MIP with infinite number of feasible solutions

Answered

Xinshuo Yang

October 25, 2022 04:47

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I 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? 

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• Official comment

  

  Simranjit Kaur

  • Gurobi Staff

  November 16, 2025 23:26

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  Jaromił Najman

  • Gurobi Staff

  October 25, 2022 08:01 Edited

  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, 
  Jaromił

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