Solutions to Home Work1

1) Imagine that you have
trained your St. Bernard, Bernie, to carry a box of three 8mm tapes

instead of a flask of brandy. (When
your disk fills up, you consider that an emergency).

These tapes each contain 7 gigabytes.
The dog can travel to your side, wherever you may be, at 18 km/hour. For what
range of distances does Bernie have a higher data rate than a transmission line
whose data rate(excluding overhead) is 150 Mbps?

Solution:

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 higher
rate than the communication line.

3) The performance of a
client-server system is influenced by two network factors: the bandwidth of the
network(how many bits/sec it can transport) and the
latency (how many seconds it takes for the first bit to get from the client to
the server). Give an example of a network that exhibits high bandwidth and high
latency. Then give an example of one with low bandwidth and low latency.

Solution:

A transcontinental fiber
link might have many gigabits/sec of bandwidth, but the latency will also be
high due to the speed of light propogation over
thousands of kilometers. In contrast, a 56-kbps modem calling a computer in the
same building will have a low bandwidth

and low latency.

11) What are two reasons for
using layered protocols?

Solution:

Among other reasons for
using layered protocols, using them leads to breaking
up the design problem into smaller, more manageable pieces, and layering means
that protocols can be changed without affecting higher or lower ones.

17) In some networks, the
data link layer handles transmission errors by requesting damaged frames to be
retransmitted. If the probability of a frame's being damaged is p, what is the
mean number of transmissions required to send a frame? Assume that
acknowledgements are never lost.

Solution:

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

∞ ∞

∑ kP_{k}_{ =
}∑ k(1-p)p^{k-1 =
} 1 / (1-p)

k=1 k=1

19) If the unit exchanged at
the data link layer is called a frame and the unit exchanged at the network
level is called a packet, do frames encapsulate packets or do packets
encapsulate frames? Explain your answer.

Solution:

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).