Question

我是古罗比的新朋友。我写了下面的代码。我保存在Sample.lp中。以

运行

gurobi_cl sample.lp

Maximize
 x1 +  x2 +  x3 + 1 x4 + 1 x5

Subject To
c1: 3 x1 + 5 x2 + 2 x3 + 5 x4 + 7 x5 - 28 <= 0
c2: 2 x1 + 0 x2 + 0 x3 + 8 x4 - 14 <= 0
c3: 4 x4 + 5 x5 - 22 <= 0
c4: 3 x2 - 2 <= 0
D1: 3 x4 -1 >= 0

Bounds
x1 <= 1
x2 <= 1
x3 <= 1
x4 <= 1
x5 <= 1

Integers
x1 x2 x3 x4 x5
End

似乎我没有得到正确的解决方案。你能帮我么。 x的值为二进制。

修改后，我得到了正确的解决方案。如果没有目标函数，我的兴趣是获得通用的二进制解。如果有很多这样的解决方案，我想得到很少（例如1000）。如何解决呢？为此，写

 Maximize
  1

我遇到错误：

Solution count 0

Model is infeasible or unbounded
Best objective -, best bound -, gap -

Answer 1

您的lp文件左侧有一个常量，古罗比将这些常量解释为变量。

 >>> m.read("sample.lp")
    Read LP format model from file sample.lp
Reading time = 0.00 seconds
: 5 rows, 10 columns, 16 nonzeros

>>> m.getVars()                                                                                                                                                                                                          
[<gurobi.Var x1>,
 <gurobi.Var x2>,
 <gurobi.Var x3>,
 <gurobi.Var x4>,
 <gurobi.Var x5>,
 <gurobi.Var 28>,
 <gurobi.Var 14>,
 <gurobi.Var 22>,
 <gurobi.Var 2>,
 <gurobi.Var 1>]

要解决此问题，请将常量向右移动。

    Maximize
 x1 +  x2 +  x3 + 1 x4 + 1 x5

Subject To
c1: 3 x1 + 5 x2 + 2 x3 + 5 x4 + 7 x5 <= 28
c2: 2 x1 + 0 x2 + 0 x3 + 8 x4 <= 14
c3: 4 x4 + 5 x5 <= 22
c4: 3 x2 <= 2
D1: 3 x4 >= 1

Bounds
x1 <= 1
x2 <= 1
x3 <= 1
x4 <= 1
x5 <= 1

Integers
x1 x2 x3 x4 x5
End

这将给出预期的答案。

>>> m.read("sample.lp")
        Read LP format model from file sample.lp
    Reading time = 0.00 seconds
    : 5 rows, 5 columns, 11 nonzeros
>>> m.optimize()
   Optimal solution found (tolerance 1.00e-04)
   Best objective 4.000000000000e+00, best bound 4.000000000000e+00, gap 0.0000%

>>> m.getVars()
   [<gurobi.Var x1 (value 1.0)>,
 <gurobi.Var x2 (value 0.0)>,
 <gurobi.Var x3 (value 1.0)>,
 <gurobi.Var x4 (value 1.0)>,
 <gurobi.Var x5 (value 1.0)>]

如何使用Gurobi解决优化问题

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