[PATCH 1/6] sched: Remove one division operation in find_busiest_queue()

[Peter Zijlstra <peterz@infradead.org>]

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From: Joonsoo Kim <iamjoonsoo.kim@lge.com>

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>
[peterz: expanded changelog]
Signed-off-by: Peter Zijlstra <peterz@infradead.org>
---
 kernel/sched/fair.c |    9 ++++-----
 1 file changed, 4 insertions(+), 5 deletions(-)

--- a/kernel/sched/fair.c
+++ b/kernel/sched/fair.c
@@ -5018,7 +5018,7 @@ static struct rq *find_busiest_queue(str
 				     struct sched_group *group)
 {
 	struct rq *busiest = NULL, *rq;
-	unsigned long max_load = 0;
+	unsigned long busiest_load = 0, busiest_power = SCHED_POWER_SCALE;
 	int i;
 
 	for_each_cpu(i, sched_group_cpus(group)) {
@@ -5049,10 +5049,9 @@ static struct rq *find_busiest_queue(str
 		 * the load can be moved away from the cpu that is potentially
 		 * running at a lower capacity.
 		 */
-		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;
 		}
 	}