3.4 Linear Programming

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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. DataTables

(per table)Chairs

(per chair)Hours

AvailableProfit Contribution $7 $5

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 > 0C > 0

Model SummaryMax 7T + 5C (profit)

Subject to the constraints:

3T + 4C < 2400 (carpentry hrs)

2T + 1C < 1000 (painting hrs)

C < 450 (max # chairs)

T > 100 (min # tables)

T, C > 0(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.

CarpentryConstraint Line

3T + 4C = 2400

Intercepts

(T = 0, C = 600)

(T = 800, C = 0)

0 800 T

C

600

0 Feasible< 2400 hrs

Infeasible> 2400 hrs

3T + 4C = 2400

PaintingConstraint Line

2T + 1C = 1000

Intercepts

(T = 0, C = 1000)

(T = 500, C = 0)

0 500 800 T

C1000

600

0

2T + 1C = 1000

0 100 500 800 T

C1000

600450

0

Max Chair Line

C = 450

Min Table Line

T = 100

FeasibleRegion

0 100 200 300 400 500 T

C

500

400

300

200

100

0

Objective Function Line

7T + 5C = Profit

7T + 5C = $4,040

Optimal Point(T = 320, C = 360)

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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 P lines for arbitrary

profit levels.

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

Labor Clay RevenuePRODUCT (hr/unit) (lb/unit) ($/unit)Bowl 1 4 40Mug 2 3 50

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

Decision variablesxb = number of bowls to producexm = number of mugs to produce

RESOURCE REQUIREMENTS

• 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?

• Xc = acres of corn• Xo = acres of oats• Maximum Profit = 40Xc + 30Xo

• Xc+Xo ≤240• 2Xc+Xo ≤320• Xc ≥0• Xo ≥0

LP Formulation: Example

Maximize Z = $40 x1 + 50 x2

Subject tox1 + 2x2 40 hr (labor constraint)

4x1 + 3x2 120 lb (clay constraint)x1 , x2 0

Solution is x1 = 24 bowls x2 = 8 mugsRevenue = $1,360

Graphical Solution: Example

4 x1 + 3 x2 120 lb

x1 + 2 x2 40 hr

50 –

40 –

30 –

20 –

10 –

0 – |10

|60

|50

|20

|30

|40 x1

x2

Extreme Corner Points

x1 = 224 bowlsx2 =8 mugsZ = $1,360 x1 = 30 bowls

x2 =0 mugsZ = $1,200

x1 = 0 bowlsx2 =20 mugsZ = $1,000

A

BC|

20|

30|

40|

10 x1

x2

40 –

30 –

20 –

10 –

0 –

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