On Thu, Mar 03, 2016 at 05:24:32PM +0100, Rafael J. Wysocki wrote:

On Thu, Mar 3, 2016 at 1:20 PM, Peter Zijlstra [off-list ref] wrote:

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On Wed, Mar 02, 2016 at 11:49:48PM +0100, Rafael J. Wysocki wrote:

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+       min_f = sg_policy->policy->cpuinfo.min_freq;
+       max_f = sg_policy->policy->cpuinfo.max_freq;
+       next_f = util > max ? max_f : min_f + util * (max_f - min_f) / max;

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In case a more formal derivation of this formula is needed, it is
based on the following 3 assumptions:

(1) Performance is a linear function of frequency.
(2) Required performance is a linear function of the utilization ratio
x = util/max as provided by the scheduler (0 <= x <= 1).

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(3) The minimum possible frequency (min_freq) corresponds to x = 0 and
the maximum possible frequency (max_freq) corresponds to x = 1.

(1) and (2) combined imply that

f = a * x + b

(f - frequency, a, b - constants to be determined) and then (3) quite
trivially leads to b = min_freq and a = max_freq - min_freq.

3 is the problem, that just doesn't make sense and is probably the
reason why you see very little selection of the min freq.

It is about mapping the entire [0,1] interval to the available frequency range.

Yeah, but I don't see why that makes sense..

I till overprovision things (the smaller x the more), but then it may
help the race-to-idle a bit in theory.

So, since we also have the cpuidle information, could we not make a
better guess at race-to-idle?

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Suppose a machine with the following frequencies:

        500, 750, 1000

And a utilization of 0.4, how does asking for 500 + 0.4 * (1000-500) =
700 make any sense? Per your point 1, it should should be asking for
0.4 * 1000 = 400.

Because, per 1, at 500 it runs exactly half as fast as at 1000, and we
only need 0.4 times as much. Therefore 500 is more than sufficient.

OK, but then I don't see why this reasoning only applies to the lower
bound of the frequency range.  Is there any reason why x = 1 should be
the only point mapping to max_freq?

Well, everything that goes over the second to last freq would end up at
the last (max) freq.

Take again the 500,750,1000 example, everything that's >750 would end up
at 1000 (for relation_l, >875 for _c).

But given the platform's cpuidle information, maybe coupled with an avg
idle est, we can compute the benefit of race-to-idle and over provision
based on that, right?

If not, then I think it's reasonable to map the middle of the
available frequency range to x = 0.5 and then we have b = 0 and a =
(max_freq + min_freq) / 2.

So I really think that approach falls apart on the low util bits, you
effectively always run above min speed, even if min is already vstly
over provisioned.