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Linear Programming Problem in Equation Form 📂Optimization

Linear Programming Problem in Equation Form

Definition 1

For matrices ARm×nA \in \mathbb{R}^{m \times n}, bRm×1\mathbf{b} \in \mathbb{R}^{m \times 1}, and cRn\mathbf{c} \in \mathbb{R}^{n}, the following linear programming problem is called in standard form or equational form. MaximizecTxsubject toAx=bx0 \begin{matrix} \text{Maximize} & \mathbf{c}^{T} \mathbf{x} \\ \text{subject to} & A \mathbf{x} = \mathbf{b} \\ & \mathbf{x} \ge \mathbf{0} \end{matrix}


Description

Conversion to Standard Form

In principle, any linear programming problem can be converted into standard form. Consider the following maximization problem, for example.

Maximize3x12x2subject to2x1x24x1+3x25x20 \begin{matrix} \text{Maximize} & 3 x_{1} - 2 x_{2} \\ \text{subject to} & 2 x_{1} - x_{2} \le 4 \\ & x_{1} + 3 x_{2} \ge 5 \\ & x_{2} \ge 0 \end{matrix}

The basic idea of conversion is ‘consolidating inequalities’. The first constraint 2x1x24 2 x_{1} - x_{2} \le 4 can in fact be represented as 2x1x2+x3=4 2 x_{1} - x_{2} + x_{3} = 4 without any issue, regarding some x30x_{3} \ge 0. Because 2x1x22 x_{1} - x_{2} supplements the amount by which 44 is smaller. Variables that serve to loosen the equation in this manner are called slack variables. The second constraint x1+3x25 x_{1} + 3 x_{2} \ge 5 can be handled by multiplying both sides by 1-1 and introducing a new slack variable x4x_{4}, just as with the first constraint, becoming x13x2+x4=5 - x_{1} - 3 x_{2} + x_{4} = - 5 . At a glance, Maximize3x12x2subject to2x1x2+x3=4x13x2+x4=5x2,x3,x40 \begin{matrix} \text{Maximize} & 3 x_{1} - 2 x_{2} \\ \text{subject to} & 2 x_{1} - x_{2} + x_{3} = 4 \\ & - x_{1} - 3 x_{2} + x_{4} = - 5 \\ & x_{2}, x_{3}, x_{4} \ge 0 \end{matrix} seems to meet the conditions for equational form, but since there’s yet no constraint for x10x_{1} \ge 0, it needs to be accommodated. Given that x1x_{1} is allowed to be any real number, breaking it into the difference x1=y1z1x_{1} = y_{1} - z_{1} of two positive numbers removes x1x_{1} from the equation, allowing it to be presented in equational form as follows. Maximize3(y1z1)2x2subject to2(y1z1)x2+x3=4(y1z1)3x2+x4=5y1,z1,x2,x3,x40 \begin{matrix} \text{Maximize} & 3 \left( y_{1} - z_{1} \right) - 2 x_{2} \\ \text{subject to} & 2 \left( y_{1} - z_{1} \right) - x_{2} + x_{3} = 4 \\ & - \left( y_{1} - z_{1} \right) - 3 x_{2} + x_{4} = - 5 \\ & y_{1}, z_{1}, x_{2}, x_{3}, x_{4} \ge 0 \end{matrix}

Geometry of Equational Form

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Equation Ax=bA \mathbf{x} = \mathbf{b} forms a general plane, a hyperplane, and taking only the part that forms a cone shape within the Euclidean space according to condition x0\mathbf{x} \ge \mathbf{0}. With the introduction of slack variables, the dimension increases, but the space to be explored for optimization is significantly simplified.

This solution space is none other than a simplex, and herein lies the idea for the simplex method.


  1. Matousek. (2007). Understanding and Using Linear Programming: p41. ↩︎