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: Ricardo Neri <ricardo.neri-calderon@linux.intel.com>
Signed-off-by: Peter Zijlstra (Intel) <peterz@infradead.org>
Tested-by: Zhang Rui <rui.zhang@intel.com>
Link: https://lore.kernel.org/r/20230406203148.19182-6-ricardo.neri-calderon@linux.intel.com

Author: ricardon

kernel/sched/fair.c

@@ -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: