First let's find out the CPU execution time for the given scenario let's call it CPU1:-
average number of cycles per instruction = total number of cycles / total number of instruction
= (0.4 * 2 + 0.2 * 4 + 0.18 * 4 + 0.22 * 6 )/1
=0.8 + 0.8 +0.72 + 1.32
=3.64 cycles/instruction
CPU1 execution time = number of instructions * average number of cycles/instruction * cycle time
= 1*3.64 * t (cycle time = t)
= 3.64t
now,let the case of CPU2:-
here,30% of the ALU ops are being replaced by new kind of instruction which takes 4 clock cycles.let's call these new instruction as ALU2 ops.
now overall percentage of this new instruction in instruction set = 30% of 40
= (30 * 40) /100 = 1200/100 = 12% in overall
so,remaining 28% is ALU ops instruction which still takes 2 cycles.
now average number of cycles per instruction = (0.28 * 2 + 0.12 * 4 + 0.2 * 4 + 0.18 * 4 + 0.22 * 8)/1
=3.1072 cycles/instructions
CPU2 execution time = number of instructions * average number of
cycles/instruction * cycle time
= 1* 3.1072 * t
=3.1072 t (cycle time doesn't change)
clearly we can see that CPU2 takes less time than CPU1 so,second approach is more faster than first approach.
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