12.7 Case Studies
Case Study 1: Product Mix Optimization
A company produces two products, A and B.
- Product A requires 2 hours of labour and 3 units of raw material per unit.
- Product B requires 3 hours of labour and 2 units of raw material per unit.
- Available resources: 100 labour hours and 90 units of raw material.
- Profit: $40 per unit of A, $30 per unit of B.
Formulate the LP model
Objective: Maximize profit:
Z = 40A + 30B
Subject to:
- Labor: 2A + 3B ≤ 100
- Raw material: 3A + 2B ≤ 90
- Non-negativity: A, B ≥ 0
Case Study 2: Graphical Solution of an LP Problem
Maximize:
Z = 3x + 2y
Subject to:
- X + y ≤ 4
- X ≤ 3
- Y ≤ 2
- x, y ≥ 0
Solution:
Plot the feasible region defined by the constraints and identify the corner points. Evaluate the objective function at each corner to find the maximum value of Z.
The plot below shows:
- The constraint lines as solid curves.
- The feasible region is shaded in grey.
- An example objective function line 3x + 2y = 12 is shown as a dashed line, which helps visualize the direction of optimization.
To find the optimal solution, you would slide the objective line parallel to itself toward the farthest point within the feasible region. The optimal value occurs at one of the corner points of the feasible region.
Corner points of the feasible region and their corresponding values for the objective function Z = 3x + 2y:
| Corner Point | Objective Value Z = 3x + 2y |
| (0,0) | 0 |
| (3,0) | 9 |
| (0,2) | 4 |
| Corner Point | Objective Value Z = 3x + 2y |
| (3,1) | 11 <- Optimal Solution |
| (2,2) | 10 |
Optimal Solution:
- Point: (3,1)
- Maximum Value of Z: 11
This means the objective function reaches its maximum value of 11 at the point x=3, y=1, within the feasible region.
Image Description
A graph titled “Feasible Region and Optimal Solution” shows the feasible region for a linear programming problem. The x-axis ranges from 0 to 4, and the y-axis ranges from 0 to 4. Three constraint lines are drawn:
Blue line:
𝑥
+
𝑦
≤
4
x+y≤4, sloping downward from (0, 4) to (4, 0).
Orange vertical line:
𝑥
≤
3
x≤3.
Green horizontal line:
𝑦
≤
2
y≤2.
The shaded gray region represents the feasible solution space, bounded below by the x-axis and the y-axis. The optimal solution is marked with a red dot at the point (3, 1), labelled “(3, 1), Z = 11.” This point lies on both the blue line and the orange line.
Case Study 3: Diet Optimization Problem
A nutritionist must design a diet that provides at least:
- 300 units of vitamin A
- 400 units of vitamin C
Two food options:
- Food X: 50 units of A, 20 units of C per ounce, costs $3
- Food Y: 20 units of A, 50 units of C per ounce, costs $4
Formulate the LP model.
Objective: Minimize cost:
Z = 3X + 4Y
Subject to:
- Vitamin A: 50X+20Y ≥ 300
- Vitamin C: 20X + 50Y ≥ 400
- Non-negativity: X,Y ≥ 0
Case Study 4: Manufacturing Resource Allocation
A fabrication department produces four parts: A, B, C, and D.
- Variable costs: A = $3, B = $2, C = $5, D = $1
- Machines: Laser Cutter (max 10 hrs), Edge Filer (8 hrs), Spot Welder (5 hrs)
Machine time requirements (in hours):
| Part | Laser Cut | Edge File | Spot Weld |
| A | 2 | 1 | 1 |
| B | 1 | 2 | 1 |
| C | 3 | 1 | 1 |
| D | 0 | 0 | 0 |
Objective: Minimize total cost:
Z = 3A + 2B + 5C + 1D
Subject to:
- Laser Cut: 2A + B + 3C ≤ 10
- Edge File: A + 2B + C ≤ 8
- Spot Weld: A + B + C ≤ 5
- Non-negativity: A, B, C, D ≥ 0
Question:
How many units of each part should be fabricated to minimize total cost?
