1、 VI Others:Know Grey: UnnecessaryChapter 1 IntroductionProblems2. An alternative to a LAN is simply a big timesharing system with terminals forall users. Give two advantages of a client-server system using a LAN.(M)使用局域网模型可以容易地增加节点。如果局域网只是一条长的电缆,且不会因个别的失效而崩溃( 例如采用镜像服务器)的情况下,使用局域网模型会更便宜。使用局域网可提供更多的计算
2、能力和更好交互式接口。3. The performance of a client-server system is influenced by two network factors:the bandwidth of the network (how many bits/sec it can transport) and the latency(how many seconds it takes for the first bit to get from the client to the server). Givean example of a network that exhibits
3、high bandwidth and high latency. Then givean example of one with low bandwidth and low latency.(E)横贯大陆的光纤连接可以有很多千兆位/秒带宽, 但是由于光速度传送要越过数千公里,时延将也高。相反,使用 56 kbps 调制解调器呼叫在同一大楼内的计算机则有低带宽和较低的时延。4. Besides bandwidth and latency, what other parameter is needed to give a goodcharacterization of the quality of
4、 service offered by a network used for digitizedvoice traffic?(M)声音的传输需要相应的固定时间,因此网络时隙数量是很重要的。传输时间可以用标准偏差方式表示。实际上,短延迟但是大变化性比更长的延迟和低变化性更糟。6. A client-server system uses a satellite network, with the satellite at a height of40,000 km. What is the best-case delay in response to a request?(E)由于请求和应答都必须通
5、过卫星,因此传输总路径长度为 160,000 千米。在空气和真空中的光速为 300,000 公里/秒, 因此最佳的传播延迟为 160,000/300,000秒,约 533 msec。8. A collection of five routers is to be connected in a point-to-point subnet.Between each pair of routers, the designers may put a high-speed line, amedium-speed line, a low-speed line, or no line. If it take
6、s 100 ms of computer timeto generate and inspect each topology, how long will it take to inspect all ofthem?将路由器称为 A,B,C,D 和 E.则有 10 条可能的线路;AB, AC, AD, AE, BC, BD, BE, CD, CE,和 DE每条线路有 4 种可能性(3 速度或者不是线路),拓扑的总数为 410 = 1,048,576。检查每个拓扑需要 100 ms,全部检查总共需要 104,857. 6 秒,或者稍微超过 29个小时。9. A group of 2n - 1 r
7、outers are interconnected in a centralized binary tree, with arouter at each tree node. Router i communicates with router j by sending a messageto the root of the tree. The root then sends the message back down to j. Derive anapproximate expression for the mean number of hops per message for large n
8、,assuming that all router pairs are equally likely.(H)这意味着,从路由器到路由器的路径长度相当于路由器到根的两倍。 若在树中,根深度为 1,深度为 n,从根到第 n 层需要 n-1 跳,在该层的路由器为 0.50。从根到 n-1 层的路径有 router 的 0.25 和 n-2 跳步。 因此,路径长度 l 为:l = 0.5*(n-1)+0.25*(n-2)+0.125*(n-3)结果化简为 ln2,平均路由路径为 2n-4。10. A disadvantage of a broadcast subnet is the capacity
9、wasted when multiplehosts attempt to access the channel at the same time. As a simplistic example,suppose that time is divided into discrete slots, with each of the n hosts attempting touse the channel with probability p during each slot. What fraction of the slots arewasted due to collisions?(H)区分
10、n-2 事件。 事件 1 到 n 由主机成功地、没有冲突地使用这条信道的事件组成。 这些可能性的事件的概率为 p(1p)n-1。事件 n+1 是一个空闲的信道,其概率为(1- p)n。事件 n+2 是一个冲突。由于事件 n+2 互斥,它们可能发生的事件必须统一合计。 冲突的可能性等于那些小部分的槽的浪费,只是1 np(1p)n-1 (1p)n11. What are two reasons for using layered protocols?通过协议分层可以把设计问题划分成较小的易于处理的片段分层意味着某一层的协议的改变不会影响高层或低层的协议13. What is the princip
11、al difference between connectionless communication andconnection-oriented communication?主要的区别有两条。其一:面向连接通信分为三个阶段,第一是建立连接,在此阶段,发出一个建立连接的请求。第二阶段,只有在连接成功建立之后,保持连接状态,才能开始数据传输。第三阶段,当数据传输完毕,必须释放连接。而无连接通信没有这么多阶段,它直接进行数据传输。其二:面向连接的通信具有数据的保序性, 而无连接的通信不能保证接收数据的顺序与发送数据的顺序一致。14. Two networks each provide reliab
12、le connection-oriented service. One of themoffers a reliable byte stream and the other offers a reliable message stream. Arethese identical? If so, why is the distinction made? If not, give an example of howthey differ.(E)不相同。在报文流中,网络保持对报文边界的跟踪;而在字节流中,网络不做这样的跟踪。例如,一个进程向一条连接写了 1024 字节,稍后又写了另外 1024 字节
13、。那么接收方共读了 2048 字节。对于报文流,接受方将得到两个报文。每个报文 1024字节。 而对于字节流,报文边界不被识别。接收方把全部的 2048 个字节当作一个整体,在此已经体现不出原先有两个报文的事实。15. What does negotiation mean when discussing network protocols? Give anexample.(E)协商就是要让双方就在通信期间将使用的某些参数或数值达成一致。最大分组长度就是一个例子。16. In Fig. 1-19, a service is shown. Are any other services implic
14、it in this figure?If so, where? If not, why not?服务是由 k 层向 k1 层提供的。服务必须由下层 k 提供,即,对层 k 的服务是由 k- 1 层提供的。17. In some networks, the data link layer handles transmission errors byrequesting damaged frames to be retransmitted. If the probability of a frames beingdamaged is p, what is the mean number of tr
15、ansmissions required to send a frame?Assume that acknowledgements are never lost.(M)假设某帧传到第 k 次才传输成功,起初 k-1 次传输皆尝试失败,概率为 pk-1 , 第 k次传输成功,概率为(1-p) ,则发送一帧成功的平均传输次数为:1. Which of the OSI layers handles each of the following:a. (a) Dividing the transmitted bit stream into frames.b. (b) Determining which route through the subnet to use.(E)把传输的比
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