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