1. Standard Minimization Problems with the Dual
Appendix simplex method

2. 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

3. 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

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

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

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

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

8. 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

9. 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

10. 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

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

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

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

14. 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

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

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

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