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Exercises
1) What advantage does a circuit-
switched network have over a
packet-switched network?
Answer: A circuit-switched
network can guarantee a certain
amount of end-to-end bandwidth
for the duration of a call. Most
packet-switched networks today
(including the Internet) cannot
make any end-to-end guarantees for
bandwidth.
2) Consider an application that
transmits data at a steady rate (for
example, the sender generates an N-
bit unit of data every k time units,
where k is small and fixed). Also,
when such an application starts, it
will continue running for a
relatively long period of time.Would a packet-switched network
or a circuit-switched network be
more appropriate for this
application? Why?
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Answer: A circuit-switched
network would be better in this
case, because the application
involves long sessions with
predictable bandwidthrequirements. Since the
transmission rate is fixed and
known, bandwidth can be reserved
for each communication. In
addition, there are no overhead
costs for setting up and tearing
down a circuit connection.
3) Consider two hosts, Hosts A and
B, connected by a single link of
rateR bps. Suppose that the two
hosts are separted bym meters, and
suppose the propagation speed
along the link iss meters/sec. Host
A is to send a packet of sizeL bits
to Host B.
a) Express the propagation
delay,d ro in terms ofm ands .
Answer: dprop= m/s
b) Determine the transmission
time of the packet,dtrans in terms
ofL andR .
Answer: dtrans= L/R
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c) Ignoring processing and queing
delays, obtain an expression for
the end-to-end delay.
Answer: m/s + L/R
d) Suppose Host A begins to
transmit the packet at timet=0 .
At timet=dtrans , where is the last
bit of the packet?
Answer: the bit is just leaving
node A
e) Supposedprop is greater
thandtrans . At timet=dtrans ,
where is the first bit of the packet?
Answer: the first bit is in the link
(it has not reached B yet)
f)) Supposedprop is less
thandtrans . At timet=dtrans ,
where is the first bit of the packet?
Answer: the first bit has reached B
g)
Supposes=2.5*108
,L=100 bits
andR=28 kbps. Find the
distancem so
thatdprop equalsdtrans
Answer: m = LR/S = 893 Km
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4) Consider sending a large file of F
bits from Host A to Host B. There
are two links and one node between
A and B. Neglect propagationdelays and queueing delays. Host A
segments the file into segments of S
bits each. Assume that F/S is an
integer. Host A adds h bits of
header to each segment forming
packets of size h+S each. Each link
has a transmission rate of R bps.
Find the end-to-end delay in
sending the file.
Answer: 2*(h+S)/R + ((F/S)-
1)*(h+S)/R = (h+S)/R * (2 + (F/S)
1) = (h+S)/R * ( (F/S) +1) seconds
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5) Consider sending voice from
host A to host B over a packet-
switched network (for example,
Internet phone). Host A converts
analog voice to a digital 64 Kbps
bit stream on the fly. Host A then
groups the bits into 48-byte packets.
There is one link between host A
and B; its transmission rate is 1
Mbps and its propagation delay is 2
ms. As soon as host A gathers a
packet, it sends it to host B. As
soon as host B receives an entire
packet, it converts the packet's bitsto an analog signal. How much time
elapses from the time a bit is
created (from the original analog
signal at host A) until the bit is
decoded (as part of the analog
signal at host B)
Answer:
Before the first bit of a packet can
be transmitted, all the other bits
belonging to the same packet need
to be generated. This requires: 48.8
/ 64x10^3 = 6 ms
The time to transmit this packet is:
48.8 / 10^6 = .384 ms
The propagation delay is 2 ms
Therefore, the delay until decoding
is: 6ms + 0.384ms + 2ms = 8.384
ms
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