A channel has a bit rate of 4 kbps and one-way propagation delay of 20 ms. The channel uses stop and wait protocol. The transmission time of the acknowledgement frame is negligible. To get a channel efficiency of at least 50%, the minimum frame size should be
(A) 80 bytes
(B) 80 bits
(C) 160 bytes
(D) 160 bits
Bit rate = 4 kbps
One-way propagation delay = 20 ms
Efficiency = Transmission time of packet/(Transmission time of packet + 2 * Propagation delay)
0.5 = x/(x + 2 * 20 * 10-3)
x = 20 * 10-3
x = 40 * 10-3
Minimum frame size / Bit rate = 40 * 10-3
Therefore, Minimum frame size = 40 * 10-3 * 4 * 103 = 160 bits
Thus, option (D) is correct.
Please comment below if you find anything wrong in the above post.
Attention reader! Don’t stop learning now. Learn all GATE CS concepts with Free Live Classes on our youtube channel.