Solving the dual problem.

Maximize z = 40w_{1} + 50w_{2}

subject to

2w_{1} + 3w_{2} ≤
3

4w_{1} + 2w_{2 }≤ 3

w_{1}, w_{2} ≥ 0

After adding slack variables, we have

Maximize z = 40w_{1} + 50w_{2} + 0x_{3} + 0x_{4}

2w_{1} + 3w_{2} + x_{3} = 3

4w_{1} + 2w2 + x_{4 }= 3

w_{1}, w_{2}, x_{3}, x_{4} ≥
0

Where x_{3} and x_{4} are slack variables.

w_{1}= 0, w_{2} = 0, z = 0

x_{3} = 3, x_{4} = 3

c_{j} |
40 | 50 | 0 | 0 | ||
---|---|---|---|---|---|---|

c_{B} |
Basic variables B |
w_{1} |
w_{2} |
x_{3} |
x_{4} |
Solution values b (=X _{B}) |

0 | x_{3} |
2 | 3 | 1 | 0 | 3 |

0 | x_{4} |
4 | 2 | 0 | 1 | 3 |

z_{j}-c_{j} |
-40 | -50 | 0 | 0 |

Table 2

c_{j} |
40 | 50 | 0 | 0 | ||
---|---|---|---|---|---|---|

c_{B} |
Basic variables B |
w_{1} |
w_{2} |
x_{3} |
x_{4} |
Solution values b (=X _{B}) |

50 | w_{2} |
2/3 | 1 | 1/3 | 0 | 1 |

0 | x_{4} |
8/3 | 0 | -2/3 | 1 | 1 |

z_{j}-c_{j} |
-20/3 | 0 | 50/3 | 0 |

Table 3

c_{j} |
40 | 50 | 0 | 0 | ||
---|---|---|---|---|---|---|

c_{B} |
Basic variables B |
w_{1} |
w_{2} |
x_{3} |
x_{4} |
Solution values b (=X _{B}) |

50 | w_{2} |
0 | 1 | 1/2 | -1/4 | 3/4 |

40 | w_{1} |
1 | 0 | -1/4 | 3/8 | 3/8 |

z_{j}-c_{j} |
0 | 0 | 15 | 5/2 |

The optimal solution is:

w_{1 }= 3/8, w_{2} = 3/4

z = 40 X 3/8 + 50 X 3/4= 105/2.

In case of **primal problem**, you noted that the values of z_{j}-c_{j} under the surplus variables x_{3} and x_{4} were 3/8
and 3/4. In case of **dual problem**, these values are the optimal values
of dual variables w_{1} and w_{2}.

Further, the values of the objective functions in both the problems are same (i.e., 105/2)

- If one problem has an unbounded optimal solution, then the other problem cannot have a feasible solution.
- Optimal solution of either problem gives complete information about the optimal solution of the other.
- If either the primal or dual problem has a finite optimal solution, then the other has also a finite optimal solution.
- The dual of a dual is primal.