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SubjectRe: [PATCH] sched/fair: Simplify the scene where both local and busiest are group_fully_busy
On Wed, 6 Apr 2022 at 13:46, zgpeng <zgpeng.linux@gmail.com> wrote:
>
> When both local and busiest group are group_fully_busy type, because
> the avg_load of the group of type group_fully_busy is not calculated, the
> avg_load value is equal to 0. In this case, load balancing will not actually
> done, because after a series of calculations in the calculate_imbalance, it
> will be considered that load balance is not required. Therefore,it is not
> necessary to enter calculate_imbalance to do some useless work.
>
> Signed-off-by: zgpeng <zgpeng@tencent.com>
> Reviewed-by: Samuel Liao <samuelliao@tencent.com>
> ---
> kernel/sched/fair.c | 12 ++++++++++++
> 1 file changed, 12 insertions(+)
>
> diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
> index 9f75303..cc1d6c4 100644
> --- a/kernel/sched/fair.c
> +++ b/kernel/sched/fair.c
> @@ -9634,6 +9634,18 @@ static struct sched_group *find_busiest_group(struct lb_env *env)
> * busiest doesn't have any tasks waiting to run
> */
> goto out_balanced;
> +

We are there because both local and busiest are not overloaded, local
is idle or newly_idle and there might be an opportunity to pull a
waiting task on local to use this local cpu. I wonder if we should not
cover this opportunity in calculate_imbalance instead of skipping it

> + if (local->group_type == group_fully_busy)
> + /*
> + * If local group is group_fully_busy, the code goes here,
> + * the type of busiest group must also be group_fully_busy.
> + * Because the avg_load of the group_fully_busy type is not
> + * calculated at present, it is actually equal to 0. In this
> + * scenario, load balance is not performed. therefore, it can
> + * be returned directly here, and there is no need to do some
> + * useless work in calculate_imbalance.
> + */
> + goto out_balanced;
> }
>
> force_balance:
> --
> 2.9.5
>

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