计算机网络第六版自顶向下教材复习题与问题解答

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《计算机网络：自顶向下方法》第六版英文教材的课后习题与问题解答是一份专门为教师设计的参考资料，包含了原书作者Jim Kurose和Keith Ross所著第五版《计算机网络：自顶向下方法》的复习问题和实践题目的详细解答。该文档的版本日期为2012年5月，仅限教学人员使用，禁止私自复制或向他人分发，包括其他教师。发布在公共网站上的行为也被明确禁止。对于寻求更新解决方案的人士，编者愿意提供最新的解决方案手册。该书强调了"host"和"end system"这两个术语在文本中的通用性，指的是个人电脑、工作站、Web服务器和邮件服务器等各类计算机系统。第一章的复习问题涉及网络基础知识，可能探讨了网络的定义、层次结构以及主机和终端系统在通信中的角色。例如，问题可能会问及为何在讨论网络时可以将这两种术语互换使用，或者它们如何共同构成网络的基本组件。答案部分可能详细解释了诸如网络协议栈、TCP/IP模型的各个层及其功能、域名系统(DNS)的工作原理，以及端口在数据传输中的作用等内容。此外，还可能涉及网络拓扑结构，如总线型、环形、星型和网状网络的区别，以及这些结构对网络性能的影响。对于网络安全和协议的理解，可能会有针对防火墙、加密技术、安全协议（如SSL/TLS）以及网络攻击防御策略的问题。这些问题旨在帮助学生深入理解计算机网络的运作原理，并通过实践应用提升他们的理论知识。请注意，由于版权原因，这些解答仅供教学人员内部参考，确保学术诚信和尊重知识产权是至关重要的。若想获取完整和最新的答案，请咨询官方渠道或直接联系作者。

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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