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”
The source IP address is 192.168.1.102 and source TCP port 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?
The IP address of gaia.cs.umass.edu is 184.108.40.206 and the port number it is using for sending and receiving TCP segments is port 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?
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 is 0. The segment is identified as a SYN segment since the flag SYN bit is set to 1.
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4. 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?
Sequence number of the SYNACK segment sent by gaia.cs.umass.edu to the client computer in reply to the SYN is 0. The value of the Acknowledgement field in the SYNACK segment is 1 and the segment is identified as a SYNACK segment since the flag SYN&ACK bits are set to 1.
5. 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.
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? 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.
The Estimated RTT was calculated using the following equation using Initial Estimated RTT = initial Sample RTT
Estimated RTT = 0.875*Estimated RTT + 0.125*Sample RTT
7. What is the length of each of the first six TCP segments?
The length of the first TCP segment is 565 bytes. The other five are 1460 bytes each.
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?
Minimum amount of available buffer space advertised at the received for the entire trace is 5840 bytes. The available buffer space is equal to the window size field of the acknowledgment segment. No, the lack of receiver buffer space never throttles the sender on inspection
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9. Are there any retransmitted segments in the trace file? What did you check for (in the trace) in order to answer this question?
No, there aren’t retransmitted segments in the trace file. This has been verified by checking the Time Sequence Graph (Stevens) of this trace. The graph doesn’t show any horizontal overlapping of points since the sequence numbers are always increasing from one segment to another.
10. What is the throughput (bytes transferred per unit time) for the TCP connection?
The throughput can be computed by the difference between the sequence number of the first TCP segment and the acknowledged sequence number of the last ACK, which in this case is 164091 bytes. Therefore the total amount of bytes transferred is 164091 – 1 = 164090 bytes. The time of transmission is the difference between the time of the first TCP segment and the time of the last ACK. Therefore, the total transmission time is 5.455830- 0.026477 = 5.429353seconds. Hence, the throughput for the TCP connection is:
164090/5.429353 = 29.514 KByte/sec.
11. 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.
The TCP slow start phase occurs in the range marked on the graph. This phase initiates as soon as a connection is established and hence when the HTTP POST is transmitted as seen in question 5. Upon arrival of an acknowledgment, if the congestion window size increases the Maximum Segment Size (MSS) by one, this means we are still in the slow start phase. In order to identify when the end of the slow start phase and the start of congestion avoidance occurs, we have to inspect how the congestion window size varies in relation to the arrival of acknowledgments. Since the slow start phase cannot continue forever, a threshold must be predetermined in order to stop this phase. After the threshold is reached, the congestion avoidance phase initiates the additive increase procedure which allows the window size to increase linearly in contrast with the exponential nature of the slow start phase. This procedure continues until a timeout occurs or the maximum window size is reached.
Whilst the idealized behavior of TCP assumes that we are aggressive when sending data, this might cause congestion of the network due to an overload of traffic. Thus we should use the Additive Increase Multiplicative Decrease (AIMD) algorithm in order to identify network congestion and therefore decrease the window size. However in practice TCP behavior is greatly dependent with regards to the application such as web based applications. In such a case several small web objects are transmitted before the slow start phase ends and hence incur huge delays. As a result when this phase ends, there would be no data left to be sent and therefore no advantage is taken of the increase in window size.
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