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

** 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 _______________________________________________ rules-users mailing listrules-users@lists.jboss.orghttps://lists.jboss.org/mailman/listinfo/rules-users -- With kind regards, Geoffrey De Smet _______________________________________________ rules-users mailing list rules-users(a)lists.jboss.org https://lists.jboss.org/mailman/listinfo/rules-users