An optimization problem $(\mathcal{X} , f)$ is called linear (a linear optimization problem) if $\mathcal{X} \subset \R ^n$ is a polyhedron and $f: \R ^n \to \Rbar$ is a linear function.
Recall that $f$ is linear means there exists
$c \in \R ^n$ such that
\[
f(x) = c^\top x \quad \text{for all } x \in \R ^n
\] \[
\mathcal{X} = \Set{x \in \R ^n}{Ax \leq b}
\]
Given data $A \in \R ^{m \times n}$, $b \in
\R ^n$, $c \in \R ^n$, we want to find $x \in
\R ^d$ to
\[
\begin{aligned}
\text{minimize} &\quad c^\top x \\
\text{subject to} &\quad Ax \leq b
\end{aligned}
\]
In the context of linear optimization,
$c^\top x$ is often abbreviated $cx$.
As usual, $x \in \R ^n$ is called
feasible (a
feasible solution) if $Ax
\leq b$.
$x^\star \in \R ^n$ is called
optimal (an
optimal solution,
optimum solution) if
$c^\top x^\star \leq c^\top x$ for all $x \in
\R ^n$.
We sometimes denote the rows of $a$ by
$\bar{a}_i^\top \in \R ^n$ for $i = 1, \dots ,
m$, i.e.,
\[
A = \bmat{- & \bar{a}_1^\top & - \\ &\vdots& \\ - &
\bar{a}_m^\top & - } \in \R ^{m \times n}
\]
The problem $(A, b, c)$ is
infeasible if the
polyhedron
\[
P = \Set{x \in \R ^n}{Ax \leq b}
\]
Here’s an infeasible instance. Define
\[
A = \bmat{1 \\ -1} \quad b = \bmat{-1 \\ -1} \quad c =
\bmat{1}
\]
Here’s an unbounded instance: drop the second inequality constraint. Then we are interested in finding $x_1$ to minimize $x_1$ subject to $x_1 \leq -1$. Given $\alpha \in \R $, if $\alpha > 0$ pick $x_1 = -1$, else pick $2\alpha $. This problem is unbounded.
Here’s a simple example with an optimal
solution.
Modify $b = \bmat{1 & 0}^\top $.
Now we want to find $x_1$ so that
\[
x_1 \leq 1 \quad \text{ and } x_1 \geq 0
\]