Simplex 单纯形算法的python

发布时间：2019-08-17 08:49:47编辑：auto阅读（1659）

相关理论知识参考 单纯形理论知识

算法可以在给定一个包含线性规划问题的标准形式的描述下，求解该线性规划问题。
例如某一个 pro.txt 文件内容如下:

6
3
3 -1 1 -2 0 0
2 1 0 1 1 0
-1 3 0 -3 0 1
-3 4 12
-7 7 -2 -1 -6 0

执行算法之后得到结果：

x_1 = 0.000000,x_2 = 0.000000,x_3 = 0.000000,x_4 = 1.500000,x_5 = 2.500000,x_6 = 16.500000
objective value is : -16.500000
1-th line constraint value is : -3.000000
2-th line constraint value is : 4.000000
3-th line constraint value is : 12.000000

代码如下：

#encoding=utf-8
__author__ = 'ysg'
import numpy as np #python 矩阵操作lib


class Simplex():

    def __init__(self):
        self._A = "" # 系数矩阵
        self._b = "" #
        self._c = '' #约束
        self._B = '' #基变量的下标集合
        self.row = 0 #约束个数

    def solve(self, filename):
        #读取文件内容，文件结构前两行分别为 变量数 和 约束条件个数
        #接下来是系数矩阵
        #然后是b数组
        #然后是约束条件c
        #假设线性规划形式是标准形式（都是等式）

        A = []
        b = []
        c = []
        with open(filename,'r') as f:
            self.var = int(f.readline())
            self.row = int(f.readline())

            for i in range(self.row):
                x =map(int, f.readline().strip().split(' '))
                A.append(x)
            b=(map(int, list(f.readline().split(' '))))
            c=(map(int, list(f.readline().split(' '))))


        self._A = np.array(A, dtype=float)
        self._b = np.array(b, dtype=float)
        self._c = np.array(c, dtype=float)
        # self._A = np.array([[3,-1,1,-2,0,0],[2,1,0,1,1,0],[-1,3,0,-3,0,1]],dtype=float)
        # self._b = np.array([-3,4,12],dtype=float)
        # self._c = np.array([-7, 7, -2, -1, -6, 0],dtype=float)
        self._B = list()
        self.row = len(self._b)
        self.var = len(self._c)
        (x,obj) = self.Simplex(self._A,self._b,self._c)
        self.pprint(x,obj,A)


    def pprint(self,x,obj,A):
        px = ['x_%d = %f'%(i+1,x[i]) for i in range(len(x))]
        print ','.join(px)
        print ('objective value is : %f'% obj)
        print '------------------------------'
        for i in range(len(A)):
            print '%d-th line constraint value is : %f' % (i+1, x.dot(A[i]))

    #添加人工变量得到一个初始解
    def InitializeSimplex(self,A,b):

        b_min, min_pos = (np.min(b), np.argmin(b))  # 得到最小bi

        #将bi全部转化成正数
        if (b_min < 0):
            for i in range(self.row):
                if i != min_pos:
                    A[i] = A[i] - A[min_pos]
                    b[i] = b[i] - b[min_pos]
            A[min_pos] = A[min_pos]*-1
            b[min_pos] = b[min_pos]*-1

        #添加松弛变量
        slacks = np.eye(self.row)
        A = np.concatenate((A,slacks),axis=1)
        c = np.concatenate((np.zeros(self.var),np.ones(self.row)),axis=1)
        # 松弛变量全部加入基,初始解为b
        new_B = [i + self.var for i in range(self.row)]

        #辅助方程的目标函数值
        obj = np.sum(b)

        c = c[new_B].reshape(1,-1).dot(A) - c
        c = c[0]
        #entering basis
        e= np.argmax(c)

        while c[e] > 0:
            theta = []
            for i in range(len(b)):
                if A[i][e] > 0:
                    theta.append(b[i]/A[i][e])
                else:
                    theta.append(float("inf"))

            l = np.argmin(np.array(theta))

            if theta[l] == float('inf'):
                print 'unbounded'
                return False

            (new_B, A, b, c, obj) = self._PIVOT(new_B, A, b, c, obj, l , e)

            e = np.argmax(c)

        #如果此时人工变量仍在基中，用原变量去替换之
        for mb in new_B:
            if mb >= self.var:
                row = mb-self.var
                i = 0
                while A[row][i] == 0 and i < self.var:
                    i+=1
                (new_B, A, b, c, obj) = self._PIVOT(new_B,A,b,c,obj,new_B.index(mb),i)

        return (new_B, A[:,0:self.var], b)

    #算法入口
    def Simplex(self,A,b,c):
        B = ''
        (B, A ,b) = self.InitializeSimplex(A,b)

        #函数目标值
        obj = np.dot(c[B],b)

        c = np.dot(c[B].reshape(1,-1), A) - c
        c = c[0]

        # entering basis
        e = np.argmax(c)
        # 找到最大的检验数，如果大于0，则目标函数可以优化
        while c[e] > 0:
            theta = []
            for i in range(len(b)):
                if A[i][e] > 0:
                    theta.append(b[i] / A[i][e])
                else:
                    theta.append(float("inf"))

            l = np.argmin(np.array(theta))

            if theta[l] == float('inf'):
                print 'unbounded'
                return False

            (B, A, b, c, obj) = self._PIVOT(B, A, b, c, obj, l, e)

            e = np.argmax(c)


        x = self._CalculateX(B,A,b,c)
        return (x,obj)


    #得到完整解
    def _CalculateX(self,B,A,b,c):

        x = np.zeros(self.var,dtype=float)
        x[B] = b
        return x

    # 基变换
    def _PIVOT(self,B,A,b,c,z,l,e):
        # main element is a_le
        # l represents leaving basis
        # e represents entering basis

        main_elem = A[l][e]
        #scaling the l-th line
        A[l] = A[l]/main_elem
        b[l] = b[l]/main_elem

        #change e-th column to unit array
        for i in range(self.row):
            if i != l:
                b[i] = b[i] - A[i][e] * b[l]
                A[i] = A[i] - A[i][e] * A[l]

        #update objective value
        z -= b[l]*c[e]

        c = c - c[e] * A[l]

        # change the basis
        B[l] = e

        return (B, A, b, c, z)


s = Simplex()
s.solve('pro.txt')

注释差不多写的比较清楚，可以参考上面的理论知识链接。

大佬请绕路。

有错欢迎指出。

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