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Unformatted text preview: 5.5 Dual problem: minimization with problem constraints of the form 5 Associated with each minimization problem with constraints is a maximization problem called the dual problem. The dual problem will be illustrated through an example. Read the textbook carefully to learn the details of this method. We wish to minimize the objective function subject to certain constraints: 5 C = 16 x1 + 9 x2 + 21x3 x1 + x2 + 3 x3 12 2 x1 + x2 + x3 16 x1 , x2 , x3 0 Initial matrix We start with an initial matrix , A, corresponds to the problem constraints: 1 1 3 12 1 1 16 2 A= 9 21 1 16 Transpose of matrix A To find the transpose of matrix A, interchange the rows and columns so that the first row of A is now the first column of A transpose. A =
T 1 2 16 1 9 1 1 21 3 12 16 1 Dual of the minimization problem is the following maximization problem: Maximize P under the following constraints. P = 12 y1 + 16 y2 y1 + 2 y2 16 y1 + y2 3 y1 + y2 y1 , y2 0 9 21 Theorem 1: Fundamental principle of Duality A minimization problem has a solution if and only if its dual problem has a solution. If a solution exists, then the optimal value of the minimization problem is the same as the optimum value of the dual problem. Forming the Dual problem with slack variables x x x
1,
2, 3 result: y1 + 2 y2 + x1 y1 + y2 + 3 y1 + y2 + x2 = 16 =9 x3 = 21 12 y1  16 y2 + p = 0 Form the simplex tableau for the dual problem and determine the pivot element The first pivot element is 2 (in red) because it is located in the column with the smallest negative number at the bottom(16) and when divided into the rightmost constants, yields the smallest quotient (16 divided by 2 is 8) y1 x1 x2 x3 P 1 1 3 y2 2 1 1 x1 x2 1 0 0 0 1 0 0 x3 0 0 1 0 P 16 9 21 0  12  16 0 Divide row 1 by the pivot element (2) and change the exiting variable to y
2 (in red) Result:
y2 x2 x3 y1 .5 1 3 y2 1 1 1 x1 .5 0 0 x2 0 1 0 x3 0 0 1 P 8 9 21 P 12 16 0 0 0 0 Perform row operations to get zeros in the column containing the pivot element. Identify the next pivot element (0.5) (in red) 1*row 1 + R2=R2 1*row1+r3 =r3 16*r1+r4 = r4 y2 x2 x3 y1 .5 y2 1 0 0 x1 .5 x2 0 x3 0 0 1 P 8 1 13 .5
2.5 .5 1 .5 0 New pivot element P 4 0 8 0 0 128 Pivot element located in this column Variable y1 becomes new entering variable Divide row 2 by 0.5 to obtain a 1 in the pivot position. y2 y1 x3 y1 .5 1 2.5 y2 1 0 0 0 x1 .5 1 .5 8 x2 0 2 0 x3 P 0 8 0 2 1 13 P 4 0 0 128 More row operations 0.5*row2 + row1=Row1 2.5*row 2 + row3=row3 4*row2+row4=row4 y2 y1 x3 P y1 0 1 0 0 y2 x1 1 1 0 1 0 2 0 4 x2 1 2 5 8 x3 P 0 8 0 2 1 8 0 136 Solution: An optimal solution to a minimization problem can always be obtained from the bottom row of the final simplex tableau for the dual problem. Minimum of P is 136. It occurs at x 1 = 4 , x 2 =8, x 3 =0 y1 y2 y1 x3 P 0 1 0 0 y2 1 0 0 0 x1 1 x2 1 x3 0 0 1 P 8 2 8 1 2 2 5 4 8 0 136 ...
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This note was uploaded on 03/27/2008 for the course MATH 150 taught by Professor Unk during the Winter '07 term at San Jacinto.
 Winter '07
 Unk

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