计算机网络6版教师专用复习题与问题解答

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《计算机网络：自顶向下方法》第六版的复习问题和问题解答文档是一个专门为教育工作者设计的教学辅助工具。这份文档包含了第五版Jim Kurose和Keith Ross所著的经典教材《计算机网络：自顶向下方法》的详细答案，旨在帮助教师在教学过程中为学生提供参考。它强调了这些解决方案仅供教师使用，禁止私自复制或分发给其他人员，包括其他教师，并且不得将内容公开发布到互联网上。文档发布日期为2012年5月，反映了当时版本的解决方案。作者对多年来参与准备和提供反馈的学生、同事表示感谢，特别提到了Hong Gang Zhang、Rakesh Kumar、Prithula Dhungel和Vijay Annapureddy等人的贡献。同时，也感谢所有读者的建议和纠错，他们共同提升了文档的质量。在第一章的复习问题中，"主机"和"端系统"这两个术语被互换使用，这表明在该书的讨论中，它们具有同等含义。端系统包括个人电脑（PC）、工作站、Web服务器以及邮件服务器等各类设备，这些是构成计算机网络的基本元素。书中采用自顶向下的方法来阐述网络体系结构，从全局视角出发，逐步分解各个层次的功能和协议，让读者更好地理解网络通信的原理。通过查阅这份解决方案，教师可以确保学生正确理解和掌握课程中的关键概念，如网络层模型、数据传输、路由选择、网络安全等。对于那些希望深入学习或准备考试的学生来说，这份资料也是一个宝贵的补充资源。然而，为了尊重版权和维护教学秩序，仅限于授权人员获取并用于教学目的。

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.

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克里斯蒂亚诺·宇

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