[AMPL 2939] Constraint formulation

[alisonthomas]

Hi,

Can someone please advise me how I should formulate a constraint that
represents the following:

IF( "decision variable A" > 0, then 0, else "decision variable B")

I also need to formulate a similar constraint, where "decision
variable B" is 1 and is therefore a binary decision variable i.e.
IF( "decision variable C" > 0, then 0, else 1)

Thank you so much.
Alison

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[Robert Fourer-2]

Alison,

Your "IF ..." looks like an expression that you would like to have in your
constraint, but it is not the whole constraint -- there is no "=" or ">=" or
"<=" in it.  You would need to state the whole constraint in order for
someone to say how it could be reformulated for a solver.  Also you would
need to specify for every decision variable whether it is integer or
continuous, and what its lower and upper bounds are.

Bob Fourer
4er@...

> -----Original Message-----
> From: AlisonT [mailto:alisonthomas@...]
> Sent: Sunday, November 01, 2009 11:09 AM
> To: AMPL Modeling Language
> Subject: [AMPL 2939] Constraint formulation
>
> Hi,
>
> Can someone please advise me how I should formulate a constraint that
> represents the following:
>
> IF( "decision variable A" > 0, then 0, else "decision variable B")
>
> I also need to formulate a similar constraint, where "decision
> variable B" is 1 and is therefore a binary decision variable i.e.
> IF( "decision variable C" > 0, then 0, else 1)
>
> Thank you so much.
> Alison
>

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[Paul A. Rubin]

I agree with Bob that your question could use some refinement, but I
think I know what you're getting at, so I'll take a shot.

I don't think you can do exactly what you want, but you may come
close.  Let M be a sufficiently large positive constant and epsilon a
sufficiently small (but strictly positive) constant, and let Y (a
variable) be the output of your first IF.  Introduce a new binary
variable Z together with the following constraints:

(1) epsilon*Z <= A <= M*Z
(2) B - M*Z <= Y <= B + M*Z
(3) -M*(1-Z) <= Y <= M*(1-Z).

A <= 0 forces Z = 0 in the left part of (1), which forces Y = B in (2)
and makes (3) vacuous.  A >= epsilon forces Z = 1 in the right side of
(1), which forces Y = 0 in (3) and makes (2) vacuous.  Besides
introducing a binary variable, there is one catch: 0 < A < epsilon has
been excised from the feasible region.  So this won't work if
arbitrarily small but positive values of A are legitimate contenders
for an optimal solution.

Incidentally, I used the same M in every constraint because I'm too
lazy to type subscripts, but you can (and probably should) use a
distinct M (the tightest value you can find that is guaranteed not to
cut off an optimal solution) in each constraint.

/Paul

On Nov 1, 12:09 pm, AlisonT <alisontho...@...> wrote:

> Hi,
>
> Can someone please advise me how I should formulate a constraint that
> represents the following:
>
> IF( "decision variable A" > 0, then 0, else "decision variable B")
>
> I also need to formulate a similar constraint, where "decision
> variable B" is 1 and is therefore a binary decision variable i.e.
> IF( "decision variable C" > 0, then 0, else 1)
>
> Thank you so much.
> Alison

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[alisonthomas]

Hi Bob and Paul,

thank you both for the replies.  I should have expressed my question
with more clarity - sorry for that.

Thank you Paul for your answer and explanation, which answers my
question perfectly.

Is there a list available of piece-meal linearisations such as this?

Thank you so much.
Alison

On Nov 5, 11:06 pm, Paul <ru...@...> wrote:

> I agree with Bob that your question could use some refinement, but I
> think I know what you're getting at, so I'll take a shot.
>
> I don't think you can do exactly what you want, but you may come
> close.  Let M be a sufficiently large positive constant and epsilon a
> sufficiently small (but strictly positive) constant, and let Y (a
> variable) be the output of your first IF.  Introduce a new binary
> variable Z together with the following constraints:
>
> (1) epsilon*Z <= A <= M*Z
> (2) B - M*Z <= Y <= B + M*Z
> (3) -M*(1-Z) <= Y <= M*(1-Z).
>
> A <= 0 forces Z = 0 in the left part of (1), which forces Y = B in (2)
> and makes (3) vacuous.  A >= epsilon forces Z = 1 in the right side of
> (1), which forces Y = 0 in (3) and makes (2) vacuous.  Besides
> introducing a binary variable, there is one catch: 0 < A < epsilon has
> been excised from the feasible region.  So this won't work if
> arbitrarily small but positive values of A are legitimate contenders
> for an optimal solution.
>
> Incidentally, I used the same M in every constraint because I'm too
> lazy to type subscripts, but you can (and probably should) use a
> distinct M (the tightest value you can find that is guaranteed not to
> cut off an optimal solution) in each constraint.
>
> /Paul
>
> On Nov 1, 12:09 pm, AlisonT <alisontho...@...> wrote:
>
>
>
> > Hi,
>
> > Can someone please advise me how I should formulate a constraint that
> > represents the following:
>
> > IF( "decision variable A" > 0, then 0, else "decision variable B")
>
> > I also need to formulate a similar constraint, where "decision
> > variable B" is 1 and is therefore a binary decision variable i.e.
> > IF( "decision variable C" > 0, then 0, else 1)
>
> > Thank you so much.
> > Alison- Hide quoted text -
>
> - Show quoted text -

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[Paul-432]

AlisonT wrote:

>
> Thank you Paul for your answer and explanation, which answers my
> question perfectly.

You're welcome.  First thing I've done perfectly in ages. ;-)
>
> Is there a list available of piece-meal linearisations such as this?

I'm not sure.  The closest I can come is a rather interesting paper by
John Hooker that you can get from his web site (http://
web.tepper.cmu.edu/jnh/milpmodeling2.pdf).  I downloaded it a while
back and have been meaning to read it, but <sigh>work keeps
interfering</sigh>.  It's somewhat mathematical, so if you are looking
for a "cheat sheet", this is definitely not the place to look.

/Paul

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