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Date: Fri, 4 Mar 2016 02:14:06 +0100
Message-ID: <CAJZ5v0i6A8EDWXjU5+ANW_n1Rpv1xaWwWV7Q9-1NgO5P8HnhUw@mail.gmail.com>
Subject: Re: [PATCH 6/6] cpufreq: schedutil: New governor based on scheduler
 utilization data
From: "Rafael J. Wysocki" <rafael@kernel.org>
To: Peter Zijlstra <peterz@infradead.org>
Cc: "Rafael J. Wysocki" <rafael@kernel.org>,
        Vincent Guittot <vincent.guittot@linaro.org>,
        "Rafael J. Wysocki" <rjw@rjwysocki.net>,
        Linux PM list <linux-pm@vger.kernel.org>,
        Juri Lelli <juri.lelli@arm.com>,
        Steve Muckle <steve.muckle@linaro.org>,
        ACPI Devel Maling List <linux-acpi@vger.kernel.org>,
        Linux Kernel Mailing List <linux-kernel@vger.kernel.org>,
        Srinivas Pandruvada <srinivas.pandruvada@linux.intel.com>,
        Viresh Kumar <viresh.kumar@linaro.org>,
        Michael Turquette <mturquette@baylibre.com>
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On Thu, Mar 3, 2016 at 5:47 PM, Peter Zijlstra <peterz@infradead.org> wrote:
> On Thu, Mar 03, 2016 at 05:37:35PM +0100, Peter Zijlstra wrote:
>> On Thu, Mar 03, 2016 at 05:24:32PM +0100, Rafael J. Wysocki wrote:
>> > >> f = a * x + b
>
>> > If not, then I think it's reasonable to map the middle of the
>> > available frequency range to x = 0.5 and then we have b = 0 and a =
>> > (max_freq + min_freq) / 2.

That actually should be a = max_freq + min_freq, because I want
(max_freq + min_freq) / 2 = a / 2.

>> So I really think that approach falls apart on the low util bits, you
>> effectively always run above min speed, even if min is already vstly
>> over provisioned.
>
> Ah nevermind, I cannot read. Yes that is worth trying I suppose. But the
> b=0,a=1 thing seems more natural still.

It is somewhat imbalanced, though.  If all of the values of x are
equally probable (or equally frequent), the probability of running
above the middle frequency is lower than the probability of running
below it.