I'm currently working on a scheduling problem using the :- use_module(library(clpq)) library. My problem lies with retrieving the full solution to my problem statement.
schedule(BestSchedule, BestTotTime) :- %BestSchedule of format [test(task,startTime,Duration Task, Core]
( findall(T,task(T),Tasks),
findall(C,core(C),Cores),
init_time, %Retract all facts of bestsofar(_,_)
const_time(Tasks,Schedule,Cores,TotTime),%Here probably goes something wrong
assign_processor(Schedule,Cores, TotTime),
minimize(TotTime),
update_time(Schedule, TotTime),
fail
; bestsofar(BestSchedule,BestTotTime)
).
assign_processor([], _, _).
assign_processor([task(T,S,D,C)|Rest], Cores, TotTime) :-
assign_processor(Rest,Cores,TotTime),
core(C), %Pick a core
process_cost(T,C,D), %Core/Task known, setting Duration (D)
const_resource(task(T,S,D,C),Rest), %Setting extra constraints on chosen core
bestsofar(_,CurrentBestTime), %Get current best time from fact bestsofar
{TotTime < CurrentBestTime}. %Set new constraint based on currentBestTime
const_resource(_,[]).
const_resource(Task,[Task2|Rest]) :-
no_data(Task,Task2),
no_conflict(Task,Task2), %No overlap on same processor
const_resource(Task, Rest).
no_conflict(task(_,S,D,C),task(_,S2,D2,C2)) :-
C \== C2,!; %Tasks not on same processor
{ S+D =< S2; %Set no overlapping start/end times
S2+D2 =< S }.
no_data(task(T,S,D,C), task(T2,S2,D2,C2)) :-
%depends_on(T,T2,_) = T2 needs to be finished before T can start
%channel(C,C2,L,_) = Data transfer between cores generated latency L
%Set no overlap including the latency if tasks are dependent
depends_on(T,T2,_),channel(C,C2,L,_),!, { S2 + D2 =< S - L };
depends_on(T2,T,_),!,channel(C2,C,L,_),!, { S + D =< S2 - L};
true.
init_time :-
retract(bestsofar(_,_)), fail
;
assert(bestsofar(foobar, 1000)).
update_time(Schedule,TotTime) :-
retract(bestsofar(_,_)),!,
assert(bestsofar(Schedule,TotTime)).
S = [task(t7, 100, 10, c1), task(t1, 0, 10, c1), task(t6, 75, 10, c1),
task(t2, _G66, 10, c1), task(t4, 50, 10, c2), task(t3, 25, 10, c1),
task(t5, 50, 10, c1)],
ET = 110.
This solution seems to be correct, but I don't have any specific value of task(t2, _G66, 10, c1) (tasknumber, start time, duration, processor core).
To my knowledge this is a local variable which values should be between 25<_G66<35 OR 60<_G66<75 but I can't seem to find a way to print these values in Prolog itself. I thought the minimize(TotTime) would force all variables be minimized which seems to happen with the rest.
Edit:
Added some more code to show where the problem should lie. No other failures are produced somewhere else. bestsofar/2 is used store the current best schedule solution and execution time. When we find a better, faster schedule we replace it using update_time/2. This searching will always fail, this way all possible solutions are tested. Once done we reach bestsofar(BestSchedule,BestTotTime) and return these values.
If I look at the debugger before returning the result.B=35-A which does support my manual test of 35<B<50 or 60<B<75. Can't really make the deduction myself because I don't know how to interpret the _ value in these constraints.
[ task(t7,100,10,c1),
task(t1,0,10,c1),
task(t6,75,10,c1),
task(t2,B,10,c1),
task(t4,50,10,c2),
task(t3,25,10,c1),
task(t5,50,10,c1)
], % with constraints
{}(_ = -55 - A ',' _ = -40 - A ',' _ = -25 + A ',' _ = -10 + A ',' _ = -30 - A ',' _ = -5 - A ',' A >= -5 ',' A =< 0 ',' _ = -55 - A ',' _ = -25 + A ',' _ = 65 + A ',' B = 35 - A)
Without no_data/2 my code does work for examples where no channel latency is used. Therefore I guess any problem should lie in that piece of code.
Runnable code if interested: http://pastebin.com/3PmRu7Aq
