| [1] | 1 | /* BPP, Bin Packing Problem */ |
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| 2 | |
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| 3 | /* Written in GNU MathProg by Andrew Makhorin <mao@gnu.org> */ |
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| 4 | |
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| 5 | /* Given a set of items I = {1,...,m} with weight w[i] > 0, the Bin |
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| 6 | Packing Problem (BPP) is to pack the items into bins of capacity c |
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| 7 | in such a way that the number of bins used is minimal. */ |
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| 8 | |
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| 9 | param m, integer, > 0; |
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| 10 | /* number of items */ |
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| 11 | |
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| 12 | set I := 1..m; |
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| 13 | /* set of items */ |
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| 14 | |
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| 15 | param w{i in 1..m}, > 0; |
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| 16 | /* w[i] is weight of item i */ |
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| 17 | |
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| 18 | param c, > 0; |
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| 19 | /* bin capacity */ |
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| 20 | |
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| 21 | /* We need to estimate an upper bound of the number of bins sufficient |
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| 22 | to contain all items. The number of items m can be used, however, it |
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| 23 | is not a good idea. To obtain a more suitable estimation an easy |
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| 24 | heuristic is used: we put items into a bin while it is possible, and |
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| 25 | if the bin is full, we use another bin. The number of bins used in |
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| 26 | this way gives us a more appropriate estimation. */ |
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| 27 | |
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| 28 | param z{i in I, j in 1..m} := |
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| 29 | /* z[i,j] = 1 if item i is in bin j, otherwise z[i,j] = 0 */ |
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| 30 | |
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| 31 | if i = 1 and j = 1 then 1 |
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| 32 | /* put item 1 into bin 1 */ |
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| 33 | |
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| 34 | else if exists{jj in 1..j-1} z[i,jj] then 0 |
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| 35 | /* if item i is already in some bin, do not put it into bin j */ |
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| 36 | |
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| 37 | else if sum{ii in 1..i-1} w[ii] * z[ii,j] + w[i] > c then 0 |
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| 38 | /* if item i does not fit into bin j, do not put it into bin j */ |
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| 39 | |
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| 40 | else 1; |
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| 41 | /* otherwise put item i into bin j */ |
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| 42 | |
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| 43 | check{i in I}: sum{j in 1..m} z[i,j] = 1; |
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| 44 | /* each item must be exactly in one bin */ |
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| 45 | |
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| 46 | check{j in 1..m}: sum{i in I} w[i] * z[i,j] <= c; |
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| 47 | /* no bin must be overflowed */ |
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| 48 | |
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| 49 | param n := sum{j in 1..m} if exists{i in I} z[i,j] then 1; |
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| 50 | /* determine the number of bins used by the heuristic; obviously it is |
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| 51 | an upper bound of the optimal solution */ |
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| 52 | |
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| 53 | display n; |
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| 54 | |
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| 55 | set J := 1..n; |
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| 56 | /* set of bins */ |
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| 57 | |
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| 58 | var x{i in I, j in J}, binary; |
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| 59 | /* x[i,j] = 1 means item i is in bin j */ |
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| 60 | |
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| 61 | var used{j in J}, binary; |
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| 62 | /* used[j] = 1 means bin j contains at least one item */ |
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| 63 | |
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| 64 | s.t. one{i in I}: sum{j in J} x[i,j] = 1; |
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| 65 | /* each item must be exactly in one bin */ |
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| 66 | |
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| 67 | s.t. lim{j in J}: sum{i in I} w[i] * x[i,j] <= c * used[j]; |
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| 68 | /* if bin j is used, it must not be overflowed */ |
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| 69 | |
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| 70 | minimize obj: sum{j in J} used[j]; |
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| 71 | /* objective is to minimize the number of bins used */ |
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| 72 | |
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| 73 | data; |
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| 74 | |
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| 75 | /* The optimal solution is 3 bins */ |
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| 76 | |
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| 77 | param m := 6; |
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| 78 | |
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| 79 | param w := 1 50, 2 60, 3 30, 4 70, 5 50, 6 40; |
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| 80 | |
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| 81 | param c := 100; |
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| 82 | |
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| 83 | end; |
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