linux-kernel - Re: [PATCH v4 1/2] sched: make __update_cpu

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Message-ID: <20151022102852.GC28741@byungchulpark-X58A-UD3R>
Date:	Thu, 22 Oct 2015 19:28:52 +0900
From:	Byungchul Park <byungchul.park@....com>
To:	Peter Zijlstra <peterz@...radead.org>
Cc:	mingo@...nel.org, linux-kernel@...r.kernel.org, fweisbec@...il.com,
	tglx@...utronix.de
Subject: Re: [PATCH v4 1/2] sched: make __update_cpu_load() handle active
 tickless case

On Tue, Oct 20, 2015 at 11:48:48AM +0200, Peter Zijlstra wrote:
> On Tue, Oct 20, 2015 at 09:49:32AM +0900, Byungchul Park wrote:
> > > >             = A^i * cpu_load(n-i) + (A^(i-1) + A^(i-2) + ... + 1) * B
> > > >             , where i = pending_updates - 1
> > > 
> > > You missed an opportunity here, if you take i==n you avoid the need for
> > > i entirely.
> > 
> > i don't think so. as i said, _n_ is the current tick -1 and _i_ is
> > pending_updates - 1. we cannot take i == n, but should keep (n-i).
> 
> Just think relative; it doesn't matter when in time we do this.
> So 0 to n is identical to any other interval.

You are right. I'm always tired these days. I was confused.
Sorry for that.

> 
> > > >             = A^i * cpu_load(n-i) + B * (A^i - 1) / (A - 1)
> > > >             , by geometric series formula for sum
> > > 
> > > That's wrong; the limited geometric series expands to:
> > 
> > NO, that's not wrong. it doesn't matter at all.
> > 
> > a * (1 - r^n) / (1 - r)
> > = a * (-1)(r^n - 1) / (-1)(r - 1)
> > = a * (r^n - 1) / (r - 1)
> > 
> > i mean these two are exactly same.
> 
> Ah indeed! Sorry for that. I clearly didn't think beyond the series
> expansion I found.
> 
> > > I've rewritten the things a little; does this look good to you?
> > 
> > however, your expressions and descriptions below look better than me,
> > except some logical errors. could you keep my logical flow unchagned?
> 
> > > + *   load[i]_n = (1 - 1/2^i) * load[i]_n-1 + (1/2^i) * load_n-1
> > > + *             = A * load[i]_n-1 + B ; A := (1 - 1/2^i), B := (1/2^i) * load
> > > + *             = A * (A * load[i]_n-2 + B) + B
> > > + *             = A * (A * (A * load[i]_n-3 + B) + B) + B
> > > + *             = A^3 * load[i]_n-3 + (A^2 + A + 1) * B
> > > + *             = A^n * load[i]_0 + (A^(n-1) + A^(n-2) + ... + 1) * B
> > > + *             = A^n * load[i]_0 + ((1 - A^n) / (1 - A)) * B
> > > + *             = (1 - 1/2^i)^n * (load[i]_0 - load) + load
> 
> That is the same logic, right?

Yes, absolutely same logic.

Thanks a lot,
byungchul

> 
> Please be more specific as to what you'd like restored.
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