計算機網(wǎng)絡(luò)自頂向下方法習(xí)題答案

1、Computer Networking: A Top-Down Approach Featuring the Internet, 4th EditionSolutions to Review Questions and ProblemsVersion Date: October 29, 2007This document contains the solutions to review questions and problems for the 4th edition of Computer Networking: A Top-Down Approach Featuring the Inte

2、rnet 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 any solutions on a publicly-available Web site. Well be happy to provide a copy (up-to-date) of this

3、solution manual ourselves to anyone who asks.All material © copyright 1996-2007 by J.F. Kurose and K.W. Ross. All rights reservedChapter 1 Review Questions2Chapter 2 Review Questions13Chapter 3 Problems32Chapter 4 Review Questions53Chapter 5 Review Questions72Chapter 1 Review Questions1. There

4、is no difference. Throughout this 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, ove

5、r for dinner. Alice doesnt simply 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” bac

6、k and forth until they agree on 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 net

7、working program usually has two 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. 1. Dial-up modem over telephone line: reside

8、ntial; 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): mobile5. HFC bandwidth is shared among the users. On the downstream channel, all packets emanat

9、e from a single source, namely, the head end. Thus, there are no collisions in the downstream channel.6. Current possibilities include: dial-up; DSL; cable modem; fiber-to-the-home. 7. Ethernet LANs have transmission rates of 10 Mbps, 100 Mbps, 1 Gbps and 10 Gbps. For an X Mbps Ethernet (where X = 1

10、0, 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.8. Ethernet most commonly runs over twisted-pair copper wire and “thin” coaxial ca

11、ble. It also can run over fibers optic links and thick coaxial cable.9. 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, bandwidth is dedicated; HFC, downstream channel is 10-

12、30 Mbps and upstream channel is usually less than a few Mbps, bandwidth is shared.10. There are two most popular wireless Internet access technologies today:a) Wireless LANIn a wireless LAN, wireless users transmit/receive packets to/from a base station (wireless access point) within a radius of few

13、 tens of meters. The base station is typically connected to the wired Internet and thus serves to connect wireless users to the wired network.b) Wide-area wireless access networkIn these systems, packets are transmitted over the same wireless infrastructure used for cellular telephony, with the base

14、 station thus being managed by a telecommunications provider. This provides wireless access to users within a radius of tens of kilometers of the base station.11. A circuit-switched network can guarantee a certain amount of end-to-end bandwidth for the duration of a call. Most packet-switched networ

15、ks today (including the Internet) cannot make any end-to-end guarantees for bandwidth. 12. In a packet switched network, the packets from different sources flowing on a link do not follow any fixed, pre-defined pattern. In TDM circuit switching, each host gets the same slot in a revolving TDM frame.

16、 13. 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 time t1, it can begin to transmit the packet to the receiving host at ti

17、me 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/R2.14. A tier-1 ISP connects to all other tier-1 ISPs; a tier-2 ISP connects to only a few of the tier-1 ISPs

18、. Also, a tier-2 ISP is a customer of one or more tier-1.15. a) 2 users can be supported because each user requires half of the link bandwidth. b) Since each user requires 1Mbps when transmitting, if two or fewer users transmit simultaneously, a maximum of 2Mbps will be required. Since the available

19、 bandwidth of the shared link is 2Mbps, there will be no queuing delay before the link. Whereas, if three users transmit simultaneously, the bandwidth required will be 3Mbps which is more than the available bandwidth of the shared link. In this case, there will be queuing delay before the link. c) P

20、robability that a given user is transmitting = 0.2 d) Probability that all three users are transmitting simultaneously = = (0.2)3 = 0.008. Since the queue grows when all the users are transmitting, the fraction of time during which the queue grows (which is equal to the probability that all three us

21、ers are transmitting simultaneously) is 0.008. 16. The delay components are processing delays, transmission delays, propagation delays, and queuing delays. All of these delays are fixed, except for the queuing delays, which are variable.17. Java Applet18. 10msec; d/s; no; no19. a) 500 kbps b) 64 sec

22、onds c) 100kbps; 320 seconds20. End system A breaks the large file into chunks. To each chunk, it adds header generating multiple packets from the file. The header in each packet includes the address of the destination: end system B. The packet switch uses the destination address to determine the ou

23、tgoing link. Asking which road to take is analogous to a packet asking which outgoing link it should be forwarded on, given the packets address.21. Java Applet22. Five generic tasks are error control, flow control, segmentation and reassembly, multiplexing, and connection setup. Yes, these tasks can

24、 be duplicated at different layers. For example, error control is often provided at more than one layer.24. Application-layer message: data which an application wants to send and passed onto the transport layer; transport-layer segment: generated by the transport layer and encapsulates application-l

25、ayer 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 link-layer header. 25. Routers process layers 1 through 3. (This is a little bit of a white lie, as modern

26、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.26. a) Virus Requires some form of human interaction to spread. Classic example: E-mail viruses.b)WormsNo user replication needed.

27、Worm in infected host scans IP addresses and port numbers, looking for vulnerable processes to infect.c) Trojan horseHidden, devious part of some otherwise useful software.27. Creation of a botnet requires an attacker to find vulnerability in some application or system (e.g. exploiting the buffer ov

28、erflow vulnerability that might exist in an application). After finding the vulnerability, the attacker needs to scan for hosts that are vulnerable. The target is basically to compromise a series of systems by exploiting that particular vulnerability. Any system that is part of the botnet can automa

29、tically scan its environment and propagate by exploiting the vulnerability. An important property of such botnets is that the originator of the botnet can remotely control and issue commands to all the nodes in the botnet. Hence, it becomes possible for the attacker to issue a command to all the nod

30、es, that target a single node (for example, all nodes in the botnet might be commanded by the attacker to send a TCP SYN message to the target, which might result in a TCP SYN flood attack at the target). 28. Trudy can pretend to be Bob to Alice (and vice-versa) and partially or completely modify th

31、e message(s) being sent from Bob to Alice. For example, she can easily change the phrase “Alice, I owe you $1000” to “Alice, I owe you $10,000”. Furthermore, Trudy can even drop the packets that are being sent by Bob to Alice (and vise-versa), even if the packets from Bob to Alice are encrypted. Cha

32、pter 1 Problems:Problem 1. There is no single right answer to this question. Many protocols would do the trick. Here's a simple answer below:Messages from ATM machine to ServerMsg namepurpose-HELO <userid>Let server know that there is a card in the ATM machineATM card transmits user ID to

33、ServerPASSWD <passwd>User enters PIN, which is sent to serverBALANCEUser requests balanceWITHDRAWL <amount>User asks to withdraw moneyBYEuser all doneMessages from Server to ATM machine (display)Msg namepurpose-PASSWDAsk user for PIN (password)OKlast requested operation (PASSWD, WITHDRAW

34、L) OKERRlast requested operation (PASSWD, WITHDRAWL) in ERRORAMOUNT <amt>sent in response to BALANCE requestBYEuser done, display welcome screen at ATMCorrect operation:client serverHELO (userid)->(check if valid userid)<-PASSWDPASSWD <passwd>->(check password)<-OK (password

35、is OK)BALANCE-><-AMOUNT <amt>WITHDRAWL <amt>->check if enough $ to cover withdrawl<-OKATM dispenses $BYE-><-BYEIn situation when there's not enough money:HELO (userid)->(check if valid userid)<-PASSWDPASSWD <passwd>->(check password)<-OK (password

36、is OK)BALANCE-><-AMOUNT <amt>WITHDRAWL <amt>->check if enough $ to cover withdrawl<-ERR (not enough funds)error msg displayedno $ given outBYE-><-BYEProblem 2. a) A circuit-switched network would be well suited to the application described, because the application invol

37、ves long sessions with predictable smooth bandwidth requirements. Since the transmission rate is known and not bursty, bandwidth can be reserved 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 d

38、own a circuit connection, which are amortized over the lengthy duration of a typical application session.b) Given such generous link capacities, the network needs no congestion control mechanism. In the worst (most potentially congested) case, all the applications simultaneously transmit over one or

39、 more particular network links. However, since each link offers sufficient bandwidth to handle the sum of all of the applications' data rates, no congestion (very little queuing) will occur.Problem 3.a) We can n connections between each of the four pairs of adjacent switches. This gives a maximu

40、m of 4n connections.b) We can n connections passing through the switch in the upper-right-hand corner and another n connections passing through the 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 tol

41、lbooth 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 to service the 10 cars. Each of these cars has a propagation delay of 60 minutes before arriving at the second tollbooth. Thus, all the cars are lined up b

42、efore the second tollbooth after 62 minutes. The whole process repeats itself for traveling between the second and third tollbooths. Thus the total 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

43、, i.e., 122 minutes and 48 seconds.Problem 5a) seconds.b) seconds.c) seconds.d) The bit is 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) Wantkm.Problem 6Consider the first bit in a packet. Before this bit can be transmitted, all

44、 of the bits in the packet must be generated. This requiressec=6msec.The time required to transmit the packet issec=sec.Propagation delay = 2 msec.The delay until decoding is6msec +sec + 2msec = 8.384msecA similar analysis shows that all bits experience a delay of 8.384 msec.Problem 7a) 10 users can

45、 be supported because each user requires one tenth of the bandwidth.b) .c) .d) .We use the central limit theorem to approximate this probability. Let be independent random variables such that .“11 or more users” when is a standard normal r.v. Thus “10 or more users”.Problem 8a) 10,000b)Problem 9The

46、first end system requires L/R1 to transmit the packet onto the first link; the packet propagates over the first link in d1/s1; the packet switch adds a processing delay of dproc; after receiving the entire packet, the packet switch requires L/R2 to transmit the packet onto the second link; the packe

47、t propagates over the second link in d2/s2. Adding these five delays givesdend-end = L/R1 + L/R2 + d1/s1 + d2/s2 + dprocTo answer the second question, we simply plug the values into the equation to get 8 + 8 + 16 + 4 + 1 = 37 msec.Problem 10Because bits are immediately transmitted, the packet switch

48、 does not introduce any delay; in particular, it does not introduce a transmission delay. Thus,dend-end = L/R + d1/s1 + d2/s2For the values in Problem 9, we get 8 + 16 + 4 = 28 msec. Problem 11The arriving packet must first wait for the link to transmit 3,500 bytes or 28,000 bits. Since these bits a

49、re transmitted at 1 Mbps, the queuing delay is 28 msec. Generally, the queuing delay is nL + (L - x)/R.Problem 12The queuing delay is 0 for the first transmitted packet, L/R for the second transmitted packet, and generally, (n-1)L/R for the nth transmitted packet. Thus, the average delay for the N p

50、ackets is(L/R + 2L/R + . + (N-1)L/R)/N = L/RN(1 + 2 + . + (N-1) = LN(N-1)/(2RN) = (N-1)L/(2R)Note that here we used the well-known fact that1 + 2 + . + N = N(N+1)/2Problem 13It takes seconds to transmit the packets. Thus, the buffer is empty when a batch of packets arrive.The first of the packets ha

51、s no queuing delay. The 2nd packet has a queuing delay of seconds. The th packet has a delay of seconds.The average delay is.Problem 14a) The transmission delay is . The total delay isb) Let .Total delay = Problem 15a) There are nodes (the source host and the routers). Let denote the processing dela

52、y at the th node. Let be the transmission rate of the th link and let. Let be the propagation delay across the th link. Then.b) Let denote the average queueing delay at node . Then.Problem 16The command:Problem 17Throughput = minRs, Rc, R/MProblem 18a) 40,000 bitsb) 40,000 bitsc) The bandwidth-delay

53、 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 football fielde) s/R Problem 1925 bpsProblem 20a) 40,000,000 bitsb) 400,000 bitsc) .25 metersProblem 21a) ttrans + tprop = 400 msec + 40 msec = 440 msecb) 10 * (ttrans + 2 tpro

54、p) = 10*(40 msec + 80 msec) = 1.2 secProblem 22a) 150 msecb) 1,500,000 bitsc) 600,000,000 bits Problem 23Lets 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

55、 the bags and ticket. This is additional information added in the Baggage 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 pass

56、es through security, and additional stamp is often added to his/her ticket, indicating that the passenger has passed through a security check. This information is used to ensure (e.g., by later checks for the security information) secure transfer of people.Problem 24a) Time to send message from source host to first packet switch = . With store-and-forward switching, the total time to move message from source host to destination h