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优质文档ANDREW S. TANENBAUM-COMPUTER NETWORKS FOURTH EDITION PROBLEM SOLUTIONSCollected and Modified By YanZhenXing, Mail To: Classify: EEasy, MMiddle, HHard,DDeleteGreen: Important Red: Master Blue: 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)运用局域网模型可以简洁地增加节点。假如局域网只是一条长的电缆,且不会因个别的失效而崩溃(例如采纳镜像效劳器)的状况下,运用局域网模型会更廉价。运用局域网可供给更多的计算实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 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,运用 56 kbps 调制解调器呼叫在同一大楼内的计算机那么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 service offered by a network used for digitizedvoice traffic?(M)声音的传输须要相应的固定时间,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。事实上,短延迟但是大变更性比更长的延迟和低变更性更糟。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)由于恳求和应答都必需通过卫星,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,因此最正确的传播延迟为 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 takes 100 ms of computer timeto generate and inspect each topology, how long will it take to inspect all ofthem?(E)将路由器称为 A,B,C,D和 E.那么有 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。检查每个拓扑须要 100 ms,全部检查总共须要 104,857. 6 秒,或者稍稍超过 29个小时。9. A group of 2n - 1 routers 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,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。假设在树中,根深度为 1,深度为 n,从根到第 n 层须要 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 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 n-2事务。事务 1 到 n 由主机胜利地、没有冲突地运用这条信道的事务组成。 这些可能性的事务的概率为 p(1p)n-1。事务 n+1 是一个空闲的信道,其概率为(1- p)n。事务 n+2 是一个冲突。由于事务 n+2 互斥,它们可能发生的事务必需统一合计。冲突的可能性等于那些小局部的槽的奢侈,只是1 np(1p)n-1 (1p)n11. 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变更不会影响高层或低层的协议13. What is the principal difference between connectionless communication andconnection-oriented communication?(E)优质文档主要的区分有两条。其一:面对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恳求。其次阶段,只有在连接胜利建立之后,保持连接状态,才能起先数据传输。第三阶段,当数据传输完毕,必需释放连接。而无连接通信没有这么多阶段,它干脆进展数据传输。其二:面对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依次和发送数据的依次一样。14. Two networks each provide reliable 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)不一样。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。那么接收方共读了 2048字节。对于报文流,承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,在此已经表达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运用的某些参数或数值达成一样。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 implicit in this figure?If so, where? If not, why not?(E)效劳是由 k 层向 k1 层供给的。效劳必需由下层 k 供给,即,对层 k 的效劳是由 k- 1 层供给的。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 transmissions required to send a frame?Assume that acknowledgements are never lost.(M)假设某帧传到第 k 次才传输胜利,起初 k-1 次传输皆尝试失败,概率为 pk-1 ,第 k次传输胜利,概率为(1-p),那么发送一帧胜利的平均传输次数为: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把传输的比特流划分为帧数据链路层确定运用哪条路径通过子网网络层.19. If the unit exchanged at the data link level is called a frame and the unitexchanged at the network level is called a packet, do frames encapsulate packets ordo packets encapsulate frames? Explain your answer.E帧封装包。 当一个包到达数据链路层时,整个数据包,包括包头、数据及全部内容,都用作帧的数据区。或者说,将整个包放进一个信封(帧)里面,(假如能装入的话)。21. List two ways in which the OSI reference model and the TCP/IP referencemodel are the same. Now list two ways in which they differ.M相像点:都是独立的协议栈的概念;层的功能也大体相像。不同点:OSI 更好的区分了效劳、接口和协议的概念,因此比 TCP/IP 具有更好的隐藏性,能够比拟简洁的进展替换;OSI 是先有的模型的概念,然后再进展协议的实现,而 TCP/IP 是先有协议,然后建立描述该协议的模型;层次数量有差异;TCP/IP没有会话层和表示层,OSI 不支持网络互连。OSI 在网络层支持无连接和面对连接的通信,而在传输层仅有面对连接的通信,而 TCP/IP 在网络层仅有一种通信模式无连接,但在传输层支持两种模式。22. What is the main difference between TCP and UDP?ETCP是面对连接的,而 UDP是一种数据报效劳。25. When a transferred between two computers, two acknowledgementstrategies are possible. In the first one, the chopped up into packets, which areindividually acknowledged by the receiver, but the as a whole is notacknowledged. In the second one, the packets are not acknowledged individually,but the entire acknowledged when it arrives. Discuss these two approaches.E优质文档- 2 - 优质文档优质文档假如网络简洁丧失分组,那么对每一个分组逐一进展确认较好,此时仅重传丧失的分组。假如网络高度牢靠,那么在不发过失的状况下,仅在整个文件传送的结尾发送一次确认,从而削减了确认的次数,节约了带宽;不过,即使有单个分组丧失,也需要重传整个文件。26. Why does ATM use small, fixed-length cells?E因为这样可以快速地经由交换机转发,并且这是在硬件上完成的。这样的设计使得制造可以同时并行处理多个 CELLS 的硬件设备更加简洁。另外,它们不会阻碍传输线路很久,更加简洁保证供给出高质量的效劳。28. An image is 1024 x 768 pixels with 3 bytes/pixel. Assume the image isuncompressed. How long does it take to transmit it over a 56-kbps modem channel?Over a 1-Mbps cable modem? Over a 10-Mbps Ethernet? Over 100-MbpsEthernet?(E)该图像大小为 1024 * 768 * 3 * 8 = 18,874,368 bits. 传输速率为 56Kbits/sec,须要 18,874,368 / 56,000 = 337.042 sec. 传输速率为 1Mbits/sec,须要 18,874,368 / 106 = 18.874 sec. 传输速率为 10Mbits/sec,须要 18,874,368 / 107 = 1.887 sec. 传输速率为 100Mbits/sec,须要 18,874,368 / 108 = 0.189 sec.29. Ethernet and wireless networks have some similarities and some differences.One property of Ethernet is that only one frame at a time can be transmitted on anEthernet. Does 802.11 share this property with Ethernet? Discuss your answer.(E)想象一下隐藏终端的问题。假设一个无线网络里有五台终端,从 A 至 E,使它们每一台都只可以联系到和其相邻的两个邻居之一,那么 A 在和 B 通讯的同时 D 可以和 E进展通讯。因此无线网络有潜在的并行性,这和以太网上不同的。30. Wireless networks are easy to install, which makes them inexpensive sinceinstallation costs usually far overshadow equipment costs. Nevertheless, they alsohave some disadvantages. Name two of them.E无线网络的缺点:一是平安性,偶然出此时此刻无线网络内的人都能监听到网络上传递的消息;再有就是牢靠性,无线网络在传输过程中会出现很多错误;另外,因为很多无线设备须要移动,电池运用寿命不长也是其缺点之一。Chapter 2 The Physical Problems2. A noiseless 4-kHz channel is sampled every 1 msec. What is the maximum datarate?E由尼奎斯特定理,无噪声信道最大数据传输率=2Hlog2V b/s。依题有带宽 H =4kHz,因此最大数据传输率确定于每次采样所产生的比特数log2V。假如每次采样产生 16bits,那么数据传输率可达 128kbps;假如每次采样产生 1024bits,那么可达 8.2Mbps。3. Television channels are 6 MHz wide. How many bits/sec can be sent iffour-level digital signals are used? Assume a noiseless channel.E依题有带宽 H = 6MHz,每次采样 log2V = 2bit由尼奎斯特定理,可发送的最大数据传输率为 2Hlog2V = 24Mbps。4. If a binary signal is sent over a 3-kHz channel whose signal-to-noise ratio is 20dB, what is the maximum achievable data rate?M由香农定理信道比为 S/N 的有噪声信道的最大数据传输率 = Hlog2(1+S/N)。依题知带宽 H = 3kHz,信噪比为 10lgS/N = 20 dB,知 S/N =100由于 log21016.658,该信道的信道容量为 3log2(1+100)=19.98kbps再依据尼奎斯特定理,发送二进制信号的 3kHz信道的最大数据传输速率为2Hlog2V = 2*3 log22 = 6kbps 综上,可以取得的最大数据传输速率为 6kbps。5. What signal-to-noise ratio is needed to put a T1 carrier on a 50-kHz line?MT1 信号的带宽 = 1.544 * 106Hz,为发送 T1信号,由香农定理,最大数据传输率= Hlog2(1+S/N) = 1.544 * 106Hz,依题知带宽 H = 50 kHz,解得 S/N = 2311 再由尼奎斯特定理 2Hlog2V = 2Hlog2S/N = 93 dB所以,在 50kHz线路上运用 T1载波须要 93dB的信噪比。7. How much bandwidth is there in 0.1 micron of spectrum at a wavelength of 1micron?(M)依题知频段为 0.1,波长为 1因此,在 0.1的频段中可以有 30THz。- 3 - 优质文档8. It is desired to send a sequence of computer screen images over an optical fiber.The screen is 480 x 640 pixels, each pixel being 24 bits. There are 60 screen imagesper second. How much bandwidth is needed, and how many microns of wavelengthare needed for this band at 1.30 microns?M12. Multipath fading is maximized when the two beams arrive 180 degrees out ofphase. How much of a path difference is required to maximize the fading for a50-km-long 1-GHz microwave link?E优质文档传输数据的速率为 480u61472X64024u61472X60bps,即 442Mbps。由公式,这里光速 c =300000km/s,依题 f=1GHz,所以微波的波长是 30cm。优质文档须要 442Mbps的带宽,对应的波长范围是。假如一个波比另一个波多行进 15cm,那么它们到达时将 180 异相。明显,答案和链路长度是 50km的事实无关。18. A simple telephone system consists of two end offices and a single toll office towhich each end office is connected by a 1-MHz full-duplex trunk. The averagetelephone is used to make four calls per 8-hour workday. The mean call duration is6 min. Ten percent of the calls are long-distance (i.e., pass through the toll office).优质文档9. Is the Nyquist theorem true for optical fiber or only for copper wire?D尼奎斯特定理是一个数学性质,不涉及技术处理。该定理说,假如你有一个函数,它的傅立叶频谱不包含高于 f的正弦和余弦,那么以 2 f的频率采样该函数,那么你就可以获得该函数所包含的全部信息。因此尼奎斯特定理适用于全部介质。10. In Fig. 2-6 the lefthand band is narrower than the others. Why?EWhat is the maximum number of telephones an end office can support? (Assume 4kHz per circuit.)E每部电话每小时做 0.5次通话,每次通话 6分钟。因此一部电话每小时占用一条电路 3分钟,60/3=20,即 20部电话可共享一条线路。由于只有 10%的呼叫是长途,所以 200部电话占用一条完全时间的长途线路。局间干线复用了 1000000/4000=250条线路,每条线路支持 200部电话,因此,一个端局可以支持的电话部数为200*250=50000。.21. The cost of a fast microprocessor has dropped to the point where it is nowpossible to put one in each modem. How does that affect the handling of telephoneline errors?E通常在物理层对于在线路上发送的比特不采纳任何过失订正措施。在每个调制解调器中都包括一个 CPU使得有可能在第一层中包含错误订正码,从而大大削减第二层所看到的错误率。由调制解调器做的错误处理可以对其次层完全透亮。此时此刻许多调制解调器都有内建的错误处理功能。优质文档由于这 3个波段的频率范围大体上相等,依据公式得小,才能保持f大约相等。,小的波段 也22. A modem constellation diagram similar to Fig. 2-25 has data points at thefollowing coordinates: (1, 1), (1, -1), (-1, 1), and (-1, -1). How many bps can a modemwith these parameters achieve at 1200 baud?E优质文档顺便指出,3 个带宽大致一样的事实是所运用的硅的种类的一个碰巧的特性反映。11. Radio antennas often work best when the diameter of the antenna is equal tothe wavelength of the radio wave. Reasonable antennas range from 1 cm to 5 metersin diameter. What frequency range does this cover?E- 4 - 每个波特有 4 个合法值,因此比特率是波特率的两倍。优质文档优质文档对应于 1200波特,数据速率是 2400bps。23. A modem constellation diagram similar to Fig. 2-25 has data points at (0, 1)and (0, 2). Does the modem use phase modulation or amplitude modulation?E相位总是 0,但运用两个振幅,因此这是干脆的幅度调制。24. In a constellation diagram, all the points lie on a circle centered on the origin.What kind of modulation is being used?E假如全部的点都在同一圆周上,那么它们有着同样的幅度,所以没有运用幅度调制。在星座图中从来就不运用频率调制,所以,这里所采纳的编码是纯相位调制。25. How many frequencies does a full-duplex QAM-64 modem use?E全双工的 QAM-64 运用了两个频率。一个给上行流,一个给下行流。调制机制本身只运用了相位和幅度调制,这里对频率不做调制。26. An ADSL system using DMT allocates 3/4 of the available data channels to thedownstream link. It uses QAM-64 modulation on each channel. What is the capacityof the downstream link?MDMT 指离散的多信道调制。这里总共有 256 条信道,减去 6 条给 POTS 以及再减少 2 条用于限制,余下的 248 条留给数据。依题其中的 3/4 即 186 条信道给下行流。ADSL 是以 4000 baud/s 进展调置。所以对 QAM-646 bits/baud可得每条信道的带宽为 24,000 bps所以下行流总的带宽为 24,000 bps*186=4.464 Mbps27. In the four-sector LMDS example of Fig. 2-30, each sector has its own36-Mbps channel. According to queueing theory, if the channel is 50% loaded, thequeueing time will be equal to the download time. Under these conditions, how longdoes it take to download a 5-KB Web page? How long does it take to download thepage over a 1-Mbps ADSL line? Over a 56-kbps modem?ELMDS 指本地多点分发效劳。一个 5-KB 的网页数据量为 40,000 bits,对于 36 Mbps的信道而言下载时间是 1.1msec,依题平均排队延迟时间也是 1.1msec,那么总下载时间是 2.2msec,对 ADSL 而言并没有排队延迟时间,所以 1 Mbps 的下载时间是 40msec,在 56 kbps 的条件下下载时间是 714 msec.28. Ten signals, each requiring 4000 Hz, are multiplexed on to a single channelusing FDM. How much minimum bandwidth is required for the multiplexedchannel? Assume that the guard bands are 400 Hz wide.E对于 10 个 4000 Hz的信号,我们须要运用 9 个防护频段来幸免可能的干扰。所须要的最小带宽为 4000 * 10 + 400 * 9 = 43,600 Hz29. Why has the PCM sampling time been set at 125 sec?E125的采样时间对应于每秒 8000次采样。一个典型的电话通道为 4kHz。依据尼奎斯特定理,为获得一个 4kHz的通道中的全部信息须要每秒 8000次的采样频率。30. What is the percent overhead on a T1 carrier; that is, what percent of the1.544 Mbps are not delivered to the end user?MT1 线路的每一帧中,端点用户运用 193位中的 1687*24位,开销占 25=193-168位,因此开销比例等于 25/193=13%。31. Compare the maximum data rate of a noiseless 4-kHz channel usinga.(a) Analog encoding (e.g., QPSK) with 2 bits per sample.b.(b) The T1 PCM system.E两种状况下均为 8000 次采样/秒。运用 2 进制编码,那么对 a 每次采样中发送 2 位数据,对 T1 线路,每次采样发送 7 位数据。所以相对的最大数据传输率为:(a)每次采样 2比特的模拟编码 2Hlog2V = 16 kbps(b) T1 PCM系统 2Hlog2V = 56 kbps32. If a T1 carrier system slips and loses track of where it is, it tries toresynchronize using the 1st bit in each frame. How many frames will have to beinspected on average to resynchronize with a probability of 0.001 of being wrong?M10 个帧。在数字通道上某些随机比特是 0101010101模式的概率是 1/1024。察看 10 个帧,假设每一帧中的第一位形成比特串 0101010101,那么判定同步胜利,而误判的概率为 1/1024,小于 0.001。33. What is the difference, if any, between the demodulator part of a modem andthe coder part of a codec? (After all, both convert analog signals to digital ones.) (M)有区分。编码器承受随意的模拟信号,并从它产生数字信号。而解调器仅仅承受调制了的正弦或余弦波,产生数字信号。优质文档- 5 - 优质文档优质文档34. A signal is transmitted digitally over a 4-kHz noiseless channel with onesample every 125 sec. How many bits per second are actually sent for each of theseencoding methods?(M)c.(a) CCITT 2.048 Mbps standard.d.(b) DPCM with a 4-bit relative signal value.e.(c) Delta modulation.aCCITT 2.048Mbps标准用 32个 8位数据样本组成一个 125的根本帧,30个信道用于传信息,2个信道用于传限制信号。在每一个 4kHz信道上发送的数据率就是 8*8000=64kbps。b差分脉码调制DPCM是一种压缩传输信息量的方法,它发送的不是每一次抽样的二进制编码值,而是两次抽样的差值的二进制编码。此时此刻相对差值是 4位,所以对应每个 4kHz信道实际发送的比特速率为 4*8000=32kbps。c增量调制的根本思想是:当抽样时间间隔 s t很短时,模拟数据在两次抽样之间的变更很小,可以选择一个适宜的量化值 v 作为阶距。把两次抽样的差异近似为不是增加一个 v 就是削减一个 v。这样只需用 1bit二进制信息就可以表示一次抽样结果,而不会引入很大误差。因此,此时对应每个 4kHz信道实际发送的数据速率为 1*8000=8kHz。35. A pure sine wave of amplitude A is encoded using delta modulation, with xsamples/sec. An output of +1 corresponds to a signal change of +A/8, and an outputsignal of -1 corresponds to a signal change of -A/8. What is the highest frequencythat can be tracked without cumulative error?(E)在波的 1/4周期内信号必需从 0上升到 A。为了能够跟踪信号,在 T/4的时间内假定波的周期是 T必需采样 8次,即每一个全波采样 32次,采样的时间间隔是 1/x,因此波的全周期必需足够的长,使得能包含 32次采样,即 T 32/x,或 f max=x/32。37. In Fig. 2-37, the user data rate for OC-3 is stated to be 148.608 Mbps. Showhow this number can be derived from the SONET OC-3 parameters.(H)根本的 SONET同步光网络帧是每 125产生 810字节。由于 SONET 是同步的,因此不管是否有实际要发送的数据,帧都存在。每秒 8000帧的速率正好符合全部数字电话系统中运用的 PCM信道的采样率。对于 810 字节的 SONET 帧,通常用 90 列乘以 9 行的矩形来描述,每个单元对应一个字节。每秒传送 8000 次,每次 8*810=6480 位,总数据传输率为 51.84Mbps。这就是根本的 SONET信道,它被称作同步传输信号 STS-1,全部的 SONET 干线都是 STS-1 的倍数。每一帧的前 3列被保存,用于管理信息系统,前 3行包含段开销,后 6行包含线路开销。剩下的87列包含 8798800050.112Mbps 的用户数据。用户数据称为同步载荷信封,即 SPE可以从帧内的任一位置起先,并不限于第 1 行,第 4 列。线路开销的第一行包含指向 SPE 的第一字节的指针,SPE 的第一列是路径开销。路径开销不是严格的 SONET构造,它在嵌入在载荷信封中。路径开销端到端的流过网络,因此把它和端到端的运载用户信息的 SPE相关联是有意义的。然而,它从可供给应终端的用户数据中的 50.112Mbps中又减去 19880000.576Mbps,使之变成 49.536Mbps。OC-3 相当于 3 个 OC-1 复用在一起,因此其用户数据传输速率是 49.546u61472X3 148.608 Mbps。39. What is the essential difference between message switching and packetswitching?E报文交换中,对于数据块的大小没有任何限制。分组交换那么对于数据块的大小有最大分组长度的限制,任何报文超出了这一限制都会被分割成小块的多个分组。40. What is the available user bandwidth in an OC-12c connection?H当一条线路例如 OC-3没有被复用,而是仅传输来自一个源的数据,那么在线路名称后面加一个字母 c表示 conactenation,即串联。因此,OC-3 表示了由 3条独立的 OC-1 线路构成的一条 155.52Mbps 线路,而 OC-3c 表示来自于单个源的155.52Mbps的数据流。OC-3c 流内的 3 个 OC-1 流被按列交替插入,首先是流 1的- 6 - 优质文档第 1列,流 2的第 1列,流 3的第 1列,随后是流 1的第 2列,流 2的第 2列,以此类推,最终形成 270列宽 9行高的帧。OC-3c 流中的用户实际数据传输速率比 OC-3 流的速率略高149.760Mbps和148.608Mbps,因为路径开销仅在 SPE 中出现一次,而不是当运用 3条单独 OC-1流时出现的 3次。换句话说,OC-3c 中 270列中的 260列可用于用户数据,而在OC-3中仅能运用 258 列。更高层次的串联帧如 OC-12c也存在这样的状况。OC-12c帧有 12*90=1080 列和 9 行。其中段开销和线路开销占 12*3=36列,这样同步载荷信封就有 1080-36=1044 列。SPE 中仅 1 列用于路径开销,结果就是 1043列用于用户数据。由于每列 9 个字节,因此一个 OC-12c 帧中用户数据比特数是 8 u61472X9104375096。每秒 8000帧,得到用户数据速率 750968000 =600768000bps,即 600.768Mbps。所以,在一条 OC-12c连接中可供给的用户带宽是 600.768Mbps。为了使分组交换比电路交换快,必需:所以:43. Suppose that x bits of user data are to be transmitted over a k-hop path in apacket-switched network as a series of packets, each containing p data bits and hheader bits, with x p + h. The bit rate of the lines is b bps and the propagationdelay is negligible. What value of p minimizes the total delay?M所须要的分组总数是 x/p,因此加上头信息的总数据量为(p+h)x/p 位。源端发送这些位须要时间为(p+h)x/pb,中间的路由器重传最终一个分组所花的总时间为(k-1)(p+h)/b,优质文档41. Three packet-switching networks each contain n nodes. The first network hasa star topology with a central switch, the second is a (bidirectional) ring, and thethird is fully interconnected, with a wire from every node to every other node. Whatare the best-, average-, and-worst case transmission paths in hops?E三种网络的特性如下:星型:最好为 2,最差为 2,平均为 2;环型:最好为 1,最差为 n/2,平均为 n/4因此我们得到的总的延迟为对该函数求 p的导数,得到,令,得到优质文档假如考虑 n为奇偶数,因为 p0,所以故时能使总的延迟最小。优质文档那么 n为奇数时,最坏为n-1/2,平均为n+1/4n为偶数时,最坏为 n/2,平均为 n2/4(n1) 全连接:最好为 1,最差为 1,平均为 1。42. Compare the delay in sending an x-bit message over a k-hop path in acircuit-switched network and in a (lightly loaded) packet-switched network. Thecircuit setup time is s sec, the propagation delay is d sec per hop, the packet size is pbits, and the data rate is b bps. Under what conditions does the packet network havea lower delay?(M)对于电路交换, t= s 时电路建立起来;ts+x/d时报文的最终一位发送完毕;t=s+x/d+kb 时报文到达目的地。而对于分组交换,最终一位在 t=x/b时发送完毕。为到达最终目的地,最终一个分组必需被中间的路由器重发 k-1 次,每次重发花时间 p/b,所以总的延迟为44. In a typical mobile phone system with hexagonal cells, it is forbidden to reusea frequency band in an adjacent cell. If 840 frequencies are available, how many canbe used in a given cell?B每个蜂窝单元都有 6 个邻居。假定中间的单元运用频率 A,它的邻居那么可以依次运用 B,C,B,C,B,C,也就是说,只需用三个相互独立的单元就足够了。所以,每个单元可以运用 280 种频率。47. Sometimes when a mobile user crosses the boundary from one cell to another,the current call is abruptly terminated, even though all transmitters and receiversare functioning perfectly. Why?E在邻近的蜂窝单元中频率不能复用。所以当一名用户从一个单元移动到另一个单元时,必需给他安排一个新的频率。假如当用户移到一个新的单元,但是当前的新单元中的全
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