Messages in this thread | | | Date | Tue, 30 Sep 2014 20:41:29 +0200 | From | Peter Zijlstra <> | Subject | Re: [PATCH] sched: fix spurious active migration |
| |
On Tue, Sep 30, 2014 at 10:41:08AM +0200, Vincent Guittot wrote: > diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c > index 2a1e6ac..adad532 100644 > --- a/kernel/sched/fair.c > +++ b/kernel/sched/fair.c > @@ -6425,13 +6425,14 @@ static struct sched_group *find_busiest_group(struct lb_env *env) > > if (env->idle == CPU_IDLE) { > /* > - * This cpu is idle. If the busiest group load doesn't > - * have more tasks than the number of available cpu's and > - * there is no imbalance between this and busiest group > - * wrt to idle cpu's, it is balanced. > + * This cpu is idle. If the busiest group is not overloaded > + * and there is no imbalance between this and busiest group > + * wrt to idle cpus, it is balanced. The imbalance becomes > + * significant if the diff is greater than 1 otherwise we > + * might end up to just move the imbalance on another group > */ > - if ((local->idle_cpus < busiest->idle_cpus) && > - busiest->sum_nr_running <= busiest->group_weight) > + if ((local->idle_cpus <= (busiest->idle_cpus + 1)) &&
So I'm thick and I don't get this one.. In fact I don't seem to understand the existing code either.
If we're idle, and busiest is overloaded, we want to have tasks. Why would we care about number of idle cpus etc..
> + !(busiest->group_type == group_overloaded))
Would not: busiest->group_type != group_overloaded, read more natural? Also, would it make sense to make this the first condition?
> goto out_balanced; > } else {
| |