Solucionario De Tamebaum

COMPUTER NETWORKS
FOURTH EDITION

PROBLEM SOLUTIONS

ANDREW S. TANENBAUM
Vrije Universiteit
Amsterdam, The Netherlands

PRENTICE HALL PTR
UPPER SADDLE RIVER, NJ 07458

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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 theLAN 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, a56-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 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 ofpropagation 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 theresponse 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 balances) built into it. Thisinertia 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 specialinterest 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
(threespeeds or no line), so the total number of topologies is 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...