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=
i
∈
K
c
i
(
a
i
+
δ
i
−
r
i
)
2
+
i
∈
K
c
i
(
a
i
−
r
i
)
2
≤
δ
i
≤
δ
≤
δ
i
−
−
/
≤
≤
/
/
≤
and bearing in mind 1
i
and
1
2
r
i
1
2 it follows that 1
2
r
i
−
δ
r
i
, therefore
i
)
≥
i
∈
K
c
i
(
a
i
+
δ
i
−
r
i
)
+
i
∈
K
c
i
(
a
i
−
r
i
)
2
2
f
(
s
−
r
≥
i
∈
K
c
i
(
a
i
+
δ
i
−
r
i
)
2
+
i
∈
K
c
i
(
a
i
−
r
i
)
2
s
−
=
(
)
.
f
r
Then, the minimum cannot be achieved in
s
. The same conclusion can be drawn for
d
<
0, which finishes the proof.
Remark 2.
Now, it is not difficult to develop a program for the cases where
|
d
|
=
0
,
1
,
2
,
or 3. Note that
|
d
|≤
[
n
/
2
]
, so this program will solve the problem whenever
n
takes values less than 3 (if
coefficients
c
i
differ) because it is easy for the values
b
i
to have different signs. The
program will, therefore, solve the problem in a wide variety of cases.
≤
7. Moreover, when
n
>
7, it is more probable that
|
d
|
Example 4.
Let us consider the optimization problem of minimizing function (
3.32
)
subject to (
3.33
) with the following data
n
=
10;
c
1
=
1
,
c
2
=
1
,
c
3
=
3
,
c
4
=
5
,
c
5
=
7
,
c
6
=
7
,
c
7
=
8
,
c
8
=
11
,
c
9
=
15
,
c
10
=
16
.
With these data and changing the value of
s
, the following values for
d
are obtained.
We will write the pairs
(
,
)
s
d
:
(
10
,
2
)
(
20
,
1
)
(
30
,
1
)(
40
,−
2
)
(
50
,
1
)
(
60
,
0
)
(
70
,
2
)(
80
,
0
)
(
90
,
0
)
(
100
,−
2
)(
110
,
0
)(
120
,
0
)
(
130
,−
1
)(
140
,
0
)(
150
,
1
)(
160
,
0
)
As can be seen
|
d
| <
3 is satisfied in all the cases. Observe also that the values 0, 1,
−
1, 2 and
−
2 appear with high irregularity.
Remark 3.
To apply the solution of the optimization problem to solve the
WIG
it
is necessary that
α
i
≤
m
for
i
=
1
,
2
,...,
n
. It is clear that this condition will be
satisfied if
m
≥
a
i
+
d
for
i
=
1
,
2
,...,
n
,where
a
i
and
d
are given by (
3.38
)and(
3.40
)
respectively.
Now we can solve some further examples.
Example 5.
Let us consider the situation described in Example
3
with the following
data:
s
=
80; a fine that must be paid for every positive result in cowshed C
i
given
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