I just have a simple question, a bit silly, but I just need some clarification for an upcoming exam so I don't make a stupid mistake. I am currently taking a class in computer organization and design and am learning about execution time, CPI, clock cycles, etc.
For a problem, I have to calculate the amount of cycles for 2 compilers and find out which one is faster and by how much given the number of instructions and the cycles for each instruction. My main problem is figuring how much faster the faster compiler is.
For example lets say their are two compilers:
Compiler 1 has 3 load instructions, 4 store instructions, and 5 add
instructions.
Compiler 2 has 5 load instructions, 4 store instructions, and 3 add
instructions
A load instruction takes 2 cycles, a store instruction takes 3 cycles and a add instruction takes 1 cycle
So what I would do this add up to the instructions (3+4+5) and (5+4+3) which both equal to 12 instructions.
I'd then calculate the cycles by multiplying the number of instructions by the cycles and adding them all together like this
Compiler 1: (3*2)+(4*3)+(5*1) = 23 cycles
Compiler 2: (5*2)+(4*3)+(3*1) = 25 cycles
So obviously compiler 1 is faster because it requires less cycles. To find out how much faster compiler 1 is against compiler 2 would I just divide the ratio of the cycles?
My calculation was 23/25 = 0.92, so compiler 1 is 0.92 times faster than compiler 2 (92% faster).
A classmate of mine was discussing this with me and claims that it would be 25/23 which would mean it is 1.08 times faster.
I know I can also calculate this by dividing the cycles by the instructions like:
23 cycles/12 instructions = 1.91
25 cycles/12 instructions = 2.08
and then 1.91/2.08 = 0.92 which is the same as the above answer.
I'm not sure which way would be correct.
I was also wondering if the amount of instructions are difference for the second compiler, let's say 15 instructions. Would calculating the ratio of the cycles be sufficient enough?
Or would I have to divide the cycles with the instructions (cycles/instructions) but put 15 instructions for both?
(ex. 23/15 and 25/15?) and then divide the quotients of both to get
the times faster? I also get the same number(0.92) in that case.
Thank you for any clarification.
The first compiler would be 1.08 times the speed of the second compiler, which is 8% faster (because 1.0 + 0.08 = 1.08).