linux-kernel - Re: [PATCH 2/5] writeback: dirty position control

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Message-ID: <20110826001846.GA6118@localhost>
Date:	Fri, 26 Aug 2011 08:18:46 +0800
From:	Wu Fengguang <fengguang.wu@...el.com>
To:	Peter Zijlstra <peterz@...radead.org>
Cc:	Vivek Goyal <vgoyal@...hat.com>,
	"linux-fsdevel@...r.kernel.org" <linux-fsdevel@...r.kernel.org>,
	Andrew Morton <akpm@...ux-foundation.org>,
	Jan Kara <jack@...e.cz>, Christoph Hellwig <hch@....de>,
	Dave Chinner <david@...morbit.com>,
	Greg Thelen <gthelen@...gle.com>,
	Minchan Kim <minchan.kim@...il.com>,
	Andrea Righi <arighi@...eler.com>,
	linux-mm <linux-mm@...ck.org>,
	LKML <linux-kernel@...r.kernel.org>
Subject: Re: [PATCH 2/5] writeback: dirty position control

On Thu, Aug 25, 2011 at 12:12:58AM +0800, Peter Zijlstra wrote:
> On Wed, 2011-08-24 at 08:12 +0800, Wu Fengguang wrote:
> > > 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
> 
> 
> The assumption here is that N is a constant.. in the above case
> pos_ratio would eventually end up at 1 and things would be good again. I
> see your argument about oscillations, but I think you can introduce
> similar effects by varying N.

Yeah, it's very possible for N to change over time, in which case
balanced_rate will adapt to new N in similar way.

Thanks,
Fengguang
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