Fill in the cost matrix of an assignment problem and click on 'Solve'. The optimal assignment will be determined and a step by step explanation of the hungarian algorithm will be given.

Fill in the cost matrix (random cost matrix):

Size: 3x3 4x4 5x5 6x6 7x7 8x8 9x9 10x10

This is the original cost matrix:

72 | 88 | 3 | 34 |

48 | 59 | 69 | 11 |

68 | 87 | 52 | 9 |

82 | 8 | 7 | 31 |

**Subtract row minima**

We subtract the row minimum from each row:

69 | 85 | 0 | 31 | (-3) |

37 | 48 | 58 | 0 | (-11) |

59 | 78 | 43 | 0 | (-9) |

75 | 1 | 0 | 24 | (-7) |

**Subtract column minima**

We subtract the column minimum from each column:

32 | 84 | 0 | 31 |

0 | 47 | 58 | 0 |

22 | 77 | 43 | 0 |

38 | 0 | 0 | 24 |

(-37) | (-1) |

**Cover all zeros with a minimum number of lines**

There are 4 lines required to cover all zeros:

32 | 84 | 0 | 31 | x |

0 | 47 | 58 | 0 | x |

22 | 77 | 43 | 0 | x |

38 | 0 | 0 | 24 | x |

**The optimal assignment**

Because there are 4 lines required, the zeros cover an optimal assignment:

32 | 84 | 0 | 31 |

0 | 47 | 58 | 0 |

22 | 77 | 43 | 0 |

38 | 0 | 0 | 24 |

This corresponds to the following optimal assignment in the original cost matrix:

72 | 88 | 3 | 34 |

48 | 59 | 69 | 11 |

68 | 87 | 52 | 9 |

82 | 8 | 7 | 31 |

The optimal value equals 68.

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