Messages in this thread | | | Date | Fri, 22 Jun 2018 16:48:40 +0200 | From | Peter Zijlstra <> | Subject | Re: [PATCH v6 04/11] cpufreq/schedutil: use rt utilization tracking |
| |
On Fri, Jun 22, 2018 at 04:11:59PM +0200, Peter Zijlstra wrote: > On Fri, Jun 22, 2018 at 03:54:24PM +0200, Vincent Guittot wrote: > > On Fri, 22 Jun 2018 at 15:26, Peter Zijlstra <peterz@infradead.org> wrote: > > > > define f (u,r,n) { return u + ((u/(1-r)) - u) * (u/(1-r))^n; }
> > I'm a bit lost with your example. > > u = 0.2 (for cfs) and r=0.7 (let say for rt) in your example and idle is 0.1 > > > > For rt task, we run 0.7 of the time at f=1 then we will select f=0.4 > > for run cfs task with u=0.2 but u is the utilization at f=1 which > > means that it will take 250% of normal time to execute at f=0.4 which > > means 0.5 time instead of 0.2 at f=1 so we are going out of time. In > > order to have enough time to run r and u we must run at least f=0.666 > > for cfs = 0.2/(1-0.7). > > Argh.. that is n=0. So clearly I went off the rails somewhere.
Aah, I think the number I've been computing is a 'corrected' u. Not an f. It made sure that 0 idle got u=1, but no more.
| |