Question

For a new version of processor, suppose the capacitive load remains, how much more energy will the processor consume if we increase voltage by 20percent and increase clock rate by 20percent?
I do not want you to answer the question for me, but please give me the step to solve this. How do I solve this problem.

Answer 1

Answer:

72.80 % more energy will be consumed.

Explanation:

First of all, let us have a look at the formula of energy for a processor.

$E = CV^2f$

Where, $E$ is the energy

$C$ is the capacitance

$V$ is the voltage and

$f$ is the clock rate.

Let $E_1$ be the energy of older processor.

$C_1$ be the capacitance of older processor.

$V_1$ be the voltage of older processor

$f_1$ be the capacitance of older processor

So, $E_1 = C_1V_1^2f_1$ ....... (1)

and

Let $E_2$ be the energy of newer processor.

$C_2$ be the capacitance of newer processor.

$V_2$ be the voltage of newer processor

$f_2$ be the capacitance of newer processor

$E_2 = C_2V_2^2f_2$ ....... (2)

Dividing equation (2) with equation (1):

$\dfrac{E_2}{E_1} = \dfrac{C_2V_2^2f_2}{C_1V_1^2f_1}$

As per given statement:

$C_1=C_2$

$V_2=1.2\times V_1$

$f_2=1.2\times f_1$

Putting the values above:

$\dfrac{E_2}{E_1} = \dfrac{C_1\times (1.2V_1)^2\times 1.2f_1}{C_1V_1^2f_1}\\\Rightarrow \dfrac{E_2}{E_1} = \dfrac{(1.2)^2\times 1.2}{1}\\\Rightarrow \dfrac{E_2}{E_1} = 1.728\\\Rightarrow \bold{E_2 = 1.728 \times E_1}$

Energy consumed by newer processor is 1.728 times the energy consumed by older processor.

OR

72.80 % more energy will be consumed by the newer processor.