Suppose we are transmitting frames between two nodes using Stop-and-Wait protocol. The frame size is 3000 bits. The transmission rate of the channel is 2000 bps (bits/second) and the propagation delay between the two nodes is 100 milliseconds. Assume that the processing times at the source and destination are negligible. Also, assume that the size of the acknowledgement packet is negligible. Which ONE of the following most accurately gives the channel utilization for the above scenario in percentage?

To find the channel utilization (efficiency $\eta$) for the Stop-and-Wait protocol:

1. Transmission time ($T_t$):

   $$T_t = \frac{\text{Frame size}}{\text{Bandwidth}} = \frac{3000\text{ bits}}{2000\text{ bps}} = 1.5\text{ seconds} = 1500\text{ ms}$$

2. Propagation delay ($T_p$):

   $$T_p = 100\text{ ms} = 0.1\text{ seconds}$$

3. Total cycle time ($T_{\text{total}}$): Since the processing time and acknowledgment packet transmission time are negligible:

   $$T_{\text{total}} = T_t + 2 \times T_p = 1.5 + 2 \times 0.1 = 1.7\text{ seconds} = 1700\text{ ms}$$

4. Channel Utilization ($\eta$):

   $$\eta = \frac{T_t}{T_t + 2 \times T_p} = \frac{1500}{1700} = \frac{15}{17} \approx 0.88235 = 88.235\%$$

Rounding to two decimal places gives 88.23%, which corresponds to option (A).