# coding=utf-8
# 單純形法的實現，只支援最簡單的實現方法
# 且我們假設約束矩陣a的最後m列是可逆的
# 這樣就必須滿足a是行滿秩的（m*n的矩陣）
import numpy as np
class
******x
(object):
def__init__
(self, c, a, b)
:# 形式 minf(x)=c.tx
# s.t. ax=b
self.c = c
self.a = a
self.b = b
defrun(self)
: c_shape = self.c.shape
a_shape = self.a.shape
b_shape = self.b.shape
assert c_shape[0]
== a_shape[1]
,"not aligned a with c shape"
assert b_shape[0]
== a_shape[0]
,"not aligned a with b shape"
# 找到初始的b，n等值
end_index = a_shape[1]
- a_shape[0]
n = self.a[:,
0:end_index]
n_columns = np.arange(
0, end_index)
c_n = self.c[n_columns,:]
# 第乙個b必須是可逆的矩陣，其實這裡應該用演算法尋找，但此處省略
b = self.a[
:, end_index:
] b_columns = np.arange(end_index, a_shape[1]
) c_b = self.c[b_columns,:]
steps =
0while
true
: steps +=
1print
("steps is {}"
.format
(steps)
) is_optim, b_columns, n_columns = self.main_******x(b, n, c_b, c_n, self.b, b_columns, n_columns)
if is_optim:
break
else
: b = self.a[
:, b_columns]
n = self.a[
:, n_columns]
c_b = self.c[b_columns,:]
c_n = self.c[n_columns,:]
defmain_******x
(self, b, n, c_b, c_n, b, b_columns, n_columns)
: b_inverse = np.linalg.inv(b)
p =(c_n.t - np.matmul(np.matmul(c_b.t, b_inverse)
, n)
).flatten(
)if p.
min(
)>=0:
is_optim =
true
print
("reach optimization."
)print
("b_columns is {}"
.format
(b_columns)
)print
("n_columns is {}"
.format
(sorted
(n_columns)))
best_solution_point = np.matmul(b_inverse, b)
print
("best solution point is {}"
.format
(best_solution_point.flatten())
)print
("best value is {}"
.format
(np.matmul(c_b.t, best_solution_point)
.flatten()[
0]))
print
("\n"
)return is_optim, b_columns, n_columns
else
:# 入基
n_i_in = np.argmin(p)
n_i = n[
:, n_i_in]
.reshape(-1
,1)# by=ni， 求出基
y = np.matmul(b_inverse, n_i)
x_b = np.matmul(b_inverse, b)
n_i_out = self.find_out_base(y, x_b)
tmp = n_columns[n_i_in]
n_columns[n_i_in]
= b_columns[n_i_out]
b_columns[n_i_out]
= tmp
is_optim =
false
print
("not reach optimization"
)print
("in base is {}"
.format
(tmp)
)print
("out base is {}"
.format
(n_columns[n_i_in]))
# 此時已經被換過去了
print
("b_columns is {}"
.format
(sorted
(b_columns)))
print
("n_columns is {}"
.format
(sorted
(n_columns)))
print
("\n"
)return is_optim, b_columns, n_columns
deffind_out_base
(self, y, x_b)
:# 找到x_b/y最小且y>0的位置
index =
min_value =
for i, value in
enumerate
(y):
if value <=0:
continue
else:/
float
(value)
) actual_index = index[np.argmin(min_value)
]return actual_index
if __name__ ==
"__main__"
:'''
c = np.array([-20, -30, 0, 0]).reshape(-1, 1)
a = np.array([[1, 1, 1, 0], [0.1, 0.2, 0, 1]])
b = np.array([100, 14]).reshape(-1, 1)
'''c = np.array([-
4,-1
,0,0
,0])
.reshape(-1
,1) a = np.array([[
-1,2
,1,0
,0],
[2,3
,0,1
,0],
[1,-
1,0,
0,1]
])b = np.array([4
,12,3
]).reshape(-1
,1) ******x = ******x(c, a, b)
******x.run(
)
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