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:

62 | 77 | 96 | 64 |

42 | 82 | 69 | 30 |

75 | 47 | 11 | 31 |

44 | 9 | 93 | 26 |

**Subtract row minima**

We subtract the row minimum from each row:

0 | 15 | 34 | 2 | (-62) |

12 | 52 | 39 | 0 | (-30) |

64 | 36 | 0 | 20 | (-11) |

35 | 0 | 84 | 17 | (-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 | 15 | 34 | 2 | x |

12 | 52 | 39 | 0 | x |

64 | 36 | 0 | 20 | x |

35 | 0 | 84 | 17 | x |

**The optimal assignment**

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

0 | 15 | 34 | 2 |

12 | 52 | 39 | 0 |

64 | 36 | 0 | 20 |

35 | 0 | 84 | 17 |

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

62 | 77 | 96 | 64 |

42 | 82 | 69 | 30 |

75 | 47 | 11 | 31 |

44 | 9 | 93 | 26 |

The optimal value equals 112.

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