1. Linear Programming: Model Formulation and Graphic Solution
Lecture 2

2. A Maximization Model Example
The Beaver Creek Pottery Company
Given these limited resources, the company desires to know how many bowls and mugs to produce each day in order to maximize profit.
There are 40 hours of labor and 120 kg of clay available each day for production.

3. Resource Requirements
Beaver Creek Pottery
Resource Requirements
Product
Labor
(hr/unit)
Clay
(kg/unit)
Profit
($/unit)
Bowl 1 4 40
Mug 2 3 50

4. Maximize Z = 40 x1 + 50 x2 Modeling Decision Variables
Objective Function
x1 : number of bowls to produce
x2 : number of mugs to produce
Total Profit = 40 x x2
40 x1 = profit from bowls
50 x2 = profit from mugs
Maximize Z = 40 x x2

5. Non-negativity Constraint
Model Constraints
Labor Constraint
Total Labor used in production = 1 x1 + 2 x2
1 x1 + 2 x2 ≤ 40 hr
Clay Constraint
Total Clay used in production = 4 x1 + 3 x2
4 x1 + 3 x2 ≤ 120 lb
Non-negativity Constraint
x1 ≥ 0, x2 ≥ 0

6. Linear Programming Model
Maximize Z = 40 x x2
Subject to:
1 x1 + 2 x2 ≤ 40
4 x1 + 3 x2 ≤ 120
x1, x2 ≥ 0

7. Feasible / Infeasible If x1 = 5, x2 = 10 Z = 40 . 5 + 50 . 10 = 700

8. Graphical Solution of Linear Programming Models10 20 30 40 50 60
Coordinates for graphical analysis

9. Graph of the labor constraint linex2 10 20 30 40 50 60
x1 + 2 x2 = 40 x1

10. The labor constraint areax2 10 20 30 40 50 60 M L
x1 + 2 x2 ≤ 40 K x1

11. Graph of the labor constraint linex2 10 20 30 40 50 60
4 x1 + 3 x2 = 120 x1

12. The clay constraint areax2 10 20 30 40 50 60 M L
4 x1 + 3 x2 ≤ 120 K x1

13. Graph of both model constraintsx2 10 20 30 40 50 60
4 x1 + 3 x2 = 120
x1 + 2 x2 = 40 x1

14. The feasible solution area constraintsx2 10 20 30 40 50 60
T: Infeasible
S: Infeasible
R: Feasible
4 x1 + 3 x2 = 120 T S R
x1 + 2 x2 = 40 x1

15. Objective function line for Z = $ 800x2 10 20 30 40 50 60
800 = 40 x x2 x1

16. Alternative objective function lines for profits, Z, of $ 800, $ 1200, $ 1600x2 40
800 = 40 x x2 30
1200 = 40 x x2 20
1600 = 40 x x2 10 10 20 30 40 x1

17. Identification of optimal solution pointx2 10 20 30 40 50 60
800 = 40 x x2
Optimal solution point B x1

18. Optimal solution coordinatesx2 40 35
4 x1 + 3 x2 = 120 30 25 A 20 15 10 B 8
x1 + 2 x2 = 40 5 C
Prof. M.A.Shouman 5 10 15 20 24 25 30 35 40 x1

19. Solutions at all corners pointsx2 40 35
4 x1 + 3 x2 = 120
x1 = 0 bowls
x2 = 20 mugs
Z = $ 1,000 30 25
x1 = 24 bowls
x2 = 8 mugs
Z = $ 1,360 A 20
x1 = 30 bowls
x2 = 0 mugs
Z = $ 1,200 15 10 B 8 5
x1 + 2 x2 = 40 C
Prof. M.A.Shouman 5 10 15 20 24 25 30 35 40 x1

20. Optimal solution point
The optimal solution with Z = 70 x x2 x2 40 35
4 x1 + 3 x2 = 120 30 25 A 20
Optimal solution point
x1 = 30 bowls
x2 = 0 mugs
Z = $ 1,200 15 10 B 5
x1 + 2 x2 = 40 C
Prof. M.A.Shouman 5 10 15 20 25 30 35 40 x1

22. A Minimization Model Example
The Farmer’s Field
The farmer’s field requires at least 16 kg. of nitrogen and 24 kg. of phosphate.
Super-gro costs $6 per bag, and Crop-quick costs $3.
The farmer wants to know how many bags of each brand to purchase in order to minimize the total cost of fertilizing.

23. Chemical Contribution
The Farmer’s Field
Chemical Contribution
Brand
Nitrogen
(kg./bag)
Phosphate
Super-gro 4 2
Crop-quick 3

24. Minimize Z = 6 x1 + 3 x2 Subject to: 2 x1 + 4 x2 ≥ 16 4 x1 + 3 x2 ≥ 24
Linear Programming Model
Minimize Z = 6 x1 + 3 x2
Subject to:
2 x1 + 4 x2 ≥ 16
4 x1 + 3 x2 ≥ 24
x1, x2 ≥ 0

25. Constraint lines for fertilizer modelx2 12 10 8
4 x1 + 3 x2 = 24 6 4 2
2 x1 + 4 x2 = 16
Prof. M.A.Shouman 2 4 6 8 10 12 14 16 x1

26. Feasible solution areax2 12 10 8
4 x1 + 3 x2 = 24 6
Feasible solution area 4 2
2 x1 + 4 x2 = 16 x1 2 4 6 8 10 12

27. Optimal solution point
The optimal solution point x2
Optimal solution point 12
x1 = 0 bags of Super-gro
x2 = 8 bags of Crop-quick
Z = $ 24 10 A 8 6
Z = 6 x1 + 3 x2 4 B 2 C x1 2 4 6 8 10 12

28. Irregular Types of Linear Programming Problems

29. Multiple Optimal Solutionsx2 40 35 30
Point B
Point C 25
x1 = 24
x2 = 8
Z = $ 1,200
x1 = 30 bowls
x2 = 0 mugs
Z = $ 1,200 A 20 15 10 B 5 C
Prof. M.A.Shouman 5 10 15 20 25 30 35 40 x1

30. An Infeasible Problem C B A x1 = 4 x2 = 6 4 x1 + 2 x2 = 8 x2 12 10 8 62 4 6 8 10 12

31. An Unbounded Problem x1 = 4 Minimize Z = 4 x1 + 2 x2 Subject to:12 10 8
Z = 4 x1 + 2 x2 6
x2 = 6 4 2 x1 2 4 6 8 10 12