linux-kernel - Re: [RFC PATCH v3 0/6] sched/cpufreq: Make schedutil energy aware

lists.openwall.net		lists / announce owl-users owl-dev john-users john-dev passwdqc-users yescrypt popa3d-users / oss-security kernel-hardening musl sabotage tlsify passwords / crypt-dev xvendor / Bugtraq Full-Disclosure linux-kernel linux-netdev linux-ext4 linux-hardening linux-cve-announce PHC
Open Source and information security mailing list archives

Hash Suite: Windows password security audit tool. GUI, reports in PDF.

[<prev] [next>] [<thread-prev] [thread-next>] [day] [month] [year] [list]

Message-ID: <2cbde0fe-c10c-0ebb-32ef-2d522986bc89@arm.com>
Date:   Fri, 18 Oct 2019 09:44:44 +0200
From:   Dietmar Eggemann <dietmar.eggemann@....com>
To:     Peter Zijlstra <peterz@...radead.org>,
        Quentin Perret <qperret@...gle.com>
Cc:     Douglas Raillard <douglas.raillard@....com>,
        linux-kernel@...r.kernel.org, linux-pm@...r.kernel.org,
        mingo@...hat.com, rjw@...ysocki.net, viresh.kumar@...aro.org,
        juri.lelli@...hat.com, vincent.guittot@...aro.org,
        qperret@...rret.net, patrick.bellasi@...bug.net, dh.han@...sung.com
Subject: Re: [RFC PATCH v3 0/6] sched/cpufreq: Make schedutil energy aware

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)

[...]