Information Technology Reference
In-Depth Information
•
Since it is reasonable to accept that 'John is old'
and
'John is old' does coincide
with 'John is old', and 'John is young'
or
'John is young' does coincide with
'John is young' , we can decide to take either
T
=
min
,
S
=
max, or
T
and
S
as
ordinal-sums.
•
Provided idempotency is avoided i.e., 'John is young'
and
'John is young' does
coincide with 'John is very young', instead of
T
prod,
that is more interactively than min. In this case, because it does not seem that
duality should be avoided, we could tak
e
S
=
min, we can take
T
=
Prod
∗
, an
d
then
=
√
6
6
35
√
6
36
Pro
d
∗
(
1
1
μ
v
er y youn
g
or mol old
(
)
=
36
,
)
=
6
(
+
)
=
.
,
45
1
0
5636
√
6
6
that is greater than the value
=
0
.
408 obtained with
=
max.
x
Example 2.2.56
In
X
=[
0
,
10
]
the predicate
P
=
big is represented by
μ(
x
)
=
10
.
In which points in
[
0
,
10
]
is the degree of 'big' less than that of 'not big'?
)
μ
(
Solution
.Given
μ
, the problem is to find for which
x
∈
X
it is
μ(
x
x
)
=
))
=
˕
−
1
)
˕
−
1
1
N
˕
(μ(
x
(
1
−
˕(μ(
x
)))
, that is,
μ(
x
(
2
)
. Then,
x
10
x
•
If
N
=
N
0
,
1
−
10
,or
x
5.
(
√
2
10
1
−
x
10
,or
x
2
•
If
N
=
N
1
,
+
20
x
−
100
0, that means
x
10
−
1
)
=
10
1
+
4
.
142
√
75
x
10
1
−
x
10
,or
x
2
•
If
N
=
N
2
,
+
10
x
−
50
0, that means
x
−
5
=
3
.
66
2
x
10
1
+
Hence,
•
If
N
=
N
0
,thesetis
[
0
,
5
]
, and the threshold (of selfcontradiction) of
big
is 5.
•
If
N
=
N
1
,thesetis
[
0
,
4
.
142
]
, and the threshold is 4
.
142
•
=
[
,
.
]
.
66
Notice that changing
big
by
not big
, the thresholds do remain but the sets are, respec-
tively,
If
N
N
2
,thesetis
0
3
66
, and the threshold is 3
[
5
,
10
]
,
[
4
.
142
,
10
]
, and
[
3
.
66
,
10
]
.
2.3 On Aggregating Imprecise Information
The kind of problems this section will deal with are like the following. An exam is
corrected by three referees
R
1
,
R
2
,
R
3
, each one with a different weight of strongness
3, such that
i
=
1
W
W
(
R
i
)
∈[
0
,
1
]
,1
i
(
R
i
)
=
1. Each referee assigns a
numerical qualification
p
i
to the exam delivered by a given student. How
these qualification can be “aggregated” to obtain final qualification for the student's
exam? A recognized usual way of doing it is by the
weighted mean
:
∈[
0
,
10
]
1
10
Q
p
1
10
·
p
2
10
·
p
3
10
·
=
W
(
R
1
)
+
W
(
R
2
)
+
W
(
R
3
),
p
i
10
with
∈[
0
,
10
]
. For example, if
W
=
(
0
.
5
,
0
.
3
,
0
.
2
)
and
P
=
(
7
,
6
,
5
)
, it follows
Search WWH ::
Custom Search