On Thu, Jan 10, 2013 at 9:00 AM, Preeti U Murthy <[email protected]> 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. _______________________________________________ Kernelnewbies mailing list [email protected] http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
