Solving Arithmetically Max c 1 *X 1 +…+ c n *X n = z s.t. a 11 *X 1 +…+ a 1n *X n b 1 … a m1 *X 1 +…+ a mn *X n b m X 1, X n 0
Convert Inequalities Max c 1 *X 1 +…+c n *X n +0*S 1 +…+0*S m = z s.t. a 11 *X 1 +…+a 1n *X n +1*S 1 +…+0*S m = b 1 … a m1 *X 1 +…+a mn *X n +0*S 1 +…+1*S m = b m X 1, X 2, S 1, S m 0 Max c 1 *X 1 +…+c n *X n = z s.t. a 11 *X 1 +…+a 1n *X n b 1 … a m1 *X 1 +…+a mn *X n b m X 1, X 2 0
Arithmetic Problem, 1 Number of unknown variables: n+m Number of equations (constraints): m Can only solve m equations for m unknown variables
Arithmetic Problem, 2 From graphical illustration, we know Solution occurs at extreme point
1 Constraint 1 Non-Zero Variable Max Objective s.t. X1 + 2*X2 + S = 10 X1, X2, S 0 X2 X1 10 5
Arithmetic Problem, 3 Extreme points have only m non-zero variables and n-m zero variables Thus, if we knew which n-m variables are zero, we could solve for the remaining m variables using our m equation system New task: Which variables should remain zero?
Arithmetic Solution 1.Pick n-m variables and set to zero 2.Solve mxm equation system 3.Compute optimality indicators 4.If optimal, done; otherwise
Optimal solution is extreme point Number of non-zero variables = number of equations Non-zero (positive) variables are called basic variables All remaining variables are called non- basic variables Non-basic variables are zero.
Decomposed LP in Matrix Notation Max C B X B + C NB X NB s.t. B X B + A NB X NB = b X B, X NB 0
Solution of LP in Matrix Notation BX B =b-AX NB BX B =b X B =B -1 b
Solver tasks Determine which variables form basis Invert coefficient matrix to calculate optimal variable values
If optimal solution is an extreme point, why not simply calculate all extreme points and choose the one with highest objective function value?
Number of Possible Extreme Points n.. Number of Variables m.. Number of Equations