---
title: "Centre of gravity method for warehouse"
subject: "Industrial Engineering and Management"
chapter: 2
key: "semester_6/industrial_engineering_and_management:centre-of-gravity-location-method"
source: https://www.bctengineer.com/q/semester_6/industrial_engineering_and_management/centre-of-gravity-location-method
site: BCT Engineering
---

# Centre of gravity method for warehouse

Industrial Engineering and Management · Chapter 2: Design of Production Systems

- Asked in: year not given, 6 marks
- Video: [Operations Management: Location Selection – Centre of Gravity Method](https://www.youtube.com/watch?v=cipzCjYhs38) — The Business Doctor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/industrial_engineering_and_management/centre-of-gravity-location-method
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_engineering_and_management/chapter-2

## Question

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.

## 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).

## Answer

#### 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:

$$
\bar{x} = \frac{\sum x_i W_i}{\sum W_i}, \qquad \bar{y} = \frac{\sum y_i W_i}{\sum W_i}
$$

#### Calculation

| Market | $x_i$ | $y_i$ | $W_i$ | $x_iW_i$ | $y_iW_i$ |
|---|---|---|---|---|---|
| M1 | 20 | 60 | 400 | 8,000 | 24,000 |
| M2 | 50 | 20 | 300 | 15,000 | 6,000 |
| M3 | 80 | 70 | 500 | 40,000 | 35,000 |
| M4 | 30 | 90 | 200 | 6,000 | 18,000 |
| **Total** | | | **1,400** | **69,000** | **83,000** |

$$
\bar{x} = \frac{69000}{1400} = 49.29\ \text{km}, \qquad \bar{y} = \frac{83000}{1400} = 59.29\ \text{km}
$$

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).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What does the centre of gravity method find?
   A. The point farthest from all markets
   B. The point that minimises total weighted distance, using shipment volumes as weights
   C. The cheapest land
   D. The point with equal distance to all markets
   Answer: B. It treats shipment volumes as weights and finds the load-distance minimising point. (Key point: Principle)

2. How is the x coordinate of the centre calculated?
   A. Sum of x_i divided by number of markets
   B. Sum of W_i divided by sum of x_i
   C. Sum of (x_i W_i) divided by sum of W_i
   D. Largest x_i
   Answer: C. x-bar = sum(x_i W_i) / sum(W_i), a weighted average. (Key point: Formula)

3. What are the coordinates found for the centre?
   A. (49.3, 59.3)
   B. (50, 60)
   C. (69.0, 83.0)
   D. (49.3, 83.0)
   Answer: A. 69000/1400 = 49.29 and 83000/1400 = 59.29. (Key point: Result)

4. Which is a limitation of the method?
   A. It ignores shipment volume
   B. It cannot use coordinates
   C. It uses curved road distances only
   D. It assumes straight-line distances and considers only transport cost
   Answer: D. Real roads, rivers and hills are ignored and costs like labour, land and taxes are not included. (Key point: Limitation)

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