Question

A 8-core machine has 4 times the performance of a single-core machine of the same frequency. Performance is proportional to frequency. Voltage decreases proportionally to frequency. To achieve the same performance, how much (in percentage) dynamic power would the 8-core system save?

Answer 1

Answer: The 8-core machine saves  87.5% of the dynamic power.

Explanation:

Let Fold = f , Vold = V , Cold = Capacitance

so

Old Dynamic power = Cold × (Vold × Vold) × f

therefore for the 8-core machine

 Fnew / Fold = 1/4

Fnew = Fold/4

we were told that Voltage decreases proportional to frequency,

so

Vnew / Vold = 1/4

Vnew = V / 4

So New Capacitance will be;

Cnew = Cold

Thus, New Dynamic power = 8 × Cnew × ( Vnew × Vnew ) ×  Fnew

= 8 × Cold × (Vold × Vold/16) × ( f/4 )

=  8 × ( Cold ) × ( Vold × Vold ) × ( f ) / 64

= (Old Dynamic Power) / 8

therefore

Old Dynamic Power / New Dynamic Power = 8

Thus, Percentage of power saved will be;

Percentage power saved = 100 × ( Old Dynamic Power - New Dynamic Power ) / Old Dynamic Power

=   100 × (8-1) / 8

= 87.5 %

Therefore The 8-core machine saves  87.5% of the dynamic power.