The utilization of a subset of the performance equation as a performance metric is a pitfall. To illustrate this, assume the following two processors. P1 has a clock rate of 4 GHz, average CPI of 0.9, and requires the execution of 5.0E9 instructions. P2 has a clock rate of 3 GHz, an average CPI of 0.75, and requi res the execution of 1.0E9 instructions. One usual fallacy is to consider the computer with the largest clock rate as having the largest performance. Check if this is true for P1 and P2.

Given dа ta:

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.