A sender using Stop-and-Wait ARQ protocol has a timeout of 2 ms. If the distance between sender and receiver is 1000 km and the speed of signal is 2 * 10^8 m/s, what is the maximum data rate?
Q1. A sender using Stop-and-Wait ARQ protocol has a timeout of 2 ms. If the distance between sender and receiver is 1000 km and the speed of signal is 2 * 10^8 m/s, what is the maximum data rate?
Answer: 2 Mbps
Explanation: Round trip time (RTT) = 2 * distance / speed = 2 * 1000 * 1000 / (2 * 10^8) = 10 ms. For Stop-and-Wait, efficiency = 1 / (1 + 2a), where a = RTT / transmission time. To maximize data rate, transmission time should be minimized.