if condition in ampl

Question

I am new in ampl and I want to use if condition in ampl with the following information:

I have a binary variable X[p,r], where {p in P, r in R}. Now I want to make a new constraint such that the variable R[p,r] is used where X[p,r]=0. I do not know how I can write it or even if the ampl can handle it or not, I tried the following constraint but they did not work:

s.t. a1{r in R, p in P and X[p,r]=0}: 
s.t. a2{r in R p in P and X[p,r]=0};
s.t. a2{r in R ,p in P, and X[p,r]=0};
s.t. a2{r in R, p in P: and X[p,r]=0};

Answer 1

You cannot include a decision variable in the "for all" part of the constraint (in AMPL, the part inside the {...}). Instead, you need build into the constraint itself the logic that says the constraint is only active if X[p,r] = 0. The way to do that depends on the type of constraint: >=, =, or <=. I'll write each case separately, and I'll do it in a generic way instead of specific to your problem.

In the explanation below, I assume that the constraint is written as

a[1]y[1] + ... + a[n]y[n] >=/=/<= b, 

where a[i] and b are constants and y[i] are decision variables. I also assume we want the constraint to hold if x = 0, where x is a binary decision variable, and we don't care whether the constraint holds if x = 1.

Let M be a new parameter (constant) that equals a large number.

Greater-than-or-equal-to constraints:

The constraint is a[1]y[1] + ... + a[n]y[n] >= b. Rewrite it as

a[1]y[1] + ... + a[n]y[n] >= b - Mx.

Then, if x = 0, the constraint holds, and if x = 1, it has no effect since the right-hand side is very negative.

(If all of the a[i] are nonnegative, you can instead use

a[1]y[1] + ... + a[n]y[n] >= bx,

which is tighter.)

Less-than-or-equal-to constraints:

The constraint is a[1]y[1] + ... + a[n]y[n] <= b. Rewrite it as

a[1]y[1] + ... + a[n]y[n] <= b + Mx.

Then, if x = 0, the constraint holds, and if x = 1, it has no effect since the RHS is very large.

Equality constraints:

The constraint is a[1]y[1] + ... + a[n]y[n] = b. Rewrite it as

a[1]y[1] + ... + a[n]y[n] <= b + Mx
a[1]y[1] + ... + a[n]y[n] >= b - Mx.

Then, if x = 0, the equality constraint holds, and if x = 1, the constraints have no effect.

Note: If your model is relatively large, i.e., it takes a non-negligible amount of time to solve, then you need to be careful with big-M-type formulations. In particular, you want M to be as small as possible while still enforcing the logic of the constraints above.