Standard Minimization Problems with the Dual Appendix simplex method

STANDARD MINIMIZATION PROBLEM A standard minimization problem is a linear programming problem with an objective function that is to be minimized. The objective function is of the form : Z= aX1 + bX2 + cX3….. where a, b, c, . . . are real numbers and X1, X2, X3, . . . are decision variables. Constraints are of the form: AX1 + BX2 + CX3+ …… ≥ M where A, B, C,... are real numbers and M is nonnegative

STANDARD MINIMIZATION PROBLEM (cont.) Example: Determine if the linear programming problem is a standard minimization problem Minimize Z = 4X1+ 8X2 Subject to 3X1 + 4X2 ≤ -9 X2 ≥ 5 X1,x2 ≥0

STANDARD MINIMIZATION PROBLEM (cont.) Solution Minimize Z = 4X1+ 8X2 Subject to 3X1 + 4X2 ≤ -9 X2 ≥ 5 X1,x2 ≥0 Multiply First constraint by -1

STANDARD MINIMIZATION PROBLEM (cont.) We got: Minimize Z = 4X1+ 8X2 Subject to 3X1 + 4X2 ≥9 X2 ≥ 5 X1,x2 ≥0

The Dual For a standard minimization problem whose objective function has nonnegative coefficients, it may construct a standard maximization problem called the dual problem

Using Duals to Solve Standard Minimization Problems Example: Minimize Z= 2X1 + 3X2 Subject to X1+ X2 ≥ 12 3X1 + 2X2 ≥ 4 X1, X2≥ 0

Using Duals to Solve Standard Minimization Problems (cont.) The solution 1- construct a matrix for the problem as: 1 12 3 2 4 X1+ X2 ≥ 12 3X1 + 2X2 ≥ 4 2X1 + 3X2= Z

Using Duals to Solve Standard Minimization Problems (cont.) 2- The transpose of the matrix is created by switching the rows and columns The dual problem is: Maximize Z= 12X1+ 4X2 ST X1+ 3X2 ≤ 2 X1 + 2X2 ≤ 3 X1,X2 ≥0 1 3 2 12 4 X1+ 3X2 ≤ 2 X1+ 2X2 ≤ 3 12X1+ 4X2 = Z

Using Duals to Solve Standard Minimization Problems (cont.) Then Adding in the slack variables and rewriting the objective function yield the system of equations: X1+ 3X2 + S1= 2 X1 + 2X2 + S2= 3 X1,X2 ≥0

Using Duals to Solve Standard Minimization Problems (cont.) The initial simplex tableau: Basis X1 X2 S1 S2 RHS 1 3 2 Z -12 -4

Basis X1 X2 S1 S2 RHS 1 3 2 -1 Z 32 12 24

Basis X1 X2 S1 S2 RHS 1 3 2 -1 Z 32 12 24 X1 value X2 value

Minimization in other case If objective function is minimization and all constraints are “<“ , the solution can be found by multiply objective function by -1 , then objective function will convert to Max and solve the problem as simplex method. Example: Min z= 3x1 – 2x2 ST X1+x2<= 12 X2<= 24 X1,X2>=0

Minimization in other case (cont.) Solution Min z= 3x1 – 2x2 ST X1+x2<= 12 X2<= 24 X1,X2>=0 Multiply by -1

Minimization in other case (cont.) Max z= -3x1 + 2x2 ST X1+x2<= 12 X2<= 24 X1,X2>=0

Minimization in other case (cont.) Basis X1 X2 S1 S2 RHS 1 12 24 Z 3 -2