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    SubjectRe: [PATCH v6 04/11] cpufreq/schedutil: use rt utilization tracking
    On Fri, Jun 22, 2018 at 03:54:24PM +0200, Vincent Guittot wrote:
    > On Fri, 22 Jun 2018 at 15:26, Peter Zijlstra <peterz@infradead.org> 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.

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