Messages in this thread | | | Date | Thu, 3 Mar 2016 17:47:51 +0100 | From | Peter Zijlstra <> | Subject | Re: [PATCH 6/6] cpufreq: schedutil: New governor based on scheduler utilization data |
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On Thu, Mar 03, 2016 at 05:37:35PM +0100, Peter Zijlstra wrote: > On Thu, Mar 03, 2016 at 05:24:32PM +0100, Rafael J. Wysocki wrote: > > >> f = a * x + b
> > 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.
Ah nevermind, I cannot read. Yes that is worth trying I suppose. But the b=0,a=1 thing seems more natural still.
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