3.4 Linear Programming

1. 3.4 Linear Programming

2. What is a Linear Program? A linear program is a mathematical model that indicates the goal and requirements of an allocation problem. It has two or more non-negativevariables. Its objective is expressed as a mathematical function. The objective function plots as a line on a two-dimensional graph. There are constraints that affect possible levels of the variables. In two dimensions these plot as lines and ordinarily define areas in which the solution must lie.

3. Properties of LP Models • Seek to minimize or maximize • Include “constraints” or limitations • There must be alternatives available • All equations are linear

4. Example LP Model Formulation:The Product Mix Problem Decision: How much to make of > 2 products? Objective: Maximize profit Constraints: Limited resources

5. Example: Flair Furniture Co. Two products: Chairs and Tables Decision: How many of each to make this month? Objective: Maximize profit

6. Flair Furniture Co. Data • Other Limitations: • Make no more than 450 chairs • Make at least 100 tables

7. Decision Variables: T = Num. of tables to make C = Num. of chairs to make Objective Function: Maximize Profit Maximize $7 T + $5 C

8. Constraints: • Have 2400 hours of carpentry time available 3 T + 4 C < 2400 (hours) • Have 1000 hours of painting time available 2 T + 1 C < 1000(hours)

9. More Constraints: • Make no more than 450 chairs C < 450 (num. chairs) • Make at least 100 tables T > 100 (num. tables) Nonnegativity: Cannot make a negative number of chairs or tables T > 0 C > 0

10. Model Summary 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)

11. 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.

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

13. C 1000 600 0 Painting Constraint Line 2T + 1C = 1000 Intercepts (T = 0, C = 1000) (T = 500, C = 0) 2T + 1C = 1000 0 500 800 T

14. C 1000 600 450 0 Max Chair Line C = 450 Min Table Line T = 100 Feasible Region 0 100 500 800 T

15. C 500 400 300 200 100 0 Objective Function Line 7T + 5C = Profit 7T + 5C = $4,040 Optimal Point (T = 320, C = 360) 0 100 200 300 400 500 T

16. Finding Most Attractive Corner • The optimal solution will always correspond to a corner point of the feasible solution region. • Because there can be many corners, the most attractive corner is easiest to find visually. • That is done by plotting two P lines for arbitrary profit levels.

17. 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

18. Because graphs can be inaccurate due to human error and often the numbers are very large or very small it is best to identify the corner point through a system of equations. Simply identify which lines intersect to form the corner point and solve them simultaneously.

19. RESOURCE REQUIREMENTS Labor Clay Revenue PRODUCT (hr/unit) (lb/unit) ($/unit) Bowl 1 4 40 Mug 2 3 50 There are 40 hours of labor and 120 pounds of clay available each day Decision variables xb= number of bowls to produce xm= number of mugs to produce LP Model: Example

20. We have 240 acres of land to plant corn and oats. We make a profit of $40 per acre of corn and $30 per acre of oats. We have 320 hours of available labor. Corn takes 2 hours to plant per acre and oats require 1 hour. How many acres of each should we plant to maximize our profits?

21. Xc = acres of corn • Xo = acres of oats • Maximum Profit = 40Xc + 30Xo • Xc+Xo ≤240 • 2Xc+Xo ≤320 • Xc ≥0 • Xo ≥0

22. LP Formulation: Example Maximize Z = $40 x1 + 50 x2 Subject to x1 + 2x2 40 hr (labor constraint) 4x1 + 3x2 120 lb (clay constraint) x1 , x2 0 Solution is x1 = 24 bowls x2 = 8 mugs Revenue = $1,360

23. 50 – 40 – 30 – 20 – 10 – 0 – x2 4 x1 + 3 x2 120 lb x1 + 2 x2 40 hr | 10 | 20 | 30 | 40 | 50 | 60 x1 Graphical Solution: Example

24. x1 = 0 bowls x2 =20 mugs Z = $1,000 x2 x1 = 224 bowls x2 =8 mugs Z = $1,360 40 – 30 – 20 – 10 – 0 – x1 = 30 bowls x2 =0 mugs Z = $1,200 A B C | 10 | 20 | 30 | 40 x1 Extreme Corner Points

25. Mixture • A rancher is mixing two types of food, Brand X and Brand Y, for his cattle. If each serving is required to have 60 grams of protein and 30 grams of fat, where Brand X has 15 grams of protein and 10 grams of fat and costs 80 cents per unit, and Brand Y contains 20 grams of protein and 5 grams of fat, and costs 50 cents per unit, how much of each type should be used to minimize the cost to the rancher? • Let X = # of units of Brand X • Let Y = # of units of Brand Y • Minimize Cost = .80X + .50Y • 15X + 20Y ≥60 • 10X + 5Y ≥ 30 • Non-negative Constraints