sched: Remove one division operation in find_busiest_queue()

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>

kernel/sched/fair.c

@@ -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;
 		}
 	}