sched/fair: Do strict inequality check for busiest misfit task group

In update_sd_pick_busiest(), when comparing two sched groups that are
both of type group_misfit_task, we currently consider the new group as
busier than the current busiest group even if the new group has the
same misfit task load as the current busiest group. We can avoid some
unnecessary writes if we instead only consider the newest group to be
the busiest if it has a higher load than the current busiest. This
matches the behavior of other group types where we compare load, such as
two groups that are both overloaded.

Let's update the group_misfit_task type comparison to also only update
the busiest group in the event of strict inequality.

Signed-off-by: David Vernet <void@manifault.com>
Signed-off-by: Ingo Molnar <mingo@kernel.org>
Reviewed-by: Vincent Guittot <vincent.guittot@linaro.org>
Reviewed-by: Valentin Schneider <vschneid@redhat.com>
Link: https://lore.kernel.org/r/20240206043921.850302-3-void@manifault.com

diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
index 41dda5311770dff5315f2c0cc755ab36acc71af8..448520f4fe83f921a3f10696cd98d6bbd070a265 100644 (file)
--- a/kernel/sched/fair.c
+++ b/kernel/sched/fair.c
@@ -10032,7 +10032,7 @@ static bool update_sd_pick_busiest(struct lb_env *env,
                 * If we have more than one misfit sg go with the biggest
                 * misfit.
                 */
-               if (sgs->group_misfit_task_load < busiest->group_misfit_task_load)
+               if (sgs->group_misfit_task_load <= busiest->group_misfit_task_load)
                        return false;
                break;