[rules-users] Continuous Planning values for task planning
Geoffrey De Smet
ge0ffrey.spam at gmail.com
Mon Aug 20 03:46:48 EDT 2012
You 'll want incremental score calculation (with delta's) for your "end
times".
http://docs.jboss.org/drools/release/5.4.0.Final/drools-planner-docs/html_single/index.html#incrementalScoreCalculation
So that naturally puts the calculation of those end times in the
scoreDRL (or IncrementalJavaCalculator if you're not using drools).
Whether or not that endTime should be a property on the model (at least
the model that Planner works with), is an open design question.
If it isn't, you can use insertLogicals in DRL, like I did in
nurserostering to calculate the number of sequential weekends being
worked etc.
If it is a property on your model, either the DRL must first "correct
it" (with higher salience rules for example),
or custom moves must "correct it" as they are being done (which is very
hard as it entails constraints functionality).
As for the cloning: it's quite simple:
Either the model's entity contains the endTime property, then clone it.
If it doesn't, then there's nothing to clone.
I don't see any problems related to the cloning.
Op 18-08-12 01:53, Josef Bajada schreef:
> Hi Geoffrey,
>
> Following your advice and after gaining some more understanding of
> planner, I think approaching the problem as a chain of tasks one of
> the other makes sense. It would have some 'wait' time between tasks
> where the end time of the previous task is smaller than the minimum
> time the task has to wait after its dependency (which could be
> different from the previous task).
>
> I've noticed that in most examples (TSP and VRP), there is some
> separation between the model and the planning entity that is being
> moved around in the chain, which also makes sense. (For instance in
> TSP, Visit is the planning entity while City is the data entity). When
> the planning entity gets cloned, the data entity gets assigned to the
> clone.
>
> I am concerned however, that with my dependency between tasks and
> their end times (which are as such a property of the planning entity
> not the data entity) I won't be able to model them in this way. (For a
> task to know whether it has violated its hard constraint it needs to
> get access to the end time of its dependency, which is in the planning
> entity). My concern is that I might end up with a whole mess when it
> comes to the cloning of tasks. I am also concerned about the
> performance of computing the end time of each node recursively based
> on its previous task and dependent tasks.
>
> What is the best approach in this case?
>
> thanks,
>
> Josef
>
>
>
> On 25 July 2012 08:30, Geoffrey De Smet <ge0ffrey.spam at gmail.com
> <mailto:ge0ffrey.spam at gmail.com>> wrote:
>
>
> Op 24-07-12 23:14, Josef Bajada schreef:
>> Hi Geoffrey,
>>
>> Thanks for your reply.
>>
>> > Does it make sense to wait longer than 7 mins after task A
>> (presuming no other task forces occupies the user at that time)?
>> > Put differently: Can we say that the starting time of B =
>> Math.max((endTime of task before B), (endTime of task A + 7
>> minutes))?
>> > If we can say that, it's pointless to investigate the solutions
>> where task B starts 8 minutes after task A and the user doing no
>> task that last minute.
>>
>> Yes, the starting time of B = Math.max((endTime of task before
>> B), (endTime of task A + 7mins)) as long as it is smaller than
>> (endTime of task A + 8mins).
>> Yes, it is pointless to investigate the solutions where task B
>> starts 8 minutes after task A and the user doing no task that
>> last minute.
>> The 8 minute is just a constraint that the task in between tasks
>> A and B cannot take longer than 7:59s.
>>
>> I am thinking that maybe instead of using time itself as the
>> planning variable, we would use time just to determine the Hard
>> and Soft scores.
>> So if Task B is scheduled after Task A + 8mins by the solver,
>> then it inflicts on the hard score. Similarly if Task B is
>> scheduled before Task A + 7 mins.
>> Does my reasoning make sense in any way?
> Yes, but personally, I 'd design it differently (although I have
> no proof that my way would be better), like this:
> "Task B is scheduled after Task A + 8mins by the solver" => make
> this a hard constraint
> "Task B is scheduled before Task A + 7 mins" => make this a
> build-in hard constraint (= not a constraint in the scoreDRL or
> ScoreCalculator, but by design, see manual).
> Each Task is assigned to a previousTaskOrPerson (and this variable
> is chained). It does not know it's startingTime directly.
> The scoreDRL or ScoreCalculator calculates the startingTime of a
> Task dynamically, by applying this logic:
> starting time of B = Math.max((endTime of previousTaskOrPerson
> of B), (endTime of task A + 7mins))
> Note: "Chained=true" guarantees that there are no cycles of Tasks
> and that no Tasks exists with a previousTaskOrPerson == null.
> Note: "(endTime of task A + 7mins)" is not hard coded in the score
> function: you won't find "7" or "A" in there.
>
>
>>
>> thanks,
>> Josef
>>
>>
>>
>> On 24 July 2012 20:46, Geoffrey De Smet <ge0ffrey.spam at gmail.com
>> <mailto:ge0ffrey.spam at gmail.com>> wrote:
>>
>>
>> Op 23-07-12 16 <tel:23-07-12%2016>:26, Josef Bajada schreef:
>>> Hi Geoffrey,
>>>
>>> Well I want to leave 'space' between tasks in the situations
>>> where there are hard constraints that require me to put this
>>> space.
>> This makes the chaining technique harder to model, but I
>> wouldn't write it off yet.
>>
>>>
>>> As a simple example:
>>>
>>> Task A: Put pasta in boiling water (duration 40 seconds)
>>> Task B: Take pasta out of boiling water (duration 50
>>> seconds, cannot start before 7 mins after Task A finishes,
>>> cannot start after 8 mins after Task A finishes)
>> Does it make sense to wait longer than 7 mins after task A
>> (presuming no other task forces occupies the user at that time)?
>> Put differently: Can we say that the starting time of B =
>> Math.max((endTime of task before B), (endTime of task A + 7
>> minutes))?
>> If we can say that, it's pointless to investigate the
>> solutions where task B starts 8 minutes after task A and the
>> user doing no task that last minute.
>> If we can say that, then chaining can calculate the the
>> starting time of a task on the fly differently.
>>
>>> Task C: Chop vegetables (duration 2 minutes).
>>>
>>> This will evidently leave some gaps. The ideal result from
>>> the solver should be:
>>>
>>> Task A: at time 0 (ends at 40s)
>>> Task C: at time 41s (ends at 2:41)
>>> Task B: at time 7:40
>>>
>>> There is a gap between C and B which is OK.
>>>
>>> If another Task is added to the story:
>>> Task D: Prepare sauce (duration 7 minutes)
>>>
>>> I would want the following result:
>>>
>>> Task A: at time 0 (ends at 40s)
>>> Task D: at time 41s (ends 7:41s)
>>> Task B: at time 8:42s (ends 9:32s)
>>> Task C: at time 9:33s (ends 11:33s)
>>>
>>> Task C can actually take place before Task A too.
>>>
>>> I still need to read and understand the chaining
>>> functionality properly. Do you think it would allow me to
>>> achieve the above?
>>>
>> I don't know.
>> But using continuous variables in a search problem such as
>> this that smells discrete with discrete constraints (A must
>> start before B, ...), could blow up the search space
>> unnecessarily.
>>
>> If you want to look into using continuous variables: the
>> support for it is limited currently.
>> you can reuse the Drools Planner metaheuristic algorithms
>> (including termination, score, ...), but there's no decent
>> generic move factory support for continuous variables yet.
>> So you 'll have to write a custom MoveFactory that creates a
>> limited subset of moves.
>> Also, construction heuristics can't handle continuous
>> variables yet, so you 'll have to write a custom
>> SolutionIntializer.
>> There are examples with a custom MoveFactory and a custom
>> SolutionIntializer where you can copy paste from, but none
>> with continuous variables at the moment.
>>
>>> thanks,
>>>
>>> Josef
>>>
>>>
>>>
>>> On 22 July 2012 20:05, Geoffrey De Smet
>>> <ge0ffrey.spam at gmail.com <mailto:ge0ffrey.spam at gmail.com>>
>>> wrote:
>>>
>>> Presuming that you don't want to leave space between
>>> tasks, you can design your model differently by using
>>> the "chained" functionality:
>>> it will be far more efficient and the planning variable
>>> won't be continuous.
>>>
>>> Let's presume you're scheduling Tasks to Persons.
>>>
>>> @PlanningEntity
>>> class Task implements TaskOrPerson {
>>>
>>> ...
>>>
>>> @PlanningVariable(chained = true)
>>> @ValueRanges({
>>> @ValueRange(type =
>>> ValueRangeType.FROM_SOLUTION_PROPERTY, solutionProperty
>>> = "taskList"),
>>> @ValueRange(type =
>>> ValueRangeType.FROM_SOLUTION_PROPERTY, solutionProperty
>>> = "personList",
>>> excludeUninitializedPlanningEntity = true)})
>>> public TaskOrPerson getPreviousTaskOrPerson() {
>>> return previousTaskOrPerson;
>>> }
>>>
>>> public int getDuration() {
>>> return duration;
>>> }
>>>
>>> public int getStartingTime() {
>>> int startingTime = 0;
>>> TaskOrPerson taskOrPerson =
>>> getPreviousTaskOrPerson();
>>> while (taskOrPerson instanceof Task) { //
>>> Every chain is guarantee to end up with an anchor (= Person)
>>> startingTime += ((Task)
>>> taskOrPerson).getDuration();
>>> taskOrPerson = ((Task)
>>> taskOrPerson).getPreviousTaskOrPerson()
>>> }
>>> return startingTime;
>>> }
>>>
>>> }
>>>
>>> class Person implements TaskOrPerson {
>>>
>>> }
>>>
>>> For a good example, take a look at the VehicleRouting
>>> example.
>>> For more info about chaining, in the manual see section
>>> 4.3.4.2.6. Chained
>>> http://docs.jboss.org/drools/release/5.4.0.Final/drools-planner-docs/html_single/index.html
>>>
>>> Op 22-07-12 18 <tel:22-07-12%2018>:00, Josef Bajada schreef:
>>>> Hi,
>>>>
>>>> I am new to Drools and Drools Planner, so apologies if
>>>> I am asking anything obvious.
>>>>
>>>> My objective is to implement a simple (for now) planner
>>>> which schedules tasks according to 2 main criteria:
>>>> - Their duration (in seconds)
>>>> - Their dependencies on other tasks (e.g. Hard
>>>> Constraint that Task B has to start between 180 and 200
>>>> seconds after Task A finishes).
>>>>
>>>> Since there are gaps between dependent tasks as part of
>>>> the hard constraints other tasks can be fitted in
>>>> between dependent tasks.
>>>> So the Solver needs to find the optimal start time for
>>>> each task that satisfies the hard constraints, and in
>>>> the shortest total timeline possible to complete all
>>>> tasks (soft constraint).
>>>>
>>>> The main problem I am finding is that this start time,
>>>> which is essentially the planning variable is a
>>>> continuous variable.
>>>> Chapter 4 of the Drools documentation mentions very
>>>> briefly (Section 4.3.4.1) that planning variables can
>>>> be continuous, but there does not seem to be any more
>>>> details about how to achieve this.
>>>>
>>>> Even if the planning variable was discrete (say bins of
>>>> 5 second intervals), there is no upper bound as such.
>>>>
>>>> How is it best to handle such planning variables in
>>>> Drools Planner?
>>>>
>>>> thanks,
>>>> josef
>>>>
>>>>
>>>>
>>>>
>>>> _______________________________________________
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>>>> https://lists.jboss.org/mailman/listinfo/rules-users
>>>
>>> --
>>> With kind regards,
>>> Geoffrey De Smet
>>>
>>>
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>>>
>>>
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>>
>> --
>> With kind regards,
>> Geoffrey De Smet
>>
>>
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>>
>>
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>
> --
> With kind regards,
> Geoffrey De Smet
>
>
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--
With kind regards,
Geoffrey De Smet
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