问题
I'm working on a problem where I use the cumulatives/[2,3]
predicate.
But I get very bad performance when I try to combine this with minimize
in labeling
I have the following demo. 10 task, all with duration 1, 4 machines, all with capacity=1. My goal is to minimize the total time, i.e. 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).
If I now run this and solve for 1 second I get:
| ?- 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] ?
I.e. all tasks has been scheduled to run on machine 1
I need to run the solver for 30 second before I see any sign of minimization:
| ?- 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] ?
If I run for 60 second I start to get acceptable results:
| ?- 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] ?
I feel that this is taking way too long time. Any comments on why it takes so long time?
回答1:
I found two standard tricks that speed up your code.
First, variable order. You are labeling all the M variables before the S variables. That takes some 51 seconds. It's much better to fix both S and M for one task at a time. I.e. variable order [S1,M1,S2,M2,S3,M3,S4,M4,S5,M5,S6,M6,S7,M7,S8,M8,S9,M9,S10,M10]. That brings the time down to some 2 seconds.
Second, your tasks are interchangeable, and so are your machines. Perhaps that will not always be the case, if your code was meant to be a toy example and not the real thing. But if you have such symmetries, you can prevent the search going down a lot of culs de sac by breaking the symmetries e.g. by:
lex_chain([[S1,M1],[S2,M2],[S3,M3],[S4,M4],[S5,M5],[S6,M6],[S7,M7],[S8,M8],[S9,M9],[S10,M10]]),
That brings the time down to 10 milliseconds.
Both these tricks are standard in the craft of constraint programming.
回答2:
By using the task_intervals(true)
option on the cumulatives
predicate the speed is really improved:
cumulatives(Tasks, Machines, [bound(upper),task_intervals(true)] )
gives solvingtime of 2ms to the code in the original question without changing anyting else.
来源:https://stackoverflow.com/questions/23649534/use-of-cumulatives