Question

Explain by providing an example, why MIPS is not an accurate measure for computing performance among computers?

Answer 1

Consider an Example
clock rate = 500 MHz = 500*10^6
Compiler 1
Instruction A count = 5 x 10^9       CPI = 1
Instruction B count = 1 x 10^9       CPI = 2
Instruction C count = 1 x 10^9       CPI = 3
Compiler 2
Instruction A count = 10 x 10^9       CPI = 1
Instruction B count = 1 x 10^9       CPI = 2
Instruction C count = 1 x 10^9       CPI = 3

Total CPU Clock cycles by compiler 1 = (5 * 1 + 1 * 2 + 1 * 3) * 10^9 = 10 * 10^9
Total CPU Clock cycles by compiler 2 = (10 * 1 + 1 * 2 + 1 * 3) * 10^9 = 15 * 10^9
CPU time1= (10 * 10^9)/ (500 * 10^6) = 20 seconds
CPU time2= (15 * 10^9)/ (500 * 10^6) = 30 seconds

MIPS1 = (5 + 1 + 1) * 10^9/ (20 * 10^6) = 350
MIPS2 = (10 + 1 + 1) * 10^9/ (30 * 10^6) = 400

From above example, we can see that
MIPS(2) is greater than MIPS(1)

But execution time of 1 is lesser than 2.
Therefore, MIPS is not an accurate measure for computing performance among computers.