Re: [PATCH RESEND] sched/fair: Only increment deadline once on yield

Peter Zijlstra
2025-09-11T11:37:56Z

On Thu, Sep 11, 2025 at 11:51:13AM +0200, Fernand Sieber wrote:
> If a task yields, the scheduler may decide to pick it again. The task in
> turn may decide to yield immediately or shortly after, leading to a tight
> loop of yields.
>
> If there's another runnable task as this point, the deadline will be
> increased by the slice at each loop. This can cause the deadline to runaway
> pretty quickly, and subsequent elevated run delays later on as the task
> doesn't get picked again. The reason the scheduler can pick the same task
> again and again despite its deadline increasing is because it may be the
> only eligible task at that point.
>
> Fix this by updating the deadline only to one slice ahead.
>
> Note, we might want to consider iterating on the implementation of yield as
> follow up:
> * the yielding task could be forfeiting its remaining slice by
> incrementing its vruntime correspondingly

This..

> diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
> index b173a059315c..d6a0d22d08d7 100644
> --- a/kernel/sched/fair.c
> +++ b/kernel/sched/fair.c
> @@ -8921,7 +8921,7 @@ static void yield_task_fair(struct rq *rq)
> */
> rq_clock_skip_update(rq);
>
> - se->deadline += calc_delta_fair(se->slice, se);
> + se->deadline = se->vruntime + calc_delta_fair(se->slice, se);

Have you tried:

se->vruntime = se->deadline;
se->deadline += calc_delta_fair(se->slice, se);

instead?