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

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Date:	Wed, 24 Aug 2011 18:12:58 +0200
From:	Peter Zijlstra <peterz@...radead.org>
To:	Wu Fengguang <fengguang.wu@...el.com>
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 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.
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