# Decision Optimization

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## Transformation into 2nd order cone

• #### 1.  Transformation into 2nd order cone

Posted Mon November 29, 2021 07:37 AM
Dear all,

I am struggling with quadratic optimization right now. I have a problem with a linear objective and several linear constraints. However, one set of constraints takes the form \$c \geq a / (a+b)\$ which obviously is a quadratic constraint. Here, \$a,b\$ are non-negative real numbers, while \$c\$ takes any value between \$0\$ and \$1\$. On the selected domain, the quadratic matrix is positive semi-definite, but not on the entire area (so if \$a,b,c\$ take negative values).

Is it possible to still work with this constraint in CPLEX?
If push comes to shove, I will replace this with a few piece-wise linear functions, but I was hoping that there was a solution without any binary/integer variables.

Thank you!

• #### 2.  RE: Transformation into 2nd order cone

Posted Mon November 29, 2021 03:19 PM
Edited by Paul Rubin Mon November 29, 2021 03:29 PM
Presumably your first step is to convert the inequality to \$ac + bc - a \ge 0\$, which is quadratic. I'm not sure what you mean by the quadratic matrix being positive semidefinite "on the selected domain", since the Hessian matrix is a constant (and is not psd). In any case, the feasible region is clearly not convex. For instance, (a, b, c) = (0, 0.1, 0) and (a, b, c) = (10, 0, 1) both satisfy the constraint but their midpoint (5, 0.05, 0.5) does not. CPLEX can solve nonconvex QPs and MIQPs (I think), but as far as I know it cannot solve nonconvex MIQCPs (i.e., where the feasible region is nonconvex). So I think you are going to have to do a PWL approximation or perhaps switch to a nonconvex solver.

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Paul Rubin
Professor Emeritus
Michigan State University
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