A | B | C | D | E | |
A | 0 | - | - | - | - |
B | 20 | 0 | - | - | - |
C | 60 | 50 | 0 | - | - |
D | 100 | 90 | 40 | 0 | - |
E | 90 | 80 | 50 | 30 | 0 |
A & B are
closest (20 units): join them
into one cluster (AB) joining at 20, and recalculate the
average distance from C, D, and E to (AB).
[For example, the distance from C to (AB) = (60 +
50)/2 = 55, and the distance from D to (AB) = (100
+ 90)/2 = 95]. This gives:
(AB) | C | D | E | |
(AB) | 0 | - | - | - |
C | 55 | 0 | - | - |
D | 95 | 40 | 0 | - |
E | 85 | 50 | 30 | 0 |
D & E are
closest (30 units): join them into one cluster (DE) joining
at 30, and recalculate the average distances between (AB),
C, and (DE).
[For
example, the distance from (AB) to (DE) = (95 +
85)/2 = 90]. This gives:
(AB) | C | (DE) | |
(AB) | 0 | - | - |
C | 55 | 0 | - |
(DE) | 90 | 45 | 0 |
C & (DE) are
closest (45 units): join them into one cluster (CDE)
joining at 45, and recalculate the average distance between (CDE)
and (AB).
This gives:
(AB) | (CDE) | |
(AB) | 0 | - |
(CDE) | 72.5 | 0 |
The two clusters join at 72.5. This completes the analysis. [Commentary on calculations]
These results may be presented as a phenogram with nodes at 20, 30, 45, and 72.5 units. The phenogram indicates that A & B are similar to each other, as are D & E, and that C is more similar to D & E :