[e2e] [Fwd: RED-->ECN]

[Vishal Misra]

On Sat, 3 Feb 2001, Alhussein Abouzeid wrote:

> 
> The I in the PI controller is "Integrator", which is some sort of
> averaging. Just like the averaging of the queue size, a PI controller does
> introduce delay in the feedback (another pole as Saverio said), and it is
> some sort of averaging, right?

The pole is at 0. The PI controller also has a zero, which is at a
frequency higher than 0.

> You can on the other hand add a D controller (a zero). In all cases,
> you will end up optimizing responsiveness and unstability (KD) versus
> sluggishness and stability (I).


The PI controller is the P controller and I controller summed up in
parallel, something like (in the Laplace domain, with gains k1 and k2)
 
                   |-------[k1 P]------|
(q-q_ref)   -------|                   +---- p
                   |-------[k2 I]------|
 
This gives the transfer function between the error signal (q-q_ref) and
the marking probability p.
 
Call q - q_ref = q~
 
so p(s) = k1*q~(s)+k2*q~(s)/s
 
you can conceptually think of that as
 
 p(s) = (s*k1(q~)+k2(q~(s))/s, which is
           ^          ^     ^
           D          P    \int
 
i.e. a D term and a P term going into an integrator (a queue). The classic
Lindley equation for the queue then gives p(t+1) (new queue) in terms of
the [total arrival+old queue]+.  The feedback control will drive the total
arrival to zero, i.e. q~-> 0, which means target buffer size is met and
(q~)' -> 0, which implies that the rates are matched (this is also the
"clear buffer, match rate" scheme of REM). At that point, the marking
probability would have converged giving p(t+1) = [0+p(t)]+

-Vishal