Suppose we are transmitting frames between two nodes using Stop-and-Wait…
2025
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?
- A.
88.23
- B.
93.75
- C.
85.44
- D.
66.67
Attempted by 101 students.
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Correct answer: A
Solution:
For Stop‑and‑Wait protocol, channel utilization U = (frame transmission time) / (frame transmission time + round‑trip propagation delay + any acknowledgement transmission time + processing times). Acknowledgement transmission time and processing times are given as negligible here, so:
Frame transmission time = frame size / transmission rate = 3000 bits / 2000 bps = 1.5 s.
Round‑trip propagation delay = 2 × 100 ms = 2 × 0.1 s = 0.2 s.
Total cycle time = 1.5 s + 0.2 s = 1.7 s.
Utilization = 1.5 / 1.7 ≈ 0.88235 = 88.235% ≈ 88.23%.
Therefore the most accurate channel utilization for the given scenario is 88.23%.
Common mistakes to avoid:
Using only one propagation delay instead of the round‑trip propagation (this overestimates utilization).
Forgetting to convert milliseconds to seconds when adding propagation delay.
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