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:

8 | 34 | 24 | 83 |

75 | 63 | 50 | 77 |

57 | 34 | 93 | 15 |

67 | 8 | 82 | 79 |

**Subtract row minima**

We subtract the row minimum from each row:

0 | 26 | 16 | 75 | (-8) |

25 | 13 | 0 | 27 | (-50) |

42 | 19 | 78 | 0 | (-15) |

59 | 0 | 74 | 71 | (-8) |

**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 | 26 | 16 | 75 | x |

25 | 13 | 0 | 27 | x |

42 | 19 | 78 | 0 | x |

59 | 0 | 74 | 71 | x |

**The optimal assignment**

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

0 | 26 | 16 | 75 |

25 | 13 | 0 | 27 |

42 | 19 | 78 | 0 |

59 | 0 | 74 | 71 |

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

8 | 34 | 24 | 83 |

75 | 63 | 50 | 77 |

57 | 34 | 93 | 15 |

67 | 8 | 82 | 79 |

The optimal value equals 81.

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