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:

12 | 42 | 51 | 68 |

64 | 6 | 63 | 2 |

26 | 6 | 65 | 29 |

19 | 20 | 9 | 27 |

**Subtract row minima**

We subtract the row minimum from each row:

0 | 30 | 39 | 56 | (-12) |

62 | 4 | 61 | 0 | (-2) |

20 | 0 | 59 | 23 | (-6) |

10 | 11 | 0 | 18 | (-9) |

**Subtract column minima**

Because each column contains a zero, subtracting column minima has no effect.

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

There are 4 lines required to cover all zeros:

0 | 30 | 39 | 56 | x |

62 | 4 | 61 | 0 | x |

20 | 0 | 59 | 23 | x |

10 | 11 | 0 | 18 | x |

**The optimal assignment**

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

0 | 30 | 39 | 56 |

62 | 4 | 61 | 0 |

20 | 0 | 59 | 23 |

10 | 11 | 0 | 18 |

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

12 | 42 | 51 | 68 |

64 | 6 | 63 | 2 |

26 | 6 | 65 | 29 |

19 | 20 | 9 | 27 |

The optimal value equals 29.

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