Nested set in OPL/CPLEX

Question

I'm trying to solve a scheduling problem using OPL/CPLEX. In the model, there are nine tasks in which each task has its specific precedence. the precedence relationship is as follows:

[{1}, {1}, {1,2}, {1,2,3}, {4,7}, {6}, {6}, {5}, {8}];​

Where, for example, tasks {1,2,3} are the precedence of task four. I have defined two ways to create this relationship. First, by using a nested tuple set:

tuple link{
   {int} task;
};
{link} relations[tasks]=
[{<{1}>}, {<{1}>}, {<{1,2}>}, {<{1,2,3}>}, {<{4,7}>}, {<{6}>}, {<{6}>}, {<{5}>}, {<{8}>}];​

Second, by nested set:

setof(int) operations[tasks] = [{1}, {1}, {1,2}, {1,2,3}, {4,7}, {6}, {6}, {5}, {8}];​

Now, I would like to use this relationship in the following constraint:

\begin{equation}\sum_{h \in H} \sum_{i \in I} i . x_{i h j} \leq \sum_{h \in H} \sum_{i \in I} i . x_{i h t} \forall j, t \in N, j<t\end{equation}​ Where, ${i \in workstation}$, ${h \in workers}$ and ${j,t \in tasks}$.

The OPL code is:

  e4: forall(j,t in tasks: j<t)
    sum(h in workers, i in workstation) i*x[i,h,j] <= 
        sum(h in workers, i in workstation) i*x[i,h,t];

I'm wondering if, which one of these methods can be applied to the constraint and how can I fix this issue?

Answer 1

Why not writing

{int} tasks=asSet(1..9);
{int} workstation={1,2};
{int} workers={1,2};
setof(int) operations[tasks]=[{1},{1}, {1,2}, {1,2,3}, {4,7}, {6}, {6}, {5}, {8}];

dvar boolean x[workers,tasks,tasks];

subject to
{
  e4: forall(j,t in tasks: j in operations[t])
sum(h in workers, i in workstation)i*x[i,h,j] <= 
sum(h in workers,i in workstation)i*x[i,h,t];}

and later on you asked for the same with tuple:

{int} tasks=asSet(1..9);
{int} workstation={1,2};
{int} workers={1,2};
//setof(int) operations[tasks]=[{1},{1}, {1,2}, {1,2,3}, {4,7}, {6}, {6}, {5}, {8}];
tuple link{
   {int} task;
};
link relations[tasks]=
[<{1}>,<{1}>, <{1,2}>, <{1,2,3}>, <{4,7}>, <{6}>, <{6}>, <{5}>, <{8}>];;
dvar boolean x[workers,tasks,tasks];

subject to
{
  e4: forall(j,t in tasks: j in relations[t].task)
sum(h in workers, i in workstation)i*x[i,h,j] <= 
sum(h in workers,i in workstation)i*x[i,h,t];}