Standard Minimization Problems with the Dual Appendix simplex

Standard Minimization Problems with the Dual Appendix simplex method 1

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= a. X 1 + b. X 2 + c. X 3…. . where a, b, c, . . . are real numbers and X 1, X 2, X 3, . . . are decision variables. § Constraints are of the form: AX 1 + BX 2 + CX 3+ …… ≥ M where A, B, C, . . . are real numbers and M is nonnegative 2

STANDARD MINIMIZATION PROBLEM (cont. ) Example: Determine if the linear programming problem is a standard minimization problem Minimize Z = 4 X 1+ 8 X 2 Subject to 3 X 1 + 4 X 2 ≤ -9 X 2 ≥ 5 X 1, x 2 ≥ 0 3

STANDARD MINIMIZATION PROBLEM (cont. ) Solution Minimize Z = 4 X 1+ 8 X 2 Subject to 3 X 1 + 4 X 2 ≤ -9 Multiply First constraint by -1 X 2 ≥ 5 X 1, x 2 ≥ 0 4

STANDARD MINIMIZATION PROBLEM (cont. ) We got: Minimize Z = 4 X 1+ 8 X 2 Subject to 3 X 1 + 4 X 2 ≥ 9 X 2 ≥ 5 X 1, x 2 ≥ 0 5

The Dual § For a standard minimization problem whose objective function has nonnegative coefficients, it may construct a standard maximization problem called the dual problem 6

Using Duals to Solve Standard Minimization Problems Example: Minimize Z= 2 X 1 + 3 X 2 Subject to X 1+ X 2 ≥ 12 3 X 1 + 2 X 2 ≥ 4 X 1, X 2≥ 0 7

Using Duals to Solve Standard Minimization Problems (cont. ) The solution 1 - construct a matrix for the problem as: 1 1 12 X 1+ X 2 ≥ 12 3 2 4 3 X 1 + 2 X 2 ≥ 4 2 3 1 2 X 1 + 3 X 2= Z 8

Using Duals to Solve Standard Minimization Problems (cont. ) 2 - The transpose of the matrix is created by switching the rows and columns 1 3 2 X 1+ 3 X 2 ≤ 2 1 2 3 X 1+ 2 X 2 ≤ 3 12 4 1 12 X 1+ 4 X 2 = Z The dual problem is: Maximize Z= 12 X 1+ 4 X 2 ST X 1+ 3 X 2 ≤ 2 X 1 + 2 X 2 ≤ 3 X 1, X 2 ≥ 0 9

Using Duals to Solve Standard Minimization Problems (cont. ) § Then Adding in the slack variables and rewriting the objective function yield the system of equations: X 1+ 3 X 2 + S 1= 2 X 1 + 2 X 2 + S 2= 3 X 1, X 2 ≥ 0 10

Using Duals to Solve Standard Minimization Problems (cont. ) The initial simplex tableau: Basis X 1 X 2 S 1 S 2 RHS S 1 1 3 1 0 2 S 2 1 2 0 1 3 Z -12 -4 0 0 0 11

Basis X 1 X 2 S 1 S 2 RHS X 1 1 3 1 0 2 S 2 0 -1 -1 1 1 Z 0 32 12 0 24

Basis X 1 X 2 S 1 S 2 RHS X 1 1 3 1 0 2 S 2 0 -1 -1 1 1 Z 0 32 12 0 24 X 1 value X 2 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= 3 x 1 – 2 x 2 ST X 1+x 2<= 12 X 2<= 24 X 1, X 2>=0

Minimization in other case (cont. ) • Solution Min z= 3 x 1 – 2 x 2 ST X 1+x 2<= 12 X 2<= 24 X 1, X 2>=0 Multiply by -1

Minimization in other case (cont. ) Max z= -3 x 1 + 2 x 2 ST X 1+x 2<= 12 X 2<= 24 X 1, X 2>=0

Minimization in other case (cont. ) Basis X 1 X 2 S 1 S 2 RHS X 1 1 0 12 S 2 0 1 24 Z 3 -2 0 0 0