Re: [PATCH 6/6] cpufreq: schedutil: New governor based on scheduler utilization data

Rafael J. Wysocki
2016-03-03T16:24:37Z

On Thu, Mar 3, 2016 at 1:20 PM, Peter Zijlstra <peterz@infradead.org> wrote:
> On Wed, Mar 02, 2016 at 11:49:48PM +0100, Rafael J. Wysocki wrote:
>> >>> + 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;
>
>> 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).
>
>> (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.

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

> 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?

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.

I'll try that and see how it goes.

> Note. we all know that 1 is a 'broken' assumption, but lacking anything
> better I think its a reasonable one to make.

Right.