Originally posted by: Sultan_Niz7755

Hello, everybody. At the moment I am dealing with the topic of planning patients into surgery. I have to apply Column Genearation and implement it in OPL. In this respect I had two questions. I have successfully written original problem in opl, but I have some difficulties in writing the master and sub problem. the questions are as follows:
1. let Ω be the set of all feasible columns. How can I write this in OPL?

2. i have the problem with inserting the dual variables in the sub problem. like this:

Let πi , λt , βt , and γt be the optimal dual variables to an RMP, associated respectively with constraints (10), (11) and (12).

I am very thankful for any feedback.

Regards, Sherzod

Here is a very simple .mod file (adapted from the opl/examples/opl_interfaces/java/javaknapsack/cutstock_ext_main.mod example in the distribution) that shows how you can add variables to a model:

// This tuple represents a variable that can be added.
// The variable has an index, an objective coefficient and a coefficient
// for the one constraint that our model has. If the model had more
// constraints in which the variable could appear, then we would of course
// need more constraints (either add more fields or change the type of the
// field to an array of floats).
tuple T {
  key int   idx;
      float obj;   // Objective coefficient of variable
      float coef;  // Coefficient of variable in constraint
}

// Each element in this set defines a variable.
{T} columns = ...;

// The variables.
dvar boolean x[columns];

maximize sum(j in columns) j.obj * x[j];

subject to {
  // A simple constraints: Pick at most 5 variables.
  sum(j in columns) j.coef * x[j] <= 5;
}

main {
  // Generate the initial model and extract required data from it.
  var masterOpl = thisOplModel;
  masterOpl.generate();
  var masterDef = masterOpl.modelDefinition;
  var masterCplex = cplex;
  var masterData = masterOpl.dataElements;

  for (var iter = 1; iter <= 9; ++iter) {
    // Solve a model with the current set of columns.
    sub = new IloOplModel(masterDef,masterCplex);
    sub.addDataSource(masterData);
    sub.generate();
    masterCplex.solve();
    writeln("## Iteration " + iter + ": " + masterCplex.getObjValue());
    for (var j in masterData.columns) {
       writeln("x[" + j.idx + "] = " + sub.x[j]);
    }

    // TODO: Find a column to add.

    // Add the new column to the master data.
    // Here we simply add a new column with objective 'iter' and coefficient
    // 1.0 in our constraint.
    masterData.columns.add(masterData.columns.size, // idx
                           iter, // obj
                           1.0   // coef
                           );

    sub.end();
  }

  0;
}

In addition to the .mod you also need this .dat:

columns = { };

Originally posted by: Sultan_Niz7755

Hi, Daniel,

Thank you so much for your reply. I'm actually new to Cplex, I'm not really that pure. I understood Column Generation to mean that I first have to create compact model, then master and subproblem.  And then to connect all the models together, I need the Main function, right?  The column-oriented formulation of the model confuses me all the time. In the Cutting Stock example, only a dual variable pi was inserted, but in my case I have to define that the above 4 variables are optimal dual variables for my restricted master problem. These optimal dual variables I need to apply to my pricing problem to have the solution to the subproblem. Hier ist my OPL Code for Master Problem:

 int  N = ...;// Number of Patients
  int j = ...; // Number of tasks
  range Patient = 1..N;
  range Task = 1..j;
  int P = ...; // Number of Porters
  range Porter = 1..P;
  int R = ...; // Number of Operating rooms
  range Room = 1..R;
  int B = ...; // Number of recovery beds
  range Bed = 1..B;
  int Period = ...;
  range H = 0..Period-1; // discrete Time horizon of H Periods, The discrete Time unit is set equal to 10 Minutes
  int A = ...;
  range K = 1..A;
   
  
  int p[Patient, Task] = ...;
  dvar float+ c[K];
  dvar boolean x[Patient, K];//1,if column k is related to patient i
  dvar int+ s[K,Task];//starting time of the task j according to column k
  dvar int+ y[K];//let y(k) for k€K be a binary decision variable that takes 1 if the feasible column k is selected and 0 otherwise
  dvar boolean u[H,K];
  dvar boolean v[H,K];
  dvar boolean w[H,K];
  
  minimize sum (k in K)c[k]*y[k];
  subject to  {
  forall (k in K) c[k] == s[k,4] + (sum (i in Patient) x[i,k]*p[i,4]);
  forall (i in Patient, k in K, t in H) u[t,k] == (((s[k, 1] <= t)  && (s[k, 1] + p[i, 1] - 1) >= t) + ((s[k, 4] <= t)  && (s[k, 4] + p[i, 4] - 1) >= t));
  forall (i in Patient, k in K, t in H) v[t,k] == ((s[k, 2] <= t)  && (s[k, 2] + p[i, 2] + p[i,5] - 1) >= t);
  forall (i in Patient, k in K, t in H) u[t,k] == ((s[k, 4] <= t)  && (s[k, 4] + p[i, 4] - 1) >= t);
  
  forall (i in Patient) sum (k in K) (x[i,k]*y[k]) == 1;
  forall (t in H, pi in Porter) sum (k in K) u[t,k]*y[k] <= pi;
  forall (t in H, r in Room) sum (k in K) v[t,k]*y[k] <= r;
  forall (t in H, b in Bed) sum (k in K) w[t,k]*y[k] <= b;
  
    
  }
  

and here pricing problem

thank you again,

regards Sherzod

Originally posted by: Sultan_Niz7755

hier is pricing problem

Sorry, I cannot dig into your particular model in detail. But I think there is some confusion about the cutstock example. Maybe I did not point to the best variant of that. You could take a look at

opl/examples/opl/cutstock/cutstock_main.mod

That is similar but a bit simpler. The dual values are used for communicating with the sub problem. However, the columns that are added to the master are added implicitly via this statement:

masterData.Patterns.add(masterData.Patterns.size,1,subOpl.Use.solutionValue);

This adds a new elements to the set

{pattern} Patterns = ...;

and since the variables are defined as

dvar float Cut[Patterns] in 0..1000000;

this results in one more variable if we regerenate the model from scratch.