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.