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In the Stop and Wait protocol, the efficiency is primarily determined by the ratio of the time taken to transmit a frame to the total time taken for the round-trip of a frame plus the time taken for the acknowledgment to come back. This can be simplified as the ratio of the transmission time (Tt) to the sum of the transmission time and twice the propagation time (since the signal has to travel to the receiver and then the acknowledgment has to travel back to the sender), i.e., Tt + 2Tp.
Given that the transmission time (Tt) for a frame is 20 ns and the propagation time (Tp) is 30 ns, the efficiency (η) can be calculated as follows:
[ η = frac{T_t}{T_t + 2 * T_p} ]
Substituting the given values:
[ η = frac{20}{20 + 2 * 30} = frac{20}{80} = frac{1}{4} ]
[ η = 0.25 ] or 25%
Thus, the efficiency of the Stop and Wait protocol under the given conditions would be 25%.