Location problems: Centre of Gravity, Median Method and Factor Rating — from load–distance optimisation to weighted multi-factor scoring of candidate locations.
A company must locate a new plant to serve a set of existing facilities — suppliers that ship raw material in, and markets that receive finished goods. Each facility sits at map coordinates (x, y) and moves a known volume of goods per period at a known transport rate per unit of distance.
The Centre of Gravity method places the plant at the load-weighted average of the facility coordinates, minimising the total load–distance cost of moving goods to and from the plant:
x̄ = Σ (Vᵢ · Rᵢ · xᵢ) / Σ (Vᵢ · Rᵢ)
ȳ = Σ (Vᵢ · Rᵢ · yᵢ) / Σ (Vᵢ · Rᵢ)
Total cost = Σ Vᵢ · Rᵢ · dᵢ, dᵢ = √((x̄ − xᵢ)² + (ȳ − yᵢ)²)
Your task: enter the facilities below — coordinates, volumes and rates — and read off the optimal plant location and its total period cost. Add or remove facilities to see how the answer moves.
Edit any value inline. Volume is the load moved per period; rate is the transport cost per unit load per unit distance.
| Facility | x | y | Volume V | Rate R | Weight V·R | |
|---|---|---|---|---|---|---|
| 2,000 | ||||||
| 1,500 | ||||||
| 3,000 | ||||||
| 3,600 | ||||||
x̄
52.97
ȳ
44.65
Total cost / period
351,659
| Facility | Weight V·R | Distance dᵢ | Cost V·R·dᵢ |
|---|---|---|---|
| Supplier A | 2,000 | 48.34 | 96,673 |
| Supplier B | 1,500 | 44.87 | 67,310 |
| Market C | 3,000 | 33.70 | 101,088 |
| Market D | 3,600 | 24.05 | 86,587 |
Circles are facilities (area ∝ weight); the cross is the optimal plant location.
Try this: add a very large market far to one corner and watch the optimal plant slide toward it. Then set its rate to zero — the pull disappears entirely.