Answer:
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
