The Simplex Tableau Layout

A linear program in standard form is

$$\text{maximize } z = c_1 x_1 + c_2 x_2 + \cdots + c_n x_n$$

subject to

$$a_{i1} x_1 + a_{i2} x_2 + \cdots + a_{in} x_n \leq b_i, \quad i = 1, \ldots, m,$$

with $x_j \geq 0$ and $b_i \geq 0$. To work with the constraints algebraically, we introduce slack variables $s_i \geq 0$ that turn each inequality into an equality:

$$a_{i1} x_1 + \cdots + a_{in} x_n + s_i = b_i.$$

The simplex tableau packages every coefficient into one table:

$$\begin{array}{c|cccccc|c} \text{Basis} & x_1 & \cdots & x_n & s_1 & \cdots & s_m & \text{RHS} \\ \hline s_1 & a_{11} & \cdots & a_{1n} & 1 & \cdots & 0 & b_1 \\ \vdots & \vdots & & \vdots & \vdots & & \vdots & \vdots \\ s_m & a_{m1} & \cdots & a_{mn} & 0 & \cdots & 1 & b_m \\ \hline z & -c_1 & \cdots & -c_n & 0 & \cdots & 0 & 0 \end{array}$$

The pieces of the tableau:

• The leftmost Basis column lists the variables currently in the basis (one per constraint row).
• The middle columns hold the coefficients of every variable -- decision and slack alike.
• The rightmost RHS column holds the constraint constants $b_i$ in the upper block and the current objective value in the bottom-right cell.
• The bottom row -- the z-row -- stores the negated objective coefficients $-c_j$ under the decision variables, with $0$ under each basic (slack) variable.

At the start of the method, the slacks form the initial basis, so the slack columns form an identity block and the z-row is exactly $[-c_1, \ldots, -c_n, 0, \ldots, 0 \,|\, 0]$.