Question

A and B which are separated by 920 Km are directly connected using some type of physical link. A is sending packets to B with a packet size of 8000 bits at a bit rate of 2 Mbps. If the first bit transmitted by A has just reached B when the last bit is being transmitted by A, calculate the velocity of propagation in the medium in two different ways:
(i) Using the concept of delay-bandwidth product
(ii) Using the concept of bit width

Answer 1

1 - Using the concept of delay-bandwidth product

distance = 920 km

Propagation time = the time taken by a bit to travel from source to destination

Round Trip Time = 2 * Propogation time.

BandWidth Product is the the amount of data that can be in transit in the network. which is

BandWidthProduct = total_available_bandwidth * roundTripTime

In this case Band WidthProduct = 8000 bits because the first bit transmitted by A has just reacched B when

the last bit is being transmitted by A.

=>   $8000 = 2 * 10^{6} * (Round Trip Time) => 8000 = 2 * 10^{6} * 2 * \frac{distance}{velocity of propagation}$

=>velocity of propagation = $\frac{2 * 10^{6} * 2 * 920 * 10^{3}}{8000}$ = m/s

2 - Using the concept of bitwidth

Propogation delay = $\frac{distance}{velocity of propogation}$

In this case as mentioned the first bit transmitted by A has just reached B when the last bit is being transmitted by A

So the transmission time at A is equal to propogation delay

=> propogation delay = $\frac{8000}{2 * 10^{6}}$ = secs = 4 milli secs

=> $4 * 10^{-3} = \frac{920 * 10 ^{3}}{ velocity}$

=> $velocty = \frac{920 * 10 ^{3}}{4 * 10^{-3}}$ => m/sec =>

that's it !

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