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1. 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, ete. 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 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 HFC: home; 4. 100 Mbps switched Ethernet: enterprise. 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 downstream 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 fibers optic links. 9. 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 b) Since cach user requires 1Mbps when transmitting, if two or fewer users transmit simultancously, a maximum of 2Mbps will be required. Since the available 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) Probability 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 users are transmitting simultaneously) is 0.008. 14. Tf the two TSPs 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 pecring with cach other directly, the two ISPs can reduce their payments to their provider ISPs. An Internct Exchange Points (LXP) (typically in a standalone building with its own switches) is a mecting 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 [XP a relatively small fee, which may depend on the amount of traffic sent to or received from the IXP. 15. Google's 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 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 expericnee, since it has to usc few intermediary ISPs. Sccond, it can save moncy by sending less traffic into provider networks. Third, if ISPs decide to charge more moncy to highly profitable content providers (in countrics where net neutrality docsn't apply), the content providers can avoid these extra payments. 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. a) 1000 km, 1 Mbps, 100 bytes b) 100 km, | Mbps, 100 bytes 18. LOmsec; d/s; no; no 19. a) 500 kbps b) 64 seconds c) 100kbps; 320 seconds 20. End system A breaks the large file into chunks. It adds header to each chunk, thereby generating multiple packets from the file. The header in cach packct 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 road to take is analogous to a packct asking which outgoing link it should be forwarded on, given the packet’s 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 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. that particular vulnerability. Any system that is part of the botnet can automatically 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 issuc 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 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 the message(s) being sent from Bob to Alice. For example. she can easily change the phrase “Alice, T 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 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 Server Msg name purpose Let server know that there is UkLO
BY Messages from Server to ATM machine (display) PASSWD BATANCF User asks to withdraw moneyuser all done purpose (PASSWD, WTTHDRAWT,) Msg name PASSWD Ask user tor PIN (password) last requested operation OKlast OK requesled operalion in RRROR ERR (PASSWD, WITHDRAWL) AMOUN'I sent in BYE Correct response to BALANCE request user dons, display welcome screen al. ATM operation: client server ann theck if valid ——o > userid) PASSWD (check g------- password) OK (password anne is OK) BATANCR AMOUNT no $ given ouLRYF, BYR Problem 2 At time N*(L/R) the first packet has reached the destination, the second packet is stored in the last router, the third packet is stored in the next-to-last router, etc. At time N*(L/R) + L/R, the second packet has reached the destination, the third packet is stored in the last router, ete. Continuing with this logic, we see that at time N*(L/R) + (P-1)*(L/R) = (N+P- 1)*(L/R) all packets have reached the destination. Problem 3 a) A circuit-switched network would be well suited to the application, 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 without significant waste. In addition, the overhead costs of setting up and tearing down connections are amortized over the lengthy duration of a typical application session. b) In the worst case, all the applications simultaneously transmit over one or more network links. However, since each link has sufficient bandwidth to handle the sum of all of the applications’ data rates, no congestion (very little queuing) will occur. Given such generous link capacities, the network does not need congestion control mechanisms. Problem 4 a) Between the switch in the upper left and the switch in the upper right we can have 4 connections. Similarly we can have four connections between each of the 3 other pairs of adjacent switches. Thus, this network can support up to 16 connections. b) We can 4 connections passing through the switch in the upper-right-hand corner and another 4 connections passing through the switch in the lower-left-hand corner, giving a total of 8 connections. c) Yes. For the connections between A and C, we route two connections through B and two connections through D. For the connections between B and D, we route two connections through A and two connections through C. In this manner, there are at most 4 connections passing through any link. Problem 5 Tollbooths are 75 km apart, and the cars propagate at 17S5kam/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 2 minutes, for the first tollbooth to service the 10 cars. Each of these cars has a propagation delay of 25.7 minutes (travel 75 km) before arriving at the second tollbooth. Thus, all the cars are lined up before the second tollbooth after 27.7 minutes. The whole process repeats itself for traveling between the second and third tollbooths. It also takes 2 minutes for the third tollbooth to service the 10 cars. Thus the total delay is 57.4 minutes. b) Delay between tollbooths is 8*12 seconds plus 25.7 minutes. The total delay is twice this amount plus 8*12 scconds, i.c., 53 minutes. Problem 6 a) seconds. b) seconds. c) seconds. 120 x 0.10.9) when is a standard normal r.v. Thus “51 of more users” . Problem 9 a) 10,000 b) Problem 10 The first end system requires 1/R1 to transmit the packet onto the first link; the packet propagates over the first link in di/s1; the packet switch adds a processing delay of dproc; after receiving the entire packet, the packet switch connecting the first and the sccond link requires L/R2 to transmit the packct onto the second link; the packet propagates over the second link in d2/s2. Sumilarly, we can find the delay caused by the second switch and the third link: L/R3, dproc, and d3/s3. Adding these five delays gives dend-end = L/R1 | L/R2 | LAR3 | difss \ d2/s2 \ d3/s3\ dproc | dproe To answer the second question, we simply plug the values into the equation to get 4.8 + 4.8 + 4.8 +20+16+4+3+3=60.4 msec. Problem 11 Because bits are immediately transmitted, the packet switch does not introduce any delay; in particular, it does not introduce a transmission delay. Thus, dend-end = L/R \ di/s1 | d2/s2\ d3/s3 For the values in Problem 10, we get 4.8 + 20+ 16+ 4=44.8 msec. Problem 12 The arriving packct must first wait for the link to transmit 4.5 *1,500 bytes — 6,750 bytes or 54,000 bits. Since thesc bits are transmitted at 2.5 Mbps, the queuing delay is 21.6 msce. Generally, the queuing delay is (nL + (L- x)\/R. Problem 13 a) The queuing delay is 0 for the first transmitted packet, /./R for the second transmitted packet, and generally, (n-1)L/R for the nm transmitted packet. Thus, the average delay for the NV packets is: (L/R + QAR F oossee. + (N-DL/RYIN — LARN) * (142 4... + (N-L) = LRN) * N(N-1)/2 = LN(N-1)/2RN) = (N-1)L/(2R) Note that here we used the well-known fact: TMD crane IN=NNID2 b) It takes seconds to transmit the packets. Thus, the buffer is empty when a each batch of packets arrive. Thus, the average delay of a packet across all batches is the average delay within one batch, ie, (N-L)Z/2R. Problem 14 a) The transmission delay is . The total delay is b) Let. Total delay = For x=0, the total delay =0; as we increase x, total delay increases, approaching infinity as x approaches I/a. Problem 15 Total delay . ‘Traceroutes between San Diego Super Computer Center and www.poly.edu a) The average (mean) of the round-trip delays at cach of the three hours is 71.18 ms, 71.38 ms and 71.55 ms, respectively. The standard deviations are 0.075 ms, 0.21 ms, 0.05 ms, respectively. b) In this example, the traceroutes have 12 routers in the path at each of the three hours. No, the paths didn’t change during any of the hours. c) Traceroute packets passed through four ISP networks from source to destination. Yes, in this experiment the largest delays occurred at peering interfaces between adjacent ISPs. Traccroutes from www.stella-net.net (France) to www.poly.cdu (USA). d) The average round-trip delays at each of the three hours are 87.09 ms, 86.35 ms and 86.48 ms, respectively. The standard deviations are 0.53 ms, 0.18 ms, 0.23 ms, respectively. In this example, there are 11 routers in the path at cach of the three hours. No, the paths didn’t change during any of the hours. Traceroute packets passed three ISP networks from source to destination. Yes, in this experiment the largest delays occurred at peering interfaces between adjacent ISPs. Problem 19 1 x (1 - 1) messages will be sent. This answer supports the Metcalfe’s law. Problem 20 Throughput — min{Rs, Re, R/M} Problem 21 If only use one path, the max throughput is given by: If usc all paths, the max throughput is given by. Problem 22 Probability of successfully receiving a packet is: ps~ (1-p)n. The number of transmissions necded to be performed until the packet is successfully received by the client is a geometric random variable with success probability ps. Thus, the average number of transmissions needed is given by: 1/ps. Then, the average number of re-transmissions needed. is given by: 1/ps-1. Problem 23 Let's call the first packet A and call the second packet B. a) If the bottleneck link is the first link, then packet B is queued at the first link waiting for the transmission of packet A. So the packet inter-arrival time at the destination is simply L/Rs. b) If the second link is the bottleneck link and both packets are sent back to back, it must be true that the second packet arrives at the input queue of the second link before the second link finishes the transmission of the first packet. That is, L/Rs + L/Rs + dprop < L/Rs + dprop + L/Re The left hand side of the above inequality represents the time needed by the second packet to arrive at the input queue of the second link (the second link has not started transmitting the second packet yet). The right hand side represents the time needed by the first packet to finish its transmission onto the second link. c) 0.5 meters Problem 28 a) tirans + tprep — 400 msec + 80 msec = 480 msec. b) 20 * (tirans + 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 29 Recall geostationary satellite is 36,000 kilometers away from earth surface. a) 150 msec b) 1,500,000 bits c) 600,000,000 bits Problem 30 Let’s 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 sccurity and additional stamp is often added to his/her ticket, indicating that the passenger has passed through a security check. This information is uscd to censure (c.g., by later checks for the security information) sccure transfer of people. Problem 31 a) Time to send message from source host to first packet switch — 106 5x106= 0. 2sec With store-and-forward switching, the total time to move message from source host to destination host =0. 2sec x 3hops = 0. 6sec b) Time 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 Is packet is received at the second switch — c) Time at which Istpacket is received at the destination host — . After this, every Smsec one packet will be received; thus time at which last (800m) packet is received — . It can be scen that delay in using message segmentation is significantly less (almost 1/3ra). d) i. Without message segmentation, if bit crrors are not tolerated, if there is a single bit crror, 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 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 32 Yes, the delays in the animation correspond to the delays in the Problem 31. The propagation delays affect the overall end-to-end delays both for packet switching and message switching equally. 
