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by which we obtain
y
6
y
4
y
5
y
2
y
3
y
1
In Step 2, if we let
p
=
q
=
r
=
2, then we have
r
1
=
(
0
.
2132
,
0
.
4666
,
0
.
3202
),
r
2
=
(
0
.
3316
,
0
.
3516
,
0
.
3168
)
r
3
=
(
0
.
2708
,
0
.
3771
,
0
.
3521
),
r
4
=
(
0
.
3140
,
0
.
2664
,
0
.
4196
)
r
5
=
(
0
.
3278
,
0
.
3259
,
0
.
3463
),
r
5
=
(
0
.
3547
,
0
.
2992
,
0
.
3461
)
Then we calculate the scores of all the alternatives:
S
(
r
1
)
=−
0
.
2534
,
S
(
r
2
)
=−
0
.
0200
,
S
(
r
3
)
=−
0
.
1064
S
(
r
4
)
=
0
.
0475
,
S
(
r
5
)
=
0
.
0019
,
S
(
r
6
)
=
0
.
0555
and thus,
y
6
y
4
y
5
y
2
y
3
y
1
In Step 2, if we use the GIFWBGM to aggregate the performances of the alterna-
tives (here, we let
p
=
q
=
r
=
1
)
, then
r
1
=
(
0
.
2401
,
0
.
4773
,
0
.
2826
),
r
2
=
(
0
.
3111
,
0
.
3542
,
0
.
3347
)
r
3
=
(
.
,
.
,
.
),
r
4
=
(
.
,
.
,
.
)
0
2546
0
3879
0
3575
0
2929
0
2681
0
4390
r
5
=
(
.
,
.
,
.
),
r
6
=
(
.
,
.
,
.
)
0
3268
0
3313
0
3419
0
3404
0
3000
0
3596
from which we calculate the scores of all the alternatives:
S
(
r
1
)
=−
0
.
2732
,
S
(
r
2
)
=−
0
.
0431
,
S
(
r
3
)
=−
0
.
1333
S
(
r
4
)
=
0
.
0248
,
S
(
r
5
)
=−
0
.
0045
,
S
(
r
6
)
=
0
.
0404
and thus,
y
6
y
4
y
5
y
2
y
3
y
1
If we let
p
=
q
=
r
=
2, then
r
1
=
(
0
.
2029
,
0
.
4861
,
0
.
3110
),
r
2
=
(
0
.
3052
,
0
.
3577
,
0
.
3371
)
r
3
=
(
0
.
2514
,
0
.
4004
,
0
.
3482
),
r
4
=
(
0
.
2875
,
0
.
2712
,
0
.
4413
)
r
5
=
(
0
.
3265
,
0
.
3395
,
0
.
3340
),
r
5
=
(
0
.
3357
,
0
.
3016
,
0
.
3627
)
and the scores of all the alternatives are:
S
(
r
1
)
=−
0
.
2832
,
S
(
r
2
)
=−
0
.
0525
,
S
(
r
3
)
=−
0
.
1490
S
(
r
4
)
=
0
.
0163
,
S
(
r
5
)
=−
0
.
0129
,
S
(
r
6
)
=
0
.
0341
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