Originally posted by: appleme

 Hi,all.

tuple  JUP
  {   int index;
    string p;
    string u;      
  } ; 
 
 {JUP} jup=...;

 
  dvar int+ jupz[jup] in 0..1;

I want to get the count from  jupz where  jup.p="T" and jupz[jup]==1 ,How to realize?I found there is no count function to do this.

Could you help me

Thanks.

Originally posted by: ChrisBr

Hello,

You could try this:
  c == count(all(j in jup : j.p=="T") jupz[j], 1);

I suggest you have a look at the documentation
IDE and OPL > Optimization Programming Language (OPL) > Language Quick Reference > OPL functions > count

I hope this helps,

Chris.
 

Originally posted by: appleme

Thanks for your help.

I tested,It works well.

Originally posted by: appleme

 I found a question.

c == count(all(j in jup : j.p=="T") jupz[j], 1);

this line is a constraint,It is not a equation.

could this function "count(all(j in jup : j.p=="T") jupz[j], 1)" set to a int variable or as condition in if ? 

I found

if(count(all(j in jup : j.p=="T") jupz[j], 1)<=5) or c=count(all(j in jup : j.p=="T") jupz[j], 1);

it doesn't work.

Originally posted by: ChrisBr

Hello,

Since the result of "count(...)" depends on decision variables, it is not possible to set it to an int as long as the values of the concerned decision variables are not known. But this is possible after the resolution when the decision variables are fixed.

You can set it to a "dexpr int" or define a "dvar int" and constrain it to be equal to the result of "count(...)".

To sum up, all the following statements are allowed:

dexpr int c1 = count(all(j in jup : j.p=="T") jupz[j], 1);
dvar int c2 in 0..card(jup);
subject to {
  c2 == count(all(j in jup : j.p=="T") jupz[j], 1);
}
int c3 = count(all(j in jup : j.p=="T") jupz[j], 1);

About your second question: it is mentioned in documentation that
        Conditions in if-else statements must be ground; that is, they must not contain decision variables.

So you cannot write:

 if(count(all(j in jup : j.p=="T") jupz[j], 1)<=5) ...

But you can use implications of constraints instead, for example:

((count(all(j in jup : j.p=="T") jupz[j], 1)) <=5) => (...);

or conditional expression, for example:

  k == (((count(all(j in jup : j.p=="T") jupz[j], 1)) <=5) ? 1000 : -1);

I hope this clarifies,

Chris.