Re: [PATCH v2 08/11] sched: get CPU's activity statistic

[Morten Rasmussen]

On Wed, Jun 04, 2014 at 10:35:15AM +0100, Vincent Guittot wrote:
> On 4 June 2014 11:23, Peter Zijlstra <peterz@xxxxxxxxxxxxx> wrote:
> > On Wed, Jun 04, 2014 at 09:55:42AM +0100, Morten Rasmussen wrote:
> >> Both running_avg and runnable_avg are affected by other tasks on the
> >> same cpus, but in different ways. They are equal if you only have one
> >> task on a cpu. If you have more, running_avg will give you the true
> >> requirement of the tasks until the cpu is fully utilized. At which point
> >> the task running_avg will drop if you add more tasks (the unweighted sum
> >> of task running_avgs remains constant).
> >>
> >> runnable_avg on the other hand, might be affected as soon as you have
> >> two task running on the same cpu if they are runnable at the same time.
> >> That isn't necessarily a bad thing for load-balancing purposes, because
> >> tasks that are runnable at the same time are likely to be run more
> >> efficiently by placing them on different cpus. You might view as at sort
> >> of built in concurrency factor, somewhat similar to what Yuyang is
> >> proposing. runnable_avg increases rapidly when the cpu is over-utilized.
> >
> > Agreed.
> >
> >> > I'm not sure I see how 100% is possible, but yes I agree that runnable
> >> > can indeed be inflated due to this queueing effect.
> >>
> >> You should only be able to get to 75% worst case for runnable_avg for
> >> that example. The total running_avg is 50% no matter if the tasks
> >> overlaps or not.
> >
> > Yes, 75% is what I ended up with.
> 
> Can you explain how you reach 75% as it depends on the runtime and a
> runtime longer than 345ms will end to a 100% load whatever the
> idletime was previously ?

If the busy period of each task is long enough that the first one to run
runs to completetion before the other task is scheduled you get 25% and
50%. But as I said in my other reply, you can get higher task runnable
if the tasks busy period is long enough that you switch between them
before the first one completes.
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