From mboxrd@z Thu Jan 1 00:00:00 1970 Return-Path: Received: (majordomo@vger.kernel.org) by vger.kernel.org via listexpand id S1755968Ab1HXANF (ORCPT ); Tue, 23 Aug 2011 20:13:05 -0400 Received: from mga03.intel.com ([143.182.124.21]:47549 "EHLO mga03.intel.com" rhost-flags-OK-OK-OK-OK) by vger.kernel.org with ESMTP id S1752730Ab1HXANC (ORCPT ); Tue, 23 Aug 2011 20:13:02 -0400 X-ExtLoop1: 1 X-IronPort-AV: E=Sophos;i="4.68,272,1312182000"; d="scan'208";a="41713267" Date: Wed, 24 Aug 2011 08:12:58 +0800 From: Wu Fengguang To: Vivek Goyal Cc: Peter Zijlstra , "linux-fsdevel@vger.kernel.org" , Andrew Morton , Jan Kara , Christoph Hellwig , Dave Chinner , Greg Thelen , Minchan Kim , Andrea Righi , linux-mm , LKML Subject: Re: [PATCH 2/5] writeback: dirty position control Message-ID: <20110824001257.GA6349@localhost> References: <20110808141128.GA22080@localhost> <1312814501.10488.41.camel@twins> <20110808230535.GC7176@localhost> <1313154259.6576.42.camel@twins> <20110812142020.GB17781@localhost> <1314027488.24275.74.camel@twins> <20110823034042.GC7332@localhost> <1314093660.8002.24.camel@twins> <20110823141504.GA15949@localhost> <20110823174757.GC15820@redhat.com> MIME-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Disposition: inline In-Reply-To: <20110823174757.GC15820@redhat.com> User-Agent: Mutt/1.5.20 (2009-06-14) Sender: linux-kernel-owner@vger.kernel.org List-ID: X-Mailing-List: linux-kernel@vger.kernel.org > You somehow directly jump to > > balanced_rate = task_ratelimit_200ms * write_bw / dirty_rate > > without explaining why following will not work. > > balanced_rate_(i+1) = balance_rate(i) * write_bw / dirty_rate Thanks for asking that, it's probably the root of confusions, so let me answer it standalone. It's actually pretty simple to explain this equation: write_bw balanced_rate = task_ratelimit_200ms * ---------- (1) dirty_rate If there are N dd tasks, each task is throttled at task_ratelimit_200ms for the past 200ms, we are going to measure the overall bdi dirty rate dirty_rate = N * task_ratelimit_200ms (2) put (2) into (1) we get balanced_rate = write_bw / N (3) So equation (1) is the right estimation to get the desired target (3). As for write_bw balanced_rate_(i+1) = balanced_rate_(i) * ---------- (4) dirty_rate Let's compare it with the "expanded" form of (1): write_bw balanced_rate_(i+1) = balanced_rate_(i) * pos_ratio * ---------- (5) dirty_rate So the difference lies in pos_ratio. Believe it or not, it's exactly the seemingly use of pos_ratio that makes (5) independent(*) of the position control. Why? Look at (4), assume the system is in a state - dirty rate is already balanced, ie. balanced_rate_(i) = write_bw / N - dirty position is not balanced, for example pos_ratio = 0.5 balance_dirty_pages() will be rate limiting each tasks at half the balanced dirty rate, yielding a measured dirty_rate = write_bw / 2 (6) Put (6) into (4), we get balanced_rate_(i+1) = balanced_rate_(i) * 2 = (write_bw / N) * 2 That means, any position imbalance will lead to balanced_rate estimation errors if we follow (4). Whereas if (1)/(5) is used, we always get the right balanced dirty ratelimit value whether or not (pos_ratio == 1.0), hence make the rate estimation independent(*) of dirty position control. (*) independent as in real values, not the seemingly relations in equation Thanks, Fengguang