Question

You have a large inventory of Intel processors that run at 3 GHz and 1 Volt, and consume 100 W (of which, 20 W is leakage) when running a given CPU-bound application. This application finishes in 100 seconds on this processor. The processor is capable of DFS and DVFS, but cannot reduce its voltage under 0.9 V. A customer places an order for motherboards that have inexpensive cooling and power delivery that can only handle a processor that runs at 40 W. How will you configure your Intel processor so that it meets these specifications and consumes as little energy as possible for the same CPU-bound application? How much energy does the processor consume to finish this application on this new system?

Answer 1

Solution

v=voltage

f=frequency

Leakage Power=20W

Dynamic Power=80W.

Leakage Power = v * c

Dynamic Power = activity * capacity * v² * f

In order to the have the least power, we set the new voltage to 0.9V

New Leakage Power

= 20W * 0.9

= 18W

New Dynamic Power

= 80W*0.9² * new f/old f

= 40W – 18W

= 22W

So we configure the processor that new frequency is 1.02 GHz, new voltage is 0.9V to meet the specification and consumes as little energy as possible.

New time = 100 * old f / new f

= 294.11

So

the total energy

= New time * power

= 40W * 294.11

= 11764.7J

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