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Subject: Re: [RFC PATCH v3 0/6] sched/cpufreq: Make schedutil energy aware
To:     Peter Zijlstra <peterz@infradead.org>,
        Quentin Perret <qperret@google.com>
Cc:     Douglas Raillard <douglas.raillard@arm.com>,
        linux-kernel@vger.kernel.org, linux-pm@vger.kernel.org,
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        juri.lelli@redhat.com, vincent.guittot@linaro.org,
        qperret@qperret.net, patrick.bellasi@matbug.net, dh.han@samsung.com
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From:   Dietmar Eggemann <dietmar.eggemann@arm.com>
Message-ID: <2cbde0fe-c10c-0ebb-32ef-2d522986bc89@arm.com>
Date:   Fri, 18 Oct 2019 09:44:44 +0200
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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]

> We then do: 'x = util - util_avg' and use that on the first
> OPP >= min_freq:
> 
> 	cost(x) = cost_j + cost_j * x ; freq_j >= min_freq
> 	        = cost_j * (1 + x)
> 
> And then we find the highest OPP that has:
> 
> 	cost_k <= cost(x)

[...]