Elementary Linear Algebra by Kenneth Kuttler

Author:Kenneth Kuttler
Language: eng
Format: epub, pdf
Publisher: Saylor Foundation

Next it is necessary to make the last two columns on the bottom left row into simple columns. Performing the row operation, this yields an initial simplex tableau,

Now the algorithm involves getting rid of the positive entries on the left bottom row. Begin with the first column. The pivot is the 2. An application of the simplex algorithm yields the new tableau

Now go to the third column. The pivot is the 3/2 in the second row. An application of the simplex algorithm yields

(11.20)

and you see there are only nonpositive numbers on the bottom left column so the process stops and yields 0 for the minimum of z = x7+ x8. As for the other variables, x1 = 5/3, x2 = 0, x3 = 1/3, x4 = 0, x5 = 0, x6 = 5. Now as explained in the above observation, this is a basic feasible solution for the original system (11.18).

Now consider a maximization problem associated with the above constraints.

Example 11.4.3 Maximize x1−x2+2x3 subject to the constraints, 2x1+x2−x3 ≥ 3, x1+x2+x3 ≥ 2, x1 + x2 + x3 ≤ 7 and x ≥ 0.

From (11.20) you can immediately assemble an initial simplex tableau. You begin with the first 6 columns and top 3 rows in (11.20). Then add in the column and row for z. This yields

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