Wireshark TCP Lab  

Introduction

This assignment will investigate the behaviour of TCP (Transmission Control Protocol). The application well suited for this assignment is Wireshark, it has the features suitable for this assignment and enables me to output the relevant graphs needed. The graph being used is a Time-Sequence-Graph (Stevens).

Methodology

In order to achieve what is needed I will need to analyse a trace of the TCP segments sent and received in transferring a 150KB file containing the text of Lewis Carrol’s Alice’s Adventures in Wonderland from my computer to a remote server. I will study TCP’s use of sequence and acknowledgement numbers for providing reliable data transfer; See TCP’s congestion control algorithm – slow start and congestion avoidance – in action; and look at TCP’s receiver-advertised flow control mechanism. I will also output graphs that wireshark will allow me to do so within one of its functions.

Wireshark will obtain a packet trace of the TCP transfer of a file from my computer to a remote server. I will do so by accessing a Web page that will allow me to enter the name of a file stored on my computer which contains the ASCII text of Alice in Wonderland, and then transfer the file to a web server using the HTTP POST method.

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

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.

1. 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)

2. The acknowledged sequence number of the last ACK (164091 bytes for No. 202 segment).

3. Therefore, the total data are 164091 - 1 = 164090 bytes.
4. 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).

5. Therefore, the total transmission time is 5.455830 - 0.026477 = 5.4294 seconds.

6. 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