Cplex c ++多维决策变量 [英] Cplex c++ multidimensional decision variable

问题描述

我是新的使用cplex，我尝试在互联网上找到一些信息，但没有找到清楚的东西，以帮助我的问题。

我有P [k] k将等于1到4并且我有一个决策变量x [i] [k]必须等于0或1（也是p [ k]）

i在1到5之间

  IloEnv env; 
 IloModel model（env）; 
 IloNumVarArray p（env）; 
 p.add（IloNumVar（env，0，1））; 
 p.add（IloNumVar（env，0，1））; 
 p.add（IloNumVar（env，0，1））; 
 IloIntVar x（env，0，1）; 
 
 model.add（IloMaximize（env，1000 * p [1] + 2000 * p [2] + 500 * p [3] + 1500 * p [4] 
 
 for（int k = 1; k <= 4; k ++）{
 for（int i = 1; i <= 5; i ++）{
 model。 x [i] [k] + x [i] [k] => 2 * p [k] + x [i] [k] ;）; 
}} 
  

循环应该这样：

x [1] [1] + x [2] [1] + x [3] [1] + x [4] [1] + x [5] * p [1];

x [1] [2] + x [2] [2] + x [3] [2] + x [4] 2] + x [5] [2] => 2 * p [2];

x [1] x [3] [3] + x [4] [3] + x [5] [3] => 2 * p [3];

] [4] + x [2] [4] + x [3] [4] + x [4] [4] + x [5] [4]

有没有人有想法？

$ b

但是我离这个结果很远。 b $ b

感谢

解决方案

您可能想要使用IloNumExpr

  for（int k = 0; k <4; k ++）{
 IloNumExpr sum_over_i（env） 
 for（int i = 0; i <5; i ++）{
 sum_over_i + = x [i] [k]; 
} 
 model.add（sum_over_i> = 2 * p [k];）; 
} 
  

您还需要将x声明为二维数组。

  IloArray x（env，4）; 
 for（int k = 0; k <4; ++ k）
 x [k] = IloIntVarArray（env，5，0，1） 
  

此外，在c ++中，数组索引从0到大小-1，而不是1到大小。您的目标应该是

  model.add（IloMaximize（env，1000 * p [0] + 2000 * p [1] + 500 * p [2] + 1500 * p [3]））; 
  

I'm new using cplex and I try to find some information on internet but didn't find clear stuff to help me in my problem.

I have P[k] k will be equal to 1 to 4

and I have a decision variable x[i][k] must be equal to 0 or 1 (also p[k])

the i is between 1 to 5

For now I do like this

  IloEnv env;
  IloModel model(env);
  IloNumVarArray p(env);
  p.add(IloNumVar(env, 0, 1));
  p.add(IloNumVar(env, 0, 1));
  p.add(IloNumVar(env, 0, 1));
  IloIntVar x(env, 0, 1);

  model.add(IloMaximize(env, 1000 * p[1] + 2000 * p[2] + 500 * p[3] + 1500 * p[4]));

   for(int k = 1; k <= 4; k++){
    for(int i = 1; i <= 5; i++){
      model.add(x[i][k] + x[i][k] + x[i][k] + x[i][k] + x[i][k] => 2 * p[k]; );
    }}

The loop should do something like this:

x[1][1] + x[2][1] + x[3][1] + x[4][1] + x[5][1] => 2 * p[1];

x[1][2] + x[2][2] + x[3][2] + x[4][2] + x[5][2] => 2 * p[2];

x[1][3] + x[2][3] + x[3][3] + x[4][3] + x[5][3] => 2 * p[3];

x[1][4] + x[2][4] + x[3][4] + x[4][4] + x[5][4] => 3 * p[4];

but I'm far away from this result.

Does anyone have an idea?

Thanks

解决方案

You probably want to use an IloNumExpr

for(int k = 0; k < 4; k++){
   IloNumExpr sum_over_i(env);
   for(int i = 0; i < 5; i++){
        sum_over_i += x[i][k];
   }
   model.add(sum_over_i >= 2 * p[k]; );
}

You also need to declare x as a 2-dimensional array.

IloArray x(env, 4);
for (int k = 0; k < 4; ++k)
      x[k] = IloIntVarArray(env, 5, 0, 1);

Also, in c++, array indices are from 0 to size-1, not 1 to size. Your objective should be written

model.add(IloMaximize(env, 1000 * p[0] + 2000 * p[1] + 500 * p[2] + 1500 * p[3]));

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