Dear
in "min hours daily for all teachers", i can't make the min hours = 1, the min hours (allowed) is 2 .
please let the minimum number allowed = 1 .
in the examination timetable there are 2 activities in the day, so i can't used 2, and fet not allowed 1 ?! :'(
regards
Min hours 1 will have no effect. You can use max hours 1, min n days between activities, max days per week or min days per week if you want to force somethink like that.
this the timetable, "max days per week" not work.
i want to let the min hours for any teacher = 1 , I do not want a blank day for any teacher.
then just use min days per week.
Quote from: Volker Dirr on November 03, 2011, 10:39:50 PM
then just use min days per week.
when i used "min days per week"
when i click "gernerate botton" i got this massege:
Cannot optimize because for teacher Abcd you have min days per week 9 and he has only 0 activities - impossible. Please correct and try again.
teacher Abcd have not enough activities. you told fet that Abcd must teach at least (min) 1 hour per day with that constraint. that is why fet complain.
solution 1:
delete teacher abcd (or give him more activities. at least 9)
solution 2:
don't add this constraint to all teachers. add it only to teachers that have at least 9 activities.
All teachers to have 10 activities in 9 days
Equal number of activities teachers.
please see attachment file in the first post.
i downloaded that file. fet complain. that is correct. there are 7 teachers in that file without any activity.
please go to fet->advanced->activity planning.
check the lower grid. there you can see that 7 teachers without activity very easy.
please delete that teachers.
you can also see them at fet->statistics->teachers statistics.
ok, i deleted the 7 teachers, and click generate button , fet :
197 out of 280 activities placed
then stop ?
maybe fet need longer.
maybe the table is impossible with such a difficult constraint.
i can't check that now. i must sleep now.
please post the new dataset (because it is difficult to read arabic letters, so i need very long to delete that teachers :-(
i will check tomorrow (in around 16 hours) if you post the latest dataset.
this the last file with English alphabet.
First of all you have 10 activities with min n day constraint, but only 9 days. You use "only" a 95% weight for this, but this waste time.
"Solution" 1:
In my optinion you should remove one activity per each min n day constraint and increase weight up to 100%.
"Solution" 2:
Since you use the min days per week constraint the min n day constraints are not any longer needed at all. So you can even delete all.
If you do this, you will see that fet sadly still hang at the same position.
Second point is this:
Watch your space constraints. you have many space constraint and all disallow the same slots. Each time you remove one of these constraints, the max number of placed activities grow.
So it look like it is impossible. I didn't see the exactly reason for this, but maybe rethink if all space constraint are needed and why that disalow to solve the dataset. I deleted a few room constraints and the timetable was possible in that case. maybe do the same and then check the timetable. it is maybe much easier for you to understand the problem if you see that timetable. then rethink why it is not possible to add the space constraint.
I saw now you have 28 teachers, but only 18 rooms.
So you must have at least 28 free slots per day if you want that eavery teacher teach 1 hour per day.
But you disallow to many rooms at several days.
You have 2 slots per day. So 2*18rooms=36 rooms (slots per day).
If all 28 teachers need one slot per day, then you have 36rooms - 28room = 8 rooms left.
So you are only allowed to disallow max 8 rooms per day! But you disallow much more rooms.
It is imposible that way.
Quote from: Volker Dirr on November 04, 2011, 06:15:01 PM
I saw now you have 28 teachers, but only 18 rooms.
So you must have at least 28 free slots per day if you want that eavery teacher teach 1 hour per day.
But you disallow to many rooms at several days.
You have 2 slots per day. So 2*18rooms=36 rooms (slots per day).
If all 28 teachers need one slot per day, then you have 36rooms - 28room = 8 rooms left.
So you are only allowed to disallow max 8 rooms per day! But you disallow much more rooms.
It is imposible that way.
so i hope you add "min hours daily for all teachers".
regards
Quote from: Silver on November 04, 2011, 09:10:49 PM
so i hope you add "min hours daily for all teachers".
No, we will not do that, because:
a) that constraint will have no effect (because days without activities are allowed with min hours daily constraint)
b) min day per week can be used as "min hours = 1". so your request is already included
c) that doesn't help your dataset, because the dataset will be never solveable. please read my previous post again and check that. YOUR DATASET IS IMPOSSIBLE WITH THAT REQUEST! i wrote the proof for that in my previous post.
i solved the problem temporarily by:
- "teachers min days per week" = 8
- "teachers max days per week" = 8
please see attachment.
and Thank you for your interest.
Yes, that is possible and a nice solution. (But that was not your request. Your request was that no teacher should have a free day. Now everybody will get a free day.)
QuoteBut that was not your request.
When I did not find a solution,
I tried to make the distribution of free days equally.
of course. but why not = 7 with both constraints. so all teachers have 2 free days. (=6 sadly doesn't work because of room problem).