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I am trying to represent the following inequality:

$$\frac{x}{1-x} \leq y \qquad\mathrm{with}\qquad 0<x<1$$

The function on the left is convex (its second derivative is always positive over the domain $0<x<1$), meaning that the described set is convex. (Without the domain constraint, however, the domain is far from convex…) Drawing it gives the same conclusion:

The set described by this inequality

However, I cannot find a DCP representation of this set (even using nonstandard cones).

I can do it with a slight modification (replacing $x$ with $x^2$, two quad-over-lin constraints do the trick, CVX also accepts the constraint as $y\geq\frac{x^2}{z}$ and $z\leq 1-x^2$), but it is not describing the same set, and the difference is really significant in my case.

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2 Answers 2

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Here is an alternative answer.

Clearly, the problem is equivalent to $$ \begin{array}{rcl} \frac{1-t}{t} & \leq & y \\ 1-x & = & t \\ \end{array} $$ which in turn is equivalent to $$ \begin{array}{rcl} \frac{1}{t} -1 & \leq & y \\ 1-x & = & t \\ \end{array} $$ This is clearly SOCP representable.

You should end up with $$ \begin{array}{rcl} 2 (1-x) (1+y) & \geq & \sqrt{2}^2 \\ 1-x & \geq & 0 \\ 1+y & \geq & 0 \\ \end{array} $$ which is the same as saying $$ \left [ \begin{array}{c} 1-x \\ 1+y \\ \sqrt{2} \\ \end{array} \right ] \in RQ $$ where RQ is a rotated quadratic cone as defined in the Mosek modelling cook book.

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  • $\begingroup$ Thanks for your detailed answer :)! $\endgroup$
    – dourouc05
    Apr 28, 2020 at 17:49
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You can model your constraint with the second order cone constraint $$\sqrt{2^2 + (x+y)^2} \le 2-x+y,$$ and a lower bound on $x$.

I found this by first multiplying both sides by $1-x$ to obtain $(1-x)y - x >= 0$. As we have a quadratic inequality with a convex feasible region, this hints to a quadratic or second order cone.

Rewriting this constraint gives $(1-x)(1+y) \ge 1$. Then I used Equation (7) in this paper to turn the constraint into a second order cone constraint.

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    $\begingroup$ Thanks for your answer :)! $\endgroup$
    – dourouc05
    Apr 28, 2020 at 17:49

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