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-puter Networking: A Top-Down Approach Featuring the Internet, 3rd EditionSolutions to Review Questions and ProblemsNote: These solutions are inplete. plete solutions will be available by 1 September 2005Version Date: July 1, 2004This document contains the solutions to review questions and problems for the 3rd edition of puter 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 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 1996-2004 by J.F. Kurose and K.W. Ross. All rights reservedChapter 1 Review Questions1. There is no difference. Throughout this te*t, 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 simply just call Bob on the phone and say, “e 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 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 e*cuses.3. A networking program usually has two programs, each running on a different host, municating with each other. The program that initiates the munication 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) 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 application 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 in 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 service are: No handshaking No guarantees of reliable data transfer No flow control or congestion control5. 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 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 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-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 fi*ed, pre-defined pattern. In TDM circuit 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 pletes 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 time t1. At time t2 = t1 + L/R2, the router pletes 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 switch 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 VCs; (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 ma*imum end-to-end packet delay when VCs are used.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: pany; 5. Wireless LAN: mobile; 6. Cellular mobile access (for e*ample, WAP): mobile13. 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. Also, a tier-2 ISP is a customer of one or more tier-114. 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 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 and 10 Gbps. For an * Mbps Ethernet (where * = 10, 100, 1,000 or 10,000), a user can continuously transmit at the rate * 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 * Mbps.17. Ethernet most monly runs over twisted-pair copper wire and “thin coa*ial cable. It also can run over fibers optic links and thick coa*ial 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, bandwidth 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 ponents are processing delays, transmission delays, propagation delays, and queuing delays. All of these delays are fi*ed, e*cept for the queuing delays, which are variable.20. Five generic tasks are error control, flow control, segmentation and reassembly, multiple*ing, and connection setup. Yes, these tasks can be duplicated at different layers. For e*ample, error control is often provided at more than one layer.21. The five layers in the Internet 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 the 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 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 ponents, and process layer four as well.)Link layer switches process layers 1 through 2. Hosts process all five layers.Chapter 1 ProblemsProblem 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 namepurpose-HELO Let server know that there is a card in the ATM machineATM card transmits user ID to ServerPASSWD User enters PIN, which is sent to serverBALANCEUser requests balanceWITHDRAWL User asks to withdraw moneyBYEuser all doneMessages from Server to ATM machine (display)Msg namepurpose-PASSWDAsk user for PIN (password)OKlast requested operation (PASSWD, WITHDRAWL) OKERRlast requested operation (PASSWD, WITHDRAWL) in ERRORAMOUNT sent in response to BALANCE requestBYEuser done, display wele screen at ATMCorrect operation:client serverHELO (userid)-(check if valid userid)-PASSWDPASSWD -(check password)-AMOUNT WITHDRAWL -check if enough $ to cover withdrawl(check if valid userid)-PASSWDPASSWD -(check password)-AMOUNT WITHDRAWL -check if enough $ to cover withdrawl-BYEProblem 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 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, 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, no congestion (very little queueing) will occur.Problem 3.a) We can n connections between each of the four pairs of adjacent switches. This gives a ma*imum 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 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 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 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, i.e., 122 minutes and 48 seconds.Problem 5.a) The time to transmit one packet onto a link is . The time to deliver the packet over Q links is. Thus the total latency is.b)c) Because there is no store-and-forward delays at the links, the total delay is.Problem 6. a) 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 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 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 e*perience a delay of 8.384 msec.Problem 8. a) 10 users can be supported because each user requires one tenth of the bandwidth.b).c).d).We use the central limit theorem to appro*imate this probability. Let be independent random variables such that .“11 or more userswhen is a standard normal r.v. Thus “10 or more users.Problem 9.a) 10,000Problem 10. It takes seconds to transmit the packets. Thus, the buffer is empty when a batch of packets arrive.The first of the packets has no queueing delay. The 2nd packet has a queueing delay of seconds. The th packet has a delay of seconds.The average delay is.Problem 11. a) The transmission delay is . The total delay isb) Let .Total delay = Problem 12. a) There are nodes (the source host and the routers). Let denote the processing delay 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 13. The mand:traceroute -q 20 .eure.frwill get 20 delay measurements from the issuing host to the host, .eure.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 ma*imum number of bits that can be in the linkd) 1 bit is 250 meters long, which is longer than a football fielde) s/RProblem 15.25 bpsProblem 16.a) 40,000,000 bitsb) 400,000 bitsc) .25 metersProblem 17.a) ttrans + tprop = 400 msec + 40 msec = 440 msecb) 10 * (ttrans + 2 tprop) = 10*(40 msec + 80 msec) = 1.2 secProblem 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 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 passes through security, and additional stamp is often added to his/her ticket, indicating the at 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 20.a) time taken 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 host = b) time taken to send 1st packet from source host to first packet switch = . . Time at which 2nd packet is received at the first switch = time at which 1st packet is received at the second switch = c) time at which 1st packet is received at the destination host = . . After this, every 1msec one packet will be received, thus time at which last (5000th) packet is received = . 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. Since header size is usually the same for packets regardless of its size, with message segmentation total amount of header bytes sent increases.Problem 21.Java Applet Problem 22.Time at which the 1st packet is received at the destination = sec. After this, one packet is received by the destination every sec. Thus delay in sending the whole file = To calculate the value of S which leads to the minimum delay,Chapter 2 Review Questions1. The Web: ; file transfer: FTP; remote login: Telnet; Network News: NNTP; : SMTP.2. Network architecture refers to organization of munication into layers (e.g., the five-layer Internet architecture). Application architecture, on the other hand, is designed by an application developer and dictates how the application is (e.g., client-server or P2P)3. Instant Messaging involves the initiator to contact a centralized server to locate the address (IP address.) of the receiver: client server model. After this, the instant messaging can be peer to peer message between the two municating parties are sent directly between them.4. The process which initiates a service request is the client; the process that waits to be contacted is the server.5. No. As stated in the te*t, all munication sessions have a client side and a server side. In a P2P file-sharing application, the peer that is receiving a file is typically the client and the peer that is sending the file is typically the server. 6. The IP address of the destination hosts and the port numbers of the destination socket.7. You probably use a browser and a mail reader on a daily basis. You may also use an FTP user agent, a Telnet user agent, an audio/video player user agent (such as a Real Networks player), an instant messaging agent, a P2P file-sharing agent, etc.8. There are no good e*amples of an application that requires no data loss and timing. If you know of one, send an to the authors.9. A protocol uses handshaking if the two municating entities first e*change control packets before sending data to each other. SMTP uses handshaking at the application layer whereas does not.10. The applications that use those protocols require that all application data is received in the correct order and without gaps. TCP provides this service whereas UDP does not.11. In both cases, the site must keep a database record for the user. With cookies, the user does not e*plicitly provide a username and password each time it visits the site. However, browser identifies the user by sending the users cookie number each time the user accesses the site.12. In persistent without pipelining, the browser first waits to receive a response from the server before issuing a new request. In persistent with pipelining, the browser issues requests as soon as it has a need to do so, without waiting for response messages from the server.13. Web caching can bring the desired content “closer to the user, perhaps to the same LAN to which the users host is connected. Web caching can reduce the delay for all objects, even objects that are not cached, since caching reduces the traffic on links. 15. FTP uses two parallel TCP connections, one connection for sending control information (such as a request to transfer a file) and another connection for actually transferring the file. Because the control information is not sent over the same connection that the file is sent over, FTP sends control information out of band.16. Message is sent from Alices host to her mail server over . Alices mail server then sends the message to Bobs mail server over SMTP. Bob then transfers the message from his mail server to his host over POP3.18. With download and delete, after a u
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