Information Technology Reference
In-Depth Information
1
2
5
7
1
/
2
1
3
4
A
=
.
4
1
/
5
1
/
3
1
2
1
/
7
1
/
4
1
/
2
1
Table 2.1 Results of stud ents' knowledge appraising
x
y
No.
i
i
1
Excellent (“A”)
0.8
2
Excellent (“A”)
0.9
3
“good” (“B”)
0.6
4
“good” (“B”)
0.7
5
"satisfactory” (“C”)
0.3
6
"satisfactory” (“C”)
0.1
7
"satisfactory” (“C”)
0.2
8
"unsatisfactory” (“E”)
0
We determine eigenvalues of this matrix:
λ
=
4
022
;
λ
=
0
010
;
2
1
λ
=
0
006
+
0
0294
i
;
λ
=
0
006
0
0294
i
. Let us select a maximum eigenvalue
3
4
(
)
ω
=
0
859
;
0
466
;
0
180
;
0
109
and the corresponding eigenvector
. Let us build
4
a line
= xy by four points (0.9; 0.859), (0.8; 0.466), (0.7; 0.180),
(0.6; 0.109) applying a method of least squares. Let us check a value
()
2
537
1
499
()
y
1
=
2
537
1
499
=
1
038
. Apparently, the condition
y
1
1
is satisfied, that
[] ()
x
0
:
μ
x
=
1
ensures
existence
at
least
one
point
.
Thus,
4
() (
)
μ
x
0
985
;
1
0
385
;
0
.
By means of membership function
4
()
μ
x
, it is possible to define explicitly the
4
()
μ
x
0
59
x
<
0
985
right boundary of membership function
, i.e. with
3
()
()
μ
x
=
1
μ
x
=
2
499
2
537
x
.
3
4
()
μ is not possible as there are no students
who have got a point “B”, and whose points are
x
Obviously, the further building of
3
y
<
0
59
. In this case we come
i
()
μ of term X . Let us consider outcomes
of the pupils who have got points "C" and “E”. Let us arrange them as per
decreasing points y . We obtain the following conditional ordinal series
No.5, No. 7, No. 6, No. 8.
to which the numerical ordinal number corresponds
0.3; 0.2; 0.1; 0.
The teacher makes paired comparisons of knowledge of students from the
conditional ordinal series using Saati scale. The following matrix of paired
comparisons is obtained:
x
to building of membership function
2
Search WWH ::




Custom Search