#### shg

##### MrExcel MVP

- Joined
- May 7, 2008

- Messages
- 21,837

- Office Version
- 2010

- Platform
- Windows

C | D | E | |

6 | x | y | Wgt |

7 | 200 | 50 | 2500 |

8 | 300 | 100 | 1300 |

9 | 100 | 150 | 5000 |

Assuming the weight is taken to mean the number of trips you need to make each year (out and back, no stops at other customers), where should you locate a distribution center to minimize total distance?

What they did is take the weighted average of x and weighted average of y ({158, 114}), just what I'd have done; it makes the weighted average distance 81.6 miles per trip.

Except it's the wrong answer. Solver would put the facility at {100, 150}, which makes the average trip 70.6 miles.

Can someone explain the disconnect?