We can model disjunctions (note I am not looking for convex hull) of $t$ unbounded convex polyhedra given by $A^{(1)}x^{(1)}\leq b^{(1)}$,$\dots$,$A^{(t)}x^{(t)}\leq b^{(t)}$ exactly with a mixed integer linear or convex program with $O(t)$ integer variables and perhaps $2^{O(t)}$ real variables?

Let the $i$th convex polyhedra be $A^{(i)}x^{(i)}\leq b^{(i)}$ where $A^{(i)}\in\mathbb R^{m_i\times n}$ and $b^{(i)}\in\mathbb R^{m_i}$ are fixed. Then if we introduce binary variables $y_1,\dots,y_t\in\{0,1\}$ and a real vector $x=\sum_{i=1}^tx^{(i)}\in\mathbb R^n$ then
$$A^{(i)}x^{(i)}\leq b^{(i)}y_i$$
$$y_1+\dots+y_t=1$$
$$x=\sum_{i=1}^tx^{(i)}\in\mathbb R^n$$
suffices.

However the trick breaks down if $b^{(i)}=0$ at $i\in\{1,\dots,t\}$. That is if $b^{(i)}$ are $0$ vectors then the trick breaks down.

>If the $t$ polytopes are given by $A^{(i)}x^{(i)}\leq0$ where $A^{(i)}\in\mathbb R^{m_i\times n}$ and $b^{(i)}\in\mathbb R^{m_i}$ are fixed then how do we model unions? Is there a standard trick with at least convex convex constraints introduced?


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This is what I am thinking for the case each entry of $x^{(i)}_j\in\mathbb R$ of vectors $x^{(i)}\in\mathbb R^n$ satisfy $0\leq x^{(i)}_j\leq1$.
$$B^{(i)}x^{(i)}\leq0$$
$$y_1+\dots+y_t=1$$
$$B^{(i)}=y_iA^{(i)}$$
$$B=B^{(1)}+\dots+B^{(t)}$$
$$x=\sum_{i=1}^tx^{(i)}\in\mathbb R^n$$
$$\mbox{//AND done with linear programming}$$
$$//\max(0,a+b-1)\le\mbox{AND}(a,b)\le\min(a,b,1)$$
$$x=\sum_{i=1}^t\mbox{AND}(y_i,x^{(i)})\in\mathbb R^n$$
seems to work if vectors $x^{(i)}$ are non-negative and each entry is in $[0,1]$.

By scaling above trick if it works then it also works for case  $0\leq x^{(i)}_j\leq\mbox{B}$ for a bound $B$ by scaling $A^{(i)}$.

1. Is my reasoning correct?

2. Is there a general way for arbitrary $x^{(i)}\in\mathbb R^{n}$?