文章目录
- 简单回顾
- 线性规划LP
- 整数规划IP
- 0-1规划
简单回顾
线性规划是数学规划中的一类最简单规划问题,常见的线性规划是一个有约束的,变量范围为有理数的线性规划。如:
使用matlab
的linprog
函数即可求解简单的线性规划问题,可以参考这篇博客:
MATLAB求解线性规划(含整数规划和0-1
规划)问题
matlab
求解线性规划LP
问题需要化为最小化问题,所有约束条件必须为≤
类型,限制较多。本文介绍使用python+gurobi
进行求解。
python+gurobi
介绍参考这篇博客:
gurobi最新下载安装教程 2023.11
线性规划LP
import gurobipy
from gurobipy import GRB# 创建模型
c = [7, 12]
a = [[9, 4],[4, 5],[3, 10]]
b = [300, 200, 300]
MODEL = gurobipy.Model("Example")# 创建变量
x = MODEL.addVars(2, lb=0, ub=gurobipy.GRB.INFINITY, name='x')# 更新变量环境
MODEL.update()# 创建目标函数
MODEL.setObjective(x.prod(c), gurobipy.GRB.MAXIMIZE)# 创建约束条件
MODEL.addConstrs(x.prod(a[i]) <= b[i] for i in range(3))# 执行线性规划模型
MODEL.optimize()
print("Obj:", MODEL.objVal)
for v in MODEL.getVars():print(f"{v.VarName}:{round(v.X,3)}")
Gurobi Optimizer version 10.0.3 build v10.0.3rc0 (win64)CPU model: Intel(R) Core(TM) i7-8565U CPU @ 1.80GHz, instruction set [SSE2|AVX|AVX2]
Thread count: 4 physical cores, 8 logical processors, using up to 8 threadsOptimize a model with 3 rows, 2 columns and 6 nonzeros
Model fingerprint: 0x6b25b35d
Coefficient statistics:Matrix range [3e+00, 1e+01]Objective range [7e+00, 1e+01]Bounds range [0e+00, 0e+00]RHS range [2e+02, 3e+02]
Presolve time: 0.01s
Presolved: 3 rows, 2 columns, 6 nonzerosIteration Objective Primal Inf. Dual Inf. Time0 3.2500000e+30 2.812500e+30 3.250000e+00 0s2 4.2800000e+02 0.000000e+00 0.000000e+00 0sSolved in 2 iterations and 0.01 seconds (0.00 work units)
Optimal objective 4.280000000e+02
Obj: 428.0
x[0]:20.0
x[1]:24.0
最终可得最优解为x = 20, y = 24, 最优值为428。
gurobi
对最大化问题、最小化问题,大于等于和小于等于约束都支持。
整数规划IP
import gurobipy
from gurobipy import GRB
import numpy as np# 创建模型
c = [3, 2]
a = [[2, 3],[4, 2]]
b = [14, 18]
MODEL = gurobipy.Model("Example")# 创建变量
#x = MODEL.addVars(2, lb=0, ub=gurobipy.GRB.INFINITY, name='x')x1 = MODEL.addVar(vtype=GRB.INTEGER,lb=0,ub=GRB.INFINITY, name='x1')
x2 = MODEL.addVar(vtype=GRB.INTEGER,lb=0,ub=GRB.INFINITY, name='x2')
# 更新变量环境MODEL.update()# 创建目标函数
MODEL.setObjective(c[0]*x1+c[1]*x2, gurobipy.GRB.MAXIMIZE)# 创建约束条件
for i in range(2):MODEL.addConstr(a[i][0]*x1 + a[i][1]*x2 <= b[i])# 执行线性规划模型
MODEL.optimize()
print("Obj:", MODEL.objVal)
for v in MODEL.getVars():print(f"{v.VarName}:{round(v.X,3)}")
Gurobi Optimizer version 10.0.3 build v10.0.3rc0 (win64)CPU model: Intel(R) Core(TM) i7-8565U CPU @ 1.80GHz, instruction set [SSE2|AVX|AVX2]
Thread count: 4 physical cores, 8 logical processors, using up to 8 threadsOptimize a model with 2 rows, 2 columns and 4 nonzeros
Model fingerprint: 0x15a6e8bd
Variable types: 0 continuous, 2 integer (0 binary)
Coefficient statistics:Matrix range [2e+00, 4e+00]Objective range [2e+00, 3e+00]Bounds range [0e+00, 0e+00]RHS range [1e+01, 2e+01]
Found heuristic solution: objective 14.0000000
Presolve time: 0.00s
Presolved: 2 rows, 2 columns, 4 nonzeros
Variable types: 0 continuous, 2 integer (0 binary)Explored 0 nodes (0 simplex iterations) in 0.00 seconds (0.00 work units)
Thread count was 8 (of 8 available processors)Solution count 1: 14 Optimal solution found (tolerance 1.00e-04)
Best objective 1.400000000000e+01, best bound 1.400000000000e+01, gap 0.0000%
Obj: 14.0
x1:4.0
x2:1.0
可得该整数规划问题的最优解为x1 = 4, x2 = 1
,最优值为14
。
如果解其对应的松弛问题:
import gurobipy
from gurobipy import GRB
import numpy as np# 创建模型
c = [3, 2]
a = [[2, 3],[4, 2]]
b = [14, 18]
MODEL = gurobipy.Model("Example")# 创建变量
x = MODEL.addVars(2, lb=0, ub=gurobipy.GRB.INFINITY, name='x')# 更新变量环境MODEL.update()# 创建目标函数
MODEL.setObjective(x.prod(c), gurobipy.GRB.MAXIMIZE)# 创建约束条件
MODEL.addConstrs(x.prod(a[i]) <= b[i] for i in range(2))# 执行线性规划模型
MODEL.optimize()
print("Obj:", MODEL.objVal)
for v in MODEL.getVars():print(f"{v.VarName}:{round(v.X,3)}")
Gurobi Optimizer version 10.0.3 build v10.0.3rc0 (win64)CPU model: Intel(R) Core(TM) i7-8565U CPU @ 1.80GHz, instruction set [SSE2|AVX|AVX2]
Thread count: 4 physical cores, 8 logical processors, using up to 8 threadsOptimize a model with 2 rows, 2 columns and 4 nonzeros
Model fingerprint: 0x15a42e7d
Coefficient statistics:Matrix range [2e+00, 4e+00]Objective range [2e+00, 3e+00]Bounds range [0e+00, 0e+00]RHS range [1e+01, 2e+01]
Presolve time: 0.01s
Presolved: 2 rows, 2 columns, 4 nonzerosIteration Objective Primal Inf. Dual Inf. Time0 5.0000000e+30 2.750000e+30 5.000000e+00 0s2 1.4750000e+01 0.000000e+00 0.000000e+00 0sSolved in 2 iterations and 0.01 seconds (0.00 work units)
Optimal objective 1.475000000e+01
Obj: 14.75
x[0]:3.25
x[1]:2.5
可以发现对应的解是x1 = 3.25, x2 = 2.5
, 最优值为14.75
。松弛问题的最优解总是优于整数规划问题的。
0-1规划
无论是matlab
的linprog
函数还是gurobi,0-1
规划实际上只需要在整数规划的基础上,让决策变量的定义域在0-1
之间即可。
仍然是上面的同一个问题:
## 0-1规划import gurobipy
from gurobipy import GRB# 创建模型
c = [3, 2]
a = [[2, 3],[4, 2]]
b = [14, 18]
MODEL = gurobipy.Model("Example")# 创建变量
#x = MODEL.addVars(2, lb=0, ub=gurobipy.GRB.INFINITY, name='x')x1 = MODEL.addVar(vtype=GRB.INTEGER,lb=0,ub=1, name='x1')
x2 = MODEL.addVar(vtype=GRB.INTEGER,lb=0,ub=1, name='x2')
# 更新变量环境MODEL.update()# 创建目标函数
MODEL.setObjective(c[0]*x1+c[1]*x2, gurobipy.GRB.MAXIMIZE)# 创建约束条件
for i in range(2):MODEL.addConstr(a[i][0]*x1 + a[i][1]*x2 <= b[i])# 执行线性规划模型
MODEL.optimize()
print("Obj:", MODEL.objVal)
for v in MODEL.getVars():print(f"{v.VarName}:{round(v.X,3)}")
Gurobi Optimizer version 10.0.3 build v10.0.3rc0 (win64)CPU model: Intel(R) Core(TM) i7-8565U CPU @ 1.80GHz, instruction set [SSE2|AVX|AVX2]
Thread count: 4 physical cores, 8 logical processors, using up to 8 threadsOptimize a model with 3 rows, 2 columns and 6 nonzeros
Model fingerprint: 0x6b25b35d
Coefficient statistics:Matrix range [3e+00, 1e+01]Objective range [7e+00, 1e+01]Bounds range [0e+00, 0e+00]RHS range [2e+02, 3e+02]
Presolve time: 0.01s
Presolved: 3 rows, 2 columns, 6 nonzerosIteration Objective Primal Inf. Dual Inf. Time0 3.2500000e+30 2.812500e+30 3.250000e+00 0s2 4.2800000e+02 0.000000e+00 0.000000e+00 0sSolved in 2 iterations and 0.01 seconds (0.00 work units)
Optimal objective 4.280000000e+02
Obj: 428.0
x[0]:20.0
x[1]:24.0
Gurobi Optimizer version 10.0.3 build v10.0.3rc0 (win64)CPU model: Intel(R) Core(TM) i7-8565U CPU @ 1.80GHz, instruction set [SSE2|AVX|AVX2]
Thread count: 4 physical cores, 8 logical processors, using up to 8 threadsOptimize a model with 3 rows, 6 columns and 18 nonzeros
Model fingerprint: 0x157f6a4a
Coefficient statistics:Matrix range [9e+00, 6e+01]Objective range [6e+00, 1e+01]Bounds range [0e+00, 0e+00]RHS range [6e+01, 2e+02]
Presolve time: 0.01s
Presolved: 3 rows, 6 columns, 18 nonzerosIteration Objective Primal Inf. Dual Inf. Time0 0.0000000e+00 4.187500e+01 0.000000e+00 0s3 2.9842520e+01 0.000000e+00 0.000000e+00 0sSolved in 3 iterations and 0.01 seconds (0.00 work units)
Optimal objective 2.984251969e+01
Obj: 29.84251968503937
x[0]:0.0
x[1]:0.433
x[2]:0.0
x[3]:4.252
x[4]:0.0
x[5]:0.0
Gurobi Optimizer version 10.0.3 build v10.0.3rc0 (win64)CPU model: Intel(R) Core(TM) i7-8565U CPU @ 1.80GHz, instruction set [SSE2|AVX|AVX2]
Thread count: 4 physical cores, 8 logical processors, using up to 8 threadsOptimize a model with 2 rows, 2 columns and 4 nonzeros
Model fingerprint: 0x15a6e8bd
Variable types: 0 continuous, 2 integer (0 binary)
Coefficient statistics:Matrix range [2e+00, 4e+00]Objective range [2e+00, 3e+00]Bounds range [0e+00, 0e+00]RHS range [1e+01, 2e+01]
Found heuristic solution: objective 14.0000000
Presolve time: 0.00s
Presolved: 2 rows, 2 columns, 4 nonzeros
Variable types: 0 continuous, 2 integer (0 binary)Explored 0 nodes (0 simplex iterations) in 0.00 seconds (0.00 work units)
Thread count was 8 (of 8 available processors)Solution count 1: 14 Optimal solution found (tolerance 1.00e-04)
Best objective 1.400000000000e+01, best bound 1.400000000000e+01, gap 0.0000%
Obj: 14.0
x1:4.0
x2:1.0
Gurobi Optimizer version 10.0.3 build v10.0.3rc0 (win64)CPU model: Intel(R) Core(TM) i7-8565U CPU @ 1.80GHz, instruction set [SSE2|AVX|AVX2]
Thread count: 4 physical cores, 8 logical processors, using up to 8 threadsOptimize a model with 2 rows, 2 columns and 4 nonzeros
Model fingerprint: 0x15a42e7d
Coefficient statistics:Matrix range [2e+00, 4e+00]Objective range [2e+00, 3e+00]Bounds range [0e+00, 0e+00]RHS range [1e+01, 2e+01]
Presolve time: 0.01s
Presolved: 2 rows, 2 columns, 4 nonzerosIteration Objective Primal Inf. Dual Inf. Time0 5.0000000e+30 2.750000e+30 5.000000e+00 0s2 1.4750000e+01 0.000000e+00 0.000000e+00 0sSolved in 2 iterations and 0.01 seconds (0.00 work units)
Optimal objective 1.475000000e+01
Obj: 14.75
x[0]:3.25
x[1]:2.5
Gurobi Optimizer version 10.0.3 build v10.0.3rc0 (win64)CPU model: Intel(R) Core(TM) i7-8565U CPU @ 1.80GHz, instruction set [SSE2|AVX|AVX2]
Thread count: 4 physical cores, 8 logical processors, using up to 8 threadsOptimize a model with 2 rows, 2 columns and 4 nonzeros
Model fingerprint: 0xdff3d373
Variable types: 0 continuous, 2 integer (0 binary)
Coefficient statistics:Matrix range [2e+00, 4e+00]Objective range [2e+00, 3e+00]Bounds range [1e+00, 1e+00]RHS range [1e+01, 2e+01]
Found heuristic solution: objective 5.0000000Explored 0 nodes (0 simplex iterations) in 0.00 seconds (0.00 work units)
Thread count was 1 (of 8 available processors)Solution count 1: 5 Optimal solution found (tolerance 1.00e-04)
Best objective 5.000000000000e+00, best bound 5.000000000000e+00, gap 0.0000%
Obj: 5.0
x1:1.0
x2:1.0
可得0-1规划的最优解是x1 = x2 = 1
, 最优值 = 5
。
当然0-1规划的典型应用场景是指派问题、运输问题、排班问题等。