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μ
α
then the ranges of
and
v
α
are as follows, respectively:
ϑ(α)
−
2
1
≤
μ
α
≤
ϑ(α),
0
≤
v
α
≤
1
−
ϑ(α)
ϑ(α)
Based on the analysis above, in what follows, we introduce a new method for
ranking IFVs (Zhang and Xu 2012):
Let
α
i
=
(μ
α
i
,
v
α
i
,π
α
i
)(
i
=
1
,
2
,...,
n
)
be a collection of IFVs, then we rank
these IFVs according to the following steps:
Step 1
Calculate the
ϑ
values of the IFVs
α
i
(
i
=
1
,
2
,...,
n
)
using Eq. (
1.15
).
Step 2
Rank the IFVs
α
i
(
i
=
1
,
2
,...,
n
)
according to the order of the
ϑ
values,
and the IFV with the larger
ϑ
value should be ranked first. If there exist some IFVs
with the same
value, then go to Step 3.
Step 3
Calculate the accuracy degrees of these IFVs using Eq. (
1.4
), and then rank
the IFVs according to the following principles:
ϑ
(1) If
ϑ(α
i
)>ϑ(α
j
)
, then
α
i
>α
j
.
(2) If
ϑ(α
i
)
=
ϑ(α
j
)
, then
(
a
)
If
H
(α
i
)>
H
(α
j
),
then
α
i
>α
j
;
(
b
)
If
H
(α
i
)<
H
(α
j
),
then
α
i
<α
j
;
(
c
)
If
H
(α
i
)
=
H
(α
j
),
then
α
i
=
α
j
,
(1.16)
which are in accordance with the basic principles introduced at the beginning of
Sect.
1.1.2
.
In the following, we give an example to illustrate the method above and compare
it with all the existing ones:
Example 1.4
(Zhang and Xu 2012) Let
α
1
=
(
0
.
6
,
0
.
1
,
0
.
3
), α
2
=
(
0
.
6
,
0
.
15
,
0
.
25
),
α
3
=
(
be five IFVs. Here, we
rank them using all the methods discussed previously. The derived results are listed
in Table
1.1
(Zhang and Xu 2012).
0
.
5
,
0
,
0
.
5
), α
4
=
(
0
.
2
,
0
.
3
,
0
.
5
)
, and
α
5
=
(
0
,
0
.
8
,
0
.
2
)
According to the data in Table
1.1
, we can get the following ranking results:
(i) By the formulas (
1.3
) and (
1.4
), we get
α
1
>α
3
>α
2
>α
4
>α
5
.
Table 1.1
The results derived by the existing methods
α
i
S
(α
i
)
d
1
(α
i
,α
∗
)
d
2
(α
i
,α
∗
)
S
(α
i
)
H
(α
i
)
L
(α
i
)
J
(α
i
)ϑ
)
i
(0.6,0.1,0.3)
0.5
0.7
0.9
0.25
0.4
0.26
6/7
9/13
(0.6,0.15,0.25)
0.45
0.75
0.85
0.275
0.4
0.25
0.8
17/25
(0.5,0,0.5)
0.5
0.5
1
0.25
0.5
0.375
1
2/3
−
.
(0.2,0.3,0.5)
0
1
0.5
0.7
0.55
0.8
0.6
0.4
7/15
(0,0.8,0.2)
−
0
.
8
0.8
0.2
0.9
1
0.6
0
1/6
μ
α
μ
α
+
v
α
μ
α
+
Note 1
. In the process of calculating
J
(α
)
, we choose
σ
=
and
θ
=
i
v
α
v
α
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