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>
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