# GATE | Gate IT 2005 | Question 72

• Last Updated : 28 Jun, 2021

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

Explanation:
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.

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