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Date:	Thu, 02 Apr 2009 07:21:26 -0400
From:	Gregory Haskins <ghaskins@...ell.com>
To:	Peter Zijlstra <peterz@...radead.org>
CC:	Ingo Molnar <mingo@...e.hu>, Steven Rostedt <rostedt@...dmis.org>,
	LKML <linux-kernel@...r.kernel.org>,
	Thomas Gleixner <tglx@...utronix.de>
Subject: Re: [PATCH] sched_rt: fix overload bug on rt group scheduling

Peter Zijlstra wrote:
> On Wed, 2009-04-01 at 20:58 -0400, Gregory Haskins wrote:
>   
>> Hi Peter,
>>
>> Peter Zijlstra wrote:
>>     
>>> Fixes an easily triggerable BUG() when setting process affinities.
>>>
>>> Make sure to count the number of migratable tasks in the same place:
>>> the root rt_rq. Otherwise the number doesn't make sense and we'll hit
>>> the BUG in set_cpus_allowed_rt().
>>>
>>> Also, make sure we only count tasks, not groups (this is probably
>>> already taken care of by the fact that rt_se->nr_cpus_allowed will be 0
>>> for groups, but be more explicit)
>>>
>>> Signed-off-by: Peter Zijlstra <a.p.zijlstra@...llo.nl>
>>> Tested-by: Thomas Gleixner <tglx@...utronix.de>
>>> CC: stable@...nel.org
>>> ---
>>>  kernel/sched_rt.c |   16 +++++++++++++++-
>>>  1 files changed, 15 insertions(+), 1 deletions(-)
>>>
>>> diff --git a/kernel/sched_rt.c b/kernel/sched_rt.c
>>> index de4469a..c1ee8dc 100644
>>> --- a/kernel/sched_rt.c
>>> +++ b/kernel/sched_rt.c
>>> @@ -10,6 +10,8 @@ static inline struct task_struct *rt_task_of(struct sched_rt_entity *rt_se)
>>>  
>>>  #ifdef CONFIG_RT_GROUP_SCHED
>>>  
>>> +#define rt_entity_is_task(rt_se) (!(rt_se)->my_q)
>>> +
>>>  static inline struct rq *rq_of_rt_rq(struct rt_rq *rt_rq)
>>>  {
>>>  	return rt_rq->rq;
>>> @@ -22,6 +24,8 @@ static inline struct rt_rq *rt_rq_of_se(struct sched_rt_entity *rt_se)
>>>  
>>>  #else /* CONFIG_RT_GROUP_SCHED */
>>>  
>>> +#define rt_entity_is_task(rt_se) (1)
>>> +
>>>  static inline struct rq *rq_of_rt_rq(struct rt_rq *rt_rq)
>>>  {
>>>  	return container_of(rt_rq, struct rq, rt);
>>> @@ -73,7 +77,7 @@ static inline void rt_clear_overload(struct rq *rq)
>>>  
>>>  static void update_rt_migration(struct rt_rq *rt_rq)
>>>  {
>>> -	if (rt_rq->rt_nr_migratory && (rt_rq->rt_nr_running > 1)) {
>>> +	if (rt_rq->rt_nr_migratory > 1) {
>>>   
>>>       
>> The rest of the patch is making sense to me, but I am a little concerned
>> about this change.
>>
>> The original logic was designed to catch the condition when you might
>> have a non-migratory task running, and a migratory task queued.   This
>> would mean nr_running == 2, and nr_migratory == 1, which is eligible for
>> overload handling.  (Of course, the opposite could be true..the
>> migratory is running and the non-migratory is queued...we cannot discern
>> the difference here and we go into overload anyway.  This is just
>> suboptimal but functionally correct).
>>
>> What can happen now is you could have that above condition but we will
>> not go into overload unless there is at least two migratory tasks
>> queued.  This will undoubtedly allow a potential scheduling latency on
>> task #2.
>>
>> I think we really need to qualify overload on both running > 1 and at
>> least one migratory task.  Is there a way to get this state, even if by
>> other means?
>>     
>
> Ah, yes, I missed that bit. I ripped out the rt_nr_running because I 1)
> didn't think of this, and 2) rt_nr_running is accounted per rt_rq, not
> per-cpu, so it doesn't match.
>
> Since rt_nr_running is also used in a per rt_rq setting, changing that
> isn't possible and we'd need to introduce another per-cpu variant is you
> want to re-instate this.
>   
Yeah, I actually don't care if its literally a nr_running stat
reinstated, or some other way to restore "correctness" ;)

Double bonus if you can solve that problem I mentioned above where I
can't tell if its really eligible for overload in all cases (but goes
into overload anyway to be conservative).  I had been thinking of doing
something like subtracting the nr_migration number when a migratory task
is put on the cpu.  But this is kind of messy because you need to handle
all the places that can manipulate nr_migratory to make sure it doesnt
break.

Thanks Peter!
-Greg




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