Is \(x\) bounded? If so \((L \le x \le U)\), you replace with it with a sum of binary variables: $$x = \sum_{i=L}^U i x_i$$ Exactly one of the \(x_i\) is nonzero: $$\sum_{i=L}^U x_i = 1$$ Therefore, you can rewrite your constraint as: $$y = \sum_{i=L}^U \frac{N}{i} x_i$$
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rschwarz |