How is a conditional summation possible in Cplex?

问题:

I want to sum all used resources among times in my model (it's rcpsp model) how can I do it in CPLEX? at first I wrote this:

forall(k in K)
  forall(t in 1..f[nAct])
    sum(i in I:f[i]-d[i]<=t-1 && t<=f[i]) r[i,k] <= aR[k]; 

(note: K is a range for resources, nAct is number of activities, f[i] is an array dvar and indicates finishing time of activity i, d[i] is duration of i,r[i,k] is required resource of k for activity i and aR[k] is available resources of k.)

The problem is that the cplex doesn't accept decision variable in sum's condition. I changed it to:

forall(k in K)
  forall(t in 1..f[nAct])
    sum(i in I) (f[i]-d[i]<=t-1 && t<=f[i])*r[i,k] <= aR[k]; 

But it didn't work. it made True constraints in Problem Browser after run(I don't know why) and it made this constraint ineffective.

Any Idea how to fix it?

回答1:

There are several ways to put your problem into an integer programming framework. There are books written on this subject. I think this is the simplest formulation.

I assume that in your problem, r[i,k] and d[i] are known and that the time horizon is broken into discrete time periods.

• on[i,t] indicator that activity i is active at time t
• start[i,t] indicator that activity i starts at the start of period t
• end[i,t] indicator that activity i finishes at the end of period t

So in[i,t] replaces the condition f[i]-d[i]<=t-1 && t<=f[i])*r[i,k] Your constraint becomes

forall(k in K)
   forall(t in 1..f[nAct])
      sum(i in I : r[i,k] = 1) on[i,t] <= aR[k]; 

You also need to add constraints to enforce the definition of on, start and off.

   forall(t in 2..f[nAct])
      forall(i in I)
         on[i,t-1] - on[i,t] = end[i,t-1] - start[i,t];

   forall(i in I)
      on[i,0] = start[i,0];

   forall(i in I)
      sum(t in 1..f[nAct]) start[i,t] = 1;
   forall(i in I)
      sum(t in 1..f[nAct]) end[i,t] = 1;
   forall(i in I)
      sum(t in 1..f[nAct]) on[i,t] = d[i];

回答2:

You can use dexpr for manipulating decision variables. Here is an example from the same resource IBM Knowledge Center.

Without dexpr

dvar int x in 0..20;
dvar int y in 0..20;
dvar int d;
dvar int s;
maximize (d);
subject to {
  d==x-y;
  s==x+y;
  s<=15;
  s<=x-2*y;
  d>=2;
  d<=y+8;
  1<=d;
}

With dexpr

dvar int x in 0..20;
dvar int y in 0..20;
dexpr int d=x-y;
dexpr int s=x+y;
maximize (d);
subject to {
  s<=15;
  s<=x-2*y;
  d>=2;
  d<=y+8;
  1<=d;
}