You are designing a system for a real-time application in which specific deadlines must be met. Finishing the computation faster gains nothing. You find that your system can execute the necessary code, in the worst case, twice as fast as necessary. How much energy do you save if you execute at the current speed and turn off the system when the computation is complete and how much energy do you save if you set the voltage and frequency to be half as much?

Answer / Explanation:

To proper understand the answers that is given to the question, we need to understand some basic terms that has been used in the question.

Energy: This can be refereed to as the quantitative property that is transferred to an object for the purpose of the object working or to heat up the object. It can also be referred to as conserved quantity that is energy can be converted from one form or state to another but cannot destroyed.

Power:  This can be defined as the rate of doing work or transferring heat per unit time from one state to another. The SI Units of power is watt which is equal to one joule per second.

Hence, the formula that links energy and power is:

Energy = Power x Time

Now. referring back to the question (a) asking how much energy do we save if we execute at the current speed and turn off the system when the computation is complete: The answer is  = 50%. That is 50% of the energy is saved.

(b) If we recall the formula for calculating energy,

we have:

Energy = 1 /2 Load x V²

Changing the frequency does not affect the energy.However, it affects the power.

So therefore, the new energy is 1 / 2 Load x ( 1/2 V)² ,

reducing it to about 1 /4 of the old energy.

twist boundary

explanation: