Calculating the MIPS Rating for a Multi-Processor System

  • Thread starter Thread starter johnnyrocket
  • Start date Start date
  • Tags Tags
    Mips Rating
Click For Summary

Discussion Overview

The discussion revolves around calculating the MIPS rating for a multi-processor system based on a problem involving instruction execution time and parallel processing capabilities. Participants explore the implications of a system with N processors executing a program with specific characteristics, including instruction count and execution time.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant outlines the problem involving a system with N processors executing 16 million instructions, noting that 60% of the code can be executed in parallel.
  • Another participant proposes a formula for calculating MIPS, suggesting that with a 4GHz processor, the MIPS rating for a single processor could be 4000 MIPS.
  • Some participants express uncertainty about the relevance of the MIPS rating and suggest focusing on formulating the problem algebraically instead.
  • A participant introduces a speedup formula, proposing that the speedup factor is related to the number of processors and the proportion of parallelizable code, leading to a calculation that suggests N equals 6.
  • Another participant confirms the calculation of N as 6 but questions the formula used, providing an alternative formulation based on runtime comparisons between single and multi-processor executions.

Areas of Agreement / Disagreement

Participants generally agree on the value of N being 6 based on the calculations presented. However, there is disagreement regarding the correct formulation of the speedup and MIPS calculations, with some participants suggesting different approaches and interpretations.

Contextual Notes

There are unresolved aspects regarding the assumptions made in the calculations, particularly concerning the definitions of parallel and non-parallel execution times and how they relate to the overall speedup and MIPS ratings.

Who May Find This Useful

This discussion may be useful for individuals interested in computer architecture, parallel processing, and performance metrics in multi-processor systems.

johnnyrocket
Messages
8
Reaction score
0
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?
 
Physics news on Phys.org
Try to convert the problem statement into an algebraic equation.
 
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?
 
You don't need to know the MIPS rating. Do what rcgldr suggested, which is to write the problem statement as an algebraic expression.
 
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?
 
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 said:
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.
 
johnnyrocket said:
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.
 
johnnyrocket said:
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
 

Similar threads

Comp Sci Instruction scheduling in multiple-issue RISCV-processor
  • · Replies 7 ·
Replies
7
Views
3K
Comp Sci DLX 5 stage pipeline processor execution and speedup
  • · Replies 3 ·
Replies
3
Views
2K
MIPS Number of Instructions Executed and Memory Accessed
  • · Replies 2 ·
Replies
2
Views
5K
Numerical Problem on Processor Pool Model
  • · Replies 1 ·
Replies
1
Views
3K
Calculations prior to determining algorithm complexity order
  • · Replies 1 ·
Replies
1
Views
2K
Insights Parallel Programming on a CPU with AVX-512
  • · Replies 2 ·
Replies
2
Views
3K
Insights Parallel Programming on an NVIDIA GPU
  • · Replies 2 ·
Replies
2
Views
3K
How to compare CPUs (for example Intel vs Apple)?
  • · Replies 38 ·
2
Replies
38
Views
8K
Interrupts, bus request operations
  • · Replies 10 ·
Replies
10
Views
3K
How Do RISC and CISC Architectures Affect Program Speed?
  • · Replies 1 ·
Replies
1
Views
2K