Math, asked by belagurpreet12, 2 months ago

Solve the following LLP by Simplex Method :-

Max Z = 3 X1 + 2 X2 + 5 X3

Subject to :

X1 + X2 + X3 ≤ 9

2 X1 + 3X2 +5X3 ≤30

2 X1 -X2 -X3 ≤8

And X1, X2, X3 ≥ 0​

Answers

Answered by CutieAaiti23
0

Answer:

Step-by-step explanation:

To solve the given linear programming problem using the simplex method, we first write the problem in standard form by introducing slack variables and expressing the objective function as a linear combination of the variables:

Maximize Z = 3X1 + 2X2 + 5X3

Subject to:

X1 + X2 + X3 + S1 = 9

2X1 + 3X2 + 5X3 + S2 = 30

2X1 - X2 - X3 + S3 = 8

X1, X2, X3, S1, S2, S3 ≥ 0

We then construct the initial simplex tableau as follows:

| X1 | X2 | X3 | S1 | S2 | S3 | RHS |

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

S1| 1 | 1 | 1 | 1 | 0 | 0 | 9 |

S2| 2 | 3 | 5 | 0 | 1 | 0 | 30 |

S3| 2 |-1 |-1 | 0 | 0 | 1 | 8 |

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

Z | 3 | 2 | 5 | 0 | 0 | 0 | 0 |

The numbers in the first row represent the coefficients of the variables and slack variables in the constraints, while the numbers in the last column represent the right-hand side values of the constraints. The numbers in the last row represent the coefficients of the variables in the objective function.

We choose X1 as the entering variable and S1 as the leaving variable, since the ratio of the RHS values to the coefficients for X1 is smallest for the first constraint. We then perform row operations to make X1 the basic variable for the first constraint and update the tableau accordingly:

| X1 | X2 | X3 | S1 | S2 | S3 | RHS |

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

X1| 1 | 1 | 1 | 1 | 0 | 0 | 9 |

S2|-1 | 1 | 3 |-2 | 1 | 0 | 12 |

S3|-4 | 3 | 3 |-2 | 0 | 1 | 1 |

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

Z | 0 |-1 | 2 | 3 | 0 | 0 | 27 |

We choose X3 as the entering variable and S2 as the leaving variable, since the ratio of the RHS values to the coefficients for X3 is smallest for the second constraint. We then perform row operations to make X3 the basic variable for the second constraint and update the tableau accordingly:

| X1 | X2 | X3 | S1 | S2 | S3 | RHS |

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

X1| 0 | 2 |-1 | 3 |-1 | 0 | 3 |

X3| 0 | 4/3| 1 |-2/3|

Similar questions