Information Technology Reference
In-Depth Information
u
j
,
i
,
G
+1
=
x
(
lo
)
if
u
j
,
i
,
G
+1
<
x
(
lo
)
(3.7)
x
(
hi
)
if
u
j
,
i
,
G
+1
>
x
(
hi
)
Each value, which is replicated, is tagged for its value and index. Only those values,
which are deemed replicated, are repaired, and the rest of the values are not manip-
ulated. A second sequence is now calculated for values, which are not present in the
solution. It stands to reason that if there are replicated values, then some feasible values
are missing. The pseudocode if given in Fig 3.3.
Three unique repairment strategies were developed to repair the replicated values:
front mutation
,
back mutation
and
random mutation
, named after the indexing used for
each particular one.
3.4.1.1 Front Mutation
Front mutation indexes the repairment from the front of the replicated array with values
randomly selected from the missing value array as shown in Fig 3.4.
Array
Solution
,
ViolateVal
,
MissingVal
;
for (int
i
= 0;
i
<
sizeo f ViolateVal
;
i
++)
Solution [
ViolateVal
[
i
]] = Random [
MissingVal
] ;
}
Fig. 3.4.
Pseudocode for front mutation
Illustration
:
In order to understand
front mutation
, assume an in
−
feasible solution of dimension
D
= 10:
x
=
.
The first step is to isolate all repetitive values in the solution. These are highlighted
in the following array:
x
=
{
3
,
4
,
2
,
1
,
3
,
5
,
6
,
7
,
10
,
5
}
{
3
,
4
,
2
,
1
,
3
,
5
,
6
,
7
,
10
,
5
}
. As shown, the values 3 and 5 are
repeated in the solution.
All
first
occurring values are now set as default:
x
=
{
3
,
4
,
2
,
1
,
3
,
5
,
6
,
7
,
10
,
5
}
.
{
3
,
4
,
2
,
1
,
5
,
5
,
So now only two positions are replicated, index 5 and 10 as given:
x
=
6
,
7
,
10
,
10
}
.
An array of missing values is now generated as
MV
=
{
8
,
9
}
, since values 8 and 9
are missing from the solution.
An insertion array is now randomly generated, which specifies the position of the
insertion of each value:
IA
=
. Since only two values were missing so only
two random numbers are generated. In this respect, the first value 2 in
IA
, outlines
that the value pointed by index 1 in
MV
which is 8 is to be placed in the sec-
ond indexed in-feasible solution and likewise for the other missing value given as:
x
=
{
2
,
1
}
{
3
,
4
,
2
,
1
,
1
,
5
,
6
,
7
,
10
,
2
}
.
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