Discussion
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
The channel utilization is about 88.23%.
In stop-and-wait the sender transmits one frame and then waits a full round trip for the acknowledgement. With negligible ACK size and processing times:
\[ \begin{aligned} T_{t} &= \frac{3000}{2000} = 1.5\ \text{s} \\ T_{p} &= 100\ \text{ms} = 0.1\ \text{s} \\ U &= \frac{T_{t}}{T_{t} + 2T_{p}} = \frac{1.5}{1.7} = 0.8824 \end{aligned} \]
That is 88.23% to two decimal places (88.235 truncated). 93.75% uses only one propagation delay (1.5/1.6), forgetting that the acknowledgement must travel back. 66.67% and 85.44% match no consistent use of these figures.
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