# Solved Exercise of Minimization of 2 variables with the Big M MethodExample of Linear Programming - Big M Method

### Problem Data

We want to Minimize the following problem:

Objective Function
Z = X1 - 2X2
Subject to the following constraints
• X1 + X2 2
• -X1 + X2 1
• 0X1 + X2 3

X1, X2 ≥ 0

Description
Solved Exercise of Minimization of 2 variables with the Big M MethodSolve the linear programming problem shown above using the Big M 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 S1 will be subtracted and the artificial variable A1 will be added. In the initial matrix, A1 will be in the base.
• Constraint 2: It has sign “” (greatest equal), therefore the excess variable S2 will be subtracted and the artificial variable A2 will be added. In the initial matrix, A2 will be in the base.
• 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:

Minimize Z = X1 - 2X2 + 0S1 + 0S2 + 0S3 + MA1 + MA2

Subject to:

• X1 + X2 - S1 + 0S2 + 0S3 + A1 + 0A2 = 2
• -X1 + X2 + 0S1 - S2 + 0S3 + 0A1 + A2 = 1
• 0X1 + X2 + 0S1 + 0S2 + S3 + 0A1 + 0A2 = 3

### Initial Matrix

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

Table 1 Cj Cb Base X1 X2 S1 S2 S3 1 -2 0 0 0 M M M 1 1 -1 0 0 1 0 2 M -1 1 0 -1 0 0 1 1 0 0 1 0 0 1 0 0 3 -1 2M + 2 -M -M 0 0 0 3M

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 = (M×1) + (M×-1) + (0×0) - (1) = -1
• Z2 = (Cb,1×X2,1) + (Cb,2×X2,2) + (Cb,3×X2,3) - Cj,2 = (M×1) + (M×1) + (0×1) - (-2) = 2M + 2
• Z3 = (Cb,1×S1,1) + (Cb,2×S1,2) + (Cb,3×S1,3) - Cj,3 = (M×-1) + (M×0) + (0×0) - (0) = -M
• Z4 = (Cb,1×S2,1) + (Cb,2×S2,2) + (Cb,3×S2,3) - Cj,4 = (M×0) + (M×-1) + (0×0) - (0) = -M
• Z5 = (Cb,1×S3,1) + (Cb,2×S3,2) + (Cb,3×S3,3) - Cj,5 = (M×0) + (M×0) + (0×1) - (0) = 0
• Z6 = (Cb,1×A1,1) + (Cb,2×A1,2) + (Cb,3×A1,3) - Cj,6 = (M×1) + (M×0) + (0×0) - (M) = 0
• Z7 = (Cb,1×A2,1) + (Cb,2×A2,2) + (Cb,3×A2,3) - Cj,7 = (M×0) + (M×1) + (0×0) - (M) = 0
• Z8 = (Cb,1×R1) + (Cb,2×R2) + (Cb,3×R3) = (M×2) + (M×1) + (0×3) = 3M

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: [-1, 2M + 2, -M, -M, 0, 0, 0]. The highest value is = 2M + 2 which corresponds to the X2 variable. This variable will enter the base and its values in the table will form our pivot column.

Note: To check the chosen value, simply replace the variable M by a large number (e.g. M = 1000000) in the vector of reduced costs.

d) Feasibility Condition (Outgoing Variable):

The feasibility condition will be verified by dividing the values of column R by the pivot column X2. 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 A1 → R1 / X2,1 = 2 / 1 = 2
• Row A2 → R2 / X2,2 = 1 / 1 = 1 (The Lowest Value)
• Row S3 → R3 / X2,3 = 3 / 1 = 3

The lowest value corresponds to the A2 row. This variable will come from the base. The pivot element corresponds to the value that crosses the X2 column and the A2 row = 1.

Enter the variable X2 and the variable A2 leaves the base. The pivot element is 1

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