Nt1310 Unit 3

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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, 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” (refer to Figure 2 in the “Getting Started with Wireshark” Lab if you’re uncertain about the Wireshark windows.
A)The IP address and TCP port number used by the client computer is:
IP address:192.168.1.102 and TCP port number is 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?
A)The IP address of gaia.cs.umass.edu is Destination: …show more content…
What is the length of each of the first six TCP segments?
A)The length of the first TCP is 62 bytes .The second is 62 bytes .Third is 54 bytes .Fourth is 619 bytes .Fifth is 1514 bytes .Sixth is 60 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?
A)The available buffer space is from the server and is 17520 bytes .The maximum buffer size is 62780 and the sender is never throttled.
10. Are there any retransmitted segments in the trace file? What did you check for (in the trace) in order to answer this question?
A)
No there are no retransmitted segments.The lines are increasing in a continous pattern. 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 247 in the text).
A)The receiver typically acknowledges the data received by the server between two ACK’s .The receiver is ACKing every other segment .For example the segment no 12 data is 1550 …show more content…
What is the throughput (bytes transferred per unit time) for the TCP connection? Explain how you calculated this value.
A) The alice.txt on the hard drive is 152,138 bytes, and the download time is 1.578736000 (First

TCP segment) ‐ 0.271257000 (last ACK) = 1.307479 second. Along these lines, the throughput for the

TCP association is figured as 152,138/1.307479=116359.803867 bytes/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 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.
A)
The slow start takes place at 0 seconds and then ends at 0.1 secons.Congestion avoidance takes place over 0.7 seconds 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
A) The slow start takes place at 0 seconds and then ends at 0.1 secons.Congestion avoidance takes place over 0.7
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