Re: [PATCH 5/5] sched: limit sched_slice if it is more than sysctl_sched_latency

[Preeti U Murthy <preeti@linux.vnet.ibm.com>]

Hi Joonsoo,

On 04/01/2013 10:39 AM, Joonsoo Kim wrote:
> Hello Preeti.
> So we should limit this possible weird situation.
>>>
>>> Signed-off-by: Joonsoo Kim <iamjoonsoo.kim@lge.com>
>>>
>>> diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
>>> index e232421..6ceffbc 100644
>>> --- a/kernel/sched/fair.c
>>> +++ b/kernel/sched/fair.c
>>> @@ -645,6 +645,9 @@ static u64 sched_slice(struct cfs_rq *cfs_rq, struct sched_entity *se)
>>>  	}
>>>  	slice = calc_delta_mine(slice, se->load.weight, load);
>>>
>>> +	if (unlikely(slice > sysctl_sched_latency))
>>> +		slice = sysctl_sched_latency;
>>
>> Then in this case the highest priority thread would get
>> 20ms(sysctl_sched_latency), and the rest would get
>> sysctl_sched_min_granularity * 10 * (1024/97977) which would be 0.4ms.
>> Then all tasks would get scheduled ateast once within 20ms + (0.4*9) ms
>> = 23.7ms, while your scheduling latency period was extended to 40ms,just
>> so that each of these tasks don't have their sched_slices shrunk due to
>> large number of tasks.
> 
> I don't know I understand your question correctly.
> I will do my best to answer your comment. :)
> 
> With this patch, I just limit maximum slice at one time. Scheduling is
> controlled through the vruntime. So, in this case, the task with nice -20
> will be scheduled twice.
> 
> 20 + (0.4 * 9) + 20 = 43.9 ms
> 
> And after 43.9 ms, this process is repeated.
> 
> So I can tell you that scheduling period is preserved as before.
> 
> If we give a long period to a task at one go, it can cause
> a latency problem. So IMHO, limiting this is meaningful.

Thank you very much for the explanation. Just one question. What is the
reason behind you choosing sysctl_sched_latency as the upper bound here?

Regards
Preeti U Murthy