# Calculating the MIPS Rating

1. Dec 7, 2011

### johnnyrocket

I need some help with this problem I am trying to solve if anyone can help.

Each instruction on a certain computer system requires 1 cycle for any type of instruction. This system includes N processors and is used to execute a program contain 16 million instructions and 60% of the code can be executed in parallel. All of the processors run at a 4GHz clock rate. If the program is executed using only one of the processors, the execution time is observed to be twice that required when using all N processors to execute the program.
--What is the value of N?
--What would be the effective MIPS rating for this program on the single-processor system and on the N-processor system?

2. Dec 7, 2011

### rcgldr

Try to convert the problem statement into an algebraic equation.

3. Dec 7, 2011

### johnnyrocket

Ok. I think the MIPS equation is:

MIPS = number of Instructions / (time to execute * 1 million)

So I would say that I can assume that if each instruction takes 1 cycle, with a 4GHz processor, I can execute 4 *10^9 instructions per second. So:

MIPS = 16million / (.0004 seconds to execute 16mil instructions * 1 million)
= 4000 MIPS rating on a single processor?

4. Dec 7, 2011

### D H

Staff Emeritus
You don't need to know the MIPS rating. Do what rcgldr suggested, which is to write the problem statement as an algebraic expression.

5. Dec 7, 2011

### johnnyrocket

I'm not exactly sure what he's suggesting. The 2nd part of the question asks for mips rating for the single and N processor systems, so that's what I was trying to solve for. Can you offer me any guidance? Am I on the right track?

6. Dec 7, 2011

### johnnyrocket

I think the correct formula is as follows, but I'm not sure:

Speedup = 1/((non parallel) +(parallel/# processors)). So given that they tell me the run time is 2x as fast on 6 processors, I'm guessing that is speedup. So I would say:

2 = 1/((.40)+(.60/N))
Which makes N = 6?

7. Dec 7, 2011

### johnnyrocket

I meant to say 2x as fast on N processors.

8. Dec 8, 2011

### D H

Staff Emeritus
Correct.

You do need the MIPS rating for the second half of the question, and you're earlier answer of 4000 MIPS for the single processor case was correct.

9. Dec 8, 2011

### rcgldr

Right answer, but wrong formula. Since the run time with N processors is 1/2 the run time with one processor:

1/2 (runtime) = 40% (runtime) / 1 + 60% (runtime) / N

.5 = .4 / 1 + .6 / N
.1 = .6 / N
N = .6 / .1 = 6