实验九：求解线性/整数/01/连续线性规划

本博文源于python基础，旨在讲解用python做线性规划。先讲解python下做线性规划的相关函数，然后python四道例题进行演示。

0.linprog()函数调用格式

optimize.linprog(C,A,B,Aeq,beq,bounds,method,X0,options)

• C 是最值向量
• A和B对应线性不等式约束
• Aeq和Beq对应线性等式约束
• bounds对应公式中的Lb和Ub,即决策向量的上界和下界
• method是求解器(方法)的类型
• X0是X的初始值
• option提供常用最值算法，如Nelder-Mead,Powell,CG,BFGS,Newtonn-CG等；部分参数可默认。

1.求解线性规划$maxf=2x_1+3x_2+4x_3$

$$s.t.\{\begin{array}{r}1.5x_1+3x_2+5x_3\le 600\\ 280x_1+250x_2+400x_3\le 60000\\ x_1,x_2,x_3\ge 0\end{array}$$

>>> from scipy import optimize # 导入库
>>> from sympy import * # 导入库
>>> C = Matrix([2,3,4]) # 确定目标函数系数矩阵C
>>> A = Matrix([[1.5,3,5],[280,250,400]]) # 确定约束条件系数矩阵A
>>> B = Matrix([[600],[60000]]) # 确定约束条件常数矩阵B
>>> res = optimize.linprog(-C,A,B,method='simplex') # 问题要求最大值，雇佣-C
>>> print(res)
     con: array([], dtype=float64)
     fun: -632.258064516129
 message: 'Optimization terminated successfully.'
     nit: 3
   slack: array([0., 0.])
  status: 0
 success: True
       x: array([ 64.51612903, 167.74193548,   0.        ])
>>>

通过观察解之的$x_1=64.52,x_2=167.74,x_3=0$时，此时线性规划最优值为$f=632.26$

2.用python求解连续线性规划问题

$$\begin{gathered} maxf=72x_1+64x_2 \\ s.t\{\begin{array}{r}{x}_1+x_2\le 50\\ 12x_1+8x_2\le 480\\ 3x_1\le 100\\ x_1\ge 0,x_2\ge 0\end{array} \end{gathered}$$

>>> from scipy import optimize
>>> from sympy import * # 第一行第二行导入库
>>> C = Matrix([72,64]) # 确定目标函数系数矩阵C
>>> A = Matrix([[1,1],[12,8],[3,0]]) # 确定约束条件系数矩阵A
>>> B = Matrix([[50],[480],[100]]) # 确定约束条件常数矩阵B
>>> res = optimize.linprog(-C,A,B,method='simplex') # 问题要求最低阿志，故用-C
>>> res # 最优值是显示结果的相反数
     con: array([], dtype=float64)
     fun: -3360.0
 message: 'Optimization terminated successfully.'
     nit: 4
   slack: array([ 0.,  0., 40.])
  status: 0
 success: True
       x: array([20., 30.])
>>>

最后得出结果是$x_1=20,x_2=30$，此时线性规划问题最优值为$f=3360$

3.求解整数规划

$$\begin{gathered} maxy \\ s.t.\{\begin{array}{r}y\le \frac{1}{2}{x}_1,y\le \frac{1}{3}{x}_2\\ 5x_1+4x_2\le 960\\ 9x_1+10x_2\le 1440\\ (5x_1+4x_2)-(9x_1+10x_2)\ge -60\\ (5x_1+4x_2)-(9x_1+10x_2)\le 60\\ x_1\ge 0,x_2\ge 0,y\ge 0,且为整数\end{array} \end{gathered}$$

import pulp

model = pulp.LpProblem("Example 8",pulp.LpMaximize)
x1 = pulp.LpVariable("x1",lowBound=0,cat='Integer') # 限制约束变量为整数
x2 = pulp.LpVariable('x2',lowBound=0,cat = 'Integer') # 限制约束变量为整数
y = pulp.LpVariable('y',lowBound=0,cat='Integer') # 限制约束变量为整数
# 目标函数
model += y
# 约束条件
model += y-0.5*x1 <= 0
model += y-1/3*x2 <= 0
model += 5*x1+4*x2 <= 960
model += 9*x1+10*x2 <= 1440
model += 4*x1+6*x2 <= 60

model.solve() # 解决问题
pulp.LpStatus[model.status]

print('x1={}'.format(x1.varValue))
print('x2={}'.format(x2.varValue)) # 打印最有解
print("最大产量为y=",pulp.value(model.objective)) # 打印最优值

$x_1与x_2$就是最优解

4.求解0-1规划

$$maxf=1000x_1+1500x_2+900x_3+2100x_{4}s.t.\{\begin{array}{r}4x_1+6x_2+6x_3+8x_4\le 20\\ x_i=0或1，i=1,2,3,4\end{array}$$

import pulp
model = pulp.LpProblem("Example 8",pulp.LpMaximize)
x1 = pulp.LpVariable("x1",lowBound=0,cat='Binary') # 限制约束变量为0-1
x2 = pulp.LpVariable('x2',lowBound=0,cat = 'Binary') # 限制约束变量为0-1
x3 = pulp.LpVariable('x3',lowBound=0,cat = 'Binary') # 限制约束变量为0-1
x4 = pulp.LpVariable('x4',lowBound=0,cat = 'Binary') # 限制约束变量为0-1

# 目标函数
model += 1000*x1+1500*x2+900*x3+2100*x4
# 约束条件
model += 4*x1+6*x2+6*x3+8*x4 <= 20

model.solve() # 解决问题
pulp.LpStatus[model.status]

print('x1={}'.format(x1.varValue))
print('x2={}'.format(x2.varValue)) # 打印最有解
print('x3={}'.format(x3.varValue))
print('x4={}'.format(x4.varValue))

print('x1={}'.format(x1.varValue))

print("最大产量为y=",pulp.value(model.objective)) # 打印最优值