Dear Friends and colleagues
I have an optimization in which I have a nonlinear term in the following form:
x(t)* a(k)
where, x and a are variables. a is a binery variable and the sets in which each of the variables are defined is not the same. Would you please suggest me a method that I can use to handle this term and transfer my model to a mixed integer linear programming?
Thank you for your suggestions.
I agree with Mr. Iago Augusto Carvalho. We need explanations about those questions.
Can you also provide information about whether each of x(t) and a(k) are linear or nonlinear?
Best regards,
Ivan K. Singgih
Dear Lago and dear Ivan
Thank you very much for your kind response. This non-linear term is a part of an equality constraint. Our objective function is linear. The other constraints even equalities or inequalities are also linear. But, we have two terms of nonlinearity such as " x(t)* a(k) " in one of the equality constraints. Also our problem is a kind of minimization. x(t) and a(k) are two variables.
@Ivan Kristianto Singgih @Iago Augusto Carvalho
Look up big-M constraints. They can be used to make a bound dependent on the value of a binary variable.
You can introduce a new variable 'z(t,k)' to hold the value of x(t)*a(k), and use that in the original equality. And to set the value of 'z', create constraints for its bounds with big-M:
"z(t,k) >= x(t) - M*(1-a(k))" and
"z(t,k) = -M*a(k)" and
"z(t,k)
You can use a new variable v(k,t) instead of the product, adding the following constraints:
v(k,t)
... I was assuming that the variables are binary. If they are not, you have to use the big M
Dear Mohammad Reza Salehizadeh, the answers of Olivér Ősz and Vladimir Marianov seem interesting to me. Based on these proposals, I would like to offer you the following, which I believe will have the advantage of not finding major problems for the implementation, as will happen with the previous two.
So a bit based on Olivér Ősz and Vladimir Marianov, I propose the following change of variable:
Y1(t, k) = x(t)*a(k) this implies that always Y1(t, k) it will be equal to x(t)
To achieve that x(t) can be zero, it is used
Y2(t, k) = M*a(k) where M is a large enough value.
Finally:
Y1(t, k) ≤ Y2(t, k) with this it guarantees that Y1(t, k) will be zero when a(k) is zero.
When a (k) is 1, Y1(t, k) will take the value of x(t), since it is being minimized.
This procedure is the one used when working with fixed cost problems, I recommend that you review the work of Gamboa (https://www.researchgate.net/publication/318198709_Decision_Support_System_to_facilitate_teaching_of_Integer_Programming_Sistema_de_apoyo_a_las_decisiones_para_facilitar_la_ensenanza_de_Programacion_Entera ).
If you have other similar combinations in your problem, you can repeat the same process using Y3 and Y4; Y5 and Y6, and so on.
I hope it will be useful and many successes in your research.
Best regards,
José Hernández.
Hi all,
Let me offer a last version (José's version seems to have a nonlinear constraint, is that right?):
y(k,t) = x(t) - M*(1-a(k)) (2)
The value of M could be = max(t) {x(t)}
Equation (1) makes y(k,t) = 0 whenever a(k) = 0; and it allows it to take any value in the range of x(t) when a(k) = 1
Equation (2) can be read as:
y(k,t) >= x(t) - M when a(k) = 0 (recall that M is the maximum value that x(t) can take, so this equation is really y(k,t) >= 0)
and
y(k,t) >= x(t) when a(k) = 1
Since it is a minimization, y(k,t) will take the value of x(t) whenever a(k) = 1.
Hope it helps.
Cheers,
Vlad
Very good, dear Vladimir Marianov, our proposals should reach similar results.
There is no problem in my approach, there is no non-linear constraint, because the change of variable is done outside the problem, it is not worked as a restriction.
When the problem is solved, if necessary, the variable change is returned, to know the value of x(t). The change of variable, is just necessary to be able to remove from the problem the non-linear restriction x(t)*a(k).
Thanks for the observation.
Best regards,
José Hernández.
Depends on why you would have to use the product.
If the product should just serve to have the production amount c set to zero if the oroduction unit is off - switching state reflected by a.
This means that x is is a so-called semi-continuous variable (available e.g. in CPLEX, where the modeling is done by the solver).
If you want to model such thing yourself, it's simple if x has a lober bound lb and an upper bound ub:
a*lb
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