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v α
1 + π α
1
1
μ α
1 + π α
(1.15)
=
=
1
α = α = (
where 0
ϑ(α)
1. We call
ϑ(α)
the
ϑ
value of the IFV. When
1
,
0
,
0
)
,
it gets the maximum
ϑ
value:
ϑ(α) =
1; and when
α = α = (
0
,
1
,
0
)
, it gets the
minimum
ϑ
value:
ϑ(α) =
0.
From Eq. ( 1.15 ), we can easily conclude that (Zhang and Xu 2012):
(1) If two IFVs have the same hesitancy degree, then the IFV which has the smaller
non-membership degree or the larger membership degree should be ranked first.
(2) If two IFVs have the same non-membership degree, then the IFV which has the
smaller hesitancy degree should be ranked first.
(3) If two IFVs have the same membership degree, then the IFVwhich has the larger
hesitancy degree should be ranked first.
When we use Eq. ( 1.15 ) to rank IFVs, the derived result is in accordance with the
first principle introduced at the beginning of Sect. 1.1.2 , i.e., the IFV which has the
larger membership degree and the smaller non-membership degree should be given
priority. Let's prove it below:
Proof Let
α 1 = α 1 ,
v
α 1 α 1 )
and
α 2 = α 2 ,
v
α 2 α 2 )
be two IFVs, and suppose
that
μ α 1 α 2 and v
α 1 <
v
α 2 , then by Eq. ( 1.15 ), we get
1
v α 1
1
v α 2
ϑ(α 1 ) =
v α 1 ,ϑ(α 2 ) =
2
μ α 1
2
μ α 2
v α 2
Let
μ α 1 μ α 2 = 1 and v α 2
v α 1 = 2 , then we have
1
v
1
v
1
v
α
α
α
1
1
1
ϑ(α 1 ) =
1 >
1 =
2
μ α 1
v
2
α 1 1 )
v
2
μ α 2
v
α
α
α
1
1
v
α 1
ϑ(α 1 )>
2
μ α 2
v
α 1
and similarly,
1
v
1
μ α 2
1
μ α 2
α 2
ϑ(α 2 ) =
α 2 =
1
α 2 <
1
2
μ α 2
v
2
μ α 2
v
2
μ α 2 (
v
α 2 2 )
1
μ α 2
1
v α 1
=
v α 1 =
1
2
μ α 2
2
μ α 2
v α 1
1
v α 1
ϑ(α 2 )<
2
μ α 2
v
α
1
Thus, we can obtain
1
v
α
1
ϑ(α 2 )<
1 <ϑ(α 1 ) ϑ(α 1 )>ϑ(α 2 )
2
μ α 2
v
α
 
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