Answer:
P2 has the highest performance expressed in instructions per second i.e. 2.5 * 10^9
Explanation:
Given:
Clock rate of P1 = 3 GHz
Clock rate of P2 = 2.5 GHz
Clock rate of P3 = 4.0 GHz
Cycles Per Instruction CPI of P1 = 1.5
Cycles Per Instruction CPI of P2 = 1.0
Cycles Per Instruction CPI of P3 = 2.2
To find:
Highest Performance expressed in instruction per second.
Solution:
Performance = clock rate / CPI
The performance of processor depends on instructions count and CPU time.
As we know that
CPU time = Instructions * Cycles Per Instruction / clock rate
Instructions per second = Instruction count / CPU time
As:
CPU time = I * CPI/ clock rate
I/CPU time = clock rate/CPI
So IPS = clock rate/ CPI
Hence
Performance = Clock rate / Cycles per Instruction = clock rate CPI
Performance for P1 = IPS = clock rate / CPI = 3 GHz / 1.5 = 2
As we know that 1 GHz = 10 ^ 9 Hz. So:
Performance of P1 expressed in instructions per second is 2 * 10^9
Performance for P2 = IPS = clock rate / CPI = 2.5 / 1.0 = 2.5
Performance of P2 expressed in instructions per second is 2.5 * 10^9
Performance for P3 = IPS = clock rate / CPI = 4 GHz / 2.2= 1.82
Performance of P3 expressed in instructions per second is 1.82 * 10^9
Hence P2 has the highest performance expressed in instructions per second.
