# Big M Method Examples – Problems with Solutions π₯

The Big M Method is a widely used approach to solving linear programming exercises using the simplex algorithm. Our post presents you with the 5 Big M Method examples.

We have included maximization and minimization problems as well as the step-by-step to reach the final solution. It is important to mention that all solutions were obtained using our premium version of the Big M method online calculator.

## Example 1:

Maximize Z = 2X1 + 5X2 + 3X3
subject to
X1 + 2X2 – X3 β€ 7
-X1 + X2 – 2X3 β€ 5
X1 + 4X2 + 3X3 β₯ 1
2X1 – X2 + 4X3 = 6
X1, X2, X3 β₯ 0

### Solution 1:

The problem will be adapted to the standard linear programming model, adding the slack, surplus and / or artificial variables in each of the constraints:

• Constraint 1: It has a sign “β€” (less than or equal) so the slack variable will be added S1.
• Constraint 2: It has a sign “β€” (less than or equal) so the slack variable will be added S2.
• Constraint 3: It has a sign “β₯” (greater than or equal) so the surplus variable S3 will be subtracted and the artificial variable A1 will be added.
• Constraint 4: It has an “=” sign (equal) so the artificial variable A2 will be added.

The problem has artificial variables so we will use the Big M method. As the problem is one of maximization, the artificial variables will be subtracted from the objective function multiplied by a very large number (represented by the letter M) in this way the simplex algorithm will penalize them and eliminate them from the base.

The problem is shown below in standard form. The coefficient 0 (zero) will be placed where it corresponds to create our table:

#### Objective Function:

Maximize: Z = 2X1 + 5X2 + 3X3 + 0S1 + 0S2 + 0S3 – MA1 – MA2

#### Subject to:

1X1 + 2X2 – 1X3 + 1S1 + 0S2 + 0S3 + 0A1 + 0A2 = 7

-1X1 + 1X2 – 2X3 + 0S1 + 1S2 + 0S3 + 0A1 + 0A2 = 5

1X1 + 4X2 + 3X3 + 0S1 + 0S2 – 1S3 + 1A1 + 0A2 = 1

2X1 – 1X2 + 4X3 + 0S1 + 0S2 + 0S3 + 0A1 + 1A2 = 6

X1, X2, X3, S1, S2, S3, A1, A2 β₯ 0

The letters in the tables correspond to:

• Cj: Cost vector. Contains the coefficients of the objective function variables.
• Cb: Coefficients of the variables that make up the solution vector.
• Base: Variables forming the solution vector.
• Z: In this row you will find the values of the reduced cost vector. It is calculated by multiplying the solution vector by the constraint coefficients and subtracting the vector of costs.
• Xn, Sn and An: Each column displays the coefficients of the corresponding variable.
• R: This column contains the independent term of each equation.
• An additional numerical subscript will be added to the equations to identify their position in the table. Example: X1,2: Coefficient of X1 in row 2. For Cj and Z the subscripts will indicate their position in the column. Example: Cj,3: I would indicate the coefficient of Cj in column 3.

### Iteration 1

For this table, the value of row Z will be calculated as follows:

Z1 = (Cb,1*X1,1) + (Cb,2*X1,2) + (Cb,3*X1,3) + (Cb,4*X1,4) – Cj1 = (0*1) + (0*-1) + (-M*1) + (-M*2) – (2) = -3M-2

Z2 = (Cb,1*X2,1) + (Cb,2*X2,2) + (Cb,3*X2,3) + (Cb,4*X2,4) – Cj2 = (0*2) + (0*1) + (-M*4) + (-M*-1) – (5) = -3M-5

Z3 = (Cb,1*X3,1) + (Cb,2*X3,2) + (Cb,3*X3,3) + (Cb,4*X3,4) – Cj3 = (0*-1) + (0*-2) + (-M*3) + (-M*4) – (3) = -7M-3

Z4 = (Cb,1*S1,1) + (Cb,2*S1,2) + (Cb,3*S1,3) + (Cb,4*S1,4) – Cj4 = (0*1) + (0*0) + (-M*0) + (-M*0) – (0) = 0

Z5 = (Cb,1*S2,1) + (Cb,2*S2,2) + (Cb,3*S2,3) + (Cb,4*S2,4) – Cj5 = (0*0) + (0*1) + (-M*0) + (-M*0) – (0) = 0

Z6 = (Cb,1*S3,1) + (Cb,2*S3,2) + (Cb,3*S3,3) + (Cb,4*S3,4) – Cj6 = (0*0) + (0*0) + (-M*-1) + (-M*0) – (0) = M

Z7 = (Cb,1*A1,1) + (Cb,2*A1,2) + (Cb,3*A1,3) + (Cb,4*A1,4) – Cj7 = (0*0) + (0*0) + (-M*1) + (-M*0) – (-M) = 0

Z8 = (Cb,1*A2,1) + (Cb,2*A2,2) + (Cb,3*A2,3) + (Cb,4*A2,4) – Cj8 = (0*0) + (0*0) + (-M*0) + (-M*1) – (-M) = 0

Z9 = (Cb,1*R,1) + (Cb,2*R,2) + (Cb,3*R,3) + (Cb,4*R,4) = (0*7) + (0*5) + (-M*1) + (-M*6) = -7M

Table 1 Cj Cb Base X1 X2 X3 S1 S2 S3 2 5 3 0 0 0 -M -M 0 1 2 -1 1 0 0 0 0 7 0 -1 1 -2 0 1 0 0 0 5 -M 1 4 3 0 0 -1 1 0 1 -M 2 -1 4 0 0 0 0 1 6 -3M-2 -3M-5 -7M-3 0 0 M 0 0 -7M
In the reduced cost vector (Z) we have negative values, so we must select the most negative one for the pivot column (maximization).

The solution vector is composed of the following numbers: [-3M-2, -3M-5, -7M-3, 0, 0, M, 0, 0]. The most negative is = -7M-3 which corresponds to the X3 variable. This variable will enter the base and its values in the table will form our pivot column.

If it is not clear to you why -7M-3 is the most negative value, you can replace the variable M by high number as 1000000 in the vector of reduced costs and re-evaluate the results.

The feasibility condition will be verified by dividing the values of column R by the pivot column X3. 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 β R1 / X3,1 = 7 / -1 = N/A
Row S2 β R2 / X3,2 = 5 / -2 = N/A
Row A1 β R3 / X3,3 = 1 / 3 = 0.(3) (The Lowest Value)
Row A2 β R4 / X3,4 = 6 / 4 = 1.5

The numbers in parentheses represent repeating decimals. Example: 1.(3) = 1.33333…

The lowest value corresponds to the A1 row. This variable will come from the base. The pivot element corresponds to the value that crosses the X3 column and the A1 row = 3.

Enter the variable X3 and the variable A1 leaves the base. The pivot element is 3

### Iteration 2

For this table, the value of row Z will be calculated as follows:

Z1 = (Cb,1*X1,1) + (Cb,2*X1,2) + (Cb,3*X1,3) + (Cb,4*X1,4) – Cj1 = (0*4/3) + (0*-1/3) + (3*1/3) + (-M*2/3) – (2) = -2/3M-1

Z2 = (Cb,1*X2,1) + (Cb,2*X2,2) + (Cb,3*X2,3) + (Cb,4*X2,4) – Cj2 = (0*10/3) + (0*11/3) + (3*4/3) + (-M*-19/3) – (5) = 19/3M-1

Z3 = (Cb,1*X3,1) + (Cb,2*X3,2) + (Cb,3*X3,3) + (Cb,4*X3,4) – Cj3 = (0*0) + (0*0) + (3*1) + (-M*0) – (3) = 0

Z4 = (Cb,1*S1,1) + (Cb,2*S1,2) + (Cb,3*S1,3) + (Cb,4*S1,4) – Cj4 = (0*1) + (0*0) + (3*0) + (-M*0) – (0) = 0

Z5 = (Cb,1*S2,1) + (Cb,2*S2,2) + (Cb,3*S2,3) + (Cb,4*S2,4) – Cj5 = (0*0) + (0*1) + (3*0) + (-M*0) – (0) = 0

Z6 = (Cb,1*S3,1) + (Cb,2*S3,2) + (Cb,3*S3,3) + (Cb,4*S3,4) – Cj6 = (0*-1/3) + (0*-2/3) + (3*-1/3) + (-M*4/3) – (0) = -4/3M-1

Z7 = (Cb,1*A1,1) + (Cb,2*A1,2) + (Cb,3*A1,3) + (Cb,4*A1,4) – Cj7 = (0*1/3) + (0*2/3) + (3*1/3) + (-M*-4/3) – (-M) = 7/3M+1

Z8 = (Cb,1*A2,1) + (Cb,2*A2,2) + (Cb,3*A2,3) + (Cb,4*A2,4) – Cj8 = (0*0) + (0*0) + (3*0) + (-M*1) – (-M) = 0

Z9 = (Cb,1*R,1) + (Cb,2*R,2) + (Cb,3*R,3) + (Cb,4*R,4) = (0*22/3) + (0*17/3) + (3*1/3) + (-M*14/3) = -14/3M+1

Each value in the table will be iterated considering the following:

• New Value Pivot Row = Current Value Pivot Row / Pivot Element
• New Value Other Rows = Current Value – (Pivot Column Row Element * New Value Pivot Row)

Pivot row calculations (Row NΒ° 3):

 Current Value Pivot Row Pivot Element New Value Pivot Row 1 4 3 0 0 -1 1 0 1 3 3 3 3 3 3 3 3 3 1 / 3 = 1/3 4 / 3 = 4/3 3 / 3 = 1 0 / 3 = 0 0 / 3 = 0 -1 / 3 = -1/3 1 / 3 = 1/3 0 / 3 = 0 1 / 3 = 1/3

Next, we'll calculate the new values for all other rows of the table:

Row 1:

 Current Value Pivot Column Row Element New Value Pivot Row New Value 1 2 -1 1 0 0 0 0 7 -1 -1 -1 -1 -1 -1 -1 -1 -1 1/3 4/3 1 0 0 -1/3 1/3 0 1/3 1 – (-1Γ1/3) = 4/3 2 – (-1Γ4/3) = 10/3 -1 – (-1Γ1) = 0 1 – (-1Γ0) = 1 0 – (-1Γ0) = 0 0 – (-1Γ-1/3) = -1/3 0 – (-1Γ1/3) = 1/3 0 – (-1Γ0) = 0 7 – (-1Γ1/3) = 22/3

Row 2:

 Current Value Pivot Column Row Element New Value Pivot Row New Value -1 1 -2 0 1 0 0 0 5 -2 -2 -2 -2 -2 -2 -2 -2 -2 1/3 4/3 1 0 0 -1/3 1/3 0 1/3 -1 – (-2Γ1/3) = -1/3 1 – (-2Γ4/3) = 11/3 -2 – (-2Γ1) = 0 0 – (-2Γ0) = 0 1 – (-2Γ0) = 1 0 – (-2Γ-1/3) = -2/3 0 – (-2Γ1/3) = 2/3 0 – (-2Γ0) = 0 5 – (-2Γ1/3) = 17/3

Row 4:

 Current Value Pivot Column Row Element New Value Pivot Row New Value 2 -1 4 0 0 0 0 1 6 4 4 4 4 4 4 4 4 4 1/3 4/3 1 0 0 -1/3 1/3 0 1/3 2 – (4Γ1/3) = 2/3 -1 – (4Γ4/3) = -19/3 4 – (4Γ1) = 0 0 – (4Γ0) = 0 0 – (4Γ0) = 0 0 – (4Γ-1/3) = 4/3 0 – (4Γ1/3) = -4/3 1 – (4Γ0) = 1 6 – (4Γ1/3) = 14/3
Table 2 Cj Cb Base X1 X2 X3 S1 S2 S3 2 5 3 0 0 0 -M -M 0 4/3 10/3 0 1 0 -1/3 1/3 0 22/3 0 -1/3 11/3 0 0 1 -2/3 2/3 0 17/3 3 1/3 4/3 1 0 0 -1/3 1/3 0 1/3 -M 2/3 -19/3 0 0 0 4/3 -4/3 1 14/3 -2/3M-1 19/3M-1 0 0 0 -4/3M-1 7/3M+1 0 -14/3M+1
In the reduced cost vector (Z) we have negative values, so we must select the most negative one for the pivot column (maximization).

The solution vector is composed of the following numbers: [-2/3M-1, 19/3M-1, 0, 0, 0, -4/3M-1, 7/3M+1, 0]. The most negative is = -4/3M-1 which corresponds to the S3 variable. This variable will enter the base and its values in the table will form our pivot column.

If it is not clear to you why -4/3M-1 is the most negative value, you can replace the variable M by high number as 1000000 in the vector of reduced costs and re-evaluate the results.

The feasibility condition will be verified by dividing the values of column R by the pivot column S3. 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 β R1 / S3,1 = 22/3 / -1/3 = N/A
Row S2 β R2 / S3,2 = 17/3 / -2/3 = N/A
Row X3 β R3 / S3,3 = 1/3 / -1/3 = N/A
Row A2 β R4 / S3,4 = 14/3 / 4/3 = 3.5 (The Lowest Value)

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 S3 column and the A2 row = 4/3.

Enter the variable S3 and the variable A2 leaves the base. The pivot element is 4/3

### Iteration 3

For this table, the value of row Z will be calculated as follows:

Z1 = (Cb,1*X1,1) + (Cb,2*X1,2) + (Cb,3*X1,3) + (Cb,4*X1,4) – Cj1 = (0*3/2) + (0*0) + (3*1/2) + (0*1/2) – (2) = -1/2

Z2 = (Cb,1*X2,1) + (Cb,2*X2,2) + (Cb,3*X2,3) + (Cb,4*X2,4) – Cj2 = (0*7/4) + (0*1/2) + (3*-1/4) + (0*-19/4) – (5) = -23/4

Z3 = (Cb,1*X3,1) + (Cb,2*X3,2) + (Cb,3*X3,3) + (Cb,4*X3,4) – Cj3 = (0*0) + (0*0) + (3*1) + (0*0) – (3) = 0

Z4 = (Cb,1*S1,1) + (Cb,2*S1,2) + (Cb,3*S1,3) + (Cb,4*S1,4) – Cj4 = (0*1) + (0*0) + (3*0) + (0*0) – (0) = 0

Z5 = (Cb,1*S2,1) + (Cb,2*S2,2) + (Cb,3*S2,3) + (Cb,4*S2,4) – Cj5 = (0*0) + (0*1) + (3*0) + (0*0) – (0) = 0

Z6 = (Cb,1*S3,1) + (Cb,2*S3,2) + (Cb,3*S3,3) + (Cb,4*S3,4) – Cj6 = (0*0) + (0*0) + (3*0) + (0*1) – (0) = 0

Z7 = (Cb,1*A1,1) + (Cb,2*A1,2) + (Cb,3*A1,3) + (Cb,4*A1,4) – Cj7 = (0*0) + (0*0) + (3*0) + (0*-1) – (-M) = M

Z8 = (Cb,1*A2,1) + (Cb,2*A2,2) + (Cb,3*A2,3) + (Cb,4*A2,4) – Cj8 = (0*1/4) + (0*1/2) + (3*1/4) + (0*3/4) – (-M) = M+3/4

Z9 = (Cb,1*R,1) + (Cb,2*R,2) + (Cb,3*R,3) + (Cb,4*R,4) = (0*17/2) + (0*8) + (3*3/2) + (0*7/2) = 9/2

Each value in the table will be iterated considering the following:

• New Value Pivot Row = Current Value Pivot Row / Pivot Element
• New Value Other Rows = Current Value – (Pivot Column Row Element * New Value Pivot Row)

Pivot row calculations (Row NΒ° 4):

 Current Value Pivot Row Pivot Element New Value Pivot Row 2/3 -19/3 0 0 0 4/3 -4/3 1 14/3 4/3 4/3 4/3 4/3 4/3 4/3 4/3 4/3 4/3 2/3 / 4/3 = 1/2 -19/3 / 4/3 = -19/4 0 / 4/3 = 0 0 / 4/3 = 0 0 / 4/3 = 0 4/3 / 4/3 = 1 -4/3 / 4/3 = -1 1 / 4/3 = 3/4 14/3 / 4/3 = 7/2

Next, we'll calculate the new values for all other rows of the table:

Row 1:

 Current Value Pivot Column Row Element New Value Pivot Row New Value 4/3 10/3 0 1 0 -1/3 1/3 0 22/3 -1/3 -1/3 -1/3 -1/3 -1/3 -1/3 -1/3 -1/3 -1/3 1/2 -19/4 0 0 0 1 -1 3/4 7/2 4/3 – (-1/3Γ1/2) = 3/2 10/3 – (-1/3Γ-19/4) = 7/4 0 – (-1/3Γ0) = 0 1 – (-1/3Γ0) = 1 0 – (-1/3Γ0) = 0 -1/3 – (-1/3Γ1) = 0 1/3 – (-1/3Γ-1) = 0 0 – (-1/3Γ3/4) = 1/4 22/3 – (-1/3Γ7/2) = 17/2

Row 2:

 Current Value Pivot Column Row Element New Value Pivot Row New Value -1/3 11/3 0 0 1 -2/3 2/3 0 17/3 -2/3 -2/3 -2/3 -2/3 -2/3 -2/3 -2/3 -2/3 -2/3 1/2 -19/4 0 0 0 1 -1 3/4 7/2 -1/3 – (-2/3Γ1/2) = 0 11/3 – (-2/3Γ-19/4) = 1/2 0 – (-2/3Γ0) = 0 0 – (-2/3Γ0) = 0 1 – (-2/3Γ0) = 1 -2/3 – (-2/3Γ1) = 0 2/3 – (-2/3Γ-1) = 0 0 – (-2/3Γ3/4) = 1/2 17/3 – (-2/3Γ7/2) = 8

Row 3:

 Current Value Pivot Column Row Element New Value Pivot Row New Value 1/3 4/3 1 0 0 -1/3 1/3 0 1/3 -1/3 -1/3 -1/3 -1/3 -1/3 -1/3 -1/3 -1/3 -1/3 1/2 -19/4 0 0 0 1 -1 3/4 7/2 1/3 – (-1/3Γ1/2) = 1/2 4/3 – (-1/3Γ-19/4) = -1/4 1 – (-1/3Γ0) = 1 0 – (-1/3Γ0) = 0 0 – (-1/3Γ0) = 0 -1/3 – (-1/3Γ1) = 0 1/3 – (-1/3Γ-1) = 0 0 – (-1/3Γ3/4) = 1/4 1/3 – (-1/3Γ7/2) = 3/2
Table 3 Cj Cb Base X1 X2 X3 S1 S2 S3 2 5 3 0 0 0 -M -M 0 3/2 7/4 0 1 0 0 0 1/4 17/2 0 0 1/2 0 0 1 0 0 1/2 8 3 1/2 -1/4 1 0 0 0 0 1/4 3/2 0 1/2 -19/4 0 0 0 1 -1 3/4 7/2 -1/2 -23/4 0 0 0 0 M M+3/4 9/2
In the reduced cost vector (Z) we have negative values, so we must select the most negative one for the pivot column (maximization).

The solution vector is composed of the following numbers: [-1/2, -23/4, 0, 0, 0, 0, M, M+3/4]. The most negative is = -23/4 which corresponds to the X2 variable. This variable will enter the base and its values in the table will form our pivot column.

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 S1 β R1 / X2,1 = 17/2 / 7/4 = 4.(857142) (The Lowest Value)
Row S2 β R2 / X2,2 = 8 / 1/2 = 16
Row X3 β R3 / X2,3 = 3/2 / -1/4 = N/A
Row S3 β R4 / X2,4 = 7/2 / -19/4 = N/A

The numbers in parentheses represent repeating decimals. Example: 1.(3) = 1.33333…

The lowest value corresponds to the S1 row. This variable will come from the base. The pivot element corresponds to the value that crosses the X2 column and the S1 row = 7/4.

Enter the variable X2 and the variable S1 leaves the base. The pivot element is 7/4

### Iteration 4

For this table, the value of row Z will be calculated as follows:

Z1 = (Cb,1*X1,1) + (Cb,2*X1,2) + (Cb,3*X1,3) + (Cb,4*X1,4) – Cj1 = (5*6/7) + (0*-3/7) + (3*5/7) + (0*32/7) – (2) = 31/7

Z2 = (Cb,1*X2,1) + (Cb,2*X2,2) + (Cb,3*X2,3) + (Cb,4*X2,4) – Cj2 = (5*1) + (0*0) + (3*0) + (0*0) – (5) = 0

Z3 = (Cb,1*X3,1) + (Cb,2*X3,2) + (Cb,3*X3,3) + (Cb,4*X3,4) – Cj3 = (5*0) + (0*0) + (3*1) + (0*0) – (3) = 0

Z4 = (Cb,1*S1,1) + (Cb,2*S1,2) + (Cb,3*S1,3) + (Cb,4*S1,4) – Cj4 = (5*4/7) + (0*-2/7) + (3*1/7) + (0*19/7) – (0) = 23/7

Z5 = (Cb,1*S2,1) + (Cb,2*S2,2) + (Cb,3*S2,3) + (Cb,4*S2,4) – Cj5 = (5*0) + (0*1) + (3*0) + (0*0) – (0) = 0

Z6 = (Cb,1*S3,1) + (Cb,2*S3,2) + (Cb,3*S3,3) + (Cb,4*S3,4) – Cj6 = (5*0) + (0*0) + (3*0) + (0*1) – (0) = 0

Z7 = (Cb,1*A1,1) + (Cb,2*A1,2) + (Cb,3*A1,3) + (Cb,4*A1,4) – Cj7 = (5*0) + (0*0) + (3*0) + (0*-1) – (-M) = M

Z8 = (Cb,1*A2,1) + (Cb,2*A2,2) + (Cb,3*A2,3) + (Cb,4*A2,4) – Cj8 = (5*1/7) + (0*3/7) + (3*2/7) + (0*10/7) – (-M) = M+11/7

Z9 = (Cb,1*R,1) + (Cb,2*R,2) + (Cb,3*R,3) + (Cb,4*R,4) = (5*34/7) + (0*39/7) + (3*19/7) + (0*186/7) = 227/7

Each value in the table will be iterated considering the following:

• New Value Pivot Row = Current Value Pivot Row / Pivot Element
• New Value Other Rows = Current Value – (Pivot Column Row Element * New Value Pivot Row)

Pivot row calculations (Row NΒ° 1):

 Current Value Pivot Row Pivot Element New Value Pivot Row 3/2 7/4 0 1 0 0 0 1/4 17/2 7/4 7/4 7/4 7/4 7/4 7/4 7/4 7/4 7/4 3/2 / 7/4 = 6/7 7/4 / 7/4 = 1 0 / 7/4 = 0 1 / 7/4 = 4/7 0 / 7/4 = 0 0 / 7/4 = 0 0 / 7/4 = 0 1/4 / 7/4 = 1/7 17/2 / 7/4 = 34/7

Next, we'll calculate the new values for all other rows of the table:

Row 2:

 Current Value Pivot Column Row Element New Value Pivot Row New Value 0 1/2 0 0 1 0 0 1/2 8 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 6/7 1 0 4/7 0 0 0 1/7 34/7 0 – (1/2Γ6/7) = -3/7 1/2 – (1/2Γ1) = 0 0 – (1/2Γ0) = 0 0 – (1/2Γ4/7) = -2/7 1 – (1/2Γ0) = 1 0 – (1/2Γ0) = 0 0 – (1/2Γ0) = 0 1/2 – (1/2Γ1/7) = 3/7 8 – (1/2Γ34/7) = 39/7

Row 3:

 Current Value Pivot Column Row Element New Value Pivot Row New Value 1/2 -1/4 1 0 0 0 0 1/4 3/2 -1/4 -1/4 -1/4 -1/4 -1/4 -1/4 -1/4 -1/4 -1/4 6/7 1 0 4/7 0 0 0 1/7 34/7 1/2 – (-1/4Γ6/7) = 5/7 -1/4 – (-1/4Γ1) = 0 1 – (-1/4Γ0) = 1 0 – (-1/4Γ4/7) = 1/7 0 – (-1/4Γ0) = 0 0 – (-1/4Γ0) = 0 0 – (-1/4Γ0) = 0 1/4 – (-1/4Γ1/7) = 2/7 3/2 – (-1/4Γ34/7) = 19/7

Row 4:

 Current Value Pivot Column Row Element New Value Pivot Row New Value 1/2 -19/4 0 0 0 1 -1 3/4 7/2 -19/4 -19/4 -19/4 -19/4 -19/4 -19/4 -19/4 -19/4 -19/4 6/7 1 0 4/7 0 0 0 1/7 34/7 1/2 – (-19/4Γ6/7) = 32/7 -19/4 – (-19/4Γ1) = 0 0 – (-19/4Γ0) = 0 0 – (-19/4Γ4/7) = 19/7 0 – (-19/4Γ0) = 0 1 – (-19/4Γ0) = 1 -1 – (-19/4Γ0) = -1 3/4 – (-19/4Γ1/7) = 10/7 7/2 – (-19/4Γ34/7) = 186/7
Table 4 Cj Cb Base X1 X2 X3 S1 S2 S3 2 5 3 0 0 0 -M -M 5 6/7 1 0 4/7 0 0 0 1/7 34/7 0 -3/7 0 0 -2/7 1 0 0 3/7 39/7 3 5/7 0 1 1/7 0 0 0 2/7 19/7 0 32/7 0 0 19/7 0 1 -1 10/7 186/7 31/7 0 0 23/7 0 0 M M+11/7 227/7
Since there are no negative values in the reduced cost vector (Z) = [31/7, 0, 0, 23/7, 0, 0, M, M+11/7], this means that we are at the optimal point (maximization).

The optimal solution is Z = 227/7

X1= 0, X2= 34/7, X3= 19/7, S1= 0, S2= 39/7, S3= 186/7, A1= 0, A2= 0