Return-Path: Received: (majordomo@vger.kernel.org) by vger.kernel.org via listexpand id S1758646AbcCDBOL (ORCPT ); Thu, 3 Mar 2016 20:14:11 -0500 Received: from mail-lb0-f195.google.com ([209.85.217.195]:34082 "EHLO mail-lb0-f195.google.com" rhost-flags-OK-OK-OK-OK) by vger.kernel.org with ESMTP id S1756879AbcCDBOJ (ORCPT ); Thu, 3 Mar 2016 20:14:09 -0500 MIME-Version: 1.0 In-Reply-To: <20160303164751.GD6375@twins.programming.kicks-ass.net> References: <2495375.dFbdlAZmA6@vostro.rjw.lan> <1842158.0Xhak3Uaac@vostro.rjw.lan> <20160303122030.GN6356@twins.programming.kicks-ass.net> <20160303163735.GS6356@twins.programming.kicks-ass.net> <20160303164751.GD6375@twins.programming.kicks-ass.net> Date: Fri, 4 Mar 2016 02:14:06 +0100 X-Google-Sender-Auth: mq5qDhFXV9HDGuKfs_py_qIfOtY Message-ID: Subject: Re: [PATCH 6/6] cpufreq: schedutil: New governor based on scheduler utilization data From: "Rafael J. Wysocki" To: Peter Zijlstra Cc: "Rafael J. Wysocki" , Vincent Guittot , "Rafael J. Wysocki" , Linux PM list , Juri Lelli , Steve Muckle , ACPI Devel Maling List , Linux Kernel Mailing List , Srinivas Pandruvada , Viresh Kumar , Michael Turquette Content-Type: text/plain; charset=UTF-8 Sender: linux-kernel-owner@vger.kernel.org List-ID: X-Mailing-List: linux-kernel@vger.kernel.org Content-Length: 1007 Lines: 23 On Thu, Mar 3, 2016 at 5:47 PM, Peter Zijlstra wrote: > 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. That actually should be a = max_freq + min_freq, because I want (max_freq + min_freq) / 2 = a / 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. It is somewhat imbalanced, though. If all of the values of x are equally probable (or equally frequent), the probability of running above the middle frequency is lower than the probability of running below it.