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I am looking for tips on how to handle a planning problem with three business requirements that cannot be broken. I could turn that into three Hard Constraints (HC).

In the score calculator, the negative value of each of the three HC can be very different at a given point in the planning process:

HC1 = -10
HC2 = -10000000
HC3 = -2222

I did some testing using BendableLongScore in calculateScore, it seems during the test that HC1 considered the most important and if it gets the value 0, the planning can end with that solution.

Before I go any further is this the best way to handle 3 HCs in a planning problem?. I was thinking that I might turn to HardMediumSoft score and then work with a new HC that is an equally weighted sum of the three HC, in principle like this:

hardscore = hardscor1+hardscore2+hardscore3

Any tips? Thanks

user2212019
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1 Answers1

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No need to use something as complex as BendableLongScore. Simply use a normal HardSoft(Long)Score: score.setHardScore(hardscor1+hardscore2+hardscore3).

See score weighting vs score levels in docs chapter 5.

Geoffrey De Smet
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