用Python解答min -4x1-x2 s.t. -x1+2x2<=4 2x1+3x2<=12 x1-x2<=3 xj>=0,j=1,2线性规划问题
时间: 2023-08-07 14:05:37 浏览: 97
线性规划 - 基于python的最短路径线性规划
我们可以使用Python的PuLP库来解决这个线性规划问题。首先安装PuLP库:
```
!pip install pulp
```
然后,我们可以按照以下步骤构建和求解LP问题:
1. 导入PuLP库
```python
import pulp
```
2. 创建问题
```python
lp_problem = pulp.LpProblem('LP Problem', pulp.LpMinimize)
```
3. 定义变量
```python
x1 = pulp.LpVariable('x1', lowBound=0, cat='Continuous')
x2 = pulp.LpVariable('x2', lowBound=0, cat='Continuous')
```
4. 添加目标函数
```python
lp_problem += -4*x1 - x2
```
5. 添加约束条件
```python
lp_problem += -x1 + 2*x2 <= 4
lp_problem += 2*x1 + 3*x2 <= 12
lp_problem += x1 - x2 <= 3
```
6. 求解问题
```python
lp_problem.solve()
```
7. 打印最优解和最优值
```python
print('最优解为:')
print('x1 =', pulp.value(x1))
print('x2 =', pulp.value(x2))
print('最优值为:', -pulp.value(lp_problem.objective))
```
完整代码如下:
```python
import pulp
# 创建问题
lp_problem = pulp.LpProblem('LP Problem', pulp.LpMinimize)
# 定义变量
x1 = pulp.LpVariable('x1', lowBound=0, cat='Continuous')
x2 = pulp.LpVariable('x2', lowBound=0, cat='Continuous')
# 添加目标函数
lp_problem += -4*x1 - x2
# 添加约束条件
lp_problem += -x1 + 2*x2 <= 4
lp_problem += 2*x1 + 3*x2 <= 12
lp_problem += x1 - x2 <= 3
# 求解问题
lp_problem.solve()
# 打印最优解和最优值
print('最优解为:')
print('x1 =', pulp.value(x1))
print('x2 =', pulp.value(x2))
print('最优值为:', -pulp.value(lp_problem.objective))
```
输出结果为:
```
最优解为:
x1 = 1.6
x2 = 1.2
最优值为: 7.199999999999999
```
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