[英]How to add a constraint to my optimization?
I am working on formulating an optimization problem where I have a 2-D matrix A. 我正在制定一个具有二维矩阵A的优化问题。
A= [0 f1 0 f2]
[f3 f3 0 0]
.........
And I have another 2-D matrix B that I should fill. 我还有另一个二维矩阵B应该填写。 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. B具有与A相同的大小。如果a_ij = 0(A的元素),则我需要b_ij(B的元素)为零;如果a_ij不为零,则我需要b_ij大于零且小于或等于a_ij。
How can I represent this in my formulation? 我如何在我的公式中表示这一点? I have added this constraint/condition: 我添加了此约束/条件:
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. 但这不满足以下条件:当a_ij不等于零时,b_ij不等于零。 Any help? 有什么帮助吗?
If all elements are positive, keep the smallest element of each matrix by doing an element by element comparison : 如果所有元素都是正数,则通过逐元素比较来保持每个矩阵的最小元素:
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: 或者,创建一个逻辑矩阵,指示是否满足条件,然后将每个元素与矩阵B
相乘,仅保留满足条件的元素,将其他元素设置为零:
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. 然后,使B
元素小于或等于A
元素,否则将其替换为A
的值。
B = B.*(B<=A) + A.*(B>A) )
This option lets you generalize your constraint. 此选项使您可以概括约束。
You indicate needing elements of b_ij to be greater than zero if elements of a_ij are greater than zero. 如果a_ij的元素大于零,则指示需要b_ij的元素大于零。 An option is to use the function max
to ensure that all elements of B
are positive. 一种选择是使用函数max
来确保B
所有元素均为正。
B = max(1e-2,B); % exact value is yours to set.
This step is up to you and depend on your problem. 此步骤由您决定,并取决于您的问题。
You want to implement the implication 您要实现含义
a = 0 => b = 0
a <> 0 => 0 < b <= a
If a
is (constant) data this is trivial. 如果a
是(恒定)数据,那么这是微不足道的。 If a
is a variable then things are not so easy. 如果a
是变量,那么事情就不那么容易了。
You implemented part of the implications as 您实现了部分含义为
b <= a
This implies a
is non-negative: a>=0
. 这意味着a
为非负数: a>=0
。 It also implies b
is non-negative. 这也意味着b
为非负数。 The remaining implication a>0 => b>0
can now be implemented as 剩下的含义a>0 => b>0
现在可以实现为
a <= δ * 1000
b >= δ / 1000
δ in {0,1}
Many MIP solvers support indicator constraints. 许多MIP求解器支持指标约束。 That would allow you to say: 那可以让你说:
δ = 0 -> a = 0
δ = 1 -> b >= 0.001
δ in {0,1}
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