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μ
α
>
Case 2
v
α
1
−
v
1
−
v
1
2
α
α
⇒
α
)
>
α
)
=
(
1
−
μ
α
)
+
(
1
−
v
(
1
−
v
α
)
+
(
1
−
v
1
2
⇒
ϑ(α) >
Case 3
μ
α
<
v
α
1
−
v
α
1
−
v
α
1
2
⇒
v
α
)
<
v
α
)
=
(
1
−
μ
α
)
+
(
1
−
(
1
−
v
α
)
+
(
1
−
1
2
⇒
ϑ(α) <
μ
α
and non-
membership degree
v
α
change in specific ranges, whichwe can see from the contours.
In the following, we shall demonstrate it:
Let
In fact, for each value derived by Eq. (
1.15
), its membership degree
α
=
(μ
α
,
v
α
,π
α
)
be an IFV, then
(1) If
ϑ(α)
≤
0
.
5, then
(a) When
π
α
=
0,
μ
α
and
v
α
get the maximums, i.e.,
π
α
=
0
⇒
μ
α
+
v
α
=
1
and
1
−
v
α
ϑ(α)
=
=
1
−
v
α
⇒
v
α
=
1
−
ϑ(α)
⇒
μ
α
=
ϑ(α)
1
+
π
α
(b) When
μ
α
=
0,
v
gets the minimum, i.e.,
α
1
−
v
1
−
v
1
−
v
1
−
2
ϑ(α)
α
α
α
ϑ(α)
=
=
=
⇒
v
α
=
1
+
π
α
1
+
1
−
μ
α
−
v
2
−
v
1
−
ϑ(α)
α
α
then the ranges of
μ
α
and
v
α
are as follows, respectively:
1
−
2
ϑ(α)
0
≤
μ
α
≤
ϑ(α),
≤
v
α
≤
1
−
ϑ(α)
1
−
ϑ(α)
(2)If0
.
5
<ϑ(α)
≤
1, then
(a) When
π
α
=
0
,μ
α
and
v
α
get the maximums, i.e.,
μ
α
=
ϑ(α)
and
v
α
=
;
(b) When
v
α
=
1
−
ϑ(α)
0
,μ
α
gets the minimum, i.e.,
1
−
v
α
1
−
v
α
1
2
ϑ(α)
−
1
ϑ(α)
=
=
=
⇒
μ
α
=
1
+
π
α
1
+
1
−
μ
α
−
v
α
2
−
μ
α
ϑ(α)
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