2022年网络复习 .pdf

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1、HW1 Problem 1 Design and describe an application-level protocol to be used between an automatic teller machine and a banks centralized computer. You protocol should allow a users card and password to be verified, the account balance (which is maintained at the centralized computer) to be queried, an

2、d an account withdrawal to be made (that is, money disbursed to the user). Your protocol entities should be able to handle the all-too-common case in which there is not enough money in the account to cover the withdrawal. Specify your protocol by listing the messages exchanged and the action taken b

3、y the automatic teller machine or the banks centralized computer on transmission and receipt of messages. Sketch the operation of your protocol for the case of a simple withdrawal with no errors, using a diagram similar to that in Figure 1.2. Explicitly state the assumption made by your protocol abo

4、ut underlying end-to-end transport service. Answer：automatic teller machinecentralized computerStart(UserID)Error-Unvalid UserIDEndError-Not Enough BalanceShowBalanceError-WrongPWDcheck UserIDcheck UserPWDcheck Enough BalanceDispenses MoneyPWDPWD(UserPWD)Right(ChooseService)BalanceWithDrawl(Amount)E

5、noughEnd名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 1 页，共 27 页 - - - - - - - - - Messages from automatic teller machine to centralized computer Start(UserID) Let server know that there is a card in the automatic teller machineautomatic teller machinetransmits User

6、ID to centralized computerPWD(UserPWD) User enters PWD, which is sent to centralized computerBalance User requests balance WithDrawl(Amount) User asks to withdraw money End user all done Messages from centralized computer to automatic teller machine PWD Ask user for UserPWd Right(ChooseService) last

7、 requested operation (PASSWD, WITHDRAWL) OK Error last requested operation (UserID,PWD, WithDrawl) in Error ShowBalance(Balance)sent in response to BALANCE request Enough sent in response to WithDrawl request End user done, display welcome screen at automatic teller machine Problem 5 Consider sendin

8、g a packet of F bits over a path of Q links. Each link transmits at R bps. The network is lightly loaded so that there are no queuing delays. Propagation delay is negligible. a. Suppose the network is a packet-switched virtual circuit network. Denote the VC set-up time by ts seconds. Suppose the sen

9、ding layers add a total of h bits of header to each packet. How long does it take to send the file from source to destination? b. Suppose the network is a packet-switched datagram network and a connectionless service is used. Now suppose each packet has 2h bits of header. How long does it take to se

10、nd the file? c. Finally, suppose that the network is a circuit-switched network. Further suppose that the transmission rate of the circuit between source and destination is R bps. Assuming ts set-up time and h bits of header appended to the entire file, how long does it take to send the file? Answer

11、：(1). The length of each packet: (F+h) bits The time ench packet takes: (F+h)/R seconds Q links total time: Q(F+h)/R secondsThe total time: st+Q(F+h)/R seconds(2) Q(F+h)/R seconds (3) st+(F+h)/R seconds Problem6 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 2 页，共 27

12、 页 - - - - - - - - - This elementary problem begins to explore propagation delay and transmission delay, two central concepts in data networking. Consider two hosts, Host A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meters, and suppose the propagat

13、ion speed along the link is s meters/sec. Host A is to send a packet of size L bits to Host B. a. Express the propagation delay, dprop in terms of m and s. b. Determine the transmission time of the packet, dtrans in terms of L and R.c. Ignoring processing and queuing delays, obtain an expression for

14、 the end-to-end delay. d. Suppose Host A begins to transmit the packet at time t = 0. At time t = dtrans, where is the last bit of the packet?e. Suppose dprop is greater than dtrans. At time t = dtrans, where is the first bit of the packet? f. Suppose dprop is less than dtrans. At time t = dtrans, w

15、here is the first bit of the packet? g. Suppose s = 2.5*108, L = 100 bits, and R = 28 kbps. Find the distance m so that dprop equals dtrans. Answer：a)smdprop/seconds. b)RLdtrans/seconds. c)/(RLsmdendtoendseconds. d) The bit is just leaving Host A. e) The first bit is in the link and has not reached

16、Host B. f) The first bit has reached Host B. g) Want 8 9 3105.2102810083SRLmkm. Problem7 In this problem we consider sending voice from Host A to Host B over a packet-switched network (for example, Internet phone). Host A converts on-the-fly analog voice to a digital 64- Kbps bit stream. Host A then

17、 groups the bits into 48-byte packets. There is one link between host A and B; its transmission rate is 1 Mbps and its propagation delay is 2 mesc. As soon as Host A gathers a packet, it sends it to Host B. As soon as Host B receives an entire packet, it converts the packets bits to an analog signal

18、. How much time elapses from the time a bit is created (from the original analog signal at A) until a bit is decoded (as part of the analog signal at B)? Answer：Before the first bit in a packet can be transmitted, all of the bits in the packet must be generated. This requires 31064848sec=6msec. 名师资料

19、总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 3 页，共 27 页 - - - - - - - - - The time required to transmit the packet is 6101848sec=384sec. Propagation delay = 2 msec. The delay until decoding is 6msec +384sec + 2msec = 8.384msec Problem 10 Consider the queuing delay in a

20、 router buffer (preceding an outbound link). Suppose all packets are L bits, the transmission rate is R bps, and that N packets arrive to the buffer every LN/R seconds. Find the average queuing delay of a packet. Answer：RNLnNRLRLnNNnNn2)1(/) 1(1101Problem 13 Perform a traceroute between source and d

21、estination on the same continent at three different hours of the day. a. Find the average and standard deviation of the round-trip delays at each of the three hours. b. Find the number of routers in the path at each of the three hours. Did the paths change during any of the hours? c. Try to identify

22、 the number of ISP networks that the Traceroute packets pass through from source to destination. Routers with similar names and/or similar IP addresses should be considered as part of the same ISP. In your experiments, do the largest delays occur at peering interfaces between adjacent ISPs? d. Repea

23、t the above for a source and destination on different continents. Compare the intracontinent and intercontinent results. Answer：名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 4 页，共 27 页 - - - - - - - - - (1) the average and standard deviation of the round-trip delays

24、: (20/3+18/3+21/3+28/3+18/3+17/3+29/3)/7+(10/3+4/3+8/3+19/3+53/3+49/3+34/3)/7)/2=7.81 ms the standard deviation of the round-trip delays: 0.3613 (2) the number of routers in first experiment: 7 the number of routers in second experiment: 7 The paths din not chang. (3) In my experiment,123.126.6.198

25、and 123.126.248.74 are similar IP address. They should be considered as part of the same ISP. The largest delays did notoccur at peering interfaces between adjacent ISPs. Problem 14 Suppose two hosts, A and B, are separated by 10,000 kilometers and are connected by a direct link of R = 1 Mbps. Suppo

26、se the propagation speed over the link is 2.5* 810meters/sec. a. Calculate the bandwidth-delay product, R* proptb. Consider sending a file of 400,000 bits from Host A to Host B. Suppose the file is sent continuously as one big message. What is the maximum number of bits that will be in the link at a

27、ny given time? c.Provide an interpretation of the bandwidth-delay product. d. What is the width (in meters) of a bit in the link? Is it longer than a football field? e. Derive a general expression for the width of a bit in terms of the propagation speed s, the transmission rate R, and the length of

28、the link m. 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 5 页，共 27 页 - - - - - - - - - Answer：a) 40000 bits b)40000 bitsc)The bandwidth-delay product of a link is the maximum number of bits that can be in the linkd)1 bit is 250 meters long, which is longer than a fo

29、otball fielde)s/RProblem17 Refer again to problem 14. a. How long does it take to send the file, assuming it is sent continuously? b. Suppose now the file is broken up into 10 packets with each packet containing 20,000bits. Suppose that each packet is acknowledged by the receiver and the transmissio

30、n time of an acknowledgment packet is negligible. Finally, assume that the sender cannot send a packet until the preceding one is acknowledged. How long does it take to send the file? c.Compare the results from (a) and (b). Answer：a) ttrans + tprop = 400 msec + 40 msec = 440 msec b) 10 * (ttrans + 2

31、 tprop) = 10*(40 msec + 80 msec) = 1.2 sec Problem 20 In modern packet-switched networks, the source host segments long, application-layer messages (for examples, an image or a music file) into smaller packets and sends the packets into the network. The receiver then reassembles the packets back int

32、o the original message. We refer to this process as message segmentation. Figure 1.21 illustrates the end-to-end transport of a message with and without message segmentation. Consider a message that is 7.5*10e6 bits long that is to be sent from source to destination in Figure1.21. Suppose each link

33、in the figure is 1.5Mbps. Ignore propagation, queueing, and processing delays a. Consider sending the message from source to destination without message segmentation. How long does it take to move the message from source to destination without message segmentation? How long does it take to move the

34、message from the source host to the first packet switch, what is the total time to move the message from source host to destination host? b. Now suppose that the message is segmented into 5,000 packets, with each packet 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第

35、6 页，共 27 页 - - - - - - - - - being 1,500 bits long. How long does it take to move the first packet from the source host to the first packet switch? When the first packet is being sent from the first switch to the second switch, the second packet is being sent from the source host to the second switc

36、h. At what time will the second packet be fully received at the first switch? c. How long does it take to move the file from source host to destination host when message segmentation is used? Compare this result with your answer in a part (a) and comment. d. Discuss the drawbacks of message segmenta

37、tion. Answer：a) Time to send message from source host to first packet switch = sec5sec105 .1105. 766. With store-and-forward switching, the total time to move message from source host to destination host = sec153sec5hopsb) Time to send 1st packet from source host to first packet switch = . sec1sec10

38、5 .1105.163m. Time at which second packet is received at the first switch = time at which first packet is received at the second switch = sec2sec12mmc) Time at which 1st packet is received at the destination host = . sec33sec1mhopsm. After this, every 1msec one packet will be received; thus time at

39、which last (5000th) packet is received = sec002.5sec1*4999sec3mm. d) Comparement: (1)Packets have to be put in sequence at the destination. (2)Message segmentation results in many smaller packets. Since header size is usually the same for all packets regardless of their size, with message segmentati

40、on the total amount of header bytes is more. Problem 22 Consider sending a large file of F bits from Host A to Host B. There are two links (and one switch) between A and B, and the links are uncongested (that is, no queuing delays). Host A segments the file into segments of S bits each and adds 40 b

41、its of header to each segment, forming packets of L = 40 + S bits. Each link has a transmission rate of R bps. Find the value of S that minimizes the delay of moving the packet from Host A to Host B. Neglect propagation delay. 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - -

42、 - - 第 7 页，共 27 页 - - - - - - - - - Answer：Time at which the first packet is received at the destination = 240RSsec. After this, one packet is received at destination every RS40sec. Thus delay in sending the whole file= ) 1(40)40()1(240SFRSRSSFRSdelayTo calculate the value of S which leads to the mi

43、nimum delay, FSRSSSRFdelaydSd4001)401(02HW2 Problem 1True or false? A.When a user requests a Web page that consists of some text and two images. For this page, the client will send one request message and receive three response messages. B.Two distinct Web pages(for example,www.mit.edu/reaearch.html

44、 and www.mit.edu/students.html)can be send over the same persistent connection. Answer A.False After the client receive the web page, the client will send three request message and receive three response messages. B.True Problem 6. Suppose within your Web browser you click on a link to obtain a Web

45、page. The IP address for the associated URL is not cached in your local host, so a DNS look-up is necessary to obtain the IP address. Suppose that n DNS servers are visited before your host receives the OP address form DNS; the successive visits incur an RTT of 1RTT ,nRTT .Further suppose that the W

46、eb page associated with the link contains exactly one object, consisting of a small amount of HTML text. Let 0RTTdenote the RTT between the local host and the server containing the object. Assuming zero transmission time of the object, how much time elapses form when the client clicks on the link un

47、til the client receives the objects? Answer:1RTT +nRTT +2*0RTTProblem 7. Referring to Problem 6, suppose the HTML file references three very small objects on the same server. Neglecting transmission times, how much time elapses with 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - -

48、 - - - - - 第 8 页，共 27 页 - - - - - - - - - a.Non-persistent HTTP with no parallel TCP connections? b.Non-persistent HTTP with no parallel connections? c.Persistent HTTP with pipelining? Answer a.1RTT +nRTT +2*0RTT +3*2*0RTT =1RTT +nRTT +8*0RTTb.It still uses TCP,So 1RTT +nRTT +2*0RTT +3*2*0RTT =1RTT

49、+nRTT +8*0RTTIf it is Non-persistent HTTP with parallel connections 1RTT +nRTT +2*0RTT +2*0RTT =1RTT +nRTT +4*0RTTc.1RTT +nRTT +2*0RTT +0RTT =1RTT +nRTT +3*0RTTProblem 9. Consider Figure 2.11,for which there is an institutional network connected to the Internet. Suppose that the average object size

50、is 900,000 bits and that the average request rate from institution s browsers to the origin servers is 1.5 requests per second. Also suppose that the amount of time it takes from when the router on the Internet side of the access link forwards an HTTP request until it receives the response in two se