Maliheh Aramon

Hi Maik,  You can define the decision variables as binary variables \(x_{ij} \) being equal to 1 if order \(i\) is delivered by product \(j\) and to 0 otherwise. Let us further define \(r_i\) to re...

Hi Andres,  The Gurobi academic license does not have any restriction. It is the same full-featured version offered to commercial users. The amount of memory that you can access depends on the hard...

Hi Thalles, I think you are mapping the blocks and the years to the items and knapsacks, respectively. In the solution you shared, we have x[1, 2] = 1 and x[5, 2] = 1 meaning that both items 1 and ...

Hi Thalles, I do not see in your code snippet how you are defining \(\texttt{items}\) and \(\texttt{knapsacks}\) variables. I am assuming that they are lists of item and knapsack strings.We need to...

Hi Arthur, The first inconsistency is that all runs except one ran out of memory. I suspect the concurrent LP optimization at the root node caused a memory spike, as the logs show slight difference...

For the model.update() to discard the lazy constraints, you would need to modify the model. You can consider adding a dummy variable, for example, and then call the update method. I also tested you...

Please see the last response to this post. To discard all lazy constraints, you can call model.reset() or make a modification to the model, for example, by adding a dummy variable. However, you can...

Hi Jose, Yes, you are right. If you modify the model, for example, by adding a new constraint or variable, or by changing coefficients, the lazy constraints generated in the callback are discarded ...

Sorry for the confusion. I was not sure why you needed to call the optimize() method again, and was just trying to make sure you meant calling it on the same model object again.

If the model object remains the same and is not disposed of, yes the lazy constraints persist if you recall model.optimize(cb). As an example, if you initially call model.optimize(cb) with TimeLimi...