On 21 December 2012 09:53, Vincent Guittot <vincent.guit...@linaro.org> wrote:
> On 21 December 2012 06:47, Namhyung Kim <namhy...@kernel.org> wrote:
>> Hi Vincent,
>>
>> On Thu, Dec 13, 2012 at 11:11:11AM +0100, Vincent Guittot wrote:
>>> On 13 December 2012 03:17, Alex Shi <alex....@intel.com> wrote:
>>> > On 12/12/2012 09:31 PM, Vincent Guittot wrote:
>>> >> +static bool is_buddy_busy(int cpu)
>>> >> +{
>>> >> + struct rq *rq = cpu_rq(cpu);
>>> >> +
>>> >> + /*
>>> >> + * A busy buddy is a CPU with a high load or a small load with a
>>> >> lot of
>>> >> + * running tasks.
>>> >> + */
>>> >> + return ((rq->avg.runnable_avg_sum << rq->nr_running) >
>>> >
>>> > If nr_running a bit big, rq->avg.runnable_avg_sum << rq->nr_running is
>>> > zero. you will get the wrong decision.
>>>
>>> yes, I'm going to do that like below instead:
>>> return (rq->avg.runnable_avg_sum > (rq->avg.runnable_avg_period >>
>>> rq->nr_running));
>>
>> Doesn't it consider nr_running too much? It seems current is_buddy_busy
>> returns false on a cpu that has 1 task runs 40% cputime, but returns true
>> on a cpu that has 3 tasks runs 10% cputime each or for 2 tasks of 15%
>> cputime each, right?
>
> Yes it's right.
>>
>> I don't know what is correct, but just guessing that in a cpu's point
>> of view it'd be busier if it has a higher runnable_avg_sum than a
>> higher nr_running IMHO.
>
sorry, the mail has been sent before i finish it
> The nr_running is used to point out how many tasks are running
> simultaneously and the potential scheduling latency of adding
The nr_running is used to point out how many tasks are running
simultaneously and as a result the potential scheduling latency.
I have used the shift instruction because it was quite simple and
efficient but it may give too much weight to nr_running. I could use a
simple division instead of shifting runnable_avg_sum
>
>>
>>
>>>
>>> >
>>> >> + rq->avg.runnable_avg_period);
>>> >> +}
>>> >> +
>>> >> +static bool is_light_task(struct task_struct *p)
>>> >> +{
>>> >> + /* A light task runs less than 25% in average */
>>> >> + return ((p->se.avg.runnable_avg_sum << 1) <
>>> >> + p->se.avg.runnable_avg_period);
>>> >
>>> > 25% may not suitable for big machine.
>>>
>>> Threshold is always an issue, which threshold should be suitable for
>>> big machine ?
>>>
>>> I'm wondering if i should use the imbalance_pct value for computing
>>> the threshold
>>
>> Anyway, I wonder how 'sum << 1' computes 25%. Shouldn't it be << 2 ?
>
> The 1st version of the patch was using << 2 but I received a comment
> saying that it was may be not enough aggressive so I have updated the
> formula with << 1 but forgot to update the comment. I will align
> comment and formula in the next version.
> Thanks for pointing this
>
> Vincent
>
>>
>> Thanks,
>> Namhyung
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