Re: [Patch v3 3/6] sched/fair: Implement prefer sibling imbalance calculation between asymmetric groups

[Tim Chen <tim.c.chen@linux.intel.com>]

On Fri, 2023-07-14 at 18:44 +0530, Shrikanth Hegde wrote:
> 
> > +	/* Take advantage of resource in an empty sched group */
> > +	if (imbalance == 0 && local->sum_nr_running == 0 &&
> > +	    busiest->sum_nr_running > 1)
> > +		imbalance = 2;
> > +
> 
> I don't see how this case would be true. When there are unequal number of cores and local->sum_nr_ruuning 
> is 0, and busiest->sum_nr_running is atleast 2, imbalance will be atleast 1. 

I think you are correct.  With at least 2 task in the busiest group,
imbalance will be at least 1.  This is the effect of doing rounding
when adding the (ncores_local + ncores_busy) rounding factor.  

Returning an imbalance value of 1 will not be correct as we
will be dividing imbalance by 2 and we will still not move task
to the empty group as intended. 

So this code should be updated as below:


diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
index 3fc8d3a3bd22..16bf75e6a775 100644
--- a/kernel/sched/fair.c
+++ b/kernel/sched/fair.c
@@ -9400,7 +9400,7 @@ static inline long sibling_imbalance(struct lb_env *env,
        imbalance /= ncores_local + ncores_busiest;
 
        /* Take advantage of resource in an empty sched group */
-       if (imbalance == 0 && local->sum_nr_running == 0 &&
+       if (imbalance <= 1 && local->sum_nr_running == 0 &&
            busiest->sum_nr_running > 1)
                imbalance = 2;
 

Tim