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。