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T
λ
,
= (
w 1 ,
w 2 ,...,
w n )
,
∈[
,
] ,
,
=
,
,...,
where
1
w
w i
0
1
x i
X
i
1
2
n , and
i = 1 w i =
1.
Eq. ( 2.1 ) can weight not only the deviation of each IFV, but also the deviations of
the correspondingmembership degree, the non-membership degree and the hesitancy
(indeterminancy) degree. It is more general than the similarity measure:
ϑ (
A 1 ,
A 2 )
n
1
2 n
2
2
2
=
1
1 ((μ A 1 (
x j ) μ A 2 (
x j ))
+ (
v A 1 (
x j )
v A 2 (
x j ))
+ A 1 (
x j ) π A 2 (
x j ))
j
=
(2.2)
and thus, Eq. ( 2.1 ) is of high flexibility. If we take
ϑ(
A 1 ,
A 2 )
as the function of w ,
then it is a bounded function. Let
n
x i ) | λ + β 2 |
x i ) | λ + β 3 | π A 1 (
d
(
w
) =
w i | μ A 1 (
x i ) μ A 2 (
v A 1 (
x i )
v A 2 (
x i )
i
=
1
x i ) | λ ),
π A 2 (
λ
1
(2.3)
then we need to solve the maximum and minimum problem of Eq. ( 2.1 ), which can
be transformed to solve the maximum and minimum problem of d
(
w
)
. Since
n
x i ) | λ + β 2 |
x i ) | λ
d
(
w
) =
w i 1 | μ A 1 (
x i ) μ A 2 (
v A 1 (
x i )
v A 2 (
i
=
1
x i ) | λ )
+ β 3 | π A 1 (
x i ) π A 2 (
x i ) | λ + β 2 |
x i ) | λ
ma i { β 1 | μ A 1 (
x i ) μ A 2 (
v A 1 (
x i )
v A 2 (
x i ) | λ }
+ β 3 | π A 1 (
x i ) π A 2 (
1
(2.4)
There must exist a positive integer k such that
x i ) | λ + β 2 |
x i ) | λ + β 3 | π A 1 (
x i ) | λ }
ma i { β 1 | μ A 1 (
x i ) μ A 2 (
v A 1 (
x i )
v A 2 (
x i ) π A 2 (
x k ) | λ + β 2 |
x k ) | λ + β 3 | π A 1 (
= β 1 | μ A 1 (
x k ) μ A 2 (
v A 1 (
x k )
v A 2 (
x k )
x k ) | λ }
π A 2 (
1
(2.5)
Hence, when w k =
1 and w i =
0, i
=
k , the equality holds. Also since
n
x i ) | λ + β 2 |
x i ) | λ
d
(
w
) =
w i 1 | μ A 1 (
x i ) μ A 2 (
v A 1 (
x i )
v A 2 (
i
=
1
x i ) | λ )
+ β 3 | π A 1 (
x i ) π A 2 (
 
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