Re: [RFC PATCH v1 8/8] sched/deadline: make bandwidth enforcement scale-invariant

Peter Zijlstra
2017-07-25T13:51:26Z

On Tue, Jul 25, 2017 at 09:03:08AM +0200, Luca Abeni wrote:

> > I'm still confused..
> >
> > So GRUB does:
> >
> > dq = Uact -dt
> >
> > right?
>
> Right. This is what the original (single processor) GRUB did. And this
> was used by the "GRUB-PA" algorithm:
> https://www.researchgate.net/profile/Giuseppe_Lipari/publication/220800940_Using_resource_reservation_techniques_for_power-aware_scheduling/links/09e41513639b2703fc000000.pdf
>
> (basically, GRUB-PA uses GRUB for reclaiming, and scales the CPU
> frequency based on Uact)
>
>
> > Now, you do DVFS using that same Uact. If we lower the clock, we need
> > more time, so would we then not end up with something like:
> >
> > dq = 1/Uact -dt
>
> Well, in the GRUB-PA algorithm GRUB reclaiming is the mechanism used to
> give more runtime to the task... Since Uact is < 1, doing
> dq = - Uact * dt
> means that we decrease the current runtime by a smaller amount of time.
> And so we end up giving more runtime to the task: instead of giving
> dl_runtime every dl_period, we give "dl_runtime / Uact" every
> dl_period... And since the CPU is slower (by a ratio Uact), this is
> equivalent to giving dl_runtime at the maximum CPU speed / frequency
> (at least, in theory :).
>
>
> > After all; our budget assignment is such that we're able to complete
> > our work at max freq. Therefore, when we lower the frequency, we'll have
> > to increase budget pro rata, otherwise we'll not complete our work and
> > badness happens.
>
> Right. But instead of increasing dl_runtime, GRUB-PA decreases the
> amount of time accounted to the current runtime.
>
>
> > Say we have a 1 Ghz part and Uact=0.5 we'd select 500 Mhz and need
> > double the time to complete.
> >
> > Now, if we fold these two together, you'd get:
> >
> > dq = Uact/Uact -dt = -dt
>
> Not sure why " / Uact"... According to the GRUB-PA algorithm, you just
> do
> dq = - Uact * dt = -0.5dt
> and you end up giving the CPU to the task for 2 * dl_runtime every
> dl_period (as expected)

Yeah, I seem to have gone off the rails there... Bah I'm terminally
confused now. Let me try and get my brain the right way up.