計算機網(wǎng)絡(luò)——自頂向下方法與Internet特色(第三版)習(xí)題答案

1、計算機網(wǎng)絡(luò)自頂向下方法與Internet特色(第三版)習(xí)題答案 導(dǎo)讀：就愛閱讀網(wǎng)友為您分享以下“計算機網(wǎng)絡(luò)自頂向下方法與Internet特色(第三版)習(xí)題答案”資訊，希望對您有所幫助，感謝您對的支持! Computer Networking: A Top-Down Approach Featuring the Internet, 3rd Edition Solutions to Review Questions and Problems Version Date: April 26, 2005 This document contains the solutions to review que

2、stions and problems for the 3rd edition of Computer Networking: A Top-Down Approach Featuring the Internet by Jim Kurose and Keith Ross. These solutions are being made available to instructors ONLY. Please do NOT copy or distribute this document to others (even other instructors). Please do not post

3、 any solutions on a publicly-available Web site. Well be happy to provide a copy (up-to-date) of this solution manual ourselves to anyone who asks. All material copyright 1996-2005 by J.F. Kurose and K.W. Ross. All rights reserved Chapter 1 Review Questions 1. There is no difference. Throughout this

4、 text, the words “host” and “end system” are used interchangeably. End systems include PCs, workstations, Web servers, mail servers, Internet-connected PDAs, WebTVs, etc. 2. Suppose Alice, an ambassador of country A wants to invite Bob, an ambassador of country B, over for dinner. Alice doesnt simpl

5、y just call Bob on the phone and say, “come to our dinner table now”. Instead, she calls Bob and suggests a date and time. Bob may respond by saying hes not available that particular date, but he is available another date. Alice and Bob continue to send “messages” back and forth until they agree on

6、a date and time. Bob then shows up at the embassy on the agreed date, hopefully not more than 15 minutes before or after the agreed time. Diplomatic protocols also allow for either Alice or Bob to politely cancel the engagement if they have reasonable excuses. 3. A networking program usually has two

7、 programs, each running on a different host, communicating with each other. The program that initiates the communication is the client. Typically, the client program requests and receives services from the server program. 4. The Internet provides its applications a connection-oriented service (TCP)

8、and a connectionless service (UDP). Each Internet application makes use of one these two services. The two services will be discussed in detail in Chapter 3. Some of the principle characteristics of the connection-oriented service are: Two end-systems first “handshake” before either starts to send a

9、pplication data to the other. Provides reliable data transfer, i.e., all application data sent by one side of the connection arrives at the other side of the connection in order and without any gaps. Provides flow control, i.e., it makes sure that neither end of a connection overwhelms the buffers i

10、n the other end of the connection by sending to many packets to fast. Provides congestion control, i.e., regulates the amount of data that an application can send into the network, helping to prevent the Internet from entering a state of grid lock. The principle characteristics of connectionless ser

11、vice are: No handshaking No guarantees of reliable data transfer No flow control or congestion control 5. Flow control and congestion control are two distinct control mechanisms with distinct objectives. Flow control makes sure that neither end of a connection overwhelms the buffers in the other end

12、 of the connection by sending to many packets to fast. Congestion control regulates the amount of data that an application can send into the network, helping to prevent congestion in the network core (i.e., in the buffers in the network routers). 6. The Internets connection-oriented service provides

13、 reliable data transfer by using acknowledgements and retransmissions. When one side of the connection doesnt receive an acknowledgement (from the other side of the connection) for a packet it transmitted, it retransmits the packet. 7. A circuit-switched network can guarantee a certain amount of end

14、-to-end bandwidth for the duration of a call. Most packet-switched networks today (including the Internet) cannot make any end-to-end guarantees for bandwidth. 8. In a packet switched network, the packets from different sources flowing on a link do not follow any fixed, pre-defined pattern. In TDM c

15、ircuit switching, each host gets the same slot in a revolving TDM frame. 9. At time t0 the sending host begins to transmit. At time t1 = L/R1, the sending host completes transmission and the entire packet is received at the router (no propagation delay). Because the router has the entire packet at t

16、ime t1, it can begin to transmit the packet to the receiving host at time t1. At time t2 = t1 + L/R2, the router completes transmission and the entire packet is received at the receiving host (again, no propagation delay). Thus, the end-to-end delay is L/R1 + L/R2. 10. In a VC network, each packet s

17、witch in the network core maintains connection state information for each VC passing through it. Some of this connection state information is maintained to a VC-number translation table. (See page 25) 11. The cons of VCs include (i) the need to have a signaling protocol to set-up and tear-down the V

18、Cs; (ii) the need to maintain connection state in the packet switches. For the pros, some researchers and engineers argue that it is easier to provide QoS services - such as services that guarantee a minimum transmission rate or services that guarantee maximum end-to-end packet delay when VCs are us

19、ed. 12. 1. Dial-up modem over telephone line: residential; 2. DSL over telephone line: residential or small office; 3. Cable to HFC: residential; 4. 100 Mbps switched Etherent: company; 5. Wireless LAN: mobile; 6. Cellular mobile access (for example, WAP): mobile 13. A tier-1 ISP connects to all oth

20、er tier-1 ISPs; a tier-2 ISP connects to only a few of the tier-1 ISPs. Also, a tier-2 ISP is a customer of one or more tier-1 14. A POP is a group of one or more routers in an ISPs network at which routers in other ISPs can connect. NAPs are localized networks at which many ISPs (tier-1, tier-2 and

21、 lower-tier ISPs) can interconnect. 15. HFC bandwidth is shared among the users. On the downstream channel, all packets emanate from a single source, namely, the head end. Thus, there are no collisions in the downstream channel. 16. Ethernet LANs have transmission rates of 10 Mbps, 100 Mbps, 1 Gbps

22、and 10 Gbps. For an X Mbps Ethernet (where X = 10, 100, 1,000 or 10,000), a user can continuously transmit at the rate X Mbps if that user is the only person sending data. If there are more than one active user, then each user cannot continuously transmit at X Mbps. 17. Ethernet most commonly runs o

23、ver twisted-pair copper wire and “thin” coaxial cable. It also can run over fibers optic links and thick coaxial cable. 18. Dial up modems: up to 56 Kbps, bandwidth is dedicated; ISDN: up to 128 kbps, bandwidth is dedicated; ADSL: downstream channel is .5-8 Mbps, upstream channel is up to 1 Mbps, ba

24、ndwidth is dedicated; HFC, downstream channel is 10-30 Mbps and upstream channel is usually less than a few Mbps, bandwidth is shared. 19. The delay components are processing delays, transmission delays, propagation delays, and queuing delays. All of these delays are fixed, except for the queuing de

25、lays, which are variable. 20. Five generic tasks are error control, flow control, segmentation and reassembly, multiplexing, and connection setup. Yes, these tasks can be duplicated at different layers. For example, error control is often provided at more than one layer. 21. The five layers in the I

26、nternet protocol stack are from top to bottom the application layer, the transport layer, the network layer, the link layer, and the physical layer. The principal responsibilities are outlined in Section 1.7.1. 22. application-layer message: data which an application wants to send and passed onto th

27、e transport layer; transport-layer segment: generated by the transport layer and encapsulates application-layer message with transport layer header; network-layer datagram: encapsulates transport-layer segment with a network-layer header; link-layer frame: encapsulates network-layer datagram with a

28、link-layer header. 23. Routers process layers 1 through 3. (This is a little bit of a white lie, as modern routers sometimes act as firewalls or caching components, and process layer four as well .) Link layer switches process layers 1 through 2. Hosts process all five layers. Chapter 1 Problems Pro

29、blem 1. There is no single right answer to this question. Many protocols would do the trick. Heres a simple answer below: Messages from ATM machine to ServerMsg name purpose- -HELO userid Let server know that there is a card in theATM machineATM card transmits user ID to ServerPASSWD passwd User ent

30、ers PIN, which is sent to server BALANCE User requests balanceWITHDRAWL amount User asks to withdraw moneyBYE user all done Messages from Server to ATM machine (display)Msg name purpose- -PASSWD Ask user for PIN (password)OK last requested operation (PASSWD, WITHDRAWL)OKERR last requested operation

31、(PASSWD, WITHDRAWL)in ERRORAMOUNT amt sent in response to BALANCE requestBYE user done, display welcome screen at ATMCorrect operation: client server HELO (userid) - (check if valid userid)- PASSWDPASSWD passwd - (check password)- OK (password is OK)BALANCE - AMOUNT amt WITHDRAWL amt - check if enou

32、gh $ to cover withdrawl - OKATM dispenses $BYE - BYE In situation when theres not enough money: HELO (userid) - (check if valid userid)- PASSWDPASSWD passwd - (check password)- OK (password is OK)BALANCE - AMOUNT amtWITHDRAWL amt - check if enough $ to cover withdrawl- ERR (not enough funds)error ms

33、g displayedno $ given outBYE - BYE Problem 2. a) A circuit-switched network would be well suited to the application described, because the application involves long sessions with predictable smooth bandwidth requirements. Since the transmission rate is known and not bursty, bandwidth can be reserved

34、 for each application session circuit with no significant waste. In addition, we need not worry greatly about the overhead costs of setting up and tearing down a circuit connection, which are amortized over the lengthy duration of a typical application session. b) Given such generous link capacities

35、, the network needs no congestion control mechanism. In the worst (most potentially congested) case, all the applications simultaneously transmit over one or more particular network links. However, since each link offers sufficient bandwidth to handle the sum of all of the applications data rates, n

36、o congestion (very little queueing) will occur. Problem 3. a) We can n connections between each of the four pairs of adjacent switches. This givesa maximum of 4n connections.b) We can n connections passing through the switch in the upper-right-hand corner andanother n connections passing through the

37、 switch in the lower-left-hand corner, giving a total of 2n connections. Problem 4. Tollbooths are 100 km apart, and the cars propagate at 100km/hr. A tollbooth services a car at a rate of one car every 12 seconds.(a) There are ten cars. It takes 120 seconds, or two minutes, for the first tollbooth

38、to service the 10 cars. Each of these cars have a propagation delay of 60 minutes before arriving at the second tollbooth. Thus, all the cars are lined up before the second tollbooth after 62 minutes. The whole process repeats itself for traveling between the second and third tollbooths. Thus the to

39、tal delay is 124 minutes. (b) Delay between tollbooths is 7*12 seconds plus 60 minutes, i.e., 61 minutes and 24 seconds. The total delay is twice this amount, i.e., 122 minutes and 48 seconds. Problem 5. a) The time to transmit one packet onto a link is R h L /) (+. The time to deliver the packet ov

40、er Q links isR h L Q /) (+. Thus the total latency is R h L Q t s /) (+.b) R h L Q /) 2(+c) Because there is no store-and-forward delays at the links, the total delay isR L h t s /) (+. Problem 6. a) s m d prop /= seconds.b) R L d trans /= seconds.c) ) /(R L s m d end to end += seconds.d) The bit is

41、 just leaving Host A.e) The first bit is in the link and has not reached Host B.f) The first bit has reached Host B.g) Want()893105. 2102810083=S R L m km.Problem 7. Consider the first bit in a packet. Before this bit can be transmitted, all of the bits in the packet must be generated. This requires

42、31064848sec=6msec.The time required to transmit the packet is6101848sec=384sec.Propagation delay = 2 msec.The delay until decoding is 6msec +384sec + 2msec = 8.384msec A similar analysis shows that all bits experience a delay of 8.384 msec. Problem 8. a) 10 users can be supported because each user r

43、equires one tenth of the bandwidth. b) 1. 0=p .c) ()n n p p n 40140. d) ()=90401401n n n p p n . We use the central limit theorem to approximate this probability. Let j X be independent random variables such that ()p X P j =1.(P “11 or more users”)=101401j j X P=9. 01. 04069. 01. 040410401401j j j j

44、 X P X P ()16. 36. 6=Z P Z P 999. 0=when Z is a standard normal r.v. Thus (P “10 or more users”)001. 0.Problem 9. a) 10,000b) ()+=M N n n M n p p n M 11Problem 10. It takes R LN / seconds to transmit the N packets. Thus, the buffer is empty when a batch of N packets arrive. The first of the N packet

45、s has no queueing delay. The 2nd packet has a queueing delay of R L / seconds. The n th packet has a delay of R L n /) 1( seconds. The average delay is2) 1(2) 1(11/) 1(1101=N R L N N N R L n N R L R L n N N n Nn .Problem 11. a) The transmission delay is R L /. The total delay isIR L R L I R IL =+1/)

46、 1( b) Let R L x /=.Total delay = axx 1Problem 12. a) There are Q nodes (the source host and the 1N routers). Let q proc d denote theprocessing delay at the q th node. Let q R be the transmission rate of the q th link and let q q trans R L d /=. Let q prop d be the propagation delay across the q th

47、link. Then =+=Q q q propq trans q proc end to end d d d d 1.b) Let q queue d denote the average queueing delay at node q . Then=+=Q q q queue q prop q trans q proc end to end d d d d d 1. Problem 13. The command: traceroute -q 20 www.eurecom.fr will get 20 delay measurements from the issuing host to

48、 the host, www.eurecom.fr. The average and standard deviation of these 20 measurements can then be collected. Do you see any differences in your answers as a function of time of day? Problem 14. a) 40,000 bitsb) 40,000 bitsc) the bandwidth-delay product of a link is the maximum number of bits that c

49、an be in the linkd) 1 bit is 250 meters long, which is longer than a football fielde) s/RProblem 15. 25 bps Problem 16. a) 40,000,000 bitsb) 400,000 bitsc) .25 meters Problem 17. a) t trans + tprop = 400 msec + 40 msec = 440 msecb) 10 * (ttrans + 2 tprop ) = 10*(40 msec + 80 msec) = 1.2 sec Problem

50、18. a) 150 msecb) 1,500,000 bitsc) 600,000,000 bits Problem 19. Lets suppose the passenger and his/her bags correspond to the data unit arriving to the top of the protocol stack. When the passenger checks in, his/her bags are checked, and a tag is attached to the bags and ticket. This is additional

51、information added in theBaggage layer if Figure 1.20 that allows the Baggage layer to implement the service or separating the passengers and baggage on the sending side, and then reuniting them(hopefully!) on the destination side. When a passenger then passes through security, and additional stamp i

52、s often added to his/her ticket, indicating the at the passenger haspassed through a security check. This information is used to ensure (e.g., by later checks for the security information) secure transfer of people. Problem 20. a) Time to send message from source host to first packet switch =sec 5se

53、c 105. 1105. 766=. With store-and-forward switching, the total time to move message from source host to destination host = sec 153sec 5=hopsb) Time to send 1st packet from source host to first packet switch = .sec 1sec 105. 1105. 163m =. Time at which 2nd packet is received at the first switch = tim

54、e at which 1st packet is received at the second switch = sec 2sec 12m m = c) Time at which 1st packet is received at the destination host = .sec 33sec 1m hops m =. After this, every 1msec one packet will be received; thus time at which last (5000th ) packet is received =sec 002. 5sec 1*4999sec 3=+m

55、m . It can be seen that delay in using message segmentation is significantly less (almost 1/3rd ).d) Drawbacks:i. packets have to be put in sequence at the destination.ii. Message segmentation results in many smaller packets. Sinceheader size is usually the same for all packets regardless of theirsi

56、ze, with message segmentation the total amount of header bytesis more. Problem 21. Java Applet Problem 22. Time at which the 1st packet is received at the destination =240+R S sec. After this, one packet is received at destination every RS 40+sec. Thus delay in sending the whole file = ) 1(4040() 1(240+=+=SF R S R S S F R S delay To calcula