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-negative variables

.

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.

Properties of LP Models

1) Seek to minimize or maximize 2) Include “constraints” or limitations 3) There must be alternatives available 4) All equations are linear

Example LP Model Formulation: The Product Mix Problem

Decision: How much to make of > 2 products?

Objective: Maximize profit Constraints: Limited resources

Example: Flair Furniture Co.

Two products: Chairs and Tables Decision: How many of each to make this month?

Objective: Maximize profit

Flair Furniture Co. Data

Tables (per table) Chairs (per chair) Profit Contribution $7 $5 Hours Available Carpentry 3 hrs 4 hrs 2400 Painting 2 hrs 1 hr 1000 Other Limitations: • Make no more than 450 chairs • Make at least 100 tables

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

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)

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

Model Summary

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

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.

C

Carpentry Constraint Line 3T + 4C = 2400

600

Intercepts (T = 0, C = 600) (T = 800, C = 0)

0 0 Feasible < 2400 hrs Infeasible > 2400 hrs 800 T

C 1000

Painting Constraint Line 2T + 1C = 1000

600

Intercepts (T = 0, C = 1000) (T = 500, C = 0)

0 0 500 800 T

Max Chair Line C = 450

C 1000

Min Table Line T = 100

600 450 Feasible 0 0 100 Region 500 800 T

C

Objective Function Line 7T + 5C = Profit

500 400 Optimal Point (T = 320, C = 360) 300 200 100 0 0 100 200 300 400 500 T

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 profit levels.

P

lines for

arbitrary

16

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

• 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.

LP Model: Example

PRODUCT Bowl Mug RESOURCE REQUIREMENTS

Labor (hr/unit)

1 2

Clay (lb/unit)

4 3

Revenue ($/unit)

40 50 There are 40 hours of labor and 120 pounds of clay available each day Decision variables

x

b = number of bowls to produce

x

m = number of mugs to produce

• 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?

• X c • X o • Maximum Profit = 40X c • X c +X o ≤240 • 2X • X • X o c c = acres of corn = acres of oats +X ≥0 ≥0 o ≤320 + 30X o

LP Formulation: Example

Maximize

Z

= $40

x

1 + 50

x

2 Subject to

x

1 4

x

1 + 2 + 3

x x

2 2

x

1 ,

x

2  40 hr (labor constraint)  120 lb (clay constraint)  0 Solution is

x

1 = 24 bowls Revenue = $1,360

x

2 = 8 mugs

Graphical Solution: Example

x

2 50 – 40 – 30 – 20 – 10 – 0 – 4

x

1 + 3

x

2  120 lb | 10 | 20 | 30 | 40

x

1 + 2

x

2  40 hr | 50 | 60

x

1

Extreme Corner Points

x

2 40 – 30 – 20 –

A

10 – 0 – | 10

x

1 = 0 bowls

x

2 = 20 mugs Z = $1,000 | 20

B

| 30

C

| 40

x

1 = 224 bowls

x

2 = 8 mugs Z = $1,360

x

1 = 30 bowls

x

2 = 0 mugs Z = $1,200

x

1

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