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Consider a channel with source and target having distance between them of 10km. Propagation delay is 10 microsec/km. What will be the data rate of the channel if the RTT is equal to the transmission delay and packet size is 125 bytes?

a) 5MBps

b)5Mbps

c)10KBps

d)15kbps

a) 5MBps

b)5Mbps

c)10KBps

d)15kbps

+2 votes

Round trip delay is the delay involved in a round trip in a channel from sender to receiver and back to sender from receiver.Hence,

R.T.T = 2 * P.T. as we know propogation time is time spent in a channel from sender to receiver or vice versa.

According to the question,

R.T.T = T.T(Transmission Time)

Also we know that transmission time = data size / data rate

And we are given P.T./km = 10 μs and distance = 10km so P.T = 100 μs , so R.T.T = 200 μs

Given data size = 125 bytes = 1000 bits .Hence we have the equation,

10^{3}/x = 200/10^{6 }which gives x = 10^{6} * 1000 / 200 = 5 * 10^{6} bps = 5 Mbps

Hence , B) is the correct option.

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