计算机网络原理习题讲解(23页).doc

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1、-计算机网络原理习题讲解-第 23 页Chapter I 1. What is the difference between a host and an end system? List the types of end systems. Is a Web server an end system?2. What is a client program? What is a server program? Does a server program request and receive services from a client program?3. List six access tec

2、hnologies. Classify each one as residential access, company access, or mobile access.4. 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 dedicated.5

3、. Describe the most popular wireless Internet access technologies today. Compare and contrast them.6. What advantage does a circuit-switched network have over a packet-switched network? What advantages does TDM have over FDM in a circuit-switched network?7. Consider sending a packet from a source ho

4、st to a destination 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?8. How long does it take a packet of length 2,000 bytes to propagate over a link of distance 2,000 km, propagation speed m/s, and transmission rat

5、e 2 Mbps? More generally, how long does it take a packet of length L to propagate over a link of distance d, propagation speed s, and transmission rate R bps? Does this delay depend on packet length? Does this delay depend on transmission rate?9. What are the five layers in the Internet protocol sta

6、ck? What are the principal responsibilities of each of these layers?10. 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?11. What is an application-layer message? A transport-layer segment? A network-la

7、yer datagram? A link-layer frame? 12. This elementary problem begins to explore propagation delay and transmission delay, two central concepts in data networking. 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

8、the propagation speed along the link is s meters/sec. Host A is to send a packet of size L bits to Host B.a. Express the propagation delay, , in terms of m and s.b. Determine the transmission time of the packet, in terms of L and R.c. Ignoring processing and queuing delays, obtain an expression for

9、the end-to-end delay.d. Suppose Host A begins to transmit the packet at time t = 0. At time ,where is the last bit of the packet?e. Suppose is greater than . At time t = ,where is the first bit of the packet? f. Suppose is less than . At time t = , where is the first bit of the packet? g. Suppose ,

10、L = 100bits, and R = 28 kbps. Find the distance m so that equals .13. In modern packet-switched networks, the source host segments long, application-layer messages (for example, an image or a music file) into smaller packets and sends the packets into the network. The receiver then reassembles the p

11、ackets back into the original message. We refer to this process as message segmentation. Figure 1.24 illustrates the end-to-end transport of a message with and without message segmentation. Consider a message that is bits long that is to be sent from source to destination in Figure 1.24. Suppose eac

12、h link in the figure is 2 Mbps. Ignore propagation, queuing, and processing delays.a. Consider sending the message from source to destination without message segmentation. How long does it take to move the message from the source host to the first packet switch? Keeping in mind that each switch uses

13、 store-and-forward packet switching, what is the total time to move the message from source host to destination host?b. Now suppose that the message is segmented into 4,000 packets, with each packet being 2,000 bits long. How long does it take to move the first packet from source host to the first s

14、witch? When the first packet is being sent from the first switch to the second switch, the second packet is being sent from the source host to the first switch. At what time will the second packet be fully received at the first switch?c. How long does it take to move the file from source host to des

15、tination host when message segmentation is used? Compare this result with your answer in part (a) and comment.d. Discuss the drawbacks of message segmentation.14. 下列说法中,正确的是( )。A.在较小范围内布置的一定是局域网,币在较大范围内布置的一定是广域网B.城域网是连接广域网而覆盖园区的网络C.城域网是为淘汰局域网和广域网而提出的一种网络技术D.局域网是基于广播技术发展起来的网络,广域网是基于交换技术发展起来的向络解答:D。通常

16、而言,局域网的覆盖范围较小,而广域网的覆盖范围较大,但这并不绝对。有时候在一个不大的范围内采用广域网,这取决于应用的需要和是否采用单一网络等多种因素。特别是局域网技术的进步,使得其覆盖范围越来越大,达到几十千米的范围。城域网是利用广域网技术、满足一定区域需求的一种网络,事实上,城域网的范围弹性非常大。最初的局域网采用广播技术,这种技术一直被沿用,而广域网最初使用的是交换技术,也一直被沿用。 15. 相对于o的7层参考模型的低4层,TCP/IP协议集内对应的层次有( )。A.传输层、互联网层、网络接口层和物理层B.传输层、互联网层、网络接口层C.传输层、互联网层、ATM层和物理层D.传输层、网络

17、层、数据链路蜃和物理层解答:B。根据TCP/P分层模型可知,其对应OSI低4层的分别是传输层、互联网层、网络接口层。16. 在C/S模式的网络中,最恰当的是( )。A.客户机提出请求,服务器响应请求、进行处理并返回结果B.服务器有时可以同时为多个客户机服务C.客户机可以将服务器的资源各份在本地,以避免向服务器请求服务D.服务器永远是网络的瓶颈解答:A。根据C/S模式的定义,选项A描述了C/S模式的基本工作流程。服务器必须总能而不是有时可以同时为多个客户机服务,否则网络就没有了存在的价值。由于服务器的资源太庞大,而且很多资源因为知识产权、保密、管理复杂等一系列的原因,使客户机不可能都把服务器的资

18、源备份到本地。从表面上看,服务器可能是网络的瓶颈, 但事实上:在多数情况下网络的主要瓶颈不在服务器，而在通信线路。17. 比较分组交换与报文交换,说明分组交换优越的原因。解答:报文交换网络与分组交换的原理都是:将用户数据加上源地址、自的地址、长度、校验码等辅助信息封装成PDU,发送给下一个节点。下一个节点收到后先暂存报文,待输出线路空闲时再转发给下一个节点,重复这一过程直到到达目的节点。每个PDU可单独选择到达目的节点的路径。这种方式也称为存储转发方式。两者的不同之处是:分组交换所生成的PDU的长度较短,而且是固定的;而报文交换的PDU的长度不是固定的。正是这一差别,使得分组交换具有独特的优点

19、:缓冲区易于管理;分组的平均延迟更小,网络中占用的平均缓冲区更少;更易标准化;更适合应用。所以现在的主流网络基本上都可以看成是分组交换网络。18. 单顶选择题【1】第一个分组交换网是(A)。A.ARPAnet B.X.25 C.以太网 D.Internet【2】在大多数网络中,数据链路层都是用请求重发已损坏了帧的办法来解决发送出错问题。如果一个帧被损坏的概率是p,而且确认信息不会丢失,则发送一帧的平均发送次数是( d )A.1+p B. I-p C.1/(1+p) D. 1/(1-p)【3】物理层的电气特性规定的特性包括（ b ）A.接插件的形式 B.信号的电压值 C.电缆的长度 D.各引脚的

20、功能【4】网卡是完成( b )的功能。A.物理层 B.数据链路层 C.物理层和数据链路层 D.数据链路层和网络层【5】通信子网不包括( d )。A.物理层 B.数据链路层 C.网络层 D.传输层【6】当数据由端系统A传至端系统B时,不参与数据封装工作的是( a )。A.物理层 B.数据链路层 C.应用层 D.表示层【7】RFC是（）。（重庆大学2007年试题）A.因特网标准的形式 B.一种网络协议 C.一种网络文件格式 D.一种网络技术解析：所有的因特网标准都是以RFC的形式在因特网上发表。RFC(reguest for comments)的意思就是“请求评论”。所有的RFC文档都可以从因特网

21、上免费下载。但应注意,并非所有的RFC文档都是因特网标准,只有一小部分RFC文档最后才能变成因特网标准。RFC接收到时间的先后从小到大编上序号(即RFCxxxx,这里xxxx是阿拉伯数字)。一个RFC文档更新后就使用一个新的编号,并在文档中指出原来老编号的RFC文档已成为陈旧的。简言之,RFC是因特网标准的形式。所以选项A为正确答案。答案:A【8】在OSI的七层模型中，工作在第三层以上的网间连接设备是( )。(华中科技大学2003年试题)A.集线器 B.网关 C.网桥 D.中继器解析:集线器属于LAN与大型机以及LAN与WAN的互连。网挢工作于数据链数据通信系统中的基础设备,应用于OSI参考模

22、型第一层。集线器的设计目标主要是优化网络布线结构，简化网络管理，主要功能是对接收到的信号进行再生整形放大，以扩大网络的传输距离，同时把所有节点集中在以它为中心的结点上。网关亦称网间协议转换器,工作于OSI/RM的传输层、会话层、表示层和应用层。网关不仅具有路由器的全部功能,同时还可以完成因操作系统差异引起的通信协议之间的转换。网关可用于LAN-LAN、路层。它要求两个互连网络在数据链路层以上采用相同或兼容的网络协议。中继器是最简单的网络互连设备,主要完成物理层的功能,负责在两个结点的物理层上按位传递信息,完成信号的复制、调整和放大功能,以此来延长网络的长度。它位于0SI参考模型中物理层。由此可

23、知,网关工作于OSI/RM的传输层、会话层、表示层和应用层。所以选项B为正确答案。答案：B【9】在OSI七层结构模型中，处于数据链路层于传输层之间的是（ ）(华中科技大学2003年试题)A 物理层 B. 网络层 C. 会话层 D. 表示层 解析：OSI/RM网络结构模型将计算机网络体系结构的通信协议规定为物理层、数据链路层、网络层、传输层、会话层、表示层、应用层，共七层。因此,网络层处于数据链路层与传输层之间。所以选项B为正确答案。答案:B【10】完成路径选择功能是在OSI参考模型的（ ）。（华中科技大学2003年试题）A.物理层 B.数据链路层 C.网络层 D.传输层解析:物理层:主要是利用

24、物理传输介质为数据链路层提供物理连接,以便透明地传递比特流。数据链路层:分为MAC和LLC,传送以帧为单位的数据,采用差错控制,流量控制方法。网络层:实现路由选择、拥塞控制和网络互连功能,使用TCP和UDP协议。传输层:是向用户提供可靠的端到端服务,透明地传送报文,使用TCP协议。由此可知,网络层具有路径选择的功能。所以选项C为正确答案。答案:C【11】在TCP/IP协议簇的层次中，解决计算机之间通信问题是在（ ）。（华中科技大学2003年试题）A.网络接口层 B.网际层 C.传输层 D.应用层解析:TCP/IP协议族把整个协议分成四个层次:(1)网络接口层:负责接收P数据报,并把该数据报发送

25、到相应的网络上。从理论上讲,该层不是TCP/IP协议的组成部分,但它是TCP/IP协议的基础,是各种网络与TCP/IP协议的接口。 .(2)网络层(也叫网际层):网络层解决了计算机到计算机通信的问题。因特网在该层的协议主要有网络互联协议IP、网间控制报文协议ICMP、地址解析协议ARP等。(3)传输层:传输层提供一个应用程序到另一个应用程序之间端到端的通信。因特网在该层的协议主要有传输控制协议TCP、用户数据报协议UDP等。(4)应用层:是TCP/IP协议的最高层,与0sI参考模型的上三层的功能类似。因特网在该层的协议主要有文件传输协议FTP、远程终端访问协议Telnet、简单邮件传输协议sM

26、TP和域名服务协议DNS等。由此可知,网际层解决了计算机到计算机通信的问题。所以选项B为正确答案。答案:B【12】TCP/IP参考模型的网际层用于实现地址转换的协议有( )。 A.ARP B.ICMP C. UDP D.TCP 解析:ARP地址解析协议就是主机在发送帧前,将目标IP地址转换成目标MAC地址的过程。ARP协议的基本功能就是通过目标设备的IP地址,查询目标设备的MAC地址,以保证通信的顺利进行。所以选项A正确。ICMP协议是TCP/IP协议集中的一个子协议,属于网络层协议,主要用于在主机与路由器之间传递控制信息,包括报告错误、交换受限控制和状态信息等。因此,ICMP协议不具备地址转

27、换功能,排除选项B。用户数据报协议(UDP)是ISO参考模型中一种无连接的传输层协议,提供面向事务的简单不可靠信息传送服务。UDP协议基本上是IP协议与上层协议的接口。由此可知,UDP不具备地址转换功能,排除选项C。TCP是面向连接的传输层协议,，它提供的是一种虚电路方式的运输服务。因此，排除选项D。19. 简述面向连接服务于面向非连接服务的特点。解析:面向连接服务是电话系统服务模式的抽象。每一次完整的数据传输都必须经过建立连接、数据传输和终止连接三个过程。无连接服务是邮政系统服务模式的抽象。在无连接服务的情况下,两个实体之间的通信不需要先建立好一个连接,因此其下层的有关资源不需要事先进行预定

28、保留。这些资源将在数据传输时动态地进行分配。答案:面向连接服务的特点是,在数据交换之前,必须先建立连接。当数据交换结束后,则应终止这个连接。在数据传输过程中,各数据报地址不需要携带目的地址,而是使用连接号。接收到的数据与发送方的数据在内容和顺序上是一致的。无连接服务的特点,每个报文带有完整的目的地址,每个报文在系统中独立传送。无连接服务不能保证报文到达的先后顺序,先发送煦报文不一定先到。无连接服务不保证报文传输的可靠性。Chapter 21. For a communication session between a pair of processes, which process is th

29、e client and which is the server?2. What is the difference between network architecture and application architecture?3. What information is used by a process running on one host to identify a process running on another host?4. Suppose you wanted to do a transaction from a remote client to a server a

30、s fast as possible. Would you use UDP or TCP? Why?5. List the four broad classes of services that a transport protocol can provide. For each of the service classes, indicate if either UDP or TCP (or both) provides such a service.6. Why do HTTP, FTP, SMTP, and POP3 run on top of TCP rather than on UD

31、P?7. What is meant by a handshaking protocol?8. Suppose Alice, with a Web-based e-mail account (such as Hotmail or gmail), sends a message to Bob, who accesses his mail from his mail server using POP3. Discuss how the message gets from Alices host to Bobs host. Be sure to list the series of applicat

32、ion-layer protocols that are used to move the message between the two hosts.9. From a users perspective, what is the difference between the download-and-delete mode and the download-and-keep mode in POP3?10. Is it possible for an organizations Web server and mail server to have exactly the same alia

33、s for a hostname (for example, foo. com)? What would be the type for the RR that contains the hostname of the mail server?11. Why is it said that FTP sends control information out-of-band?12. True or false?a. A user requests a Web page that consists of some text and three images. For this page, the

34、client will send one request message and receive four response messages.b. Two distinct Web pages (for example, and www.mit.edu/students. html) can be sent over the same persistent connection.c. With nonpersistent connections between browser and origin server, it is possible for a single TCP segment

35、 to carry two distinct HTTP request messages.d. The Date: header in the HTTP response message indicates when the object in the response was last modified.13. Suppose within your Web browser you click on a link to obtain a Web page. The IP address for the associated URL is not cached in your local ho

36、st, so a DNS lookup is necessary to obtain the IP address. Suppose that n DNS servers are visited before your host receives the IP address from DNS; the successive visits incur an RTT of . Further suppose that the Web page associated with the link contains exactly one object, consisting of a small a

37、mount of HTML text. Let denote the RTT between the local host and the server containing the object. Assuming zero transmission time of the object, how much time elapses from when the client clicks on the link until the client receives the object?14. Referring to Problem P7, suppose the HTML file ref

38、erences three very small objects on the same server. Neglecting transmission times, how much time elapses witha. Non-persistent HTTP with no parallel TCP connections?b. Non-persistent HTTP with parallel connections?c. Persistent HTIP?通知第二次试验时间：11月8日，上午10:00 12:00地点：九教北401试验内容：实验指南实验四、实验五、实验六；参考GBN源代

39、码资料存储在： 邮箱：net_class_test 密码：123456abcdChapter 3 1. （教材R3）Describe why an application developer might choose to run an application over UDP rather than TCP.A: An application developer may not want its application to use TCPs congestion control, which can throttle the applications sending rate at tim

40、es of congestion. Often, designers of IP telephony and IP videoconference applications choose to run their applications over UDP because they want to avoid TCPs congestion control. Also, some applications do not need the reliable data transfer provided by TCP.2. （教材R4）Why is it that voice and video

41、traffic is often sent over TCP rather than UDP in todays Internet. (Hint: The answer we are looking for has nothing to do with TCPs congestion-control mechanism.)A: Since most firewalls are configured to block UDP traffic, using TCP for video and voice traffic lets the traffic though the firewalls.3

42、. （教材R5）Is it possible for an application to enjoy reliable data transfer even when the application runs over UDP? If so, how? A: Yes. The application developer can put reliable data transfer into the application layer protocol. This would require a significant amount of work and debugging, however.

43、4. （教材R6）Consider a TCP connection between Host A and Host B. Suppose that the TCP segments traveling from Host A to Host B have source port number X and destination port number y. What are the source and destination port numbers for the segments traveling from Host B to Host A? A: Source port numbe

44、r y and destination port number x.5. （教材R7）Suppose a process in Host C has a UDP socket with port number 6789. Suppose both Host A and Host B each send a UDP segment to Host C with destination port number 6789. Will both of these segments be directed to the same socket at Host C? If so, how will the

45、 process at Host C know that these two segments originated from two different hosts? A: Yes, both segments will be directed to the same socket. For each received segment, at the socket interface, the operating system will provide the process with the IP addresses to determine the origins of the indi

46、vidual segments.6. （教材R9）In our rdt protocols, why did we need to introduce sequence numbers? A: Sequence numbers are required for a receiver to find out whether an arriving packet contains new data or is a retransmission.7. （教材10）In our rdt protocols, why did we need to introduce timers? A: To hand

47、le losses in the channel. If the ACK for a transmitted packet is not received within the duration of the timer for the packet, the packet (or its ACK or NACK) is assumed to have been lost. Hence, the packet is retransmitted.8. （教材R14）Suppose Host A sends two TCP segments back to back to Host B over a TCP connection. The first segment has sequence number 90; the second has sequence number 110. a. How much data is in the first segment?b. Suppose that the first segment is lost but the second