U
R
U
G
R
TO
TU
, U
R
U
G
R
TO
TU
,2 PROBLEM SOLUTIONS FOR CHAPTER 2
5. 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.
6. A transcontinental fiber link might have many gigabits/sec of bandwidth, but
TU
the latency will also be high due to the speed of light propagation over thou-
sands of kilometers. Similarly, a satellite link may run at megabits/sec but have
a high latency to send a signal into orbit and back. In contrast, a 56-kbps
modem calling a computer in the same building has low bandwidth and low
latency. So do low-end local and personal area wireless technologies such as
Zigbee.
TO
7. No. The speed of propagation is 200,000 km/sec or 400 meters/sec. In 20
sec, the signal travels 4 km. Thus, each switch adds the equivalent of 4 km
of extra cable. If the client and server are separated by 5000 km, traversing
even 50 switches adds only 200 km to the total path, which is only 4%. Thus,
switching delay is not a major factor under these circumstances.
8. The delay is 1% of the total time, which means
100 s n
R
= 0. 01
29, 700 km
+ 100 s n
300, 000 km/s
G
, where n is the number of satellites.
29, 700 km
+ 100 s n = 100 100 s n
300, 000 km/s
U
29, 700 km
= 99 100 s n
300, 000 km/s
29, 700 km
= 10 = n
R
300, 000 km/s 99 100 s
This means the signal must pass 10 satellites for the switching delay to be 1%
of the total delay.
U
9. 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.
10. Traveling at 2/3 the speed of light means 200,000 km/sec. The signal travels
for 100 milliseconds, or 0.1 seconds. This means the signal traversed a dis-
tance of 200, 000 0. 1 = 4000 km.
, All Chapters Included
All Answers Included
PROBLEM SOLUTIONS FOR CHAPTER 1 3
11. 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
TU
reasoned opinions be to written into law may be undesirable. The potential ef-
fects of advertising campaigns by special interest groups of one kind or anoth-
er 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 re-
sults.
12. Call the routers A, B, C, D, and E. There are ten potential lines: AB, AC, AD,
TO
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 410 = 1, 048, 576. At 50
ms each, it takes 52,428.8 sec, or about 14.6 hours to inspect them all.
13. 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.
R
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) + ...
G
or
infinity infinity
l= n (0. 5)i − i(0. 5)i
U
i=1 i=1
This expression reduces to l = n − 2. The mean router-router path is thus
2n − 4.
14. Distinguish n + 2 events. Events 1 through n consist of the corresponding host
R
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 chan-
nel, with probability (1 − p)n. Event n + 2 is a collision. Since these n + 2
U
events are exhaustive, their probabilities must sum to unity. The probability of
a collision, which is equal to the fraction of slots wasted, is then just
1 − np(1 − p)n − 1 − (1 − p)n.
15. Instead of trying to foresee bad things and avoid them from happening, suc-
cessful networks are fault-tolerant. They allow bad things to happen but iso-
late or hide them from the rest of the system. Examples include error correc-
tion, error detection, and network routing.
R
U
G
R
TO
TU
, U
R
U
G
R
TO
TU
,2 PROBLEM SOLUTIONS FOR CHAPTER 2
5. 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.
6. A transcontinental fiber link might have many gigabits/sec of bandwidth, but
TU
the latency will also be high due to the speed of light propagation over thou-
sands of kilometers. Similarly, a satellite link may run at megabits/sec but have
a high latency to send a signal into orbit and back. In contrast, a 56-kbps
modem calling a computer in the same building has low bandwidth and low
latency. So do low-end local and personal area wireless technologies such as
Zigbee.
TO
7. No. The speed of propagation is 200,000 km/sec or 400 meters/sec. In 20
sec, the signal travels 4 km. Thus, each switch adds the equivalent of 4 km
of extra cable. If the client and server are separated by 5000 km, traversing
even 50 switches adds only 200 km to the total path, which is only 4%. Thus,
switching delay is not a major factor under these circumstances.
8. The delay is 1% of the total time, which means
100 s n
R
= 0. 01
29, 700 km
+ 100 s n
300, 000 km/s
G
, where n is the number of satellites.
29, 700 km
+ 100 s n = 100 100 s n
300, 000 km/s
U
29, 700 km
= 99 100 s n
300, 000 km/s
29, 700 km
= 10 = n
R
300, 000 km/s 99 100 s
This means the signal must pass 10 satellites for the switching delay to be 1%
of the total delay.
U
9. 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.
10. Traveling at 2/3 the speed of light means 200,000 km/sec. The signal travels
for 100 milliseconds, or 0.1 seconds. This means the signal traversed a dis-
tance of 200, 000 0. 1 = 4000 km.
, All Chapters Included
All Answers Included
PROBLEM SOLUTIONS FOR CHAPTER 1 3
11. 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
TU
reasoned opinions be to written into law may be undesirable. The potential ef-
fects of advertising campaigns by special interest groups of one kind or anoth-
er 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 re-
sults.
12. Call the routers A, B, C, D, and E. There are ten potential lines: AB, AC, AD,
TO
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 410 = 1, 048, 576. At 50
ms each, it takes 52,428.8 sec, or about 14.6 hours to inspect them all.
13. 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.
R
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) + ...
G
or
infinity infinity
l= n (0. 5)i − i(0. 5)i
U
i=1 i=1
This expression reduces to l = n − 2. The mean router-router path is thus
2n − 4.
14. Distinguish n + 2 events. Events 1 through n consist of the corresponding host
R
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 chan-
nel, with probability (1 − p)n. Event n + 2 is a collision. Since these n + 2
U
events are exhaustive, their probabilities must sum to unity. The probability of
a collision, which is equal to the fraction of slots wasted, is then just
1 − np(1 − p)n − 1 − (1 − p)n.
15. Instead of trying to foresee bad things and avoid them from happening, suc-
cessful networks are fault-tolerant. They allow bad things to happen but iso-
late or hide them from the rest of the system. Examples include error correc-
tion, error detection, and network routing.
