sched/rt: Improve pick_next_highest_task_rt()

Avoid extra work by continuing on to the next rt_rq if the highest
prio task in current rt_rq is the same priority as our candidate
task.

More detailed explanation:  if next is not NULL, then we have found a
candidate task, and its priority is next->prio.  Now we are looking
for an even higher priority task in the other rt_rq's.  idx is the
highest priority in the current candidate rt_rq.  In the current 3.3
code, if idx is equal to next->prio, we would start scanning the tasks
in that rt_rq and replace the current candidate task with a task from
that rt_rq.  But the new task would only have a priority that is equal
to our previous candidate task, so we have not advanced our goal of
finding a higher prio task.  So we should avoid the extra work by
continuing on to the next rt_rq if idx is equal to next->prio.

Signed-off-by: Michael J Wang <mjwang@broadcom.com>
Acked-by: Steven Rostedt <rostedt@goodmis.org>
Reviewed-by: Yong Zhang <yong.zhang0@gmail.com>
Signed-off-by: Peter Zijlstra <a.p.zijlstra@chello.nl>
Link: http://lkml.kernel.org/r/2EF88150C0EF2C43A218742ED384C1BC0FC83D6B@IRVEXCHMB08.corp.ad.broadcom.com
Signed-off-by: Ingo Molnar <mingo@kernel.org>

1 files changed, 1 insertions, 1 deletions

diff --git a/kernel/sched/rt.c b/kernel/sched/rt.c
index b60dad720173..44af55e6d5d0 100644
--- a/kernel/sched/rt.c
+++ b/kernel/sched/rt.c
@@ -1428,7 +1428,7 @@ static struct task_struct *pick_next_highest_task_rt(struct rq *rq, int cpu)
 next_idx:
 		if (idx >= MAX_RT_PRIO)
 			continue;
-		if (next && next->prio < idx)
+		if (next && next->prio <= idx)
 			continue;
 		list_for_each_entry(rt_se, array->queue + idx, run_list) {
 			struct task_struct *p;