Re: [PATCH] sched/fair: Simplify the scene where both local and busiest are group_fully_busy

[Vincent Guittot]

On Thu, 7 Apr 2022 at 04:38, 彭志刚 <zgpeng.linux@xxxxxxxxx> wrote:
>
> When the type of the local group is group_has_spare, it does not affect it, and it also has the opportunity
>
> to pull the process to the local cpu.
>

My point is :

local group is group_fully_busy
busiest group is group_fully_busy

but

local cpu is idle or newly idle otherwise we would have already returned

Currently, calculate_imbalance returns imbalance=0 because it is based
on avg_load which is not set for busiest group. Instead of skipping
calculate_imbalance, we might compute an imbalance similarly to what
is done for spare_capacity because local cpu being idle is an
opportunity to run something else

>
>
> This patch deals with scenarios where both the local group and the busiest group type are group_fully_busy.
>
> In this scenario, because the avg_load of the group of type group_fully_busy is not calculated, this value is 0.
>
> Therefore, the condition of local->avg_load >= busiest->avg_load in calculate_imbalance is satisfied, so the
>
> imbalance will be set to 0; Therefore, in this scenario, the original logic has no chance to pull the process to
>
> the local cpu for execution.  I think it can be judged at the upper level, and there is no need to go into
>
> calculate_imbalance to do some useless work.
>
>
> Vincent Guittot <vincent.guittot@xxxxxxxxxx> 于2022年4月6日周三 23:41写道：
>>
>> On Wed, 6 Apr 2022 at 13:46, zgpeng <zgpeng.linux@xxxxxxxxx> 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@xxxxxxxxxxx>
>> > Reviewed-by: Samuel Liao <samuelliao@xxxxxxxxxxx>
>> > ---
>> >  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
>> >