Students min hours continuously

[Horatiu]

I've been "battling" gaps for students, and there is no way of avoiding them, given the restrictions that I am facing.

However, after discussing with students, it emerged that they are not bothered by gaps, as long as they have at least 3 hours continuously. Even if they have 3 hours in the morning and 3 more in the evening (thus 4 consecutive gaps).

But I have no idea how to implement that in fet. ant help would be appreciated.

[Volker Dirr]

You mean he need at leat min 3 hours gaps (not only 1 or 2)?
i am not sure if that is realy a good idea, but you can maybe do that easy this way:
add pseudoactivities with only students set and 3 hours duration.
add constraint max 0 gaps per students.

[Horatiu]

No. What I meant is this:
Students are OK with gaps (no matter how long - even 3 hours), as long as they have at least 3 hours continuously
Students don't want to come to school for only 1 hour, then leave, then come back (maybe for only 1 hour again).

[Volker Dirr]

ah. ok. i understand now.

currently you can't enter that exactly in fet.

i don't know your dataset. so maybe my ideas are not very good:
i think you should think about this solution:
maybe add a min hours per day constraint for the students.
maybe add max gaps per day for students.


i unnderstand your request, but don't forget that your request is "only" a psychological problem and not depending on a real time problem.
example:
maybe you have 9 timeslots (hours) per day.
student A get activities on Monday: 1st, 2nd, 3rd, 7th, 8th, 9th.
so this fit your request (at least 3 hours continiuesly).

an other students get this:
Monday: 1st, 2nd, 3rd, 4th, 8,th, 9th.
so this doesn't fit your request, because there are only 2 hours continiously (in the morning).

but now lets check the difference.
Both leave home at the same time (because school starts with 1st hour) -> so  no difference.
Both will come back home at the same time (because school ends with 9th hour) -> so  no difference.
both work 6 hours per day -> so  no difference.
both have 3 hours continiously a big gap -> so no difference.

so if the students stay at school during the gap, both must wait 3 hours.
if they drive home during the gap: it doens't matter if you have the first or the second timetable. the way to travell is the same.

so you see, it is a "only" a psychological if you think you need to come only for 2 hours.

in my opinion you should think about solving/optimising in an other way, not the way you requested. (but i can't do that without knowing more about your dataset.)

[Horatiu]

I already work with min_3_hours daily and max_1_gap daily for students.
Problem is I get a lot of:
- hour-gap-hour-hour-hour (and students have the tendency to not attend the first one),
or
- hour-hour-hour-gap-hour (students have the tendency to not attend the last one).

Worst case scenario:
- hour-gap-hour-gap-hour-hour (this is why i use max_1gap_daily; this situation is to be avoided under any circumstances).

- 3 hours continuously (my example from the first post) is the ideal situation.
- 2 hours continuously (your example) is also acceptable
I tried to think a way, using dummy activities, but with no results. That is way I asked on the forum;)

[Liviu Lalescu]

Unfortunately, this constraint is easy to express in words, but too difficult for me to implement.

[Horatiu]

So no dummy activities trick here, right?

[Liviu Lalescu]

I am not sure, I didn't think of tricks, I was just talking about the constraint.

Probably there are no possible tricks, but I am not sure.

[Volker Dirr]

[...]
- 2 hours continuously (your example) is also acceptable

I fear you didn't understodd 100% my idea.
I can tell the same story with 1 hours continiously:
a third studendt get:
Monday: 1st, 2nd, 3rd, 4th, 5,th, 9th.
so this doesn't fit your request, because there are only 1 hours continiously.

but now lets check the difference.
All 3 students leave home at the same time (because school starts with 1st hour) -> so  no difference.
All will come back home at the same time (because school ends with 9th hour) -> so  no difference.
All work 6 hours per day -> so  no difference.
All have 3 hours continiously a big gap -> so no difference.

so if the students stay at school during the gap, both must wait 3 hours.
if they drive home during the gap: it doens't matter if you have the first or the second timetable. the way to travell is the same.

so you see, it is a "only" a psychological if you think you need to come only for 1 hours.

in my opinion you should think about solving/optimising in an other way, not the way you requested. (but i can't do that without knowing more about your dataset.)

- hour-gap-hour-hour-hour (and students have the tendency to not attend the first one),
- hour-hour-hour-gap-hour (students have the tendency to not attend the last one).

But the students want to learn and want to get a certification at the end, so it is easy in my opinion:
there are at least 3 possibilities:
1. if students miss an hour, then they didn't write such a good test at the end, so they get a bad mark. (if they learn at home and write a good test at the end of semester -> where is the problem?)
2. at end of each hour the teacher just do a small test. if a students is missing, he get 0 point, so a bad mark at the end of semester.
3. if a students is missing 3 or more times a year, then he didn't get a certification about this subject.
(at my university some teacher didn't allowed the students to write the final test if they miss more then 2 times.)

So no [...] trick here, right?

I don't know your dataset, but maybe an easy one it so limit the max hours per day at the same time, because that mean there will not be days with many hours. if there are no days with many hours it also mean the other days of (maybe) more continiusly hours per day (if you still limit the max gaps per day).

I tried to think a way, using dummy activities, but with no results.

Yes, maybe: (it is possible, but maybe - i don't know your dataset - to much work to add this).
count the number of free timeslots of an student.
add as many dummy activities to the student as he has free timesolts.
allow this dummy activities only at special time.
for example if you have 9 hours a day do it like this: (a=allowed, n=not allowed)
annnannna
so you can see he get at least 3 hours continiously.

[Horatiu]

Dear Volker, it seems that my students and your students are not alike. Neither my teachers and your teachers;)

Anyway, thanks for your suggestions, I will toy a bit with dummies and preferred time slots. This might be useful for the next semester, so plenty of time to explore...

Thanks again!

[Horatiu]

Dear Volker, it seems that my students and your students are not alike.

The Romanian students are staging a protest today: by the new Education Law, attending classes is mandatory; they are protesting against that! They do not want to attend classes!

[Volker Dirr]

i don't think our students are much different.

1. [...] if they learn at home and write a good test at the end of semester -> where is the problem?

i don't know about the students protest in romania, but what is the reason for this new law if the old law was good? (maybe i missunderstood, but i thought you was talking about guys that are over 18 years old. i can't belive the law force them to study. if they don't want to study, why don't stay that guys at home or search an other job? Or is that new law for pupil under 18 years? in that case i understand that they must forced to study and i also understand that some kids have psychological problem with that. sadly you must try the dummy trick in that case.)

what is your position? do you like the new law or not? then rethink about your request. does it realy change something?
you said:

students have the tendency to not attend

What is the reason for this? (the guys want to study - but they also have tendency to not attend - that is not very logical - except there are other reasons like: a bad teacher, to full rooms (not seat), ... . But your can't solve that problem with your request. you need an other solution for that problems.)

Maybe they just didn't think logically. Tell them my sample. What did they answer?

Or is there an other reason why the do not attend?

If they do not attend? What is the (your) problem?
Who has tendency to not attend? Does it make sence to care about that? Maybe some poor students are glade if the activity is not so full and the teacher can care more about them. Maybe some poor students miss, but they are anyway not clever enough for the job they are studying for.

Do they still have this unlogical psychological problem "i must come just for an hour?". I know myself there are also some students thinking like that in germany. But answer this question yourself before you spend to much time into something that is not worth to care about, because the problem is maybe at an other site.

[Volker Dirr]

hmmm...

how many hours has a student max per week?
how many hours has he min per week?

There might be an other solution.

If you answer for example he has exact 30 hours per week (5 days week), then you can maybe do this:
- add constraint max gaps per day = 1
- add constraint max hours per day = 6.
- add constraint max hours continiusly = 3.

that mean he will never have 1 or 2 hours continiously, but it also mean he will have one gap per day.


if 2 hours continiously is also ok:
a) if you answer: he has min 25 and max 30 hours:
- add max gaps per day = 1
- add max hours per day = 6
- add constraint max hours continiusly = 4
- add min hours per day = 2

b) if you answert: he has min 30 and max 35 hours:
- add max gaps per day = 1
- add max hours per day = 7
- add constraint max hours continiusly = 4

...

of course that mean he has at least 1 gap per day (except last smaple. it might happen just 2 hours continiously at a single day.)

so what is the min and max number of hours? maybe you can solve it this way easier? But it depend on your dataset.

[Horatiu]

When I say "students", I mean University. So 18+ years old. Let's not go into any further details about this. I studied both in Romania and in Italy, and students' attitude towards learning was totally different. Why...? Well, the answer is a complex one, and we risk going off-topic here. We could chat in private if you want...

On topic now;)

My students have 10-12 "hours" per week (1 "hour" = 90 mins). However, they can have free days.
- max gaps per day = 1
- min hours daily = 2
- max hours daily = ? (no more than 4)
- max hours continuously=?
Is your idea valid for my case?

[Volker Dirr]

How many timeslots (hours) do you have per day? 5 slots?
Is there a student with 11 hours per week?