linux-kernel - Re: [PATCH v6 04/11] cpufreq/schedutil: use rt utilization tracking

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Message-ID: <CAKfTPtC5HLD=3W-w0XgwMrzhLfgE8onE4RmGTXyCWx6F=-uVhw@mail.gmail.com>
Date:   Fri, 22 Jun 2018 14:23:22 +0200
From:   Vincent Guittot <vincent.guittot@...aro.org>
To:     Peter Zijlstra <peterz@...radead.org>
Cc:     Quentin Perret <quentin.perret@....com>,
        Ingo Molnar <mingo@...nel.org>,
        linux-kernel <linux-kernel@...r.kernel.org>,
        "Rafael J. Wysocki" <rjw@...ysocki.net>,
        Juri Lelli <juri.lelli@...hat.com>,
        Dietmar Eggemann <dietmar.eggemann@....com>,
        Morten Rasmussen <Morten.Rasmussen@....com>,
        viresh kumar <viresh.kumar@...aro.org>,
        Valentin Schneider <valentin.schneider@....com>,
        Patrick Bellasi <patrick.bellasi@....com>,
        Joel Fernandes <joel@...lfernandes.org>,
        Daniel Lezcano <daniel.lezcano@...aro.org>,
        Ingo Molnar <mingo@...hat.com>
Subject: Re: [PATCH v6 04/11] cpufreq/schedutil: use rt utilization tracking

On Fri, 22 Jun 2018 at 13:37, Peter Zijlstra <peterz@...radead.org> wrote:
>
> On Fri, Jun 22, 2018 at 08:58:53AM +0100, Quentin Perret wrote:
> > Say we have 50% of the capacity stolen by RT, and a 25% CFS task
> > running. If we re-scale, we'll end up with a 50% request for CFS
> > (util==512 for your code above). But if we want to see a little bit
> > of idle time in the system, we should really request an OPP for 75%+ of
> > capacity no ? Or am I missing something ?
>
> That is true.. So we could limit the scaling to the case where there is
> no idle time, something like:
>
>         util = sg_cpu->util_cfs;
>
>         cap_cfs = (1024 - (sg_cpu->util_rt + ...));
>         if (util == cap_cfs)
>                 util = sg_cpu->max;
>
> That specifically handles the '0% idle -> 100% freq' case, but I don't
> realy like edge behaviour like that. If for some reason it all doesn't
> quite align you're left with bits.
>
> And the linear scaling is the next simplest thing that avoids the hard
> boundary case.
>
> I suppose we can make it more complicated, something like:
>
>              u           u
>   f := u + (--- - u) * (---)^n
>             1-r         1-r
>
> Where: u := cfs util
>        r := \Sum !cfs util
>        f := frequency request
>
> That would still satisfy all criteria I think:
>
>   r = 0      -> f := u
>   u = (1-r)  -> f := 1
>
> and in particular:
>
>   u << (1-r) -> f ~= u
>
> which casuses less inflation than the linear thing where there is idle
> time.
>
> In your specific example that ends up with:
>
>              .25           .25
>   f = .25 + (--- - .25) * (---)^n = .25 + .0625 (for n=2)
>              .5            .5     = .25 + .125  (for n=1)
>
> But is that needed complexity?

And we are not yet at the right value for quentin's example as we need
something around 0.75 for is example
The non linearity only comes from dl so if we want to use the equation
above, u should be (cfs + rt) and r = dl
But this also means that we will start to inflate the utilization to
get higher OPP even if there is idle time and lost the interest of
using dl bw