Relationship between Optimal Solutions of a Linear Program and its Integer Counterpart with Binary Constraint Matrix

Question

Consider a matrix $\mathbf{A} \in \{0, 1\}^{K \times N}$ and a vector $\mathbf{b} \in \mathbb{Z}_{>0}^{K \times 1}$ (i.e., elements of $\mathbf{A}$ are either $0$ or $1$, and elements of $\mathbf{b}$ are positive integers). We define two optimization problems:

Problem 1 (Linear Program) \begin{align*} & \text{Minimize:} & & \sum_{i=1}^{N} x_i \\ & \text{Subject to:} & & \mathbf{A} \mathbf{x} \ge \mathbf{b} \\ & & & x_i \geq 0 \quad \forall i \in \{1, \ldots, N\} \end{align*} Let the optimal solution be $\mathbf{x}^* = [x_1^*, \ldots, x_N^*]^{\mathrm{T}}$, and define $R_1 := \sum_{i=1}^{N} x_i^*$.

Problem 2 (Integer Program) \begin{align*} & \text{Minimize:} & & \sum_{i=1}^{N} y_i \\ & \text{Subject to:} & & \mathbf{A} \mathbf{y} \ge \mathbf{b} \\ & & & y_i \in \mathbb{N} \quad \forall i \in \{1, \ldots, N\} \end{align*} Let the optimal solution be $\mathbf{y}^* = [y_1^*, \ldots, y_N^*]^{\mathrm{T}}$, and define $R_2 := \sum_{i=1}^{N} y_i^*$.

Question:

Is it always true that $R_2 = \lceil R_1 \rceil$?

I have conducted extensive numerical simulations with randomly generated matrices $\mathbf{A} \in \{0,1\}^{K \times N}$ and vectors $\mathbf{b} \in \mathbb{Z}_{>0}^K$, and I have yet to find a counterexample. Any rigorous proof or a counterexample would be greatly appreciated.

This question has also been posted on [Operations Research Stack Exchange]https://or.stackexchange.com/questions/13181/relationship-between-optimal-solutions-of-a-linear-program-and-its-integer-count

Answer 1

For a simple example where the ILP and its linear relaxation can be arbitrarily far apart, consider the following problem with $3N$ variables:

\begin{align*} & \text{Minimize:} & & \sum_{i=1}^N (x_i + y_i + z_i) \\ & \text{Subject to:} & & x_i + y_i \ge 1 & \quad \forall i \in \{1, \ldots, N\}\\ & & & x_i + z_i \ge 1 & \quad \forall i \in \{1, \ldots, N\}\\ & & & y_i + z_i \ge 1 & \quad \forall i \in \{1, \ldots, N\}\\ & & & x_i, y_i, z_i \geq 0 & \quad \forall i \in \{1, \ldots, N\} \end{align*}

Here, $R_1 = \frac32 N$ (obtained by setting $x_i = y_i = z_i = \frac12$ for all $i$) but $R_2 = 2N$ (since at least two of $x_i, y_i, z_i$ must be at least $1$ for each $i$.) As soon as $N\ge 2$, we have $R_2 > \lceil R_1 \rceil$: when $N=2$, we have $R_1 = \lceil R_1 \rceil = 3$ but $R_2 = 4$.

In general, as soon as there is any gap between $R_1$ and $R_2$, we can combine many independent copies of the same problem to make that gap arbitrarily large, until eventually it exceeds $1$ and so $R_2$ must be bigger than $\lceil R_1 \rceil$. Here, this is done with a three-variable problem, which is the smallest case for which a gap occurs.