Calculating the MIPS Rating for a Multi-Processor System

[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?

 

[rcgldr]

Try to convert the problem statement into an algebraic equation.

 

[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?

 

[D H]

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

 

[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?

 

[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?

 

[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?

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

 

[D H]

Which makes N = 6?

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.

 

[rcgldr]

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

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