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

72.80 % more energy will be consumed.

Step-by-step 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):

`(E_2)/(E_1) = (C_2V_2^2f_2)/(C_1V_1^2f_1)`

As per given statement:

`C_1=C_2`

`V_2=1.2* V_1`

`f_2=1.2* f_1`

Putting the values above:

`(E_2)/(E_1) = (C_1* (1.2V_1)^2* 1.2f_1)/(C_1V_1^2f_1)\\\Rightarrow (E_2)/(E_1) = ((1.2)^2* 1.2)/(1)\\\Rightarrow (E_2)/(E_1) = 1.728\\\Rightarrow \bold{E_2 = 1.728 * 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.