# Constraint Binding

Extreme Points and Vertex of Polyhedron ^{↵}

x ∈ P is an extreme point of P if

`x ̸= λ · y + (1 − λ) · z for all y, z ∈ P \ {x}, 0 ≤ λ ≤ 1,`

`i. e., x is not a convex combination of two other points in P.`

`x ∈ P is a vertex of P if`

`there is some c ∈ R n such that c T · x < cT · y for all y ∈ P \ {x},`

`i. e., x is the unique optimal solution to the LP min{c T · z | z ∈ P}.`

Maxiumum subset (x) of Linear Indepedent Vectors

`Amount of Matrices -1 = Dimension(x) ~ `

`Max no. Linearly Indepedent Vectors in space = x`

`Try combination [Top Left + Row(i++)] → [Elements =! Col (Diagonal)] until sum /= 0`

\

`Meaning Linearly Indepedent Vectors (Non-Singular).`

M^{−1}

`[(x)(x)(x)(x) | (1)(0)(0)(0)]`

`[(x)(x)(x)(x) | (0)(1)(0)(0)]`

`[(x)(x)(x)(x) | (0)(0)(1)(0)]`

`[(x)(x)(x)(x) | (0)(0)(0)(1)]`

`1`

^{st}`(x)`

→ `2`

^{nd}` correspondent(1) for operations.`

`Rx → Rx (+) (-) (x) Ry`

`Col++ until matches right side`

Primal Solution

`If ! (A · x = b) then convert to standard form.`

`Ax = b`

`(A`

^{−1}`)Ax = (A`

^{−1}`)b`

`Ix (Identity) = (A`

^{−1}`)b`

`x = (A`

^{−1}`)b → Xb [Basic (!0)] = (B`

^{−1}`)b`

`~ (B`

^{−1}`)b ~`

`4 x 4 [Matrix] 4 x 1 [Vector]`

`4 x 1`

`[(x)(x)(x)(x) | (y)]`

`[(x)(x)(x)(x) | (y)]`

`[(x)(x)(x)(x) | (y)]`

`[(x)(x)(x)(x) | (y)]`

`Primal → x(0, 0, x, x, x, x)`

Tableau Approach of Simplex Method

`If there is a (-) value as a minimization problem then improvement iteration continues:`

`Left Side is (0, 0, 0, x, x ,x) therefore Non-Basic.`

`Choose 1 Column and Swap Rows`

`Blands Rule → Which Non-Basic will become Basic, Which Basic will become Basic`

`Lowest Subscript [x`

_{1}`]`

`Constraint Rows (L / R) → [x`

_{1}` ≤ (L / R)]`

`Minimum that satisfies conditions is Pivot`

`Correspond to [1] Column`