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Pertemuan 2 – Menyelesaikan Formulasi Model Dengan Metode Grafik

Pertemuan 2 – Menyelesaikan Formulasi Model Dengan Metode Grafik. Riset Operasiomal - dewiyani. Methods to Solve LP Problems. Graphical Method Simplex Method. Problem Definition A Maximization Model Example (1 of 3). Product mix problem - Beaver Creek Pottery Company

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Pertemuan 2 – Menyelesaikan Formulasi Model Dengan Metode Grafik

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  1. Pertemuan 2 – MenyelesaikanFormulasi Model DenganMetodeGrafik RisetOperasiomal- dewiyani

  2. Methods to Solve LP Problems • Graphical Method • Simplex Method

  3. Problem Definition A Maximization Model Example (1 of 3) • Product mix problem - Beaver Creek Pottery Company • How many bowls and mugs should be produced to maximize profits given labor and materials constraints? • Product resource requirements and unit profit:

  4. Problem Definition A Maximization Model Example (2 of 3) Resource 40 hrs of labor per day Availability: 120 lbs of clay Decision x1 = number of bowls to produce per day Variables: x2 = number of mugs to produce per day Objective Maximize Z = $40x1 + $50x2 Function: Where Z = profit per day Resource 1x1 + 2x2 40 hours of labor Constraints: 4x1 + 3x2 120 pounds of clay Non-Negativity x1  0; x2  0 Constraints:

  5. Problem Definition A Maximization Model Example (3 of 3) Complete Linear Programming Model: Maximize Z = $40x1 + $50x2 subject to: 1x1+ 2x2  40 4x2+ 3x2  120 x1, x2  0

  6. Feasible Solutions • A feasible solution does not violate any of the constraints: • Example x1 = 5 bowls • x2 = 10 mugs • Z = $40x1 + $50x2 = $700 • Labor constraint check: • 1(5) + 2(10) = 25 < 40 hours, within constraint • Clay constraint check: • 4(5) + 3(10) = 70 < 120 pounds, within constraint

  7. Infeasible Solutions • An infeasible solution violates at least one of the constraints: • Example x1 = 10 bowls • x2 = 20 mugs • Z = $1400 • Labor constraint check: • 1(10) + 2(20) = 50 > 40 hours, violates constraint

  8. Graphical Solution of Linear Programming Models • Graphical solution is limited to linear programming models containing only two decision variables (can be used with three variables but only with great difficulty). • Graphical methods provide visualization of how a solution for a linear programming problem is obtained.

  9. Coordinate Axes Graphical Solution of Maximization Model (1 of 13) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 • After defining the axes, begin drawing the constraints • What is the x or y intercept? • What is the slope? Figure 2.1 Coordinates for Graphical Analysis

  10. Labor Constraint Graphical Solution of Maximization Model (2 of 13) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.1 Graph of Labor Constraint

  11. Labor Constraint Area Graphical Solution of Maximization Model (3 of 13) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 • After the labor constraint, let’s move onto the clay constraint • What is the x or y intercept? • What is the slope? Figure 2.3 Labor Constraint Area

  12. Clay Constraint Area Graphical Solution of Maximization Model (4 of 13) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.4 Clay Constraint Area

  13. Both Constraints Graphical Solution of Maximization Model (5 of 13) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.5 Graph of Both Model Constraints

  14. Don’t forget the non-negative constraints Graphical Solution of Maximization Model (6 of 13) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 X2 = 0 X1 = 0 Figure 2.6 Feasible Solution Area

  15. Feasible Solution Area Graphical Solution of Maximization Model (7 of 13) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.6 Feasible Solution Area

  16. Objective Solution = $800 Graphical Solution of Maximization Model (8 of 13) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Arbitrarily set the objective function as $800 X2 = 0 X1 = 0 Figure 2.7 Objection Function Line for Z = $800

  17. Alternative Objective Function Solution Lines Graphical Solution of Maximization Model (9 of 13) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 X2 = 0 X1 = 0 Figure 2.8 Alternative Objective Function Lines

  18. Optimal Solution Graphical Solution of Maximization Model (10 of 13) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Above and Beyond What if we forgot to include the non-negative constraints? Figure 2.9 Identification of Optimal Solution

  19. Optimal Solution Coordinates Graphical Solution of Maximization Model (11 of 13) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.10 Optimal Solution Coordinates

  20. Corner Point Solutions Graphical Solution of Maximization Model (12 of 13) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Why are corner points important? Figure 2.11 Solution at All Corner Points

  21. Optimal Solution for New Objective Function Graphical Solution of Maximization Model (13 of 13) Maximize Z = $70x1+ $20x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.12 Optimal Solution with Z = 70x1 + 20x2

  22. Problem Definition A Minimization Model Example (1 of 7) • Two brands of fertilizer available - Super-Gro, Crop-Quick. • Field requires at least 16 pounds of nitrogen and 24 pounds of phosphate. • Super-Gro costs $6 per bag, Crop-Quick $3 per bag. • Problem: How much of each brand to purchase to minimize total cost of fertilizer given following data ?

  23. Problem Definition A Minimization Model Example (2 of 7) Decision Variables: x1 = bags of Super-Gro x2 = bags of Crop-Quick The Objective Function: Minimize Z = $6x1 + 3x2 Where: $6x1 = cost of bags of Super-Gro $3x2 = cost of bags of Crop-Quick Model Constraints: 2x1 + 4x2 16 lb (nitrogen constraint) 4x1 + 3x2 24 lb (phosphate constraint) x1, x2 0 (non-negativity constraint)

  24. Model Formulation and Constraint Graph A Minimization Model Example (3 of 7) Minimize Z = $6x1 + $3x2 subject to: 2x1 + 4x2  16 4x1+ 3x2  24 x1, x2  0 Figure 2.14 Graph of Both Model Constraints

  25. Feasible Solution Area A Minimization Model Example (4 of 7) Minimize Z = $6x1 + $3x2 subject to: 2x1 + 4x2  16 4x1 + 3x2  24 x1, x2  0 After we determine the feasible solution area, what do we do to determine the optimal solution? Figure 2.15 Feasible Solution Area

  26. Optimal Solution Point A Minimization Model Example (5 of 7) Minimize Z = $6x1 + $3x2 subject to: 2x1 + 4x2  16 4x1 + 3x2  24 x1, x2  0 Figure 2.16 Optimum Solution Point

  27. Restaurant Case Study • Angela and Zooey decided to open a French restaurant, a small town near the University they graduated from • Before they offer a full variety menu, they would like to do a “test run” to learn more about the market • Offering two four-course meals each night, one with beef and one with fish as the main course • Experiment with different appetizers, soups, side dishes and desserts • Need to determine how many meals to prepare each night to maximize profit • Impact ingredients purchasing, staffing and work schedule • Cannot afford too much waste • Estimated about 60 meals per night. Available kitchen labor = 20 hours • Initial market analysis indicates that • For health reasons, at least three fish dinners for every two beef dinners • Regardless at least 10% of their customers will order beef dinners

  28. Restaurant Case StudyUnderstand the Problem Objective Constraints Parameters Need to determine how many meals to prepare each night to maximize profit Estimated about 60 meals per night. Available kitchen labor = 20 hours For health reasons, at least three fish dinners for every two beef dinners Regardless at least 10% of their customers will order beef dinners

  29. Restaurant Case Study Formulate the Problem Objective Constraints Parameters Maximize Z = 12*x1 + 16*x2 subject to x1 + x2 ≤ 60 0.25*x1 + .50*x2 ≤ 20 Or 2*x1 – 3*x2 ≥ 0 Or – 0.10*x1 + 0.90*x2 ≥ 0 x1 / x2 ≥ 3 / 2 x2 / (x1 + x2) >= .10 x1, x2 ≥ 0 Need to determine how many meals to prepare each night to maximize profit Estimated about 60 meals per night. Available kitchen labor = 20 hours For health reasons, at least three fish dinners for every two beef dinners Regardless at least 10% of their customers will order beef dinners

  30. Restaurant Case Study Graphical Solution # of meals Fish to Beef Ratio Minimal Beef entrée % Obj. Fct. Labor constraint

  31. Restaurant Case Study What-if Analysis (1 of 2) Angela and Zooey wonders what would be the impact of increasing fish dish price to match profit with beef dish? Based on the initial market survey estimates, the demand for beef will increase from 10% of the sales to 20%. Maximize Z = 16*x1 + 16*x2 What are the slopes? subject to x1 + x2 ≤ 60 0.25*x1 + .50*x2 ≤ 20 Or 2*x1 – 3*x2 ≥ 0 Or – 0.20*x1 + 0.80*x2 ≥ 0 x1 / x2 ≥ 3 / 2 x2 / (x1 + x2) >= 0.20 x1, x2 ≥ 0 At least 20% of their customers will order beef dinners

  32. Irregular Types of Linear Programming Problems • For some linear programming models, the general rules do not apply. • Special types of problems include those with: • Multiple optimal solutions • Infeasible solutions • Unbounded solutions

  33. Multiple Optimal Solutions Beaver Creek Pottery Example Objective function is parallel to a constraint line. Maximize Z=$40x1 + 30x2 subject to: 1x1 + 2x2  40 4x2 + 3x2  120 x1, x2  0 Where: x1 = number of bowls x2 = number of mugs Figure 2.18 Example with Multiple Optimal Solutions

  34. An Infeasible Problem Every possible solution violates at least one constraint: Maximize Z = 5x1 + 3x2 subject to: 4x1 + 2x2 8 x1 4 x2 6 x1, x2 0 Figure 2.19 Graph of an Infeasible Problem

  35. An Unbounded Problem Value of objective function increases indefinitely: Maximize Z = 4x1 + 2x2 subject to: x1 4 x2 2 x1, x2 0 Figure 2.20 Graph of an Unbounded Problem

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