Re: [PATCH 6/6] cpufreq: schedutil: New governor based on scheduler utilization data
From: Rafael J. Wysocki
Date: Wed Mar 09 2016 - 18:42:01 EST
On Wed, Mar 9, 2016 at 11:15 AM, Juri Lelli <juri.lelli@xxxxxxx> wrote:
> sorry if I didn't reply yet. Trying to cope with jetlag and
> talks/meetings these days :-). Let me see if I'm getting what you are
> discussing, though.
> On 08/03/16 21:05, Rafael J. Wysocki wrote:
>> On Tue, Mar 8, 2016 at 8:26 PM, Peter Zijlstra <peterz@xxxxxxxxxxxxx> wrote:
>> > On Tue, Mar 08, 2016 at 07:00:57PM +0100, Rafael J. Wysocki wrote:
>> >> On Tue, Mar 8, 2016 at 12:27 PM, Peter Zijlstra <peterz@xxxxxxxxxxxxx> wrote:
>> a = max_freq gives next_freq = max_freq for x = 1, but with that
>> choice of a you may never get to x = 1 with frequency invariant
>> because of the feedback effect mentioned above, so the 1/n produces
>> the extra boost needed for that (n is a positive integer).
>> Quite frankly, to me it looks like linear really is a better
>> approximation for "raw" utilization. That is, for frequency invariant
>> x we should take:
>> next_freq = a * x * max_freq / current_freq
>> (and if x is not frequency invariant, the right-hand side becomes a *
>> x). Then, the extra boost needed to get to x = 1 for frequency
>> invariant is produced by the (max_freq / current_freq) factor that is
>> greater than 1 as long as we are not running at max_freq and a can be
>> chosen as max_freq.
> Expanding terms again, your original formula (without the 1.1 factor of
> the last version) was:
> next_freq = util / max_cap * max_freq
> and this doesn't work when we have freq invariance since util won't go
> over curr_cap.
Can you please remind me what curr_cap is?
> What you propose above is to add another factor, so that we have:
> next_freq = util / max_cap * max_freq / curr_freq * max_freq
> which should give us the opportunity to reach max_freq also with freq
> This should actually be the same of doing:
> next_freq = util / max_cap * max_cap / curr_cap * max_freq
> We are basically scaling how much the cpu is busy at curr_cap back to
> the 0..1024 scale. And we use this to select next_freq. Also, we can
> simplify this to:
> next_freq = util / curr_cap * max_freq
> and we save some ops.
> However, if that is correct, I think we might have a problem, as we are
> skewing OPP selection towards higher frequencies. Let's suppose we have
> a platform with 3 OPPs:
> freq cap
> 1200 1024
> 900 768
> 600 512
> As soon a task reaches an utilization of 257 we will be selecting the
> second OPP as
> next_freq = 257 / 512 * 1200 ~ 602
> While the cpu is only 50% busy in this case. And we will go at max OPP
> when reaching ~492 (~64% of 768).
> That said, I guess this might work as a first solution, but we will
> probably need something better in the future. I understand Rafael's
> concerns regardin margins, but it seems to me that some kind of
> additional parameter will be probably needed anyway to fix this.
> Just to say again how we handle this in schedfreq, with a -20% margin
> applied to the lowest OPP we will get to the next one when utilization
> reaches ~410 (80% busy at curr OPP), and so on for the subsequent ones,
> which is less aggressive and might be better IMHO.
Well, Peter says that my idea is incorrect, so I'll go for
next_freq = C * current_freq * util_raw / max
where C > 1 (and likely C < 1.5) instead.
That means C has to be determined somehow or guessed. The 80% tipping
point condition seems reasonable to me, though, which leads to C =