Re: Status of X86_P4_CLOCKMOD?

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On Sat, Feb 25, 2006 at 01:53:26PM +0100, Adrian Bunk wrote:
> > > Doesn't less heat imply less power consumption?
> > 
> > Not in this case no.
> >...
> 
> Sorry for the dumb question, but how could this work physically?
> 
> If a computer produces less heat with the same power consumption, what 
> happens with the other energy?

No. Let's do the math (again), and (again) for the actual values of an Intel
Pentium(R) M Processor, 1400 MHz @ 1.484 V, even though the same rules of
physics, logic and mathematics apply to _all_ processors.

Power consumption in idle state C2 (Stop-Grant state)	 7.3 W
Power consumption when "skipping instructions"
	because of throttling (Stop-Grant state)	 7.3 W

Power consumption when doing work			22.0 W


This means that if the processor idle percentage is _larger_ than (1 -
throttling rate), throttling has no effect at all.


Now, let's assume there is some work for the CPU to do which keeps it busy
for one second @ 1.4 GHz. How much energy is needed to get this work done?

0% throttling:		22 Ws	(1s)
25% throttling:		24 Ws	(1.3s)
50% throttling:		29 Ws	(2s)
75% throttling:		44 Ws	(4s)


Now let's also assume there is nothing else to do during a span of four
seconds: then, independent of the throttling setting, the CPU power
consumption is 44 Ws for these four seconds.


However: for the 75% throttling state, the CPU only produces 11 W of heat
_all the time_ -- this means, the fan or air conditioning must only consider
11 W. For 0%, the CPU may produce 44 W of heat in a second -- and to cool
that sufficiently, the fan _may_ need to run faster, which consumes more
energy than is saved by only having to cool 7.3 W (instead of 11W) the other
three seconds.


So: P4-clockmod style throttling only makes sense if either

a) the idle handler does not enter the Stop-Grant state (C2) efficiently, or

b) the load varies significantly over time in a manner which has effect on
   the fan, and where the latency induced by throttling doesn't matter.


	Dominik
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