cpufreq: intel_pstate: Use average P-State instead of current P-State

The result returned by pid_calc() is subtracted from current_pstate
(which is the P-State requested during the last period) in order to
obtain the target P-State for the current iteration.

However, current_pstate may not reflect the real current P-State of
the CPU. In particular, that P-State may be higher because of the
frequency sharing per module.

The theory is:
 - The load is the percentage of time spent in C0 and is related to
   the average P-State during the same period.
 - The last requested P-State can be completely different than the
   average P-State (because of frequency sharing or throttling).
 - The P-State shift computed by the pid_calc is based on the load
   computed at average P-State, so the shift must be relative to
   this average P-State.

Using the average P-State instead of current P-State improves power
without significant performance penalty in cases when a task migrates
from one core to other core sharing frequency and voltage.

Performance and power comparison with this patch on Cherry Trail
platform using Android:

Benchmark               ?Perf    ?Power
FishTank                10.45%    3.1%
SmartBench-Gaming       -0.1%   -10.4%
SmartBench-Productivity -0.8%   -10.4%
CandyCrush                n/a   -17.4%
AngryBirds                n/a    -5.9%
videoPlayback             n/a   -13.9%
audioPlayback             n/a    -4.9%
IcyRocks-20-50           0.0%   -38.4%
iozone RR               -0.16%  -1.3%
iozone RW                0.74%  -1.3%

Signed-off-by: Philippe Longepe <philippe.longepe@linux.intel.com>
Signed-off-by: Srinivas Pandruvada <srinivas.pandruvada@linux.intel.com>
Signed-off-by: Rafael J. Wysocki <rafael.j.wysocki@intel.com>

0 files changed, 0 insertions, 0 deletions