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>

diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
index f918635efe092073eb698d9a662353438cbbd5d2..8aa217f62a9e5443b1221788d64473f3eb6a26c6 100644 (file)
--- a/kernel/sched/fair.c
+++ b/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;
                }
        }