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

From: Peter Zijlstra
Date: Thu Aug 11 2011 - 18:57:05 EST


On Tue, 2011-08-09 at 19:20 +0200, Peter Zijlstra wrote:
> So going by:
>
> write_bw
> ref_bw = dirty_ratelimit * pos_ratio * --------
> dirty_bw
>
> pos_ratio seems to be the feedback on the deviation of the dirty pages
> around its setpoint. So we adjust the reference bw (or rather ratelimit)
> to take account of the shift in output vs input capacity as well as the
> shift in dirty pages around its setpoint.
>
> From that we derive the condition that:
>
> pos_ratio(setpoint) := 1
>
> Now in order to create a linear function we need one more condition. We
> get one from the fact that once we hit the limit we should hard throttle
> our writers. We get that by setting the ratelimit to 0, because, after
> all, pause = nr_dirtied / ratelimit would yield inf. in that case. Thus:
>
> pos_ratio(limit) := 0
>
> Using these two conditions we can solve the equations and get your:
>
> limit - dirty
> pos_ratio(dirty) = ----------------
> limit - setpoint
>
> Now, for some reason you chose not to use limit, but something like
> min(limit, 4*thresh) something to do with the slope affecting the rate
> of adjustment. This wants a comment someplace.

Ok, so I think that pos_ratio(limit) := 0, is a stronger condition than
your negative slope (df/dx < 0), simply because it implies your
condition and because it expresses our hard stop at limit.

Also, while I know this is totally over the top, but..

I saw you added a ramp and brake area in future patches, so have you
considered using a third order polynomial instead?

The simple:

f(x) = -x^3

has the 'right' shape, all we need is move it so that:

f(s) = 1

and stretch it to put the single root at our limit. You'd get something
like:

s - x 3
f(x) := 1 + (-----)
d

Which, as required, is 1 at our setpoint and the factor d stretches the
middle bit. Which has a single (real) root at:

x = s + d,

by setting that to our limit, we get:

d = l - s

Making our final function look like:

s - x 3
f(x) := 1 + (-----)
l - s

You can clamp it at [0,2] or so. The implementation wouldn't be too
horrid either, something like:

unsigned long bdi_pos_ratio(..)
{
if (dirty > limit)
return 0;

if (dirty < 2*setpoint - limit)
return 2 * SCALE;

x = SCALE * (setpoint - dirty) / (limit - setpoint);
xx = (x * x) / SCALE;
xxx = (xx * x) / SCALE;

return xxx;
}

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