could you please explain the your logic with binary variable delta, for more general case, for instance, if y={a: x>0, b: x<0}.

If we assume \(y\) can be either \(a\) or \(b\) (constants) when \(x=0\), and \(x \in [L,U]\) (with \(L<0\) and \(U>0\)) then we can write

\[\begin{matrix} y = \begin{cases}a&x\ge 0\\b&x\le 0 \end{cases} & \Longrightarrow & \boxed{\begin{align}&L(1-\delta) \le x \le U \delta\\&y=a\delta + b (1-\delta)\\ &\delta \in \{0,1\} \end{align}} \end{matrix} \] |

I am not sure if the left part is formulated mathematically correctly. May be I should say:

\[y = \begin{cases} a&x> 0\\ b&x< 0 \\ a \text{ or } b & x=0 \end{cases}\] |

If we want to exclude \(x=0\) then we need to add some small numbers:

\[\begin{matrix} y = \begin{cases}a&x> 0\\b&x < 0 \end{cases} & \Longrightarrow & \boxed{\begin{align}&\varepsilon + (L-\varepsilon)(1-\delta) \le x \le –\varepsilon +(U+\varepsilon)\delta\\&y=a\delta + b (1-\delta)\\&\delta \in \{0,1\}\\ &\varepsilon=0.001\end{align}} \end{matrix} \] |

If \(a\) and \(b\) are variables (instead of constants) things become somewhat more complicated. I believe the following is correct:

\[\boxed{\begin{align} &\varepsilon + (L-\varepsilon)(1-\delta) \le x \le –\varepsilon +(U+\varepsilon)\delta\\ &a+M_1(1-\delta)\le y \le a+ M_2(1-\delta)\\ &b+M_3\delta\le y \le b+M_4\delta\\ &\delta \in \{0,1\}\\ &\varepsilon=0.001\\ &M_1 = b^{LO}-a^{UP}\\ &M_2 = b^{UP}-a^{LO}\\ &M_3 = a^{LO}-b^{UP}\\ &M_4 = a^{UP}-b^{LO} \end{align}}\] |

I used here \(a \in [a^{LO},a^{UP}]\) and \(b \in [b^{LO},b^{UP}]\).

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