I spent much time striving to solve an impossible timetable
I wonder if FET can automatically report it impossible then abort
-Fisrt case: a teacher having only 5 hours per week: 1+1+1+2 (Min days between activities=1)
With Min hours per day for teachers = 2
This lead for an impossible timetable, but FET desperately trying to solve it for hours and hours
-Second case:
A teacher has 15 hours
With a maximum of 4 hours per day, given a 6 days week, this would allow him 1 free day or even 2 free days (4+4+4+4)
But activities were split in a rather strange fashion
Group G1: 3+3
Group G2: 2+2+2+3 (Min days between activities = 1)
Activities from G1 alone would occupy him for 2 days
Then come activities form G2 which require additional 4 days
Total is 6 days !!
I set Max days per week = 5
This leads to yet another impossible timetable
My question: can FET automatically report it impossible then abort?
Yes, it is of course possible to detect such things, but I think there are too many possibilities of mistakes like that for me to take care of all of them. I'll add your request in the TODO and I'll think about it.