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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