Re: [RFC PATCH v1] sched/fair: limit load balance redo times at the same sched_domain level

[Vincent Guittot]

On Tue, 23 Feb 2021 at 06:41, Li, Aubrey <aubrey.li@xxxxxxxxxxxxxxx> wrote:
>
> Hi Vincent,
>
> Sorry for the delay, I just returned from Chinese New Year holiday.
>
> On 2021/1/25 22:51, Vincent Guittot wrote:
> > On Mon, 25 Jan 2021 at 15:00, Li, Aubrey <aubrey.li@xxxxxxxxxxxxxxx> wrote:
> >>
> >> On 2021/1/25 18:56, Vincent Guittot wrote:
> >>> On Mon, 25 Jan 2021 at 06:50, Aubrey Li <aubrey.li@xxxxxxxxx> wrote:
> >>>>
> >>>> A long-tail load balance cost is observed on the newly idle path,
> >>>> this is caused by a race window between the first nr_running check
> >>>> of the busiest runqueue and its nr_running recheck in detach_tasks.
> >>>>
> >>>> Before the busiest runqueue is locked, the tasks on the busiest
> >>>> runqueue could be pulled by other CPUs and nr_running of the busiest
> >>>> runqueu becomes 1, this causes detach_tasks breaks with LBF_ALL_PINNED
> >>>
> >>> We should better detect that when trying to detach task like below
> >>
> >> This should be a compromise from my understanding. If we give up load balance
> >> this time due to the race condition, we do reduce the load balance cost on the
> >> newly idle path, but if there is an imbalance indeed at the same sched_domain
> >
> > Redo path is there in case, LB has found an imbalance but it can't
> > move some loads from this busiest rq to dest rq because of some cpu
> > affinity. So it tries to fix the imbalance by moving load onto another
> > rq of the group. In your case, the imbalance has disappeared because
> > it has already been pulled by another rq so you don't have to try to
> > find another imbalance. And I would even say you should not in order
> > to let other level to take a chance to spread the load
> >
> >> level, we have to wait the next softirq entry to handle that imbalance. This
> >> means the tasks on the second busiest runqueue have to stay longer, which could
> >> introduce tail latency as well. That's why I introduced a variable to control
> >> the redo loops. I'll send this to the benchmark queue to see if it makes any
> >
> > TBH, I don't like multiplying the number of knobs
>
> Sure, I can take your approach, :)
>
> >>>
> >>> --- a/kernel/sched/fair.c
> >>> +++ b/kernel/sched/fair.c
> >>> @@ -7688,6 +7688,16 @@ static int detach_tasks(struct lb_env *env)
> >>>
> >>>         lockdep_assert_held(&env->src_rq->lock);
> >>>
> >>> +       /*
> >>> +        * Another CPU has emptied this runqueue in the meantime.
> >>> +        * Just return and leave the load_balance properly.
> >>> +        */
> >>> +       if (env->src_rq->nr_running <= 1 && !env->loop) {
>
> May I know why !env->loop is needed here? IIUC, if detach_tasks is invoked

IIRC,  my point was to do the test only when trying to detach the 1st
task. A lot of things can happen when a break is involved but TBH I
can't remember a precise UC. It may be over cautious

> from LBF_NEED_BREAK, env->loop could be non-zero, but as long as src_rq's
> nr_running <=1, we should return immediately with LBF_ALL_PINNED flag cleared.
>
> How about the following change?
>
> diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
> index 04a3ce20da67..1761d33accaa 100644
> --- a/kernel/sched/fair.c
> +++ b/kernel/sched/fair.c
> @@ -7683,8 +7683,11 @@ static int detach_tasks(struct lb_env *env)
>                  * We don't want to steal all, otherwise we may be treated likewise,
>                  * which could at worst lead to a livelock crash.
>                  */
> -               if (env->idle != CPU_NOT_IDLE && env->src_rq->nr_running <= 1)
> +               if (env->idle != CPU_NOT_IDLE && env->src_rq->nr_running <= 1) {

IMO, we must do the test before:  while (!list_empty(tasks)) {

because src_rq might have become empty if waiting tasks have been
pulled by another cpu and the running one became idle in the meantime

> +                       /* Clear the flag as we will not test any task */
> +                       env->flag &= ~LBF_ALL_PINNED;
>                         break;
> +               }
>
>                 p = list_last_entry(tasks, struct task_struct, se.group_node);
>
> Thanks,
> -Aubrey