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# How to add a constraint to my optimization?

Jhon Kami
1#
Jhon Kami Published in 2017-12-07 09:42:07Z
 I am working on formulating an optimization problem where I have a 2-D matrix A.  A= [0 f1 0 f2] [f3 f3 0 0] .........  And I have another 2-D matrix B that I should fill. B has the same size of A. I need b_ij (element of B) to be zero if a_ij=0 (element of A) and I need b_ij to be greater than zero and less than or equal to a_ij if a_ij is not zero. How can I represent this in my formulation? I have added this constraint/condition:  b_ij<=a_ij  But this does not satisfy the condition that states that b_ij is not equal zero when a_ij is not equal zero. Any help?
Laure
2#
 If all elements are positive, keep the smallest element of each matrix by doing an element by element comparison : B2 = min(A,B)  Alternatively, create a logical matrix indicating if a condition is answered and multiply element by element with the matrix B , only elements who satisfy the condition remain, others are set to zero: B = B.*(A~=0)  Then keep elements of B that are smaller or equal to elements of A, and replace them by the value of A otherwise. B = B.*(B<=A) + A.*(B>A) )  This option lets you generalize your constraint. You indicate needing elements of b_ij to be greater than zero if elements of a_ij are greater than zero. An option is to use the function max to ensure that all elements of B are positive. B = max(1e-2,B); % exact value is yours to set.  This step is up to you and depend on your problem.
 You want to implement the implication a = 0 => b = 0 a <> 0 => 0 < b <= a  If a is (constant) data this is trivial. If a is a variable then things are not so easy. You implemented part of the implications as b <= a  This implies a is non-negative: a>=0. It also implies b is non-negative. The remaining implication a>0 => b>0 can now be implemented as a <= δ * 1000 b >= δ / 1000 δ in {0,1}  Many MIP solvers support indicator constraints. That would allow you to say: δ = 0 -> a = 0 δ = 1 -> b >= 0.001 δ in {0,1}