Question

Suppose we are transmitting frames between two nodes using the 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?

• A. 88.23
• B. 93.75
• C. 85.44
• D. 66.67

Hint:

Channel utilization in Stop-and-Wait protocol is calculated as the ratio of transmission time to total cycle time, considering both transmission and round-trip delay.

Answer 1

The Correct Option is A

The time to transmit the frame is given by: \[ T_t = \frac{{Frame Size}}{{Transmission Rate}} = \frac{3000}{2000} = 1.5 { seconds} \] The total round-trip time (RTT) includes the propagation delay: \[ RTT = 2 \times {Propagation Delay} = 2 \times 0.1 = 0.2 { seconds} \] Total time required for one cycle (transmission + RTT) is: \[ T_{{total}} = T_t + RTT = 1.5 + 0.2 = 1.7 { seconds} \] Channel utilization is given by: \[ {Utilization} = \frac{T_t}{T_{{total}}} \times 100 = \frac{1.5}{1.7} \times 100 = 88.23% \] Thus, the correct answer is (A) 88.23%.