On Thu, Jan 10, 2013 at 9:00 AM, Preeti U Murthy
<preeti@linux.vnet.ibm.com> wrote:
> d1's 'groups',both the sd0s.Here is
> the next advantage.It needs information about the sched group alone and
> will not bother about the individual cpus in it.it checks if
> load(sd0[cpu2,cpu3]) > load(sd0[cpu0,cpu1])
> Only if this is true does it go on to see if cpu2/3 is more loaded.If
> there were no scheduler domain or groups,we would have to see the states
> of cpu2 and cpu3 in two iterations instead of 1 iteration like we are
> doing now.

Thanks Peter and preeti, I had seen that intel link and had read but
was not very clear with it,
with both explanations and new links I am clear.