《计算机网络：自顶向下方法》官方解答(英文版)

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"《计算机网络：自顶向下方法》第六版官方解答(英文版)" 《计算机网络：自顶向下方法》是一本广泛用于教学和自学的权威教材，它由Jim Kurose和Keith Ross合著。书中采用了一种独特的自顶向下的方法来教授计算机网络的知识，即从应用层开始，逐渐深入到网络的底层机制。这份官方解答是针对第五版教材的，包含了书中的复习问题和习题的解答，旨在帮助教师和学生更好地理解和掌握课程内容。解答集的版本日期为2012年5月，仅供教师使用，禁止复制或分发给他人，也不应将其发布在公共网站上。作者鼓励有需要的人直接联系他们获取最新版本的解答手册。该书的作者对多年来提供帮助的学生和同事表示感谢，特别是Hong Gang Zhang、Rakesh Kumar、Prithula Dhungel和Vijay Annapureddy。同时，他们也感谢所有提出建议和纠正错误的读者。所有材料的版权自1996年至2012年由J.F. Kurose和K.W. Ross持有，所有权利保留。在第一章的复习问题中，提到了"主机"（host）和"端系统"（end system）这两个术语。在本书中，这两个词被互换使用，它们涵盖了个人电脑、工作站、Web服务器和邮件服务器等不同类型的设备。这反映了网络中与之交互的各种终端系统的概念。本章的讨论可能涵盖网络的基本原理，如OSI模型和TCP/IP协议栈，以及主机在互联网上的通信方式。问题可能涉及如何封装数据包，从应用层协议（如HTTP、FTP）到传输层的TCP和UDP，再到网络层的IP，最后通过数据链路层和物理层在物理介质上传输比特流。此外，还可能涉及端系统的地址（如IP地址和MAC地址）以及端到端的通信过程。通过解答这些问题，学生能够深入理解网络的各个层次，包括它们的功能、工作原理以及如何协同工作以实现可靠的数据传输。这将为后续章节的学习打下坚实的基础，包括网络的性能、安全性和新兴技术等内容。

b) In this example of traceroutes from one city in France and fro m 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.

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