计算机网络习题2解析

第一章——计算机网络概述一、填空题1.计算机网络的网络资源包括________、________和________。

2.1969年12月，Internet的前身________的投入运行，标志着计算机网络的兴起。

3.国际标准化组织（英文简称____）在1984年正式颁布了________________使计算机网络体系结构实现了标准化。

4.________________是计算机网络最基本的功能之一。

5.计算机网络是________技术与________技术结合的产物。

6.Internet的应用有________、信息发布、电子商务、远程音频、视频应用。

7.计算机网络是由________系统和________系统构成的；从逻辑功能上看，则是由________和________组成的；从拓扑结构看是由一些________和________构成的。

8.________________又称网络单元，一般可分为三类：________、________和________。

9.________是指两个网络节点之间承载信息和数据的线路，可分为______________和____________。

10.__________提供访问网络和处理数据的能力，由主机系统、终端控制器和终端组成；__________是计算机网络中负责数据通信的部分，主要完成数据的传输、交换以及通信控制，由________、________组成。

11.网络硬件系统是指构成计算机网络的硬件设备，包括各种____________、_______及________；网络软件主要包括____________、____________和____________。

12.__________是计算机网络的主体，按其在网络中的用途和功能的不同，可分为________和________两大类。

13.____________是网络中用户使用的计算机设备，又称______；____________是通过网络操作系统为网上工作站提供服务及共享资源的计算机设备。

第二章1. (Q2)For a communication session between a pair of processes, which process is the client and which is the server?Answer:The process which initiates the communication is the client; the process that waits to be contacted is the server..2. (Q3) What is the difference between network architecture and application architecture?Answer:Network architecture refers to the organization of the communication processintolayers (e.g., the five-layer Internet architecture). Application architecture, on the other hand, is designed by an application developer and dictates the broadstructure of the application (e.g., client-server or P2P)3. (Q4) What information is used by a process running on one host to identify a process running on another host?Answer: The IP address of the destination host and the port number of the destinationsocket.4. (Q6) Referring to Figure 2.4, we see that none of the application listed in Figure 2.4 requires both no data loss and timing. Can you conceive of an application that requires no data loss and that is also highly time-sensitive?Answer: There are no good example of an application that requires no data loss and timing.If you know of one, send an e-mail to the authors5. (Q9) Why do HTTP, FTP, SMTP, and POP3 run on top of TCP rather than on UDP?Answer: The applications associated with those protocols require that all application databe received in the correct order and without gaps. TCP provides this servicewhereas UDP does not.6. (Q11) What is meant by a handshaking protocol?Answer: A protocol uses handshaking if the two communicating entities first exchangecontrol packets before sending data to each other. SMTP uses handshaking at theapplication layer whereas HTTP does not.7. (Q13) Telnet into a Web server and send a multiline request message. Include in the request message the If-modified-since: header line to force a response message with the 304 Not Modified status code.Answer: Issued the following command (in Windows command prompt) followed by theHTTP GET message to the “” web server:> telnet 80Since the index.html page in this web server was not modified since Fri, 18 May2007 09:23:34 GMT, the following output was displayed when the abovecommands were issued on Sat, 19 May 2007. Note that the first 4 lines are theGET message and header lines input by the user and the next 4 lines (startingfrom HTTP/1.1 304 Not Modified) is the response from the web server.8. (Q14) Consider an e-commerce site that wants to keep a purchase record for each of its customers. Describe how this can be done with cookies.Answer: When the user first visits the site, the site returns a cookie number. This cookie number is stored on the user’s host and is managed by the browser. During each subsequent visit (and purchase), the browser sends the cookie number back to the site. Thus the site knows when this user (more precisely, this browser) is visiting the site.9. (Q15) 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 Alice’s host to Bob’s host. Be sure to list the series of application-layer protocols that are used to move the message between the two hosts.Answer: Message is sent from Alice’s host to her mail server over HTTP. Alice’s mail server then sends the message to Bob’s mail server over SMTP. Bob then transfers the message from his mail server to his host over POP3.10. (Q10) Recall that TCP can be enhanced with SSL to provide process-to-process securityservices, including encryption. Does SSL operate at the transport layer or the application layer? If the application developer wants TCP to be enhanced with SSL, what does the developer have to do?Answer: SSL operates at the application layer. The SSL socket takes unencrypted data fromthe application layer, encrypts it and then passes it to the TCP socket. If theapplication developer wants TCP to be enhanced with SSL, she has to include theSSL code in the application.11. (Q16) Print out the header of an e-mail message you have recently received. How manyReceived: header lines are there? Analyze each of the header lines in the message.Answer: from 65.54.246.203 (EHLO )Received：(65.54.246.203) by with SMTP; Sat, 19 May 2007 16:53:51 -0700from ([65.55.135.106]) by Received: with Microsoft SMTPSVC(6.0.3790.2668); Sat, 19 May 2007 16:52:42 -0700 Received: from mail pickup service by with Microsoft SMTPSVC; Sat,19 May 2007 16:52:41 -0700Message-ID: <*******************************************>Received: from 65.55.135.123 by with HTTP; Sat, 19 May 2007 23:52:36 GMTFrom: "prithuladhungel"<***************************>To: ******************Bcc:Subject: Test mailDate: Sat, 19 May 2007 23:52:36 +0000Mime-Version:1.0Content-Type: Text/html; format=flowedReturn-Path: ***************************Figure: A sample mail message headerReceived: This header field indicates the sequence in which the SMTP serverssend and receive the mail message including the respective timestamps.In this example there are 4 “Received:” header lines. This means the mailmessage passed through 5 different SMTP servers before being delivered to thereceiver’s mail box. The la st (forth) “Received:” header indicates the mailmessage flow from the SMTP server of the sender to the second SMTP server inthe chain of servers. The sender’s SMTP server is at address 65.55.135.123 andthe second SMTP server in the chain is .The third “Received:” header indicates the mail message flow from the secondSMTP server in the chain to the third server, and so on.Finally, the first “Received:” header indicates the flow of the mail message fromthe forth SMTP server to t he last SMTP server (i.e. the receiver’s mail server) inthe chain.Message-id: The message has been given this*************************************************(bybay0-omc3-s3.bay0.hotmail.com. Message-id is a unique string assigned by the mail systemwhen the message is first created.From: This indicates the email address of the sender of the mail. In the givenexample,**************************************To: This field indicates the email address of the receiver of the mail. In theexample, the ****************************Subject: This gives the subject of the mail (if any specified by the sender). In theexample, the subject specified by the sender is “Test mail”Date: The date and time when the mail was sent by the sender. In the example,the sender sent the mail on 19th May 2007, at time 23:52:36 GMT.Mime-version: MIME version used for the mail. In the example, it is 1.0.Content-type: The type of content in the body of the mail message. In theexample, it is “text/html”.Return-Path: This specifies the email address to which the mail will be sent if thereceiver of this mail wants t o reply to the sender. This is also used by the sender’smail server for bouncing back undeliverable mail messages of mailer-daemonerror messages. In the example, the return path is“***************************”.12. (Q18) Is it possible for an organizat ion’s Web server and mail server to have exactly thesame alias for a hostname (for example, )? What would be the type for the RR that contains the hostname of the mail server?Answer: Yes an organization’s mail server and Web server can have the sa me alias for ahost name. The MX record is used to map the mail server’s host name to its IPaddress.13. (Q19) Why is it said that FTP sends control information “out-of-band”?Answer:FTP uses two parallel TCP connections, one connection for sending controlinformation (such as a request to transfer a file) and another connection foractually transferring the file. Because the control information is not sent over thesame connection that the file is sent over, FTP sends control information out ofband.14. (P6) Consider an HTTP client that wants to retrieve a Web document at a given URL. The IPaddress of the HTTP server is initially unknown. What transport and application-layer protocols besides HTTP are needed in this scenario?Answer:Application layer protocols: DNS and HTTPTransport layer protocols: UDP for DNS; TCP for HTTP15. (P9) Consider Figure2.12, for which there is an institutional network connected to theInternet. Suppose that the average object size is 900,000 bits and that the average request rate from the institution’s browsers to the origin servers is 10 requests per second. Also suppose that the amount of time it takes from when the router on the Internet side of the access link forwards an HTTP request until it receives the response is two seconds on average (see Section 2.2.5).Model the total average response times as the sum of the average access delay (that is, the delay from Internet router to institution router) and the average Internet delay. For the average access delay, use △/(1-△β), where △is the average time required to send an object over the access link and βis the arrival rate of objects to the access link.a. Find the total average response time.b. Now suppose a cache is installed in the institutional LAN. Suppose the hit rate is 0.6. Findthe total response time.Answer:a.The time to transmit an object of size L over a link or rate R is L/R. The average timeisthe average size of the object divided by R:Δ= (900,000 bits)/(1,500,000 bits/sec) = 0.6 secThe traffic intensity on the link is (1.5 requests/sec)(0.6 sec/request) = 0.9. Thus, theaverage access delay is (0.6 sec)/(1 - 0.9) = 6 seconds. The total average response timeis therefore 6 sec + 2 sec = 8 sec.b.The traffic intensity on the access link is reduced by 40% since the 40% of therequestsare satisfied within the institutional network. Thus the average access delayis(0.6 sec)/[1–(0.6)(0.9)] = 1.2 seconds. The response time is approximately zero iftherequest is satisfied by the cache (which happens with probability 0.4); the averageresponse time is 1.2 sec + 2 sec = 3.2 sec for cache misses (which happens 60% of thetime). So the average response time is (0.4)(0 sec) + (0.6)(3.2 sec) = 1.92 seconds.Thusthe average response time is reduced from 8 sec to 1.92 sec.16. (P12) What is the difference between MAIL FROM: in SMTP and From: in the mail messageitself?Answer: The MAIL FROM: in SMTP is a message from the SMTP client that identifies the senderof the mail message to the SMTP server. TheFrom: on the mail message itself is NOTanSMTP message, but rather is just a line in the body of the mail message.17. (P16) Consider distributing a file of F = 5 Gbits to N peers. The server has an upload rate ofu s = 20 Mbps, and each peer has a download rate of d i =1 Mbps and an upload rate of u. For N = 10, 100, and 1,000 and u = 100 Kbps, 250 Kbps, and 500 Kbps, prepare a chart giving the minimum distribution time for each of the combinations of N and u for both client-server distribution and P2P distribution.Answer:For calculating the minimum distribution time for client-server distribution, we use thefollowing formula:D cs = max {NF/u s , F/d min }Similarly, for calculating the minimum distribution time for P2P distribution, we use thefollowing formula:D P 2P = max {F /u s ,F /d min ,NF /( u s + u i n i =1 )} Where,F = 5 Gbits = 5 * 1024 Mbits u s = 20 Mbps d min = d i = 1 MbpsClient Server:N 10 100 1000 200 Kbps10240 51200 512000 u 600 Kbps10240 51200 512000 1Mbps10240 51200 512000Peer to Peer:N 10 100 1000 200 Kbps10240 25904.3 47559.33 U 600 Kbps10240 13029.6 16899.64 1 Mbps10240 10240 10240。

计算机网络(第二版)课后习题答案计算机网络(第二版)课后习题答案一、绪论计算机网络是指在多个计算机之间传输数据和共享资源的系统。

随着互联网的普及和发展，计算机网络已经成为现代社会中不可或缺的组成部分。

本文将回答《计算机网络(第二版)》课后习题，并提供详细的解答。

二、物理层1. 什么是物理层？物理层的任务是什么？物理层是计算机网络模型中的第一层，主要负责传输比特流。

其任务包括确定物理传输媒介的规范、数据的编码认证、物理连接的建立和维护等。

2. 传输媒介可分为哪几种类型？各有什么特点？传输媒介可分为有线传输媒介和无线传输媒介两种类型。

有线传输媒介包括双绞线、同轴电缆和光纤等，其特点是传输速度快、传输距离较长、抗干扰能力强。

无线传输媒介包括无线电波和红外线等，其特点是灵活性高、易于扩展和部署，但传输速度和距离受到限制。

3. 什么是调制和解调？其作用是什么？调制是将数字信号转换为模拟信号的过程，解调是将模拟信号转换为数字信号的过程。

调制和解调的作用是在发送端将数字数据转换为适合在传输媒介上传输的模拟信号，然后在接收端将模拟信号转换为可被计算机理解的数字数据。

4. 什么是信道复用？常见的信道复用技术有哪些？信道复用是指通过合理地利用通信线路，将不同用户的数据流合并在一起传输的技术。

常见的信道复用技术包括频分复用（FDM）、时分复用（TDM）和码分复用（CDM）等。

三、数据链路层1. 数据链路层有哪些基本的功能？数据链路层的基本功能包括封装成帧、物理寻址、错误检测和流量控制等。

封装成帧将网络层交付的数据分成适当的数据帧进行传输；物理寻址通过物理地址标识源和目的设备；错误检测使用帧检验序列等方法检测传输中的错误；流量控制通过控制数据的发送速率来保证接收端能够正确接收数据。

2. 什么是差错控制？常见的差错检测技术有哪些？差错控制是指在数据传输过程中采取一定的机制来检测和纠正传输中发生的差错。

常见的差错检测技术有纵向奇偶校验、循环冗余检验（CRC）和海明码等。

1．请举例说明信息、数据与信号之间的关系。

答：数据通信中被传输的二进制代码称为数据，数据是信息的载体。

数据涉及对事物的表示形式，信息涉及对数据所表示内容的解释。

数据通信的任务是要传输二进制代码的比特序列，而不需要解释代码所表示的内容。

例如，二进制编码按从高位到低位的顺序排列，英文单词“NETWORK”的ASCII码编码的二进制比特序列应该是“1001110 1000101 1010100 1010111 1001111 1010010 1001011”。

如果要从一个主机将这个比特序列准确地传送到另一个主机，并且两个主机都使用ASCII编码，那么另一个主机就可以将接收的比特序列解释为“NETWORK”。

2.通过比较说明双绞线、同轴电缆与光纤等3种常用传输介质的特点。

答：传输介质是网络中连接收发双方的物理通路，也是通信中实际传送信息的载体。

网络中常用的传输介质有：双绞线、同轴电缆和光缆。

(1) 双绞线由按规则螺旋结构排列的2根、4根或8根绝缘导线组成。

各线对螺旋排列的目的是使线对之间的电磁干扰最小。

双绞线分为屏蔽双绞线与非屏蔽双绞线两种。

双绞线的主要特点是：连接距离比较短，抗干扰性比较好，支持点—点连接与多点连接，价格低于其他传输介质，安装与维护比较方便。

(2) 同轴电缆由内导体、外屏蔽层、绝缘层及外部保护层组成。

同轴介质的特性参数由内、外导体及绝缘层的电参数与机械尺寸决定。

同轴电缆可以分为基带同轴电缆和宽带同轴电缆两种。

同轴电缆的主要特点是：连接距离比较长，抗干扰性很好，支持点—点连接与多点连接，价格介于双绞线与光缆之间，安装与维护比较方便。

(3) 光缆是网络传输介质中性能最好、应用前途最广泛的一种。

光纤是一种直径为50～100μm、能传导光波的柔软介质。

在折射率较高的单根光纤外面，用折射率较低的包层包裹起来构成一条光纤通道；由多条光纤组成一束就构成一条光缆。

光纤分为单模光纤与多模光纤两种。

光缆的主要特点是：低损耗和宽频带、高数据传输速率、低误码率和安全保密性好，价格高于双绞线与同轴电缆，安装与维护比较困难。