Question

A processor has a clock rate of 2.5 GHz. It’s dynamic power is 80 W and static power is 20 watt. It executes a certain program in 25 seconds.

a. What is the power consumption if clock frequency is slowed down by 30%?

b. What will be the new execution time for the program which earlier took 25 seconds after slowing down the clock as in part a?

c. What is the power consumption if voltage is decreased by 15%?

Answer 1

Part a.)

The static power consumption will remain the same as it is (largely) independent of frequency

The dynamic power consumption is given by the following formula:

Pd = C*f*V^2

where C is a constant.

As C & V are the same and only frequency is slowed down by 30%, so the ratio of new dynamic power to old dynamic power will be:

Pd(new) / Pd(Old) = (C*(f*(1-0.3))*V^2) / C*f*V^2

= 0.7

Thus Pd(new) = 0.7 * Pd(Old)

= 0.7 * 80

= 56 Watt

So, Total power consumption = 20 + 56

= 76 Watts

Part b.)

Since execution time is inversely proportional to clock frequency, so we have the following relationship

T(new)/T(old) = f(old)/f(new)

T(new) = 25 * (2.5 / (2.5 *(1-0.3)))

T(new) = 35.7 seconds

Thus the new execution time will be 35.7 (~36) seconds

Part c.)

We can use the same relation as in part a. but this time only the voltage will change as:

Pd(new) / Pd(Old) =    (C*f*(V*(1-0.15))^2) / (C*f*V^2)

Pd(new) / Pd(Old) = (0.85)^2

Pd(new) = 0.7225 * Pd(Old)

= 0.7225 * 80

= 57.8 Watts

Thus, total power consumption = 20 + 57.8

= 77.8 Watts