Minimum Frame Size for 50% Utilization on a Stop-and-Wait Link

Given: bit rate $R = 64$ kbps, one-way propagation delay $T_p = 20$ ms, ACK transmission time and node processing delay are negligible, target utilization $U \ge 50\%$.

Step 1: Time for one stop-and-wait cycle

One cycle = transmit the frame + propagate to receiver + propagate the ACK back:

$$T_{cycle} = T_t + T_p + T_p = T_t + 2T_p$$

Step 2: Utilization formula

$$U = \frac{\text{time the link is busy sending data}}{\text{total cycle time}} = \frac{T_t}{T_t + 2T_p}$$

Step 3: Apply the 50% condition

$$0.5 \le \frac{T_t}{T_t + 0.040}$$

$$0.5\,(T_t + 0.040) \le T_t \;\Longrightarrow\; 0.5T_t + 0.020 \le T_t$$

$$0.020 \le 0.5\,T_t \;\Longrightarrow\; T_t \ge 0.040\ \text{s} = 40\ \text{ms}$$

So the frame must occupy the link for at least 40 ms, i.e. $T_t \ge 2T_p$.

Step 4: Convert transmission time into frame size

$$L = R \times T_t = 64{,}000\ \text{bits/s} \times 0.040\ \text{s} = 2560\ \text{bits}$$

$$L = \frac{2560}{8} = 320\ \text{bytes}$$

Check

$T_t = 320 \times 8 / 64000 = 0.04$ s, so $U = 0.04/(0.04 + 0.04) = 0.5$, i.e. exactly 50%.

Final answer: the minimum frame size is 320 bytes. Any frame of 320 bytes or more yields a utilization of at least 50%.