What will be the efficiency of a Stop and Wait protocol, if the transmission time for a frame is 20 ns and the propagation time is 30 ns?
- A. 20%
- B. 25%
- C. 40%
- D. 66%
Answer: B. 25%
Explanation: Efficiency = Tt / (Tt + 2Tp) = 20 / (20 + 2*30) = 20 / 80 = 1/4 = 25%.