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:

92 | 18 | 85 | 21 |

93 | 43 | 96 | 18 |

57 | 73 | 5 | 83 |

33 | 95 | 77 | 51 |

**Subtract row minima**

We subtract the row minimum from each row:

74 | 0 | 67 | 3 | (-18) |

75 | 25 | 78 | 0 | (-18) |

52 | 68 | 0 | 78 | (-5) |

0 | 62 | 44 | 18 | (-33) |

**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:

74 | 0 | 67 | 3 | x |

75 | 25 | 78 | 0 | x |

52 | 68 | 0 | 78 | x |

0 | 62 | 44 | 18 | x |

**The optimal assignment**

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

74 | 0 | 67 | 3 |

75 | 25 | 78 | 0 |

52 | 68 | 0 | 78 |

0 | 62 | 44 | 18 |

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

92 | 18 | 85 | 21 |

93 | 43 | 96 | 18 |

57 | 73 | 5 | 83 |

33 | 95 | 77 | 51 |

The optimal value equals 74.

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