Originally posted by: MachexC

Hi Alex,

I have a constraint as following, which don't be a constraint Big M.

B(i) = a*X(i) + b*Y(i) -c*F (i);

I want to convert to a logic constraint in OPL.

If F(i) = 0 then B(i) = a*X(i) + b*Y(i);

If F(i) = 1 then B(i) = a*X(i) + b*Y(i) - c;

Thank you for your help.

Mac

Hi,

you could try

B[i] == a*X[i] + b*Y[i] - c * (F[i]==1);

regards

https://www.linkedin.com/pulse/making-decision-optimization-simple-alex-fleischer/

Originally posted by: MachexC

Hi Alex,

For F(i) =1, we have B[i] == a*X[i] + b*Y[i] - c * (F[i]==1);

In this case, F(i) = 0, how is?

Thank you,

Mac

Hi,

if you write

B[i] == a*X[i] + b*Y[i] - c * (F[i]==1);

then when you have F[i] equals 0, you get

B[i] == a*X[i] + b*Y[i]

regards

Originally posted by: MachexC

Hi Alex,

With such way, it is the same, we write as follow. Right?

forall i = 1..n {

F[i] = 1 => B[i] == a*X[i] + b*Y[i] - c ||

F[i] = 0 => B[i] == a*X[i] + b*Y[i] ;

}

Thank you,

Bests and regards,

Mac

Hi,

you may write

If F(i) = 0 then B(i) = a*X(i) + b*Y(i);

If F(i) = 1 then B(i) = a*X(i) + b*Y(i) - c;

as

(F[i]==0) => (B[i]==a*X[i] + b*Y[i] );

(F[i]==1) => (B[i]==a*X[i] + b*Y[i] - c);

regards

Originally posted by: MachexC

Hi Alex,

Thank you for your suggestion.

When I use case 1, then model run faster but the results don't see any F[i] = 0. This is model reads only values F[v] =1. Meanwhile, the case 2 is run with the very slow runtime, the engine log can not see. Could you explain for me know?

Case 1:

forall i = 1..n {

F[i] = 1 => B[i] == a*X[i] + b*Y[i] - c;

}

Case 2:

forall i = 1..n {

(F[i]==1) => (B[i]==a*X[i] + b*Y[i] - c);

F[i]==0) => (B[i]==a*X[i] + b*Y[i] );

}

Bests,

Mac

Hi,

that is why I would write

B[i] == a*X[i] + b*Y[i] - c * (F[i]==1);

Your case 1 and 2 are not equivalent with each other. Case 2 is the same as what I suggest

regards