CE/CS/EE 330 Computer Networks
CE/CS/EE 330 Computer NetworksFall Semester 2006
Assignment 1_Solution1. Consider two hosts, A and B, connected
by a single link of rate R bps. Suppose that the two hosts are
separated by m meters, and suppose the propagation speed along the
link is s meters/sec. Host A is to send a packet of size L bits to
host B.a. Express the propagation delay, dprop, in terms of m and
s.
b. Determine the transmission time of the packet, dtrans, in
terms of L and R.
c. Ignoring processing and queuing delays, obtain an expression
for the end-to-end delay.
d. Suppose host A begins to transmit the packet at time t = 0.
At time t = dtrans, where is the last bit of the packet?
e. Suppose dprop is greater than dtrans. At time t = dtrans,
where is the first bit of the packet?
f. Suppose dprop is less than dtrans. At time t = dtrans, where
is the first bit of a packet?
g. Suppose s = 2.5 x 108, L = 100 bits, and R = 28 Kbps. Find
the distance m so that dprop equals dtrans.
2. Consider the queuing delay in router buffer (preceding an
outbound link). Suppose all packets are L bits, the transmission
rate is R bps, and that N packets simultaneously arrive at the
buffer every LN/R seconds. Find the average queuing delay of a
packet. (Hint: The queuing delay of the first packet is zero; for
the second packet L/R; for the third packet 2L/R. The N-th packet
has already been transmitted when the second batch of packets
arrives.)Solutions:
Problem 1:
a) dprop = m / s seconds.b) dtrans = R / L seconds.
c) dend-to-end = ( m / s + L / R ) seconds.
d) The bit is just leaving Host A.
e) The first bit is in the link and has not reached host B.
f) The first bit has reached Host B.
g) m = L/R*S = 100/28x103 (2.5x108) = 893 km
Problem 2:
It takes LN/R seconds to transmit the N packets. Thus, the
buffer is empty when a batch of N packet arrives.
The first of the N packets has no queuing delay. The 2nd packet
has a queuing delay of L/R seconds. The nth packet has a delay of
(n 1)L/R seconds.
The average delay is:-
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