A link has a transmission speed of 10^6 bits/sec. It uses data packets of size…
2015
A link has a transmission speed of 10^6 bits/sec. It uses data packets of size 1000 bytes each. Assume that the acknowledgment has negligible transmission delay, and that its propagation delay is the same as the data propagation delay. Also assume that the processing delays at nodes are negligible. The efficiency of the stop-and-wait protocol in this setup is exactly 25%. The value of the one-way propagation delay (in milliseconds) is ____________.
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Correct answer: 12
Key idea: for stop-and-wait, the sender transmits a packet (transmission time) and then waits for the round-trip propagation before sending the next packet. Acknowledgment transmission time is negligible.
Compute transmission time: packet size = 1000 bytes = 8000 bits; link speed = 10^6 bits/s, so transmission time = 8000 / 10^6 = 0.008 s = 8 ms.
Write efficiency formula: efficiency = transmission time / (transmission time + 2 × one-way propagation delay). Given efficiency = 0.25, so 0.25 = 8 ms / (8 ms + 2 × t_prop).
Solve for one-way propagation delay: 0.25(8 + 2t_prop) = 8 → 2 + 0.5·(2t_prop) = 8 simplifies to 0.75·8 = 0.5·(2t_prop) → t_prop = 1.5 × 8 ms = 12 ms.
Answer: 12 ms (one-way propagation delay).
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