11, 12 Optimization Based Methods
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Transcript of 11, 12 Optimization Based Methods
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Optimization based methods
.
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1
n
ij k
k
s b
1 1
m m
ij ij
i j
s X
1
1
m
ij
j
X i
1
m
jjj
X p
ij jjX X
Let the similarity between two machines i and j be defined as
where bk=1 if a
ik= a
jkand 0 otherwise.
Let Xjj
= 1 if point (or machine) j is a median.Let Xij= 1 if point (or machine) i is attached to median j
The p-median formulation is as follows:
Maximize
Subject to
Xij =0,1
1
n
ij ik jk
k
d a a
dij= n
sij.
Minimize
1 1
m m
ij iji j
d X
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1 2 3 4 5 6 7 81 1 1 1 12 1 1 1 13 1 1 14 1 1 15 1 1 1 1 16 1 1 1 1 1 1
1 2 3 4 5 61
0
6
1
1
7
6
2 6 0 5 5 3 43 1 5 0 2 8 54 1 5 2 0 6 75 7 3 8 6 0 36 6 4 5 7 3 0
12 1 0 0 1 0 0 1 0 1 0 1 1 0 1 0 1 6d
Also d12
= 8s12
= 6.
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The optimum solution is
X11
= X31
= X41
= X22
= X52
= X62
= 1; Y11
= Y41
= Y51
= Y61
= Y22
= Y32
= Y72
= Y82
= 1;
Z262
= Z261
= Z551
= Z552
= Z611
= Z612
= Z641
= Z642
= 1.
This means that there are four intercell moves M2-P6, M5-P5, M6-P1 and M6-P4.
This is the same solution obtained using the p median model for p = 2.
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Assignment model
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1 2 3 4 5 6 7 81 1 1 1 12 1 1 1 13 1 1 14 1 1 15 1 1 1 1 16 1 1 1 1 1 1
1 2 3 4 5 61
0
6
1
1
7
6
2 6 0 5 5 3 43 1 5 0 2 8 54 1 5 2 0 6 75 7 3 8 6 0 36 6 4 5 7 3 0
12 1 0 0 1 0 0 1 0 1 0 1 1 0 1 0 1 6d
Also d12
= 8s12
= 6.
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The solution to the assignment problem is
X13
= X26
= X34
= X41
= X52
= X65
= 1 with Z = 14.
From this solution we observe that machine 1 is assigned to machine 3,
which is assigned to machine 4. Machine 4 is assigned to machine 1.
We observe a subtour 1-3-4-1 out of the assignments.
Therefore the three machines {1, 3, 4} form a group.
Similarly the machine cell {2, 6, 5} is formed.
Applying the maximum density rule the corresponding part families
are {1, 4, 5, 6} and {2, 3, 7, 8}. There are four intercell moves.
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10
1111111111222222222233333333334444
1234567890123456789012345678901234567890123
1 1 1
2 1 1 1 1 11 1 1
3 1 1 111
4 1 1 1 1 1 1 1
5 1 11 111 1 1 1 1 1 1 1
6 11 111 111 1 1 1 111 1 11 11
7 1 1 1
8 111 11 11 111 11 11 1 11 1 1
9 1 1 1 1 1 1 11 1 1
10 1 11 11 1 1
11 1 1 1 1 1 1
12 1 1 1 1 1
13 1 1
14 1 1 1 1
15 1 1 1 1 1 1 1
16 1 1 1 1 11 1
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12
An Assignment model Solution
X(1,13) = X(13,1) = 1
X(2,9) = X(9,16) = X(16,2) = 1
X(3,14) = X(14,3) = 1
X(4,5) = X(5,15) = x(15,4) = 1
X(6,8) = X(8,6) = 1
X(7,10) = X(10,7) = 1
X(11,12) = X(12,11) = 1
Seven machine groups
{1,13}, {2,9,16}, {3,14}, {4,5,15}, {6,8}, {7,10} and {11,12}.
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13
Part groups
{1,12}, {2,42,37}, {3,24}, {4,10,18}, {5,14,19,23,8,15,16,29},
{6,17}, {7,34}, {9,20}, {11,22}, {13,25}, {21,41}, {26,31,39},
{27,30}, {28,38}, {32,40}, {33,43}, {35,36}.
Initial Solution
Group Machines Components
123
4567
1,132,9,163,14
4,5,156,87,1011,12
32,42,37,4,10,18,28,38,32,406,17,7,34,35,36
5,14,19,15,16,29,9,21,41,331,8,11,12,20,23,31,39,3,4313,25,2624,22,27,30
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Group Components Machines1 2,42,37,4,10,18,28,38,32,40 1,2,9,16
2 6,17,7,34,35,36 3,14
3 5,14,19,15,16,29,9,21,41,33 4,5,15
4 1,8,11,12,20,23,31,39,3,43 6,8,10,75 13,25,26 7
6 24,22,27,30 11,12,13
Intermediate Solution
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11233344 1333 1111222344 112233 122223
2408827802677456589456913931312356193102470
1 1 1
2 1 1 111111
9 111111111
16 1 11 111 1
3 11111
14 1 1 1 1
4 1 11 11111
5 1111111111111
6 1 11 111111 1 11 1 1 1 111 1
8 1 1 11 111 1 111 111 1 111 11
15 1 1 11 111
7 111
10 111111
11 1 1 1 111
12 1 1111
13 1 1
Final block diagonal structure
Intercell move after duplicating 6 and 8
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Measures of performance
Minimize E
Minimize E + B
Grouping efficiency Grouping Efficacy
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1 2 3 4 5 6 7 86 1 1 1 1 1 11 1 1 1 13 1 1 14 1 1 15 1 1 1 1 12 1 1 1 1
2 7 8 3 1 4 5 66 1 1 1 1 1 15 1 1 1 1 12 1 1 1 11 1 1 1 13 1 1 14 1 1 1
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Search Algorithms
Genetic Algorithms
Simulated Annealing