- Year not given · 3+7 marks
Cluster analysis applications; K-means
Short answer
Cluster analysis is used in marketing segmentation, biology, image processing, fraud detection, etc. K-means iteratively assigns points to the nearest centroid (Euclidean distance) and recomputes means until stable. Example with points (2,2)… yields Cluster 1 = {1,2,5} centroid (2.67,2.33); Cluster 2 = {3,4} centroid (8.5,7.5).
Explain the application areas of Cluster Analysis. Describe K-means clustering algorithm with example.
Application areas of cluster analysis
- Marketing: customer segmentation by buying behaviour for targeted campaigns.
- Biology/medicine: grouping genes and proteins with similar function, taxonomy of plants and animals, disease subtypes.
- Image processing: image segmentation, colour quantisation, pattern recognition.
- Information retrieval/web: grouping documents, news and search results; recommender systems.
- Urban planning and earth science: grouping houses by type/value, earthquake zones, land-use mapping.
- Fraud and intrusion detection: objects that fit no cluster are outliers.
- Insurance, finance, social networks (community detection) and as a data compression and preprocessing step.
K-means algorithm
1. Choose k and select k initial centroids (random objects).
2. Repeat
a. Assign each object to the nearest centroid
(Euclidean distance) -> k clusters.
b. Recompute each centroid as the mean of its cluster.
3. Until no object changes cluster (centroids stable).
The objective is to minimise SSE = Σᵢ Σ_{x∈Cᵢ} ‖x − cᵢ‖².
Example (k = 2)
Points: 1 (2, 2), 2 (3, 2), 3 (8, 7), 4 (9, 8), 5 (3, 3). Initial centroids: C1 = point 1 (2, 2), C2 = point 3 (8, 7).
Iteration 1
| Point | d to C1 | d to C2 | Cluster |
|---|---|---|---|
| 1 (2, 2) | 0.00 | 7.81 | C1 |
| 2 (3, 2) | 1.00 | 7.07 | C1 |
| 3 (8, 7) | 7.81 | 0.00 | C2 |
| 4 (9, 8) | 9.22 | 1.41 | C2 |
| 5 (3, 3) | 1.41 | 6.40 | C1 |
New centroids: C1 = ((2 + 3 + 3)/3, (2 + 2 + 3)/3) = (2.67, 2.33); C2 = ((8 + 9)/2, (7 + 8)/2) = (8.5, 7.5).
Iteration 2: every point is nearest to the same centroid as before (e.g. point 5: 0.75 vs 7.1), so there is no change and the algorithm stops.
Result: Cluster 1 = {1, 2, 5} with centroid (2.67, 2.33) and Cluster 2 = {3, 4} with centroid (8.5, 7.5).
✓ Answer checked by Claude Opus 5.5 (high effort)
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