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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    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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    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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    16

    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