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.
