Python中不可行的解决方案,使用投币或

Python中不可行的解决方案,使用投币或,python,pulp,coin-or-cbc,Python,Pulp,Coin Or Cbc,我正在尝试使用投币或python解决一个线性编程问题。 我已经做了所有的工作,但似乎有一个错误,我无法确定。 很多时候,这个解决方案要么不可行,要么不正确,因为我正试图强制执行这些约束。 二进制变量的值也不正确,即~0.1^10或~0.999 请帮助我找到错误或尝试指导解决问题 我附上 zip文件(代码+excel中的输入数据文件) word文档。对于数学公式 Model1主类接受输入并创建新的输出文件,然后两个类方法构建距离矩阵和Njg矩阵 Model1类的构造函数方法中似乎存在错误,该类

我正在尝试使用投币或python解决一个线性编程问题。 我已经做了所有的工作,但似乎有一个错误,我无法确定。 很多时候,这个解决方案要么不可行,要么不正确,因为我正试图强制执行这些约束。 二进制变量的值也不正确,即~0.1^10或~0.999

请帮助我找到错误或尝试指导解决问题

我附上

  • zip文件(代码+excel中的输入数据文件)
  • word文档。对于数学公式
Model1主类接受输入并创建新的输出文件,然后两个类方法构建距离矩阵和Njg矩阵

Model1类的构造函数方法中似乎存在错误,该类用于构建TLP并解决LP

声明变量和约束的代码为:

M=100000*prob.schoolNum

    sModel = []
    for i in range(prob.schoolNum):
        sModel.append(i)
    gModel = []
    for i in xrange(prob.gradeNum):
        gModel.append(i)
    Beta = []

    for i in xrange(prob.schoolNum):
        temp = 0
        for j in xrange(prob.gradeNum):
            temp = temp + prob.Njg[i][j]
        if temp < prob.Nmax:
            Beta.append(0)
        else:
            Beta.append(1)


    # x = students of grade g transfer from school i to j
    x = LpVariable.matrix("x_igj_", (sModel, gModel, sModel), 0, 1, LpBinary)
    y = LpVariable.matrix("status of school", (sModel), 0, 1, LpBinary)
    # o = resulting students in grade in school
    o = LpVariable.matrix("o", (sModel, gModel), 0, None, LpInteger)

    # oHelper  = summation of o for all g rades
    oHelper = LpVariable.matrix("oH", (gModel), 0, None, LpInteger)
    #Njg_Helper = Total students in a particular school
    Njg_helper = LpVariable.matrix("NH", (sModel), 0, None, LpInteger)






    formulation = LpProblem("School Consolidation Model", LpMinimize)

    formulation += lpSum(((prob.Njg[i][g] * x[i][g][j] for j in sModel) for g in gModel) for i in sModel)

    for i in sModel:
        for j in sModel:
            for g in gModel:
                formulation += x[i][g][j] * prob.D[i][j] <= prob.d1

    for i in sModel:
        for j in sModel:
            for g in gModel:
                formulation += x[i][g][j] <= y[j]
    for i in sModel:
        for g in gModel:
            formulation += lpSum(x[i][g][j] for j in sModel) <= 1 - y[i]

    for j in sModel:
        formulation += ((lpSum(prob.Njg[j][g] for g in gModel) - prob.Nmax) * (1 - y[j])) <= 0

    for i in sModel:
        for j in sModel:
            if i != j and Beta[i] * Beta[j] != 1:
                formulation += (prob.D[i][j] - prob.d2) >= (y[i] + y[j] - 2) * M

    for g in gModel:
        formulation += lpSum(o[j][g] for j in sModel) == oHelper[g]
        formulation += lpSum(prob.Njg[i][g] for i in sModel) == oHelper[g]
        for j1 in sModel:
            formulation += lpSum(prob.Njg[i1][g]*x[i1][g][j1] for i1 in sModel) == o[j1][g]-prob.Njg[j1][g]*y[j1]

    formulation.solve()
sModel=[]
对于范围内的i(prob.schoolNum):
sModel.append(i)
gModel=[]
对于X范围内的i(prob.gradeNum):
gModel.append(i)
Beta=[]
对于xrange中的i(prob.schoolNum):
温度=0
对于X范围内的j(prob.gradeNum):
温度=温度+概率Njg[i][j]
如果温度公式+=x[i][g][j]*prob.D[i][j]模型中可能存在错误。我会尝试以下方法:

  • 构建一个包含所有特性的最小模型。也就是说,首先构建您可以构建的最小模型(一个玩具示例),以便生成的模型可以为人类可读。有一个或两个学生,只有两个位置,等等。一个模型你可以手工计算解决方案
  • 使用
    plup
    函数,创建包含模型的LP文件
  • 使用其他解算器运行模型,以检查是否得到相同的不可行结果(例如,如果您在非商业项目中,请尝试SCIP)
  • 读取lp文件以检查/查找模型中的错误

如果您将发现错误的代码复制到问题本身,而不是链接到zip文件,您可能会得到更多有用的输入,希望这会有所帮助。谢谢您的建议,完成。谢谢你的帮助和考虑。我们终于找到了解决办法
import wx
import wx.xrc

class WelcomeScreen(wx.Frame):
    def __init__(self, parent):
        wx.Frame.__init__(self, parent, id=wx.ID_ANY, title=u"Stock Cutting", pos=wx.DefaultPosition,
                          size=wx.Size(448, 395), style=wx.DEFAULT_FRAME_STYLE | wx.TAB_TRAVERSAL)

        self.Bind(wx.EVT_CLOSE, self.OnClose)

        # prevent resizing of the window
        self.SetSizeHintsSz(wx.Size(448, 394), wx.Size(448, 395))

        bSizer1 = wx.BoxSizer(wx.VERTICAL)

        #create image object using the jpg file stored in the root directory
        self.m_bitmap1 = wx.StaticBitmap(self, wx.ID_ANY, wx.Bitmap(u"test.jpg", wx.BITMAP_TYPE_ANY),
                                         wx.DefaultPosition, wx.DefaultSize, 0)

        bSizer1.Add(self.m_bitmap1, 0, wx.ALL, 5)

        self.m_button1 = wx.Button(self, wx.ID_ANY, u"START", wx.DefaultPosition, wx.DefaultSize, 0)
        self.m_button1.SetFont(wx.Font(15, 71, 90, 92, False, wx.EmptyString))
        self.m_button1.Bind(wx.EVT_BUTTON, self.next, self.m_button1)
        bSizer1.Add(self.m_button1, 1, wx.ALL | wx.EXPAND, 5)

        self.SetSizer(bSizer1)
        self.Layout()
        self.Centre(wx.BOTH)
        self.Show(True)

    def OnClose(self, event):
        dlg = wx.MessageDialog(self,
                               "Do you really want to close this application?",
                               "Confirm Exit", wx.OK | wx.CANCEL | wx.ICON_QUESTION)
        result = dlg.ShowModal()
        dlg.Destroy()
        if result == wx.ID_OK:
            self.Destroy()

    # function to go to the next screen to enter the data
    def next(self, event):
        self.Show(False)
#        start = home.HomeScreen(None, prob)

# main method to start the application
if __name__ == "__main__":
    app = wx.App()
    WelcomeScreen = WelcomeScreen(None)
    app.MainLoop()