Linear programming Dr. Arturo S. Leon, BSU (Spring 2010)

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1 Linear programming (Adapted from Chapter 13 Supplement, Operations and Management, 5 th edition by Roberta Russell & Bernard W. Taylor, III., Copyright 2006 John Wiley & Sons, Inc. This presentation also contains material of Pearson, Prentice hall Dr. Arturo S. Leon, BSU (Spring 2010) 1 Arturo S. Leon, BSU, Spring 2010

2 Lecture Outline S up pl e m en t 13-2 Model Formulation Graphical Solution Method Linear Programming Model Solution Solving Linear Programming Problems with Excel Sensitivity Analysis Copyright 2006 John Wiley & Sons, Inc.

3 Linear Programming (LP) A model consisting of linear relationships representing a firm s objective and resource constraints LP is a mathematical modeling technique used to determine a level of operational activity in order to achieve an objective, subject to restrictions called constraints Copyright 2006 John Wiley & Sons, Inc. Supplement 13-3

4 Linear Programming A model consisting of linear relationships representing a firm s objective and resource constraints LP is a mathematical modeling technique used to determine a level of operational activity in order to achieve an objective, subject to restrictions called constraints Copyright 2006 John Wiley & Sons, Inc. Supplement 13-4

5 Video of Linear Programming (LP) d= Copyright 2006 John Wiley & Sons, Inc. Supplement 13-5

6 LP Model Formulation Decision variables mathematical symbols representing levels of activity of an operation Objective function a linear relationship reflecting the objective of an operation most frequent objective of business firms is to maximize profit most frequent objective of individual operational units (such as a production or packaging department) is to minimize cost Constraint a linear relationship representing a restriction on decision making Copyright 2006 John Wiley & Sons, Inc. Supplement 13-6

7 LP Model Formulation (cont.) Max/min z = c 1 x 1 + c 2 x c n x n subject to: a 11 x 1 + a 12 x a 1n x n (, =, ) b 1 a 21 x 1 + a 22 x a 2n x n (, =, ) b 2 : a m1 x1 + a m2 x a mn x n (, =, ) b m x j = decision variables b i = constraint levels c j = objective function coefficients a ij = constraint coefficients Copyright 2006 John Wiley & Sons, Inc. Supplement 13-7

8 LP Model: Example PRODUCT Labor (hr/unit) Clay (lb/unit) Revenue ($/unit) Bowl Mug There are 40 hours of labor and 120 pounds of clay available each day Decision variables x 1 = number of bowls to produce x 2 = number of mugs to produce RESOURCE REQUIREMENTS Copyright 2006 John Wiley & Sons, Inc. Supplement 13-8

9 LP Formulation: Example Maximize Z = $40 x x 2 Subject to x 1 + 2x 2 40 hr (labor constraint) 4x 1 + 3x lb (clay constraint) x 1, x 2 0 Quick solution with Excel solver Initial conditions for Solver Using solver for the Bowl/mug example Decision variables X1-5 Maximize -250 x2-1 constraint 1: -7 <= 40 constraint 2: -23 <= 120 constraint 3: -5 >= 0 constraint 4: -1 >= 0

10 LP Example (Cont.) Solution Using solver for the Bowl/mug example Decision variables X1 24 Maximize 1360 x2 8 constraint 1: 40 <= 40 constraint 2: 120 <= 120 constraint 3: 24 >= 0 constraint 4: 8 >= 0 Solution is x 1 = 24 bowls x 2 = 8 mugs Revenue or benefit = $1,360

11 Graphical Solution Method 1. Plot model constraint on a set of coordinates in a plane 2. Identify the feasible solution space on the graph where all constraints are satisfied simultaneously 3. Plot objective function to find the point on boundary of this space that maximizes (or minimizes) value of objective function Copyright 2006 John Wiley & Sons, Inc. Supplement 13-11

12 LP Formulation: Example Maximize Z = $40 x x 2 Subject to x 1 + 2x 2 40 hr (labor constraint) 4x 1 + 3x lb (clay constraint) x 1, x 2 0 Solution is x 1 = 24 bowls x 2 = 8 mugs Revenue = $1,360 Copyright 2006 John Wiley & Sons, Inc. Supplement 13-12

13 Graphical Solution: Example x x x lb Area common to both constraints x x 2 40 hr x 1 Copyright 2006 John Wiley & Sons, Inc. Supplement 13-13

14 Computing Optimal Values x x x lb x x 2 40 hr x 1 + 2x 2 = 40 4x 1 + 3x 2 = 120 4x 1 + 8x 2 = 160-4x 1-3x 2 = x 2 = 40 x 2 = 8 x 1 + 2(8) = 40 x 1 = x 1 Z = $50(24) + $50(8) = $1,360 Copyright 2006 John Wiley & Sons, Inc. Supplement 13-14

15 Extreme Corner Points x A x 1 = 0 bowls x 2 = 20 mugs Z = $1,000 x 1 = 224 bowls x 2 = 8 mugs Z = $1,360 x 1 = 30 bowls x 2 = 0 mugs Z = $1, B C x 1 Copyright 2006 John Wiley & Sons, Inc. Supplement 13-15

16 Objective Function x x 1 + 3x lb A Z = 70x x 2 Optimal point: x 1 = 30 bowls x 2 = 0 mugs Z = $2, B C 30 x 1 + 2x 2 40 hr 40 Copyright 2006 John x 1 Wiley & Sons, Inc. Supplement 13-16

17 Minimization Problem CHEMICAL CONTRIBUTION Brand Nitrogen (lb/bag) Phosphate (lb/bag) Gro-plus 2 4 Crop-fast 4 3 Minimize Z = $6x 1 + $3x 2 subject to 2x 1 + 4x 1 + 4x 2 16 lb of nitrogen 3x 2 24 lb of phosphate x 1, x 2 0 Copyright 2006 John Wiley & Sons, Inc. Supplement 13-17

18 Graphical Solution x x 1 = 0 bags of Gro-plus x 2 = 8 bags of Crop-fast Z = $ A Z = 6x 1 + 3x B 6 C x 1 Copyright 2006 John Wiley & Sons, Inc. Supplement 13-18

19 Copyright 2006 John Wiley & Sons, Inc. Supplement 13-19

20 Another example: Max 7T + 5C (profit) Subject to the constraints: 3T + 4C < 2400 (carpentry hrs) 2T + 1C < 1000 (painting hrs) C < 450 (max # chairs) T > 100 (min # tables) T, C > 0 (nonnegativity)

21 Graphical Solution Graphing an LP model helps provide insight into LP models and their solutions. While this can only be done in two dimensions, the same properties apply to all LP models and solutions.

22 Carpentry Constraint Line 3T + 4C = 2400 Intercepts C 600 Infeasible > 2400 hrs (T = 0, C = 600) (T = 800, C = 0) 0 Feasible < 2400 hrs T

23 Painting Constraint Line 2T + 1C = 1000 C Intercepts (T = 0, C = 1000) (T = 500, C = 0) T

24 Max Chair Line C = 450 C 1000 Min Table Line T = Feasible 0 Region T

25 Objective Function Line 7T + 5C = Profit C Optimal Point (T = 320, C = 360) T

26 Additional Constraint Need at least 75 more chairs than tables C > T + 75 Or C New optimal point T = 300, C = 375 T = 320 C = 360 No longer feasible C T > T

27 LP Characteristics Feasible Region: The set of points that satisfies all constraints Corner Point Property: An optimal solution must lie at one or more corner points Optimal Solution: The corner point with the best objective function value is optimal

28 Special Situation in LP 1. Redundant Constraints - do not affect the feasible region Example: x < 10 x < 12 The second constraint is redundant because it is less restrictive.

29 Special Situation in LP 2. Infeasibility when no feasible solution exists (there is no feasible region) Example: x < 10 x > 15

30 Special Situation in LP 3. Alternate Optimal Solutions when there is more than one optimal solution Max 2T + 2C Subject to: T + C < 10 T < 5 C < 6 T, C > 0 C All points on Red segment are optimal T

31 Special Situation in LP 4. Unbounded Solutions when nothing prevents the solution from becoming infinitely large Max 2T + 2C Subject to: 2T + 3C > 6 T, C > 0 C T