Question

Lp model example and solution

Answer 1

Linear Programming (LP) is a mathematical technique used to optimize a certain objective function, subject to a set of constraints. Below, you'll find a simple example of an LP model along with its solution.

### Example Problem

**Objective:**
Maximize profit from producing two products, Product A and Product B.

**Data:**

- Each unit of Product A gives a profit of $5.
- Each unit of Product B gives a profit of $3.
- Product A requires 2 hours of labor and Product B requires 1 hour of labor.
- The total available labor hours are 100.
- The production limits are:
- At most 30 units of Product A.
- At most 50 units of Product B.

### LP Model

**Decision Variables:**
- Let \( x_1 \) = number of units of Product A produced.
- Let \( x_2 \) = number of units of Product B produced.

**Objective Function:**
Maximize Profit = \( 5x_1 + 3x_2 \)

**Constraints:**
1. Labor constraint: \( 2x_1 + x_2 \leq 100 \) (Total labor hours)
2. Production limit for Product A: \( x_1 \leq 30 \)
3. Production limit for Product B: \( x_2 \leq 50 \)
4. Non-negativity constraints: \( x_1 \geq 0 \), \( x_2 \geq 0 \)

### Complete Model

Maximize:
\[ Z = 5x_1 + 3x_2 \]
Subject to:
\[ 2x_1 + x_2 \leq 100 \]
\[ x_1 \leq 30 \]
\[ x_2 \leq 50 \]
\[ x_1 \geq 0 \]
\[ x_2 \geq 0 \]

### Solving the LP Problem

Using the graphical method, we can plot the constraints and find the feasible region.

1. **Plot the constraints:**
- For \( 2x_1 + x_2 = 100 \):
- If \( x_1 = 0 \), then \( x_2 = 100 \).
- If \( x_2 = 0 \), then \( x_1 = 50 \).

- For \( x_1 = 30 \) (vertical line).
- For \( x_2 = 50 \) (horizontal line).

2. **Find the feasible region:**
The feasible region will be bounded by the intersection of the above lines while also satisfying the non-negativity constraints.

3. **Identify corner points:**
After plotting, the corner points of the feasible region can be determined:
- Intersection of \( 2x_1 + x_2 = 100 \) and \( x_1 = 30 \):
\[
2(30) + x_2 = 100 \rightarrow x_2 = 40 \quad (30, 40)
\]
- Intersection of \( 2x_1 + x_2 = 100 \) and \( x_2 = 50 \):
\[
2x_1 + 50 = 100 \rightarrow 2x_1 = 50 \rightarrow x_1 = 25 \quad (25, 50)
\]
- Intersection of \( x_1 = 30 \) and \( x_2 = 50 \):
\[
(30, 50)
\]

4. **Evaluate the objective function at each corner point:**
- At \( (30, 40) \): \( Z = 5(30) + 3(40) = 150 + 120 = 270 \)
- At \( (25, 50) \): \( Z = 5(25) + 3(50) = 125 + 150 = 275 \)
- At \( (30, 50) \): \( Z = 5(30) + 3(50) = 150 + 150 = 300 \)

### Solution

The maximum profit occurs at the corner point \( (30, 50) \).

- **Optimal Solution:**
- Produce 30 units of Product A and 50 units of Product B.
- Maximum Profit = $300.

This is a simplified example, and more complex scenarios can be handled using various LP-solving techniques, including the Simplex method or software solutions like Excel Solver, R, or Python libraries.