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Message-ID: <20170330194118.GE16440@ydu19desktop>
Date: Fri, 31 Mar 2017 03:41:18 +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 Fri, Mar 31, 2017 at 03:13:55AM +0800, Yuyang Du wrote:
> 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).
>
Oh, what I meant is when p != 0, actually p>=1.
And thinking about it for a while, it's really what you want, brilliant :)
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