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:

quoted

quoted

quoted

f = a * x + b

quoted

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.