Messages in this thread | | | Subject | Re: [PATCH 2/5] writeback: dirty position control | From | Peter Zijlstra <> | Date | Fri, 12 Aug 2011 00:56:06 +0200 |
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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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