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There are two real variables $x$ and $y$. The conditions are such that:

  • if $y\le 0$, then $x=0$
  • if $y>0$, then $x=y$

How to write linear equations or inequalities to satisfy both the conditions?

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  • $\begingroup$ Note this is also sometimes noted $x = y^+$ or $x = \max(0, y)$, and outloud "x is the positive part of y" $\endgroup$
    – Stef
    Apr 25 at 9:25

2 Answers 2

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Let $L_x,U_x$ denote a lower and upper bound for $x$, and $L_y,U_y$ denote a lower and upper bound for $y$.

Define an additional binary variable $\delta \in \{0,1\}$ and enforce the following:

$$ y \le 0 \implies \delta = 1 \implies x=0 \tag{1} $$

$$ y> 0 \implies \delta = 0 \implies x=y \tag{2} $$

Or equivalently via contrapositive:

$$ \delta = 0 \implies x=y \quad \text{and} \quad y> 0 \tag{3} $$

$$ \delta = 1 \implies x=0 \quad \text{and} \quad y\le 0 \tag{4} $$

$(3)$ can be modeled as follows: \begin{align} y+L_x\cdot\delta&\le x\le y+U_x\cdot\delta \tag{5}\\ \epsilon+L_y\cdot \delta &\le y \tag{6} \end{align}

$\epsilon$ is a small positive constant (tolerance).

And $(4)$ as follows: \begin{align} 0+L_x\cdot(1-\delta)\le x&\le 0+ U_x\cdot(1-\delta) \tag{7}\\ y&\le 0+U_y \cdot(1-\delta) \tag{8} \end{align}

As per RobPratt's comments:

  • $\epsilon$ can be set to $0$ (or omitted) as $(0,0)$ is a breakpoint
  • $L_x=0$ ; $L_y=:L$
  • $U_x=U_y=:U$
  • $(8)$ is redundant with $(5)$ and $(7)$

These simplifications yield:

\begin{align} L\cdot \delta &\le y \\ y&\le x \\ x&\le y+U\cdot\delta \\ x&\le U\cdot(1-\delta) \\ x&\ge 0 \end{align}

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    $\begingroup$ You did not define $\epsilon$, but you can omit it here because $(0,0)$ is a breakpoint. $\endgroup$
    – RobPratt
    Apr 24 at 12:55
  • $\begingroup$ You can simplify further by noting that $L_x=0$ and $U_x=U_y$. $\endgroup$
    – RobPratt
    Apr 24 at 16:16
  • $\begingroup$ You have some sign errors for $L_x$ and $L_y$, and the constraints that involve $y$ but not $x$ are linear combinations of the other constraints. $\endgroup$
    – RobPratt
    Apr 24 at 17:36
  • $\begingroup$ Thanks @RobPratt for the useful comments. $\endgroup$
    – Kuifje
    Apr 24 at 18:59
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    $\begingroup$ Glad to help. Looks better now, but you can strengthen $x \le y + U \cdot \delta$ to $x \le y - L \cdot \delta$. $\endgroup$
    – RobPratt
    Apr 24 at 19:31
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It is worth noting that the desired relationship can be expressed as $x=\max(y,0)$. You want to model a disjunction of two rays from the origin. Impose finite bounds \begin{align} 0 \le x \le U \tag1\\ L \le y \le U \tag2 \end{align} and model a disjunction of two line segments (from $(0,0)$ to $(U,U)$ and from $(0,0)$ to $(0,L)$). For both, we must have $$y \le x \tag3$$ Now introduce a binary decision variable $\delta$ and enforce \begin{align} \delta = 0 &\implies y \ge x\\ \delta = 1 &\implies x \le 0 \end{align} either by using indicator constraints or by explicitly imposing big-M constraints \begin{align} x - y &\le -L\delta \tag4\\ x &\le U(1-\delta) \tag5 \end{align}

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    $\begingroup$ For some models, it suffices to require $x \geq {\rm max}(y,0)$, which is easily linearized as $x \geq y$, $x \geq 0$. But if also $x \leq {\rm max}(y,0)$ must be enforced explicitly, then the binary $\delta$ must be introduced. $\endgroup$
    – 4er
    Apr 25 at 15:25

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