Is this correct?

3) A ring topology with a total circular length of 2D, with the 2 stations a distance D apart. Acknowledgment is achieved by the allowing a packet to circulate past the destination station, back to the source station. There are N repeaters on the ring, each of which introduces a delay of one bit time.
Solve for : D=1km, B=10Mbps, P=256bits, N=10

Is this correct >

So that one way propagation delay is,
1000m = 200m/microsec x t
t = 5 microsec

Round trip propagation delay is,
2t = 10 microseconds

Time to transmit a frame is,
number of bits = data rate x time
256 = 10,000,000bps x t
t = 25.6 microsec

Ack transmission time is,
number of bits = datarate x time
88 = 10,000,000bps x t
t = 8.8 microsec

Total repeaters delay = 10 x (1/10 x 10^6) = 1microsec

The total time cycle for a packet of 256 bits = (25.6 microsec + 10 microsec) + (8.8 microsec + 10 microsec) + 1mircosec = 55.4 microsec

We multiply total time cycle for a packet of 256 bits with 31250.
or,
55.4 microsec x 31250 = 1.73 seconds

Therefore, the total elapsed time for sending 1 million characters is 1.73 seconds


Actual frame size is 256
There is no overhead

So the effective throughput is,
256/ (55.4 x 10^(-6))sec = 4.621 Mbps