Computer Sciences > GATE 2025 SET-2 > Sliding Window Protocols
Suppose we are transmitting frames between two nodes using Stop-and-Wait protocol. The frame size is 3000 bits. The transmission rate of the channel is 2000 bps (bits/second) and the propagation delay between the two nodes is 100 milliseconds. Assume that the processing times at the source and destination are negligible. Also, assume that the size of the acknowledgement packet is negligible. Which ONE of the following most accurately gives the channel utilization for the above scenario in percentage?
Correct : a
The correct answer is Option A - 88.23%.
The formula for channel utilization in Stop-and-Wait protocol is:
Utilization = Tt / (Tt + 2 × Tp)
Where Tt is transmission time and Tp is propagation delay.
Transmission Time (Tt) = Frame Size / Transmission Rate = 3000 / 2000 = 1.5 seconds
Propagation Delay (Tp) = 100 ms = 0.1 seconds
Plugging in:
Utilization = 1.5 / (1.5 + 2 × 0.1) = 1.5 / 1.7 ≈ 0.8823 = 88.23%
The 2 × Tp accounts for the round-trip propagation — signal going to the receiver and ACK coming back. Since ACK size is negligible, it doesn't add to the total time. The sender is actually transmitting only for 1.5 out of every 1.7 seconds, giving us that ~88% efficiency.
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