1. 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.
The text lists examples of the listed applications.

3. Properties of LP Models
Seek to minimize or maximize
Include “constraints” or limitations
There must be alternatives available
All equations are linear

4. Example LP Model Formulation: The Product Mix Problem
Decision: How much to make of > 2 products?
Objective: Maximize profit
Constraints: Limited resources

5. Example: Flair Furniture Co.
Two products: Chairs and Tables
Decision: How many of each to make this month?
Objective: Maximize profit

6. Flair Furniture Co. Data
Tables
(per table)
Chairs
(per chair)
Hours Available
Profit 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

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

8. Constraints:
Have 2400 hours of carpentry time available
3 T C < (hours)
Have 1000 hours of painting time available
2 T C < (hours)

9. Make no more than 450 chairs C < 450 (num. chairs)
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

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

11. 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.

12. Carpentry Constraint Line 3T + 4C = 2400 Intercepts (T = 0, C = 600)
Carpentry
Constraint Line
3T + 4C = 2400
Intercepts
(T = 0, C = 600)
(T = 800, C = 0)
Infeasible
> 2400 hrs
3T + 4C = 2400
Feasible
< 2400 hrs T

13. Painting Constraint Line 2T + 1C = 1000 Intercepts (T = 0, C = 1000)
600
Painting
Constraint Line
2T + 1C = 1000
Intercepts
(T = 0, C = 1000)
(T = 500, C = 0)
2T + 1C = 1000 T

14. Max Chair Line C = 450 Min Table Line T = 100 Feasible Region C 1000
600
450
Max Chair Line
C = 450
Min Table Line
T = 100
Feasible
Region T

15. Objective Function Line
500
400
300
200
100
Objective Function Line
7T + 5C = Profit
7T + 5C = $4,040
Optimal Point
(T = 320, C = 360) T

16. 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.

17. 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

18. 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.

19. LP Model: Example RESOURCE REQUIREMENTS Labor Clay Revenue
PRODUCT (hr/unit) (lb/unit) ($/unit)
Bowl
Mug
There are 40 hours of labor and 120 pounds of clay available each day
Decision variables
xb = number of bowls to produce
xm = number of mugs to produce
RESOURCE REQUIREMENTS

20. We have 240 acres of land to plant corn and oats
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?

21. Xc = acres of corn
Xo = acres of oats
Maximum Profit = 40Xc + 30Xo
Xc+Xo ≤240
2Xc+Xo ≤320
Xc ≥0
Xo ≥0

22. LP Formulation: Example
Maximize Z = $40 x x2
Subject to
x1 + 2x2 40 hr (labor constraint)
4x1 + 3x2 120 lb (clay constraint)
x1 , x2 0
Solution is x1 = 24 bowls x2 = 8 mugs
Revenue = $1,360

23. 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

24. Extreme Corner Points x1 = 0 bowls x2 =20 mugs Z = $1,000 x2A B C | 20 30 40 10 x1 x2
40 –
30 –
20 –
10 –
0 –

25. 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