sched: reduce calculation effort in fix_small_imbalance

Date	Mon, 17 Jun 2013 21:00:22 +0800

Actually all below item could be repalced by scaled_busy_load_per_task
	(sds->busiest_load_per_task * SCHED_POWER_SCALE)
		/sds->busiest->sgp->power;

Signed-off-by: Lei Wen <leiwen@marvell.com>
Signed-off-by: zwliew <zhaoweiliew@gmail.com>
sched: scale the busy and this queue's per-task load before compare

Date	Mon, 17 Jun 2013 21:00:23 +0800

Since for max_load and this_load, they are the value that already be
scaled. It is not reasonble to get a minimum value between the scaled
and non-scaled value, like below example.
	min(sds->busiest_load_per_task, sds->max_load);

Also add comment over in what condition, there would be cpu power gain
in move the load.

Signed-off-by: Lei Wen <leiwen@marvell.com>
Signed-off-by: zwliew <zhaoweiliew@gmail.com>
sched: scale cpu load for judgment of group imbalance

Date	Mon, 17 Jun 2013 21:00:24 +0800

We cannot compare two load directly from two cpus, since the cpu power
over two cpu may vary largely.

Suppose we meet such two kind of cpus.
CPU A:
	No real time work, and there are 3 task, with rq->load.weight
being 512.
CPU B:
	Has real time work, and it take 3/4 of the cpu power, which
makes CFS only take 1/4, that is 1024/4=256 cpu power. And over its CFS
runqueue, there is only one task with weight as 128.

Since both cpu's CFS task take for half of the CFS's cpu power, it
should be considered as balanced in such case.

But original judgment like:
        if ((max_cpu_load - min_cpu_load) >= avg_load_per_task &&
            (max_nr_running - min_nr_running) > 1)
It makes (512-128)>=((512+128)/4), and lead to imbalance conclusion...
Make the load as scaled, to avoid such case.

Change-Id: I0ed4fc29cc1d24fe253bde84c94a383cdb8eb4e8
Signed-off-by: Lei Wen <leiwen@marvell.com>
Signed-off-by: zwliew <zhaoweiliew@gmail.com>