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Y
,
Y
,...,
Y
,
(
)
(
)
(
)
++
to which, as we see above , the numerical ordered series corresponds
j
1
j
2
j
+
v
y
,
y
,...,
y
,
(
)
(
)
(
)
++
Let us perform paired comparisons of objects of this series using Saati scale,
determine
j
1
j
2
j
+
v
a
matrix
of
paired
comparisons
and
its
eigenvector
(
)
ω
=
ω
,
...,
ω
corresponding to a maximum eigenvalue. Let us
m
−
1
m
−
1
m
−
1
v
[]
y
∈
0
X
consider that evaluations
,
i
=
j
+
1
j
+
v
belong to term
with
i
m
−
ω
membership degrees
,
, accordingly. To obtain membership function
i
=
1
v
m
−
1
()
y
=
a
x
+
b
μ
x
, with the left boundary in the form
, let us use a method
m
1
−
m
−
1
m
−
1
of least squares:
v
(
)
∑
=
2
F
=
a
y
+
b
−
ω
→
min
.
(
)
m
−
1
m
−
1
j
+
1
m
−
1
m
−
1
i
i
1
From system of normal equations
∂
F
⎧
m
−
1
=
0
⎪
⎪
⎨
∂
a
m
−
1
∂
F
⎪
⎪
m
−
1
=
0
∂
b
⎩
m
−
1
We obtain unknown coefficients
v
⎛
v
⎞
⎟
⎠
v
⎞
∑
∑
∑
v
y
ω
−
y
ω
⎜
⎝
⎜
⎝
⎟
⎠
(
)
(
)
j
+
i
m
−
1
i
j
+
i
m
−
1
i
i
=
1
i
=
1
i
=
1
b
=
;
m
−
1
2
v
⎛
v
⎞
∑
∑
2
v
y
−
⎜
⎝
y
⎟
⎠
(
)
(
)
j
+
i
j
+
i
i
=
1
i
=
1
v
v
v
⎛
⎞
⎟
⎠
⎞
∑
∑
∑
v
y
ω
−
⎜
⎝
y
⎜
⎝
ω
⎟
⎠
(
)
(
)
j
+
i
m
−
1
i
j
+
i
m
−
1
i
v
v
1
1
∑
∑
i
=
1
i
=
1
i
=
1
a
=
ω
−
y
.
(
)
m
−
1
m
−
1
i
j
+
i
2
v
v
v
⎛
v
⎞
i
=
1
i
=
1
∑
∑
2
v
y
−
⎜
⎝
y
⎟
⎠
(
)
(
)
j
+
i
j
+
i
i
=
1
i
=
1
If the condition
⎛
b
⎞
b
y
⎜
⎜
⎝
−
m
⎟
⎟
⎠
=
−
a
m
+
b
>
1
m
−
1
m
−
1
a
a
is satisfied, we obtain membership function of
T
-number:
m
m
