The values of parameters for the Stop-and-Wait ARQ protocol are as given…
2017
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 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
Key assumptions: propagation delay is one-way (0.75 ms), the processing time (0.25 ms) is at the receiver before sending the ACK, the information frame size includes payload + overhead (1980 + 20 = 2000 bytes), and bitrate = 1 Mbps.
Transmit time for information frame = 2000 bytes × 8 bits/byte ÷ 1e6 b/s = 16 ms (0.016 s).
Transmit time for ACK = 20 bytes × 8 ÷ 1e6 = 0.16 ms (0.00016 s).
One-way propagation delay = 0.75 ms (0.00075 s); receiver processing time = 0.25 ms (0.00025 s).
Cycle time for one successful frame (Stop-and-Wait) = transmit data + propagation + processing + transmit ACK + propagation = 0.016 + 0.00075 + 0.00025 + 0.00016 + 0.00075 = 0.01791 s.
Useful time per cycle (time carrying payload bits) = payload bits ÷ bitrate = (1980 bytes × 8) ÷ 1e6 = 0.01584 s.
Efficiency = (useful time / cycle time) × 100 = (0.01584 / 0.01791) × 100 ≈ 88.45% (rounded to 2 decimals).
Answer: 88.45%.
Note: Different interpretations (for example, applying the 0.25 ms processing time at both sender and receiver or using other timing conventions) change the numeric result. The provided official value (86.50%) does not match the standard calculation above under the stated assumptions.
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