8: A product consists of two components A and B. These components require two
 different raw materials X and Y. 100 units of X and 200 units of Y are available for production. The
 materials are processed in three departments.  The requirement of production time in hours and materials
 in units are given in the table below.
 Departments
 Raw material input
 per run in unit
 Output of components
 per run (units)
 X Y A B
 1 7 5 6 4
 2 4 8 5 8
 3 2 7 7 3
 Formulate a progrmme to determine the number of production runs for each department, which
 will maximise the total number of components A and B for the product.

To formulate a linear programming model to maximize the total number of components \( A \) and \( B \), we'll define the decision variables, objective function, and constraints.

### Decision Variables
Let:
- \( x_1 \) be the number of production runs for Department 1.
- \( x_2 \) be the number of production runs for Department 2.
- \( x_3 \) be the number of production runs for Department 3.

### Objective Function
We want to maximize the total number of components \( A \) and \( B \) produced. The output of components per run for each department is given as follows:

- Department 1: Produces 6 units of \( A \) and 4 units of \( B \).
- Department 2: Produces 5 units of \( A \) and 8 units of \( B \).
- Department 3: Produces 7 units of \( A \) and 3 units of \( B \).

Thus, the objective function is:
\[
\text{Maximize } Z = 6x_1 + 5x_2 + 7x_3 + 4x_1 + 8x_2 + 3x_3
\]
Simplifying:
\[
\text{Maximize } Z = (6 + 4)x_1 + (5 + 8)x_2 + (7 + 3)x_3
\]
\[
\text{Maximize } Z = 10x_1 + 13x_2 + 10x_3
\]

### Constraints
The constraints are based on the availability of raw materials \( X \) and \( Y \). Given that 100 units of \( X \) and 200 units of \( Y \) are available:

1. **Material \( X \) constraint:**
   - Department 1 uses 7 units of \( X \) per run.
   - Department 2 uses 4 units of \( X \) per run.
   - Department 3 uses 2 units of \( X \) per run.
   
   Therefore, the constraint for material \( X \) is:
   \[
   7x_1 + 4x_2 + 2x_3 \leq 100
   \]

2. **Material \( Y \) constraint:**
   - Department 1 uses 5 units of \( Y \) per run.
   - Department 2 uses 8 units of \( Y \) per run.
   - Department 3 uses 7 units of \( Y \) per run.
   
   Therefore, the constraint for material \( Y \) is:
   \[
   5x_1 + 8x_2 + 7x_3 \leq 200
   \]

3. **Non-negativity constraint:**
   - \( x_1 \geq 0, x_2 \geq 0, x_3 \geq 0 \)

### Complete Linear Programming Model
\[
\text{Maximize } Z = 10x_1 + 13x_2 + 10x_3
\]
Subject to:
\[
7x_1 + 4x_2 + 2x_3 \leq 100
\]
\[
5x_1 + 8x_2 + 7x_3 \leq 200
\]
\[
x_1 \geq 0, x_2 \geq 0, x_3 \geq 0
\]

This linear programming model can now be solved using methods like the Simplex algorithm to find the optimal number of production runs for each department that maximizes the total output of components \( A \) and \( B \).

Would you like further details on how to solve this problem or any other questions?

### Further Related Questions:
1. How do we apply the Simplex algorithm to solve this LP problem?
2. What is the significance of the shadow price in this context?
3. How can we interpret the dual problem of this linear program?
4. What if the availability of raw materials changes? How would that affect the optimal solution?
5. How can sensitivity analysis be used to assess the robustness of the solution?

### Tip:
Ensure that you thoroughly check the constraints and objective function for accuracy before solving a linear programming problem, as even small errors can lead to incorrect results.

Learn how to maximize the production runs for components A and B using linear programming. Solve the problem using constraints on raw materials X and Y. Suitable for undergraduate level students and professionals.

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