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:

6 | 11 | 94 | 77 |

10 | 71 | 60 | 18 |

36 | 34 | 83 | 82 |

70 | 14 | 13 | 54 |

**Subtract row minima**

We subtract the row minimum from each row:

0 | 5 | 88 | 71 | (-6) |

0 | 61 | 50 | 8 | (-10) |

2 | 0 | 49 | 48 | (-34) |

57 | 1 | 0 | 41 | (-13) |

**Subtract column minima**

We subtract the column minimum from each column:

0 | 5 | 88 | 63 |

0 | 61 | 50 | 0 |

2 | 0 | 49 | 40 |

57 | 1 | 0 | 33 |

(-8) |

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

There are 4 lines required to cover all zeros:

0 | 5 | 88 | 63 | x |

0 | 61 | 50 | 0 | x |

2 | 0 | 49 | 40 | x |

57 | 1 | 0 | 33 | x |

**The optimal assignment**

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

0 | 5 | 88 | 63 |

0 | 61 | 50 | 0 |

2 | 0 | 49 | 40 |

57 | 1 | 0 | 33 |

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

6 | 11 | 94 | 77 |

10 | 71 | 60 | 18 |

36 | 34 | 83 | 82 |

70 | 14 | 13 | 54 |

The optimal value equals 71.

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