These two cycles can be compared on the basis of either the same compression ratio or the same maximum pressure and temperature.

1 - 2 - 3 - 5 = Otto Cycle,
for the same heat rejection Q2 the higher the 1 - 2 -4 - 5 = Diesel Cycles, heat given Q1, the higher is the cycle efficiency.
So from T-S diagram for cycle 1 - 2 - 3 - 5, Q1 is more than that for 1 - 2 - 7 - 5 (area under the curve represents Q,).
Hence ηOtto > ηDiesel For the same heat rejection by both otto and diesel cycles. Again both can be compared on the basis of same maximum pressure and temperature.

1 - 2 – 3 – 4 = Otto Cycle; Here area under the curve
1 - 2′ - 3 - 4 = Diesel Cycle
1 - 2′ - 3 - 4 is more than 1 - 2 – 3 – 4
So ηdiese l > ηotto; for the same Tmax and Pmax