Re: Which cpufreq driver is best for ...

[Dominik Brodowski <linux@brodo.de>]

Hi,

On Thu, Sep 04, 2003 at 11:57:04AM +0200, Norbert Preining wrote:
> OK, me again. Just for completeness, I now see the point and want to add
> some comments:
> 
> On Mit, 03 Sep 2003, Dominik Brodowski wrote:
> > Unfortunately, I don't have the information for your CPU. But, as a good
> > example, let's take the 1.0 GHz Pentium-III-M. 
> > 
> > 100% CPU-usage, 1.0 GHz, Speedstep-High, no throttling:		34.0 W
> > 	and let's say the CPU consuming task takes 1 minute to complete.
> > 	==> for the task you'll need 2.0 kWs [=kJ?] of energy.
> > 
> > 100% CPU-usage,	700 MHz, Speedstep-Low, no throttling:		16.1 W
> > 	--> the CPU consuming task will take 1.43 minutes to complete.
> > 	==> for the task you'll need 1.4 kWs of energy.
> 
> I guess these two values have been taken from some specification (the
> values 34.0 and 16.1)?

Yes: they are taken from Table 33 "Power Specifications for Mobile Pentium
III Processor With Intel SpeedStep Technology" on page 64 of Datasheet
283653 rev. 002 "Mobile Intel Pentium III Processor in BGA2 and Micro- PGA2
Packages", available at developer.intel.com. 

> Assume that we are running cpuXXX daemon which switches the cpu to the
> highest frequency as soon as the load is reasonable high, i.e. someting
> is done. So the power consumption in the times where cpu is used is the
> same in p4-clockmod and speedstep. I will call this T and S (for
> throttling and SpeedStep from now on). Thus, only the `idle' times make
> a difference.
> 
> Let t be the throttling factor, in our case 0.125, and s be the
> speedstep factor (top to low), in our case 0.7. For idle times 
> let c_t the cpu usage level in throttling state, and c_s the cpu usage 
> level in speedstep low state. In our case, we would have
> (1)	c_t : c_s = s : t = 700 : 125 = 5.6
> obviously, because in T we have a lower frequency, thus the cpu usage
> will be higher (is this wrong?) compared to the S state with higher
> frequency.

This is right.

> Now for the energy in S and T states:
> 	S (speedstep low):	W_s = 1.2 + ((16.1-1.2)*s*c_s)
> 	T (throttling):		W_t = 2.2 + ((34.0-2.2)*t*c_t)
> >From (1) we know that
> 	s*c_s = t*c_t
> which gives for W_s:
> 	W_s = 1.2 + ((16.1-1.2)*t*c_t)
> Now we compare W_t with W_s. First note that 
> 	t*c_t always is in [0,1]
> since both of them are in [0,1], thus for the border cases
> t*c_t = 0 and = 1, we obtain:
> 	W_t : W_s [0] = 2.2 : 1.2 = 1.833
> and
> 	W_t : W_s [1] = 34.0 : 16.1 = 2.111
> which gives, that the power consumption is indeed much higher in
> throttling state compared to speedstep low state.
> 
> Reason for all this, to make it short: SpeedStep reduces the voltage and
> the power consumption of the processor, which saves more energy than
> letting the processor run at lower frequency but on the full
> voltage/power consumption.

Indeed. But I think its worth adding that
a) if proper idling is done and CPU usage is below the throttling level, no
	power is saved at all.
b) the voltage and thus the power usage needed for "idling" is lower when
	the processor is in speedstep_low state than if it is in
	speedstep_high state.

> Now one last question: Is this computation complete rubbish? And where
> did you get the values for your processor?

In the datasheet above. For pentium4-ms I couldn't find as complete
documentation, but maybe I didn't look hard enough.

	Dominik