我如何在Cplex脚本(流控制)中添加新的约束到现有模型?

Question

在我的启发式算法中，我用不同的数据多次求解模型，每次求解后，我需要向现有模型添加一些新的约束并再次求解。我想在Cplex脚本(流量控制)中做到这一点。

Answer 1

我不确定这是否是一种最佳的方法来做到这一点(但我没有找到一个更好的)，但如果你的约束有已知的结构，你可以编辑(在运行时)一组边界在一个。dat文件，让CPLEX生成约束。

dvar float k;
{float} bounds = ...; //data.dat: bounds = {4.0};
main {
var model = thisOplModel
model.generate()
cplex.solve()
writeln(cplex.getObjValue())
var model2 = new IloOplModel(model.modelDefinition, cplex)
var data = model.dataElements
data.bounds.add(3.0)
model2.addDataSource(data)
model2.generate()
cplex.solve()
writeln(cplex.getObjValue())
}
maximize k;
subject to {
forall (b in bounds) k <= b;
}

则输出为

4
3

Answer 2

/*通常，在求解模型之后，我们需要添加新的变量和约束。我们可以通过编写新模型来实现，但我们也可以增量地实现。假设我们首先需要知道如果我们需要拥有相同数量的40座和30座公交车，然后在模型中允许50座公交车，会发生什么。我们可以写int nbKids = 300;浮动costBus40 = 500;浮动costBus30 = 400;浮动costBus50 = 700;

dvar int+ nbBus40;
dvar int+ nbBus30;
dvar int+ nbBus50;

minimize
costBus40*nbBus40  +nbBus30*costBus30+nbBus50*costBus50;

subject to
{
ctKids:40*nbBus40+nbBus30*30+nbBus50*50>=nbKids;

nbBus30==nbBus40;
nbBus50==0;
}
execute DISPLAY_After_SOLVE
{
writeln("The minimum cost is ",costBus40*nbBus40  +nbBus30*costBus30+costBus50*nbBus50);
writeln("We will use ",nbBus40," 40 seats buses ",nbBus30," 30 seats buses and ",
nbBus50," buses 50 seats");
}
which will give

The minimum cost is 4500
We will use 5 40 seats buses 5 30 seats buses and 0 buses 50 seats
and then if we comment
nbBus50==0;
which means we allow 50 seats buses then we ll get

The minimum cost is 4100
We will use 3 40 seats buses 3 30 seats buses and 2 buses 50 seats
But to do that we generate 3 times the matrix which can take time.

We can do that in a main and be incremental. And then we do not generate new matrixes but rely on incremental changes.
The technique is simply to have spare decision variables and constraints:
*/

int nbKids=300;
float costBus40=500;
float costBus30=400;

dvar int+ nbBus40;
dvar int+ nbBus30;
dvar int+ emptyVar;

minimize
costBus40*nbBus40  +nbBus30*costBus30;

subject to
{
ctKids:40*nbBus40+nbBus30*30>=nbKids;

ctEmpty:0<=0;
}
execute DISPLAY_After_SOLVE
{
writeln("The minimum cost is ",costBus40*nbBus40  +nbBus30*costBus30);
writeln("We will use ",nbBus40," 40 seats buses and ",nbBus30," 30 seats buses ");
writeln();
}
main
{
thisOplModel.generate();
writeln("Basic model");
cplex.solve();
thisOplModel.postProcess();
writeln("Let us add a row : saying that nbBus40 and nbBus30 should be equal");
thisOplModel.ctEmpty.setCoef(thisOplModel.nbBus40,1);
thisOplModel.ctEmpty.setCoef(thisOplModel.nbBus30,-1);
thisOplModel.ctEmpty.setBounds(0,0);
cplex.solve();
thisOplModel.postProcess();
writeln("Let us add a column : saying that nbBus50 could also be used and their cost is 700");
cplex.setObjCoef(thisOplModel.emptyVar,700);
thisOplModel.ctKids.setCoef(thisOplModel.emptyVar,50);
cplex.solve();
writeln("The minimum cost is ",
thisOplModel.costBus40*thisOplModel.nbBus40.solutionValue  +thisOplModel.nbBus30.solutionValue*thisOplModel.costBus30
+700*thisOplModel.emptyVar.solutionValue);
writeln("We will use ",thisOplModel.nbBus40.solutionValue," 40 seats buses ",thisOplModel.nbBus30.solutionValue,
" 30 seats buses and "+thisOplModel.emptyVar.solutionValue," 50 seats buses");
}
/*
gives

Basic model
The minimum cost is 3800
We will use 6 40 seats buses and 2 30 seats buses
Let us add a row : saying that nbBus40 and nbBus30 should be equal
The minimum cost is 4500
We will use 5 40 seats buses and 5 30 seats buses
Let us add a column : saying that nbBus50 could also be used and their cost is 700
The minimum cost is 4100
We will use 3 40 seats buses 3 30 seats buses and 2 50 seats buses
Of course we get the same results. 
*/

从https://www.linkedin.com/pulse/making-decision-optimization-simple-alex-fleischer/

增加行和列