PuLPs

Question

我正在尝试用puLP（Python）来解决MILP问题，并不断出现以下错误：

Traceback (most recent call last):
  File "main_lp.py", line 63, in <module>
    ans = solve_lp(C)
  File "/home/ashwin/Documents/Williams/f2014/math317_or/project/solve_lp.py", line 36, in solve_lp
    prob.solve()
  File "/usr/local/lib/python2.7/dist-packages/PuLP-1.5.6-py2.7.egg/pulp/pulp.py", line 1619, in solve
    status = solver.actualSolve(self, **kwargs)
  File "/usr/local/lib/python2.7/dist-packages/PuLP-1.5.6-py2.7.egg/pulp/solvers.py", line 1283, in actualSolve
    return self.solve_CBC(lp, **kwargs)
  File "/usr/local/lib/python2.7/dist-packages/PuLP-1.5.6-py2.7.egg/pulp/solvers.py", line 1346, in solve_CBC
    raise PulpSolverError("Pulp: Error while executing "+self.path)
pulp.solvers.PulpSolverError: Pulp: Error while executing /usr/local/lib/python2.7/dist-packages/PuLP-1.5.6-py2.7.egg/pulp/solverdir/cbc-32

对于我的线性规划问题，我试图把不同向量的和作为约束，我想我一定是做得不对，因为一个简单得多的问题工作时没有牵绊。我附加了代码（C是一个N×Nnumpy数组）。

def solve_lp(C):
    N = len(C)
    prob=LpProblem('Scheduling',LpMinimize)

    X = [[LpVariable('X' + str(i+1) + str(j+1), 0, C[i,j],LpBinary)
          for j in range(N)] for i in range(N)]
    X = np.array(X)
    X_o = [LpVariable('X0' + str(i), 0, None, LpBinary) for i in range(N)]
    X_t = [LpVariable('X' + str(i) + 't', 0, None, LpBinary) for i in range(N)]

    # Objective Function                                                                                                                                                
    ones_vec = list(np.ones(len(X_o)))
    prob += lpDot(ones_vec,X_o), 'Minimize Buses'

    # Constraints                                                                                                                                                       
    for i in range(N):
        row = list(X[i,:]) + [X_t[i]]
        ones_vec = list(np.ones(len(row)))
        prob += lpDot(ones_vec, row) == 1, 'Only one destination for ' + str(i)

    for j in range(N):
        col = list(X[:,j]) + [X_o[j]]
        ones_vec = list(np.ones(len(col)))
        prob += lpDot(ones_vec,col) == 1, 'Only one source for ' + str(j)

    prob.solve()
    return X, value(prob.objective)

Answer 1

由于模型中的Nan输入，我最近遇到了类似的问题。我把数据放在一个数据帧中，其中一些单元格不应该被转换成变量来提高性能。然而，在创建目标函数和约束条件时，我注意到了Nan的存在，当我改变它们时，它工作得很好。

Answer 2

请确保没有重复的LpVariable名称，并注意LpVariable名称中有不支持的字符-+[] ->/，因为所有这些字符都会自动转换为下划线_

在调用prob.solve()之前设置LpSolverDefault.msg = 1可能有助于将解算器输出打印到控制台。

Answer 3

我认为你有重复的LpVariable名称。我也有同样的问题，多亏了levis501's answer我才看到。这里：

X = [[LpVariable('X' + str(i+1) + str(j+1), 0, C[i,j],LpBinary)
      for j in range(N)] for i in range(N)]

X包含一些同名的变量。例如，对于i=0和j=10，得到“X111”；对于i=10和j=0，也得到“X111”。