# Cohen Chemicals, Inc. Minimization problemExample of the Simplex Method - Two-Phase

### Problem Data

We want to Minimize the following problem:

- Objective Function
- Z = 2500X
_{1}+ 3000X_{2} - Subject to the following constraints
- X
_{1}+ 0X_{2}≥ 30 - 0X
_{1}+ X_{2}≥ 20 - X
_{1}+ X_{2}≥ 60

X

_{1}, X_{2}≥ 0- X
- Description
**Cohen Chemicals, Inc. Minimization problem**Cohen Chemicals, Inc., produces two types of photo-developing fluids. The first, a black-and-white picture chemical, costs Cohen $2,500 per ton to produce. The second, a color photo chemical, costs $3,000 per ton. Based on an analysis of current inventory levels and outstanding orders, Cohen’s production manager has specified that at least 30 tons of the black-and-white chemical and at least 20 tons of the color chemical must be produced during the next month. In addition, the manager notes that an existing inventory of a highly perishable raw material needed in both chemicals must be used within 30 days. To avoid wasting the expensive raw material, Cohen must produce a total of at least 60 tons of the photo chemicals in the next month. Solve this problem using the two-phase method.

## Solution

To solve the problem, the iterations of the **simplex method** will be performed until the optimal solution is found.

Next, the problem will be adapted to the standard linear programming model, adding the slack, excess and/or artificial variables in each of the constraints and converting the inequalities into equalities:

**Constraint 1:**It has sign “**≥**” (greatest equal), therefore the excess variable**S**will be subtracted and the artificial variable_{1}**A**will be added. In the initial matrix,_{1}**A**will be in the base._{1}**Constraint 2:**It has sign “**≥**” (greatest equal), therefore the excess variable**S**will be subtracted and the artificial variable_{2}**A**will be added. In the initial matrix,_{2}**A**will be in the base._{2}**Constraint 3:**It has sign “**≥**” (greatest equal), therefore the excess variable**S**will be subtracted and the artificial variable_{3}**A**will be added. In the initial matrix,_{3}**A**will be in the base._{3}

Since the problem has artificial variables, the **two-phase method** will be used. In the first phase, the objective function will be modified to **minimize the sum of the artificial variables**.

Now we will show the problem in the standard form. We will place the coefficient 0 (zero) where appropriate to create our initial matrix:

Objective Function:

Minimize Z = 0X_{1} + 0X_{2} + 0S_{1} + 0S_{2} + 0S_{3} + A_{1} + A_{2} + A_{3}

Subject to:

- X
_{1}+ 0X_{2}- S_{1}+ 0S_{2}+ 0S_{3}+ A_{1}+ 0A_{2}+ 0A_{3}= 30 - 0X
_{1}+ X_{2}+ 0S_{1}- S_{2}+ 0S_{3}+ 0A_{1}+ A_{2}+ 0A_{3}= 20 - X
_{1}+ X_{2}+ 0S_{1}+ 0S_{2}- S_{3}+ 0A_{1}+ 0A_{2}+ A_{3}= 60

### Initial Matrix Phase One

We will use the coefficients of the equations to elaborate our first table:

Table 1 | C_{j} | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | |
---|---|---|---|---|---|---|---|---|---|---|

C_{b} | Base | X_{1} | X_{2} | S_{1} | S_{2} | S_{3} | A_{1} | A_{2} | A_{3} | R_{} |

1 | A_{1} | 1 | 0 | -1 | 0 | 0 | 1 | 0 | 0 | 30 |

1 | A_{2} | 0 | 1 | 0 | -1 | 0 | 0 | 1 | 0 | 20 |

1 | A_{3} | 1 | 1 | 0 | 0 | -1 | 0 | 0 | 1 | 60 |

Z | 2 | 2 | -1 | -1 | -1 | 0 | 0 | 0 | 110 |

**a) Reduced Cost Vector Calculation (Z):**

The values recorded in row Z were obtained as follows:

- Z
_{1}= (C_{b,1}×X_{1,1}) + (C_{b,2}×X_{1,2}) + (C_{b,3}×X_{1,3}) - C_{j,1}= (1×1) + (1×0) + (1×1) - (0) = 2 - Z
_{2}= (C_{b,1}×X_{2,1}) + (C_{b,2}×X_{2,2}) + (C_{b,3}×X_{2,3}) - C_{j,2}= (1×0) + (1×1) + (1×1) - (0) = 2 - Z
_{3}= (C_{b,1}×S_{1,1}) + (C_{b,2}×S_{1,2}) + (C_{b,3}×S_{1,3}) - C_{j,3}= (1×-1) + (1×0) + (1×0) - (0) = -1 - Z
_{4}= (C_{b,1}×S_{2,1}) + (C_{b,2}×S_{2,2}) + (C_{b,3}×S_{2,3}) - C_{j,4}= (1×0) + (1×-1) + (1×0) - (0) = -1 - Z
_{5}= (C_{b,1}×S_{3,1}) + (C_{b,2}×S_{3,2}) + (C_{b,3}×S_{3,3}) - C_{j,5}= (1×0) + (1×0) + (1×-1) - (0) = -1 - Z
_{6}= (C_{b,1}×A_{1,1}) + (C_{b,2}×A_{1,2}) + (C_{b,3}×A_{1,3}) - C_{j,6}= (1×1) + (1×0) + (1×0) - (1) = 0 - Z
_{7}= (C_{b,1}×A_{2,1}) + (C_{b,2}×A_{2,2}) + (C_{b,3}×A_{2,3}) - C_{j,7}= (1×0) + (1×1) + (1×0) - (1) = 0 - Z
_{8}= (C_{b,1}×A_{3,1}) + (C_{b,2}×A_{3,2}) + (C_{b,3}×A_{3,3}) - C_{j,8}= (1×0) + (1×0) + (1×1) - (1) = 0 - Z
_{9}= (C_{b,1}×R_{1}) + (C_{b,2}×R_{2}) + (C_{b,3}×R_{3}) = (1×30) + (1×20) + (1×60) = 110

**b) Optimality Condition (Incoming Variable):**

In the **reduced cost vector (Z)** we have positive values, so we must select the **highest value** for the pivot column (minimization).

In the vector Z (excluding the last value), we have the following numbers: **[2, 2, -1, -1, -1, 0, 0, 0]**. The highest value is = **2** which corresponds to the **X _{1}** variable. This variable will enter the base and its values in the table will form our pivot column.

**d) Feasibility Condition (Outgoing Variable):**

The feasibility condition will be verified by dividing the values of **column R** by the **pivot column X _{1}**. To process the division, the denominator must be strictly positive (If it's negative or zero, it'll display N/A = Not applicable). The

**lowest value**will define the variable that will exit from the base:

**Row A**→ R_{1}_{1}/ X_{1,1}= 30 / 1 =**30 (The Lowest Value)****Row A**→ R_{2}_{2}/ X_{1,2}= 20 / 0 = NA**Row A**→ R_{3}_{3}/ X_{1,3}= 60 / 1 = 60

The **lowest value** corresponds to the **A _{1}** row. This variable will come from the base. The pivot element corresponds to the value that crosses the

**X**column and the

_{1}**A**row =

_{1}**1**.

Enter the variable **X _{1}** and the variable

**A**leaves the base. The pivot element is

_{1}**1**

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