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SubjectRe: [PATCH 1/2] sched: push rt tasks only if newly activated tasks have been added
Hi Dmitry,

(Disclaimer: I am sick with a fever today, so hopefully I'm groking your email properly and not about to say something stupid ;)

>>> On Tue, Apr 22, 2008 at 11:30 AM, in message
<b647ffbd0804220830h6524e788n1467b027bc5bc4d2@mail.gmail.com>, "Dmitry
Adamushko" <dmitry.adamushko@gmail.com> wrote:
> Hi Gregory,
>
>
> consider the following 2-cpu system: cpu0 and cpu1.
>
> cpu0: is idle --> in such a state, it never pulls RT tasks on its own.
>
> T0 and T1 are RT tasks
>
>
> square#0:
>
> cpu1: T0 is running
>
> T1 is of the same prio as T0 (shouldn't really matter but to get the
> same result it would require altering the flow of events slightly)
>
> T1's affinity allows it to be run only on cpu1.
> T0 can run on both.
>
> try_to_wake_up() is called for T1.
> |
> --> select_task_rq_rt() => gives cpu1
> |
> --> task_wake_up_rt()
> |
> ---> push_rt_tasks() -> rq->rt.pushed = 1
>
> now, neither T1 (due to its affinity), nor T0 (it's running) can be
> pushed away to cpu0.
>
> [ btw., (1) I'd expect that this task_wake_up_rt() thing should be
> redundant, logically-wise... I'll check once more and comment later
> on.

They are both necessary, but the key is that the select_task_rq() is a best-effort route attempt, whereas the task_wake_up() routine is the authoritative router. By doing the push after activation, it allowed us to utilize a very clever and significant optimization on the pull side that Steven came up with. The details of the optimization escape me now, but I do remember it was substantial to the design. Then later we put the select_task_rq() logic in (see git-id 318e0893) to further optimize the routing by finding a likely good home before the activation takes place (saving an activation/deactivation cycle), but it still needs the post-router to protect against race conditions since its just best-effort.

> (2) any example when (p->prio >= rq->rt.highest_prio) is not true in
> task_wake_up_rt() ?

Hmm...good catch. Looks like it should be "p->prio >= rq->curr->prio" since we only need be concerned with pushing here if the task is not going to preempt current. Do you agree Steven, or am I missing something?


> ]
>
> as a result, rq->rt.pushed == 1.
>
> Now, post_schedule_rt() won't call push_rt_tasks().
>
> T0 and T1 are both running for some time on cpu1 (possibly
> context-switching if they are both of SCHED_RR type).
>
> Then they both block, _first_ T1 and then T0.
>
> After some interval of time, they wake up (let's say they are
> periodic) in the following order: _first_ T0 and then T1.
>
> rq->rt.pushed becomes 0 and here we are back to square#0. The whole
> story repeats again.
>
> cpu0 is idle so it won't pull T0. Both T0 and T1 are competing for the
> same cpu. Not good.
>
> am I missing smth?

No, I think you are indeed correct. However, I would consider the root cause of the problem to have existed prior to the "pushed" flag, so perhaps we need to address this at a different level. The case you present would have always been problematic for FIFO, and would have "worked" for RR eventually prior to the "pushed" patch. But I dont know if I like relying on how it worked before to fix up the system. At the very best, T1 would have experienced a latency equal to the remainder of T0's timeslice.

Rather, I think we need to address the preemptive behavior for the case where a migratory task is on the cpu and a non-migratory task tries to wake up. If they are equal in numerical priority, perhaps we need to treat "non-migratory" as the tie breaker. In this case, T1 would preempt T0 from cpu1, and then we would push T0 to cpu0. I don't quite have all the details about how this would work thought through yet. Perhaps I should wait until my fever lifts. ;) Thoughts?

-Greg





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