1. 1 Simplex Method for Bounded Variables Linear programming problems with lower and upper bounds Generalizing simplex algorithm for bounded variables Reference: Chapter 5 in Bazaraa, Jarvis, and Sherali

2. 2 LP Problems with Bounded Variables In most practical Problems, variables are bounded from below as well as above: We can handle the upper bound constraints implicitly and thus reduce the size of the basis matrix substationally.

3. 3 Basic Feasible Solution Consider the system: Partition A into [B,N 1,N 2 ] such that B is an invertible matrix. We call it a basis structure. The basic solution corresponding to this basis structure is determined in the following manner: If for all, we say that the solution is a basic feasible solution.

4. 4 Basic Feasible Solution (cont’d) Consider the solution space given by: Which is equivalent to:

5. 5 Basic Feasible Solution (cont’d) Suppose that x 2 and x 4 are made basic variables, then: 1) Suppose we set x 1 =0 and x 3 =0. Then x 2 =5 and x 4 =-6. The resulting solution is not a basic feasible solution. 2) Suppose we set x 1 =4 and x 3 =0. Then x 2 =1 and x 4 =6, which is a basic feasible solution. In general, what is the maximum number of basic feasible solutions we can have?

6. 6 Optimality Conditions for Minimization Problem Basic Feasible Solution: Optimality Conditions: zxBxB x N1 x N2 RHS z10c B B -1 N 1 -c N1 c B B -1 N 2 -c N2 z xBxB 0IB -1 N 1 B -1 N 2 b

7. 7 Handling Entering Variable If the nonbasic variable entering the basis is at its lower bound, then its value is increased. If the nonbasic variable entering the basis is at its upper bound, then its value is decreased. The amount of increase or decrease is determined by the constraint that the values of all basic (and nonbasic) variables remains within their lower and upper bounds.

8. 8 Example zx1x1 x2x2 x3x3 x4x4 x5x5 RHS z124100 x4x4 0211109 x5x5 011 015 The initial simplex tableau: Entering variable: x 2 Leaving variable: x 5

9. 9 Example (cont’d) zx1x1 x2x2 x3x3 x4x4 x5x5 RHS z1-2050-4-21 x4x4 010214 x2x2 011 015 Next tableau: Entering variable: x 3 Leaving variable: x 2

10. 10 Example (cont’d) zx1x1 x2x2 x3x3 x4x4 x5x5 RHS z135001-26 x4x4 0320112 x3x3 0 10 2 Next tableau: Entering variable: x 1 Leaving variable: x 4

11. 11 Example (cont’d) zx1x1 x2x2 x3x3 x4x4 x5x5 RHS z10300-28 x1x1 012/301/3 2/3 x3x3 00-1/311/3-2/38/3 The current solution is an optimal solution. Final tableau: