Question

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 acknowledge 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 in _____ (correct to 2 decimal places).

Answer 1

Data:

Number of bytes in the information frame = 1980

Number of bytes in the acknowledge frame = 20

Number of overhead bytes in the information frame = 20

Bit rate of the transmission channel = 1Mbps

Propagation delay from sender to receiver = 0.75 ms.

Time to process a frame = 0.25 ms.

Formula

Efficiency = \(\frac{{Transmission\;time}}{{transmission\;time + process\;time\; + \;RTT\; + AckFrame\;transmit\;time}}\)

Calculation:

So, useful information bytes in a frame = number of bytes in information frame – overhead bytes

 = 1980 – 20 = 1960

Transmission time (without overhead) = \(\frac{{Frame\;bits}}{{bandwidth}} = \frac{{1960\; \times 8}}{{{{10}^6}}}\) = 15.68 millisecond

Total transmission time (with overhead) = \(\frac{{1980\; \times 8}}{{{{10}^6}}}\) = 15.84 millisecond

Acknowledgement time = \(\frac{{20 \times 8}}{{{{10}^6}}}\) = 0.16 millisecond

RTT = 2 × Propagation time = 2 × 0.75 = 1.5 millisecond

Process time = 0.25 (info) + 0.25 (ack) = 0.5 millisecond

Efficiency = \(\frac{{15.68}}{{\left( {15.84 + 0.5 + 1.5 + 0.16} \right)}}\) = 0.871111 = 87.11