What is the efficiency of a stop-and-wait protocol with a bandwidth of 1 Mbps and a round-trip time of 20 ms?
Q1. What is the efficiency of a stop-and-wait protocol with a bandwidth of 1 Mbps and a round-trip time of 20 ms?
Answer: 0.05
Explanation: Efficiency = (frame size / (frame size + 2 * RTT * bandwidth)), for a 1-bit frame, efficiency = 1 / (1 + 2 * 0.02 * 10^6) = 1 / 40001 = 0.025 or 0.05 approx for a small frame.