Closed Chickenpowerrr closed 3 weeks ago
In multi-objective model checking, we restrict to policies that yield finite reward for all objectives. In particular, this means that we drop solutions that are infinitely "bad" for one or more objectives.
For the given example, this means that selecting the 'simp' action is the only valid policy for the given set of objectives. Consequently, there is no valid policy under which R{"simp"} <=0 [C]
holds. We (correctly) get false
as an answer.
Note that this assumption avoids some nasty technicalities within the algorithm (e.g. what would be the convex hull of the two points (0,1) and (inf,0) and how to represent this?).
I discussed this situation and possible workarounds also a bit in my dissertation (Section 3.3), in case you are interested.
Please also note that outputting There is no Pareto optimal scheduler that yields finite reward for all objectives
would not be accurate: Selecting simp
in s=0 is Pareto optimal and yields finite rewards.
Hi Tim,
Thanks the great explanation and reference, I read the relevant section and understand the output now. I will close the issue.
Kind regards,
Mark
When evaluating
multi(R{"inf"}min=? [C], R{"simp"}<=0 [C])
on the following MDP, the output isfalse
, while the actual result should be infinity. Hence I would expect to get the output:There is no Pareto optimal scheduler that yields finite reward for all objectives. There is no Pareto optimal scheduler that yields finite reward for all objectives
. Moreover,multi(R{"simp"}<=0 [C])
is considered achievable, somulti(R{"inf"}min=? [C], R{"simp"}<=0 [C])
should not be unachievable.