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 {