Linear Programming Econ 6000

Outline  Review the basic concepts of Linear Programming  Illustrate some problems which can be solved by linear programming

Linear Programming 1. LP- an optimization technique for determining the maximum or minimum level of allocation of resources among competing products under conditions of several inequality constraints

Advantages  allows managers to find optimal solutions when several inequality constraints are in effect  can be applied when constraints may not be binding i.e. when resources are less than fully employed(X+Y≤C for max; or X+Y≥C for min.) where C= fixed amount of the choice variable

2. Some Applications  product mix problem –Tomato product processing case  diet problem  product distribution problem  allocation of advertising budget  personnel assignment problems

3. Assumptions of LP  the objective function and constraint equations must be linear  alternatives must be available to the firm or organization  optimization of the objective function is subject to some type of restrictions related to the supply of resources and capacity of firm plants  Constant resource prices are assumed  Constant returns to scale and nonnegative values

4. Steps in the formulation of LP Problem  formulate the objective function  formulate the inequality constraints  find the optimal solutions using alternative methods

5. Methods of Solving LP Problems  Graphical method  Algebraic method  Simplex method using such software as: Lindo; or Maple in Math; solver in Excel

6. Maximization and minimization examples  Profit maximization by an electric company by producing two types of light bulbs  Cost minimization by UN in producing two types of food

6.a. Profit Maximization Case  Suppose a certain electric company produces two brands of bulbs (x 1,x 2 ) and the contribution margins per unit of x 1 and x 2 to the firm profits are $20 and $30, respectively. A summary of company data shows :

LP: Maximization Example

Steps in LP Solution  Formulate the objective function and the constraint equations.  Using the graphical method, finding the optimal values of X 1 and X 2. Solution : Maximize :  = 20X X 2 S.t. : X 1 + 2X 2 < 16 2X 1 + 2/3X 2 < 12 X 1 > 0 ; X 2 > 0

X1X1 X2X2 X 1 = 4 X 2 = A(0,16) B(0,6) C(6,4) D(8,0) E(18,0) Feasible solution space. B,C,D are called feasible solution points. Only one of this points yields an optimal solution.

Evaluating the objective function at B,C,D corner points yields :  = 20X X 2  (B) = 20(6) + 30(0) = $120  (C) = 20(4) + 30(6) = $260  (D) = 20(0) + 30(8) = $240  Point C yields the optimal feasible solution, i.e. X 1 = 4 and X 2 = 6 maximize the company’s profit.

6.b. Minimization Example  Suppose a nutritionist for a United Nations food distribution agency is concerned with developing a minimum-cost-per-day balanced diet from two basic foods, cereal and dried milk, that meets or exceeds certain nutritional requirements.  The information concerning the two food items and the requirements is summarized in the table.

LP Minimization Example *Define X 1 as the number of cereal and X 2 as the number of ounces of dried milk to be included in the diet.

Questions a. Determine the objective function. b. Determine the constraint relationships. c. Using graphical methods, determine the the optimal quantities of cereal and dried milk to include in the diet. d. How is the optimal solution in part (c) affected if the price of cereal increases to 2.5 cents per ounce ?

Solution (a) Objective : Minimize the cost of the diet Minimize C = 1.5X X 2 (b) Constraints (1) 2X 1 + 5X 2 > 100 Protein constraint (2) 100X X 2 > 500 Calorie constraint (3) 10X X 2 > 400 Vitamin constraint (4) 1X X 2 > 20 Iron constraint (5) X 1 > 0, X 2 > 0 Non-negativity - constraint

(10,20)=>optimal (25,10) X2X2 X1X Feasible solution space 2X 1 + 5X 2 > X X 2 > X X 2 > 400 1X X 2 > 20