Hello Livi!
If I leave the decay for a subject 1+1+1, how much weight will the FET change itself to 2+1 and not 3
Hello, ngo,
If it is 1+1+1 and you have a constraint min days between activities for all these 3 subactivities, FET won't put 3 on the same day.
like Liviu said. maybe a bit more detailed:
"weight" is in fact only something like: "how many times should i retry if the constraints break" (see https://www.timetabling.de/manual/FET-manual.en.html#id_35 )
It doesn't mean a percentage value of how often the constraint is broken.
with weight 0% the min n day of an activity with 2 subactivities in a 5 days week is probably broken 1/5=20% broken. While the same 0% min n day of the same activitiy with the same 2 subactivities in a 4 days week is probably broken around 1/4=25%.
If you don't set a constraint, then it is like a 0% constraint.
But a min n day constraint with weight 0% and 3 subactivities 1+1+1 will be never 3 on on a single day, since it will have max 2 subactivities consecutive
Thanks, Volker! But even if not consecutive FET won't place 3 on the same day (since FET-6.4.0).
a subject, for example math, has 3=1+1+1 so how much weight it will change to 3=2+1. I mean for small weights the FET will run faster to 100%
@Liviu
Ok. last sentence is a bit unclear. I was referring to the sentence before: comparing a min n day constraint with 0% with adding no constraint at all.
@ngoctp29121982:
I don't understand you question. A lower weight will be (in normal case) always faster then a higher weight. Since a higher weight mean that FET will try to fix a broken constraint with a higher probability; and fixing a broken constraints needs time.
I mean is there a way to make 3=1+1+1 if you put it out of 50% it will turn into 3=2+1
hmm.. maybe i misunderstand you, but that is the normal min n day constraint.
Please have a look here: (scroll down a bit and look at the pictures.)
https://www.timetabling.de/manual/FET-manual.en.html#id_15