Método de WARD Individuo X1 X2 A 10 5 B 20 20 C 30 10 D 30 15 E 5 10 Nivel 1 Combinaciones posibles: Partición (A, B), C, D, E (A, C), B, D, E (A, D), B, C, E (A, E), B, C, D (B, C), A, D, E (B, D), A, C, E (B, E), A, C, D (C, D), A, B, E (C, E), A, B, D (D, E), A, B, C 5 2 = 10 Ek EAB = 162.5 CAB = (15, 12.5) EC = ED = EE = 0 EAC = 212.5 CAC = (20, 7.5) EB = ED = EE = 0 EAD = 250 CAD = (20, 10) EB = EC = EE = 0 EAE = 25 CAE = (7.5, 7.5) EB = EC = ED = 0 EBC = 100 CBC = (25, 15) EA = ED = EE = 0 EBD = 62.5 CBD = (25, 17.5) EA = EC = EE = 0 EBE = 162.5 CBE = (12.5, 15) EA = EC = ED = 0 ECD = 12.5 CCD = (30, 12.5) EA = EB = EE = 0 ECE = 312.5 CCE = (17.5; 10) EA = EB = ED = 0 EDE = 325 CDE = (17.5; 12.5) EA = EB = EC = 0 Centroides E ΔE 162.5 162.5 212.5 212.5 250 250 25 25 100 100 62.5 62.5 162.5 162.5 12.5 12.5 312.5 312.5 325 325 Cluster (C, D), A, B, E EAB = (10 − 15)2 + (5 − 12.5)2 + (20 − 15)2 + (20 − 12.5)2 = 162.5 EAB = 102 + 52 + 202 + 202 − 2(152 + 12.52 ) = 162.5 Nivel 2 Combinaciones posibles: Partición ((C, D), A), B, E ((C, D), B), A, E ((C, D), E), A, B (A, B), (C, D), E (A, E), (C, D), B (B, E), (C, D), A 4 2 =6 Ek ECDA = 316.66 CCDA = (23.33, 10) EB = EE = 0 ECDB = 116.66 CCDB = (26.66, 15) EA = EE = 0 ECDE = 433.33 CCDE = (21.66, 11.66) EA = EB = 0 EAB = 162.5 CAB = (15, 12.5) ECD = 12.5 CCD = (30, 12.5) EE = 0 EAE = 25 CAE = (7.5, 7.5) ECD = 12.5 CCD = (30, 12.5) EB = 0 EBE = 162.5 CBE = (12.5, 15) ECD = 12.5 CCD = (30, 12.5) EA = 0 Centroides E ΔE 316.66 304.16 116.66 104.16 433.33 420.83 175 162.5 37.5 25 175 162.5 Cluster (A, E), (C, D), B Cálculos para la partición ((C, D), A), B, E: m (C,D),A E(C,D),A 2(mCD ) + mA 2(30, 12, 5) + (10, 5) = = = (23.33, 10) 3 3 3 2 (C,D),A (C,D),A 2 = (xil − ml ) = i=1 l=1 = (30 − 23.33)2 + (10 − 10)2 + (30 − 23.33)2 + (15 − 10)2 + (10 − 23.33)2 + (5 − 10)2 = 316.66 3 2 2 (C,D),A 2 (C,D),A 2 E(C,D),A = (xil ) −3 (ml ) = i=1 l=1 l=1 = 302 + 102 + 302 + 152 + 102 + 52 − 3(23.332 + 102 ) = 316.66 ΔE(C,D),A = E(C,D),A − ECD − EA = 316.66 − 12.5 − 0 = 304.16 2 nCD nA CD 1 · 2 2 ΔE(C,D),A = (ml − mA (30 − 10)2 + (12.5 − 5)2 = 304.16 l ) = nCD + nA l=1 3 1 ΔE(C,D),A = (nC + nA )ΔECA + (nD + nA )ΔEDA − nA ΔECD = nCD + nA 1 = (2 · 212.5 + 2 · 250 − 12.5) = 304.16 2+1 Nivel 3 Combinaciones posibles: Partición 3 2 =3 Centroides ((A, E), (C, D)), B CAECD = (18.75, 10) ((A, E), B), (C, D) ((C, D), B), (A, E) CAEB = (11.66, 11.66) CCD = (30, 12.5) CCDB = (26.66, 15) CAE = (7.5, 7.5) Ek EAECD = 568.75 EB = 0 EAEB = 233.33 ECD = 12.5 ECDB = 116.66 EAE = 25 E 568.75 531.25 245.8 Nivel 4 Partición Centroide E ΔE (((C, D), B), (A, E)) CCDBAE = (19, 12) 650 508.34 A E C 104.16 508.34 Ward D B 208.3 141.66 104.16 Cluster ((C, D), B), (A, E) 12.5 25 ΔE