`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=pIt 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`