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:

40 | 58 | 92 | 11 |

44 | 9 | 23 | 86 |

5 | 18 | 86 | 67 |

67 | 75 | 45 | 76 |

**Subtract row minima**

We subtract the row minimum from each row:

29 | 47 | 81 | 0 | (-11) |

35 | 0 | 14 | 77 | (-9) |

0 | 13 | 81 | 62 | (-5) |

22 | 30 | 0 | 31 | (-45) |

**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:

29 | 47 | 81 | 0 | x |

35 | 0 | 14 | 77 | x |

0 | 13 | 81 | 62 | x |

22 | 30 | 0 | 31 | x |

**The optimal assignment**

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

29 | 47 | 81 | 0 |

35 | 0 | 14 | 77 |

0 | 13 | 81 | 62 |

22 | 30 | 0 | 31 |

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

40 | 58 | 92 | 11 |

44 | 9 | 23 | 86 |

5 | 18 | 86 | 67 |

67 | 75 | 45 | 76 |

The optimal value equals 70.

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