# 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 = 2X1 + X2
Subject to the following constraints
• X1 + X2 10
• X1 - X2 0
• X1 + 0X2 4

X1, X2 ≥ 0

Description
Jack Ulern University Maximization Exercise - Simplex MethodJack 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 S1. This variable will be placed at the base in the initial matrix.
• Constraint 2: It has sign “” (least equal), therefore the slack variable S2. This variable will be placed at the base in the initial matrix.
• Constraint 3: It has sign “” (least equal), therefore the slack variable S3. This variable will be placed at the base in the initial matrix.

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 = 2X1 + X2 + 0S1 + 0S2 + 0S3

Subject to:

• X1 + X2 + S1 + 0S2 + 0S3 = 10
• X1 - X2 + 0S1 + S2 + 0S3 = 0
• X1 + 0X2 + 0S1 + 0S2 + S3 = 4

### Initial Matrix

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

Table 1 Cj Cb Base X1 X2 S1 2 1 0 0 0 0 1 1 1 0 0 10 0 1 -1 0 1 0 0 0 1 0 0 0 1 4 -2 -1 0 0 0 0

a) Reduced Cost Vector Calculation (Z):

The values recorded in row Z were obtained as follows:

• Z1 = (Cb,1×X1,1) + (Cb,2×X1,2) + (Cb,3×X1,3) - Cj,1 = (0×1) + (0×1) + (0×1) - (2) = -2
• Z2 = (Cb,1×X2,1) + (Cb,2×X2,2) + (Cb,3×X2,3) - Cj,2 = (0×1) + (0×-1) + (0×0) - (1) = -1
• Z3 = (Cb,1×S1,1) + (Cb,2×S1,2) + (Cb,3×S1,3) - Cj,3 = (0×1) + (0×0) + (0×0) - (0) = 0
• Z4 = (Cb,1×S2,1) + (Cb,2×S2,2) + (Cb,3×S2,3) - Cj,4 = (0×0) + (0×1) + (0×0) - (0) = 0
• Z5 = (Cb,1×S3,1) + (Cb,2×S3,2) + (Cb,3×S3,3) - Cj,5 = (0×0) + (0×0) + (0×1) - (0) = 0
• Z6 = (Cb,1×R1) + (Cb,2×R2) + (Cb,3×R3) = (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 X1 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 X1. 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 S1 → R1 / X1,1 = 10 / 1 = 10
• Row S2 → R2 / X1,2 = 0 / 1 = 0 (The Lowest Value)
• Row S3 → R3 / X1,3 = 4 / 1 = 4

The lowest value corresponds to the S2 row. This variable will come from the base. The pivot element corresponds to the value that crosses the X1 column and the S2 row = 1.

Enter the variable X1 and the variable S2 leaves the base. The pivot element is 1

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