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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



asked in CO & Architecture by Active (2.2k points) | 79 views

1 Answer

+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,

103/x = 200/106     which gives x = 106 * 1000 / 200 = 5 * 106 bps = 5 Mbps

Hence , B) is the correct option.

answered by Veteran (100k points)

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