Minimization case problems of school management with three variables solved by simplex techniques.

To solve a minimization case problem of school management with three variables using simplex techniques, we need to set up the linear programming model and then apply the simplex method to find the optimal solution.

Let's consider the following minimization case problem with three variables:

Minimize Z = 3x1 + 2x2 + 4x3
Subject to:
2x1 + x2 + 3x3 ≥ 10
x1 + 2x2 + x3 ≥ 8
x1, x2, x3 ≥ 0

Step 1: Convert the problem into standard form by introducing slack variables:
2x1 + x2 + 3x3 + x4 = 10
x1 + 2x2 + x3 + x5 = 8
Z = 3x1 + 2x2 + 4x3

Step 2: Set up the initial simplex tableau:
| x1 | x2 | x3 | x4 | x5 | RHS |
|----|----|----|----|----|-----|
| 2 | 1 | 3 | 1 | 0 | 10 |
| 1 | 2 | 1 | 0 | 1 | 8 |
| -3 | -2 | -4 | 0 | 0 | 0 |

Step 3: Choose the entering variable (most negative coefficient in the objective function) and the leaving variable (minimum ratio test):
Entering variable: x1
Leaving variable: x4

Step 4: Perform row operations to pivot on the entering and leaving variables:
| x1 | x2 | x3 | x4 | x5 | RHS |
|----|----|----|----|----|-----|
| 1 | 0 | 1 | 1/2| 0 | 5 |
| 0 | 2 | 0 | -1/2| 1 | 3 |
| 0 | 0 | 0 | 3/2 | 0 | 15 |

Step 5: Repeat steps 3 and 4 until all coefficients in the objective function are non-negative:
Entering variable: x2
Leaving variable: x5

Step 6: Perform row operations to pivot on the entering and leaving variables:
| x1 | x2 | x3 | x4 | x5 | RHS |
|----|----|----|----|----|-----|
| 1 | 0 | 1 | 1/2| 0 | 5 |
| 0 | 1 | 0 | -1/4| 1/2| 3/2 |
| 0 | 0 | 0 | 3/2 | 0 | 15 |

Step 7: The optimal solution is x1 = 5, x2 = 3/2, x3 = 0 with Z = 23/2.

Therefore, the optimal solution for the minimization case problem of school management with three variables using simplex techniques is Z = 23/2 with x1 = 5, x2 = 3/2, and x3 = 0.

Case Problem:

Objective: Minimize the total cost of running a school

Variables:

- x1: Number of teachers
- x2: Number of classrooms
- x3: Amount spent on supplies

Constraints:

- Number of students must be accommodated: 1000 students
- Maximum budget: $1,000,000
- Minimum number of teachers: 50

Data:

- Cost of each teacher: $50,000
- Cost of each classroom: $200,000
- Cost of supplies: $200 per student

Solution using Simplex Techniques:

Step 1: Construct the Tableau

| Variable | Coefficient | x1 | x2 | x3 | RHS |
|---|---|---|---|---|---|
| Objective | 1 | -50,000 | -200,000 | -200 | 0 |
| Constraint 1 | 1 | 1 | 0 | 0 | 1000 |
| Constraint 2 | 1 | 0 | 1 | 0 | 1,000,000 |
| Constraint 3 | 1 | -1 | 0 | 0 | -50 |

Step 2: Introduce Slack Variables

| Variable | Coefficient | x1 | x2 | x3 | s1 | s2 | RHS |
|---|---|---|---|---|---|---|---|
| Objective | 1 | -50,000 | -200,000 | -200 | 0 | 0 | 0 |
| Constraint 1 | 1 | 1 | 0 | 0 | 1 | 0 | 1000 |
| Constraint 2 | 1 | 0 | 1 | 0 | 0 | 1 | 1,000,000 |
| Constraint 3 | 1 | -1 | 0 | 0 | 0 | 0 | -50 |

Step 3: Find the Initial Feasible Solution

- Set all non-basic variables to 0.
- The initial feasible solution is: (x1 = 0, x2 = 0, x3 = 0, s1 = 1000, s2 = 1,000,000).

Step 4: Perform Pivot Operations

- Select the most negative coefficient in the objective function row.
- In this case, it is -50,000 in the x1 column.
- Choose the row with the smallest non-negative ratio of the RHS to the coefficient of the entering variable.
- In this case, it is row 1.
- Perform pivot operations and update the tableau.

Step 5: Repeat Steps 4-5 until Optimality is Achieved

| Variable | Coefficient | x1 | x2 | x3 | s1 | s2 | RHS |
|---|---|---|---|---|---|---|---|
| Objective | 1 | 0 | -100,000 | -100 | 0 | 90,000 | 50,000 |
| Constraint 1 | 1 | 0 | 1 | 0 | 1 | 0 | 250 |
| Constraint 2 | 1 | 0 | 0 | 1 | 0 | 1 | 500,000 |
| Constraint 3 | 1 | 1 | 0 | 0 | 0 | 0 | 2000 |

- Since the objective function coefficient is non-positive, optimality is achieved.

Optimal Solution:

The optimal solution is to hire 50 teachers, have 10 classrooms, and spend $200,000 on supplies. The total cost of running the school is $50,000,000.