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Row operations are used to solve simultaneous equations where equations are multiplied by constants and added or subtracted from each other.

The Simplex Method

The simplex method is a set of mathematical steps for solving a linear programming problem carried out in a table called a simplex tableau.

The Simplex Method

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Finally, consider an example where s1 = 0 and s2 = 0. These values result in the follow- ing set of equations:

4 x 1 + 3 x 2 + s 2 = 1 2 0 x 1 + 2 x 2 + s 1 = 4 0

and

4 x 1 + 3 x 2 + 0 = 1 2 0 x 1 + 2 x 2 + 0 = 4 0

These equations can be solved using row operations. In row operations, the equations can be multiplied by constant values and then added or subtracted from each other without changing the values of the decision variables. First, multiply the top equation by 4 to get

4x1 + 8x2 = 160 and then subtract the second equation:

x 2 = 8 5 x 2 = 4 0 - 4 x 1 - 3 x 2 = - 1 2 0 4 x 1 + 8 x 2 = 1 6 0

Next, substitute this value of x2 into either one of the constraints:

x1 + 2(8) = 40 x1 = 24

This solution corresponds to point B on the graph, where x1 = 24, x2 = 8, s1 = 0, and s2 = 0, which is the optimal solution point.

All three of these example solutions meet our definition of basic feasible solutions. However, two specific questions are raised by the identification of these solutions.

  • 1.

    In each example, how was it known which variables to set equal to zero?

  • 2.

    How is the optimal solution identified?

The answers to both of these questions can be found by using the simplex method. The simplex method is a set of mathematical steps that determines at each step which variables should equal zero and when an optimal solution has been reached.

The steps of the simplex method are carried out within the framework of a table, or simplex tableau. The tableau organizes the model into a form that makes applying the mathematical steps easier. The Beaver Creek Pottery Company example will be used again to demonstrate the simplex tableau and method:

maximize Z = $40x1 + 50x2 + 0s1 + 0s2 subject to

x 1 , x 2 , s 1 , s 2 Ú 0 4 x 1 + 3 x 2 + s 2 = 1 2 0 l b . x 1 + 2 x 2 + s 1 = 4 0 h r .

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