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Date:   Fri, 18 Oct 2019 09:59:57 +0200
From:   Peter Zijlstra <peterz@infradead.org>
To:     Dietmar Eggemann <dietmar.eggemann@arm.com>
Cc:     Quentin Perret <qperret@google.com>,
        Douglas Raillard <douglas.raillard@arm.com>,
        linux-kernel@vger.kernel.org, linux-pm@vger.kernel.org,
        mingo@redhat.com, rjw@rjwysocki.net, viresh.kumar@linaro.org,
        juri.lelli@redhat.com, vincent.guittot@linaro.org,
        qperret@qperret.net, patrick.bellasi@matbug.net, dh.han@samsung.com
Subject: Re: [RFC PATCH v3 0/6] sched/cpufreq: Make schedutil energy aware
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On Fri, Oct 18, 2019 at 09:44:44AM +0200, Dietmar Eggemann wrote:
> On 17/10/2019 16:11, Peter Zijlstra wrote:
> > On Thu, Oct 17, 2019 at 12:11:16PM +0100, Quentin Perret wrote:
> 
> [...]
> 
> > It only boosts when 'rq->cfs.avg.util' increases while
> > 'rq->cfs.avg.util_est.enqueued' remains unchanged (and util > util_est
> > obv).
> > 
> > This condition can be true for select_task_rq_fair(), because that is
> > ran before we do enqueue_task_fair() (for obvious raisins).
> > 
> >>> I'm still thinking about the exact means you're using to raise C; that
> >>> is, the 'util - util_est' as cost_margin. It hurts my brain still.
> >>
> >> +1 ...
> > 
> > cost_i = capacity_i / power_i ; for the i-th OPP
> 
> I get confused by this definition. efficiency=capacity/power but the
> cs->cost value used in em_pd_get_higher_freq() is defined as
> 
> cs_cost = cs->power * cpu_max_freq / cs->freq [energy_model.h]

Well, chalk that up to confusion inspired by the Changelog of patch 1.

Let me redo it with that formula then.