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 :)