MIP with infinite number of feasible solutions
AnsweredI have a MIP where two continuous variables x and y are equivalent. So in the solution the sum (denoted by s) of x and y is determined/fixed, but there are infinite many choices of feasible solutions. I have some prior knowledge about the model and thus seek for a solution where x is close to s and y is close to 0, I will set both Start and VarHintVal to guide the algorithm to converge to the desired solution. However, the returned solution is always x = 0 while y = s.
Did I misunderstand the meaning of Start and VarHintVal? What is the best way to achieve this?

Did I misunderstand the meaning of Start and VarHintVal? What is the best way to achieve this?
As you mentioned yourself, Start and VarHintVal only guide the solver to the solution. They do not guarantee that the final solution is anywhere near the values provided by Start and VarHintVal.
You could introduce a penalty term to your objective for \(x\) being far away from \(s\) and \(y\) being far away from \(0\). Or alternatively, if you know that your assumption definitely holds for the optimal solution point, you could introduce bounds \(s  \delta \leq x \leq s + \delta \) and \( \delta \leq y \leq \delta\) with \(\delta > 0\).
Best regards,
Jaromił0
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