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sched: Remove one division operation in find_busiest_queue()

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Remove one division operation in find_busiest_queue() by using
crosswise multiplication:

	wl_i / power_i > wl_j / power_j :=
	wl_i * power_j > wl_j * power_i

Signed-off-by: Joonsoo Kim <iamjoonsoo.kim@lge.com>
[ Expanded the changelog. ]
Signed-off-by: Peter Zijlstra <peterz@infradead.org>
Link: http://lkml.kernel.org/r/1375778203-31343-2-git-send-email-iamjoonsoo.kim@lge.com
Signed-off-by: Ingo Molnar <mingo@kernel.org>
This commit is contained in:
Joonsoo Kim 2013-08-06 17:36:41 +09:00 committed by Ingo Molnar
parent a4f61cc03e
commit 95a79b805b
1 changed files with 9 additions and 5 deletions
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14
kernel/sched/fair.c
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@ -4968,7 +4968,7 @@ static struct rq *find_busiest_queue(struct lb_env *env,
struct sched_group *group)
{
struct rq *busiest = NULL, *rq;
unsigned long max_load = 0;
unsigned long busiest_load = 0, busiest_power = 1;
int i;
for_each_cpu(i, sched_group_cpus(group)) {
@ -4998,11 +4998,15 @@ static struct rq *find_busiest_queue(struct lb_env *env,
* the weighted_cpuload() scaled with the cpu power, so that
* the load can be moved away from the cpu that is potentially
* running at a lower capacity.
*
* Thus we're looking for max(wl_i / power_i), crosswise
* multiplication to rid ourselves of the division works out
* to: wl_i * power_j > wl_j * power_i; where j is our
* previous maximum.
*/
wl = (wl * SCHED_POWER_SCALE) / power;
if (wl > max_load) {
max_load = wl;
if (wl * busiest_power > busiest_load * power) {
busiest_load = wl;
busiest_power = power;
busiest = rq;
}
}