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4

I assume that $y$ is constrained to the interval $[0,1]$. (You did not state this explicitly.) Let's assume that you have selected values $r_i$ such that $0=r_1 < r_2 < \dots < r_n = 1.$ If your solver supports SOS2 constraints, you can make $w_1, \dots, w_n$ nonnegative variables with the constraint $\sum_i w_i = 1$ and declare $\lbrace w_1,\dots,...


4

Adding the last constraint is required to guarantee that only one of the $r_i$ values is selected for $y$. However, you need an additional constraint to make the relationship between $y$, its piecewise linearisation variables and the remaining of the problem constraints (especially $y=f(x)$), such as: $$y = \sum_{i=1}^{n} r_i \times w_i$$


1

Have you tried solving the problem directly? This is simply a quadratic non-convex constraint: $$ y = \frac{x^2}{1-x}\implies y - yx = x^2 $$ So e.g. using Gurobi and Python, you can write: import gurobipy as gp m = gp.Model() x = m.addVar(0, x_ub) y = m.addVar(0, y_ub) m.addConstr(y - y*x == x*x)


3

I assume $X\le1$, since the formulation is infeasible for $x > 1$. What you are doing is a form of piecewise linear approximation of the function $f(x) = x^2/(1-x)$. Choosing $n$ is a matter of trial and error. Obviously, the larger $n$ is the more accurate the approximation is. Also obvious is that larger $n$ means longer solution time (and more memory ...


5

The two denominators are equal to $1$, so just omit the denominators, yielding a linear constraint.


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