#7

Simultaneous Equations

Example:

Beaver Creek Pottery Company

X1 = # of bowls to produce each day

X2 = # of mugs to produce each day

Zmax = \$40X1 + \$50X2

The company needs to know how many bowls and mugs to produce each day in order to maximize profit.

Subject to the following constraints:

X1 + 2X2 <= 40 labor, hr

4X1 + 3X2 <= 120 clay, lb

X1, X2 >= 0 nonnegativity

Graph the feasible region and locate the optimal solution.

Simplex is a matrix algebra method that requires all mathematical relationships to be equations. Therefore …

Convert inequalities to equalities. Use a slack variable to do this. This represents unused resources.

S1 = unused labor

S2 = unused clay

Constraints are now written as:

______ + ______ + ______ = 40 labor, hr

______ + ______ + ______ = 120 clay, lb

____ , ____ , ____ , ____ >= 0 nonnegativity

Question: If no bowls and mugs are produced …

… show the equation for determining the value of S1

… show the equation for determining the value of S2

… show the equation for determining the profit

Simultaneous Equations

When you have the same number of unknown variables as equations, it is possible to solve for the value of the variables.

But, as in this case, when there are 4 unknown variables (i.e. two decision variables and 2 slack variables) and 2 equations, you can let nm variables equal zero (where n equals the number of variables and m equals the number of constraints).

What if: X1 = 0 and S1 = 0

X1 + 2X2 + S1 = 40

0 + 2X2 + 0 = 40

X2 = _________

4X1 + 3X2 + S2 = 120

0 + 3X2 + S2 = 120

S2 = _________

X1 = 0, X2 = 20, S1 = 0, S2 = 60

This result, obtained through simultaneous equations, should correspond to one of the potential optimal solutions on the graph.

This solution is referred to as a basic feasible solution.

What if: X2 = 0 and S2 = 0

Solve for X1 and S1.

This result, also obtained through simultaneous equations, should correspond to another one of the potential optimal solutions on the graph.

Again, this solution is referred to as a basic feasible solution.

Row Operations

Simultaneous equations can also 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.

What if: We use 4 as the constant.

First, multiply the top equation (labor constraint) by 4

______ + ______ = ______

And then subtract the second equation to solve for X2

______ + ______ = 160

minus ______ - ______ = -120

5X2 = 40

X2 = 8

Next, substitute this value of X2 into either one (or both) of the constraints.

labor constraint clay constraint

X1 + 2(8) = 40 4X1 + 3(8) = 120

X1 = ______ X1 = ______

X1 = 24, X2 = 8

This result, also obtained through simultaneous equations, should correspond to another one of the potential optimal solutions on the graph.

Again, this solution is referred to as a basic feasible solution.

All three of these solutions are feasible solutions. But …

… in each example, how did we know which variables to set to zero?

… how was the optimal solution found in the 3rd example?

The answers to both of these questions can be found with the simplex method. Stay tuned.