wireshark实验一答案

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? Ip address 192.168.1.36

TCP port number：1957

2. What is the IP address of gaia.cs.umass.edu? On what port number

is it sending and receiving TCP segments for this connection? the IP address of gaia.cs.umass.edu：128.119.245.12 port number：80

3. 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? sequence number:0

syn 被设置为1说明是syn段

4. What is the sequence number of the SYNACK segment sent by

gaia.cs.umass.ed to 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 is it in the segment that identifies the segment as a SYNACK segment? 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.

5. What is the sequence number of the TCP segment containing the

HTTP POST command?

第11号报文段是包含 HTTP POST 命令的TCP segment。且报文段的序列号为152299

6. 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 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 segments? 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.

前6个报文段为No.11，12，14，15，17，18

序列号为：1，619，2027，3435，4843，6251.

Segment 1 Segment 2 Send time 2.513203 2.513393 ACK received time 2.799560 2.803157 RTT seconds 0.286357 0.289764 Segment 3 Segment 4 Segment 5 Segment 6 2.799582 2.799592 2.803168 2.803176 2.803157 3.088572 3.088579 3.088589 0.003575 0.28898 0.285411 0.285413 EstimatedRTT=0.875* EstimatedRTT+0.125*SampleRTT 接受到报文段1之后的EstimatedRTT为：

EstimatedRTT=RTT for segment 1=0.286157 second 接受到报文段2之后的EstimatedRTT为：

EstimatedRTT=0.875*0.286357+0.125*289764=0.286783 sencond 接受到报文段3之后的EstimatedRTT为：

EstimatedRTT=0.875*0.286783+0.125*0.003575=0.251382 second 接受到报文段4之后的EstimatedRTT为：

EstimatedRTT=0.875*0.251382+0.125*0.28898=0.256082 second 接受到报文段5之后的EstimatedRTT为：

EstimatedRTT=0.875*0.256082 +0.125*0.285411= 0.259748 second 接受到报文段6之后的EstimatedRTT为：

EstimatedRTT=0.875*0.259748+0.125*0.285413= 0.262956 second 7. What is the length of each of the first six TCP segments? 如图

答：前6个段的长度分别为：618,1408,1408,1408,1408,1408字节

8. 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 the sender?

答：接收方通知给发送方的最低窗口大小为6798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6798逐步增加到65535，窗口大小始终大于发送方发送的分组的容量

9.. Are there any retransmitted segments in the trace file? What did you check for (in the 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。 10. 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

如图可知，接收方在一个ACK确认的数据大小一般为1408字节。

The Acknowledged sequence number and the Acknowledged data:

ACK 1 ACK 2 ACK 3 ACK 4 ACK 5 ACK 6 ACK 7 ACK 8 ACK 9 ACK 10 ACK 11 Acknowledged sequence number 618 2026 3434 4842 6250 7658 9066 10218 11626 13034 14442 Acknowledged data 618 1408 1408 1408 1408 1408 1152 1408 1408 1408 1408 报文段确认数据为2816bytes=1408*2 bytes，即129131-126315

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