# Thread: Can someone help me with this

1. ## 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

2. Originally Posted by anderson
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.
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
a) Chose one of the machines for a reference, say C1, then the relative performance on program i on machine j is:

$\tau_{i,j}= \dfrac{t_{i,1}}{t_{i,j}}$

Now what difficulties are you having with the others?

CB

3. 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.
$\frac{8}{5}$, $\frac{6}{4}$, $\frac{3}{7}$
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

4. Originally Posted by anderson
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.
$\frac{8}{5}$, $\frac{6}{4}$, $\frac{3}{7}$
No that is not what you were told, relative to machine 1 the performances on the two programs are: $1,1;\frac{8}{6}, \frac{4}{5}; \frac{8}{3}, \frac{4}{7}$

CB

5. 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)?

6. Originally Posted by anderson
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)?
Thry explaining what you think you are doing

CB