数学建模--数学规划模型Python实现

线性规划scipy.optimize.linprog

from scipy.optimize import linprog

一般形式

官方文档：https://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.linprog.html

例一：

c = [-5, -4, -6]  # 目标函数系数
A = [[1, -1, 1],  # 不等式约束系数
     [3, 2, 4],
     [3, 2, 0]]
b = [20, 42, 30]  # 不等式约束值
# 上下界默认为[0, +inf]
res = linprog(c, A_ub=A, b_ub=b, method='revised simplex')

import numpy as np
np.set_printoptions(suppress=True)  # 不使用用科学计数法

res

     con: array([], dtype=float64)
     fun: -78.0
 message: 'Optimization terminated successfully.'
     nit: 2
   slack: array([32.,  0.,  0.])
  status: 0
 success: True
       x: array([ 0., 15.,  3.])

例二：

c = [0.04, 0.15, 0.1, 0.125]
A = [[-0.03, -0.3, 0, -0.15],
     [0.14, 0, 0, 0.07]]
A_eq = [[0.05, 0, 0.2, 0.1]]  # 等式约束系数矩阵，只有一行的情况下必须写两个[]
b = [-32, 42]  # 不等式约束值
b_eq = 24  # 等式约束值
res = linprog(c, A_ub=A, A_eq=A_eq, b_ub=b, b_eq=b_eq, method='revised simplex')

res

     con: array([0.])
     fun: 28.0
 message: 'Optimization terminated successfully.'
     nit: 0
   slack: array([ 0., 42.])
  status: 0
 success: True
       x: array([  0.        , 106.66666667, 120.        ,   0.        ])

无解情况，观察message中的信息

c = [1, 1]
A = [[1, 2]]
b = 8
A_eq = [[1, 1]]
b_eq = 10
res = linprog(c, A_ub=A, b_ub=b, A_eq=A_eq, b_eq=b_eq)
res

     con: array([8.98932503])
     fun: 1.0106749652571962
 message: 'The algorithm terminated successfully and determined that the problem is infeasible.'
     nit: 4
   slack: array([6.1035428])
  status: 2
 success: False
       x: array([0.12489273, 0.88578224])

整数线性规划pulp

from pulp import *

例一

# 创建LpProblem对象，并指定类型：LpMaximize
my_MipProblem = LpProblem('test1', LpMaximize)

# 指明参数范围、类型，LpInterger代表整型
x1 = LpVariable('x1', lowBound=0, cat=LpInteger)
x2 = LpVariable('x2', lowBound=0, cat=LpInteger)

# 添加目标函数以及约束
my_MipProblem += 20 * x1 + 10 * x2, 'obj'
my_MipProblem += 5 * x1 + 4 * x2 <= 24, 'c1'
my_MipProblem += 2 * x1 + 5 * x2 <= 13, 'c2'

# 返回状态，1代表成功
my_MipProblem.solve()

1

my_MipProblem.status

1

my_MipProblem.variables

<bound method LpProblem.variables of test1:
MAXIMIZE
20*x1 + 10*x2 + 0
SUBJECT TO
c1: 5 x1 + 4 x2 <= 24

c2: 2 x1 + 5 x2 <= 13

VARIABLES
0 <= x1 Integer
0 <= x2 Integer
>

# 打印求解得到的参数值
for v in my_MipProblem.variables():
    print(v.varValue)

4.0
1.0

my_MipProblem.objective

20*x1 + 10*x2 + 0

# 打印求解得到的目标函数值
value(my_MipProblem.objective)

90.0

例二

m1 = LpProblem('test2', LpMinimize)

x1 = LpVariable('x1', lowBound=0)
x2 = LpVariable('x2', lowBound=0)
x3 = LpVariable('x3', lowBound=0, cat=LpInteger)

m1 += 18*x1 + 23*x2 + 5*x3, 'obj'
m1 += 72*x1 +121*x2 + 65*x3 <= 2250, 'c1'
m1 += 72*x1 +121*x2 + 65*x3 >= 2000, 'c2'
m1 += 107*x1 + 500*x2 <= 50000, 'c3'
m1 += 107*x1 + 500*x2 >= 500, 'c4'

m1.solve()

1

m1.variables

<bound method LpProblem.variables of test2:
MINIMIZE
18*x1 + 23*x2 + 5*x3 + 0
SUBJECT TO
c1: 72 x1 + 121 x2 + 65 x3 <= 2250

c2: 72 x1 + 121 x2 + 65 x3 >= 2000

c3: 107 x1 + 500 x2 <= 50000

c4: 107 x1 + 500 x2 >= 500

VARIABLES
x1 Continuous
x2 Continuous
0 <= x3 Integer
>

value(m1.objective)

168.0

for v in m1.variables():
    print(v.name,'=',v.varValue)

x1 = 0.0
x2 = 1.0
x3 = 29.0

例三

m2 = LpProblem('test3', LpMinimize)

x1 = LpVariable('x1', lowBound=0)
x2 = LpVariable('x2', lowBound=0)
x3 = LpVariable('x3', cat=LpBinary)  # 0-1变量

m2 += -3*x1 - 2*x2 - x3, 'obj'
m2 += x1 + x2 + x3 <= 7, 'c1'
m2 += 4*x1 + 2*x2 + x3 == 12, 'c2'

m2.solve()

1

for v in m2.variables():
    print(v.name,'=',v.varValue)

x1 = 0.0
x2 = 6.0
x3 = 0.0

value(m2.objective)

-12.0

应用

0-1背包问题

from pulp import *

my_package = pulp.LpProblem('package', LpMaximize)
x1 = LpVariable('x1', cat=LpBinary)
x2 = LpVariable('x2', cat=LpBinary)
x3 = LpVariable('x3', cat=LpBinary)
x4 = LpVariable('x4', cat=LpBinary)
x5 = LpVariable('x5', cat=LpBinary)
x6 = LpVariable('x6', cat=LpBinary)
x7 = LpVariable('x7', cat=LpBinary)
x8 = LpVariable('x8', cat=LpBinary)
x9 = LpVariable('x9', cat=LpBinary)
x10 = LpVariable('x10', cat=LpBinary)

my_package += 540*x1+200*x2+180*x3+350*x4+60*x5+150*x6+280*x7+450*x8+320*x9+120*x10, 'obj'
my_package += 6*x1+3*x2+4*x3+5*x4+x5+2*x6+3*x7+5*x8+4*x9+2*x10<=30, 'c1'

my_package.solve()

1

value(my_package.objective)

2410.0

for v in my_package.variables():
    print(v.name,'=',v.varValue)

x1 = 1.0
x10 = 1.0
x2 = 1.0
x3 = 0.0
x4 = 1.0
x5 = 0.0
x6 = 1.0
x7 = 1.0
x8 = 1.0
x9 = 1.0

指派问题scipy.optimize.linear_sum_assignment

import numpy as np
from scipy.optimize import linear_sum_assignment

# 损耗矩阵
cost = np.array([[66.8, 75.6, 87, 58.6],
                 [57.2, 66, 66.4, 53],
                 [78, 67.8, 84.6, 59.4],
                 [70, 74.2, 69.6, 57.2],
                 [67.4, 71, 83.8, 62.4]])
row_ind, col_ind = linear_sum_assignment(cost)

row_ind

array([0, 1, 2, 3], dtype=int64)

col_ind

array([3, 0, 1, 2])

# 目标成员坐标
list(zip(row_ind,col_ind))

[(0, 3), (1, 0), (2, 1), (3, 2)]

# 目标函数值
cost[row_ind, col_ind].sum()

253.20000000000002

钢管切割问题

my_tube = pulp.LpProblem('tube', LpMinimize)

x1 = LpVariable('x1', lowBound=0, cat=LpInteger)
x2 = LpVariable('x2', lowBound=0, cat=LpInteger)
x3 = LpVariable('x3', lowBound=0, cat=LpInteger)
x4 = LpVariable('x4', lowBound=0, cat=LpInteger)
x5 = LpVariable('x5', lowBound=0, cat=LpInteger)
x6 = LpVariable('x6', lowBound=0, cat=LpInteger)
x7 = LpVariable('x7', lowBound=0, cat=LpInteger)

my_tube += x1+x2+x3+x4+x5+x6+x7, 'obj'
my_tube += x1 + 2*x2 + x7 >=100, 'c1'
my_tube += 3*x3+2*x4+x5+x7 >=100, 'c2'
my_tube += 4*x1+x2+2*x4+4*x5+6*x6+x7 >=100

my_tube.solve()

1

value(my_tube.objective)

91.0

for v in my_tube.variables():
    print(v.name,'=',v.varValue)

x1 = 8.0
x2 = 46.0
x3 = 26.0
x4 = 11.0
x5 = 0.0
x6 = 0.0
x7 = 0.0

非线性规划scipy.optimize.minimize

import numpy as np
from scipy.optimize import minimize

一般形式

例一

# 定义目标函数
def func1(x, sign=-1):
    return sign*(x[0]**2 + x[1]**2 - x[0]*x[1] - 2*x[0] - 5*x[1])
# 定义约束，ineq代表不等式约束
cons = ({'type': 'ineq',
         'fun': lambda x: -(x[0]-1)**2 + x[1]},
        {'type': 'ineq',
         'fun': lambda x: 2*x[0] - 3*x[1] + 6})
# 设定初始值
x0 = [0, 0]
res = minimize(func1, x0, constraints=cons, method='SLSQP')
print(res.x)
print(-res.fun)

[1.00023831 0.00000004]
-1.0000001710927542

例二

def func2(x, args=8):
    return x[0]**2 + x[1]**2 + x[2]**2 + args

cons = ({'type':'ineq', 'fun': lambda x: x[0]**2 - x[1] + x[2]**2},
        {'type':'ineq', 'fun': lambda x: -(x[0] + x[1]**2 + x[2]**2) + 20},
        {'type':'eq', 'fun': lambda x: x[0] + x[1]**2 - 2},
        {'type':'eq', 'fun': lambda x: x[1] + 2*x[2]**2 - 3})
# 设定参数上下界
bounds = ((0, None), (0, None), (0, None))
x0 = [0, 0, 0]
res = minimize(func2, x0, bounds=bounds, constraints=cons, method='SLSQP')
print(res.x)
print(res.fun)

[0.55216734 1.20325918 0.94782404]
10.651091840572583

最大最小化模型

一般形式：

from scipy.optimize import minimize

def func(x):
    for i in range(10):
        FUNC[i] = abs(x[0]-a[i]) + abs(x[1]-b[i])
    return max(FUNC)

bounds = [(3,8), (4,10)]
x0 = [6, 6]
res = minimize(func, x0, bounds=bounds, method="SLSQP")
print(res.x)
print(res.fun)

[7.75000019 8.75000039]
13.500000579039686

多目标规划

对多目标函数进行加权组合，使问题变为单目标规划

• 统一为最大或最小问题
• 标准化去量纲
• 确定权重