A link has a transmission speed of 106 bits/sec. It uses data packets of size…
2021
A link has a transmission speed of 106 bits/sec. It uses data packets of size 1000 bytes each. Assume that the acknowledgement has negligible transmission delay and that its propagation delay is the same as the data propagation delay. Also assume that the processing delays at the nodes are negligible. The efficiency of the stop-and-wait protocol in this setup is exactly 25%. The value of the one-way propagation delay (in milliseconds) is
- A.
10
- B.
24
- C.
12
- D.
05
- E.
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Correct answer: C
First, calculate the transmission delay (Tt). The packet size is 1000 bytes or 8000 bits. Dividing by the speed of 10^6 bits per second gives Tt = 8 milliseconds.
Next, use the efficiency formula for stop-and-wait: eta = Tt / (Tt + 2Tp). Substitute the known values where efficiency is 0.25 and Tt is 8 ms.
Solving the equation 0.25 = 8 / (8 + 2Tp) yields 8 + 2Tp = 32. Subtracting 8 gives 2Tp = 24, so the one-way propagation delay Tp is exactly 12 milliseconds.