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83 lines
2.2 KiB
83 lines
2.2 KiB
/* BPP, Bin Packing Problem */
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/* Written in GNU MathProg by Andrew Makhorin <mao@gnu.org> */
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/* Given a set of items I = {1,...,m} with weight w[i] > 0, the Bin
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Packing Problem (BPP) is to pack the items into bins of capacity c
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in such a way that the number of bins used is minimal. */
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param m, integer, > 0;
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/* number of items */
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set I := 1..m;
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/* set of items */
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param w{i in 1..m}, > 0;
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/* w[i] is weight of item i */
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param c, > 0;
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/* bin capacity */
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/* We need to estimate an upper bound of the number of bins sufficient
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to contain all items. The number of items m can be used, however, it
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is not a good idea. To obtain a more suitable estimation an easy
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heuristic is used: we put items into a bin while it is possible, and
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if the bin is full, we use another bin. The number of bins used in
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this way gives us a more appropriate estimation. */
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param z{i in I, j in 1..m} :=
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/* z[i,j] = 1 if item i is in bin j, otherwise z[i,j] = 0 */
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if i = 1 and j = 1 then 1
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/* put item 1 into bin 1 */
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else if exists{jj in 1..j-1} z[i,jj] then 0
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/* if item i is already in some bin, do not put it into bin j */
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else if sum{ii in 1..i-1} w[ii] * z[ii,j] + w[i] > c then 0
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/* if item i does not fit into bin j, do not put it into bin j */
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else 1;
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/* otherwise put item i into bin j */
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check{i in I}: sum{j in 1..m} z[i,j] = 1;
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/* each item must be exactly in one bin */
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check{j in 1..m}: sum{i in I} w[i] * z[i,j] <= c;
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/* no bin must be overflowed */
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param n := sum{j in 1..m} if exists{i in I} z[i,j] then 1;
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/* determine the number of bins used by the heuristic; obviously it is
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an upper bound of the optimal solution */
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display n;
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set J := 1..n;
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/* set of bins */
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var x{i in I, j in J}, binary;
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/* x[i,j] = 1 means item i is in bin j */
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var used{j in J}, binary;
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/* used[j] = 1 means bin j contains at least one item */
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s.t. one{i in I}: sum{j in J} x[i,j] = 1;
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/* each item must be exactly in one bin */
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s.t. lim{j in J}: sum{i in I} w[i] * x[i,j] <= c * used[j];
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/* if bin j is used, it must not be overflowed */
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minimize obj: sum{j in J} used[j];
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/* objective is to minimize the number of bins used */
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data;
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/* The optimal solution is 3 bins */
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param m := 6;
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param w := 1 50, 2 60, 3 30, 4 70, 5 50, 6 40;
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param c := 100;
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end;
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