纸浆求解器错误

我正在尝试在 puLP (Python) 中求解 MILP，但不断收到以下错误：

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，我试图将不同向量的总和作为约束，并且我认为我一定以某种方式做错了，因为一个更简单的问题可以顺利进行。我已附上代码（C是一个 N × Nnumpy大批）。

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)

确保没有重复的 LpVariable 名称，并注意带有不受支持的字符的 LpVariable 名称-+[] ->/因为所有这些字符都会默默地转换为下划线_

Setting LpSolverDefault.msg = 1打电话之前prob.solve()可以通过将解算器输出打印到控制台来提供帮助。