《计算机网络：自顶向下方法》(第6版)解答

需积分: 9 50 浏览量更新于2024-07-20 收藏 2.86MB PDF 举报

"计算机网络自顶向下方法答案(英文第六版)" 是一本针对Jim Kurose和Keith Ross合著的《计算机网络：自顶向下方法》第五版的解答指南，适用于教师使用，禁止复制或分发。该解答文档包含了书中复习问题和习题的答案。截至2012年5月的版本，提供了对计算机网络理论和实践的深入理解。作者对那些多年来帮助编撰解答手册的学生和同事表示感谢，特别是Hong Gang Zhang、Rakesh Kumar、Prithula Dhungel和Vijay Annapureddy，以及所有提出建议和纠正错误的读者。所有的内容均受1996年至2012年J.F.Kurose和K.W.Ross的版权保护，保留所有权利。第一章复习问题涉及到计算机网络的基础概念： 1. "Host" 和 "End System" 这两个词在本书中是同义的。它们都指的是网络中的终端系统，包括个人电脑、工作站、Web服务器和邮件服务器等。这些系统是网络通信的起点和终点，能够发起或接收数据传输。计算机网络自顶向下的方法是一种教学和理解网络原理的方式，它从应用层开始，逐步深入到更低层次的协议和技术。这种教学方式使学生能够从用户的角度出发，更好地理解网络如何支持各种应用程序。在第一章中，可能涵盖了网络的基本组件、OSI模型的介绍、网络的分类（局域网、广域网）、TCP/IP协议栈的概述，以及端系统之间的通信基础。复习问题可能会涉及以下知识点： - 应用层的作用，如HTTP、FTP、SMTP等协议的功能。 - 传输层的主要协议TCP和UDP，及其各自的特点，例如连接导向与无连接服务。 - 网络层的IP协议，包括IP地址的结构和子网掩码的概念。 - 数据链路层的MAC地址和以太网协议。 - 物理层的基本概念，如信号类型、传输介质和接口标准。解答这些问题将帮助学生巩固对计算机网络基本概念的理解，为后续章节的学习打下坚实基础。

b) In this example of traceroutes from one city in France and from another city in

Germany to the same host in United States, three links are in common including the

transatlantic link.

Problem 28

a) t

trans

+ t

prop

= 400 msec + 80 msec = 480 msec.

b) 20 * (t

trans

+ 2 t

prop

) = 20*(20 msec + 80 msec) = 2 sec.

c) Breaking up a file takes longer to transmit because each data packet and its

corresponding acknowledgement packet add their own propagation delays.

Problem 29

Recall geostationary satellite is 36,000 kilometers away from earth surface.

a) 150 msec

b) 1,500,000 bits

c) 600,000,000 bits

Problem 30

Let’s suppose the passenger and his/her bags correspond to the data unit arriving to the

top of the protocol stack. When the passenger checks in, his/her bags are checked, and a

tag is attached to the bags and ticket. This is additional information added in the

Baggage layer if Figure 1.20 that allows the Baggage layer to implement the service or

separating the passengers and baggage on the sending side, and then reuniting them

(hopefully!) on the destination side. When a passenger then passes through security and

additional stamp is often added to his/her ticket, indicating that the passenger has passed

through a security check. This information is used to ensure (e.g., by later checks for the

security information) secure transfer of people.

Problem 31

a) Time to send message from source host to first packet switch =

sec4sec

102

108





With store-and-forward switching, the total time to move message from source host

to destination host =

sec123sec4  hops

b) Time to send 1

packet from source host to first packet switch = .

sec5sec

102

101

m



. Time at which 2

packet is received at the first switch = time

at which 1

packet is received at the second switch =

sec10sec52 mm 

c) Time at which 1

packet is received at the destination host =

sec153sec5 mhopsm 

. After this, every 5msec one packet will be received; thus

time at which last (800

) packet is received =

sec01.4sec5*799sec15  mm

. It

can be seen that delay in using message segmentation is significantly less (almost

1/3

i. Without message segmentation, if bit errors are not tolerated, if there is a

single bit error, the whole message has to be retransmitted (rather than a single

packet).

ii. Without message segmentation, huge packets (containing HD videos, for

example) are sent into the network. Routers have to accommodate these huge

packets. Smaller packets have to queue behind enormous packets and suffer

unfair delays.

i. Packets have to be put in sequence at the destination.

ii. Message segmentation results in many smaller packets. Since header size is

usually the same for all packets regardless of their size, with message

segmentation the total amount of header bytes is more.

Problem 32

Yes, the delays in the applet correspond to the delays in the Problem 31.The propagation

delays affect the overall end-to-end delays both for packet switching and message

switching equally.

Problem 33

There are F/S packets. Each packet is S=80 bits. Time at which the last packet is received

at the first router is



 80

sec. At this time, the first F/S-2 packets are at the

destination, and the F/S-1 packet is at the second router. The last packet must then be

transmitted by the first router and the second router, with each transmission taking

S 80

sec. Thus delay in sending the whole file is

)2(







delay

To calculate the value of S which leads to the minimum delay,

FSdelay

400 

Problem 34

The circuit-switched telephone networks and the Internet are connected together at

"gateways". When a Skype user (connected to the Internet) calls an ordinary telephone, a

circuit is established between a gateway and the telephone user over the circuit switched

network. The skype user's voice is sent in packets over the Internet to the gateway. At the

gateway, the voice signal is reconstructed and then sent over the circuit. In the other

direction, the voice signal is sent over the circuit switched network to the gateway. The

gateway packetizes the voice signal and sends the voice packets to the Skype user.

剩余153页未读，继续阅读

CClusXX

粉丝: 0
资源: 1

《计算机网络：自顶向下方法》(第6版)解答

计算机网络自顶向下方法答案 英文第六版

计算机网络自顶向下方法第六版答案 英文版

计算机网络自顶向下方法答案(英文第六版).pdf.zip_stairsrhz_答案_自顶向下

计算机网络自顶向下方法答案(英文第六版)方法答案(英文第六版)

计算机网络自顶向下方法答案（英文第六版）

计算机网络自顶向下方法答案(英文第六版）

计算机网络自顶向下方法答案(英文第六版适用)

计算机网络自顶向下方法答案(英文第六版)1

计算机网络自顶向下方法答案(英文第六版).pdf

计算机网络自顶向下方法英文第六版 课后习题答案

最新资源

计算机网络自顶向下方法答案英文第六版

计算机网络自顶向下方法第六版答案英文版

计算机网络自顶向下方法英文第六版课后习题答案