《自顶向下(第三版)》计算机网络习题解答

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"这是一份全面的《自顶向下(第三版)》计算机网络课程的课后习题英文答案，适合教师参考。" 在计算机网络领域，《自顶向下》是一本广泛使用的经典教材，它通过互联网的角度来讲解网络原理，采用自顶向下的方法，即从应用层开始逐层深入到物理层。这本书由Jim Kurose和Keith Ross共同编写，其第二版的解决方案包含了对复习问题和习题的详细解答。这份解答文档是专为教师提供的，旨在帮助他们更好地理解和教授课程内容，不应复制或分发给他人。教师可以向作者请求最新的解决方案手册。所有材料受到1996年至2002年间的版权保护，由J.F.Kurose和K.W.Ross所有，保留所有权利。第一章的复习问题涉及到网络基础知识。例如： 1. “主机”和“端系统”这两个术语在本书中是互换使用的。端系统包括个人电脑、工作站、Web服务器、邮件服务器、连接互联网的PDA、WebTV等设备，这些都是网络中的主要参与者。 2. 这个问题以一个外交情境为例，展示了网络通信中的协议概念。Alice邀请Bob参加晚餐，不会简单地直接说“现在就来吃饭”，而是会提议一个具体的时间和日期。这与网络通信中的TCP/IP模型类似，通信双方需要协商好参数（如时间即对应于网络中的时序）才能进行有效交流。这些复习问题和解答有助于读者巩固和深化对网络层次结构、网络设备角色以及通信协议的理解。通过解答，学生能更好地掌握如何将网络原理应用于实际场景，而教师则能更有效地评估学生的学习进度和理解程度。在后续章节中，可能会涵盖更多关于网络协议、传输层、网络层、数据链路层和物理层的具体问题，涉及IP地址、路由选择、错误检测与纠正、数据编码技术等内容。这份详尽的习题答案是学习和教学计算机网络的宝贵资源，能够帮助学生和教师深入理解网络的工作机制，并提升他们在实际问题中的分析和解决能力。

access the messages from many different machines (office PC, home PC, etc.). In the

download and keep configuration, messages are not deleted after the user retrieves

the messages. This can also be inconvenient, as each time the user retrieves the stored

messages from a new machine, all of non-deleted messages will be transferred to the

new machine (including very old messages).

16. See online Java applet.

17. A host‟s authoritative name server provides the “official” translation of a hostname to

an IP address (among other things). A host‟s local name server is a proxy to the DNS

system: the host first queries its local name server for a mapping; the local name

server then obtains the mapping on the behalf of the querying host.

18. Yes an organization‟s mail server and Web server can have the same alias for a host

name. The MX record is used to map the mail server‟s host name to its IP address.

19. You can find a link to nslookup at the online site.

20. With the UDP server, there is no welcoming socket, and all data from different clients

enters the server through this one socket. With the TCP server, there is a welcoming

socket, and each time a client initiates a connection to the server, a new socket is

created. Thus, to support n simultaneous connections, the server would need n+1

sockets.

21. For the TCP application, as soon as the client is executed, it attempts to initiate a TCP

connection with the server. If the TCP server is not running, then the client will fail to

make a connection. For the UDP application, the client does not initiate connections

(or attempt to communicate with the UDP server) immediately upon execution.

22. Web caching can bring the desired content “closer” to the user, perhaps to the same

LAN to which the user‟s host is connected. Web caching can reduce the delay for all

objects, even objects that are not cached, since caching reduces the traffic on links.

23. Web caches can (and often do!) use the conditional GET for checking freshness. The

idea is exactly the same as for user-host caching.

24. DNS is often used for redirection in a CDN. When a host receives an HTML page

from a Web server, the page can contain references to objects in the CDN. The host

will then do a DNS lookup to get the IP address of the CDN. DNS can then redirect

the host to best CDN server for the host.

25. The overlay network in a P2P file sharing system consists of the nodes participating

in the file sharing system and the logical links between the nodes. There is a logical

directed link (a “directed edge” in graph theory terms) from node A to node B if node

A sends datagrams that are explicitly addressed to node B‟s IP address. An overlay

network does not include routers. With Gnutella, when a node wants to join the

b) UDPClient doesn't establish a TCP connection with the server. Thus, everything

should work fine if you first run UDPClient, then run UDPServer, and then type some

input into the keyboard.

c) If you use different port numbers, then the client will attempt to establish a TCP

connection with the wrong process or a non-existent process. Errors will occur.

Problem 12.

See Web-server programming assignment for this chapter for guidance.

Problem 13.

a) The time to transmit an object of size L over a link or rate R is L/R. The average time

is the average size of the object divided by R:

= (900,000 bits)/(1,500,000 bits/sec) = .6 sec

The traffic intensity on the link is (1.5 requests/sec)(.6 msec/request) = .9. Thus, the

average access delay is (.6 sec)/(1 - .9) = 6 seconds. The total average response time is

therefore 6 sec + 2 sec = 8 sec.

b) The traffic intensity on the access link is reduced by 40% since the 40% of the

requests are satisfied within the institutional network. Thus the average access delay

is (.6 sec)/[1 – (.6)(.9)] = 1.2 seconds. The response time is approximately zero if the

request is satisfied by the cache (which happens with probability .4); the average

response time is 1.2 sec + 2 sec = 3.2 sec for cache misses (which happens 60% of the

time). So the average response time is (.4)(0 sec) + (.6)(3.2 sec) = 1.92 seconds. Thus

the average response time is reduced from 8 sec to 1.92 sec.

Problem 14.

a) umass.edu is directed to cache-1; eurecom.fr is directed to cache-2; upenn.edu is

directed to cache-1; wimba.com is directed to cache-1; w3c.org is directed to cache-0;

ietf.org is directed to cache-2; awl.com is directed to cache-1.

b) This hashing scheme only takes into account the hostname part of the URL. Thus all

browser requests for these objects will be directed to cache-1. For each of these

objects, the first request to cache-1 will be a cache miss; cache-1 will retrieve the

document from the origin server (at UPenn) and then forward the document to the

requesting browser. The subsequent requests for the same object will be cache hits,

and will be satisfied directly from cache-1.

c) It turns out that the www.upenn.edu objects still belong in cache-1. However, the

cache in which other objects belong may change. For example, a www.eurecom.fr

object now belongs in a different cache. This results in a cache miss for the first

request for the www.eurecom.fr object. Although subsequent requests for the same

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