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Centre of gravity method for warehouse

Short answer

Use shipment volumes as weights to find weighted average coordinates: x̄ = Σ(x_i W_i)/ΣW_i, ȳ = Σ(y_i W_i)/ΣW_i. The distribution centre is at (49.3 km, 59.3 km).

A distribution centre is to serve four markets whose coordinates (in km, on a map grid) and monthly shipments are: M1 (20, 60) with 400 tonnes; M2 (50, 20) with 300 tonnes; M3 (80, 70) with 500 tonnes; M4 (30, 90) with 200 tonnes. Assuming transport cost is proportional to tonnes shipped and the same for all routes, find the location of the centre using the centre of gravity method. Explain the principle and its limitation.

Principle

The centre of gravity (load-distance) method places the facility at the weighted average of the market coordinates, using the shipment volumes as weights. This point keeps the total load-distance low and is a good first approximation to the minimum-transport-cost location:

xˉ=∑xiWi∑Wi,yˉ=∑yiWi∑Wi\bar{x} = \frac{\sum x_i W_i}{\sum W_i}, \qquad \bar{y} = \frac{\sum y_i W_i}{\sum W_i}

Calculation

Marketxix_iyiy_iWiW_ixiWix_iW_iyiWiy_iW_i
M120604008,00024,000
M2502030015,0006,000
M3807050040,00035,000
M430902006,00018,000
Total1,40069,00083,000
xˉ=690001400=49.29 km,yˉ=830001400=59.29 km\bar{x} = \frac{69000}{1400} = 49.29\ \text{km}, \qquad \bar{y} = \frac{83000}{1400} = 59.29\ \text{km}

The centre is at about (49.3, 59.3) on the grid, pulled towards M3 because it has the largest shipment.

Limitations

  • Straight-line distances are assumed; real roads, rivers and hills are ignored.
  • The result is only an approximation: the point that exactly minimises total load-distance can differ slightly from the centre of gravity.
  • The calculated point may be unavailable or very costly land, so a nearby suitable site is chosen.
  • Only transport cost is considered, not labour, land or taxes.
  • Cost per tonne-km is assumed equal for all routes and constant with volume.

Answer: Centre of gravity = (49.3 km, 59.3 km).

✓ Answer checked by Claude Opus 5.5 (high effort)

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