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

From: Juri Lelli
Date: Wed Mar 09 2016 - 23:30:46 EST


On 10/03/16 00:41, Rafael J. Wysocki wrote:
> On Wed, Mar 9, 2016 at 11:15 AM, Juri Lelli <juri.lelli@xxxxxxx> wrote:
> > Hi,
> >
> > 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?
>

The capacity at current frequency.

> > 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
> > invariance.
> >
> > 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 =
> 1.25.
>

Right. So, when using freq. invariant util we have:

next_freq = C * curr_freq * util / curr_cap

as

util_raw = util * max / curr_cap

What Vincent is saying makes sense, though. If we use
arch_scale_freq_capacity() as denominator instead of max, we can use a
single formula for both cases.

Best,

- Juri