Intuitionistic Fuzzy Aggregation Techniques - Intuitionistic Fuzzy Aggregation and Clustering - page 59

Table 1.11 Intuitionistic
fuzzy decision matrix B

G 1 G 2 G 3

y 1 (0.3,0.4,0.3) (0.7,0.2,0.1) (0.5,0.3,0.2)

y 2 (0.5,0.2,0.3) (0.4,0.1,0.5) (0.7,0.1,0.2)

y 3 (0.4,0.5,0.1) (0.7,0.2,0.1) (0.4,0.4,0.2)

y 4 (0.2,0.6,0.2) (0.8,0.1,0.1) (0.8,0.2,0.0)

y 5 (0.9,0.1,0.0) (0.6,0.3,0.1) (0.2,0.5,0.3)

into consideration in the installation problem, the weight vector of the attributes
G j

T . Assume that the characteristics of the

( j = 1 , 2 , 3 ) is w = ( 0 . 3 , 0 . 5 , 0 . 2 )

alternatives y i ( i = 1 , 2 , 3 , 4 , 5 ) with respect to the attributes G j ( j = 1 , 2 , 3 ) are

represented by the IFVs b ij = (μ ij , v ij ,π ij ) , and all b ij ( i = 1 , 2 , 3 , 4 , 5 ; j = 1 , 2 , 3 )

are contained in the intuitionistic fuzzy decision matrix B = ( b ij ) 5 × 3 (see Table 1.11 )

(Xia et al. 2012a).

Step 1 Considering all the attributes G j ( j = 1 , 2 , 3 ) are the benefit attributes,

the performance values of the alternatives y i ( i = 1 , 2 , 3 , 4 , 5 ) do not need

normalization.
Step 2 Utilize the WIFGBM (here we take p = q = 1) to aggregate all the

performance values b ij (

= , , )

of the i th line, and get the overall performance
value b i corresponding to the alternative y i :

j 1 2 3

b 1 = ( 0 . 8084 , 0 . 1034 , 0 . 0882 ), b 2 = ( 0 . 8101 , 0 . 0438 , 0 . 1461 )

b 3 = ( 0 . 8112 , 0 . 1269 , 0 . 0619 ), b 4 = ( 0 . 8561 , 0 . 0915 , 0 . 0524 )

b 5 = ( 0 . 8381 , 0 . 1006 , 0 . 0613 )

Step 3 Calculate the scores of all the alternatives:

S ( b 1 ) = 0 . 7050 , S ( b 2 ) = 0 . 7664 , S ( b 3 ) = 0 . 6843

S ( b 4 ) = 0 . 7647 , S ( b 5 ) = 0 . 7375

Since

S ( b 2 )> S ( b 4 )> S ( b 5 )> S ( b 1 )> S ( b 3 )

then the ranking of the alternatives y i ( i = 1 , 2 , 3 , 4 , 5 ) is

y 2 y 4 y 5 y 1 y 3

IFGBM 2 , 2
w

If we take p = q = 2, then by b i = (μ i , v i ,π i ) = ( b i 1 , b i 2 , b i 3 ) ,

we get

b 1 = (

. , . , . ),

b 2 = (

. , . , . )

0 8039 0 1056 0 0905 0 7924 0 0465 0 1611

b 3 = ( 0 . 8100 , 0 . 1290 , 0 . 0610 ), b 4 = ( 0 . 8406 , 0 . 0971 , 0 . 0623 )

b 5 = ( 0 . 8092 , 0 . 1130 , 0 . 0778 )