#### Transcript Document

### 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**