On Thu, Mar 03, 2016 at 05:24:32PM +0100, Rafael J. Wysocki wrote: > On Thu, Mar 3, 2016 at 1:20 PM, Peter Zijlstra <peterz@xxxxxxxxxxxxx> wrote: > > On Wed, Mar 02, 2016 at 11:49:48PM +0100, Rafael J. Wysocki wrote: > >> >>> + min_f = sg_policy->policy->cpuinfo.min_freq; > >> >>> + max_f = sg_policy->policy->cpuinfo.max_freq; > >> >>> + next_f = util > max ? max_f : min_f + util * (max_f - min_f) / max; > > > >> In case a more formal derivation of this formula is needed, it is > >> based on the following 3 assumptions: > >> > >> (1) Performance is a linear function of frequency. > >> (2) Required performance is a linear function of the utilization ratio > >> x = util/max as provided by the scheduler (0 <= x <= 1). > > > >> (3) The minimum possible frequency (min_freq) corresponds to x = 0 and > >> the maximum possible frequency (max_freq) corresponds to x = 1. > >> > >> (1) and (2) combined imply that > >> > >> f = a * x + b > >> > >> (f - frequency, a, b - constants to be determined) and then (3) quite > >> trivially leads to b = min_freq and a = max_freq - min_freq. > > > > 3 is the problem, that just doesn't make sense and is probably the > > reason why you see very little selection of the min freq. > > It is about mapping the entire [0,1] interval to the available frequency range. Yeah, but I don't see why that makes sense.. > I till overprovision things (the smaller x the more), but then it may > help the race-to-idle a bit in theory. So, since we also have the cpuidle information, could we not make a better guess at race-to-idle? > > Suppose a machine with the following frequencies: > > > > 500, 750, 1000 > > > > And a utilization of 0.4, how does asking for 500 + 0.4 * (1000-500) = > > 700 make any sense? Per your point 1, it should should be asking for > > 0.4 * 1000 = 400. > > > > Because, per 1, at 500 it runs exactly half as fast as at 1000, and we > > only need 0.4 times as much. Therefore 500 is more than sufficient. > > OK, but then I don't see why this reasoning only applies to the lower > bound of the frequency range. Is there any reason why x = 1 should be > the only point mapping to max_freq? Well, everything that goes over the second to last freq would end up at the last (max) freq. Take again the 500,750,1000 example, everything that's >750 would end up at 1000 (for relation_l, >875 for _c). But given the platform's cpuidle information, maybe coupled with an avg idle est, we can compute the benefit of race-to-idle and over provision based on that, right? > If not, then I think it's reasonable to map the middle of the > available frequency range to x = 0.5 and then we have b = 0 and a = > (max_freq + min_freq) / 2. So I really think that approach falls apart on the low util bits, you effectively always run above min speed, even if min is already vstly over provisioned. -- To unsubscribe from this list: send the line "unsubscribe linux-acpi" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
