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Date: Tue, 13 May 2014 04:34:16 +0800
From: Yuyang Du <yuyang.du@intel.com>
To: Stratos Karafotis <stratosk@semaphore.gr>
Cc: Dirk Brandewie <dirk.brandewie@gmail.com>,
        "Rafael J. Wysocki" <rjw@rjwysocki.net>,
        Viresh Kumar <viresh.kumar@linaro.org>,
        Dirk Brandewie <dirk.j.brandewie@intel.com>,
        "linux-pm@vger.kernel.org" <linux-pm@vger.kernel.org>,
        LKML <linux-kernel@vger.kernel.org>,
        Doug Smythies <dsmythies@telus.net>
Subject: Re: [RFC PATCH] cpufreq: intel_pstate: Change the calculation of
 next pstate
Message-ID: <20140512203416.GB10676@intel.com>
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On Tue, May 13, 2014 at 07:16:24AM +0300, Stratos Karafotis wrote:
> On 12/05/2014 11:01 μμ, Yuyang Du wrote:
> > On Tue, May 13, 2014 at 06:59:42AM +0300, Stratos Karafotis wrote:
> >> Hi,
> >>
> >> On 12/05/2014 10:34 μμ, Yuyang Du wrote:
> >>> On Mon, May 12, 2014 at 11:30:03PM +0300, Stratos Karafotis wrote:
> >>>> On 09/05/2014 05:56 μμ, Stratos Karafotis wrote:
> >>>>
> >>>> Next performance state = min_perf + (max_perf - min_perf) * load / 100
> >>>>
> >>> Hi,
> >>>
> >>> This formula is fundamentally broken. You need to associate the load with its
> >>> frequency.
> >>
> >> Could you please explain why is it broken? I think the load should be
> >> independent from the current frequency.
> > 
> > Why independent? The load not (somewhat) determined by that?
> > 
> > 
> 
> Maybe, in some cases yes. But not always.
> For example, please consider a CPU running a tight "for" loop in 100MHz
> for a couple of seconds. This produces a load of 100%.
> It will produce the same load (100%) in any other frequency.
 
Still fundamentally wrong, because you are not making a fair
comparison ("load" in 100MHz vs. any other freq).

Yuyang