Suppose that the stop-and-wait protocol is used on a link with a bit rate of 64 kilobits per second and 20 milliseconds propagation delay. Assume that the transmission time for the acknowledgement and the processing time at nodes are negligible. Then the minimum frame size in bytes to achieve a link utilization of at least 50% is ____________.

Given parameters:

• Bandwidth / Bit rate, $B = 64\text{ kbps} = 64 \times 10^3\text{ bps}$

• Propagation delay, $T_p = 20\text{ ms} = 20 \times 10^{-3}\text{ s}$

In a stop-and-wait protocol, link utilization (efficiency) $\eta$ is given by:

For $\eta \ge 50\% = 0.5$:

Substituting $T_p = 20\text{ ms}$:

Frame size in bits $L = B \times T_t = 64000 \times 0.04 = 2560\text{ bits}$.

In bytes, frame size $= \frac{2560}{8} = 320\text{ bytes}$.

*(Note: Following the official key where $T_t = T_p = 20\text{ ms}$ was evaluated, $L = \frac{64 \times 10^3 \times 20 \times 10^{-3}}{8} = 160\text{ bytes}$. The official key value is 160).*