# Jack Ulern University Maximization Exercise - Simplex MethodExample of the Simplex Method - Two-Phase

### Problem Data

We want to Maximize the following problem:

- Objective Function
- Z = 2X
_{1}+ X_{2} - Subject to the following constraints
- X
_{1}+ X_{2}≤ 10 - X
_{1}- X_{2}≤ 0 - X
_{1}+ 0X_{2}≤ 4

X

_{1}, X_{2}≥ 0- X
- Description
**Jack Ulern University Maximization Exercise - Simplex Method**Jack is an inspiring freshman at Ulern University. He realized that "all work and no play makes Jack a dull boy. As a result, Jack wants to apportion his available time of about 10 hours a day between work and play. He estimates that play is twice as much fun as work. He also wants to study at least as much as he plays. However, Jack realizes that if he is going to get all his homework assignments done, he cannot play more than 4 hours a day. How should Jack allocate his time to maximize his pleasure from both work and play? Solve the problem with the Simplex 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 “**≤**” (least equal), therefore the slack variable**S**. This variable will be placed at the base in the initial matrix._{1}**Constraint 2:**It has sign “**≤**” (least equal), therefore the slack variable**S**. This variable will be placed at the base in the initial matrix._{2}**Constraint 3:**It has sign “**≤**” (least equal), therefore the slack variable**S**. This variable will be placed at the base in the initial matrix._{3}

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:

Maximize Z = 2X_{1} + X_{2} + 0S_{1} + 0S_{2} + 0S_{3}

Subject to:

- X
_{1}+ X_{2}+ S_{1}+ 0S_{2}+ 0S_{3}= 10 - X
_{1}- X_{2}+ 0S_{1}+ S_{2}+ 0S_{3}= 0 - X
_{1}+ 0X_{2}+ 0S_{1}+ 0S_{2}+ S_{3}= 4

### Initial Matrix

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

Table 1 | C_{j} | 2 | 1 | 0 | 0 | 0 | |
---|---|---|---|---|---|---|---|

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

0 | S_{1} | 1 | 1 | 1 | 0 | 0 | 10 |

0 | S_{2} | 1 | -1 | 0 | 1 | 0 | 0 |

0 | S_{3} | 1 | 0 | 0 | 0 | 1 | 4 |

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

**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}= (0×1) + (0×1) + (0×1) - (2) = -2 - Z
_{2}= (C_{b,1}×X_{2,1}) + (C_{b,2}×X_{2,2}) + (C_{b,3}×X_{2,3}) - C_{j,2}= (0×1) + (0×-1) + (0×0) - (1) = -1 - Z
_{3}= (C_{b,1}×S_{1,1}) + (C_{b,2}×S_{1,2}) + (C_{b,3}×S_{1,3}) - C_{j,3}= (0×1) + (0×0) + (0×0) - (0) = 0 - Z
_{4}= (C_{b,1}×S_{2,1}) + (C_{b,2}×S_{2,2}) + (C_{b,3}×S_{2,3}) - C_{j,4}= (0×0) + (0×1) + (0×0) - (0) = 0 - Z
_{5}= (C_{b,1}×S_{3,1}) + (C_{b,2}×S_{3,2}) + (C_{b,3}×S_{3,3}) - C_{j,5}= (0×0) + (0×0) + (0×1) - (0) = 0 - Z
_{6}= (C_{b,1}×R_{1}) + (C_{b,2}×R_{2}) + (C_{b,3}×R_{3}) = (0×10) + (0×0) + (0×4) = 0

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

In the **reduced cost vector (Z)** we have negative values, so we must select the **most negative** for the pivot column (maximization).

In the vector Z (excluding the last value), we have the following numbers: **[-2, -1, 0, 0, 0]**. The most negative 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 S**→ R_{1}_{1}/ X_{1,1}= 10 / 1 = 10**Row S**→ R_{2}_{2}/ X_{1,2}= 0 / 1 =**0 (The Lowest Value)****Row S**→ R_{3}_{3}/ X_{1,3}= 4 / 1 = 4

The **lowest value** corresponds to the **S _{2}** row. This variable will come from the base. The pivot element corresponds to the value that crosses the

**X**column and the

_{1}**S**row =

_{2}**1**.

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

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

_{2}**1**

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