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R
k
=27 into the first position of partial solution for Case A
Ta b l e 5 . 5 .
Insertion of node π
j
1
2
3
4
5
6
7
8
9
10
11
j
π
27
1
22
20
50
10
33
44
41
25
24
d
π
j
π
j
+1
d
27
,
1
d
1
,
22
d
22
,
20
d
20
,
50
d
50
,
10
d
10
,
33
d
33
,
44
d
44
,
41
d
41
,
25
d
25
,
27
d
24
,
27
174
8
7
15
21
17
12
17
20
21
14
22
F
(
)=
d
1
,
22
+
d
22
,
20
+
d
20
,
50
+
d
50
,
10
+
d
10
,
33
+
d
33
,
44
+
d
44
,
41
+
d
41
,
25
+
d
25
,
24
+
d
24
,
1
+
d
27
,
1
+
d
24
,
27
−
π
d
24
,
1
F
(
)=
d
1
,
22
+
d
22
,
20
+
d
20
,
50
+
d
50
,
10
+
d
10
,
33
+
d
33
,
44
+
d
44
,
41
+
d
41
,
25
+
d
25
,
24
+
d
27
,
1
+
d
24
,
27
π
k
in the first position of the partial tour
D
B
Insertion of node
π
π
a
Remove
=
d
1
π
m
π
b
Add
=
d
k
+
d
m
π
k
π
1
π
π
D
are fit-
ness function values of the tour after insertion and the partial tour,
respectively.
)=
F
π
D
+
Add
) and
F
π
b
F
(
π
−
Remove
,where
F
(
π
Example B:
Remove
=
d
1
π
m
π
Remove
=
d
10
π
1
π
Remove
=
d
24
,
1
Add
=
d
+
d
R
k
R
k
π
1
π
m
π
π
Add
=
d
1
+
d
10
π
1
π
1
π
π
Add
=
d
24
,
27
+
d
27
,
1
F
(
)=
F
π
D
+
Add
π
−
Remove
F
(
)=
d
1
,
22
+
d
22
,
20
+
d
20
,
50
+
d
50
,
10
+
d
10
,
33
+
d
33
,
44
+
d
44
,
41
+
d
41
,
25
+
d
25
,
24
+
d
24
,
1
+
d
24
,
27
+
d
27
,
1
−
π
d
24
,
1
F
(
)=
d
1
,
22
+
d
22
,
20
+
d
20
,
50
+
d
50
,
10
+
d
10
,
33
+
d
33
,
44
+
d
44
,
41
+
d
41
,
25
+
d
25
,
24
+
d
24
,
27
+
d
27
,
1
π
k
between an edge
π
v
u
,
C
Insertion of node
π
π
a
Remove
=
d
π
u
v
π
b
Add
=
d
k
+
d
u
k
π
v
π
π
π
D
are fit-
ness function values of the tour after insertion and the partial tour,
respectively.
)=
F
π
D
+
Add
) and
F
π
b
F
(
π
−
Remove
,where
F
(
π
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