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Subject: Re: [PATCH] sched: Avoid side-effect of tickless idle on 
 update_cpu_load (v2)
From: Peter Zijlstra <peterz@infradead.org>
To: Venkatesh Pallipadi <venki@google.com>
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Date: Mon, 17 May 2010 19:35:17 +0200
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On Mon, 2010-05-17 at 09:52 -0700, Venkatesh Pallipadi wrote:

> >  load_i = ((2^i)-1)/(2^i) * load_i + 1/(2^i) * load_(i-1)

> > So because we're in no_hz, current load == 0 and we could approximate
> > the thing by:
> >
> >  load_i = ((2^i)-1)/(2^i) * load_i
> >
> > Because for i ~ 1, there is no new input, and for i >> 1 the fraction is
> > small.
> 
> Something like that. But, with total_updates = n and missed_updates = n - 1
> We do this for (n - 1)
> load_i = ((2^i)-1)/(2^i) * load_i
> And do this once.
> load_i = ((2^i)-1)/(2^i) * load_i + 1/(2^i) * cur_load

> That way we do not differentiate between whether we are in tickless or
> not and we use the same code path.

But by the above, that's not the same as without, because that does

load_i = ((2^i)-1)/(2^i) * load_i + 1/(2^i) * load_(i-1)

not

load_i = ((2^i)-1)/(2^i) * load_i + 1/(2^i) * cur_load

> > But why then do we precalculate these factors? It seems to me
> > ((2^i)-1)/(2^i) is something that is trivial to compute and doesn't
> > warrant a lookup table?
> >
> 
> Yes. Initially I had a for loop running for missed_updates to calculate
>  ((2^i)-1)/(2^i) * load_i
> in a loop.

Ah, right! So you want to calculate:

 (((2^i)-1)/(2^i))^n

Which ends up being a nasty binomial sum: 1/(2^ni) * \Sum_k^n (n choose
k) * 2^k, so yeah, I don't see a fancy way to quickly compute that.


OK, could you summarize our discussion into that comment?