Consider two processors \(P_1\) and \(P_2\) executing the same instruction…
20142015
Consider two processors \(P_1\) and \(P_2\) executing the same instruction set. Assume that under identical conditions, for the same input, a program running on \(P_2\) takes 25% less time but incurs 20% more CPI (clock cycles per instruction) as compared to the program running on \(P_1\). If the clock frequency of \(P_1\) is 1GHZ, then the clock frequency of \(P_2\) (in GHz) is ______.
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Correct answer: 1.6
Key formula: Execution time = (Instruction count × CPI) / clock frequency.
Let the CPI of the first processor be c and its clock frequency be 1 GHz. The second processor has CPI = 1.2c and its execution time is 25% less, so time2 = 0.75 × time1.
Using time = (CPI × instruction count) / frequency and cancelling the common instruction count, we get: 1.2c / f2 = 0.75 × (c / f1). Cancelling c gives 1.2 / f2 = 0.75 / f1.
Solve for f2: f2 = (1.2 / 0.75) × f1 = 1.6 × f1. With f1 = 1 GHz, f2 = 1.6 GHz.
Answer: 1.6 GHz
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