A channel has a bit rate of 4 kbps and one-way propagation delay of 20 ms. The…
2005
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
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Correct answer: D
Given: bit rate = 4 kbps (4000 bits/s), one-way propagation delay = 20 ms (0.02 s), stop-and-wait protocol, acknowledgement transmission time negligible.
Key equation: Efficiency = transmission time / (transmission time + round-trip propagation delay).
Round-trip propagation delay = 2 × 20 ms = 40 ms = 0.04 s.
Require efficiency ≥ 0.5, so let t = transmission time. Then t / (t + 0.04) ≥ 0.5.
Solve: t ≥ 0.04 s.
Frame size (bits) = bit rate × t = 4000 × 0.04 = 160 bits.
Therefore the minimum frame size to achieve at least 50% channel efficiency is 160 bits. Smaller sizes (for example 80 bits) give less than 50% efficiency; larger sizes also meet the requirement but are not minimal.
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