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

On Tue, Mar 08, 2016 at 09:05:50PM +0100, Rafael J. Wysocki wrote:

quoted

quoted

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
non-linear.

quoted

quoted

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
	  = 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
non-invariant).

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.