计算机Light-PathProvisioningoftheWDMOpticalNetworks第2期

4.51 Calculate the pulse spreading caused by chromatic dispersion for BF04431-02 multimode grade-index fiber from SpecTran, (See Figure 3.20) operating at 1300 nm. Assume = 70 nm. S: Taking 0 = 1342.5 nm, S0 = 0.097 ps
t GI
L NA 4 L NA 4 32cN 13 32cn13
1 BR 4t
where N1 core group index of refraction
Hence, BRGI
8cn13 8 3 10 8 1.486 3 4.92 10 9 bit / s 4.92Gbit / s L NA4 1 10 3 0.200 4
P 4.34
2
4
t BR
2
2
PD=2.17 dB ( realizable!)
4.56 What is the bit-rate length limitation caused by chromatic disperson for the Spectran fiber referred to in Problem 4.51? S: (BR x L)max = 1 / [4 D()] 1 4( 2 ) D
4.55 What is meant by the term dispersion power penalty? Calculate the disperson power penalty for the pulse spread obtained in Problem 4.51. The BR is 2.5 Gbit/s. S: The amount of transmitting power required for compensating the increase bit error rate (BER) cause by attenuation is called dispersion power penalty. In Problem 4.51, ttotal = tmat = 18 ns (if fiber length is 100 km) PD(dB) = -10 log10 {exp[-(1/4 ) (ttotal)2 (BR)2]} = -10 log10 {exp[-1/4 x (18 x 10-9)2 ( x 2.5 x 109)2]} = 2.17 x 104 dB, enough to burn any type of fiber, impossible. if the fiber length is 1 km:

damaged is p, what is the mean number of transmissions 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：
medium-speed line, a low-speed line, or no line. If it takes 100 ms of computer time to generate and inspect each topology, how long will it take to inspect all of them?(E) 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 routers are interconnected in a centralized binary tree, with a router at each tree node. Router i communicates with router j by sending a message to the root of the tree. The root then sends the message back down to j. Derive an approximate expression for the mean number of hops per message for large n, 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 l＝n－2，平均路由路径为 2n-4。 10. A disadvantage of a broadcast subnet is the capacity wasted when multiple hosts 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 to use the channel with probability p during each slot. What fraction of the slots are wasted due to collisions?(H) 区分 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(1－p)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(1－p)n-1 － (1－p)n 11. What are two reasons for using layered protocols?(M) 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 principal difference between connectionless communication and connection-oriented communication?(E)

ANDREW S. TANENBAUM 秒，约533 msec.----- COMPUTER NETWORKS FOURTH EDITION PROBLEM SOLUTIONS 8. A collection of five routers is to be conn ected in a poi nt-to-poi nt sub net.Collected and Modified By Yan Zhe nXing, Mail To: Betwee n each pair of routers, the desig ners may put a high-speed line, aClassify: E aEasy, M ^Middle, H Hard , DaDeleteGree n: Importa nt Red: Master Blue: VI Others:Know Grey:—Unnecessary ----------------------------------------------------------------------------------------------ML V Chapter 1 In troductio nProblems2. An alter native to a LAN is simply a big timeshari ng system with termi nals forall users. Give two adva ntages of a clie nt-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。

一、二、三、汉译英1、时分多址：TDMA （Time Division Multiple Address/ Time Division Multiple Access）2、通用无线分组业务：GPRSGeneral Packet Radio Service3、国际电报电话咨询委员会：CCITT4、同步数字体系：SDH Synchronous Digital Hierarchy （同步数字序列）5、跳频扩频：FHSS frequency hopping spread spectrum6、同步转移模块：STM synchronous transfer module7、综合业务数字网：ISDNIntegrated Services Digital Network8、城域网：MAN Metropolitan Area Network9、传输控制协议/互联网协议：TCP/IPTransmission Control Protocol/Internet Protocol10、服务质量：QOS Quality of Service11、中继线：trunk line12、传输速率：transmission rate13、网络管理：network management14、帧结构：frame structure15、移动手机：Mobile Phone 手机Handset16、蜂窝交换机：（Cellular switches）（电池开关cell switch）(cell 蜂房)17、天线：Antenna18、微处理器：microprocessor 19、国际漫游：International roaming20、短消息：short message21、信噪比：SNR(Signal to Noise Ratio)22、数字通信：Digital communication23、系统容量：system capacity24、蜂窝网：cell network（cellular network）（Honeycomb nets）25、越区切换：Handover26、互联网：internet27、调制解调器：modem28、频谱：spectrum29、鼠标：Mouse30、电子邮件：electronic mail E-mail31、子网：subnet32、软件无线电：software defined radios33、网络资源：network resources四、英译汉1、mobile communication：移动通信2、Computer user：计算机用户3、Frame format：帧格式4、WLAN：wireless local area network 无线局域网络5、Communication protocol：通信协议6、Transmission quality：传输质量7、Remote terminal：远程终端Mode8、International standard：国际标准9、GSM：全球移动通信系统Global System for Mobile Communications10、CDMA：码分多址Code Division Multiple Access11、ITU：国际电信联盟International Telecommunication Union12、PCM：pulse code modulation 脉冲编码调制13、WDM：波分复用Wavelength Division Multiplex14、FCC：联邦通信委员会Federal communications commission15、PSTN：公用电话交换网Public Switched Telephone Network16、NNI：网络节点借口Network Node Interface17、WWW：万维网World Wide Web18、VOD：视频点播Video-On-Demand19、VLR：访问位置寄存器Visitor Location Register20、MSC：移动交换中心Mobile Switching Centre21、HLR：原籍位置寄存器Home Location Register22、VLSI：超大规模集成电路Very Large Scale Integrated Circuits23、Bluetooth technology：蓝牙技术24、Matched filter：匹配滤波器25、ADSL：非对称数字用户环路AsymmetricalDigital Subscriber Loop非对称数字用户线路(Asymmetric Digital Subscriber Line)26、GPS：全球定位系统Global Position System27、ATM：异步传输模式Asynchronous Transfer三、汉译英1、脉冲编码调制（PCM）依赖于三个独立的操作：抽样、量化和编码。