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]
Date:	Thu, 3 Mar 2016 13:20:30 +0100
From:	Peter Zijlstra <peterz@...radead.org>
To:	"Rafael J. Wysocki" <rafael@...nel.org>
Cc:	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 Wed, Mar 02, 2016 at 11:49:48PM +0100, Rafael J. Wysocki wrote:
> >>> +       min_f = sg_policy->policy->cpuinfo.min_freq;
> >>> +       max_f = sg_policy->policy->cpuinfo.max_freq;
> >>> +       next_f = util > max ? max_f : min_f + util * (max_f - min_f) / max;

> In case a more formal derivation of this formula is needed, it is
> based on the following 3 assumptions:
> 
> (1) Performance is a linear function of frequency.
> (2) Required performance is a linear function of the utilization ratio
> x = util/max as provided by the scheduler (0 <= x <= 1).

> (3) The minimum possible frequency (min_freq) corresponds to x = 0 and
> the maximum possible frequency (max_freq) corresponds to x = 1.
> 
> (1) and (2) combined imply that
> 
> f = a * x + b
> 
> (f - frequency, a, b - constants to be determined) and then (3) quite
> trivially leads to b = min_freq and a = max_freq - min_freq.

3 is the problem, that just doesn't make sense and is probably the
reason why you see very little selection of the min freq.

Suppose a machine with the following frequencies:

	500, 750, 1000

And a utilization of 0.4, how does asking for 500 + 0.4 * (1000-500) =
700 make any sense? Per your point 1, it should should be asking for
0.4 * 1000 = 400.

Because, per 1, at 500 it runs exactly half as fast as at 1000, and we
only need 0.4 times as much. Therefore 500 is more than sufficient.



Note. we all know that 1 is a 'broken' assumption, but lacking anything
better I think its a reasonable one to make.

Powered by blists - more mailing lists

Powered by Openwall GNU/*/Linux Powered by OpenVZ