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Example 1
Maximize f(x) = 22x1 + 25x2 subject to the constraints:
x1 + x2 ≤ 14
x1 + 2x2 ≤ 22
x1 ≥ 0
x2 ≥ 0
We define the matrix
Chose the first column and perform Gaussian elimination and we note of (26/2 = 13, 14/1 = 14 22/1 = 22) that the 1st entry is the smallest positive value, so we perform Gaussian elimination on the 1st column using the 1st row as a pivot to get
Next, we choose the second column and note that of (13/(1/2) = 26, 1/(1/2) = 2, 9/(3/2) = 6), the second is the smallest, and therefore we continue to use Gaussian elimination on the 2nd column using the 2nd row as a pivot.
The third column still has a negative number, and therefore we check that of (12/1 = 12, 2/(−1) = −2, 6/1 = 6), the third has the smallest positive value, and therefore we perform Gaussian elimination based on that variable:
As both variables were visited, we read off from row one that x1 = 6 and from row two x2 = 8. The value of the objective function is 332.
Example 2
Use the same system of constraints as in Example 1, but use the objective function f(x) = x1 + 25x2.
In this case, only the 2nd column is in row-reduced echelon form, and therefore x2 = 11 and x1 = 0. The value of the objective function is 275 (= 11 × 25).
Example 3
Use the same system of constraints as in Example 1, but use the objective function f(x) = 22x1 + x2.
A this point, all indicators are positive, and therefore only one column is reduced, namely, that one corresponding to x1, and therefore our solution is x1 = 13 and x2 = 0, or x = (13, 0)T.
Copyright ©2005 by Douglas Wilhelm Harder. All rights reserved.
