the S

6、YNACK segment? How did gaia.cs.umass.edu determine that value? What is it in the segment that identifies the segment as a SYNACK segment? ANS: The sequence number: 0 ACKnowledgement number : 1 which is sequence number plus 1 Both the sequence flag and the ACKnowledgement flag been set as 1, identifi

7、es the segment as SYNACK segment. 6. What is the sequence number of the TCP segment containing the HTTP POST command? Note that in order to find the POST command, youll need to dig into the packet content field at the bottom of the Wireshark window, looking for a segment with a “POST ” within its DA

8、TA field. Ans: The sequence number : 1 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 2 页，共 7 页 - - - - - - - - - 7. Consider the TCP segment containing the HTTP POST as the first segment in the TCP connection. What are the sequence numbers of the first six segments

9、in the TCP connection (including the segment containing the HTTP POST)? At what time was each segment sent? When was the ACK for each segment received? Given the difference between when each TCP segment was sent, and when its acknowledgement was received, what is the RTT value for each of the six se

10、gments? What is the EstimatedRTT value (see page 249 in text) after the receipt of each ACK? Assume that the value of the EstimatedRTT is equal to the measured RTT for the first segment, and then is computed using the EstimatedRTT equation on page 249 for all subsequent segments. Note: Wireshark has

11、 a nice feature that allows you to plot the RTT for each of the TCP segments sent. Select a TCP segment in the “listing of captured packets” window that is being sent from the client to the gaia.cs.umass.edu server. Then select: Statistics-TCP Stream Graph- Round Trip Time Graph. 名师资料总结 - - -精品资料欢迎下

12、载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 3 页，共 7 页 - - - - - - - - - Segment 1 Segment 2 Segment 3 Segment 4 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 4 页，共 7 页 - - - - - - - - - Segment 5 Segment 6 ANS: Sequence number Sent time ACK received

13、 time RTT EstimatedRTT value Length /bytes Segment 1 1 0.026477 0.053937 0.02746 0.02746 565 Segment 2 566 0.041737 0.077294 0.035557 0.028472 1460 Segment 3 2026 0.054026 0.124085 0.070059 0.033670 1460 Segment 4 3486 0.054690 0.169118 0.11443 0.043765 1460 Segment 5 4946 0.077405 0.217299 0.13989

14、0.055781 1460 Segment 6 6406 0.078157 0.267802 0.18964 0.072513 1460 EstimatedRTT = 0.875 * Last EstimatedRTT + 0.125 * sample RTT After Segment 1 : EstimatedRTT = 0.02746 After Segment 2 : EstimatedRTT = 0.875 * 0.02746 + 0.125*0.035557 = 0.028472 After Segment 3 : EstimatedRTT = 0.875 * 0.028472 +

15、 0.125*0.070059 = 0.033670 After Segment 4 : EstimatedRTT = 0.875 * 0.033670 + 0.125*0.11443 = 0.043765 After Segment 5 : EstimatedRTT = 0.875 * 0.043765 + 0.125*0.13989 = 0.055781 After Segment 6 : EstimatedRTT = 0.875 * 0.055781 + 0.125*0.18964 = 0.072513 8. What is the length of each of the first

16、 six TCP segments?(see Q7) 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 5 页，共 7 页 - - - - - - - - - 9. What is the minimum amount of available buffer space advertised at the received for the entire trace? Does the lack of receiver buffer space ever throttle thesend

17、er? ANS:The minimum amount of buffer space (receiver window) advertised at gaia.cs.umass.edu for the entire trace is 5840 bytes; This receiver window grows steadily until a maximum receiver buffer size of 62780 bytes. The sender is never throttled due to lacking of receiver buffer space by inspectin

18、g this trace. 10. Are there any retransmitted segments in the trace file? What did you check for (in the trace) in order to answer this question? ANS: There are no retransmitted segments in the trace file. We can verify this by checking the sequence numbers of the TCP segments in the trace file. All

19、 sequence numbers are increasing.so there is no retramstmitted segment.11. How much data does the receiver typically acknowledge in an ACK? Can you identify cases where the receiver is ACKing every other received segment (see Table 3.2 on page 257 in the text). 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - -

20、 - - - - - - - - - 名师精心整理 - - - - - - - 第 6 页，共 7 页 - - - - - - - - - ANS: According to this screenshot, the data received by the server between these two ACKs is 1460bytes. there are cases where the receiver is ACKing every other segment 2920 bytes = 1460*2 bytes. For example 64005-61085 = 2920 12.

21、 What is the throughput (bytes transferred per unit time) for the TCP connection? Explain how you calculated this value. ANS: total amount data = 164091 - 1 = 164090 bytes #164091 bytes for NO.202 segment and 1 bytes for NO.4 segment Total transmission time = 5.455830 0.026477 = 5.4294 So the throug

22、hput for the TCP connection is computed as 164090/5.4294 = 30.222 KByte/sec.13. Use the Time-Sequence-Graph(Stevens) plotting tool to view 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

23、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.ANS: Slow start begins when HTTP POST segment begins. But we can t identify where TCPs slow start phase ends, and where congestion

24、avoidance takes over. 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 ANS: Slow start begins when HTTP POST segment begins. But we can t identify where TCPs slow start phase ends, and where congestion avoidance takes over. 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 7 页，共 7 页 - - - - - - - - -