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- August 28th 2010, 07:47 AM #1

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## Question on MIPS, clock cycle & clock per instruction

Hi everyone

I am doing COmputer Organisation, I cant find any hints to solve this question.Can someone help me?

TWO programs, X and Y were executed in THREE different machines, C1, C2 and C3 with the following outcome of execution times in second respectively; 8, 5; 6, 4; 3, 7. Calculate and determine:

a) Relative performance of the machines for each program.

any hint/formula for the next questions, can;t find anything related to this..thank you for your kind help & support

b) MIPS of C1, C2 and C3 given the clock rate 250 MHz, 180 MHz and 150 MHz respectively

c) Clock cycle for program X in C1

d) Clock per instruction for program Y on C2

e) Which machine is the fastest for each of the program?

Appreciate all your kind help & support.thank you & regards

- August 29th 2010, 12:08 AM #2

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- August 29th 2010, 08:50 AM #3

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Hi everyone

I am doing COmputer Organisation, I cant find any hints to solve this question.Can someone help me?

TWO programs, X and Y were executed in THREE different machines, C1, C2 and C3 with the following outcome of execution times in second respectively; 8, 5; 6, 4; 3, 7. Calculate and determine:

a) Relative performance of the machines for each program.

Answer is:

, ,

any hint/formula for the next questions, can;t find anything related to this..thank you for your kind help & support

b) MIPS of C1, C2 and C3 given the clock rate 250 MHz, 180 MHz and 150 MHz respectively

c) Clock cycle for program X in C1

d) Clock per instruction for program Y on C2

e) Which machine is the fastest for each of the program?

Appreciate all your kind help & support.thank you & regards

- August 29th 2010, 09:19 AM #4

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- August 29th 2010, 09:30 AM #5

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Hi

Thank you for replying, i think there is some mistake in your answer.THis is what i did so far, need your help to comment & advise.Thank you for allkind help & support.

1.) Relative performance:

Program X: 8/4/3 = 0.667

Program Y : 5/3/7 = 0.238

2) MIPS of C1,C2,

MIPS of C1: 250M/8= 31.25MIPS

MIPSfor C2:180M/6 = 30MIPS

MIPS for C3: 150M/3=50MIPS

3) Clock cycle for program X

Clock cycle time for

C1= 1/250M = 4ns

CPU execution time for a program=CPU clock cycles for a program x clock cycle time

For program A in C1

8seconds = Clock cycles x 4ns

Clock cycle for program X in C1 = 8/4ns = 2 x10^9

4) Clock per instruction for program Y on C2

Clock cycle time for

C2= 1/180M = 5.56ns

Clock per instruction = clock cycle time/number of instruction

= 5.56ns/180M = 3.086x10^-17

5)?

- August 29th 2010, 12:24 PM #6

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