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>
struct sched_group *group)
{
struct rq *busiest = NULL, *rq;
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)) {
int i;
for_each_cpu(i, sched_group_cpus(group)) {
* 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.
* 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;