Re: [PATCH v2] sched/fair: update scale invariance of PELT

[Vincent Guittot <vincent.guittot@linaro.org>]

Le Tuesday 11 Apr 2017 à 10:53:05 (+0200), Peter Zijlstra a écrit :
> On Tue, Apr 11, 2017 at 09:52:21AM +0200, Vincent Guittot wrote:
> > Le Monday 10 Apr 2017 à 19:38:02 (+0200), Peter Zijlstra a écrit :
> > > 
> > > Thanks for the rebase.
> > > 
> > > On Mon, Apr 10, 2017 at 11:18:29AM +0200, Vincent Guittot wrote:
> > > 
> > > Ok, so let me try and paraphrase what this patch does.
> > > 
> > > So consider a task that runs 16 out of our 32ms window:
> > > 
> > >    running   idle
> > >   |---------|---------|
> > > 
> > > 
> > > You're saying that when we scale running with the frequency, suppose we
> > > were at 50% freq, we'll end up with:
> > > 
> > >    run  idle
> > >   |----|---------|
> > > 
> > > 
> > > Which is obviously a shorter total then before; so what you do is add
> > > back the lost idle time like:
> > > 
> > >    run  lost idle
> > >   |----|----|---------|
> > > 
> > > 
> > > to arrive at the same total time. Which seems to make sense.
> > 
> > Yes
> 
> OK, bear with me.
> 
> 
> So we have:
> 
> 
>   util_sum' = utilsum * y^p +
> 
>                                  p-1
>               d1 * y^p + 1024 * \Sum y^n + d3 * y^0
> 	                         n=1
> 
> For the unscaled version, right?

Yes for the running state.

For sleeping state, it's just util_sum' = utilsum * y^p

>
>
> 
> Now for the scaled version, instead of adding a full 'd1,d2,d3' running
> segments, we want to add partially running segments, where r=f*d/f_max,
> and lost segments l=d-r to fill out the idle time.
> 
> But afaict we then end up with (F=f/f_max):
> 
> 
>   util_sum' = utilsum * y^p +
> 
>                                          p-1
>               F * d1 * y^p + F * 1024 * \Sum y^n + F * d3 * y^0
> 	                                 n=1

you also have a longer running time as it runs slower. We make the assumption that
d1+d2+d3 = (d1'+d2'+d3') * F

If we consider that we cross a decay window, we still have the d1 to complete
the past one but then p'*F= p and d'3 will be the remaining part of the
current window and most probably not equal to d3

so we have with current invariance:

   util_sum' = utilsum * y^(p/F) + 
                                              (p/F - 1)
               F * d1 * y^(p/F) + F * 1024 * \Sum y^n + F * d'3 * y^0
 	                                          n=1

with the new invariance we have

   util_sum' = utilsum * y^(F*p/F) + 
                                        (F*p/F - 1)
               d1 * y^(F*p/F) + 1024 * \Sum y^n + d3 * y^0
 	                                    n=1

For a sleeping time of d at max capacity, we have
a sleeping time d'=d-l, with l the lost time of the previous running time
With the current implementation: 
  util_sum' = utilsum * y^(p')
  util_sum' = utilsum * y^(p-l)

With the new invaraince, we have 
  util_sum' = utilsum * y^(p'+l)
  util_sum' = utilsum * y^(p-l+l)

> 
> And we can collect the common term F:
> 
>   util_sum' = utilsum * y^p +
> 
>                                       p-1
>               F * (d1 * y^p + 1024 * \Sum y^n + d3 * y^0)
> 	                              n=1
> 
> 
> Which is exactly what we already did.

In the new invariance scale, the F is applied on p not on the contribution
value

>
> 
> So now I'm confused. Where did I go wrong?
> 
> 
> Because by scaling the contribution we get the exact result of doing the
> smaller 'running' + 'lost' segments.