Answer:
Given Data:
Clock rate of P1 = 4 GHz
Clock rate of P2 = 3 GHz
Average CPI of P1 = 0.9
Number of Instructions = 5.0E9 = 5 × 10^9
Clock rate of P2 = 3 GHz
Average CPI of P2 = 0.75
Number of Instructions = 1.0E9 = 10^9
To find: If the computer with largest clock rate has the largest performance?
Explanation:
Solution:
As given in the question, clock rate of P1 = 4 GHz which is greater than clock rate of P2 = 3 GHz
According to the performance equation:
CPU Time = instruction count * average cycles per instruction/ clock rate
CPU Time = I * CPI / clock rate
Where instruction count refers to the number of instructions.
Performance of P1:
CPU Time (P1) = 5 * 10^9 * 0.9 / (4 * 10^9 )
= 5000000000 * 0.9 / 4000000000
= 4500000000 / 4000000000
= 1.125s
Performance of P2:
CPU Time (P2) = 10^9 * 0.75/ (3 * 10^9 )
= 750000000 / 3000000000
= 0.25s
So the Performance of P2 is larger than that of P1,
performance (P2) > performance (P1)
0.25 is better than 1.125
But clock rate of P1 was larger than P2
clock rate of P1 > clock rate of P2
4 GHz > 3 GHz
So this is a misconception about P1 and P2.
It is not true that computer with the largest clock rate as having the largest performance.
