Foundations-1 The Theory of the Simplex Method

Foundations-2 The Essence Simplex method is an algebraic procedure However, its underlying concepts are geometric Understanding these geometric concepts helps before going into their algebraic equivalents With two decision variables, the geometric concepts are easy to visualize Should understand the generalization of these concepts to higher dimensions

Foundations-3 Standard (Canonical) Form of an LP Model Maximize Z = c 1 x 1 + c 2 x 2 + … + c n x n subject to a 11 x 1 + a 12 x 2 + … + a 1n x n ≤ b 1 a 21 x 1 + a 22 x 2 + … + a 2n x n ≤ b 2 … a m1 x 1 + a m2 x 2 + … + a mn x n ≤ b m x 1 ≥ 0, x 2 ≥ 0, …, x n ≥ 0

Foundations-4 Extended Terminology max cxn variables s.to Ax  bm inequalities x  0n inequalities max cxn+m variables s.to Ax + Ix s = bm equations x, x s  0n+m inequalities Constraint boundary equation –For any constraint, obtain by replacing its , =,  by = –It defines a “flat” geometric shape: hyperplane Corner-point solution –Simultaneous solution of n constraint boundary equations ()

Foundations-5 Extended Terminology max cx + 0x s n+m variables s.to Ax + Ix s = bm equations x, x s  0n+m inequalities Indicating variables (in the augmented form) Original Constraint in Augmented Form Constraint Boundary Equation Indicating Variable x j  0 (j=1,2,…,n) x j = 0xjxj (i=1,2,…,m) x s n+i

Foundations-6

Foundations-7 Original Constraint Original Constraint in Augmented Form Constraint Boundary Equation Indicating Variable x j  0 x j = 0xjxj (i=1,2,…,m1) (i=1,2,…,m2) (i=1,2,…,m3) Indicating variables, overall

Foundations-8 Properties of CPF (BF) Solutions Property 1: a.If there is exactly one optimal solution, then it must be a corner-point feasible solution b.If there are multiple optimal solutions (and a bounded feasible region), then at least two must be adjacent corner-point feasible solutions Proof of (a) by contradiction

Foundations-9 Properties of CPF (BF) Solutions Property 2: There are only a finite number of corner-point feasible solutions max cxn variables s.to Ax  bm inequalities x  0n inequalities CP solution is defined as the intersection of n constraint boundary equations (n+m) choose n: Upper bound on # of CPF solutions

Foundations-10 Properties of CPF (BF) Solutions Property 3: a.If a corner-point feasible solution has no adjacent CPF solution that are better, then there are no better CPF solutions elsewhere b.Hence, such a CPF solution is an optimal solution (assuming LP is feasible and bounded)

Foundations-11 Simplex in Matrix (Product) Form max cxn variables s.to Ax  bm inequalities x  0n inequalities max cxm+n variables s.to Ax + Ix s = bm equations x, x s  0m+n inequalities Initial Tableau: () B.V. Original VariablesSlack Variables r.h.s. x1x1 x2x2 …xnxn x s n+1 …x s n+m Z -c00 x s n+1 AI b … x s n+m XB=XB=

Foundations-12 Simplex in Matrix (Product) Form Intermediate iterations: Let B denote the square matrix that contains the columns from [ A | I ] that correspond to the current set of basic variables, x B (in order) Similarly, let c B be the vector of elements in c that correspond to x B Then at any intermediate step, the simplex tableau is given by B.V. Original VariablesSlack Variables r.h.s. x1x1 x2x2 …xnxn x s n+1 …x s n+m Z c B B -1 A-cc B B -1 c B B -1 b xBxB B -1 AB -1 B -1 b

Foundations-13 Revised Simplex Method Initialization Find an initial BFS (if one not immediately available, do Phase 1. If Phase 1 terminates with z 1 >0, LP is infeasible) Optimality test Calculate c B B -1 A-c for basic, and c B B -1 for nonbasic variables. If all ≥ 0, stop with optimality Iterative step –Determine the entering variable as before (steepest ascent) –Let x k be the entering variable –Determine the leaving variable as before (minimum ratio test), but now only need to calculate the coefficients of the pivot column (from B -1 A or B -1 ) and the updated rhs (B -1 b) –Let x l : the leaving variable r : the equation number of the leaving variable a’ rk : coefficient of x k in r th equation (the pivot element) –Determine B -1 new =E B -1 old where E is I expect for r th column replaced with –Can calculate the new BFS using x B =B -1 b and z=c B B -1 b –Back to optimality test

Foundations-14 Revised Simplex Example Original data: Maximize z = 2x 1 +3x 2 subject tox 1 +2x 2 +s 1 = 10 3x 1 +x 2 +s 2 = 15 x 2 +s 3 = 4 x 1,x 2, s 1, s 2, s 3 ≥ 0 Maximize z = 2x 1 +3x 2 subject tox 1 +2x 2 ≤ 10 3x 1 + x 2 ≤ 15 x 2 ≤ 4 x 1,x 2 ≥0

Foundations-15 Iteration 0

Foundations-16 Iteration 0 Calculate row-0 coefficients: c B B -1 A-c = c B B -1 = Optimal? Entering variable is Calculate coefficients in pivot column using B -1 A or B -1 Calculate updated rhs using B -1 b, do the ratio test Leaving variable is Update B -1

Foundations-17 Iteration 1

Foundations-18 Iteration 1 Calculate row-0 coefficients: c B B -1 A-c = c B B -1 = Optimal? Entering variable is Calculate coefficients in pivot column using B -1 A or B -1 Calculate updated rhs using B -1 b, do the ratio test Leaving variable is Update B -1

Foundations-19 Iteration 2

Foundations-20 Iteration 2 Calculate row-0 coefficients: c B B -1 A-c = c B B -1 = Optimal? Entering variable is Calculate coefficients in pivot column using B -1 A or B -1 Calculate updated rhs using B -1 b, do the ratio test Leaving variable is Update B -1

Foundations-21 Iteration 3

Foundations-22 Iteration 3 Calculate row-0 coefficients: c B B -1 A-c = c B B -1 = Optimal? Entering variable is Calculate coefficients in pivot column using B -1 A or B -1 Calculate updated rhs using B -1 b, do the ratio test Leaving variable is Update B -1