Suppose one machine, A, executes a program with an average CPI of 1.9. Suppose another machine, B (with the same instruction set and an enhanced compiler), executes the same program with 20% less instructions and with a CPI of 1.1 at 800MHz. In order for the two machines to have the same performance, what does the clock rate of the first machine need to be

the clock rate of the first machine need to be 1.7 GHz

Explanation:

Given:

CPI of A = 1.9

CPI of B = 1.1

machine, B executes the same program with 20% less instructions and with a CPI of 1.1 at 800MHz

To find:

In order for the two machines to have the same performance, what does the clock rate of the first machine need to be

Solution:

CPU execution time = Instruction Count * Cycles per Instruction/ clock rate

CPU execution time = (IC * CPI) / clock rate

(IC * CPI) (A) / clock rate(A) =  (IC * CPI)B / clock rate(B)

(IC * 1.9) (A) / clock rate(A) = (IC * (1.1 * (1.0 - 0.20)))(B) / 800 * 10⁶ (B)

Notice that 0.20 is basically from 20% less instructions

(IC * 1.9)  / clock rate = (IC * (1.1 * (1.0 - 0.20))) / 800 * 10⁶

(IC * 1.9)  / clock rate =  (IC*(1.1 * ( 0.8))/800 * 10⁶

(IC * 1.9)  / clock rate =  (IC * 0.88) / 800 * 10⁶

clock rate (A) =  (IC * 1.9) / (IC * 0.88) / 800 * 10⁶

clock rate (A) =  (IC * 1.9) (800 * 10⁶) /  (IC * 0.88)

clock rate (A) = 1.9(800)(1000000)  / 0.88

clock rate (A) =  (1.9)(800000000)  / 0.88

clock rate (A) = 1520000000  / 0.88

clock rate (A) = 1727272727.272727

clock rate (A) = 1.7 GHz

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