Wireshark tcp
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The document analyzes a Wireshark trace file of a TCP connection between a client and a server. It identifies the IP addresses and port numbers used, examines the sequence numbers and flags in the TCP segments, calculates the round-trip times and estimated RTT for segments, and determines that no segments were retransmitted. It also finds that the receiver acknowledges about 1460 bytes of data at a time, and calculates the overall throughput of the connection.
Wireshark tcp
- 1. 1 Wireshark Lab: TCP SOLUTION Supplement to Computer Networking: A Top-Down Approach, 6th ed., J.F. Kurose and K.W. Ross © 2005-21012, J.F Kurose and K.W. Ross, All Rights Reserved The answers below are based on the trace file tcp-ethereal-trace-1 in in http://gaia.cs.umass.edu/wireshark-labs/wireshark-traces.zip TCP Basics Answer the following questions for the TCP segments: 1. What is the IP address and TCP port number used by your client computer (source) to transfer the file to gaia.cs.umass.edu? 2. What is the IP address and port number used by gaia.cs.umass.edu to receive the file. Solution: Client computer (source) IP address: 192.168.1.102 TCP port number: 1161 Destination computer: gaia.cs.umass.edu IP address: 128.119.245.12 TCP port number: 80 3. If you did this problem on your own computer, you’ll have your own solution ©2013 Pearson Education, Inc. Upper Saddle River, NJ. All Rights Reserved.
- 2. 2 Figure 1: IP addresses and TCP port numbers of the client computer (source) and gaia.cs.umass.edu 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? Solution: 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 SYN flag is set to 1 and it indicates that this segment is a SYN segment. ©2013 Pearson Education, Inc. Upper Saddle River, NJ. All Rights Reserved.
- 3. 3 Figure 2: Sequence number of the TCP 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? Solution: 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 value of the ACKnowledgement field in the SYNACK segment is determined by gaia.cs.umass.edu by adding 1 to the initial sequence number of SYN segment from the client computer (i.e. the sequence number of the SYN segment initiated by the client computer is 0.). The SYN flag and Acknowledgement flag in the segment are set to 1 and they indicate that this segment is a SYNACK segment. ©2013 Pearson Education, Inc. Upper Saddle River, NJ. All Rights Reserved.
- 4. 4 Figure 3: Sequence number and Acknowledgement number of the 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. Solution: No. 4 segment is the TCP segment containing the HTTP POST command. The sequence number of this segment has the value of 1. ©2013 Pearson Education, Inc. Upper Saddle River, NJ. All Rights Reserved.
- 5. 5 Figure 4: Sequence number of the TCP segment containing the HTTP POST command 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 237 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 237 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. Solution: 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 ©2013 Pearson Education, Inc. Upper Saddle River, NJ. All Rights Reserved.
- 6. 6 Segment 5 sequence number: 4946 Segment 6 sequence number: 6406 The sending time and the received time of ACKs are tabulated in the following table. Sent time ACK received time RTT (seconds) Segment 1 0.026477 0.053937 0.02746 Segment 2 0.041737 0.077294 0.035557 Segment 3 0.054026 0.124085 0.070059 Segment 4 0.054690 0.169118 0.11443 Segment 5 0.077405 0.217299 0.13989 Segment 6 0.078157 0.267802 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 second ©2013 Pearson Education, Inc. Upper Saddle River, NJ. All Rights Reserved.
- 7. 7 Figure 5: Segments 1 – 6 Figure 6: ACKs of segments 1 - 6 ©2013 Pearson Education, Inc. Upper Saddle River, NJ. All Rights Reserved.
- 8. 8 Figure 7: Round Trip Time Graph 8. What is the length of each of the first six TCP segments? Solution: Length of the first TCP segment (containing the HTTP POST): 565 bytes Length of each of the other five TCP segments: 1460 bytes (MSS) Figure 8: Lengths of segments 1 - 6 ©2013 Pearson Education, Inc. Upper Saddle River, NJ. All Rights Reserved.
- 9. 9 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? Solution: The minimum amount of buffer space (receiver window) advertised at gaia.cs.umass.edu for the entire trace is 5840 bytes, which shows in the first acknowledgement from the server. This receiver window grows steadily until a maximum receiver buffer size of 62780 bytes. The sender is never throttled due to lacking of receiver buffer space by inspecting this trace. Figure 9: Minimum receive window advertised at gaia.cs.umass.edu (packet No. 2) 10. Are there any retransmitted segments in the trace file? What did you check for (in the trace) in order to answer this question? Solution: There are no retransmitted segments in the trace file. We can verify this by checking the sequence numbers of the TCP segments in the trace file. In the Time- Sequence-Graph (Stevens) of this trace, 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 neighboring segments. ©2013 Pearson Education, Inc. Upper Saddle River, NJ. All Rights Reserved.
- 10. 10 Figure 10: Sequence numbers of the segments from the source (192.168.1.102) to the destination (128.119.245.12) 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). Solution: The acknowledged sequence numbers of the ACKs are listed as follows. acknowledged sequence number acknowledged data ACK 1 566 566 ACK 2 2026 1460 ACK 3 3486 1460 ACK 4 4946 1460 ACK 5 6406 1460 ACK 6 7866 1460 ACK 7 9013 1147 ACK 8 10473 1460 ACK 9 11933 1460 ACK 10 13393 1460 ACK 11 14853 1460 ACK 12 16313 1460 … The difference between the acknowledged sequence numbers of two consecutive ACKs indicates the data received by the server between these two ACKs. By inspecting the amount of acknowledged data by each ACK, there are cases where the receiver is ©2013 Pearson Education, Inc. Upper Saddle River, NJ. All Rights Reserved.
- 11. 11 ACKing every other segment. For example, segment of No. 80 acknowledged data with 2920 bytes = 1460*2 bytes. Figure 8: Cumulative ACKs (No. 80, 87, 88, etc) where the receiver is ACKing every other received segment. 12. What is the throughput (bytes transferred per unit time) for the TCP connection? Explain how you calculated this value. Solution: The computation of TCP throughput largely depends on the selection of averaging time period. As a common throughput computation, in this question, we 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) and 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. ©2013 Pearson Education, Inc. Upper Saddle River, NJ. All Rights Reserved.