Bruno: I’m not an expert on that approach, someone else around here can help you better, but:

Is b(i) a parameter or a variable ?

If it is a variable: I read from your last mail that b(i)=0 => a(i)=0. If that’s the case, I guess you’ll be fine (if the model is not large) introducing binary variables and use something like a “big M” formulation. Like I said before, I’m not an expert. If I remember correctly, EMP has some facilities for that kind of reformulations, so perhaps lloking through its manual will give you some ideas.

Best Regards

Claudio

On Mon, Feb 15, 2016 at 7:26 AM, Bruno Hannud wrote:

Dear Claudio, in my model, if b(i)=0, I attribute a(i)/b(i) = 0, otherwise, the answer is the normal one. Does this still classify in disjunctive programming?

Thanks.

Bruno

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On Feb 14, 2016, at 8:11 PM, Claudio Delpino wrote:
Hi Bruno: You need to think what values of b are allowed (feasible). If you are asking for a(i)/b(i) to be equal to a scalar (R), then b cannot be zero for any index. If the solver takes you there, you can use a small artificial bound (b.lo(i) =1E-6).
If you want different equations in the model according to the value of variables (if b=0 is allowed, and eq1 doesn't apply to those indexes ) , that is called "disjunctive programming", look it up.
Best regards!
Claudio
El 07/02/2016 16:38, "Bruno Hannud" escribiÃ³:
Dear all.
I have an equation which looks like this:
eq1.. R=E= a(i)/b(i);
And an objective function:
obj.. z =E= sum(i, a(i)/b(i));
Is there a way to sum only the terms with b(i) not equal to zero? I tried using conditionals on the R equation, but when I try to run I get an
***Error 52 Endogenous $-control not allowed.
Any help is appreciated.
Tks
Bruno
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