Saturday, November 26, 2011

Wireshark Lab: TCP














  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?

Client IP : 192.168.1.102
TCP port: health-polling (1161)

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?

Ip address : 128.119.245.12
TCP Port: 80

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?  


192.168.2.3
Source port:(1742)

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?

Sequence number: 0    (relative sequence number)
The syn segement can be seen to be set to 1 (.... .... ..1. = Syn: Set) This shows it is a syn segment

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 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: Acknowledgement number: 1    (relative ack number)
value of the ACKnowledgement field is 1
gaia.cs.umass.edu determined that value by adding 1 to the sequence number of the previous segement.

This segment is identified as a synack segment acknowledgement and syn bits are both set.





6. 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.

The sequence number of the TCP segment containing the HTTP POST
Command is 1






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 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.
Note: Wireshark has 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.


segmentRelative Segment numberSegment numberTime sentAcknowledgement ReceivedRTTEstimated RTT
110dd601f.026.054.028.028
25660dd6042.042.077.035.035
320260dd609d.054.124.070.70
434860dd60f9.055.169.114.114
549460dd60f9.077.217.140.140
664060dd61af.078.268.190.190


Estimated RTT packet 1 : 0.875 * .028 + 0.125 * .028= .028
Estimated RTT packet 2 : 0.875 *  .042+ 0.125 * .035 =  .035
Estimated RTT packet 3 : 0.875 * .054 + 0.125 * .070 = .070
Estimated RTT packet 4 : 0.875 *  .055+ 0.125 * .114 = .114
Estimated RTT packet 5 : 0.875 *  .077+ 0.125 * .140 =  .140
Estimated RTT packet 6 : 0.875 *  .078+ 0.125 * .190 = .190



8. What is the length of each of the first six TCP segments?


Segment 1 = 565 bytes
Segment 2 = 1460 bytes
Segment 3 = 1460 bytes
Segment 4 = 1460 bytes
Segment 5 = 1460 bytes
Segment 6 = 1460 bytes


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 the
sender


The minimum amount of available buffer space is advertised as (Calculated window size): 5840 bytes. The lack of receiver buffer space does not ever throttle the sender.

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

There are no retransmitted segments. To check this, I looked for any repeating segment numbers.

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).

The receiver typically acknowledges 1460 bytes in an ack. If the data is doubled then that segement is acking every other





12. What is the throughput (bytes transferred per unit time) for the TCP connection?  
Explain how you calculated this value.

The file is 177851 bytes dive that by the total time 7.596 seconds and average throughput is  23413.77 bytes per second



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 TCP’s slowstart 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

The slowstart phase begins at about zero and ends at about .15 seconds in according to the graph then congestion takes over. The measured data is only using a fraction of the window size instead of the idealized 1/3 to a half.



7 comments:

  1. This comment has been removed by the author.

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  2. can you please post the explanation on Question 7. how did you make that table ?

    ReplyDelete
  3. where did you get the numbers like 0.875 and 0.125?

    0.875 * .028 + 0.125 * .028= .028

    ReplyDelete
  4. bastard you copy from the internet and you don't explain when you are asked. idiot!

    ReplyDelete
  5. Amazing post.Thanks for your details and explanations..I want more information from your side.Thank you
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    ReplyDelete
  6. I think you made a mistake when you were calculating the estimated round trip time for all of those, in question 7.
    I'm looking specifically at the calculation (though I think you made the mistake for each line):
    Estimated RTT packet 2 : 0.875 * .042+ 0.125 * .035 = .035

    It should be:
    0.875 * "what the EstimatedRoundTripTime already was" + 0.125 * "what the RoundTripTime was for this packet".
    Instead you did it as:
    0.875 * "the Time the packet was sent" + 0.125 * "what the RoundTripTime was for this packet"
    So in other words, it should be
    Estimated RTT packet 2 : 0.875 * .028 + 0.125 * .035 = 0.25= 0.289

    Great work overall though, thanks for posting this just to help us out.

    ReplyDelete