[rules-users] Multiplying the count of each score rule [planner]

[Geoffrey De Smet]

Op 14-12-11 18:40, Patrik Dufresne schreef:
> Hi,
>
> I'm still in process to model my planning problem and I have some 
> difficulties in defining the correct weight of soft constraints. I 
> have soft constraints with different priorities : C1, C2, C3, ..., Cn 
> where C1 are higher then C2. Currently, I've tried to set different 
> weight for each of them : C1 get higher weight. But it's not 
> sufficient since multiple C2 may balance one C1. What I really need is 
> to set the priority to every soft constraints.
>
> Solution #1 :
> My first thought it to implement a new score definition 
> (HardAndSoftPriorityScoreDefinition) having separate soft score for 
> each priority. The rule may insert ConstraintOccurence by defining the 
> weight and the priority. This solution seems elegant but require 
> effort to implement the score definition, the score calculation, and 
> other things I don't even know about.
That is the perfect solution to your problem. Start by copy pasting 
DefaultHardAndSoftScoreDefinition and work your way from there.
You'll need to create at least a ScoreDefinition, ScoreCalculator and Score.

I've been thinking about adding such a "dynamic" score definition to 
planner's build-in scores,
but so far every use case where the developers said they needed this, it 
turned out end-users meant it differently:
when you break a 100 C2's, then it's better to break 1 C1 instead...
>
> Solution #2 :
> The other solution is stated in the Drools Planner User Guide :
>
>     "Most use cases will also weigh their constraints differently, by
>     multiplying the count of each score rule with its weight. For
>     example in freight routing, you can make 5 broken "avoid
>     crossroads" soft constraints count as much as 1 broken "avoid
>     highways at rush hour" soft constraint. This allows your business
>     analysts to easily tweak the score function as they see fit."
>
> Even tough I don't know how to implement this, it's seems much easier 
> to achieve since it's only a rule. Compare to solution #1, it's lack 
> the support of soft constraints with same priority but different weights.
That text describes plain-old weighting. Say C1 weights 100 and C2 
weights 2, then you can break 50 C2's for every 1 broken C1.

This is far easier and most of the time end-users actually mean this.
Make exaggerated examples (1000 C2's broken vs 1 C1 broken) and make 
your end-users decide what they prefer. If they still prefer 1000 C2's 
broken, then you need #1.

Tip: Sometimes, taking the square of a weight is a neat trick.
In bin packing, say you got
Solution A with 3 CPU and 3 CPU too little = 3² + 3² = 18
and Solution B with 4 CPU and 2 CPU too little = 4² + 2² = 20
So the second is worse even though they both miss 6 CPU.

>
> What is your opinion about both solution.
>
> Is one faster then the other ?
>
> Is it hard to create a new score definition ?
>
> Did anyone ever did this ?
>
> -- 
> Patrik Dufresne
>
>
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-- 
With kind regards,
Geoffrey De Smet

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