On Fri, Oct 18, 2019 at 09:44:44AM +0200, Dietmar Eggemann wrote: > On 17/10/2019 16:11, Peter Zijlstra wrote: > > On Thu, Oct 17, 2019 at 12:11:16PM +0100, Quentin Perret wrote: > > [...] > > > It only boosts when 'rq->cfs.avg.util' increases while > > 'rq->cfs.avg.util_est.enqueued' remains unchanged (and util > util_est > > obv). > > > > This condition can be true for select_task_rq_fair(), because that is > > ran before we do enqueue_task_fair() (for obvious raisins). > > > >>> I'm still thinking about the exact means you're using to raise C; that > >>> is, the 'util - util_est' as cost_margin. It hurts my brain still. > >> > >> +1 ... > > > > cost_i = capacity_i / power_i ; for the i-th OPP > > I get confused by this definition. efficiency=capacity/power but the > cs->cost value used in em_pd_get_higher_freq() is defined as > > cs_cost = cs->power * cpu_max_freq / cs->freq [energy_model.h]
Well, chalk that up to confusion inspired by the Changelog of patch 1. Let me redo it with that formula then.
