Excluding Zeroes

GAMS
1

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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2

Hi Bruno

You could define a third variable m(i) and an equation eq_m(i):



Eq_m(i)… m(i)*b(i) =E= a(i);



Obj… z= E= sum(I, m(i));

Cheers

Renger



From: gamsworld@googlegroups.com [mailto:gamsworld@googlegroups.com] On Behalf Of Bruno Hannud
Sent: Samstag, 6. Februar 2016 00:19
To: gamsworld
Subject: Excluding Zeroes



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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3

Tks a lot for your reply Renger.

I have a query. If b(i)=0, isn’t m(i) undefined? Or will GAMS automatically attribute the value zero to m(i)?

Rgds

Bruno

On Sunday, February 7, 2016 at 6:59:10 PM UTC-2, Renger van Nieuwkoop wrote:

Hi Bruno

You could define a third variable m(i) and an equation eq_m(i):



Eq_m(i)… m(i)*b(i) =E= a(i);



Obj… z= E= sum(I, m(i));

Cheers

Renger



From: gams...@googlegroups.com [mailto:gams...@googlegroups.com] On Behalf Of Bruno Hannud
Sent: Samstag, 6. Februar 2016 00:19
To: gamsworld
Subject: Excluding Zeroes



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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4

Renger, I think I made a mistake.

If b(i)=0, the equation might be undefined (if a(i)=0) or will have no solution (if a(i) 0).

Rgds

Bruno

On Sunday, February 7, 2016 at 6:59:10 PM UTC-2, Renger van Nieuwkoop wrote:

Hi Bruno

You could define a third variable m(i) and an equation eq_m(i):



Eq_m(i)… m(i)*b(i) =E= a(i);



Obj… z= E= sum(I, m(i));

Cheers

Renger



From: gams...@googlegroups.com [mailto:gams...@googlegroups.com] On Behalf Of Bruno Hannud
Sent: Samstag, 6. Februar 2016 00:19
To: gamsworld
Subject: Excluding Zeroes



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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5

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” escribió:

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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6

Thanks Claudio.


Sent with Unibox

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” escribió:

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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7

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


Sent with Unibox

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” escribió:

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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8

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


Sent with Unibox

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"  escribió:

    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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9

Claudio thanks for your answer. In my problem:

b(i) is a variable.

If b(i) = 0, a(i) is not necessarily zero, but I must impose that a(i)/b(i) =0

meaning that in the case b(i)=0, we exlude these values of a(i)/b(i) from the sum(i, a(i)/b(i)).

Bruno

Sent with Unibox

On Feb 15, 2016, at 1:07 PM, Claudio Delpino wrote:

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


Sent with Unibox

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"  escribió:

    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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