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Linear Programming Graphical Solution Procedure

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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.

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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.

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Graphing the Feasible Region X X1 X 1, X 2 ≥ 0 2X 1 + 1X 2 ≤ X 1 + 4X 2 ≤ X 1 + 1X 2 ≤ 700 1X 1 - 1X 2 ≤ 350 X 1, X 2 ≥ 0

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Graphing the Feasible Region X X1 2X 1 + 1X 2 ≤ 1000 X 1, X 2 ≥ 0 2X 1 + 1X 2 ≤ X 1 + 4X 2 ≤ X 1 + 1X 2 ≤ 700 1X 1 - 1X 2 ≤ 350 X 1, X 2 ≥ 0 2X 1 + 1X 2 ≤ 1000

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Graphing the Feasible Region X X1 2X 1 + 1X 2 ≤ X 1 + 4X 2 ≤ 2400 X 1, X 2 ≥ 0 2X 1 + 1X 2 ≤ X 1 + 4X 2 ≤ X 1 + 1X 2 ≤ 700 1X 1 - 1X 2 ≤ 350 X 1, X 2 ≥ 0 2X 1 + 1X 2 ≤ X1 + 4X2 ≤ 2400

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Graphing the Feasible Region X X1 2X 1 + 1X 2 ≤ X 1 + 4X 2 ≤ X 1 + 1X 2 ≤ 700 X 1, X 2 ≥ 0 2X 1 + 1X 2 ≤ X 1 + 4X 2 ≤ X 1 + 1X 2 ≤ 700 1X 1 - 1X 2 ≤ 350 X 1, X 2 ≥ 0 2X 1 + 1X 2 ≤ X1 + 4X2 ≤ X1 + 1X2 ≤ 700 Redundant Constraint

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Graphing the Feasible Region 2X 1 + 1X 2 ≤ X 1 + 4X 2 ≤ X 1 + 1X 2 ≤ 700 1X 1 - 1X 2 ≤ 350 X 1, X 2 ≥ 0 2X 1 + 1X 2 ≤ X1 + 4X2 ≤ X1 + 1X2 ≤ 700 1X1 - 1X2 ≤ 350 X X1 Feasible Region

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Characterization of Points X X1 3X1 + 4X2 ≤ X 1 + 1X 2 ≤ X1 - 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

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Graphing the Objective Function X X1 Set the objective function value equal to any number and graph it X 1 + 5X 2 = 2000

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Identifying the Optimal Point X X1 Move the objective function line parallel to itself until it touches the last point of the feasible region X 1 + 5X 2 = X 1 + 5X 2 = X 1 + 5X 2 = X 1 + 5X 2 = 4360 OPTIMAL POINT

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Determining the Optimal Point X X1 Solve the 2 equations in 2 unknowns that determines the optimal point. OPTIMAL POINT 2X1 + 1X2 = X1 + 4X2 = X 1 + 1X 2 = X 1 + 4X 2 = X 1 + 4X 2 = X 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)

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Determining the Optimal Objective Function Value X X1 Substitute x-values into objective function OPTIMAL POINT (320, 360) MAX 8 X X 2 (320) (360) + = 4360

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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

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