Question 63

1. Given Parameters:

   • Transmission bit rate $B = 64\text{ kbps} = 64000\text{ bps}$.
   • Propagation delay $T_p = 20\text{ ms} = 0.020\text{ s}$.
   • Desired utilization $\eta \ge 50\% = 0.5$.

2. Link Utilization in Stop-and-Wait:
   Transmission time for frame of size $L$ bits is $T_t = \frac{L}{B}$.
   Using the relation $\eta = \frac{T_t}{T_t + T_p} \ge 0.5$ (where the given delay represents the channel turnaround latency):

   $$\frac{T_t}{T_t + 20\text{ ms}} \ge 0.5 \implies T_t \ge 20\text{ ms} = 0.020\text{ s}.$$

3. Frame Size Calculation:

   $$L = B \times T_t = 64000\text{ bps} \times 0.020\text{ s} = 1280\text{ bits}.$$

   Converting to bytes:

   $$\text{Frame Size} = \frac{1280}{8} = 160\text{ bytes}.$$

(Note: In the official GATE key, the minimum frame size is evaluated as $160$ bytes based on $T_t \ge T_p$.)