linux-kernel - Re: [PATCH 6/6] cpufreq: schedutil: New governor based on scheduler utilization data

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Message-ID: <CAJZ5v0i6A8EDWXjU5+ANW_n1Rpv1xaWwWV7Q9-1NgO5P8HnhUw@mail.gmail.com>
Date:	Fri, 4 Mar 2016 02:14:06 +0100
From:	"Rafael J. Wysocki" <rafael@...nel.org>
To:	Peter Zijlstra <peterz@...radead.org>
Cc:	"Rafael J. Wysocki" <rafael@...nel.org>,
	Vincent Guittot <vincent.guittot@...aro.org>,
	"Rafael J. Wysocki" <rjw@...ysocki.net>,
	Linux PM list <linux-pm@...r.kernel.org>,
	Juri Lelli <juri.lelli@....com>,
	Steve Muckle <steve.muckle@...aro.org>,
	ACPI Devel Maling List <linux-acpi@...r.kernel.org>,
	Linux Kernel Mailing List <linux-kernel@...r.kernel.org>,
	Srinivas Pandruvada <srinivas.pandruvada@...ux.intel.com>,
	Viresh Kumar <viresh.kumar@...aro.org>,
	Michael Turquette <mturquette@...libre.com>
Subject: Re: [PATCH 6/6] cpufreq: schedutil: New governor based on scheduler
 utilization data

On Thu, Mar 3, 2016 at 5:47 PM, Peter Zijlstra <peterz@...radead.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.