lemon-project-template-glpk
comparison deps/glpk/examples/bpp.mod @ 9:33de93886c88
Import GLPK 4.47
author | Alpar Juttner <alpar@cs.elte.hu> |
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date | Sun, 06 Nov 2011 20:59:10 +0100 |
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-1:000000000000 | 0:575deda93e45 |
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1 /* BPP, Bin Packing Problem */ | |
2 | |
3 /* Written in GNU MathProg by Andrew Makhorin <mao@gnu.org> */ | |
4 | |
5 /* Given a set of items I = {1,...,m} with weight w[i] > 0, the Bin | |
6 Packing Problem (BPP) is to pack the items into bins of capacity c | |
7 in such a way that the number of bins used is minimal. */ | |
8 | |
9 param m, integer, > 0; | |
10 /* number of items */ | |
11 | |
12 set I := 1..m; | |
13 /* set of items */ | |
14 | |
15 param w{i in 1..m}, > 0; | |
16 /* w[i] is weight of item i */ | |
17 | |
18 param c, > 0; | |
19 /* bin capacity */ | |
20 | |
21 /* We need to estimate an upper bound of the number of bins sufficient | |
22 to contain all items. The number of items m can be used, however, it | |
23 is not a good idea. To obtain a more suitable estimation an easy | |
24 heuristic is used: we put items into a bin while it is possible, and | |
25 if the bin is full, we use another bin. The number of bins used in | |
26 this way gives us a more appropriate estimation. */ | |
27 | |
28 param z{i in I, j in 1..m} := | |
29 /* z[i,j] = 1 if item i is in bin j, otherwise z[i,j] = 0 */ | |
30 | |
31 if i = 1 and j = 1 then 1 | |
32 /* put item 1 into bin 1 */ | |
33 | |
34 else if exists{jj in 1..j-1} z[i,jj] then 0 | |
35 /* if item i is already in some bin, do not put it into bin j */ | |
36 | |
37 else if sum{ii in 1..i-1} w[ii] * z[ii,j] + w[i] > c then 0 | |
38 /* if item i does not fit into bin j, do not put it into bin j */ | |
39 | |
40 else 1; | |
41 /* otherwise put item i into bin j */ | |
42 | |
43 check{i in I}: sum{j in 1..m} z[i,j] = 1; | |
44 /* each item must be exactly in one bin */ | |
45 | |
46 check{j in 1..m}: sum{i in I} w[i] * z[i,j] <= c; | |
47 /* no bin must be overflowed */ | |
48 | |
49 param n := sum{j in 1..m} if exists{i in I} z[i,j] then 1; | |
50 /* determine the number of bins used by the heuristic; obviously it is | |
51 an upper bound of the optimal solution */ | |
52 | |
53 display n; | |
54 | |
55 set J := 1..n; | |
56 /* set of bins */ | |
57 | |
58 var x{i in I, j in J}, binary; | |
59 /* x[i,j] = 1 means item i is in bin j */ | |
60 | |
61 var used{j in J}, binary; | |
62 /* used[j] = 1 means bin j contains at least one item */ | |
63 | |
64 s.t. one{i in I}: sum{j in J} x[i,j] = 1; | |
65 /* each item must be exactly in one bin */ | |
66 | |
67 s.t. lim{j in J}: sum{i in I} w[i] * x[i,j] <= c * used[j]; | |
68 /* if bin j is used, it must not be overflowed */ | |
69 | |
70 minimize obj: sum{j in J} used[j]; | |
71 /* objective is to minimize the number of bins used */ | |
72 | |
73 data; | |
74 | |
75 /* The optimal solution is 3 bins */ | |
76 | |
77 param m := 6; | |
78 | |
79 param w := 1 50, 2 60, 3 30, 4 70, 5 50, 6 40; | |
80 | |
81 param c := 100; | |
82 | |
83 end; |