Re: [RFC PATCH 08/14] sched: normalize tg load contributions against runnable time

[Peter Zijlstra <a.p.zijlstra@chello.nl>]

On Wed, 2012-02-01 at 17:38 -0800, Paul Turner wrote:
> Entities of equal weight should receive equitable distribution of cpu time.
> This is challenging in the case of a task_group's shares as execution may be
> occurring on multiple cpus simultaneously.
> 
> To handle this we divide up the shares into weights proportionate with the load
> on each cfs_rq.  This does not however, account for the fact that the sum of
> the parts may be less than one cpu and so we need to normalize:
>   load(tg) = min(runnable_avg(tg), 1) * tg->shares
> Where runnable_avg is the aggregate time in which the task_group had runnable
> children.


>  static inline void __update_group_entity_contrib(struct sched_entity *se)
>  {
>         struct cfs_rq *cfs_rq = group_cfs_rq(se);
>         struct task_group *tg = cfs_rq->tg;
> +       int runnable_avg;
>  
>         se->avg.load_avg_contrib = (cfs_rq->tg_load_contrib * tg->shares);
>         se->avg.load_avg_contrib /= atomic64_read(&tg->load_avg) + 1;
> +
> +       /*
> +        * Unlike a task-entity, a group entity may be using >=1 cpu globally.
> +        * However, in the case that it's using <1 cpu we need to form a
> +        * correction term so that we contribute the same load as a task of
> +        * equal weight. (Global runnable time is taken as a fraction over 2^12.)
> +        */
> +       runnable_avg = atomic_read(&tg->runnable_avg);
> +       if (runnable_avg < (1<<12)) {
> +               se->avg.load_avg_contrib *= runnable_avg;
> +               se->avg.load_avg_contrib /= (1<<12);
> +       }
>  } 

This seems weird, and the comments don't explain anything.

Ah,.. you can count runnable multiple times (on each cpu), this also
means that the number you're using (when below 1) can still be utter
crap.

Neither the comment nor the changelog mention this, it should, it should
also mention why it doesn't matter (does it?).