I need to find whether an integer variable is an odd or even number. The problem is as follows.

For a set i /1:5/;
Integer variable x(i);
Variable a(i), OF;

if x(i) is odd, then a(i) => 10;
if x(i) is even, then a(i) <= 10;

x.lo(i) = 0;
x.up(i) = 10;

I need to minimize OF = sum(i,a(i));

I tried this but I am unable to figure it out. Since we can’t use variables in conditional (logical) statements using $. I don’t know how to do it. COuld anyone please help me with this?

Set i 'buses' /1:5/;
Parameters
b value /10/;
display b;
Integer Variables x;
Variable a, OF;
Equation cons1, cons2;
cons1(i).. a(i)$(mod(x(i),2)=0) =g= b;
cons2(i).. a(i)$(mod(x(i),2)=1) =l= b;
Equation obj;
obj.. OF =g= sum((i), a(i));
x.up(i) = 10;
x.lo(i) = 0;
Model OddEven / all /;
OPTION MINLP = BARON;
solve OddEven using MINLP minimizing OF;
display OF.l, x.l, a.l;

The trick is to represent x as 2*y + z where y is another integer variable and z a binary. x can even be continuous but will automatically assume integer values. z=1 tells you that x is odd, z=0 tell you that x is even. You can use the binary variable z in some logic constraints to enforce your other constraint. I used a bigM formulation. You find many other formulation, just search the forum for logical constraint.

Here is the full model (I had to correct a few syntactical issues):

Set i 'buses' /1*5/;
Parameters
b value /10/;
display b;
Integer Variables y;
binary variable z;
Variable x, a, OF;
Equation defoddeven, cons1, cons2;
defoddeven(i).. x(i) =e= 2*y(i) + z(i);
cons1(i).. a(i) =g= b - (1-z(i))*100;
cons2(i).. a(i) =l= b + z(i)*100;
Equation obj;
obj.. OF =e= sum((i), a(i));
x.up(i) = 10;
x.lo(i) = 0;
a.lo(i) = -10;
Model OddEven / all /;
solve OddEven using MIP minimizing OF;
display OF.l, x.l, a.l;