1.第一章课后习题及答案

1、第一章1(QI) What is the difference between a host and an end system List the types of end systems. Is a Web server an end systemAnswer: There is no d iff ere nee. Throughout this text, the words host and end system are used interchangeabl y. End systems in elude PCs, workstations, Web servers, mail ser

2、vers, 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 just c

3、all Bob on the phone and say, come to our dinner table now. In stead, she calls Bob and suggests a date and time Bob may resp ond 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 t

4、ime 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 excuses3. (Q3) What is a client program What is a server progr

5、am Does a server program request and receive services from a client programAnswer: 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 requests and re

6、ceives services from the server program4. (Q4) List six access technologies Classify each one as residential access, company access, or mobile accessAnswer: 1. Dial-up modem over telephone line: residential; 2 DSL over telephone line: residential or small office; 3. Cable to HFC: residential; 4. 100

7、 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 priceAnswer: Current pos

8、sibilities in elude: dial-up (up to 56kbps); DSL (up to 1 Mbps upstream, up to 8 Mbps downstream); cable modem (up to 30Mbps downstream，2 Mbps upstream6. (Q7) What are some of the physical media that Ethernet can run overAnswer: Ether net most commonly runs over twisted-pair copper wire and thin coa

9、xial cable It also can run over fibers optic links and thick coaxial cable7. (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 shared or dedicatedAnswer:

10、Dial up modems: up to 56 Kbps, bandwidth is dedicated; ISDN: up to 128 kbps, bandwidth is dedicated; ADSL: downstream channel is 58 Mbps, upstream channel is up to 1 Mbps, bandwidth is dedicated; HFCZ downstream channel is 10-30 Mbps and upstream channel is usually less than a few Mbps, bandwidth is

11、 shared8. 一,(Q13) Why is it said that packet switching employs statistical multiplexing Contraststatistical 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 any fixed, predefined patter

12、n In TDM circuit switching, each host gets the same slot in a revolving TDM frame.10. (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 multiplexing in Section .)a. Whe

13、n circuit switching is used, how many users can be supportedb. For the remainder of this problem, suppose packet switching is used Why will there be essentially no queuing delay before the link 讦 two or fewer users transmit at the same time Why will there be a queuing delay 讦 three users transmit at

14、 the same timec. 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 growsAnswer:a. 2 users can be supported because e

15、ach user requires half of the link bandwidthb 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 no queuing delay before the link Whereas, if thre

16、e 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 linkc. Probability that a given user is transmitting =d Probabilitythatallthreeusersaretran smittingsimulta neously

17、 = (3)p3(l - p) = 0.2、= 0.008, Since thequeue grows when all the usersare transmitting, the fraction of time during which the queue grows (which is equal to the probability that all three users are transmitting simultaneously) is 11. (Q16) Consider sending a packet from a source host to a destinatio

18、n host over a fixed route. List the delay components in the end-to-end delay Which of these delays are constant and which are variableAnswer: The delay comp on ents are processing delays, tran smissio n delays, propagation delays, and queuing delays All of these delays are fixed, except for the queu

19、ing delays, which are variable12. (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 Ri = 250 kbps, R2 = 500 kbps, and R3 = 1 Mbps.a Assuming no other traffic in the network, what is the throughput for the file transfer.b. Suppose the f

20、ile is 2 million bytes Roughly, how long will it take to transfer the file to Host Bc. Repeat (a) and (b), but now with R2 reduced to 200 kbps.Answer:a. 250 kbpsb 64 secondsc. 200 kbps; 80 seconds13. (P2) Consider the circuit-switched network in Figure Recall that there are n circuits on each linka.

21、 What is the maximum number of simultaneous connections that can be in progress at any one time in this networkb Suppose that all connections are between the switch in the upper-left-hand corner and the switch in the lower-right-hand corner. What is the maximum number of simultaneous connections tha

22、t can be in progressAnswer:a. We can n conn ections between each of the four pairs of adjacent switches This gives a maximum of 4n conn ections b We can n connections passing through the switch in the upper-right-hand corner and another n connec廿ons passing through the switch in the lower-left-hand

23、corner; giving a total of 2n conn ections 14. (is. (P4) Review the car-caravan analogy in Section Assume a propagation speed of 50 km/hour.a. Suppose the caravan travels 150 km, beginning in front of one tollbooth, passing through a second tollbooth, and finishing just before a third tollbooth What

24、is the end-to-end delayb Repeat (a), now assuming that there are five cars in the caravan instead of tenAnswer: Tollbooths are 150 km apart, and the cars propagate at 50 km/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 minute

25、s, 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 182 minutes. The whole process repeats itself for traveling between the second and thi

26、rd tollbooths Thus the total delay is 364 minutesb. Delay between tollbooths is 5*12 seconds plus 180 minutes, 181minutes. The total delay is twice this amount,362 minutes.16. (P5) This elementary problem begins to explore propagation delay and transmission delay, two central concepts in data networ

27、king Consider two hosts, A and B，connected by a sin创e 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 A is to send a packet of size L bits to Host Ba Express the propagation delay, d, in terms of m and s.

28、c. Determine the transmission time of the packet, d, in terms of L and Rd Ignoring processing and queuing delays, obtain an expression for the end-to-end delay.e Suppose Host A begins to transmit the packet at time t = 0. At time t = d , where is the last bit of the packetf Suppose d is greater than

29、 d At time t = d , where is the first bit of the packetg Suppose d is less than d At time t = d, where is the first bit of the packeth Suppose s = *108# L = lOObits, and R = 28kbps. Find the distance m so that d equals d Answer:a. d = m/s secondsbc. d = L/R seconds.d d 二(m/s + L/R) secondsg. The bit

30、 is just leaving Host A.f The first bit is in the link and has not reached Host Bg The first bit has reached Host Bh. Want100m=RS =28 * 10*2.5 * 108) = 893 km17. (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 anal

31、og voice to a digital 64 kbps bit stream on the fly. Host A then groups the bits into 56-Byte 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

32、 entire packet, it converts the packets bits to an analog signal. How much time elapses from 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 transm

33、itted, all of the bits in the packet must be generated This requires56*8ec = 7 msec64 103The time required to transmit the packet is56*8tscc = 896 usee500 衣 103Propagation delay = 2 msecThe delay until decoding is7msec + 896 u sec + 2msec = msecA similar analysis shows that all bits experience a del

34、ay of msec18.is. (P9) Consider a packet of length L which begins at end system A, travels over one link to a packet switch, and travels from the packet switch over a second link to a destination end system Let d& s, and denote the length, propagation speed, and the transmission rate of link i, for /

35、 = 1, 2 The packet switch delays each packet by dproc. Assuming no queuing delays, in terms of d/z s/z R初(/ = 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 * 108 m/s, the transmission rates of both

36、 links is 1 Mbps, the packet switch processing delay is 2 msec, the length of the first 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 delayAnswer: The first end system requires L/Ri to transmit the packet onto the first link; the packet propag

37、ates over the first link in di/si； 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 packet propagates over the second link in djs? Adding these five delays givesdend-end = L/Ri + L/

38、R2 + di/Si + 62/2 + dprocTo answer the second question, we simply plug the values into the equation to get 8 + 8 +24 + 12 + 2 = 54 msec.20. (PIO) In the above problem, suppose Ri = R2 = R and dproc = 0. Further suppose the packet switch does not store-and-forward packets but instead immediately tran

39、smits each bit it receivers before waiting for the packet to arrive What is the end-to-end delayAnswer: 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/Si + d(S2For the values

40、 in Problem 9, we get 8 + 24 + 12 = 44 msec.21. (Pll) Suppose N packets arrive 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 packetsAnswer: The queuing

41、 delay is 0 for the first transmitted packet, L/R for the second transmitted packet, and gen erally, (n-1)L/R for the nth tran smitted packet Thus, the average delay for the N packets is(L/R + 2L/R + + (N-1)L/R)/N = L/RN(1 +2 + + (N-l) = LN(N1)/(2RN) = (N1)L/(2R) Note that here we used the well-know

42、n fact that1+2 +N二N(N+l)/222. (P14) Consider the queuing delay in a router buffer. Let I denote traffic intensity; that is, / = La/R Suppose that the queuing delay takes the form IL/R (1-1) for IxTotal delay = :1 - CK23. (P16) Perform a Traceroute between source and destination on the same continent

43、 at threedifferent 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 number of routers in the path at each of the three hours Did the paths change during any of the hoursc Try to identify the number of ISP networks that the Tr

44、aceroute packets pass through from source to destination Routers with similar names and/or similar IP addresses should be considered as part of the same ISP. In your experiments, do the largest delays occur at the peering interfaces between adjacent ISPsd Repeat the above for a source and destinatio

45、n on different continents Compare the intra-continent and inter-continent resultsAnswer: Experiments24. (P18) Suppose two hosts, A and 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 *108 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 continuously as one large message What is the maximum number of bits that will be in the