Question

我正在使用Coin-Or的rehearse来实现线性编程。

我需要模约束。例如：x应该是3的倍数。

OsiCbcSolverInterface solver;
CelModel model(solver);
CelNumVar x;
CelIntVar z;

unsigned int mod = 3;

// Maximize

solver.setObjSense(-1.0);

model.setObjective(x);

model.addConstraint(x <= 7.5);

// The modulo constraint:

model.addConstraint(x == z * mod);

x的结果应为6。但是，z设置为2.5，这在我声明为CellIntVar时是不可能的。

如何强制z为整数？

Answer 1

我从没有使用过那个lib，但是我认为您应该遵循tests。

核心信息来自自述文件：

如果希望某些变量为整数，请使用CelIntVar而不是CelNumVar。 您还必须将求解器绑定到Integer Linear Programming求解器，例如Coin-cbc。

查看Rehearse/tests/testRehearse.cpp -> exemple4()（此处显示：代码不完整；没有复制粘贴）：

OsiClpSolverInterface *solver = new OsiClpSolverInterface();

CelModel model(*solver);

...
CelIntVar x1("x1");
...
solver->initialSolve();       // this is the relaxation (and maybe presolving)!
...
CbcModel cbcModel(*solver);   // MIP-solver
cbcModel.branchAndBound();    // Use MIP-solver

printf("Solution for x1 : %g\n", model.getSolutionValue(x1, *cbcModel.solver()));
printf("Solution objvalue = : %g\n", cbcModel.solver()->getObjValue());

这种用法（使用Osi获取LP-solver；在该Osi提供的LP-solver之上构建MIP-solver并调用brandAndBound）基本上遵循Cbc的内部接口（使用python的cylp）类似）。

仅供参考：这是来自here的CoinOR Cbc官方示例（无需排练）：

// Copyright (C) 2005, International Business Machines
// Corporation and others.  All Rights Reserved.

#include "CbcModel.hpp"

// Using CLP as the solver
#include "OsiClpSolverInterface.hpp"

int main (int argc, const char *argv[])
{
  OsiClpSolverInterface solver1;

  // Read in example model in MPS file format
  // and assert that it is a clean model
  int numMpsReadErrors = solver1.readMps("../../Mps/Sample/p0033.mps","");
  assert(numMpsReadErrors==0);

  // Pass the solver with the problem to be solved to CbcModel 
  CbcModel model(solver1);

  // Do complete search
  model.branchAndBound();

  /* Print the solution.  CbcModel clones the solver so we
     need to get current copy from the CbcModel */
  int numberColumns = model.solver()->getNumCols();

  const double * solution = model.bestSolution();

  for (int iColumn=0;iColumn<numberColumns;iColumn++) {
    double value=solution[iColumn];
    if (fabs(value)>1.0e-7&&model.solver()->isInteger(iColumn)) 
      printf("%d has value %g\n",iColumn,value);
   }
  return 0;
}

线性规划：模约束

1 个答案: