Graphical Solution Procedure Linear Programming Graphical Solution Procedure

Two Variable Linear Programs When a linear programming model consists of only two variables, a graphical approach can be employed to solve the model. There are few, if any, real life linear programming models with two variables. But the graphical illustration of this case helps develop the terminology and approaches for solving larger models.

5-Step Graphical Solution Procedure Graph the constraints. The resulting set of possible or feasible points is called the feasible region. Set the objective function value equal to any number and graph the resulting objective function line. Move the objective function line parallel to itself until it touches the last point(s) of the feasible region. This is the optimal solution. Solve for the values of the variables of the optimal solution. This involves solving 2 equations in 2 unknowns. Determin the optimal value of the objective function. This is done by substituting the variable values into the objective function formula.

Graphing the Feasible Region X2 1000 900 800 700 600 500 400 300 200 100 X1 2X1 + 1X2 ≤ 1000 3X1 + 4X2 ≤ 2400 1X1 + 1X2 ≤ 700 1X1 - 1X2 ≤ 350 X1, X2 ≥ 0

Graphing the Feasible Region X2 1000 900 800 700 600 500 400 300 200 100 X1 2X1 + 1X2 ≤ 1000 2X1 + 1X2 ≤ 1000 3X1 + 4X2 ≤ 2400 1X1 + 1X2 ≤ 700 1X1 - 1X2 ≤ 350 X1, X2 ≥ 0

Graphing the Feasible Region X2 1000 900 800 700 600 500 400 300 200 100 X1 2X1 + 1X2 ≤ 1000 3X1 + 4X2 ≤ 2400 1X1 + 1X2 ≤ 700 1X1 - 1X2 ≤ 350 X1, X2 ≥ 0 2X1 + 1X2 ≤ 1000 3X1 + 4X2 ≤ 2400

Graphing the Feasible Region X2 1000 900 800 700 600 500 400 300 200 100 X1 2X1 + 1X2 ≤ 1000 3X1 + 4X2 ≤ 2400 1X1 + 1X2 ≤ 700 1X1 - 1X2 ≤ 350 X1, X2 ≥ 0 2X1 + 1X2 ≤ 1000 1X1 + 1X2 ≤ 700 Redundant Constraint 3X1 + 4X2 ≤ 2400

Graphing the Feasible Region X2 1000 900 800 700 600 500 400 300 200 100 X1 2X1 + 1X2 ≤ 1000 3X1 + 4X2 ≤ 2400 1X1 + 1X2 ≤ 700 1X1 - 1X2 ≤ 350 X1, X2 ≥ 0 2X1 + 1X2 ≤ 1000 1X1 + 1X2 ≤ 700 1X1 - 1X2 ≤ 350 Feasible Region 3X1 + 4X2 ≤ 2400

Characterization of Points X2 1000 900 800 700 600 500 400 300 200 100 X1 (200,700) is an infeasible point 3X1 + 4X2 ≤ 2400 (200,450) is a boundary point 2X1 + 1X2 ≤ 1000 (200,200) is an interior point (450,100) is an extreme point 1X1 - 1X2 ≤ 350

Graphing the Objective Function X2 1000 900 800 700 600 500 400 300 200 100 X1 Set the objective function value equal to any number and graph it. 8X1 + 5X2 = 2000

Identifying the Optimal Point X2 1000 900 800 700 600 500 400 300 200 100 X1 Move the objective function line parallel to itself until it touches the last point of the feasible region. 8X1 + 5X2 = 4360 8X1 + 5X2 = 4000 8X1 + 5X2 = 3000 8X1 + 5X2 = 2000 OPTIMAL POINT

Determining the Optimal Point X2 1000 900 800 700 600 500 400 300 200 100 X1 Solve the 2 equations in 2 unknowns that determines the optimal point. 2X1 + 1X2 = 1000 3X1 + 4X2 = 2400 2X1 + 1X2 = 1000 3X1 + 4X2 = 2400 Multiply top equation by 4 8X1 + 4X2 = 4000 3X1 + 4X2 = 2400 Subtract second equation from first 5X1 = 1600 or X1 = 320 Substituting in first equation 2(320) + 1X2 = 1000 or X2 = 360 OPTIMAL POINT (320, 360)

Determining the Optimal Objective Function Value X2 1000 900 800 700 600 500 400 300 200 100 X1 Substitute x-values into objective function MAX 8 X1 + 5 X2 (320) (360) + = 4360 OPTIMAL POINT (320, 360)

Review 5 Step Graphical Solution Procedure Redundant Constraints Graph constraints Graph objective function line Move the objective function line parallel until it touches the last point of the feasible region Solve for optimal point Determin the optimal objective function value Redundant Constraints Identification of points Infeasible Interior Boundary Extreme