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1、Chapter 1 Review Questions1. There is no difference. Tluougliout this text, the words “host" and “end system" are used lnteicliangeablv. End systems include PCs, workstations, W eb seiveis, mail seiveis, Inteinet-coiinected PDAs. W ebTVs, etc.3. A networking progiam usually lias two piogia

2、ms, each running on a different host, conuiiuiiicating with each other. The piogiani that lmtiates the conunumcation is the clien匚 Typically, the client progiam requests and receives seivices fiom the seivei progiam 22. application-layer message: data wluch an application wants to send and passed on

3、to the tianspoit layer; tianspoit-layei segment: generated by the tianspoit layer and encapsulates application-layei message with tianspoit layer header; nemroik-layei datagiam: encapsulates tianspoit-lavei segment with a nemroik-layei header; lniklayei fiame: encapsulates nemroik-layei datagram wit

4、h a lnik-laver header23. Routers process layers 1 tluougli 3(This is a little bit of a white lie, as modern routers sometnnes act as filewalls oi caclimg components, and process layer four as well. Link layer switches process layers 1 tluougli 2. Hosts process all five layersChapter 1 ProblemsProble

5、m 6.a d m s prop = / secondsb d L R trans = / secondsc d (111/ sL/Rend to end = + ? ? seconds.d The bit is just leaving Host A.e The fiist bit is in the lmk and lias not reached Host B.f The fiist bit has reaclied Host B.g W ant(2.5 10 89328 10100 83/ =工=S =Rm L km.Problem 10.It takes LN / R seconds

6、 to transmit the N packets Thus, the buffer is empty when abatch ofN packets amveThe fiist of the N packets lias no queueing delay The 2nd packet lias a queueingdelayof L/R seconds. The n th packet lias a delay of (n ?1L/R seconds. The averagedelay is1(1/111(11 0?卜？=卜=?NRNNLRN11LRN11LRLPioblem 11.a

7、The tiansnussion delay is L/R The total delay isILRRLRIIL?+ =? 1/(1b Let x = L/R.Total delay =JaxPioblem 14.a 40,000 bitsb 40,000 bitsc the bandwidth-delay product of a lnik is the maxmium nunibei of bits that can be111the luikd 1 bit is 250 meters long, which is longer than a football fielde s/RPro

8、blem 18.a 150 msecb 1,500,000 bitsc 600,000,000 bitsChapter 2 Review Questions1. The W eb: HTTP; file tiansfer: FTP; remote login: Teliiet; Network News: NNTP; email: SMTP.5. No. As stated in the text, all conuiiunication sessions have a client side and a seiver side hi a P2P file-shaimg application

9、, the peer that is receiving a file is Typically the client and the peer that is sending the file is typically the seivei 6. The IP address of the destination host and the port number of the destination socket.10. The applications associated with those protocols lequire that all application data be

10、leceived in the collect oidei and without gaps. TCP provides this senice whereas UDP does not.22. With the UDP serve】，there is no welcoming socket, and all data fiom different clients enteis the serve】 tluough this one socke匚 With the TCP serve： there is a welcommg socket, and each tune a client mit

11、iates a connection to the seivei, a new socket is created. Thus, to support n simultaneous connections, the serve】 would need n+1 sockets. Chaptei 2 ProblemsProblem 1.aFbTcFdFProblem 4.Application layer protocols: DNS and HTTPTianspoit layer piotocols: UDP foi DNS; TCP for HTTPProblem 9.a The tune t

12、o transmit an object of size L over a link or rate R is L. R. The average tune is the average size of the object divided by R:-=(900,000 bits/( 1,500.000 bits/sec = .6 secThe traffic intensity on the link is (1.5 requests/sec(.6 msec/iequest = .9. Thus, theaverage access delay is (.6 sec/(l -.9 = 6

13、seconds The total average response tune is tlieiefoie 6 sec + 2 sec = 8 sec.b The traffic intensity on the access link is reduced by 40% smce the 40% of the requests are satisfied within the mstitutional netwoik. Thus the average access delayis (.6 sec/l 韶(.6(.9 = 1.2 seconds. The response tune is a

14、pproximately zero if the request is satisfied by the cache (which happens with probability .4; the average response tune is 1.2 sec + 2 sec = 3.2 sec for cache nusses (wlucli happens 60% of the tmie. So the average response tune is (.4(0 sec 十(.6(3.2 sec = 1.92 seconds. Thus the average response tun

15、e is reduced fiom 8 sec to 1.92 sec. Chapter 3 Review Questions1. Source port numbei y and destination port number x.2. An application developer may not want its application to use TCP's congestion contro 1, which can throttle the application's sending late at times of congestion. Often, des

16、igners of IP telephony and IPvideoconfeience applications choose to nm then applications over UDP because they want to avoid TCP's congestion control.Also, some applications do not need the reliable data tiansfei provided by TCP.4. a false b false c tnie d false e true f false g false5. a 20 byt

17、es b ack number = 90Chapter 3 ProblemsPioblem 1. source port numbers destination port nunibeis a A U S 467 23bB U S513 23c S U A 23 467dS U B23 513e Y es.fNo.Problem 3.110001010111000001010101十000100010100110011000101十One's complement =1110 1110.To detect enois, the leceivei adds the four words

18、(the three ongnial words and the checksum If the sum contams a zero, the receiver knows there lias been an eiior. All one-bit enois will be detected, but two-bit enois can be undetected (e.g., if the last digit of the fiist word is converted to a 0 and the last digit of the second word is convened t

19、o a 1.Problem 19.a Tme. Suppose the sender has a window size of 3 and sends packets 1, 2, 3 at 10 At tl (tl > tO the leceivei ACKS 1, 2, 3. At t2 (t2 > tl the sender tunes out and resends 1,2、3 At t3 the leceivei receives the duplicates and le-ackiiowledges 1， 2, 3. At t4 the sender receives t

20、heACKs that the leceivei sent at tl and advances its nwindow to 4. 5, 6. At t5 the sender receives the ACKs 1, 2, 3 the leceivei sent at t2 These ACKs are outside its window. b True. By essentially the same scenano as in (a. c Tme. d True. Note that with a window size of 1, SR, GBN, and the alternat

21、ing bit protocol are fiinctionally equivalent. The window size of 1 precludes the possibility of out-ofoidei packets (within the window. A cumulative ACK is just ail ordinaiy ACK in this situation, since it can only refer to the single packet within the window. Chapter 4 Review Questions 3. Fonvardm

22、g is about moving a packet from a router's input link to the appropriate output link Routing is about deternuiiing the end-to-routes between sources and destmations 8. switching via memoiy; switching via a bus; switching via an liiteiconnection netwoik 12. Yes. They have one address foi each int

23、erface. 16. 50% oveiliead 34. False Chapter 4 Problems Pioblem 7. a Prefix Match Link Interface 11100000 0 11100001 00000000 1 11100001 2 othenvise 3 b Prefix match for fiist address is 4th entiy: link mterface 3 Prefix match for second address is 2nd entiy: link mteiiace 1 Prefix match fbi fiist ad

24、dress is 3rd entiy: link mteiface 2 Problem 8. Destination Address Range Link Liteiface 00000000 tluough 0 00111111 01000000tluough 1 01111111 10000000 tluough 2 10111111 11000000 tluough 3 11111111 liumbei of addresses m each range = 26 = 64 Pioblem 13. Any IP address in range 5 to 101

25、.101.101.127 Four equal size subnets: 4/2& 0/28, 6/2& 12/28 Problem 16. MP3 file size = 4 million bytes. Assume the data is earned in TCP segments, with each TCP segment also having 20 bytes of header. Then each datagram can cany 1500- 40=1

26、460 bytes of the MP3 file Number of fragments requii ed = 2740 1460 4 106 = ? ?/ = All but the last fragment will be 1,500 bytes; the last fiagment will be 1060+40 = 1100 bytes. The offsets will be multiples of 185 (as in example m Section 4.4.1. Chapter 5 Review Questions 1 Although each link guarantees that an IP datagram sent over the link will be received at the othei end of the lnik without enois, it is not guaranteed that IP datagrams will ainve at the ultmiate destination m the piopei older. With IP. datagiains m the same TCP comiection can take different