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计算机网络自顶向下方法第四版英文答案解析
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更新于2024-07-26
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《计算机网络:自顶向下方法第四版》英文版答案文档于2007年10月29日发布,主要包含了该教材第四版中的复习问题和练习题解答。该书由Jim Kurose和Keith Ross合著,强调互联网的应用,是一本深入探讨计算机网络原理的经典教材。作者明确指出,这些答案仅提供给授课教师使用,禁止学员私自复制或分发,也不鼓励在公共网站上分享解题内容。
第1章复习问题部分,作者指出"host"和"end system"在本书中的用法是等同的,指的是包括个人电脑、工作站、Web服务器、邮件服务器、互联网连接的PDA、WebTV等各种设备。例如,当提到大使Alice邀请大使Bob赴晚宴的例子时,并非简单通过电话邀请,而是暗示了更复杂的网络通信过程,可能涉及到电子邮件、在线邀请系统或者专用的外交通信平台。
在学习计算机网络时,第四版的内容与第五版相比,可能会有新的理论发展、案例分析或者技术更新,但核心概念和原则基本保持一致。对于理解网络架构、协议栈、数据传输、网络服务和安全等内容,解答部分提供了重要的实践指导和验证理解的工具。通过对比阅读答案,学生可以更好地掌握课堂讲授的知识,加深对网络协议如TCP/IP的理解,以及网络设计、配置和故障排查的实际操作技巧。
这份解决方案手册是教学过程中不可或缺的参考资料,它帮助教师检查学生的理解程度,同时对于自我学习者来说,也是检验自己理论知识和技能水平的重要途径。然而,使用时需确保尊重作者版权,合法合规地获取和使用教育资源。
s3.bay0.hotmail.com. Message-id is a unique string assigned by the mail system
when the message is first created.
From: This indicates the email address of the sender of the mail. In the given
example, the sender is “prithuladhungel@hotmail.com”
To: This field indicates the email address of the receiver of the mail. In the
example, the receiver is “prithula@yahoo.com”
Subject: This gives the subject of the mail (if any specified by the sender). In the
example, 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 19
th
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 the example,
it is “text/html”.
Return-Path: This specifies the email address to which the mail will be sent if the
receiver of this mail wants to reply to the sender. This is also used by the sender‟s
mail server for bouncing back undeliverable mail messages of mailer-daemon
error messages. In the example, the return path is
“prithuladhungel@hotmail.com”.
18. With download and delete, after a user retrieves its messages from a POP server,
the messages are deleted. This poses a problem for the nomadic user, who may
want to 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).
19. 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.
20. It is not necessary that Bob will also provide chunks to Alice. Alice has to be in
the top 4 neighbors of Bob for Bob to send out chunks to her; this might not occur
even if Alice is provides chunks to Bob throughout a 30-second interval.
21. Alice will get her first chunk as a result of she being selected by one of her
neighbors as a result of an “optimistic unchoke,” for sending out chunks to her.
22. 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 link (an “edge” in graph theory terms) from node A to node B if
there is a semi-permanent TCP connection between A and B. An overlay network
does not include routers. With Gnutella, when a node wants to join the Gnutella
network, it first discovers (“out of band”) the IP address of one or more nodes
already in the network. It then sends join messages to these nodes. When the node
receives confirmations, it becomes a member of the of Gnutella network. Nodes
maintain their logical links with periodic refresh messages.
23. It is a hybrid of client server and P2P architectures:
a) There is a centralized component (the index) like in the case of a client
server system.
b) Other functions (except the indexing) do not use any kind of central server.
This is similar to what exists in a P2P system.
24. Whenever a user, say Alice wants to locate another user in the instant messaging
system, say Bob, her node would send out a query message asking for the (unique)
username (Bob) into the overlay network, which would be flooded into the
network in the same manner a query requesting the location of a resource (eg: a
file in Gnutella network) is flooded. If the user with the username (Bob) is online,
as the query reaches the node where the user is online from, the node sends back
the query response to the original query source (Alice). In this way, Alice locates
the IP address of the node Bob is currently at in the instant messaging system.
The advantage of such a design is that there is no centralized component in the
system and hence no single point of failure. The disadvantage however, is that the
flooding of each such request into the overlay network results in huge traffic in
the network. One option is to have limited scope query as in Gnutella. However,
with this approach, the user might not be located even if she is online.
25. a) User location
b) NAT traversal
26. a) File Distribution
b) Instant Messaging
c) Video Streaming
d) Distributed Computing
27. 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.
28. 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
Chapter 2 Problems
Problem 1
a) F
b) T
c) F
d) F
Problem 2
Access control commands:
USER, PASS, ACT, CWD, CDUP, SMNT, REIN, QUIT.
Transfer parameter commands:
PORT, PASV, TYPE STRU, MODE.
Service commands:
RETR, STOR, STOU, APPE, ALLO, REST, RNFR, RNTO, ABOR, DELE,
RMD, MRD, PWD, LIST, NLST, SITE, SYST, STAT, HELP, NOOP.
Problem 3
Application layer protocols: DNS and HTTP
Transport layer protocols: UDP for DNS; TCP for HTTP
Problem 4
a) The document request was http://gaia.cs.umass.edu/cs453/index.html. The Host : field
indicates the server's name and /cs453/index.html indicates the file name.
b) The browser is running HTTP version 1.1,as indicated just before hte first <cr><lf>
pair.
c) The browser is requesting a persistent connection, as indicated by the Connection:
keep-alive.
d) This is a trick question. This information is not contained in an HTTP message
anywhere. So there is no way to tell this from looking at the exchange of HTTP messages
alone. One would need information from the IP datagrams (that carried the TCP segment
that carried the HTTP GET request) to answer this question.
Problem 5
a) The status code of 200 and the phrase OK indicate that the server was able to locate the
document successfully. The reply was provided on Tuesday, 07 Mar 2006 12:39:45
Greenwich Mean Time.
b) The document index.html was last modified on Saturday 10 Dec 2005 18:27:46 GMT.
c) There are 3874 bytes in the document being returned.
d) The first five bytes of the returned document are : <!doc. The server agreed to a
persistent connection, as indicated by the Connection: Keep-Alive field
Problem 6
a) Persistent connections are discussed in section 8 of RFC 2616 (the real goal of
this question was to get you to retrieve and read an RFC). Sections 8.1.2 and
8.1.2.1 of the RFC indicate that either the client or the server can indicate to the
other that it is going to 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.
b) HTTP does not provide any encryption services.
Problem 7
The total amount of time to get the IP address is
n
RTTRTTRTT
21
.
Once the IP address is known,
O
RTT
elapses to set up the TCP connection and another
O
RTT
elapses to request and receive the small object. The total response time is
no
RTTRTTRTTRTT
21
2
Problem 8
a)
oon
RTTRTTRTTRTT 232
1
no
RTTRTTRTT
1
8
.
b)
oon
RTTRTTRTTRTT 22
1
no
RTTRTTRTT
1
4
.
c)
oon
RTTRTTRTTRTT 2
1
no
RTTRTTRTT
1
3
.
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 R:
= (900,000 bits)/(15,000,000 bits/sec) = .06 sec
The traffic intensity on the link is (15 requests/sec)(.06 msec/request) = .9. Thus, the
average access delay is (.06 sec)/(1 - .9) = .6 seconds. The total average response time
is therefore .6 sec + 2 sec = 2.6 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 (.06 sec)/[1 – (.6)(.9)] = .12 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 .12 sec + 2 sec = 2.12 sec for cache misses (which happens 60% of
the time). So the average response time is (.4)(0 sec) + (.6)(2.12 sec) = 1.272 seconds.
Thus the average response time is reduced from 2.6 sec to 1.272 sec.
Problem 10
Parallel download would only share the 100K bandwidth among the 10 connections (each
getting just 10K bits/sec) thus, there is no significant advantage here. With persistent
HTTP we avoid the SYN and SYNACK exchange but that only requires 2 seconds (1
second to send the 100 bit SYN message over the 100 bps link, and 1 second to receive
the ACK). Given that each object takes 101 seconds to send and receive the ACK, the use
of pipelining gives only a 2 percent gain.
Problem 11
TCPServer.java
import java.io.*;
import java.net.*;
class TCPServer {
public static void main(String argv[]) throws Exception
{
String clientSentence;
ServerSocket welcomeSocket = new ServerSocket(6789);
while(true) {
Socket connectionSocket = welcomeSocket.accept();
BufferedReader inFromClient = new BufferedReader(new
InputStreamReader(connectionSocket.getInputStream( ) ) );
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