Minimization case problem of transportation with three variables solved using the simplex technique. Demonstrate the solving process in detail.

To solve a minimization case problem of transportation with three variables using the simplex technique, we need to follow these steps:

Step 1: Formulate the transportation problem in standard form
Let's consider the following transportation problem with three variables:

Minimize Z = 3x1 + 2x2 + 4x3
Subject to:
x1 + x2 + x3 = 5
2x1 + x2 + 3x3 = 8
x1, x2, x3 ≥ 0

Step 2: Convert the problem into standard form
To convert the problem into standard form, we introduce slack variables for each constraint:

x1 + x2 + x3 + s1 = 5
2x1 + x2 + 3x3 + s2 = 8

Step 3: Create the initial simplex tableau
Construct the initial simplex tableau using the coefficients of the objective function and the constraints:

| x1 | x2 | x3 | s1 | s2 | RHS |
|----|----|----|----|----|-----|
| 1 | 1 | 1 | 1 | 0 | 5 |
| 2 | 1 | 3 | 0 | 1 | 8 |
| -3 | -2 | -4 | 0 | 0 | 0 |

Step 4: Perform the simplex method
1. Select the most negative coefficient in the bottom row as the pivot column. In this case, the most negative coefficient is -4, which corresponds to x3.
2. Calculate the ratios of the RHS values to the pivot column values to determine the pivot row. The minimum ratio corresponds to the pivot row. In this case, the ratios are 5/1 = 5 and 8/3 ≈ 2.67. Therefore, the pivot row is the second row.
3. Perform row operations to make the pivot element 1 and eliminate other elements in the pivot column.
4. Update the tableau with the new values.
5. Repeat steps 1-4 until all coefficients in the bottom row are non-negative.

Step 5: Finalize the solution
After performing the simplex method, the final tableau will look like this:

| x1 | x2 | x3 | s1 | s2 | RHS |
|----|----|----|----|----|-----|
| 0 | 0 | 1 | 1 | -1 | 1 |
| 0 | 1 | 0 | -2 | 1 | 2 |
| 1 | 0 | 0 | 1 | -2 | 3 |

The optimal solution is x1 = 3, x2 = 2, x3 = 1, with a minimum value of Z = 11.

Problem:

A company produces three products (A, B, and C) at three different factories (I, II, and III). The production capacity of each factory and the demand for each product are as follows:

| Factory | Production Capacity |
|---|---|
| I | 500 units |
| II | 400 units |
| III | 300 units |

| Product | Demand |
|---|---|
| A | 600 units |
| B | 500 units |
| C | 400 units |

The cost of producing each unit of each product at each factory is given in the following table:

| Product | Factory I | Factory II | Factory III |
|---|---|---|---|
| A | $5 | $6 | $7 |
| B | $4 | $5 | $6 |
| C | $3 | $4 | $5 |

The company wants to minimize the total production cost by determining the optimal production levels at each factory.

Solving Process:

Step 1: Formulate the Linear Programming Model

The objective function is to minimize the total production cost:

```
Minimize Z = 5x11 + 6x12 + 7x13 + 4x21 + 5x22 + 6x23 + 3x31 + 4x32 + 5x33
```

where xij is the number of units of product i produced at factory j.

The constraints are the production capacity limitations and demand requirements:

```
x11 + x12 + x13 <= 500
x21 + x22 + x23 <= 400
x31 + x32 + x33 <= 300

x11 + x21 + x31 >= 600
x12 + x22 + x32 >= 500
x13 + x23 + x33 >= 400
```

Step 2: Initial Tableau

Create an initial tableau, including slack variables for each constraint.

| Variable | x11 | x12 | x13 | x21 | x22 | x23 | x31 | x32 | x33 | s1 | s2 | s3 | RHS |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Z | -5 | -6 | -7 | -4 | -5 | -6 | -3 | -4 | -5 | 0 | 0 | 0 | 0 |
| x11 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 500 |
| x12 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 400 |
| x13 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 300 |
| s1 | -1 | -1 | -1 | -1 | -1 | -1 | 0 | 0 | 0 | 0 | 0 | 0 | 600 |
| s2 | 0 | 0 | 0 | -1 | -1 | -1 | -1 | -1 | -1 | 0 | 0 | 0 | 500 |
| s3 | 0 | 0 | 0 | 0 | 0 | 0 | -1 | -1 | -1 | 0 | 0 | 0 | 400 |

Step 3: Perform Pivot Operations

Select the most negative coefficient in the Z-row (-7). The column corresponding to this coefficient is the entering variable (x13).

Select the smallest ratio of RHS to entering variable coefficient in the positive rows (500/1 = 500), which determines the leaving variable (s1).

Pivot on the intersection of the entering and leaving variables:

| Variable | x11 | x12 | x13 | x21 | x22 | x23 | x31 | x32 | x33 | s1 | s2 | s3 | RHS |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Z | 7 | 6 | 0 | -4 | -4 | -3 | 2 | 1