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:

48 | 3 | 94 | 61 |

71 | 32 | 6 | 69 |

9 | 48 | 90 | 23 |

64 | 92 | 24 | 17 |

**Subtract row minima**

We subtract the row minimum from each row:

45 | 0 | 91 | 58 | (-3) |

65 | 26 | 0 | 63 | (-6) |

0 | 39 | 81 | 14 | (-9) |

47 | 75 | 7 | 0 | (-17) |

**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:

45 | 0 | 91 | 58 | x |

65 | 26 | 0 | 63 | x |

0 | 39 | 81 | 14 | x |

47 | 75 | 7 | 0 | x |

**The optimal assignment**

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

45 | 0 | 91 | 58 |

65 | 26 | 0 | 63 |

0 | 39 | 81 | 14 |

47 | 75 | 7 | 0 |

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

48 | 3 | 94 | 61 |

71 | 32 | 6 | 69 |

9 | 48 | 90 | 23 |

64 | 92 | 24 | 17 |

The optimal value equals 35.

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