【数学建模】Python+Gurobi——零基础学优化建模线性规划模型（LP）

目录

1 概述

2 算例及Matlab代码实现

2.1 算例

2.2 Python代码实现

2.3 求解结果

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1 概述

规划问题的数学模型一般由三个因素构成

决策变量 目标函数 约束条件

数学规划是运筹学的一个重要分支，线性规划是数学规划的一个重要分支
线性规划即以线性函数为目标函数，线性条件为约束条件 

建立线性规划模型的基本步骤
（1）分析问题，找出决策变量
（2）根据问题，找出决策变量必须满足的一组线性等式或者不等式约束，即为约束条件
（3）根据问题的目标，构造关于决策变量的一个线性函数，即为目标函数

2 算例及Matlab代码实现

2.1 算例

2.2 Python代码实现

from gurobipy import * # 在Python中调用gurobi求解包

Max_Obj=1  # 用于判定目标函数求最大值还是求最小值

M_LP=Model("LP")  #线性规划问题

# 变量声明
OP =M_LP.addVar(lb=-GRB.INFINITY,ub=GRB.INFINITY, name="OP")
x1 =M_LP.addVar(lb=0,ub=5, name="x1")
x2 =M_LP.addVar(lb=0,ub=3, name="x2")
x3 =M_LP.addVar(lb=0,ub=2, name="x3")

# 对变量边界进行限定的第二种写法：以x1为例
# M_LP.addConstr(x1>=0,"Bound_Con11")
# M_LP.addConstr(x1<=5,"Bound_Con12")

# 设置目标函数
if Max_Obj==0:
    M_LP.setObjective(x1+2*x2-3*x3,GRB.MINIMIZE)
else:
    M_LP.setObjective(x1+2*x2-3*x3,GRB.MAXIMIZE)

# 添加约束
M_LP.addConstr(x1+2*x2<=3,"Con1")
M_LP.addConstr(x2+x3<=2,"Con2")
M_LP.addConstr(x1+x2+x3==4,"Con3")

# Optimize model
M_LP.optimize()

M_LP.write("LP.lp")

print('**************')
print(' The optimal solution ')
print('**************')
print('OP is :',M_LP.ObjVal) # 输出目标值
print('x1 is :',x1.x) # .x 用于输出 X1 的值
print('x2 is :',x2.x)
print('x3 is :',x3.x)

from gurobipy import * # 在Python中调用gurobi求解包

Max_Obj=1  # 用于判定目标函数求最大值还是求最小值

M_LP=Model("LP")  #线性规划问题

# 变量声明
OP =M_LP.addVar(lb=-GRB.INFINITY,ub=GRB.INFINITY, name="OP")
x1 =M_LP.addVar(lb=0,ub=5, name="x1")
x2 =M_LP.addVar(lb=0,ub=3, name="x2")
x3 =M_LP.addVar(lb=0,ub=2, name="x3")

# 对变量边界进行限定的第二种写法：以x1为例
# M_LP.addConstr(x1>=0,"Bound_Con11")
# M_LP.addConstr(x1<=5,"Bound_Con12")

# 设置目标函数
if Max_Obj==0:
    M_LP.setObjective(x1+2*x2-3*x3,GRB.MINIMIZE)
else:
    M_LP.setObjective(x1+2*x2-3*x3,GRB.MAXIMIZE)

# 添加约束
M_LP.addConstr(x1+2*x2<=3,"Con1")
M_LP.addConstr(x2+x3<=2,"Con2")
M_LP.addConstr(x1+x2+x3==4,"Con3")

# Optimize model
M_LP.optimize()

M_LP.write("LP.lp")

print('**************')
print(' The optimal solution ')
print('**************')
print('OP is :',M_LP.ObjVal) # 输出目标值
print('x1 is :',x1.x) # .x 用于输出 X1 的值
print('x2 is :',x2.x)
print('x3 is :',x3.x)

2.3 求解结果

Gurobi Optimizer version 9.5.2 build v9.5.2rc0 (win64)
Thread count: 8 physical cores, 16 logical processors, using up to 16 threads
Optimize a model with 3 rows, 4 columns and 7 nonzeros
Model fingerprint: 0xdcaef29a
Coefficient statistics:
  Matrix range     [1e+00, 2e+00]
  Objective range  [1e+00, 3e+00]
  Bounds range     [2e+00, 5e+00]
  RHS range        [2e+00, 4e+00]
Presolve removed 1 rows and 2 columns
Presolve time: 0.00s
Presolved: 2 rows, 2 columns, 4 nonzeros

Iteration    Objective       Primal Inf.    Dual Inf.      Time
       0    1.0100000e+00   1.004000e+00   0.000000e+00      0s
       1   -0.0000000e+00   0.000000e+00   0.000000e+00      0s

Solved in 1 iterations and 0.00 seconds (0.00 work units)
Optimal objective -0.000000000e+00
**************
 The optimal solution 
**************
OP is : -0.0
x1 is : 3.0
x2 is : 0.0
x3 is : 1.0

Process finished with exit code 0

Gurobi Optimizer version 9.5.2 build v9.5.2rc0 (win64)
Thread count: 8 physical cores, 16 logical processors, using up to 16 threads
Optimize a model with 3 rows, 4 columns and 7 nonzeros
Model fingerprint: 0xdcaef29a
Coefficient statistics:
  Matrix range     [1e+00, 2e+00]
  Objective range  [1e+00, 3e+00]
  Bounds range     [2e+00, 5e+00]
  RHS range        [2e+00, 4e+00]
Presolve removed 1 rows and 2 columns
Presolve time: 0.00s
Presolved: 2 rows, 2 columns, 4 nonzeros

Iteration    Objective       Primal Inf.    Dual Inf.      Time
       0    1.0100000e+00   1.004000e+00   0.000000e+00      0s
       1   -0.0000000e+00   0.000000e+00   0.000000e+00      0s

Solved in 1 iterations and 0.00 seconds (0.00 work units)
Optimal objective -0.000000000e+00
**************
 The optimal solution 
**************
OP is : -0.0
x1 is : 3.0
x2 is : 0.0
x3 is : 1.0

Process finished with exit code 0