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Andre45 [30]
3 years ago
11

For a new version of processor, suppose the capacitive load remains, how much more energy will the processor consume if we incre

ase voltage by 20% and increase clock rate by 20%?
Computers and Technology
1 answer:
erastovalidia [21]3 years ago
4 0

Answer:

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

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Answer:

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- In the third for loop, find the sales people who exceeded this amount.

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