Solution:
Given:
Bit rate (R) = 4 kb/s = 4 × 10³ bits/s
Propagation delay (Tp) = 10 ms = 10 × 10⁻³ s
Required efficiency (η) ≥ 50% = 1/2
ACK transmission time = 0
Processing time = 0
Step 1: Efficiency of Stop-and-Wait
For Stop-and-Wait:
η = Tt / (Tt + 2Tp)
We need:
η ≥ 1/2
Therefore,
Tt / (Tt + 2Tp) ≥ 1/2
2Tt ≥ Tt + 2Tp
Tt ≥ 2Tp
Step 2: Find transmission time
Tt ≥ 2(10 ms)
Tt ≥ 20 ms
Step 3: Find frame size
Transmission time is:
Tt = L / R
Therefore,
L = Tt × R
L ≥ (20 × 10⁻³) × (4 × 10³)
L ≥ 80 bits
Final Answer:
Frame size ≥ 80 bits
