Advanced Communication Networks
Lab - 07
Iqbal Uddin Khan
Lab – 07 TCP Uploading of “alice.txt “wireshark after retrieving from http://gaia.cs.umass.edu/wiresharklabs/alice.txt, following are the screen shots.
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Advanced Communication Networks
Lab - 07
Iqbal Uddin Khan
Q1. 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, it’s probably easiest to select an HTTP message and explore the details of the TCP packet used to carry this HTTP message, using the “details of the selected packet header window”. Ans. Observing the screenshot present in Lab of TCP the source’s IP: 192.168.0.102 and Port: 1161.
Q2. What is the IP address of http://gaia.cs.umass.edu/ ? On what port number is it sending and receiving TCP segments for this connection? Ans. As http://gaia.cs.umass.edu/ is destination so its’ IP: 128.119.245.12 and Port: 80. Q3. 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. Source is my personal PC so its’ IP: 192.168.0.109 and Port: 40317. Q4. 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. The sequence number is 0, it is packet number 2 in my personal captured file. It is used to initiate the establishment of TCP connection. Furthermore as syn flag is set to 1, this shows that it is a syn segment.
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Advanced Communication Networks
Lab - 07
Iqbal Uddin Khan
Q5. 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 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? Ans. In my captured packet the 6th packet is the SYNACK in response to TCP SYN, the packet 2. The Acknowledgement is set i.e. 1. The sequence is 0. Q6. What is the sequence number of the TCP segment containing the HTTP POST command? Note that in order to find the POST command; you’ll need to dig into the packet content field at the bottom of the Wireshark window, looking for a segment with a “POST” within its DATA field. Ans. In my captured packets, 201 is the TCP packet that contains HTTP POST, The 148517 is the relative sequence number.
Q7. 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? Ans. If we consider HTTP POST packet (201), after digging it deep packet re-assembled details, our text was delivered in total 114 packets from 8 to 201. We consider packet 8, 9, 10, 11, 16 and 17 for further details. The answer of question query in Q7 is present in chart on next page.
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Advanced Communication Networks
Lab - 07
Iqbal Uddin Khan
S.No.
Segment Number
Packet Type
Sequence Number
Acknowledge Number
1
8
Data
1
1
2
13
Acknowledgment
0
1
3
9
Data
1389
1
4
15
Acknowledgment
1
1389
5
10
Data
2777
1
6
19
Acknowledgment
1
2777
7
11
Data
4165
1
8
23
Acknowledgment
1
4165
9
16
Data
5553
1
10
27
Acknowledgment
1
5553
11
17
Data
6941
1
12
31
Acknowledgment
1
6941
Time Dec 29, 2012 03:25:15.268983000 Dec 29, 2012 03:25:15.512339000 Dec 29, 2012 03:25:15.269005000 Dec 29, 2012 03:25:15.577564000 Dec 29, 2012 03:25:15.269017000 Dec 29, 2012 03:25:15.612703000 Dec 29, 2012 03:25:15.269029000 Dec 29, 2012 03:25:15.645253000 Dec 29, 2012 03:25:15.577624000 Dec 29, 2012 03:25:15.680372000 Dec 29, 2012 03:25:15.577665000 Dec 29, 2012 03:25:15.883541000
Q8. 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 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 for all subsequent segments. Ans. The equation mention in book is “EstimatedRTT = 0.875 * EstimatedRTT + 0.125 * SampleRTT “. RTT for each segment can be observed in [SEQ/ACK Analysis] section. 1. 2. 3. 4. 5. 6.
For Segment 8 is: 0.243356 Sec. For Segment 9 is: 0.251506 Sec. For Segment 10 is: 0.263028 Sec. For Segment 11 is: 0.277177 Sec. For Segment 16 is: 0.255373 Sec. For Segment 17 is: 0.261685 Sec. Page 4
Advanced Communication Networks
Lab - 07
Iqbal Uddin Khan
Q9. What is the length of each of the first six TCP segments? Ans. The length of each segment is 1388 bytes and segment contains HTTP POST is of767 bytes. Q10. 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? Ans. The minimum amount is exhibited in the 6th packet of the trace is 5792. The amount of buffer varies and in the last packet (246) I observed 116096, as the window size is increased it means sender was not throttled. Q11. Are there any retransmitted segments in the trace file? What did you check for (in the trace) in order to answer this question? Ans. I haven’t observed and repetition of sequence number in my trace and even the ACKs are in sequence, which shows that there is no re-transmitting of any packet, but duplicate ACKs are observed. Q12. How much data does the receiver typically acknowledge in an ACK? Can you identify cases where the receiver is ACKing every other received segment. Ans. The data is acknowledged is 1388 bytes, I observed this by subtracting the sequence ACK of previous from recent ACK, and checked all 6 ACK I mentioned early. Q13. What is the throughput (bytes transferred per unit time) for the TCP connection? Ans. The throughput of TCP connection can be find out by dividing the total bytes transferred in the total time elapsed, So the transferring starts from segment 8 and end at segment 201. The total bytes transferred are 148517. Now to calculate the total elapsed time, subtracting the time of segment 8 from the time of segment 201 we obtain 18.656306 Sec – 15.268983 Sec = 3.387323 Sec. Now 148517/3.387323 = 43844.94 Bytes/Sec or 43.844 Kbyte/sec. Hence the throughput is 350.75 kbps. Q14. 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 TCP’s 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 we’ve studied in the text. Ans. Screen shot of graph of packets from tcp-etherealtrace-1 from http://gaia.cs.umass.edu/wiresharklabs/wireshark-traces.zip is on next page. Q15. 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. Screen shot of graph of packet captured by me is the last image of this assignment.
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Advanced Communication Networks
Lab - 07
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Iqbal Uddin Khan
