BASIC FEASIBLE SOLUTIONS CONTENTS Polyhedral Sets and Polytopes Basic Feasible Solutions and Polytopes Reference: Chapters 2 and 3 in BJS book.

Hyperplanes A hyperplane in En generalizes the notion of a straight line in E2 and the notion of a plane in E3. A hyperplane H in En is a set of the form {x:px=k} where p is a nonzero vector in En and k is a scalar. If x0  H, then px0 = k, and for any other point x  H, px = k. Hence, p(x – x0) = 0. Both hyperplanes and halfspaces are convex sets. Why?

Rays A ray is a collection of points of the form {x0 + d:   0}, where d is a nonzero vector. x0 d A convex cone C is a convex set with the additional property that x  C for each x  C and for each   0.

Polyhedron A hyperplane divides En into two regions, called halfspaces: S1 = {x: px  k} S2 = {x: px  k} A polyhedral set or a polyhedron is the intersection of a finite number of halfspaces. A bounded polyhedral set is called a polytope.

Polyhedral Sets Consider the polyhedral set defined by the following inequalities: -2x1 + x2  4 x1 + x2  3 x1  2 x1  0 x2  0 A polyhedral set can be represented by the set {x: Ax  b} where A is an mxn matrix.

Faces, Edges, Adjacent Extreme Points of a Polyhedra

An Important Theorem Theorem: For any polytope S, every solution x can be represented by a convex combinations of the extreme points of the polytope. Let x1, x2, x3, … , xk denote the extreme points of a polytope S. Then, for any x  S, there exist 1, 2, 3,.., k x = 1x1 + 2x2 + 3x3 + …. + kxk, where

Another Important Theorem Consider the following linear programming problem: Minimize cx subject to Ax = b x  0 Let x1, x2, x3, … , xk be the extreme points of the constraint set S. This linear program can be transformed into the following linear program: Minimize 1, 2, 3, … , k  0 How to solve this linear program?

Another Important Theorem (contd.) Theorem. If a linear programming problem has a bounded feasible solution space, then there must exist an optimal extreme point solution. Is every optimal solution of a linear programming problem an extreme point solution? It can be shown that extreme point solutions correspond to basic feasible solutions (BFS) which are defined algebraically.

Standard Form of a Linear Program Condition 1: All right-hand side coefficients in the constraints are nonnegative.   Condition 2 : All constraints are in the equality form. Condition 3 : All variables must take nonnegative values. Maximize c1x1 + c2x2 + c3x3 + …. + cnxn subject to a11x1 + a12x2 + a13x3 + …. + a1nxn= b1 a21x1 + a22x2 + a23x3 + …. + a2nxn= b2 : am1x1 + am2x2 + am3x3 + …. + amnxn = bm x1, x2, x3 , …., xn  0

Restoring Condition 1 Condition 1: All right-hand side coefficients in the constraints are nonnegative.   Constraint: 2x1 - 3x2 + x3  - 10 Alternative constraint: -2x1 + 3x2 - x3  10

Restoring Condition 2 Condition 2 : All constraints are in the equality form.   max z = 4x1+ 3x2 max z = 4x1+ 3x2 s.t. x1 + x2  40 x1 + x2 + s1 = 40 2x1+ x2  60 2x1+ x2 + s2 = 60 x1, x2  0 x1, x2, s1, s2  0 s.t. x1 + x2  40 x1 + x2 - e1 = 40 2x1+ x2  60 2x1+ x2 - e2 = 60 x1, x2  0 x1, x2, e1, e2  0 si = Slack variables ei = Surplus variables

Restoring Condition 3 Condition 3 : All variables must take nonnegative values.   max z = 30x1 - 4x2 + 8x3 s.t. 5x1 - x2 + x3  30 x1  5 x1  0, x2 unrestricted, x3  0 Variable transformation: x2 = , = - x3 max z = 30x1 - 4( ) - 8 5x1 - -  30 x1  5 x1, , ,  0

Basic Feasible Solutions Consider the linear program with the constraints: Ax=b and x  0 Where A is an mxn matrix and b is an m vector. Let rank (A) = m. Let A= [B,N] where B is an m x m invertible matrix, and N is an m x (n-m) matrix. Then the solution of the system BxB = b is a basic solution of the linear program.

Basic Feasible Solutions (contd.) Alternatively, the solution x = [xB, xN] xB = B-1 b xN= 0 is the basic solutions of the system. If xB  0, then x is called a basic feasible solution (BFS) of the system. A basic feasible solution x is called a degenerate basic feasible solution if some component of xB is zero. B is called the basis matrix. Variables in xB are called basic variables. Variables in xN are called nonbasic variables.

Example 1 Consider a polyhedral set: Standard form: x1 + x2  6 x1 + x2 + x3 = 6 x2  3 x2 + x4 = 3 x1 , x2  0 x1 , x2 , x3 , x4  0 Constraint Matrix: A= [a1, a2, a3, a4] =

Example 1 (contd.) The solution space:

Example 1 (contd.) B = [a1,a2] = xB = xN =

Example 1 (contd.) B = [a1,a4] = xB = xN =

Example 1 (contd.) B = [a2,a3] = xB = xN =

Example 1 (contd.) B = [a2,a4] = xB = xN =

Example 1 (contd.) B = [a3,a4] = xB = xN =

Example 1 (contd.) Have we considered all the basic solutions? How many basic solutions there are? How many basic feasible solutions there are?

Example 1 (contd.) Four Basic Feasible Solutions:

Basic Feasible Solutions The possible number of basic feasible solutions is bounded by the number of ways of extracting m columns out of n columns. The maximum number of BFS =

Example 2 Consider a polyhedral set: Standard form: x1 + x2  6 x1 + x2 + x3 = 6 x2  3 x2 + x4 = 3 x1 + 2x2  9 x1 + 2x2 + x5 = 9 x1 , x2  0 x1 , x2 , x3 , x4, x5  0 Constraint Matrix: A= [a1, a2, a3, a4] =

Example 2 (contd.) Feasible Solution Space:

Example 2 (contd.) B = [a1, a2, a3]; xB = xN = B = [a1, a2, a4]; xB = Observe that several basis correspond to the same basic feasible solution. Also observe that the basic feasible solution must be degenerate for it to happen.

Summary: Basic and Basic Feasible Solutions Identify m linearly independent (basic) variables (BV). Set the remaining n - m (nonbasic) variables to zero value (NBV). Solve the equations Ax = b in terms of the basic variables. (Observe that there is a unique solution.) The resulting solution is a basic solution (BS). A basic solution is a basic feasible solution (BFS) if xj  0 for each j = 1, 2, … , N. There are at most n!/(n-m)!m! BFS. Hence there are finite number of BFS.

Adjacent Basic Feasible Solutions For any LP with m constraints, two basic feasible solutions are said to be adjacent if their sets of basic variables have m-1 basic variables in common. Basic Feasible Solution: (x1, x2, x3, x4) Adjacent Basic Feasible Solutions: (x1, x2, x3, x5) (x1, x2, x6, x4) (x1, x7, x3, x4) (x8, x2, x3, x4)