COMPUTER NETWORKS
FIFTH EDITION
PROBLEM SOLUTIONS
ANDREW S. TANENBAUM
Vrije Universiteit
Amsterdam, The Netherlands
and
DAVID WETHERALL
University of Washington
Seattle, WA
PRENTICE HALL
Upper Saddle River, NJ
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.
(i) If dog’s speed is doubled, maximum value of x is also doubled.
(ii) If tape capacity is doubled, value of x is also doubled.
(iii) If data rate of the transmission line is doubled, value of x is halved.
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 as well as video, 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 ac-
tually worse than a somewhat longer delay and low variability. For financial
transaction traffic, reliability and security are very important.
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
2 PROBLEM SOLUTIONS FOR CHAPTER 1
effects of advertising campaigns by special interest groups of one kind or an-
other 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.
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. 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
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
.
10. 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.
One possible disadvantage is the performance of a layered system is likely to
be worse than the performance of a monolithic system, although it is
extremely difficult to implement and manage a monolithic system.
11. In the ISO protocol model, physical communication takes place only in the
lowest layer, not in every layer.
12. 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 ex-
ample, 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
each. With a byte stream, the message boundaries do not count and the re-
ceiver will get the full 2048 bytes as a single unit. The fact that there were
originally two distinct messages is lost.
13. 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.
14. 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.
PROBLEM SOLUTIONS FOR CHAPTER 1 3
15. The probability, P
k
, of a frame requiring exactly k transmissions is the
probability 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
16. 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).
17. TCP is connection oriented, whereas UDP is a connectionless service.
18. 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.
19. 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 38.4 billion hosts. That sounds like a lot,
but if every television, cellphone, camera, car, and appliance in the world is
online, maybe it is plausible. The average person may have dozens of hosts
by then.
20. If the network tends to lose packets, it is better to acknowledge each one sep-
arately, so the lost packets can be retransmitted. On the other hand, if the net-
work is highly reliable, sending one acknowledgement at the end of the entire
transfer saves bandwidth in the normal case (but requires the entire file to be
retransmitted if even a single packet is lost).
21. Having mobile phone operators know the location of users lets the operators
learn much personal information about users, such as where they sleep, work,
travel and shop. This information might be sold to others or stolen; it could let
the government monitor citizens. On the other hand, knowing the location of
the user lets the operator send help to the right place in an emergency. It
might also be used to deter fraud, since a person who claims to be you will
usually be near your mobile phone.
22. The speed of light in coax is about 200,000 km/sec, which is 200 meters/μsec.
At 10 Mbps, it takes 0.1 μsec to transmit a bit. Thus, the bit lasts 0.1 μsec in
time, during which it propagates 20 meters. Thus, a bit is 20 meters long
here.
23. The image is 1600 × 1200 × 3 bytes or 5,760,000 bytes. This is 46,080,000
bits. At 56,000 bits/sec, it takes about 822.857 sec. At 1,000,000 bits/sec, it
takes 46.080 sec. At 10,000,000 bits/sec, it takes 4.608 sec. At 100,000,000
