Meaning of the bar over $\bf{c}'$ in $\bf{\bar{c}}'=\bf c' -\bf c'_B \bf B^{-1} \bf A\geq \bf 0$?

Question

I am trying to understand the page 87 Bertimas about Linear Programming. The author uses bolding and bars -- now I am starting to think that the bar means something else to vector, bolding apparently means vector and bar means something else. Please, explain the notation in the equation below.

$$\bf{\bar{c}}'=\bf c' -\bf c'_B \bf B^{-1} \bf A\geq \bf 0$$

Book on the page 87 ("Introduction to Linear Optimization")

Definition 3.3 A basis matrix $\bf B$ is said to be optimal if

$\bf{(a)}\text{ } \bf{B^{-1}}\bf{b}\geq \bf 0$, and

$\bf{(b)}\text{ } \bf{\bar{c}}'=\bf{c}'-\bf{c}'_B\bf{B}^{-1}\bf{A}\geq \bf{0}'.$

Answer 1

It means Reduced cost.

Let's consider just one element $\bar{\bf{c}}'$ where the notation is more apparent. So $\bf{\bar{c_j}}'=c_j' -\bf c'_B \bf B^{-1} \bf A_j$ where $c_j'\in\mathbb R$, a scalar. This means $\bar{\bf{c_j}}'$ must be a scalar ie $\bar{\bf{c_j}}'\in\mathbb R$.

In Summary, the bar over $c$ does not mean a vector! For example, the $\bf{\bar{c_j}}'$ means a reduced cost, a scalar, for the variable $x_j$. The bolding means a vector but with a subscript a scalar i.e. taking a value out of a vector. The other vector notation is a "half arrow" over c, like: $\vec{c}$. Maybe that was the confusion.

Examples

1. Reduced cost vector in the phase I of the Two-phase simplex?

2. Reduced cost in the Phase II of the two-phase Simplex?

3. Example about the Reduced cost in the Big-M method?