《计算机网络》自顶向下方法与Internet特色(第3版)习题解答

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"《计算机网络——自顶向下方法与Internet特色(第三版)习题答案》是针对英文版书籍的一份详细解答文档，由Jim Kurose和Keith Ross合著。该答案集仅供教师使用，禁止复制或分发给他人，也不能在公开网站上发布。所有内容受1996-2005年J.F.Kurose和K.W.Ross的版权保护，并保留所有权利。" 在《计算机网络——自顶向下方法》一书中，作者采用了一种自顶向下的教学方法，即从应用层开始，逐层深入到网络的底层机制。这种方法有助于读者理解网络如何为用户提供服务，然后逐渐揭示其工作原理。书中的习题答案涵盖了从基础概念到复杂问题的各个层面，旨在帮助学生巩固和深化对计算机网络的理解。第1章的复习问题涉及了基本的网络术语和概念。例如，问题1指出"主机"和"端系统"这两个词在本书中可以互换使用，它们包括个人电脑、工作站、Web服务器、邮件服务器、连接到互联网的PDA、WebTV等设备，这些都是网络通信中的重要终端。问题2通过一个生动的情境来解释网络通信的过程。Alice（来自国家A的大使）想要邀请Bob（国家B的大使）参加晚宴。在这个情境中，Alice不能简单地打电话告诉Bob立即前来，因为这涉及到跨越地理距离的信息传递，需要通过网络进行。这引出了网络通信的基本需求：如何在两个远程的端系统之间安全、可靠地传递信息，这正是TCP/IP协议栈的主要任务。在书中，读者会学习到网络的五层模型（应用层、传输层、网络层、数据链路层和物理层）及其各自的功能，如应用层的HTTP、FTP等协议，传输层的TCP和UDP，以及网络层的IP协议。此外，还会涉及路由选择、拥塞控制、错误检测与纠正、网络安全和隐私保护等多个主题。通过解答这些问题，学生将能够掌握网络通信的基本原理，了解Internet的架构和特性，为后续深入研究网络技术打下坚实基础。这份习题答案文档是教师评估学生理解和应用网络知识的重要工具，也是学生自我检验学习效果的有效资源。

close the persistent connection. It does so by including the including the connection-

token "close" in the Connection-header field of the http request/reply.

Problem 6.

The total amount of time to get the IP address is

RTTRTTRTT

Once the IP address is known,

RTT elapses to set up the TCP connection and another

RTT elapses to request and receive the small object. The total response time is

RTTRTTRTTRTT

Problem 7.

oon

RTTRTTRTTRTT 232

⋅

+Λ

RTTRTTRTT

oon

RTTRTTRTTRTT 22

+Λ

RTTRTTRTT

oon

RTTRTTRTTRTT

+ 2

RTTRTTRTT

Problem 8.

HTTP/1.0: GET, POST, HEAD.

HTTP/1.1: GET, POST, HEAD, OPTIONS, PUT, DELETE, TRACE, CONNECT.

See RFCs for explanations.

Problem 9.

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

∆= (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

Problem 13.

For a given input of domain name (such as ccn.com), IP address or network

administrator name,

whois database can be used to locate the corresponding registrar,

whois server, DNS server, and so on.

f) An attacker can use the whois database and nslookup tool to determine the IP address

ranges, DNS server addresses, etc., for the target institution.

g) By analyzing the source address of attack packets, the victim can use whois to obtain

information about domain from which the attack is coming and possibly inform the

administrators of the origin domain.

Problem 14.

Because the link is full duplex, you have 128 kbps in each direction, and the uploading

does not interfere with the downloading. However, for an asymmetric link, uploading can

significantly reduce the download rate due to metered acks. (The explanation of this

somewhat involved and beyond the scope of the chapter.)

Problem 15.

There are N nodes in the overlay network. There are N(N-1)/2 edges.

Problem 16.

a) In this case, each of the five Gnutella clients immediately learns that it has one less

neighbor. Consider one of these five clients, called, Bob. Suppose Bob has only three

neighbors after X drops out. Then Bob needs to establish a TCP connection with another

peer. Bob should have a fresh list of active peers; he sequentially contacts peers on this

list until one accepts his TCP connection attempt.

b) In this case, Bob does not immediately know that X has departed. Bob will only learn

about X’s departure when it attempts to send a message (query or ping) to X. When Bob

attempts to send a message, Bob’s TCP will make several unsuccessful attempts to send

the message to B. Bob’s TCP will then inform the Gnutella client that X is down. Bob

will then try to establish a TCP connection with a new peer (see part (a)) to rebuild a fifth

connection.

Problem 17.

a) The advantage of sending the QueryHit message directly over a TCP connection from

Bob to Alice is that the QueryHit message is routed by the underlying Internet without

passing through intermediate peers; thus, the delay in sending the message from Bob to

Alice should be substantially less. The disadvantage is that each peer that has a match

would ask Alice to open a TCP connection; Alice may therefore have to opens tens or

hundreds of TCP connections for a given query. Furthermore, there will be additional

complications if Alice is behind a NAT (see Chapter 4).

b) When a QueryHit message enters a peer, the peer records in a table the MessageID

along with an identifier of the TCP socket from which the message arrived. When the

same peer receives a QueryHit message with the same MessageID, it indexes the table

and determines the socket to which it should forward the message.

c) When the Query message reaches Bob, it contains an ordered list of all the IP

addresses of the peers the message passed through between Alice and Bob. When Bob

sends back a QueryHit message, it would copy the ordered list into the message. When a

peer receives the QueryHit message, it can use the list to determine the next peer in the

reverse path.

Problem 18.

Answer is unchanged for Ping/Pong messages. Replace Query message with Ping

message and QueryHit message with Pong message

Problem 19.

a) Each super-group leader is responsible for roughly 200

= 40,000 nodes. Therefore, we

would need about 100 super-group leaders to support 4 million nodes.

b) Each group leader might store the meta-data for all of the files its children are sharing.

A super-group leader might store all of the meta-data that its group-leader children store.

An ordinary node would first send a query to its group leader. The group leader would

respond with matches and then possibly forward the message to its super-group leader.

The super-group leader would respond (through the overlay network) with its matches.

The super-group leader may further forward the query to other super-group leaders.

Problem 20.

Alice sends her query to at most N neighbors. Each of these neighbors forwards the query

to at most M = N-1 neighbors. Each of those neighbors forwards the query to at most M

neighbors. Thus the maximum number of query messages is

N + NM + NM

+ … + NM

(K-1)

= N(1 + M + M

+ … + M

(K-1)

)

= N(1-M

)/(1-M)

= N[(N-1)

- 1]/(N-2) .

Problem 21.

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