1、第一章1. (Q1) What is the difference between a host and an end system? List the types of end systems. Is a Web server an end system?Answer: There is no difference. Throughout this text, the words “host” and “end system” are used interchangeably. End systems include PCs, workstations, Web servers, mail
2、servers, Internet-connected PDAs, WebTVs, etc.2. (Q2) The word protocol is often used to describe diplomatic relations. Give an example of a diplomatic protocol.Answer: Suppose Alice, an ambassador of country A wants to invite Bob, an ambassador of country B, over for dinner. Alice doesnt simply jus
3、t 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 a date
4、 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. (Q3) What is a client program? What is a se
5、rver program? Does a server program request and receive services from a client program?Answer: A networking 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 req
6、uests and receives services from the server program.4. (Q4) List six access technologies. Classify each one as residential access, company access, or mobile access.Answer: 1. Dial-up modem over telephone line: residential; 2. DSL over telephone line: residential or small office; 3. Cable to HFC: res
7、idential; 4. 100 Mbps switched Etherent: company; 5. Wireless LAN: mobile; 6. Cellular mobile access (for example, 3G/4G): mobile5. (Q5) List the available residential access technologies in your city. For each type of access, provide the advertised downstream rate, upstream rate, and monthly price.
8、Answer: Current possibilities include: dial-up (up to 56kbps); DSL (up to 1 Mbps upstream, up to 8 Mbps downstream); cable modem (up to 30Mbps downstream, 2 Mbps upstream.6. (Q7) What are some of the physical media that Ethernet can run over?Answer: Ethernet most commonly runs over twisted-pair copp
9、er wire and “thin” coaxial cable. It also can run over fibers optic links and thick coaxial cable.7. (Q8) Dial-up modems, HFC, and DSL are all used for residential access. For each of these access technologies, provide a range of transmission rates and comment on whether the transmission rate is sha
10、red or dedicated.Answer: 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 th
11、an a few Mbps, bandwidth is shared.8. (Q13) Why is it said that packet switching employs statistical multiplexing? Contrast statistical multiplexing with the multiplexing that takes place in TDM.Answer: In a packet switched network, the packets from different sources flowing on a link do not follow
12、any fixed, pre-defined pattern. In TDM circuit switching, each host gets the same slot in a revolving TDM frame.9. (Q14) Suppose users share a 2Mbps link. Also suppose each user requires 1Mbps when transmitting, but each user transmits only 20 percent of the time. (See the discussion of statistical
13、multiplexing in Section 1.3.)a. When circuit switching is used, how many users can be supported?b. For the remainder of this problem, suppose packet switching is used. Why will there be essentially no queuing delay before the link if two or fewer users transmit at the same time? Why will there be a
14、queuing delay if three users transmit at the same time?c. Find the probability that a given user is transmitting.d. Suppose now there are three users. Find the probability that at any given time, all three users are transmitting simultaneously. Find the fraction of time during which the queue grows.
15、Answer: 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 will be
16、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 = 0.2d. Pr
17、obability that all three users are transmitting simultaneously. Since the queue grows when all the users are =(33)3(1)0=0.23=0.008transmitting, the fraction of time during which the queue grows (which is equal to the probability that all three users are transmitting simultaneously) is 0.008.10. (Q16
18、) Consider sending a packet from a source host to a destination host over a fixed route. List the delay components in the end-to-end delay. Which of these delays are constant and which are variable?Answer: The delay components are processing delays, transmission delays, propagation delays, and queui
19、ng delays. All of these delays are fixed, except for the queuing delays, which are variable.11. (Q19) Suppose Host A wants to send a large file to Host B. The path from Host A to Host B has three links, of rates R1 = 250 kbps, R2 = 500 kbps, and R3 = 1 Mbps.a. Assuming no other traffic in the networ
20、k, what is the throughput for the file transfer.b. Suppose the file is 2 million bytes. Roughly, how long will it take to transfer the file to Host B?c. Repeat (a) and (b), but now with R2 reduced to 200 kbps. Answer:a. 250 kbpsb. 64 secondsc. 200 kbps; 80 seconds12. (P2) Consider the circuit-switch
21、ed network in Figure 1.8. Recall that there are n circuits on each link.a. What is the maximum number of simultaneous connections that can be in progress at any one time in this network?b. Suppose that all connections are between the switch in the upper-left-hand corner and the switch in the lower-r
22、ight-hand corner. What is the maximum number of simultaneous connections that can be in progress?Answer: a. We can n connections between each of the four pairs of adjacent switches. This gives a maximum of 4n connections.b. We can n connections passing through the switch in the upper-right-hand corn
23、er and another n connections passing through the switch in the lower-left-hand corner, giving a total of 2n connections.13. (P4) Review the car-caravan analogy in Section 1.4. Assume a propagation speed of 50 km/hour.a. Suppose the caravan travels 150 km, beginning in front of one tollbooth, passing
24、 through a second tollbooth, and finishing just before a third tollbooth. What is the end-to-end delay?b. Repeat (a), now assuming that there are five cars in the caravan instead of ten.Answer: Tollbooths are 150 km apart, and the cars propagate at 50 km/hr, A tollbooth services a car at a rate of o
25、ne 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 180 minutes before arriving at the second tollbooth. Thus, all the cars are lined up before the second tollbooth after 18
26、2 minutes. The whole process repeats itself for traveling between the second and third tollbooths. Thus the total delay is 364 minutes.b. Delay between tollbooths is 5*12 seconds plus 180 minutes, i.e., 181minutes. The total delay is twice this amount, i.e., 362 minutes.14. (P5) This elementary prob
27、lem begins to explore propagation delay and transmission delay, two central concepts in data networking. Consider two hosts, A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meters, and suppose the propagation speed along the link is s meters/sec. Host
28、 A is to send a packet of size L bits to Host B.a. Express the propagation delay, d prop , in terms of m and s.b. Determine the transmission time of the packet, d trans , in terms of L and R.c. Ignoring processing and queuing delays, obtain an expression for the end-to-end delay.d. Suppose Host A be
29、gins to transmit the packet at time t = 0. At time t = d trans , where is the last bit of the packet?e. Suppose d prop is greater than d trans . At time t = d trans , where is the first bit of the packet?f. Suppose d prop is less than d trans . At time t = d trans , where is the first bit of the pac
30、ket?g. Suppose s = 2.5*108, L = 100bits, and R = 28kbps. Find the distance m so that d prop equals d trans .Answer:a. d prop = m/s seconds.b. d trans = L/R seconds.c. d end-to-end = (m/s + L/R) seconds.d. The bit is just leaving Host A.e. The first bit is in the link and has not reached Host B.f. Th
31、e first bit has reached Host B.g. Want= = 10028103(2.5108)=893 .15. (P6) In this problem we consider sending real-time voice from Host A to Host B over a packet-switched network (VoIP). Host A converts analog voice to a digital 64 kbps bit stream on the fly. Host A then groups the bits into 56-Byte
32、packets. There is one link between Host A and B; its transmission rate is 500 kbps and its propagation delay is 2 msec. As soon as Host A gathers a packet, it sends it to Host B. As soon as Host B receives an entire packet, it converts the packets bits to an analog signal. How much time elapses from
33、 the time a bit is created (from the original analog signal at Host A) until the bit is decoded (as part of the analog signal at Host B)?Answer: Consider the first bit in a packet. Before this bit can be transmitted, all of the bits in the packet must be generated. This requires56864103sec=7 The tim
34、e required to transmit the packet is568500103sec=896 Propagation delay = 2 msec.The delay until decoding is7msec + 896sec + 2msec = 9.896 msecA similar analysis shows that all bits experience a delay of 9.896 msec.16. (P9) Consider a packet of length L which begins at end system A, travels over one
35、link to a packet switch, and travels from the packet switch over a second link to a destination end system. Let di, si, and Ri denote the length, propagation speed, and the transmission rate of link i, for i = 1, 2. The packet switch delays each packet by dproc. Assuming no queuing delays, in terms
36、of di, si, Ri, (i = 1, 2), and L, what is the total end-to-end delay for the packet? Suppose now the packet Length is 1,000 bytes, the propagation speed on both links is 2.5 * 108 m/s, the transmission rates of both links is 1 Mbps, the packet switch processing delay is 2 msec, the length of the fir
37、st link is 6,000 km, and the length of the last link is 3,000 km. For these values, what is the end-to-end delay?Answer: The 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 dpr
38、oc; after receiving the entire packet, the packet switch requires L/R2 to transmit the packet onto the second link; the packet 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 value
39、s into the equation to get 8 + 8 + 24 + 12 + 2 = 54 msec.17. (P10) In the above problem, suppose R1 = R2 = R and dproc = 0. Further suppose the packet switch does not store-and-forward packets but instead immediately transmits each bit it receivers before waiting for the packet to arrive. What is th
40、e end-to-end delay?Answer: 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 + d1/s1 + d2/s2For the values in Problem 9, we get 8 + 24 + 12 = 44 msec.18. (P11) Suppose N packets arr
41、ive simultaneously to a link at which no packets are currently being transmitted or queued. Each packet is of length L and the link has transmission rate R. What is the average queuing delay for the N packets?Answer: The queuing delay is 0 for the first transmitted packet, L/R for the second transmi
42、tted packet, and generally, (n-1)L/R for the nth transmitted packet. Thus, the average delay for the N packets 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)/219. (P14) Consider the queuing d
43、elay in a router buffer. Let I denote traffic intensity; that is, I = La/R. Suppose that the queuing delay takes the form IL/R (1-I) for I1.a. Provide a formula for the total delay, that is, the queuing delay plus the transmission delay.b. Plot the total delay as a function of L/R.Answer:a. The tran
44、smission delay is L / R . The total delay is(1)+ = /1b. Let x = L / R. = 120. (P16) Perform a Traceroute between source and destination on the same continent at three different hours of the day.a. Find the average and standard deviation of the round-trip delays at each of the three hours.b. Find the
45、 number of routers in the path at each of the three hours. Did the paths change during any of the hours?c. Try to identify the number of ISP networks that the Traceroute packets pass through from source to destination. Routers with similar names and/or similar IP addresses should be considered as pa
46、rt of the same ISP. In your experiments, do the largest delays occur at the peering interfaces between adjacent ISPs?d. Repeat the above for a source and destination on different continents. Compare the intra-continent and inter-continent results.Answer: Experiments.21. (P18) Suppose two hosts, A an
47、d B, are separated by 10,000 kilometers and are connected by a direct link of R = 2 Mbps. Suppose the propagation speed over the link is 2.5108 meters/sec.a. Calculate the bandwidth-delay product, R dprop.b. Consider sending a file of 400,000 bits from Host A to Host B. Suppose the file is sent cont
48、inuously as one large message. What is the maximum number of bits that will be in the link at any given time?c. Provide an interpretation of the bandwidth-delay product.d. What is the width (in meters) of a bit in the link? Is it longer than a football field?e. Derive a general expression for the wi
49、dth of a bit in terms of the propagation speed s, the transmission rate R, and the length of the link m.Answer: a. dprop = 107 / 2.5108 = 0.04 sec; so R dprop = 80,000bitsb. 80,000bitsc. The bandwidth-delay product of a link is the maximum number of bits that can be in the link.d. 1 bit is 125 meters long, which is longer than a football fielde. m / (R dprop ) = m / (R * m / s) = s/R22. (P20) Consider problem P18 but now with a link of R = 1 Gbps.a. Calculate the bandwidth-delay product, Rd prop
