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 — Given: Bit rate R = 4 kbps = 4000 bps One-way propagation delay = 20 ms ⇒ RTT = 40 ms For Stop-and-Wait: Efficiency = Tt / (Tt + 2Tp) For efficiency ≥ 50%: Tt…
- A.
80 bytes
- B.
80 bits
- C.
160 bytes
- D.
160 bits
- E.
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Correct answer: D
Given:
Bit rate R = 4 kbps = 4000 bps
One-way propagation delay = 20 ms ⇒ RTT = 40 ms
For Stop-and-Wait:
Efficiency = Tt / (Tt + 2Tp)
For efficiency ≥ 50%:
Tt ≥ 2Tp
⇒ Tt ≥ 40 ms
Now,
Frame size = R × Tt = 4000 × 0.04 = 160 bits