# Linear Programming Graphical Solution Procedure. Two Variable Linear Programs When a linear programming model consists of only two variables, a graphical.

## Presentation on theme: "Linear Programming Graphical Solution Procedure. Two Variable Linear Programs When a linear programming model consists of only two variables, a graphical."— Presentation transcript:

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 1.Graph the constraints. feasible regionThe resulting set of possible or feasible points is called the feasible region. 2.Set the objective function value equal to any number and graph the resulting objective function line. 3.Move the objective function line parallel to itself until it touches the last point(s) of the feasible region. optimal solutionThis is the optimal solution. 4.Solve for the values of the variables of the optimal solution. This involves solving 2 equations in 2 unknowns. 5.Solve for 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 0 100200 300400500600 700800 X1 X 1, X 2 ≥ 0 2X 1 + 1X 2 ≤ 1000 3X 1 + 4X 2 ≤ 2400 1X 1 + 1X 2 ≤ 700 1X 1 - 1X 2 ≤ 350 X 1, X 2 ≥ 0

Graphing the Feasible Region X2 1000 900 800 700 600 500 400 300 200 100 0 100200 300400500600 700800 X1 2X 1 + 1X 2 ≤ 1000 X 1, X 2 ≥ 0 2X 1 + 1X 2 ≤ 1000 3X 1 + 4X 2 ≤ 2400 1X 1 + 1X 2 ≤ 700 1X 1 - 1X 2 ≤ 350 X 1, X 2 ≥ 0 2X 1 + 1X 2 ≤ 1000

Graphing the Feasible Region X2 1000 900 800 700 600 500 400 300 200 100 0 100200 300400500600 700800 X1 2X 1 + 1X 2 ≤ 1000 3X 1 + 4X 2 ≤ 2400 X 1, X 2 ≥ 0 2X 1 + 1X 2 ≤ 1000 3X 1 + 4X 2 ≤ 2400 1X 1 + 1X 2 ≤ 700 1X 1 - 1X 2 ≤ 350 X 1, X 2 ≥ 0 2X 1 + 1X 2 ≤ 1000 3X1 + 4X2 ≤ 2400

Graphing the Feasible Region X2 1000 900 800 700 600 500 400 300 200 100 0 100200 300400500600 700800 X1 2X 1 + 1X 2 ≤ 1000 3X 1 + 4X 2 ≤ 2400 1X 1 + 1X 2 ≤ 700 X 1, X 2 ≥ 0 2X 1 + 1X 2 ≤ 1000 3X 1 + 4X 2 ≤ 2400 1X 1 + 1X 2 ≤ 700 1X 1 - 1X 2 ≤ 350 X 1, X 2 ≥ 0 2X 1 + 1X 2 ≤ 1000 3X1 + 4X2 ≤ 2400 1X1 + 1X2 ≤ 700 Redundant Constraint

Graphing the Feasible Region 2X 1 + 1X 2 ≤ 1000 3X 1 + 4X 2 ≤ 2400 1X 1 + 1X 2 ≤ 700 1X 1 - 1X 2 ≤ 350 X 1, X 2 ≥ 0 2X 1 + 1X 2 ≤ 1000 3X1 + 4X2 ≤ 2400 1X1 + 1X2 ≤ 700 1X1 - 1X2 ≤ 350 X2 1000 900 800 700 600 500 400 300 200 100 0 100200 300400500600 700800 X1 Feasible Region

Characterization of Points X2 1000 900 800 700 600 500 400 300 200 100 0 100200 300400500600 700800 X1 3X1 + 4X2 ≤ 2400 2X 1 + 1X 2 ≤ 1000 1X1 - 1X2 ≤ 350 interior (200,200) is an interior point infeasible (200,700) is an infeasible point boundary (200,450) is a boundary point extreme (450,100) is an extreme point

Graphing the Objective Function X2 1000 900 800 700 600 500 400 300 200 100 0 100200 300400500600 700800 X1 Set the objective function value equal to any number and graph it. 2000 8X 1 + 5X 2 = 2000

Identifying the Optimal Point X2 1000 900 800 700 600 500 400 300 200 100 0 100200 300400500600 700800 X1 Move the objective function line parallel to itself until it touches the last point of the feasible region. 2000 8X 1 + 5X 2 = 2000 3000 8X 1 + 5X 2 = 3000 4000 8X 1 + 5X 2 = 4000 4360 8X 1 + 5X 2 = 4360 OPTIMAL POINT

Determining the Optimal Point X2 1000 900 800 700 600 500 400 300 200 100 0 100200 300400500600 700800 X1 Solve the 2 equations in 2 unknowns that determines the optimal point. OPTIMAL POINT 2X1 + 1X2 = 1000 3X1 + 4X2 = 2400 2X 1 + 1X 2 = 1000 3X 1 + 4X 2 = 2400 8X 1 + 4X 2 = 4000 3X 1 + 4X 2 = 2400 Multiply top equation by 4 Subtract second equation from first 5X 1 = 1600 or X 1 = 320 Substituting in first equation 2(320) + 1X 2 = 1000 or X 2 = 360 (320, 360)

Determining the Optimal Objective Function Value X2 1000 900 800 700 600 500 400 300 200 100 0 100200 300400500600 700800 X1 Substitute x-values into objective function OPTIMAL POINT (320, 360) MAX 8 X 1 + 5 X 2 (320) (360) + = 4360

Review 5 Step Graphical Solution Procedure –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 – Solve for the optimal objective function value Redundant Constraints Identification of points –Infeasible –Interior –Boundary –Extreme

Download ppt "Linear Programming Graphical Solution Procedure. Two Variable Linear Programs When a linear programming model consists of only two variables, a graphical."

Similar presentations