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

Advantagesallows 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 CaseSuppose a certain electric company produces

two brands of bulbs (x1,x2) and the contribution margins per unit of x1 and x2 to the firm profits are $20 and $30, respectively. A summary of company data shows :

Product Casting Polishing

Std. bulb(x1) 1 2

Deluxe(x2) 2 2/3

Constraints : < 16 < 12

LP: Maximization Example

Steps in LP SolutionFormulate the objective function and the

constraint equations.Using the graphical method, finding the

optimal values of X1 and X2.

Solution :

Maximize : = 20X1 + 30X2

S.t. : X1 + 2X2 < 16

2X1 + 2/3X2 < 12

X1> 0 ; X2 > 0

X1

X2

X1= 4

X2 = 60

16

188

6

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 :

= 20X1 + 30X2

(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. X1 = 4 and X2 = 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 ExampleNutrient Fortified

Cereal X1 (units of

nutrient perounce)

Fortified Dried Milk X2 (units of nutrient per

ounce)

Minimum Requirements

(units)

Protein 2 5 100

Calories 100 40 500

Vitamin D 10 15 400

Iron 1 0.5 20

Cost(cents per ounce)

1.5 1.0

*Define X1as the number of cereal and X2 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 = 1.5X1 + 1.0X2

(b) Constraints (1) 2X1 + 5X2 > 100 Protein constraint (2) 100X1 + 40X2 > 500 Calorie constraint (3) 10X1 + 15X2 > 400 Vitamin

constraint (4) 1X1 + 0.5X2 > 20 Iron constraint (5) X1 > 0, X2 > 0 Non-negativity - constraint

(10,20)=>optimal

(25,10)

X2

X1

5040

200

40

20

Feasible solution space

2X1 + 5X2 > 100

100X1 + 40X2 > 500

10X1 + 15X2 > 400

1X1 + 0.5X2 > 20