计算机网络第一周作业.doc

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1、如有侵权，请联系网站删除，仅供学习与交流计算机网络第一周作业【精品文档】第 13 页Review1. The word protocol is often used to describe diplomatic relations. Give an example of a diplomatic protocol.答：比如说联合国宪章、世界贸易组织协定。2. What is the difference between a host and an end system? List the types of end systems. Is a Web server an end system?答

2、没有不同，主机系统和终端设备可以互换。终端系统包括PC、工作站、web服务器、邮件服务器等等。WEB服务器也是端系统。3. List six access technologies. Classify each one as residential access, company access, or mobile access.答：住宅接入：数字用户线DSL、混合光纤同轴电缆 公司接入：无线网接入、网线接入 无线接入：无线局域网、广域无线局域网4. What is a client program? What is a server program? Does a server progra

3、m request and receive services from a client program?答： 端系统或者主机，他们都连接在Internet上，根据不同的服务又分为客户机、服务器。客户机程序是这样一种运行在端系统上的程序上的程序，它发出请求，并且从运行在其他端设备上的程序接受服务。而服务器程序就是给那些客户机程序，提供服务的程序。5. List the available residential access technologies in your city. For each type of access, provide the advertised downstream

4、 rate, upstream rate, and monthly price.答：光纤接入（HFC） 100Mbits/s 20Mbits/s 20RMB/月6. What are some of the physical media that Ethernet can run over?答：同轴电缆、双绞线都可以作为以太网的传输媒介。8.Describe the most popular wireless Internet access technologies today. Compare and contrast them.答：主要有两种接入方式： * IEEE 802.11技术无线L

5、AN接入，用户距离接入点的几十米的范围内。通过连接到有线网的基站间接将用户连接到有线网上。 * 3G与LET网络。在这些系统中通过用于拨打电话的网络硬件设施，为在基站范围内的用户提供无线网络接入。9.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

6、 dedicated.答：拨号调制解调器的速率通常是56K左右，这个是很早的技术与速率，带宽是独占的。DSL的速率是上下行不对称的，根据费用不同分为512K、1M、2M三种，还有不限速的上网，带宽是专用的。HFC的速率优于10M、100M、1G、10G的速率，带宽是共享的。10.What is the transmission rate of Ethernet LANs？.答：以太网拥有10Mbps、100Mbps、1Gbps和10Gbps几种。不能，带宽是在该线路上的用户共享。11.Suppose there is exactly one packet switch between a se

7、nding host and a receiving host答：延时为两段延时之和，L/R1+L/R2。14. Suppose users share a 2Mbps link. Also suppose each user requires 1Mbps when transmitting。答 a：2个，一人分1Mbps的专有带宽；b：当有两个人或一个人使用网络时，最高带宽占用为2Mbps，所以绝对不会有排队延时出现。但是，当3个及以上人使用时，最高需要3Mbps，但是最高提供2Mbps，所以会出现延时；c：0.2；d：3人同时传输的概率是0.008；所以时间增长的时间出现也是0.00815

8、.Why is it said that packet switching employs statistical multiplexing?.答：统计复用是对通信链路的一种高效的利用方法。所有链路上被按需分配，是一种按照概率统计的方法假设在一个时间段内，只有某几个用户在使用链路。TDM中的统计复用是对所有用户的事先分配了链路上的所有资源，传输速率是恒定的。按照预定分配，标记适合于处理恒定传输速率的业务，比如电话电路。而分组交换是按需分配，对链路的利用效率更高，资源共享，速率不恒定。17.Consider sending a packet from a source host to a des

9、tination 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?答： 节点总时延由处理时延、传输时延、传播时延和排队时延组成。所有这些时延除了排队时延都是固定的。18. Suppose Host A wants to send a large file to Host B. The path from Host A to Host B has three links, of

10、rates R1 = 250 kbps, R2 = 500 kbps, and R3 = 1 Mbps.答：a:R = minR1,R2,R3 = min500kbp,2Mbps,1Mbps = 500kbps b:T = 4000KB / (500Kbps / 8) = 64s c:R = minR1,R2,R3 = min500kbps,100kbps,1Mbps = 100Kpbs; T = 4000KB / (100Kbps / 8) = 320s20. How long does it take a packet of length 1,000 bytes to propagate

11、over a link of distance 2,500 km, propagation speed 2.5 108 m/s.答：10 -5 s; L/s; 否；否21.Suppose end system A wants to send a large file to end system B. At a very high level, describe how end system A creates packets from the file. When one of these packets arrives to a packet switch, what information

12、 in the packet does the switch use to determine the link onto which the packet is forwarded? Why is packet switching in the Internet analogous to driving from one city to another and asking directions along the way?答：将文件拆成多个小块。并为每个块添加标头，从而从文件生成多个数据包。每个数据包中的标头包括目的地址的IP。分组交换机使用该IP地址确定输出链路。在给定数据包的目的地址情

13、况下，询问哪条路要类似于这个的数据包，询问它应该转发哪个输出链路。22. Which layers in the Internet protocol stack does a router process? Which layers does a link-layer switch process? Which layers does a host process?答: 路由器处理第1层到第3层。链路层交换机处理1到2.主机处理所有5各层。23. List five tasks that a layer can perform. Is it possible that one (or more

14、) of these tasks could be performed by two (or more) layers?答： 五个通用任务是错误控制、流控制、分段和重组、多路复用和连接设置。这些任务可以在不同的层重复。 25. What are the five layers in the Internet protocol stack? What are the principal responsibilities of each of these layers? 答： 应用层：应用层是网络应用层协议存留的地方；运输层：因特网的运输层在应用程序端点之间传输报文；网络层：因特网的网络层负责将成

15、为数据报的网络层分组从一台主机移动到另一台主机；链路层：因特网的网络层通过源和目的地之间的一系列路由器路由数据报；物理层：将帧的一个个比特从一个结点转移到另一个结点。 Problem1 Consider the circuit-switched network in Figure 1.8. Recall that there are n circuits on each link答 a：4n；b：2n。2Consider an application that transmits data at a steady rate (for example, the sender generates

16、an N-bit unit of data every k time units, where k is small and fixed). Also, when such an application starts, it will continue running for a relatively long period of time. Answer the following questions, briefly justifying your answer: 答：a：电路交换非常适合这个应用，因为传输速率已知并且会话为长会话，不会造成带宽的大量浪费。b：在最坏的情况下，所有应用程序都

17、通过一条链路传输数据，因为还未达到链路最大带宽所以不会发生拥堵，所以网络不需要拥塞控制机制。3. Review the car-caravan analogy in Section 1.4. Assume a propagation speed of 100 km/hour. a. Suppose the caravan travels 150 km, beginning in front of one tollbooth, passing through a second tollbooth, and finishing just after a third tollbooth. What

18、is the end-to-end delay? b. Repeat (a), now assuming that there are eight cars in the caravan instead of ten.答： 2min+60min+2min+60min=124min；1.4min+60min+1.4min+60min=122.8min4. (未留) 5. This elementary problem begins to explore propagation delay and transmission delay, two central concepts in data n

19、etworking. 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 A is to send a packet of size L bits to Host B.答：dprop=m/s；dtrans=L/R；end-to-end delay = m/s+

20、L/R；刚刚离开host A；还在链路上，未到达host B；已经到达了host B；m=L/R*s = 892.86km6. Suppose users share a 1 Mbps link. Also suppose each user requires 100 kbps when transmitting, but each user transmits only 10 percent of the time. (See the discussion of packet switching versus circuit switching in Section 1.3.) 答：a：1M

21、bps/100kbps = 10；b：p=0.1；c：(C40,n)*(p)n*(1-p)(40-n)；d：P(11人或以上人使用)=7. In this problem, we consider sending real-time voice from Host Ato Host B over a packet-switched network (VoIP). Host Aconverts analog voice to a digital 64 kbps bit stream on the fly. Host Athen groups the bits into 48-byte packe

22、ts. There is one link between Hosts Aand B; its transmission rate is 1 Mbps and its propagation delay is 2 msec. As soon as Host Agathers 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 the tim

23、e 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)?答：考虑包中的第一个bit,在这个比特被传输之前，包中的所有比特都必须被生成。这需要传输数据包所需的时间是传播延迟为2 msec。解码之前的延迟是7 msec + 896sec + 2msec= 9.896msec因此延迟为9.896 msec。 Consider the discussion in Section 1.3 of packe

24、t switching versus circuit switching in which an example is provided with a 1 Mbps link. Users are generating data at a rate of 100 kbps when busy, but are busy generating data only with probability p = 0.1. Suppose that the 1 Mbps link is replaced by a 1 Gbps link. a. What is N, the maximum number

25、of users that can be supported simultaneously under circuit switching? b. Now consider packet switching and a user population of M users. Give a formula (in terms of p, M, N) for the probability that more than N users are sending data.答：a).10000 b). Cnm * Pn *(1-q)(m-n)9. In the above problem, suppo

26、se R1 = R2 = R3 = R and dproc = 0. Further suppose the packet switch does not store-and-forward packets but instead immediately transmits each bit it receives before waiting for the entire packet to arrive. What is the end-to-end delay?答：由于bits是立即传输的，包交换不引入任何延迟;因此不引入传输延迟。D（end-end） = L/R + d1/s1 + d

27、2/s2-8 + 24 + 12 = 44 msec。10. Consider the queuing delay in router buffer (preceding an outbound link). Suppose all packets are L bits, the transmission rate is R bps, and that N packets simultaneously arrive at the buffer every LN/R seconds. Find the average queuing delay of a packet. (Hint:The qu

28、euing delay for the first packet is zero; for the second packet L/R; for the third packet 2L/R.The Nth packet has already been transmitted when the second batch of packets arrives.)答：总延迟=(0+1+ n -1) L/R平均延迟= (n -1)L/2R11. Suppose N packets arrive simultaneously to a link at which no packets are curr

29、ently 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?答 (L/R + 2L/R + . + (N-1)L/R)/N = L/(RN) * (1 + 2 + . + (N-1) = L/(RN) * N(N-1)/2 = LN(N-1)/(2RN) = (N-1)L/(2R)12. Consider a packet of length L whi

30、ch 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 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

31、dproc. Assuming no queuing delays, in terms of di, si, Ri, (i = 1, 2), and L, what is the total end-to-end delay for the packet? 答：第一个端系统需要L/R1时间将数据包传输到第链路上，包花费d1/s1在一个链路上，交换机在接收到整个数据包后处理延迟为dproc，包交换机需要L/R2将数据包传输到第二个链路上;包花费d2/s2在第二个链路上传播，d（end-end） = L/R1+ L/R2 + d1/s1 + d2/s2 + dproc。代入方程，得到8 + 8 +

32、 16 + 4+ 1 = 37msec。13. Consider the queuing delay 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 I =60x=6*10823. Suppose two hosts, A and B, are separated by 10,000 kilometers and are connected by a direct link of

33、R = 2 Mbps. Suppose the propagation speed over the link is 2.5108 meters/sec.答：a. Rd.=R*m/s= 1Mbps * 10000*/(2.5*) = 40000bitsb.40000bitsc. 带宽延迟取决于链路中的最大包的bitsd. width = 10000*/40000 = 500m比足球场大e. width = m+(R*m/s) = s/R24. In modern packet-switched networks, the source host segments long, applicati

34、on-layer messages (for example, an image or music file) into smaller packets and sends the packets into the network. The receiver then reassembles the packets back into the original message. We refer to this process as message segmentation. Figure 1.24 illustrates the end-to-end transport of a messa

35、ge with and without message segmentation. Consider a message is 810 bits long that is to be sent from source to destination in Figure 1.24. Suppose each link in the figure is 2 Mbps. Ignore propagation, queuing, and processing delays.答：第一个交换机接收第二个包的时间= 2*1msec = 2m seca 7.5/1.5 = 5 sec, 5sec * 3 = 15secb.3*1+4999*1= 5.002msc从源主机发送第一个包到第一个交换机的时间为1.5*103/1.5*106 = 1msd信息包必须按正确的顺序到达目的地。消息分割导致许多更小的数据包。包头大小对所有数据包都是一样的，因此总字节数更多。25. Consider sending a