Messages in this thread | | | Date | Sat, 02 Feb 2013 18:50:05 +0100 | From | Sebastian Andrzej Siewior <> | Subject | Re: [RFC 2/2] sched/fair: prefer a CPU in the "lowest" idle state |
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On 01/31/2013 03:12 AM, Michael Wang wrote: > I'm not sure, but just concern about this case: > > group 0 cpu 0 cpu 1 > least idle 4 task > > group 1 cpu 2 cpu 3 > 1 task 1 task > > The previous logical will pick group 1 and now it will take group 0, and > that cause more imbalance, doesn't it?
That depends on load of CPU 0 + 1 vs CPU 2 + 3. If the four tasks on CPU1 are idle then the previous code should return group 0. If the four tasks are running at 100% each then two of them should be migrated to CPU0 and this point the idle state does not matter :)
> May be check that state in find_idlest_cpu() will be better?
You say to move this from find_idlest_group() to find_idlest_cpu()?
> Regards, > Michael Wang
Sebastian
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