% RUNS ON mzn20_fd % RUNS ON mzn-fzn_fd % RUNS ON mzn20_fd_linear % RUNS ON mzn20_mip %------------------------------------------------------------------------------% % cutstock.mzn % Jakob Puchinger % December 2007 % vim: ft=zinc ts=4 sw=4 et tw=0 % % The cutting stock problem: Kantorovitch formulation for colgen. % %------------------------------------------------------------------------------% int: L; % stock length int: K; % upper bound on number of stock pieces int: N; % number of stock pieces array[1..N] of int: i_length; array[1..N] of int: i_demand; array[1..K] of var 0..1: pieces; array[1..K, 1..N] of var int: items; constraint forall(k in 1..K, i in 1..N)( items[k,i] >=0 ); solve ::int_search(array1d(1..K*N,items)++pieces,input_order,indomain_min,complete) minimize sum([ pieces[k] | k in 1..K]); constraint forall(i in 1..N) ( sum( [ items[k, i] | k in 1..K] ) >= i_demand[i] ); constraint forall( k in 1..K ) ( sum(i in 1..N)( items[k,i] * i_length[i] ) <= pieces[k] * L ); % required for showing the objective function var int: obj; constraint obj = sum([ pieces[k] | k in 1..K]); output [ "Cost = ", show( obj ), "\n" ] ++ [ "Pieces = \n\t" ] ++ [show(pieces)] ++ [ "\n" ] ++ [ "Items = \n\t" ] ++ [ show(items[k, i]) ++ if k = K then "\n\t" else " " endif | i in 1..N, k in 1..K ] ++ [ "\n" ]; % data: N = 3; K = sum(i_demand); L = 10; i_length = [7, 5, 3]; i_demand = [2, 2, 4];