计算机网络课后习题习题1、2.doc

第一章 习题5.A factor in the delay of a store-and-forward packet-switching system is how long it takes to store and forward a packet through a switch. If switching time is 10 sec, is this likely to be a major factor in the response of a client-server system where the client is in New York and the server is in California? Assume the propagation speed in copper and fiber to be 2/3 the speed of light in vacuum.（问：交换时间是否是延迟的主要因素）Solution：5. No. The speed of propagation is 200,000 km/sec or 200 meters/sec. In 10sec the signal travels 2 km. Thus, each switch adds the equivalent of 2 kmof extra cable. If the client and server are separated by 5000 km, traversingeven 50 switches adds only 100 km to the total path, which is only 2%. Thus,switching delay is not a major factor under these circumstances.13. What is the principal difference between connectionless communication and connection-oriented communication?（13在无连接通讯和面向连接通讯两者之间，最主要的区别是什么？）Solution：13. Connection-oriented communication has three phases. In the establishmentphase a request is made to set up a connection. Only after this phase has beensuccessfully completed can the data transfer phase be started and data transported.Then comes the release phase. Connectionless communication does not have these phases. It just sends the data.解答：面向连接通信有三个阶段，在使用网络服务之前必须建立一个连接，只有在成功建立连接后，才可以通信。通信完成后还要进入释放连接阶段。无连接通信在发送报文前并不建立连接，它没有面向连接通信的三个阶段。18.Which of the OSI layers handles each of the following:a. Dividing the transmitted bit stream into frames.b. Determining which route through the subnet to use.（18OSI模型中哪一层处理以下问题：（A）把传输的位流分成帧；（B）在通过子网的时候决定使用哪条路由路径。）Solution：18. (a) Data link layer. (b) Network layer.解答：（A）数据链路层（B）网络层21.List two ways in which the OSI reference model and the TCP/IP reference model are the same. Now list two ways in which they differ.Solution：21. Both models are based on layered protocols. Both have a network, transport, and application layer. In both models, the transport service can provide a reliable end-to-end byte stream. On the other hand, they differ in several ways. The number of layers is different, the TCP/IP does not have session or presentation layers, OSI does not support internetworking, and OSI has both connection-oriented and connectionless service in the network layer.22.What is the main difference between TCP and UDP?（22TCP 和UDP 之间最主要的区别是什么?）Solution：22. TCP is connection oriented, whereas UDP is a connectionless service.解答：最主要的区别在于它们的可靠性与是否建立连接。TCP 是一个可靠的，面向连接的协议；而UDP则是一个不可靠的，无连接的协议。26.Why does ATM use small, fixed-length cells?（26ATM 为什么使用小的，固定长度的信元？）Solution：26. Small, fixed-length cells can be routed through switches quickly, and completely in hardware. Small, fixed-size cells also make it easier to build hardware that handles many cells in parallel. Also, they do not block transmission lines for very long, making it easier to provide quality-of-service guarantees.解答：因为信元是由硬件来完成的，之所以采用固定长度的信元，主要原因有：（1）容易设计出由硬件路由器来处理的短的，固定长度的信元。（2）最小的信元不会阻塞线路很长时间。（3）保证不会按错序交换信元。28. An image is 1024 x 768 pixels with 3 bytes/pixel. Assume the image is uncompressed. 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-Mbps Ethernet?（28一幅图像分辨率为1024*768 像素，每个像素用三字节表示。假设该图像没有没压缩。请问，通过56K 的调制解调器信道来传输这幅图像需要多长时间？通过1M 的电缆调制解调器？通过10M的以太网呢？通过100M 的以太网呢？）28. The image is 1024 768 3 bytes or 2,359,296 bytes. This is 18,874,368bits. At 56,000 bits/sec, it takes about 337.042 sec. At 1,000,000 bits/sec, ittakes about 18.874 sec. At 10,000,000 bits/sec, it takes about 1.887 sec. At100,000,000 bits/sec, it takes about 0.189 sec.解答：该图像占 1024*768*3=2359296 个字节。传输时间如下：56,000 bits/s 337.042 s1,000,000 bits/s 18.874 s10,000,000 bits/s 1.887 sec.100,000,000 bits/s 0.189 sec.第二章：6.What is the difference between a passive star and an active repeater in a fiber network?Solution：6. A passive star has no electronics. The light from one fiber illuminates anumber of others. An active repeater converts the optical signal to an electricalone for further processing.9.Is the Nyquist theorem true for optical fiber or only for copper wire?Solution：9. The Nyquist theorem is a property of mathematics and has nothing to do with technology. It says that if you have a function whose Fourier spectrum does not contain any sines or cosines above f, then by sampling the function at a frequency of 2f you capture all the information there is. Thus, the Nyquist theorem is true for all media.28.Ten signals, each requiring 4000 Hz, are multiplexed on to a single channel using FDM. How much minimum bandwidth is required for the multiplexed channel? Assume that the guard bands are 400 Hz wide.（28、有10 个信号，每个都要求4000Hz，现在用FDM 将它们复用在一条信道上。对于被复用的信道，最小要求多少带宽？假设防护频段为400Hz 宽。）Solution：28. There are ten 4000 Hz signals. We need nine guard bands to avoid anyinterference. The minimum bandwidth required is 4000 10 + 400 9 =43,600 Hz.解答：最小带宽=10*4000Hz+9*400Hz=43600Hz.注意：要分隔开10 个信道，只需要个400Hz 的防护频带29.Why has the PCM sampling time been set at 125 sec?（ 29、为什么PCM 采样时间被设置为125s？）Solution：29. A sampling time of 125 sec corresponds to 8000 samples per second.According to the Nyquist theorem, this is the sampling frequency needed tocapture all the information in a 4 kHz channel, such as a telephone channel.(Actually the nominal bandwidth is somewhat less, but the cutoff is notsharp.)解答：一条普通的电话线路带宽为 4kHz。根据尼奎斯特定理，只要每秒4000*2=8000 次的采样频率就能获取一个4kHz 的信道中的全部信息，再高的采样率也无意义。所以PCM 的采样时间定为1/8000=125s。30. What is the percent overhead on a T1 carrier; that is, what percent of the 1.544 Mbps are not delivered to the end user?（T1 线路上额外开销的百分比为多少？也就是说，1.544Mbps 中有百分之多少没有被递交给最终用户？）Solution：30. The end users get 7 24 = 168 of the 193 bits in a frame. The overhead is therefore 25/193 = 13%.解答：T1 线路中，有24 条被复用在一起的语音信道，每条信道按顺序依次在输出流中插入8 位，其中7 位用于数据，1 位用于控制（即开销）