Originally posted by: Filipe Costa

Hi all,

What is the best way to implement the logic behind the alternative opl function without the absent restriction: "If a is absent, then all b intervals are absent."

Regards,

Hi,

why not having interval a not optional. Then as you said "If a is absent, then all b intervals are absent."  will never have any impact

regards

Originally posted by: Filipe Costa

Hi Alex,

Interval a needs to be optional in my case.

Regards

Hi,

if the alternative is not what you need, you can write your own constraint with the presenceOf values of all needed intervals.

(presenceOf(A)==1) => ( presenceOf(B)+presenceOf(C)<=1);

for instance

regards

Originally posted by: PhilippeLaborie

Hi Filipe, could you formalise the constraint you would like to model?

If master interval 'a' is absent, what do you want the 'b_i' intervals to be? Completely unconstrained? Or like in the alternative: one and only one must be present? Or maybe something else ...

Originally posted by: Filipe Costa

Hi Philippe,

So my master interval A is the following:

dvar interval transformation[o in Ops] optional size ftoi(ceil(operations[o].order.volume/operations[o].machine.production_rate));

operations is an array of tuples that represents all possible combinations between Orders, Raw materials and Machines:

tuple Operations{
    Order order;
    Machine machine;
    Raw_material raw material;
}

Then I have the following constraint: Each order must be transformed once.

I am representing this constraint like this:

forall(m in orders){
        sum(o in Ops: operations[o].order==m) presenceOf(transformation[o])==1; // Guarantee the transformation of each mill order
    }

But I think this constraint does not propagate well.

Any ideas to represent this constraint in other way?

Regards,

Originally posted by: PhilippeLaborie

Why don't you have an interval variable (present) that represents the interval of time an order is produced (no matter on which machine and with which material).

Say:

dvar interval order[m in orders] ...;

And then some alternative enforcing that one and only one pair machine/raw material must be selected to perform this order, something like:

forall(m in orders){
    alternative(order[m],  all(o in Ops: operations[o].order==m) transformation[o]);
}

This will ensure that one and only one pair machine/raw material is selected to perform this order m.

And you can use interval variables  order[m] to post other constraints of your model (ex: release dates, precedence constraints between orders if any, tardiness costs, etc.)

If there is no dependence between the allocated machines and the raw materials, you can even split the alternative as two alternatives, one on interval variables that represent the allocation of orders to machines. Another the allocation of orders to raw materials.