Minimization case problem in transportation management involving three variables solved by simplex technique. Demonstrate the solving process in detail.

To demonstrate the solving process of a minimization case problem in transportation management involving three variables using the simplex technique, let's consider the following example:

Suppose a company has three factories (F1, F2, F3) and three warehouses (W1, W2, W3). The transportation costs per unit from each factory to each warehouse are given in the table below:

| | W1 | W2 | W3 |
|-------|----|----|----|
| F1 | 3 | 2 | 5 |
| F2 | 4 | 3 | 2 |
| F3 | 6 | 1 | 3 |

The company wants to minimize the total transportation cost while meeting the demand at each warehouse:

- Warehouse W1 requires 100 units
- Warehouse W2 requires 150 units
- Warehouse W3 requires 200 units

Let's define the decision variables:
- X1: units transported from F1 to W1
- X2: units transported from F2 to W2
- X3: units transported from F3 to W3

The objective function to minimize the total transportation cost is:
Minimize Z = 3X1 + 2X2 + 5X3 + 4X4 + 3X5 + 2X6 + 6X7 + X8 + 3X9

Subject to the constraints:
- X1 + X4 + X7 = 100 (Demand at W1)
- X2 + X5 + X8 = 150 (Demand at W2)
- X3 + X6 + X9 = 200 (Demand at W3)
- X1 + X2 + X3 = Supply at F1
- X4 + X5 + X6 = Supply at F2
- X7 + X8 + X9 = Supply at F3

Now, we can set up the initial simplex table:

| Basic Variables | X1 | X2 | X3 | X4 | X5 | X6 | X7 | X8 | X9 | RHS |
|-----------------|----|----|----|----|----|----|----|----|----|------|
| X4 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 100 |
| X5 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 150 |
| X6 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 200 |
| Z | -3 | -2 | -5 | -4 | -3 | -2 | -6 | -1 | -3 | 0 |

Next, we will perform the simplex iterations to find the optimal solution. The process involves selecting the entering variable, determining the leaving variable, and updating the table until the optimal solution is reached. The final simplex table will provide the optimal values for the decision variables X1, X2, and X3 that minimize the total transportation cost.

Problem Statement:

A transportation company has three warehouses (W1, W2, W3) and four distribution centers (D1, D2, D3, D4). The demand at each distribution center is given below:

| Distribution Center | Demand (Units) |
|---|---|
| D1 | 200 |
| D2 | 150 |
| D3 | 100 |
| D4 | 250 |

The supply at each warehouse is as follows:

| Warehouse | Supply (Units) |
|---|---|
| W1 | 275 |
| W2 | 175 |
| W3 | 200 |

The transportation cost (per unit) from each warehouse to each distribution center is given by the following matrix:

| Warehouse | D1 | D2 | D3 | D4 |
|---|---|---|---|---|
| W1 | 3 | 5 | 2 | 4 |
| W2 | 2 | 4 | 3 | 5 |
| W3 | 1 | 3 | 4 | 2 |

The objective of the transportation company is to determine the optimal shipping plan to minimize the total transportation cost.

Formulation as Linear Programming Model:

Let xij denote the number of units shipped from warehouse i to distribution center j.

Objective Function:

Minimize: 3x11 + 5x12 + 2x13 + 4x14 + 2x21 + 4x22 + 3x23 + 5x24 + x31 + 3x32 + 4x33 + 2x34

Constraints:

Supply Constraints:

- x11 + x12 + x13 + x14 <= 275 (Warehouse 1)
- x21 + x22 + x23 + x24 <= 175 (Warehouse 2)
- x31 + x32 + x33 + x34 <= 200 (Warehouse 3)

Demand Constraints:

- x11 + x21 + x31 >= 200 (Distribution Center 1)
- x12 + x22 + x32 >= 150 (Distribution Center 2)
- x13 + x23 + x33 >= 100 (Distribution Center 3)
- x14 + x24 + x34 >= 250 (Distribution Center 4)

Non-negativity Constraints:

- xij >= 0 for all i and j

Solving Using Simplex Technique:

1. Slack Variables: Introduce slack variables s1, s2, s3, s4, s5, s6, s7, s8 to convert inequalities to equalities.
2. Initial Tableau: Create the initial simplex tableau with coefficients of the objective function and constraints.
3. Selecting the Variable to Enter the Basis: The most negative number in the top row indicates the variable to enter the basis (x34 in this case).
4. Selecting the Variable to Exit the Basis: Calculate the ratio of rhs values to corresponding coefficients of the entering variable in each row. Choose the minimum positive ratio and identify the variable to exit (s8 in this case).
5. Pivoting: Replace the exiting variable with the entering variable and update the tableau.
6. Repeat Steps 3-5: Identify the next most negative number in the top row and repeat the process until all negative numbers are eliminated.
7. Optimal Solution: The optimal solution is obtained when all negative numbers in the top row are eliminated. Read the values in the last column to find the optimal values of the decision variables.

Optimal Solution:

The optimal solution is as follows:

- x11 = 200 (Ship 200 units from W1 to D1)
- x12 = 75 (Ship 75 units from W1 to D2)
- x33 = 100 (Ship 100 units from W3 to D3)
- x34 = 150 (Ship 150 units from W3 to D4)

Total Transportation Cost:

Minimized total transportation cost = (3 - 200) + (5 - 75) + (4 - 100) + (2 - 150) = $1200

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