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dekker.one
1/***
2!Test
3expected:
4- !Result
5 solution: !Solution
6 items:
7 - [0, 0, 0]
8 - [0, 0, 0]
9 - [0, 0, 0]
10 - [0, 0, 0]
11 - [0, 0, 2]
12 - [0, 2, 0]
13 - [1, 0, 1]
14 - [1, 0, 1]
15 obj: 4
16 objective: 4
17 pieces: [0, 0, 0, 0, 1, 1, 1, 1]
18- !Result
19 solution: !Solution
20 items:
21 - [0, 0, 3]
22 - [0, 2, 0]
23 - [1, 0, 1]
24 - [1, 0, 1]
25 - [0, 0, 0]
26 - [0, 0, 0]
27 - [0, 0, 0]
28 - [0, 0, 0]
29 obj: 4
30 objective: 4
31 pieces: [1, 1, 1, 1, 0, 0, 0, 0]
32***/
33
34%------------------------------------------------------------------------------%
35% cutstock.mzn
36% Jakob Puchinger <jakobp@cs.mu.oz.au>
37% December 2007
38% vim: ft=zinc ts=4 sw=4 et tw=0
39%
40% The cutting stock problem: Kantorovitch formulation for colgen.
41%
42%------------------------------------------------------------------------------%
43
44int: L; % stock length
45int: K; % upper bound on number of stock pieces
46int: N; % number of stock pieces
47
48array[1..N] of int: i_length;
49array[1..N] of int: i_demand;
50
51array[1..K] of var 0..1: pieces;
52array[1..K, 1..N] of var int: items;
53
54constraint
55 forall(k in 1..K, i in 1..N)( items[k,i] >=0 );
56
57solve ::int_search(array1d(1..K*N,items)++pieces,input_order,indomain_min,complete) minimize
58 sum([ pieces[k] | k in 1..K]);
59
60constraint
61 forall(i in 1..N)
62 (
63 sum( [ items[k, i] | k in 1..K] ) >= i_demand[i]
64 );
65
66constraint
67 forall( k in 1..K ) (
68 sum(i in 1..N)( items[k,i] * i_length[i] ) <= pieces[k] * L
69 );
70
71 % required for showing the objective function
72var int: obj;
73constraint
74 obj = sum([ pieces[k] | k in 1..K]);
75
76output
77[ "Cost = ", show( obj ), "\n" ] ++
78[ "Pieces = \n\t" ] ++ [show(pieces)] ++ [ "\n" ] ++
79[ "Items = \n\t" ] ++
80[ show(items[k, i]) ++ if k = K then "\n\t" else " " endif |
81 i in 1..N, k in 1..K ] ++
82[ "\n" ];
83
84
85% data:
86N = 3;
87K = sum(i_demand);
88L = 10;
89i_length = [7, 5, 3];
90i_demand = [2, 2, 4];
91