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>
Showing
Please register or sign in to comment