Suppose that the stop-and-wait protocol is used on a link with a bit rate of…

Key formula: For stop-and-wait, link utilization U = transmission time / (transmission time + 2 × propagation delay) (ACK and processing time negligible).

• Step 1: Let L be the frame size in bits. R = 64,000 bps. Propagation delay = 20 ms = 0.02 s. Transmission time t_tx = L / R.

• Step 2: Require U ≥ 0.5, so t_tx / (t_tx + 2·0.02) ≥ 0.5 ⇒ t_tx ≥ 2·0.02 = 0.04 s.

• Step 3: L ≥ 0.04 × 64,000 = 2,560 bits = 320 bytes. Therefore, if 20 ms is the one-way propagation delay, the minimum frame size is 320 bytes.

Note on ambiguity: The problem statement can be interpreted in two ways. If the given 20 ms is intended as the round-trip propagation delay (or if a single propagation delay is used in the cycle), the requirement becomes t_tx ≥ 0.02 s, giving L ≥ 1,280 bits = 160 bytes. The answer 160 bytes matches that interpretation.

Final: 320 bytes if 20 ms is the one-way propagation delay; 160 bytes if 20 ms is treated as the round-trip (or only one propagation delay is used).