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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
))
+
(π
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
(β
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
}
,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
}
,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
(β
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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