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?
Answer
(A)
Explanation
In Stop-and-Wait the sender transmits one frame and then waits idle until the acknowledgement comes back, so one cycle lasts \(T_t + 2T_p\) (the ACK and processing times are negligible).
- Transmission time: \(T_t = \frac{3000}{2000} = 1.5\) s.
- Round-trip propagation: \(2T_p = 2 \times 0.1 = 0.2\) s.
Utilization \(= \frac{T_t}{T_t + 2T_p} = \frac{1.5}{1.7} \approx 0.8824\), about 88.2%, which is option A. Option B (93.75) comes from counting the propagation delay only one way: \(\frac{1.5}{1.6} = 0.9375\).
Answer: A