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 | 20 | 32 | 4 |

25 | 50 | 99 | 62 |

75 | 17 | 20 | 57 |

97 | 58 | 20 | 48 |

**Subtract row minima**

We subtract the row minimum from each row:

4 | 16 | 28 | 0 | (-4) |

0 | 25 | 74 | 37 | (-25) |

58 | 0 | 3 | 40 | (-17) |

77 | 38 | 0 | 28 | (-20) |

**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:

4 | 16 | 28 | 0 | x |

0 | 25 | 74 | 37 | x |

58 | 0 | 3 | 40 | x |

77 | 38 | 0 | 28 | x |

**The optimal assignment**

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

4 | 16 | 28 | 0 |

0 | 25 | 74 | 37 |

58 | 0 | 3 | 40 |

77 | 38 | 0 | 28 |

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

8 | 20 | 32 | 4 |

25 | 50 | 99 | 62 |

75 | 17 | 20 | 57 |

97 | 58 | 20 | 48 |

The optimal value equals 66.

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