我正在解决一个问题,我使用cumulatives/[2,3]
谓词。
但是当我尝试将其与minimize
in labeling
我有以下演示。 10 个任务,全部持续时间为 1,4 台机器,全部容量=1。我的目标是尽量减少总时间,即minimize(maximum(Es))
:
:- use_module(library(clpfd)).
:- use_module(library(lists)).
go( Ss, Es, Ms, Tm, Lab ) :-
Ss = [S1, S2, S3, S4,S5,S6,S7,S8,S9,S10], %Starttimes
Es = [E1, E2, E3, E4,E5,E6,E7,E8,E9,E10], %Endtimeds
Ms = [M1, M2, M3, M4,M5,M6,M7,M8,M9,M10], %MachineIds
domain(Ss, 1, 20),
domain(Es, 1, 20),
domain(Ms, 1, 10),
%All task has duration = 1
Tasks = [
task( S1, 1, E1, 1, M1 ),
task( S2, 1, E2, 1, M2 ),
task( S3, 1, E3, 1, M3 ),
task( S4, 1, E4, 1, M4 ),
task( S5, 1, E5, 1, M5 ),
task( S6, 1, E6, 1, M6 ),
task( S7, 1, E7, 1, M7 ),
task( S8, 1, E8, 1, M8 ),
task( S9, 1, E9, 1, M9 ),
task( S10, 1, E10, 1, M10 )
],
%All machines has resource capacity = 1
Machines = [
machine( 1, 1 ),
machine( 2, 1 ),
machine( 3, 1 ),
machine( 4, 1 )
],
cumulatives(Tasks, Machines, [bound(upper)] ),
maximum( MaxEndTime, Es ),
%Make the list of options to pass to the labeling predicate
append( [ [minimize(MaxEndTime)], [time_out( Tm, _)], Lab ], LabOpt ),
%The variables to lable:
append([Ms, Ss ], Vars),
labeling( LabOpt, Vars).
如果我现在运行这个并解决 1 秒,我会得到:
| ?- go( S, E, M, 1000, []).
E = [2,3,4,5,6,7,8,9,10,11],
M = [1,1,1,1,1,1,1,1,1,1],
S = [1,2,3,4,5,6,7,8,9,10] ?
IE。所有任务已安排在机器 1 上运行
我需要运行求解器 30 秒才能看到任何最小化的迹象:
| ?- go( S, E, M, 30000, []).
E = [2,3,4,5,6,7,8,9,10,2],
M = [1,1,1,1,1,1,1,1,1,2],
S = [1,2,3,4,5,6,7,8,9,1] ?
如果我跑 60 秒,我开始得到可接受的结果:
| ?- go( S, E, M, 60000, []).
E = [2,3,4,2,3,4,2,3,4,2],
M = [1,1,1,2,2,2,3,3,3,4],
S = [1,2,3,1,2,3,1,2,3,1] ?
我觉得这花费了太长的时间。
对于为什么需要这么长时间有什么评论吗?