Re: [11/11] system 1: Saving energy using DVFS

[Pavel Machek]

On Mon 2014-01-20 17:17:52, Catalin Marinas wrote:
> On Mon, Jan 20, 2014 at 05:10:29PM +0000, Catalin Marinas wrote:
> > On Mon, Jan 20, 2014 at 04:49:26PM +0000, Pavel Machek wrote:
> > > > To save energy, the higher frequencies should be avoided and only used
> > > > when the application performance requirements can not be satisfied
> > > > otherwise (e.g. spread tasks across more cpus if possible).
> > > 
> > > I argue this is untrue for any task where user waits for its
> > > completion with screen on. (And that's quite important subset).
> > > 
> > > Lets take Nokia n900 as an example. 
> > > 
> > > (source http://wiki.maemo.org/N900_Hardware_Power_Consumption)
> > > 
> > > Sleeping CPU: 2mA
> > > Screen on: 230mA
> > > CPU loaded: 250mA
> > > 
> > > Now, lets believe your numbers and pretend system can operate at 33%
> > > of speed with 11% power consumption.
> > > 
> > > Lets take task that takes 10 seconds on max frequency:
> > > 
> > >       ~ 10s * 470mA     	     	    = 4700mAs
> > > 
> > > You suggest running at 33% speed, instead; that means 30 seconds on
> > > low requency.
> > > 
> > > CPU on low: 25mA (assumed).
> > > 
> > >      ~ 30s * 255mA			    = 7650mAs
> > > 
> > > Hmm. So race to idle is good thing on Intel machines, and it is good
> > > thing on ARM design I have access to.
> > 
> > Race to idle doesn't mean that the screen goes off as well. Let's say
> > the screen stays on for 1 min and the CPU needs to be running for 10s
> > over this minute, in the first case you have:
> > 
> > 	10s & 250mA + 60s * 230mA = 16300mAs
> > 
> > in the second case you have:
> > 
> > 	30s * 25mA + 60s * 230mA = 14550mAs
> > 
> > That's a 1750mAs difference. There are of course other parts drawing
> > current but simple things like the above really make a difference in the
> > mobile space, both in terms of battery and thermal budget.
> 
> BTW, the proper way to calculate this is to use the energy rather than
> current x time. This would be J = Ohm * A^2 * s = V^2 / Ohm * s (so the
> impact of the current is even bigger).

You are claiming that energy is proportional to current squared?

I stand by numbers. Energy is proportional to values I quoted,
provided constant voltage.

									Pavel
-- 
(english) http://www.livejournal.com/~pavelmachek
(cesky, pictures) http://atrey.karlin.mff.cuni.cz/~pavel/picture/horses/blog.html
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