1、Computer Networking: A Top-Down Approach, 6th EditionSolutions to Review Questions and ProblemsVersion Date: May 2012This document contains the solutions to review questions and problems for the 5th edition of Computer Networking: A Top-Down Approach by Jim Kurose and Keith Ross. These solutions are
2、 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.Acknowled
3、gments: Over the years, several students and colleagues have helped us prepare this solutions manual. Special thanks goes to HongGang Zhang, Rakesh Kumar, Prithula Dhungel, and Vijay Annapureddy. Also thanks to all the readers who have made suggestions and corrected errors.All material copyright 199
4、6-2012 by J.F. Kurose and K.W. Ross. All rights reservedChapter 1 Review Questions1. There is no difference. Throughout this text, the words “host” and “end system” are used interchangeably. End systems include PCs, workstations, Web servers, mail servers, PDAs, Internet-connected game consoles, etc
5、.2. From Wikipedia: Diplomatic protocol is commonly described as a set of international courtesy rules. These well-established and time-honored rules have made it easier for nations and people to live and work together. Part of protocol has always been the acknowledgment of the hierarchical standing
6、 of all present. Protocol rules are based on the principles of civility.3. Standards are important for protocols so that people can create networking systems and products that interoperate.4. 1. Dial-up modem over telephone line: home; 2. DSL over telephone line: home or small office; 3. Cable to HF
7、C: home; 4. 100 Mbps switched Ethernet: enterprise; 5. Wifi (802.11): home and enterprise: 6. 3G and 4G: wide-area wireless.5. 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 down
8、stream channel.6. In most American cities, the 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. 8. Today, Ethernet most commonly runs over twisted-pair copper wire. It also can run over fibe
9、rs optic links.9. Dial up modems: up to 56 Kbps, bandwidth is dedicated; ADSL: up to 24 Mbps downstream and 2.5 Mbps upstream, bandwidth is dedicated; HFC, rates up to 42.8 Mbps and upstream rates of up to 30.7 Mbps, bandwidth is shared. FTTH: 2-10Mbps upload; 10-20 Mbps download; bandwidth is not s
10、hared.10. There are two popular wireless Internet access technologies today:a) Wifi (802.11) In a wireless LAN, wireless users transmit/receive packets to/from an base station (i.e., wireless access point) within a radius of few tens of meters. The base station is typically connected to the wired In
11、ternet and thus serves to connect wireless users to the wired network.b) 3G and 4G wide-area wireless access networks. In these systems, packets are transmitted over the same wireless infrastructure used for cellular telephony, with the base station thus being managed by a telecommunications provide
12、r. This provides wireless access to users within a radius of tens of kilometers of the base station.11. 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 route
13、r 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 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/R2.12. A c
14、ircuit-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. FDM requires sophisticated analog hardware to shift signal into appropriate fre
15、quency bands.13. 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 bandwidth of the shared link is 2Mbps, there
16、 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) Probability that a given user is transmitting =
17、 0.2d) Probability that all three users are transmitting simultaneously = 331p= (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 users are transmitting simultaneously) is 0.0
18、08. 14. If the two ISPs do not peer with each other, then when they send traffic to each other they have to send the traffic through a provider ISP (intermediary), to which they have to pay for carrying the traffic. By peering with each other directly, the two ISPs can reduce their payments to their
19、 provider ISPs. An Internet Exchange Points (IXP) (typically in a standalone building with its own switches) is a meeting point where multiple ISPs can connect and/or peer together. An ISP earns its money by charging each of the the ISPs that connect to the IXP a relatively small fee, which may depe
20、nd on the amount of traffic sent to or received from the IXP.15. Googles private network connects together all its data centers, big and small. Traffic between the Google data centers passes over its private network rather than over the public Internet. Many of these data centers are located in, or
21、close to, lower tier ISPs. Therefore, when Google delivers content to a user, it often can bypass higher tier ISPs. What motivates content providers to create these networks? First, the content provider has more control over the user experience, since it has to use few intermediary ISPs. Second, it
22、can save money by sending less traffic into provider networks. Third, if ISPs decide to charge more money to highly profitable content providers (in countries where net neutrality doesnt apply), the content providers can avoid these extra payments. 16. The delay components are processing delays, tra
23、nsmission delays, propagation delays, and queuing delays. All of these delays are fixed, except for the queuing delays, which are variable.17. a) 1000 km, 1 Mbps, 100 bytesb) 100 km, 1 Mbps, 100 bytes18. 10msec; d/s; no; no19. a) 500 kbpsb) 64 secondsc) 100kbps; 320 seconds20. End system A breaks th
24、e large file into chunks. It adds header to each chunk, thereby generating multiple packets from the file. The header in each packet includes the IP address of the destination (end system B). The packet switch uses the destination IP address in the packet to determine the outgoing link. Asking which
25、 road to take is analogous to a packet asking which outgoing link it should be forwarded on, given the packets destination address.21. The maximum emission rate is 500 packets/sec and the maximum transmission rate is 350 packets/sec. The corresponding traffic intensity is 500/350 =1.43 1. Loss will
26、eventually occur for each experiment; but the time when loss first occurs will be different from one experiment to the next due to the randomness in the emission process.22. Five generic tasks are error control, flow control, segmentation and reassembly, multiplexing, and connection setup. Yes, thes
27、e tasks can be duplicated at different layers. For example, error control is often provided at more than one layer.23. 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 prin
28、cipal responsibilities are outlined in Section 1.5.1.24. 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-layer message with transport layer header; network-l
29、ayer 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 network, link and physical layers (layers 1 through 3). (This is a little bit of a white lie, as modern routers sometime
30、s act as firewalls or caching components, and process Transport layer as well.) Link layer switches process link and physical layers (layers 1 through2). Hosts process all five layers.26. a) VirusRequires some form of human interaction to spread. Classic example: E-mail viruses.b) WormsNo user repli
31、cation needed. Worm in infected host scans IP addresses and port numbers, looking for vulnerable processes to infect.27. Creation of a botnet requires an attacker to find vulnerability in some application or system (e.g. exploiting the buffer overflow vulnerability that might exist in an application
32、). 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 automatically scan its environment and propagate by exploitin
33、g 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 nodes, that target a single node (for example, all nodes i
34、n 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 the message(s) being sent from Bob to Alice. For example,
35、 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. Chapter 1 ProblemsChapter1-Problem 1There is no single rig
36、ht answer to this question. Many protocols would do the trick. Heres a simple answer below:Messages from ATM machine to ServerMsg name purpose- -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 serverBALANCE User
37、 requests balanceWITHDRAWL User asks to withdraw moneyBYE user all doneMessages 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 (PASSWD, WITHDRAWL) in ERRORAMOUNT sent in response
38、to BALANCE requestBYE user done, display welcome screen at ATMCorrect operation:client serverHELO (userid) - (check if valid userid)- (check password)WITHDRAWL - check if enough $ to cover withdrawl(check if valid userid)- (check password)WITHDRAWL - check if enough $ to cover withdrawl= L/Rs + dpro
39、p + L/RcThus, the minimum value of T is L/Rc L/Rs . Problem 2440 terabytes = 40 * 1012 * 8 bits. So, if using the dedicated link, it will take 40 * 1012 * 8 / (100 *106 ) =3200000 seconds = 37 days. But with FedEx overnight delivery, you can guarantee the data arrives in one day, and it should cost
40、less than $100. Problem 25a) 160,000 bitsb) 160,000 bitsc) The bandwidth-delay product of a link is the maximum number of bits that can be in the link.d) the width of a bit = length of link / bandwidth-delay product, so 1 bit is 125 meters long, which is longer than a football fielde) s/R Problem 26
41、s/R=20000km, then R=s/20000km= 2.5*108/(2*107)= 12.5 bpsProblem 27a) 80,000,000 bitsb) 800,000 bits, this is because that the maximum number of bits that will be in the link at any given time = min(bandwidth delay product, packet size) = 800,000 bits.c) .25 metersProblem 28a) ttrans + tprop = 400 ms
42、ec + 80 msec = 480 msec.b) 20 * (ttrans + 2 tprop) = 20*(20 msec + 80 msec) = 2 sec.c) Breaking up a file takes longer to transmit because each data packet and its corresponding acknowledgement packet add their own propagation delays.Problem 29Recall geostationary satellite is 36,000 kilometers away
43、 from earth surface.a) 150 msecb) 1,500,000 bitsc) 600,000,000 bits Problem 30Lets 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
44、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 add
45、itional 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 31a) Time to send message from source host to first packet swit
46、ch = With store-and-forward switching, the total time to move message sec40286from source host to destination host = sec123sec4hopb) Time to send 1st packet from source host to first packet switch = . . Time at which 2nd packet is received at the first switch = time sec510264mat which 1st packet is
47、received at the second switch = sec10s52mc) Time at which 1st packet is received at the destination host = . After this, every 5msec one packet will be received; thus sec53sec5hoptime at which last (800th) packet is received = . It sec.4s*79sec1can be seen that delay in using message segmentation is
48、 significantly less (almost 1/3rd). d)i. Without message segmentation, if bit errors are not tolerated, if there is a single bit error, the whole message has to be retransmitted (rather than a single packet).ii. Without message segmentation, huge packets (containing HD videos, for example) are sent
49、into the network. Routers have to accommodate these huge packets. Smaller packets have to queue behind enormous packets and suffer unfair delays. e)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 all packets regardless of their size, with message segmentation the total amount of header bytes is more.Problem 32Yes, the delays in the applet correspond to the de
