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:

14 | 71 | 21 | 99 |

64 | 91 | 24 | 97 |

72 | 20 | 37 | 3 |

99 | 11 | 8 | 39 |

**Subtract row minima**

We subtract the row minimum from each row:

0 | 57 | 7 | 85 | (-14) |

40 | 67 | 0 | 73 | (-24) |

69 | 17 | 34 | 0 | (-3) |

91 | 3 | 0 | 31 | (-8) |

**Subtract column minima**

We subtract the column minimum from each column:

0 | 54 | 7 | 85 |

40 | 64 | 0 | 73 |

69 | 14 | 34 | 0 |

91 | 0 | 0 | 31 |

(-3) |

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

There are 4 lines required to cover all zeros:

0 | 54 | 7 | 85 | x |

40 | 64 | 0 | 73 | x |

69 | 14 | 34 | 0 | x |

91 | 0 | 0 | 31 | x |

**The optimal assignment**

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

0 | 54 | 7 | 85 |

40 | 64 | 0 | 73 |

69 | 14 | 34 | 0 |

91 | 0 | 0 | 31 |

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

14 | 71 | 21 | 99 |

64 | 91 | 24 | 97 |

72 | 20 | 37 | 3 |

99 | 11 | 8 | 39 |

The optimal value equals 52.

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