Paper 3, Section II, H

Optimization | Part IB, 2019

(a) Suppose that A∈Rm×nA \in \mathbb{R}^{m \times n} and b∈Rmb \in \mathbb{R}^{m}, with n⩾mn \geqslant m. What does it mean for x∈Rnx \in \mathbb{R}^{n} to be a basic feasible solution of the equation Ax=b?A x=b ?

Assume that the mm rows of AA are linearly independent, every set of mm columns is linearly independent, and every basic solution has exactly mm non-zero entries. Prove that the extreme points of X(b)={x⩾0:Ax=b}\mathcal{X}(b)=\{x \geqslant 0: A x=b\} are the basic feasible solutions of Ax=bA x=b. [Here, x⩾0x \geqslant 0 means that each of the coordinates of xx are at least 0 .]

(b) Use the simplex method to solve the linear program

max⁡4x1+3x2+7x3 s.t. x1+3x2+x3⩽144x1+3x2+2x3⩽5−x1+x2−x3⩾−2x1,x2,x3⩾0\begin{array}{cl} \max & 4 x_{1}+3 x_{2}+7 x_{3} \\ \text { s.t. } & x_{1}+3 x_{2}+x_{3} \leqslant 14 \\ & 4 x_{1}+3 x_{2}+2 x_{3} \leqslant 5 \\ & -x_{1}+x_{2}-x_{3} \geqslant-2 \\ & x_{1}, x_{2}, x_{3} \geqslant 0 \end{array}

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