Formulation for IP of large OR statement which gives a good linear relaxation Let $N$ be a very large number. I want a good way to program that $x$ should be one if and only if one of $x_i$ is equal to one.We can write the following Integer Programming problem:
\begin{align*}
\max x\\
x \leq \sum_{i=1}^N x_i\\
x_i \in \{0,1\}\\
x\in \{0,1\}
\end{align*}
Then clearly, we have that $x$ will be one whenever at least one $x_i$ is equal to one.
My problem with this is that when we take the Linear relaxation of this problem we obtain:
\begin{align*}
\max x\\
x \leq \sum_{i=1}^N x_i,\\
0 \leq x_i\\
0\leq x
\end{align*}
if now all $x_i = \frac{1}{N}$ we still obtain $x=1$ while for very large $N$ this is extremely inaccurate to what we actually want. I am thus looking for a way to rewrite the IP problem such that the linear relaxation behaves in a way that at least one $x_i$ needs to be large in order for $x$ to be large. 
Notes :


*

*This is only a part of a much larger IP problem, therefore I can not simply change the max with a min etc.

*This $x$ is used as follows : I have in my problem an $x$ and a $y$ : the $x$ is $1$ when any of the $x_i$ is equal to $1$ while the $y$ is equal to $1$ if any of the $y_i$ is equal to one. Then we further have a $z$ which is one if both $x$ and $y$ are equal to $1$, we have:
$z\leq x$
$z\leq y$
and we want to maximize $z$ (indeed then we get $z=1$ iff $x=1$ AND $y=1$).
 A: Approach #1 :
Use Dantzig Wolfe decomposition, which is always at least as tight as the initial formulation. To do this, define the set $\Omega$ of combinations that define your "master problem" :
$$
\Omega := \{(x_1,x_2,...,x_N,x) \in \mathbb{B}^{N+1} \; | \; x=1 \Leftrightarrow \; \exists i\in | x_i=1 \}
$$
For example $(0,...,0) \in \Omega$, as well as $(1,...,1)$, or $(1,0,1,...,1)$.
And let $\lambda_i$ be a binary variable that takes value $1$ if and only if combination $i \in \Omega$ is selected.
Your problem can then be formulated as follows :
$$
\max \; \sum_{i\in \Omega | x =1} \lambda_i
$$
subject to 
$$
\sum_{i\in \Omega } \lambda_i = 1 \\
\lambda_i \in \{0,1\}
$$
You will certainly have to add the other constraints (that you have not explicitey written in your question).
You can easily generate $\Omega$ explicitely beforehand, or dynamically with column generation.
Approach #2 :
Transform the problem into the following minimization problem
$$
\min z
$$
subject to
\begin{align*}
&x_i \le x \quad \forall i=1,...,N \\
&y_i \le x \quad \forall i=1,...,N \\
&x +y  \le 2z \\
&x,y,z \in \mathbb{B} \\
&x_i,y_i  \in \mathbb{B}
\end{align*}
The constraint $x+y \le 2z$ makes sure that when $x=y=1$, $z$ takes value $1$. Otherwise, since you are minimizing $z$ it will take value $0$.
This formulation is interesting as the solution with $x_i=1/N$ and $x=1$ is not optimal when relaxing integrality constraints. Indeed, since you are minimizing $z$, if $x_i=1/N$, $x$ will also take value $1/N$ (and not $1$), in order for $z$ to be minimized in the constraint $x+y\le 2z$.
A: You can enforce the relationship without depending on the objective or introducing additional variables.  Rewriting your logical proposition in conjunctive normal form somewhat automatically yields linear constraints:
\begin{equation}
x \iff \bigvee_i x_i \\
\left(x \implies \bigvee_i x_i\right) \bigwedge \left(\bigvee_i x_i \implies x\right) \\
\left(\neg x \lor \bigvee_i x_i\right) \bigwedge \left(\neg \bigvee_i x_i \lor x\right) \\
\left(\neg x \lor \bigvee_i x_i\right) \bigwedge \left(\bigwedge_i \neg x_i \lor x\right) \\
\left(\neg x \lor \bigvee_i x_i\right) \bigwedge \left(\bigwedge_i (\neg x_i \lor x)\right) \\
\left(1 - x + \sum_i x_i \ge 1\right) \bigwedge \left(\bigwedge_i (1 - x_i + x \ge 1)\right) \\
\left(x \le \sum_i x_i\right) \bigwedge \left(\bigwedge_i (x_i \le x)\right)
\end{equation}
That is,
\begin{align}
x &\le \sum_i x_i\\
x_i &\le x &&\text{for all $i$}
\end{align}
