Re: [PATCH v6 04/11] cpufreq/schedutil: use rt utilization tracking

[Peter Zijlstra]

On Fri, Jun 22, 2018 at 03:54:24PM +0200, Vincent Guittot wrote:
> On Fri, 22 Jun 2018 at 15:26, Peter Zijlstra <peterz@xxxxxxxxxxxxx> wrote:

> > $ bc -l
> > define f (u,r,n) { return u + ((u/(1-r)) - u) * (u/(1-r))^n; }
> > f(.2,.7,0)
> > .66666666666666666666
> > f(.2,.7,2)
> > .40740740740740740739
> > f(.2,.7,4)
> > .29218106995884773661
> >
> > So at 10% idle time, we've only inflated what should be 20% to 40%, that
> > is entirely reasonable I think. The linear case gave us 66%.  But feel
> > free to increase @n if you feel that helps, 4 is only one mult more than
> > 2 and gets us down to 29%.
> 
> I'm a bit lost with your example.
> u = 0.2 (for cfs) and r=0.7 (let say for rt) in your example and idle is 0.1
> 
> For rt task, we run 0.7 of the time at f=1 then we will select f=0.4
> for run cfs task with u=0.2 but u is the utilization at f=1 which
> means that it will take 250% of normal time to execute at f=0.4 which
> means 0.5  time instead of 0.2 at f=1 so we are going out of time. In
> order to have enough time to run r and u we must run at least  f=0.666
> for cfs = 0.2/(1-0.7). 

Argh.. that is n=0. So clearly I went off the rails somewhere.