Defining an activity with multiple students' sets

Started by aliponte, February 13, 2010, 10:41:46 PM

Previous topic - Next topic

0 Members and 1 Guest are viewing this topic.

aliponte

What is the better approach:

defining an activity like this:
Activity with id=xx (T:A, S:E, St:5a,5b,5c,5d)

which means a set of students' sets is assigned to this activity.
Or is it better like this with a predefined union of these students' sets (established by a group-subgroup relationship):

5abcd = union(5a, 5b, 5c, 5d)

Activity with id=xx (T:A, S:E, St:5abcd)

Thank you for responding
aliponte

Volker Dirr

the generated timetable will be the same.
just the look of the timetable is different.

maybe some room constraint will be easier to add.
in you first sample a home room constraint of students will do nothing, because there are to many students.
so you need to care about room with an other constraint.

in your second sample you can set a home room for students set "5abcd". (but i guess you will not use it, because rooms will be to small)

you can also just add the whole year "5" as students set, if 5a - 5d are all classes of that year. (so if there is no class 5e. Of course other courses in that year are allowed.) So you don't need to define a union "5abcd" if it contain all students of the year.

Liviu Lalescu

QuoteWhat is the better approach:

defining an activity like this:
Activity with id=xx (T:A, S:E, St:5a,5b,5c,5d)

which means a set of students' sets is assigned to this activity.
Or is it better like this with a predefined union of these students' sets (established by a group-subgroup relationship):

5abcd = union(5a, 5b, 5c, 5d)

Activity with id=xx (T:A, S:E, St:5abcd)

Thank you for responding
aliponte

Like Volker said, it is almost identical. Only look of timetable and indeed, home rooms for first variant is ignored, because there are >1 students sets.