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more than 3 weeks ago

To use the simplex method, we first need to convert the inequalities into equations by introducing slack variables:

1. X1 + 3X2 + 2X3 + X4 = 10

2. -X1 - 5X2 - X3 - X5 = -8

Now we can write the initial simplex tableau:

| X1 | X2 | X3 | X4 | X5 | RHS |

|----|----|----|----|----|-----|

| 1 | 3 | 2 | 1 | 0 | 10 |

| -1 | -5 | -1 | 0 | 1 | -8 |

| -1 | -10| -7 | 0 | 0 | 0 |

Next, we need to choose the entering variable. Since X1 has the most negative coefficient in the objective function, we choose X1 as the entering variable. To determine the leaving variable, we calculate the minimum ratio test for each constraint:

10/1 = 10

-8/-1 = 8

The minimum ratio is 8, so X5 leaves the basis. Perform row operations to pivot X1 into the basis:

| X1 | X2 | X3 | X4 | X5 | RHS |

|----|----|----|----|----|-----|

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

| 0 | -3 | 0 | 1 | 1 | 0 |

| 1 | 10 | 7 | 0 | 0 | 0 |

Now, X2 enters the basis. Calculate the minimum ratio test:

2/-2 = -1

0/-3 = N/A

Since the minimum ratio is negative, the solution is unbounded. Therefore, the optimal solution is not bounded and the problem is infeasible.

1. X1 + 3X2 + 2X3 + X4 = 10

2. -X1 - 5X2 - X3 - X5 = -8

Now we can write the initial simplex tableau:

| X1 | X2 | X3 | X4 | X5 | RHS |

|----|----|----|----|----|-----|

| 1 | 3 | 2 | 1 | 0 | 10 |

| -1 | -5 | -1 | 0 | 1 | -8 |

| -1 | -10| -7 | 0 | 0 | 0 |

Next, we need to choose the entering variable. Since X1 has the most negative coefficient in the objective function, we choose X1 as the entering variable. To determine the leaving variable, we calculate the minimum ratio test for each constraint:

10/1 = 10

-8/-1 = 8

The minimum ratio is 8, so X5 leaves the basis. Perform row operations to pivot X1 into the basis:

| X1 | X2 | X3 | X4 | X5 | RHS |

|----|----|----|----|----|-----|

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

| 0 | -3 | 0 | 1 | 1 | 0 |

| 1 | 10 | 7 | 0 | 0 | 0 |

Now, X2 enters the basis. Calculate the minimum ratio test:

2/-2 = -1

0/-3 = N/A

Since the minimum ratio is negative, the solution is unbounded. Therefore, the optimal solution is not bounded and the problem is infeasible.

Google Gemini AI

more than 3 weeks ago

Maximize Z' = -Z = -X1 - 10X2 - 7X3

Subject to

X1 + 3X2 + 2X3 + S1 = 10

-X1 - 5X2 - X3 + S2 = -8

X1, X2, X3, S1, S2 ≥ 0

| | X1 | X2 | X3 | S1 | S2 | RHS |

|---|---|---|---|---|---|---|

| Z' | -1 | -10 | -7 | 0 | 0 | 0 |

| X1 | 1 | 3 | 2 | 1 | 0 | 10 |

| X2 | -1 | -5 | -1 | 0 | 1 | -8 |

The most negative coefficient in the Z' row is -10, which corresponds to X2. So, X2 is the entering variable.

To determine the leaving variable, divide each positive number in the RHS column by the corresponding coefficient of the entering variable in the pivot row.

| Variable | RHS | Coefficient | Ratio |

|---|---|---|---|

| X1 | 10 | 1 | 10 |

| S2 | -8 | -1 | 8 |

The smallest ratio is 8, so S2 is the leaving variable.

Divide the pivot row (row 2) by the pivot element (-1).

Subtract 3 times row 2 from row 1.

Subtract 5 times row 2 from row 3.

| | X1 | X2 | X3 | S1 | S2 | RHS |

|---|---|---|---|---|---|---|

| Z' | -1 | 0 | -23 | 3 | 0 | -10 |

| X1 | 4 | 1 | 4 | 1 | 0 | 2 |

| X2 | 0 | 1 | 4 | 0 | 1 | -2 |

There are no negative coefficients in the Z' row, so the current solution is optimal.

The optimal solution is X1 = 2, X2 = -2 (but it is non-negative, so it's set to 0), X3 = 0.

Z' = -10 = -Z = Z