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计算机网络英文美国原版本
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根据给定文件的信息,我们可以提炼出以下IT领域的关键知识点,主要围绕计算机网络教材中的概念、原理及实际应用展开。
### 计算机网络教材:第四版问题解答
#### 作者与出版信息
- **作者**:Andrew S. Tanenbaum,Vrije Universiteit Amsterdam,荷兰阿姆斯特丹自由大学教授。
- **出版商**:Prentice Hall PTR,位于美国新泽西州上萨德尔河(Upper Saddle River)。
- **版权**:所有权利保留,未经出版社书面许可,任何部分均不得以任何形式或手段复制。
- **出版年份**:2003年。
- **ISBN编号**:0-13-046002-8。
#### 第一章问题解答概览
- **狗的数据传输速率计算**:假设一只狗能够携带21GB(168Gbits)数据,在18km/h的速度下移动,其数据传输速率为840/x Mbps(x为距离,单位为千米)。当距离小于5.6公里时,狗的传输速率高于通信线路。
- **局域网模型优势**:局域网可以逐步扩展,即使只是一条长电缆,单点故障也不会导致整个网络瘫痪(若服务器有冗余)。相比其他网络,局域网可能成本更低,提供更强大的计算能力和更好的交互界面。
- **跨大陆光纤链路与本地调制解调器对比**:尽管跨大陆光纤链路具有很高的带宽,但传播延迟较高,因为光在数千公里的距离上传播需要时间。相比之下,同一建筑内的56kbps调制解调器虽然带宽低,但延迟也低。
- **语音通信对网络延时的要求**:语音通信需要一致的传输时间,因此网络抖动(传输时间的标准差)非常重要。短延迟但大波动实际上比稍长的延迟和小波动更糟糕。
- **信号传播速度与交换延迟**:信号的传播速度约为每秒20万公里,即每微秒200米。在10微秒内,信号可以传播2公里。因此,每个交换机相当于增加了2公里的额外电缆长度。如果客户端和服务器相隔5000公里,即使经过50个交换机,总路径仅增加100公里,占总距离的2%,表明在这些情况下,交换延迟不是主要因素。
- **请求与响应路径**:请求必须上行并下行,而响应亦然。总路径长度包括往返的时间和距离。
### 综合分析与应用
- **数据传输效率**:通过比较不同媒介和条件下的数据传输速率,如狗的物理运输与通信线路的电子传输,理解了实际环境中的数据传输效率和限制。
- **网络设计考量**:局域网的设计优势,如可扩展性、容错能力、成本效益和用户界面,展示了网络架构选择的关键考虑因素。
- **网络性能指标**:了解网络性能不仅取决于带宽,还受延迟和抖动等参数的影响,强调了在设计实时通信系统时的综合考量。
- **信号传播理论**:通过计算信号传播速度和额外电缆长度对总路径的影响,揭示了网络设计中信号延迟和交换机数量的实际意义。
本章节的知识点涵盖了计算机网络设计的基本原则、性能指标、信号传播理论以及实际应用场景中的考虑因素,为深入理解和应用计算机网络技术提供了坚实的基础。
COMPUTER NETWORKS
FOURTH EDITION
PROBLEM SOLUTIONS
ANDREW S. TANENBAUM
Vrije Universiteit
Amsterdam, The Netherlands
PRENTICE HALL PTR
UPPER SADDLE RIVER, NJ 07458
© 2003 Pearson Education, Inc.
Publishing as Prentice Hall PTR
Upper Saddle River, New Jersey 07458
All rights reserved. No part of this book may be reproduced, in any form or by any means,
without permission in writing from the publisher.
Printed in the United States of America
10987654321
ISBN 0-13-046002-8
Pearson Education LTD.
Pearson Education Australia PTY, Limited
Pearson Education Singapore, Pte. Ltd.
Pearson Education North Asia Ltd.
Pearson Education Canada, Ltd.
Pearson Educación de Mexico, S.A. de C.V.
Pearson Education — Japan
Pearson Education Malaysia, Pte. Ltd.
PROBLEM SOLUTIONS 1
SOLUTIONS TO CHAPTER 1 PROBLEMS
1. The dog can carry 21 gigabytes, or 168 gigabits. A speed of 18 km/hour
equals 0.005 km/sec. The time to travel distance x km is x/0.005 = 200x sec,
yielding a data rate of 168/200x Gbps or 840/x Mbps. For x<5.6 km, the
dog has a higher rate than the communication line.
2. The LAN model can be grown incrementally. If the LAN is just a long cable.
it cannot be brought down by a single failure (if the servers are replicated) It
is probably cheaper. It provides more computing power and better interactive
interfaces.
3. A transcontinental fiber link might have many gigabits/sec of bandwidth, but
the latency will also be high due to the speed of light propagation over
thousands of kilometers. In contrast, a 56-kbps modem calling a computer in
the same building has low bandwidth and low latency.
4. A uniform delivery time is needed for voice, so the amount of jitter in the net-
work is important. This could be expressed as the standard deviation of the
delivery time. Having short delay but large variability is actually worse than
a somewhat longer delay and low variability.
5. No. The speed of propagation is 200,000 km/sec or 200 meters/µsec. In 10
µsec the signal travels 2 km. Thus, each switch adds the equivalent of 2 km
of extra cable. If the client and server are separated by 5000 km, traversing
even 50 switches adds only 100 km to the total path, which is only 2%. Thus,
switching delay is not a major factor under these circumstances.
6. The request has to go up and down, and the response has to go up and down.
The total path length traversed is thus 160,000 km. The speed of light in air
and vacuum is 300,000 km/sec, so the propagation delay alone is
160,000/300,000 sec or about 533 msec.
7. There is obviously no single correct answer here, but the following points
seem relevant. The present system has a great deal of inertia (checks and bal-
ances) built into it. This inertia may serve to keep the legal, economic, and
social systems from being turned upside down every time a different party
comes to power. Also, many people hold strong opinions on controversial
social issues, without really knowing the facts of the matter. Allowing poorly
reasoned opinions be to written into law may be undesirable. The potential
effects of advertising campaigns by special interest groups of one kind or
another also have to be considered. Another major issue is security. A lot of
people might worry about some 14-year kid hacking the system and falsifying
the results.
2 PROBLEM SOLUTIONS FOR CHAPTER 1
8. Call the routers A, B, C, D, and E. There are ten potential lines: AB, AC,
AD, AE, BC, BD, BE, CD, CE, and DE. Each of these has four possibilities
(three speeds or no line), so the total number of topologies is 4
10
= 1,048,576.
At 100 ms each, it takes 104,857.6 sec, or slightly more than 29 hours to
inspect them all.
9. The mean router-router path is twice the mean router-root path. Number the
levels of the tree with the root as 1 and the deepest level as n. The path from
the root to level n requires n − 1 hops, and 0.50 of the routers are at this level.
The path from the root to level n − 1 has 0.25 of the routers and a length of
n − 2 hops. Hence, the mean path length, l, is given by
l = 0.5 × (n − 1) + 0.25 × (n − 2) + 0.125 × (n − 3) +
...
or
l =
i =1
Σ
∞
n (0.5)
i
−
i =1
Σ
∞
i(0.5)
i
This expression reduces to l = n − 2. The mean router-router path is thus
2n − 4.
10. Distinguish n + 2 events. Events 1 through n consist of the corresponding
host successfully attempting to use the channel, i.e., without a collision. The
probability of each of these events is p(1 − p)
n − 1
. Event n + 1 is an idle
channel, with probability (1 − p)
n
. Event n + 2 is a collision. Since these
n + 2 events are exhaustive, their probabilities must sum to unity. The proba-
bility of a collision, which is equal to the fraction of slots wasted, is then just
1 − np(1 − p)
n − 1
− (1 − p)
n
.
11. Among other reasons for using layered protocols, using them leads to break-
ing up the design problem into smaller, more manageable pieces, and layering
means that protocols can be changed without affecting higher or lower ones,
12. No. In the ISO protocol model, physical communication takes place only in
the lowest layer, not in every layer.
13. Connection-oriented communication has three phases. In the establishment
phase a request is made to set up a connection. Only after this phase has been
successfully completed can the data transfer phase be started and data trans-
ported. Then comes the release phase. Connectionless communication does
not have these phases. It just sends the data.
14. Message and byte streams are different. In a message stream, the network
keeps track of message boundaries. In a byte stream, it does not. For exam-
ple, suppose a process writes 1024 bytes to a connection and then a little later
writes another 1024 bytes. The receiver then does a read for 2048 bytes.
With a message stream, the receiver will get two messages, of 1024 bytes
PROBLEM SOLUTIONS FOR CHAPTER 1 3
each. With a byte stream, the message boundaries do not count and the
receiver will get the full 2048 bytes as a single unit. The fact that there were
originally two distinct messages is lost.
15. Negotiation has to do with getting both sides to agree on some parameters or
values to be used during the communication. Maximum packet size is one
example, but there are many others.
16. The service shown is the service offered by layer k to layer k + 1. Another
service that must be present is below layer k, namely, the service offered to
layer k by the underlying layer k − 1.
17. The probability, P
k
, of a frame requiring exactly k transmissions is the proba-
bility of the first k − 1 attempts failing, p
k − 1
, times the probability of the k-th
transmission succeeding, (1 − p). The mean number of transmission is then
just
k =1
Σ
∞
kP
k
=
k =1
Σ
∞
k(1 − p)p
k − 1
=
1 − p
1
33333
18. (a) Data link layer. (b) Network layer.
19. Frames encapsulate packets. When a packet arrives at the data link layer, the
entire thing, header, data, and all, is used as the data field of a frame. The
entire packet is put in an envelope (the frame), so to speak (assuming it fits).
20. With n layers and h bytes added per layer, the total number of header bytes
per message is hn, so the space wasted on headers is hn. The total message
size is M + nh, so the fraction of bandwidth wasted on headers is
hn /(M + hn).
21. Both models are based on layered protocols. Both have a network, transport,
and application layer. In both models, the transport service can provide a
reliable end-to-end byte stream. On the other hand, they differ in several
ways. The number of layers is different, the TCP/IP does not have session or
presentation layers, OSI does not support internetworking, and OSI has both
connection-oriented and connectionless service in the network layer.
22. TCP is connection oriented, whereas UDP is a connectionless service.
23. The two nodes in the upper-right corner can be disconnected from the rest by
three bombs knocking out the three nodes to which they are connected. The
system can withstand the loss of any two nodes.
24. Doubling every 18 months means a factor of four gain in 3 years. In 9 years,
the gain is then 4
3
or 64, leading to 6.4 billion hosts. My intuition says that is
much too conservative, since by then probably every television in the world
and possibly billions of other appliances will be on home LANs connected to
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