计算机网络实验报告3-tcp,计算机网络实验报告3 TCP
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1、SHAANXI NORMAL UNIffFtSITY計算機網絡實驗報告專業:計算機科學與技術年級:班級:姓名:學號:計算機科學學院TCP協議分析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, i t?s probably easiest to select an HTTP message and explore the deta。
2、ils of the TCPLab ifpacket used to carry this HTTP message, using the“ details of the selected packetheader win dow ” (refer to Figure 2 in the“ Getti ng Started with Wiresharkyou?re un certa in about the Wireshark wi ndows).Src: 192.168,1,102 (192.168.1.102)Src Port: healrh-polling (1161J2、What is 。
3、the IP address of gaia.cs.umass.edu? On what port number is it sending and receivi ng TCP segme nts for this connection?4、What is the sequenee number of the TCP SYN segment that is used to initiate the TCP connection betwee n the clie nt computer and gaia.cs.umass.edu? What is itin the segment that 。
4、identifies the segment as a SYN segment?Sequence number: 0(relative sequence number).1. Syn: Set5、What is the sequenee number of the SYNACK segment sent by gaia.cs.umass.eduto the clie nt computer in reply to the SYN?What is the value of theACK no wledgeme nt field in the SYNACK segme nt?How did gai。
5、a.cs.umass.edudetermine that value? What is it in the segment that identifies the segment as a SYNACK segme nt?sequence number: 0(re1 ative sequence number)Ackrtowledgemenx number: 1(rel arive ack number)SYNACK = ACK + Len.1=Acknowledgement: Set6、What is the sequenee number of the TCP segment contai。
6、ning the HTTP POSTcomma nd? Note that in order to find the POST comma nd, you?ll n eed 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.sequence number: 1 (relative sequence number7、Consider the TCP segment containin。
7、g the HTTP POST as the first segment in the TCP connection. What are the seque nee nu mbers of the first six segme nts in the TCP co nn ectio n (in cludi ng the segme nt con ta ining the HTTP POST)? At what time was each segment sent? When was the ACK for each segment received? Give n the differe ne。
8、e betwee n whe n each TCP segme nt was sent, and whe n itsack no wledgeme nt 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 tothe measured RTT fo。
9、r the first segme nt, and the n is computed using the EstimatedRTT equati on on page 249 for all subseque nt segme nts.8、What is the length of each of the first six TCP segments?序號1seqnu mber Len56525661460320261460434861460549461460664061460Sen t-time6 02 的 77To54&Z60.0546900.077400.078157Received-。
10、time0.0539370.L240850.L6911S0.2172&?0.267802SimpleRTT0.0274400.0355570.0700590.1144280.1398940.189645EstimatedRTT0.0274400000 0284545ERTT ,= 0.875*ERTT + 0.125*SRTT;9、What is the minimum amount of available buffer space advertised at the receiv for the en tire trace? Does the lack of receiver buffer。
11、 space ever throttle thesen der?window size value:NO!10、 Are there any retransmitted segments in the trace file? What did you check for (in the trace) in order to an swer this questi on?NO ! Because every seque nee nu mber of ACK are differe nt.11、How much data does the receiver typically ack no wle。
12、dge in an ACK? Can you iden tify cases where the receiver is ACKi ng every other received segme nt (see Table 3.2 on page 257 in the text).12、What is the throughput (bytes transferred per unit time) for the TCP connection? Expla in how you calculated this value.35714. 28513、Use the Time-Sequence-Gra。
13、ph(Stevens) plotting tool to view the sequenee nu mber versus time plot of segme nts being sent from the clie nt to the gaia.cs.umass.edu server. Can you ide ntify where TCP?s slowstart phase beg ins and ends, and where congestion avoidanee takes over? Comment on ways in which the measured data differs from the idealized behavior of TCP that we?ve studied in the text.SequencenunibrBTin4/Squnc Grph Stvanz)150000 100000 50000 0 51 52.02 53 04 55.05 5。
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