A channel has a data rate of 64 kbps and a propagation delay of 20 ms. For what packet size (in bits) will the stop-and-wait protocol achieve 50% efficiency?
Q1. A channel has a data rate of 64 kbps and a propagation delay of 20 ms. For what packet size (in bits) will the stop-and-wait protocol achieve 50% efficiency?
Answer: 2560 bits
Explanation: Efficiency = packet size / (packet size + 2 * bandwidth-delay product). Bandwidth-delay product = 64 * 10³ * 20 * 10^-3 = 1280 bits. For 50% efficiency, 0.5 = packet size / (packet size + 2 * 1280), solving for packet size gives 2560 bits.