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