You are watching: What is it in the segment that identifies the segment as a syn segment?
2.What is the IP address of gaia.cs.umass.edu? ~ above what port number is the sending and receiving TCP segments because that this connection?
4.What is the sequence variety of the TCP SYN segment the is offered to begin the TCP connection in between the customer computer and also gaia.cs.umass.edu? What is it in the segment that identifies the segment together a SYN segment?
Answer:The sequence variety of the TCP SYN segment is 0 since it is offered to imitate the TCP connection between the customer computer and gaia.cs.umass.edu. Follow to the screenshot below, in the Flags section, the SYN flag is collection to 1 which suggests that this segment is a SYN segment.
5.What is the sequence variety of the SYNACK segment sent out by gaia.cs.umass.edu come the client computer in answer to the SYN? What is the worth of the Acknowledgement field in the SYNACK segment? how did gaia.cs.umass.edu identify that value? What is the in the segment that identifies the segment together a SYNACK segment?
Answer:According come the screenshot below, the sequence number of the SYN_ACK segment sent by gaia.cs.umass.edu to the client computer in answer to the SYN is 0. The value of the acknowledgement field in the SYN_ACK segment is identified by the server gaia.cs.umass.edu. The server to add 1 come the early stage sequence variety of the SYN segment from the customer computer. Because that this case, the early sequence variety of the SYN segment indigenous the client computer is 0, hence the value of the acknowledgement ar in the SYN_ACK segment is 1. A segment will certainly be figured out as a SYN_ACK segment if both SYN flag and also ACKnowledgement flag in the segment are set to 1.
6.What is the sequence variety of the TCP segment comprise the HTTP short article command? keep in mind that in order to uncover the article command, you’ll need to dig into the packet content ar at the bottom that the Wireshark window, in search of a segment with a “POST” in ~ its DATA field.
Answer:ACK because that segment 1 was got at 2.026105 s and ACK for segment 2 is got at 2.311435 s.
d.Given the difference between when each TCP segment was sent, and also when its acknowledgement was received, what is the RTT worth for each of the 6 segments?
RTT for segment 1 is 0.287551 seconds, RTT for segment 2 is 0.285274 seconds, RTT for segment 3 is 0.285261 seconds, RTT because that segment 4 is 0.285369 seconds, RTT because that segment 5 is 0.000176 seconds, RTT because that segment 6 is 0.286297 seconds.
9.What is the minimum lot of accessible buffer room advertised in ~ the got for the entire trace? does the lack of recipient buffer room ever throttle the sender?
10.Are there any kind of retransmitted segment in the map file? What did you examine for (in the trace) in order to answer this question?
No over there is no retransmitted segments in the trace file. This deserve to be explained by packets with very same sequence number at various time is not found.
11.How much data does the receiver frequently acknowledge in an ACK? have the right to you identify instances where the receiver is ACKing every various other received segment (see Table 3.2 on page 247 in the text).
Answer:According come the screenshot below, we can see that the ACK numbers boost in the succession of 1401, 2801, 4201, and also so on. The ACK numbers rises by 1400 every time, indicating that the receiver is acknowledging 1400 bytes.
12.What is the throughput (bytes moved per unit time) because that the TCP connection? explain how friend calculated this value.
13.Use the Time-Sequence-Graph(Stevens) plotting device to see the sequence number versus time plot of segment being sent from the customer to the gaia.cs.umass.edu server. Have the right to you identify where TCP’s slow-moving start phase begins and also ends, and where jam avoidance take away over? discuss ways in i m sorry the measured data different from the idealized action of TCP the we’ve learned in the text.
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By observing the plot, we have the right to see that the slow-start phase only lasts for first 1-1.5 second. Afterwards, it appears that the TCP session is constantly in jam avoidance state. In this case, we execute not watch the supposed linear increase behaviour, i.e. The TCP transmit window does not thrive linearly during this phase. In fact, it appears that the sender transmits packets in batches of 6. This does not seem come be resulted in by circulation control since the receiver advertised home window is significantly larger than 5 packets. The reason for this behaviour might be because of the reality that the HTTP server has applied a rate-limit of part sort.
Posted inComputer Networks, Fundamentals that Data Communication and also Networking, Lab, Solution, TCP, Wireshark