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Date:   Fri, 31 Mar 2017 03:13:55 +0800
From:   Yuyang Du <yuyang.du@...el.com>
To:     Peter Zijlstra <peterz@...radead.org>
Cc:     Paul Turner <pjt@...gle.com>, Ingo Molnar <mingo@...nel.org>,
        LKML <linux-kernel@...r.kernel.org>,
        Benjamin Segall <bsegall@...gle.com>,
        Morten Rasmussen <morten.rasmussen@....com>,
        Vincent Guittot <vincent.guittot@...aro.org>,
        Dietmar Eggemann <dietmar.eggemann@....com>,
        Matt Fleming <matt@...eblueprint.co.uk>,
        umgwanakikbuti@...il.com
Subject: Re: [RESEND PATCH 2/2] sched/fair: Optimize __update_sched_avg()

On Thu, Mar 30, 2017 at 04:14:28PM +0200, Peter Zijlstra wrote:
> On Thu, Mar 30, 2017 at 02:16:58PM +0200, Peter Zijlstra wrote:
> > On Thu, Mar 30, 2017 at 04:21:08AM -0700, Paul Turner wrote:
> 
> > > > +
> > > > +       if (unlikely(periods >= LOAD_AVG_MAX_N))
> > > >                 return LOAD_AVG_MAX;
> 
> > > 
> > > Is this correct in the iterated periods > LOAD_AVG_MAX_N case?
> > > I don't think the decay above is guaranteed to return these to zero.
> > 
> > Ah!
> > 
> > Indeed, so decay_load() needs LOAD_AVG_PERIOD * 63 before it truncates
> > to 0, because every LOAD_AVG_PERIOD we half the value; loose 1 bit; so
> > 63 of those and we're 0.
> > 
> > But __accumulate_sum() OTOH returns LOAD_AVG_MAX after only
> > LOAD_AVG_MAX_N, which < LOAD_AVG_PERIOD * 63.
> > 
> > So yes, combined we exceed LOAD_AVG_MAX, which is bad. Let me think what
> > to do about that.
> 
> 
> So at the very least it should be decay_load(LOAD_AVG_MAX, 1) (aka
> LOAD_AVG_MAX - 1024), but that still doesn't account for the !0
> decay_load() of the first segment.
> 
> I'm thinking that we can compute the middle segment, by taking the max
> value and chopping off the ends, like:
> 
> 
>              p
>  c2 = 1024 \Sum y^n
>             n=1
> 
>               inf        inf
>     = 1024 ( \Sum y^n - \Sum y^n - y^0 )
>               n=0        n=p
 
It looks surprisingly kinda works :)
 
> +	c2 = LOAD_AVG_MAX - decay_load(LOAD_AVG_MAX, periods) - 1024;
                            ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
But, I'm not sure               this is what you want (just assume p==0).

Thanks,
Yuyang

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