A channel has a bit rate of 4 kbps and one-way propagation delay of 20 ms. The…
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
Answer: D. 160 bits — CONCEPTIn stop-and-wait, the sender transmits one frame and then waits for its acknowledgement. When acknowledgement transmission time is negligible, the link…
- A.
80 bytes
- B.
80 bits
- C.
160 bytes
- D.
160 bits
- E.
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Correct answer: D
CONCEPT
In stop-and-wait, the sender transmits one frame and then waits for its acknowledgement. When acknowledgement transmission time is negligible, the link efficiency is η = Tₜ/(Tₜ + 2Tₚ), where Tₜ is frame-transmission time and Tₚ is one-way propagation delay.
For a target efficiency of one-half, frame transmission must occupy at least as much time as the round-trip propagation interval.
APPLICATION
The one-way propagation delay is Tₚ = 20 ms, so the round-trip propagation term is 2Tₚ = 40 ms.
Apply the target η ≥ 1/2: Tₜ/(Tₜ + 40 ms) ≥ 1/2. Multiplying by the positive denominator gives 2Tₜ ≥ Tₜ + 40 ms, hence Tₜ ≥ 40 ms.
Frame size L equals bit rate times transmission time: L = R × Tₜ = 4,000 bit/s × 0.040 s = 160 bits.
Because the question asks for the minimum size, the boundary value is the required result.
CROSS-CHECK
A 160-bit frame takes 160/4,000 = 0.040 s = 40 ms to transmit, so η = 40/(40 + 40) = 0.50. Any shorter frame has Tₜ < 40 ms and therefore efficiency below 50%.
Result: 160 bits