Re: [PATCH 02/18] sched: fix find_idlest_group mess logical

[Alex Shi <alex.shi@intel.com>]

On 12/11/2012 01:08 PM, Preeti U Murthy wrote:
> Hi Alex,
> 
> On 12/10/2012 01:52 PM, Alex Shi wrote:
>> There is 4 situations in the function:
>> 1, no task allowed group;
>> 	so min_load = ULONG_MAX, this_load = 0, idlest = NULL
>> 2, only local group task allowed;
>> 	so min_load = ULONG_MAX, this_load assigned, idlest = NULL
>> 3, only non-local task group allowed;
>> 	so min_load assigned, this_load = 0, idlest != NULL
>> 4, local group + another group are task allowed.
>> 	so min_load assigned, this_load assigned, idlest != NULL
>>
>> Current logical will return NULL in first 3 kinds of scenarios.
>> And still return NULL, if idlest group is heavier then the
>> local group in the 4th situation.
>>
>> Actually, I thought groups in situation 2,3 are also eligible to host
>> the task. And in 4th situation, agree to bias toward local group.
>> So, has this patch.
>>
>> Signed-off-by: Alex Shi <alex.shi@intel.com>
>> ---
>>  kernel/sched/fair.c |   12 +++++++++---
>>  1 files changed, 9 insertions(+), 3 deletions(-)
>>
>> diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
>> index df99456..b40bc2b 100644
>> --- a/kernel/sched/fair.c
>> +++ b/kernel/sched/fair.c
>> @@ -2953,6 +2953,7 @@ find_idlest_group(struct sched_domain *sd, struct task_struct *p,
>>  		  int this_cpu, int load_idx)
>>  {
>>  	struct sched_group *idlest = NULL, *group = sd->groups;
>> +	struct sched_group *this_group = NULL;
>>  	unsigned long min_load = ULONG_MAX, this_load = 0;
>>  	int imbalance = 100 + (sd->imbalance_pct-100)/2;
>>  
>> @@ -2987,14 +2988,19 @@ find_idlest_group(struct sched_domain *sd, struct task_struct *p,
>>  
>>  		if (local_group) {
>>  			this_load = avg_load;
>> -		} else if (avg_load < min_load) {
>> +			this_group = group;
>> +		}
>> +		if (avg_load < min_load) {
>>  			min_load = avg_load;
>>  			idlest = group;
>>  		}
>>  	} while (group = group->next, group != sd->groups);
>>  
>> -	if (!idlest || 100*this_load < imbalance*min_load)
>> -		return NULL;
>> +	if (this_group && idlest != this_group)
>> +		/* Bias toward our group again */
>> +		if (100*this_load < imbalance*min_load)
>> +			idlest = this_group;
> 
> If the idlest group is heavier than this_group(or to put it better if
> the difference in the loads of the local group and idlest group is less
> than a threshold,it means there is no point moving the load from the
> local group) you return NULL,that immediately means this_group is chosen
> as the candidate group for the task to run,one does not have to
> explicitly return that.

In situation 4, this_group is not NULL.
> 
> Let me explain:
> find_idlest_group()-if it returns NULL to mark your case4,it means there
> is no idler group than the group to which this_cpu belongs to, at that
> level of sched domain.Which is fair enough.
> 
> So now the question is under such a circumstance which is the idlest
> group so far.It is the group containing this_cpu,i.e.this_group.After
> this sd->child is chosen which is nothing but this_group(sd hierarchy
> moves towards the cpu it belongs to). Again here the idlest group search
> begins.
> 
>> +
>>  	return idlest;
>>  }
>>  
>>
> Regards
> Preeti U Murthy
> 


-- 
Thanks
    Alex