July 15, 2017

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By Michael R. Greenberg

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Example text

If, however, we change either constraints (3) or (4), the optimal solution will change. For example, if we change Eq. (4) from 3X1 + X2 + X5 = 15 to 3Xl + X2 + X5 = 16, the value of the optimal solution will increase from lOf to lOf, an increase off. If the right-hand-side value goes from 15 to 18, the value of the solution changes from 10f to 12. In short, for every unit 58 / 3: BEYOND THE GLOBAL O P T I M U M 15 12 9 * i 6 A \B~ 3 W) —Optimum value, Ζ = 1 θ | \m X""^^! l increase in the Eq. (4) resource, the value of the solution increases * of a unit until the solution changes.

Again, deviation from this guideline could save a few minutes work, but it might cause hours of additional work looking for errors. If you follow these three guidelines, the maximum number of row transformations that are required equals the number of elements in the matrix. In the matrix Ä case presented above, seven transformations were necessary to convert Ä into /. Two steps below the maximum were possible because in the column 1 operations (between steps 2 and 3), the element a31 was 0 in the original Ä matrix.

The element a 33 is already 1 and does not need further transformation. 26 / 2: ALGEBRAIC METHODS Step 6 Multiply row 3 by — 1 and add it to row 1. row 3 Step 7 -1[0 0 l][-2 3 1] = [0 transformed row 3 0 + row 1 1 0 new row 1 1 0 0 -1 0 -1][2 2 -3 -1 1 1 -1 0 0 3 -4 -3 -1] -1 Add row 3 to row 2. transformed row 3 from step 5 0 0 1 -2 + row 2 0 1 -1 -2 new row 2 0 1 0 -4 3 1 3 0 6 1 The final matrices A3 and B2 are as follows: 0 ol A, = 0 1 0 L° 0 lj [l 3 ß,. = -4 -2 -4 6 3 Matrix A 3 is an identity matrix, and B2 should be A À ■ 5 , = /.

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