Resolucion problemas tannenbaum redes

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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 inwriting from the publisher. Printed in the United States of America

10 9 8 7 6 5 4 3 2 1

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

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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 longcable. 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 computerin 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 network 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 200meters/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 totalpath 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 balances) 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 tobe considered. Another major issue is security. A lot of people might worry about some 14-year kid hacking the system and falsifying the results.

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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 topologiesis 410 = 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...
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