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

From: Rafael J. Wysocki
Date: Wed Mar 09 2016 - 18:29:27 EST

On Wed, Mar 9, 2016 at 5:39 PM, Peter Zijlstra <peterz@xxxxxxxxxxxxx> wrote:
> On Tue, Mar 08, 2016 at 09:05:50PM +0100, Rafael J. Wysocki wrote:
>> >> This means that on platforms where the utilization is frequency
>> >> invariant we should use
>> >>
>> >> next_freq = a * x
>> >>
>> >> (where x is given by (2) above) and for platforms where the
>> >> utilization is not frequency invariant
>> >>
>> >> next_freq = a * x * current_freq / max_freq
>> >>
>> >> and all boils down to finding a.
>> >
>> > Right.
>>
>> However, that doesn't seem to be in agreement with the Steve's results
>> posted earlier in this thread.
>
> I could not make anything of those numbers.
>
>> Also theoretically, with frequency invariant, the only way you can get
>> to 100% utilization is by running at the max frequency, so the closer
>> to 100% you get, the faster you need to run to get any further. That
>> indicates nonlinear to me.
>
> I'm not seeing that, you get that by using a > 1. No need for
> non-linear.

OK

>> >> Now, it seems reasonable for a to be something like (1 + 1/n) *
>> >> max_freq, so for non-frequency invariant we get
>> >>
>> >> nex_freq = (1 + 1/n) * current_freq * x

(*) (see below)

>> > This seems like a big leap; where does:
>> >
>> > (1 + 1/n) * max_freq
>> >
>> > come from? And what is 'n'?
>
>> a = max_freq gives next_freq = max_freq for x = 1,
>
> next_freq = a * x * current_freq / max_freq
>
> [ a := max_freq, x := 1 ] ->
>
> = max_freq * 1 * current_freq / max_freq
> = current_freq
>
> != max_freq
>
> But I think I see what you're saying; because at x = 1,
> current_frequency must be max_frequency. Per your earlier point.

Correct.

>> but with that choice of a you may never get to x = 1 with frequency
>> invariant because of the feedback effect mentioned above, so the 1/n
>> produces the extra boost needed for that (n is a positive integer).
>
> OK, so that gets us:
>
> a = (1 + 1/n) ; n > 0
>
> [ I would not have chosen (1 + 1/n), but lets stick to that ]

Well, what would you choose then? :-)

> So for n = 4 that gets you: a = 1.25, which effectively gets you an 80%
> utilization tipping point. That is, 1.25 * .8 = 1, iow. you'll pick the
> next frequency (assuming RELATION_L like selection).
>
> Together this gets you:
>
> next_freq = (1 + 1/n) * max_freq * x * current_freq / max_freq
> = (1 + 1/n) * x * current_freq

That seems to be what I said above (*), isn't it?

> Again, with n = 4, x > .8 will result in a next_freq > current_freq, and
> hence (RELATION_L) pick a higher one.

OK

>> Quite frankly, to me it looks like linear really is a better
>> approximation for "raw" utilization. That is, for frequency invariant
>> x we should take:
>>
>> next_freq = a * x * max_freq / current_freq
>
> (its very confusing how you use 'x' for both invariant and
> non-invariant).
>
> That doesn't make sense, remember:
>
> util = \Sum_i u_i * freq_i / max_freq (1)
>
> Which for systems where freq_i is constant reduces to:
>
> util = util_raw * current_freq / max_freq (2)
>
> But you cannot reverse this. IOW you cannot try and divide out
> current_freq on a frequency invariant metric.

I see.

> So going by:
>
> next_freq = (1 + 1/n) * max_freq * util (3)

I think that should be

next_freq = (1 + 1/n) * max_freq * util / max

(where max is the second argument of cpufreq_update_util) or the
dimensions on both sides don't match.

> if we substitute (2) into (3) we get:
>
> = (1 + 1/n) * max_freq * util_raw * current_freq / max_freq
> = (1 + 1/n) * current_freq * util_raw (4)
>
> Which gets you two formula with the same general behaviour. As (2) is
> the only approximation of (1) we can make.

OK

So since utilization is not frequency invariant in the current
mainline (or linux-next for that matter) AFAIC, I'm going to use the
following in the next version of the schedutil patch series:

next_freq = 1.25 * current_freq * util_raw / max

where util_raw and max are what I get from cpufreq_update_util().

1.25 is for the 80% tipping point which I think is reasonable.