Commit 5fd6d7f4 authored by Ricardo Neri's avatar Ricardo Neri Committed by Peter Zijlstra
Browse files

sched/fair: Keep a fully_busy SMT sched group as busiest 



When comparing two fully_busy scheduling groups, keep the current busiest
group if it represents an SMT core. Tasks in such scheduling group share
CPU resources and need more help than tasks in a non-SMT fully_busy group.

Signed-off-by: default avatarRicardo Neri <ricardo.neri-calderon@linux.intel.com>
Signed-off-by: default avatarPeter Zijlstra (Intel) <peterz@infradead.org>
Tested-by: default avatarZhang Rui <rui.zhang@intel.com>
Link: https://lore.kernel.org/r/20230406203148.19182-6-ricardo.neri-calderon@linux.intel.com
parent 18ad3453

kernel/sched/fair.c

+14 −2
Original line number Diff line number Diff line
@@ -9619,10 +9619,22 @@ static bool update_sd_pick_busiest(struct lb_env *env, 
		 * contention when accessing shared HW resources.

		 *

		 * XXX for now avg_load is not computed and always 0 so we

		 * select the 1st one.

		 * select the 1st one, except if @sg is composed of SMT

		 * siblings.

		 */

		if (sgs->avg_load <= busiest->avg_load)



		if (sgs->avg_load < busiest->avg_load)

			return false;



		if (sgs->avg_load == busiest->avg_load) {

			/*

			 * SMT sched groups need more help than non-SMT groups.

			 * If @sg happens to also be SMT, either choice is good.

			 */

			if (sds->busiest->flags & SD_SHARE_CPUCAPACITY)

				return false;

		}



		break;



	case group_has_spare: