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sched/rt: Do not throttle when PI boosting
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When a runqueue has rt_runtime_us = 0 then the only way it can
accumulate rt_time is via PI boosting. That causes the runqueue
to be throttled and replenishing does not change anything due to
rt_runtime_us = 0. So avoid that situation by clearing rt_time and
skip the throttling alltogether.

Signed-off-by: Peter Zijlstra <a.p.zijlstra@chello.nl>
[ Changelog ]
Signed-off-by: Thomas Gleixner <tglx@linutronix.de>
Link: http://lkml.kernel.org/n/tip-7x70cypsotjb4jvcor3edctk@git.kernel.org
Signed-off-by: Ingo Molnar <mingo@elte.hu>
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Peter Zijlstra authored and Ingo Molnar committed Mar 1, 2012
1 parent 42c62a5 commit 7abc63b
Showing 1 changed file with 18 additions and 2 deletions.
20 changes: 18 additions & 2 deletions kernel/sched/rt.c
Original file line number Diff line number Diff line change
Expand Up @@ -857,8 +857,24 @@ static int sched_rt_runtime_exceeded(struct rt_rq *rt_rq)
return 0;

if (rt_rq->rt_time > runtime) {
rt_rq->rt_throttled = 1;
printk_once(KERN_WARNING "sched: RT throttling activated\n");
struct rt_bandwidth *rt_b = sched_rt_bandwidth(rt_rq);

/*
* Don't actually throttle groups that have no runtime assigned
* but accrue some time due to boosting.
*/
if (likely(rt_b->rt_runtime)) {
rt_rq->rt_throttled = 1;
printk_once(KERN_WARNING "sched: RT throttling activated\n");
} else {
/*
* In case we did anyway, make it go away,
* replenishment is a joke, since it will replenish us
* with exactly 0 ns.
*/
rt_rq->rt_time = 0;
}

if (rt_rq_throttled(rt_rq)) {
sched_rt_rq_dequeue(rt_rq);
return 1;
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