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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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