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= 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
2
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
3
STANDARD MINIMIZATION
PROBLEM (cont.)
Solution
Minimize Z = 4X1+ 8X2
Subject to
3X1 + 4X2 ≤ -9 Multiply First constraint by -1
X2 ≥ 5
X1,x2 ≥0
4
STANDARD MINIMIZATION
PROBLEM (cont.)
We got:
Minimize Z = 4X1+ 8X2
Subject to
3X1 + 4X2 ≥9
X2 ≥ 5
X1,x2 ≥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= 2X1 + 3X2
Subject to
X1+ X2 ≥ 12
3X1 + 2X2 ≥ 4
X1, X2≥ 0
7
Using Duals to Solve Standard
Minimization Problems (cont.)
The solution
1- construct a matrix for the problem as:
1
1
12
X1+ X2 ≥ 12
3
2
4
3X1 + 2X2 ≥ 4
2
3
1
2X1 + 3X2= 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
X1+ 3X2 ≤ 2
1
2
3
X1+ 2X2 ≤ 3
12
4
1
12X1+ 4X2 = Z
The dual problem is:
Maximize Z= 12X1+ 4X2
ST
X1+ 3X2 ≤ 2
X1 + 2X2 ≤ 3
X1,X2 ≥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:
X1+ 3X2 + S1= 2
X1 + 2X2 + S2= 3
X1,X2 ≥0
10
Using Duals to Solve Standard
Minimization Problems (cont.)
The initial simplex tableau:
Basis
X1
X2
S1
S2
RHS
S1
1
3
1
0
2
S2
1
2
0
1
3
Z
-12
-4
0
0
0
11
Basis
X1
X2
S1
S2
RHS
X1
1
3
1
0
2
S2
0
-1
-1
1
1
Z
0
32
12
0
24
Basis
X1
X2
S1
S2
RHS
X1
1
3
1
0
2
S2
0
-1
-1
1
1
Z
0
32
12
0
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
X1
1
1
1
0
12
S2
0
1
0
1
24
Z
3
-2
0
0
0