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

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Message-ID: <20110826095356.GA5124@localhost>
Date:	Fri, 26 Aug 2011 17:53:56 +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 Fri, Aug 26, 2011 at 04:56:11PM +0800, Peter Zijlstra wrote:
> On Fri, 2011-08-26 at 09:56 +0800, Wu Fengguang wrote:
> >         /*
> >          * A linear estimation of the "balanced" throttle rate. The theory is,
> >          * if there are N dd tasks, each throttled at task_ratelimit, the bdi's
> >          * dirty_rate will be measured to be (N * task_ratelimit). So the below
> >          * formula will yield the balanced rate limit (write_bw / N).
> >          *
> >          * Note that the expanded form is not a pure rate feedback:
> >          *      rate_(i+1) = rate_(i) * (write_bw / dirty_rate)              (1)
> >          * but also takes pos_ratio into account:
> >          *      rate_(i+1) = rate_(i) * (write_bw / dirty_rate) * pos_ratio  (2)
> >          *
> >          * (1) is not realistic because pos_ratio also takes part in balancing
> >          * the dirty rate.  Consider the state
> >          *      pos_ratio = 0.5                                              (3)
> >          *      rate = 2 * (write_bw / N)                                    (4)
> >          * If (1) is used, it will stuck in that state! Because each dd will be
> >          * throttled at
> >          *      task_ratelimit = pos_ratio * rate = (write_bw / N)           (5)
> >          * yielding
> >          *      dirty_rate = N * task_ratelimit = write_bw                   (6)
> >          * put (6) into (1) we get
> >          *      rate_(i+1) = rate_(i)                                        (7)
> >          *
> >          * So we end up using (2) to always keep
> >          *      rate_(i+1) ~= (write_bw / N)                                 (8)
> >          * regardless of the value of pos_ratio. As long as (8) is satisfied,
> >          * pos_ratio is able to drive itself to 1.0, which is not only where
> >          * the dirty count meet the setpoint, but also where the slope of
> >          * pos_ratio is most flat and hence task_ratelimit is least fluctuated.
> >          */ 
> 
> I'm still not buying this, it has the massive assumption N is a
> constant, without that assumption you get the same kind of thing you get
> from not adding pos_ratio to the feedback term.

The reasoning between (3)-(7) actually assumes both N and write_bw to
be some constant. It's documenting some stuck state..

> Also, I've yet to see what harm it does if you leave it out, all
> feedback loops should stabilize just fine.

That's a good question. It should be trivial to try out equation (1)
and see how it work out in practice. Let me collect some figures..

Thanks,
Fengguang
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