IP address of gaia.cs.umass.edu: 128.119.245.12
Port number: 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?
IP address: 192.168.1.102
TCP port number: 1161
Destination IP address: 128.119.245.12
TCP port number: 80
TCP Basics
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 of the TCP SYN segment is used to initiate the TCP connection between the client computer and gaia.cs.umass.edu. The value is 0 in this trace. The segment that identifies the segment as a SYN segment is set to 1. Both illustrated below.
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?
Sequence number of the SYNACK segment from gaia.cs.umass.edu to the client computer in reply to the SYN has the value of 0 in this trace.
The value of the ACKnowledgement field in the SYNACK segment is 1.
The SYN flag and Acknowledgement flag in the segment are set to 1 and they indicate that this segment is a SYNACK segment.
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.
No. 4 segment is the TCP segment containing the HTTP POST command. The sequence number of this segment has the value of 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.
The HTTP POST segment is considered as the first segment. Segments 1 – 6 are No. 4, 5, 7, 8, 10, and 11 in this trace respectively. The ACKs of segments 1 – 6 are No. 6, 9, 12, 14, 15, and 16 in this trace.
Segment 1 sequence number: 1
Segment 2 sequence number: 566
Segment 3 sequence number: 2026
Segment 4 sequence number: 3486
Segment 5 sequence number: 4946
Segment 6 sequence number: 6406
Below shows how the RTT is calculated:
ACK received time – Sent time (Seconds) = RTT (Seconds)
Segment 1: 0.053937 - 0.026477 = 0.02746
Segment 2: 0.077294 - 0.041737 = 0.035557
Segment 3: 0.124085 - 0.054026 = 0.070059
Segment 4: 0.169118 - 0.054690 = 0.11443
Segment 5: 0.217299 - 0.077405 = 0.13989
Segment 6: 0.267802 - 0.078157 = 0.18964
EstimatedRTT = 0.875 * EstimatedRTT + 0.125 * SampleRTT
EstimatedRTT after the receipt of the ACK of segment 1:
EstimatedRTT = RTT for Segment 1 = 0.02746 second
EstimatedRTT after the receipt of the ACK of segment 2:
EstimatedRTT = 0.875 * 0.02746 + 0.125 * 0.035557 = 0.0285
EstimatedRTT after the receipt of the ACK of segment 3:
EstimatedRTT = 0.875 * 0.0285 + 0.125 * 0.070059 = 0.0337
EstimatedRTT after the receipt of the ACK of segment 4:
EstimatedRTT = 0.875 * 0.0337+ 0.125 * 0.11443 = 0.0438
EstimatedRTT after the receipt of the ACK of segment 5:
EstimatedRTT = 0.875 * 0.0438 + 0.125 * 0.13989 = 0.0558
EstimatedRTT after the receipt of the ACK of segment 6:
EstimatedRTT = 0.875 * 0.0558 + 0.125 * 0.18964 = 0.0725
Wireshark has a feature that will allow me to plot the RTT for each of the TCP segments sent. In order to do so, I will select Statistics-> TCP Stream Graph -> Round Trip Time Graph. The graph is illustrated below.
8. What is the length of each of the first six TCP segments?
The length of the first TCP segment is 565 and the other five TCP segments is 1460.
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 reciever buffer space never throttles the sender because the reciever window grows throughtout the trace. The minimum amount of available buffer space is 17520bytes, which shows in the first acknowledgement from the server.
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 in the trace file. I checked the sequence numbers of the TCP segments in the trace file.
The graph below illustrates sequence number versus time. The graph allows us to see whether there are any implications caused by packet loss or long delays. It shows steady transmissions even though at the start it had a slow start. All sequence numbers from the source 192.168.1.102 to the destination 128.119.245.12 are increasing monotonically with respect to time. If there is a retransmitted segment, the sequence number of this retransmitted segment should be smaller than those of its neighbouring segments.
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).
12. What is the throughput (bytes transferred per unit time) for the TCP connection?
Explain how you calculated this value.
The computation of TCP throughput largely depends on the selection of averaging time period. As a common throughput computation, I will select the average time period as the whole connection time. Then, the average throughput for this TCP connection is computed as the ratio between the total amount data and the total transmission time.
- The total amount data transmitted can be computed by the difference between the sequence number of the first TCP segment (i.e. 1 byte for No. 4 segment)
- The acknowledged sequence number of the last ACK (164091 bytes for No. 202 segment).
- Therefore, the total data are 164091 - 1 = 164090 bytes.
- The whole transmission time is the difference of the time instant of the first TCP segment (i.e., 0.026477 second for No.4 segment) and the time instant of the last ACK (i.e., 5.455830 second for No. 202 segment).
- Therefore, the total transmission time is 5.455830 - 0.026477 = 5.4294 seconds.
- Hence, the throughput for the TCP connection is computed as 164090/5.4294 = 30.222 KByte/sec.
TCP congestion control in action
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.
References
- Kurose, J.F. and Ross, K.W. (2008). Computer Networking: A Top-Down Approach (Pearson Education). ISBN-10: 0321513258; ISBN-13: 978-0321513250
