For a new version of processor, suppose the capacitive load remains, how much more energy will the processor consume if we increase voltage by 20% and increase clock rate by 20%?

The answer is below

Explanation:

The amount of power dissipated by a processor is given by the formula:

P = fCV²

Where f = clock rate, C = capacitance and V = Voltage

For the old version of processor with a clock rate of f, capacitance C and voltage of V, the power dissipated is:

P(old) = fCV²

For the new version of processor with a clock rate of 20% increase = (100% + 20%)f = 1.2f, capacitance is the same = C and voltage of 20% increase = 1.2V, the power dissipated is:

P(new) = 1.2f × C × (1.2V)² = 1.2f × C × 1.44V² =1.728fCV² = 1.728 × Power dissipated by old processor

Hence, the new processor is 1.728 times (72.8% more) the power of the old processor