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:

89 | 36 | 6 | 69 |

72 | 47 | 34 | 31 |

40 | 33 | 23 | 7 |

8 | 80 | 8 | 50 |

**Subtract row minima**

We subtract the row minimum from each row:

83 | 30 | 0 | 63 | (-6) |

41 | 16 | 3 | 0 | (-31) |

33 | 26 | 16 | 0 | (-7) |

0 | 72 | 0 | 42 | (-8) |

**Subtract column minima**

We subtract the column minimum from each column:

83 | 14 | 0 | 63 |

41 | 0 | 3 | 0 |

33 | 10 | 16 | 0 |

0 | 56 | 0 | 42 |

(-16) |

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

There are 4 lines required to cover all zeros:

83 | 14 | 0 | 63 | x |

41 | 0 | 3 | 0 | x |

33 | 10 | 16 | 0 | x |

0 | 56 | 0 | 42 | x |

**The optimal assignment**

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

83 | 14 | 0 | 63 |

41 | 0 | 3 | 0 |

33 | 10 | 16 | 0 |

0 | 56 | 0 | 42 |

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

89 | 36 | 6 | 69 |

72 | 47 | 34 | 31 |

40 | 33 | 23 | 7 |

8 | 80 | 8 | 50 |

The optimal value equals 68.

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