Originally posted by: Janisch

Hello,

I want to tie a condition to the sum. And that should only be summed if the end times of both stations are the same, the product instances differ and each overload is greater than zero. Unfortunately, an error appears on the says that I can not use intervall variables. how can I fix the mistake?

definitions:

• dvar interval station[1..N][s in 1..S][g in 1..G]
• dexpr int overload [i in 1..N][s in 1..S][g in 1..G]
• int binaerueberlast[1..N][s in 1..S][g in 1..G]

dexpr int wieoftgleichzeitig[s in 1..S][g in 1..G]= sum(i in 1..N,f in 1..N: endOf(station[i][s][g])==endOf(station[f][s][g])&&i!=f&&overload[i][s][g]>0&&overload[f][s][g]>0)binaerueberlast[i][s][g];

regards

Hi

the condition you use in the slicing should be bound

but it is not!

what you could try is to change

sum(I in set : condition) I

into 

sum(I in set) (condition)*I

regards

Originally posted by: Janisch

Thank you.
But now another problem arises:

"Decision variable (or expression)" station "not allowed"

dexpr int wieoftgleichzeitig[s in 1..S][g in 1..G]= sum(i in 1..N,f in 1..N: (endOf(station[i][s][g])==endOf(station[f][s][g])&&i!=f&&overload[i][s][g]>0&&overload[f][s][g]>0))binaerueberlast[i][s][g];

How can I fix this?

regards

Originally posted by: Petr Vilím

Hello,

what Alex meant is that decision variables cannot be used in the sum condition (they are not bound at the time the sum expression is created). If you want to exclude some operands from the sum depending on the values of decision variables then you have to modify the operand to be 0 when necessary.

In your case only condition i != f does not depend on decision variables so it could be used as an argument to the sum. So your expression will start by:

dexpr int wieoftgleichzeitig[s in 1..S][g in 1..G]= sum(i in 1..N,f in 1..N: i!=f) 

All the remaining conditions must be moved into the sum operands. You can use boolean expressions as 0/1 expressions. And using multiplication the result will be 0 whenever the codition is false. So the rest of the expression:

((endOf(station[i][s][g])==endOf(station[f][s][g]) && overload[i][s][g]>0 && overload[f][s][g]>0) * binaerueberlast[i][s][g])

Note that in general long expressions like this do not propagate very well though.

Petr