GATE 2017 CS – Question 55
The values of parameters for the Stop-and-Wait ARQ protocol are as given below:
Bit rate of the transmission channel = 1 Mbps.
Propagation delay from sender to receiver = 0.75 ms.
Time to process a frame = 0.25 ms.
Number of bytes in the information frame = 1980.
Number of bytes in the acknowledgement frame = 20.
Number of overhead bytes in the information frame = 20.
Assume that there are no transmission errors. Then, the transmission efficiency (expressed in percentage) of the Stop-and-Wait ARQ protocol for the above parameters is ________ (correct to 2 decimal places).
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Correct answer: 86.5 to 89.5
Explanation
Sending the frame takes $\frac{1980 \times 8}{10^6} = 15.84$ ms and the acknowledgement takes $\frac{20 \times 8}{10^6} = 0.16$ ms. One full cycle is $15.84 + 0.75 + 0.25 + 0.16 + 0.75 = 17.75$ ms. The useful data is $1980 - 20 = 1960$ bytes, which takes $15.68$ ms to send. The efficiency is $\frac{15.68}{17.75} = 0.8834$, which is 88.34 percent.