 Pertemuan 2 – Menyelesaikan Formulasi Model Dengan Metode Grafik Riset Operasiomal- dewiyani 1.

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Pertemuan 2 – Menyelesaikan Formulasi Model Dengan Metode Grafik Riset Operasiomal- dewiyani 1

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:

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

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

7 A feasible solution does not violate any of the constraints: Example x 1 = 5 bowls x 2 = 10 mugs Z = \$40x 1 + \$50x 2 = \$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 Feasible Solutions

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

9  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. Graphical Solution of Linear Programming Models

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

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

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

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

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

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

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

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

18 Alternative Objective Function Solution Lines Graphical Solution of Maximization Model (9 of 13) Figure 2.8 Alternative Objective Function Lines Maximize Z = \$40x 1 + \$50x 2 subject to:1x 1 + 2x 2  40 4x 1 + 3x 2  120 x 1, x 2  0 X 2 = 0 X 1 = 0

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

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

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

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

23 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 ?

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25 Problem Definition A Minimization Model Example (2 of 7) Decision Variables: x 1 = bags of Super-Gro x 2 = bags of Crop-Quick The Objective Function: Minimize Z = \$6x 1 + 3x 2 Where: \$6x 1 = cost of bags of Super-Gro \$3x 2 = cost of bags of Crop-Quick Model Constraints: 2x 1 + 4x 2  16 lb (nitrogen constraint) 4x 1 + 3x 2  24 lb (phosphate constraint) x 1, x 2  0 (non-negativity constraint)

26 Model Formulation and Constraint Graph A Minimization Model Example (3 of 7) Minimize Z = \$6x 1 + \$3x 2 subject to:2x 1 + 4x 2  16 4x 1 + 3x 2  24 x 1, x 2  0 Figure 2.14 Graph of Both Model Constraints

27 Feasible Solution Area A Minimization Model Example (4 of 7) Minimize Z = \$6x 1 + \$3x 2 subject to:2x 1 + 4x 2  16 4x 1 + 3x 2  24 x 1, x 2  0 Figure 2.15 Feasible Solution Area After we determine the feasible solution area, what do we do to determine the optimal solution?

28 Optimal Solution Point A Minimization Model Example (5 of 7) Minimize Z = \$6x 1 + \$3x 2 subject to:2x 1 + 4x 2  16 4x 1 + 3x 2  24 x 1, x 2  0 Figure 2.16 Optimum Solution Point

Restaurant Case Study 29  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 Main DishTime to PrepareProfit Fish15 minutes\$12 Beef30 minutes\$16

Restaurant Case Study Understand the Problem 30  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 Main DishTime to PrepareProfit Fish (x1)15 minutes (1/4 hour)\$12 Beef (x2)30 minutes (1/2 hour)\$16 Objective Constraints Parameters

Restaurant Case Study Formulate the Problem 31 Maximize Z = 12*x1 + 16*x2 Or 2*x1 – 3*x2 ≥ 0 Or – 0.10*x1 + 0.90*x2 ≥ 0 subject tox1 + x2 ≤ 60 0.25*x1 +.50*x2 ≤ 20 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 Main DishTime to PrepareProfit Fish (x1)15 minutes (1/4 hour)\$12 Beef (x2)30 minutes (1/2 hour)\$16 Objective Constraints Parameters

Restaurant Case Study Graphical Solution 32 # of meals Fish to Beef Ratio Minimal Beef entrée % Labor constraint Obj. Fct.

Restaurant Case Study What-if Analysis (1 of 2) 33 Maximize Z = 16*x1 + 16*x2 Or 2*x1 – 3*x2 ≥ 0 Or – 0.20*x1 + 0.80*x2 ≥ 0 subject tox1 + x2 ≤ 60 0.25*x1 +.50*x2 ≤ 20 x1 / x2 ≥ 3 / 2 x2 / (x1 + x2) >= 0.20 x1, x2 ≥ 0  At least 20% of their customers will order beef dinners Main DishTime to PrepareProfit Fish (x1)15 minutes (1/4 hour)\$16 Beef (x2)30 minutes (1/2 hour)\$16 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%. What are the slopes?

34  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 Irregular Types of Linear Programming Problems

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

36 An Infeasible Problem Every possible solution violates at least one constraint: Maximize Z = 5x 1 + 3x 2 subject to: 4x 1 + 2x 2  8 x 1  4 x 2  6 x 1, x 2  0 Figure 2.19 Graph of an Infeasible Problem

37 Value of objective function increases indefinitely: Maximize Z = 4x 1 + 2x 2 subject to: x 1  4 x 2  2 x 1, x 2  0 An Unbounded Problem Figure 2.20 Graph of an Unbounded Problem

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