Re: [PATCH 6/6] cpufreq: schedutil: New governor based on scheduler utilization data
From: Peter Zijlstra
Date: Wed Mar 09 2016 - 11:39:46 EST
On Tue, Mar 08, 2016 at 09:05:50PM +0100, Rafael J. Wysocki wrote:
> >> This means that on platforms where the utilization is frequency
> >> invariant we should use
> >> next_freq = a * x
> >> (where x is given by (2) above) and for platforms where the
> >> utilization is not frequency invariant
> >> next_freq = a * x * current_freq / max_freq
> >> and all boils down to finding a.
> > Right.
> However, that doesn't seem to be in agreement with the Steve's results
> posted earlier in this thread.
I could not make anything of those numbers.
> Also theoretically, with frequency invariant, the only way you can get
> to 100% utilization is by running at the max frequency, so the closer
> to 100% you get, the faster you need to run to get any further. That
> indicates nonlinear to me.
I'm not seeing that, you get that by using a > 1. No need for
> >> Now, it seems reasonable for a to be something like (1 + 1/n) *
> >> max_freq, so for non-frequency invariant we get
> >> nex_freq = (1 + 1/n) * current_freq * x
> > This seems like a big leap; where does:
> > (1 + 1/n) * max_freq
> > come from? And what is 'n'?
> a = max_freq gives next_freq = max_freq for x = 1,
next_freq = a * x * current_freq / max_freq
[ a := max_freq, x := 1 ] ->
= max_freq * 1 * current_freq / max_freq
But I think I see what you're saying; because at x = 1,
current_frequency must be max_frequency. Per your earlier point.
> 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).
OK, so that gets us:
a = (1 + 1/n) ; n > 0
[ I would not have chosen (1 + 1/n), but lets stick to that ]
So for n = 4 that gets you: a = 1.25, which effectively gets you an 80%
utilization tipping point. That is, 1.25 * .8 = 1, iow. you'll pick the
next frequency (assuming RELATION_L like selection).
Together this gets you:
next_freq = (1 + 1/n) * max_freq * x * current_freq / max_freq
= (1 + 1/n) * x * current_freq
Again, with n = 4, x > .8 will result in a next_freq > current_freq, and
hence (RELATION_L) pick a higher one.
> 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
(its very confusing how you use 'x' for both invariant and
That doesn't make sense, remember:
util = \Sum_i u_i * freq_i / max_freq (1)
Which for systems where freq_i is constant reduces to:
util = util_raw * current_freq / max_freq (2)
But you cannot reverse this. IOW you cannot try and divide out
current_freq on a frequency invariant metric.
So going by:
next_freq = (1 + 1/n) * max_freq * util (3)
if we substitute (2) into (3) we get:
= (1 + 1/n) * max_freq * util_raw * current_freq / max_freq
= (1 + 1/n) * current_freq * util_raw (4)
Which gets you two formula with the same general behaviour. As (2) is
the only approximation of (1) we can make.