COPT下载及基础使用¶
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COPT介绍¶
下载流程¶
- 前往官网申请求解器
- 审核通过后，会有安装包和licence发送到邮箱
- 安装好后将邮件附带的
license.dat和license.key拷贝到COPT的安装目录下，也就是拷贝到C:\Program Files\COPT下 - 大功告成
如碰到问题可以参考用户手册
COPT基础使用方法¶
以这个问题为例： $$ min \ Z = 3x + 5y $$
\[ s.t. \begin{cases} 2x+y \ge 8 \\ x+2y \ge 6 \end{cases} \]
类导入以及实例创建¶
Envr类：创建一个COPT环境model类：代表优化模型，包含所有变量，约束与目标函数
# 导入
import coptpy as cp
from coptpy import COPT
# 创建实例
env = cp.Envr()
model = env.createModel(name = 'model_name')Cardinal Optimizer v8.0.2. Build date Dec 1 2025
Copyright Cardinal Operations 2025. All Rights Reservedmodel类基本信息¶
model.status：模型解的状态- COPT.Status.OPTIMAL (5)：线性规划找到全局最优解
- COPT.Status.MIP_OPTIMAL (10)：整数规划找到全局整数最优解
- COPT.Status.FEASIBLE (7)：找到可行解（非最优）
- COPT.Status.MIP_FEASIBLE (11)：整数规划找到可行解（非最优）
- COPT.Status.INFEASIBLE (2)：模型无解（约束冲突）
- COPT.Status.UNBOUNDED (3)：模型无界解（目标函数无限优化）
- COPT.Status.TIMEOUT (15)：模型求解超时
- COPT.Status.LICENSEERROR (17)：许可证失效，无法求解
model.objval：目标函数值（存储模型的最优目标函数值）
添加决策变量¶
model.addVar(lb,ub,vtype= ,name)：添加一个单独的决策变量- lb：下界
- ub：上界
- vtype：变量的类型
- COPT.CONTINUOUS (连续变量)
- COPT.INTEGER (整数变量)
- COPT.BINARY (二进制变量，即 0 或 1)
- name：变量名称
model.addVars(*indices, lb, ub, obj, vtype= , nameprefix="")：添加一组类型为tuple的变量，返回tuplelist
# 添加一个决策变量
x = model.addVar(lb=0.0, ub=COPT.INFINITY, vtype=COPT.CONTINUOUS, name='x')
# 等价写法
y = model.addVar(lb=0.0, ub=cp.COPT.INFINITY, vtype=cp.COPT.CONTINUOUS, name='y')
print(x,y)<coptpy.Var: x> <coptpy.Var: y># 添加一组决策变量(2 * 3,下标从(0,0) 到(1,2))
x = model.addVars(2, 3, vtype=COPT.INTEGER, nameprefix='x')
print(x.select()) [<coptpy.Var: x(0,0)>, <coptpy.Var: x(0,1)>, <coptpy.Var: x(0,2)>, <coptpy.Var: x(1,0)>, <coptpy.Var: x(1,1)>, <coptpy.Var: x(1,2)>]设置目标函数¶
Model.setObjective(expr, sense=None)expr：表达式（必填，无默认值）sense：指定优化方向（默认最小）cp.COPT.MINIMIZE（求最小值）cp.COPT.MAXIMIZE（求最大值）
# 例如求3x+5y的最小值
model.setObjective(3*x + 5*y, sense=cp.COPT.MINIMIZE)添加约束¶
- Model.addConstr(lhs, sense=None, rhs=None, name="")
lhs：约束的左侧表达式sense：关系运算符cp.COPT.LESS_EQUAL：小于等于cp.COPT.EQUAL：等于cp.COPT.GREATER_EQUAL：大于等于
rhs：约束的右侧表达式
- Model.addConstr(表达式)
# 添加约束1
model.addConstr(2*x + 1*y,cp.COPT.GREATER_EQUAL, 8)
# 添加约束2，这种写法更简单
model.addConstr(1*x + 2*y >= 6)<coptpy.Constraint: >求解参数设置：¶
Model.setParam(paramname, newval)model.setParam(COPT.Param.TimeLimit, 3600)：设置求解时间限制model.setParam(COPT.Param.RelGap, 0.1)：设置求解MIP的求解Gapmodel.setParam(COPT.Param.LazyConstraints, 1)：关闭延迟约束（1开启）model.setParam(COPT.Param.Threads, -1)：调用全部CPU线程求解（\(\ge 1\)调用固定线程数目（2/4/6））model.setParam(COPT.Param.Logging, 0)：关闭日志打印（1开启，2详细日志）model.setParam(COPT.Param.Presolve, 2)：开启高级预处理（0关闭，1低价预处理）
求解模型¶
model.solve：求解模型，会输出一个日志
## 求解模型
model.solve()Model fingerprint: 392e914b
Using Cardinal Optimizer v8.0.2 on macOS (aarch64)
Hardware has 10 cores and 10 threads. Using instruction set ARMV8 (30)
Minimizing a MIP problem
The original problem has:
2 rows, 29 columns and 4 non-zero elements
3 binaries and 18 integers
Starting the MIP solver with 10 threads and 32 tasks
Presolving the problem
The presolved problem has:
2 rows, 2 columns and 4 non-zero elements
2 integers
Problem info:
Range of matrix coefficients: [1e+00,2e+00]
Range of rhs coefficients: [6e+00,8e+00]
Range of bound coefficients: [6e+00,8e+00]
Range of cost coefficients: [3e+00,5e+00]
Density of cost: 100.0%
Nodes Active LPit/n IntInf BestBound BestSolution Gap Time
0 1 -- 0 0.000000e+00 -- Inf 0.01s
H 0 1 -- 0 0.000000e+00 5.800000e+01 100.00% 0.01s
H 0 1 -- 0 0.000000e+00 2.300000e+01 100.00% 0.01s
H 0 1 -- 0 0.000000e+00 1.700000e+01 100.00% 0.01s
0 1 -- 2 1.666667e+01 1.700000e+01 1.961% 0.01s
1 0 0.0 2 1.700000e+01 1.700000e+01 0.000% 0.01s
1 0 0.0 2 1.700000e+01 1.700000e+01 0.000% 0.02s
Best solution : 17.000000000
Best bound : 17.000000000
Best gap : 0.0000%
Solve time : 0.02
Solve node : 1
MIP status : solved
Solution status : integer optimal (relative gap limit 0.0001)
Violations : absolute relative
bounds : 0 0
rows : 0 0
integrality : 0输出解的值和变量名¶
model.objval：输出目标值Model.getVars():获得模型的所有变量；var.index：获得变量的index，这个不同于变量名，只是一个序号；var.x: 获得变量var在最优解中的取值；var.getName(): 获得变量名。
model.getVars()
print(f"x的取值为：{x.x}，y的取值为{y.x}，最小值为{model.objval}")
```plaintext
x的取值为：4.0，y的取值为1.0，最小值为17.0
### 完整示例
```python
# 完整求解示例：min 3x+5y，s.t. 2x+y≥8, x+2y≥6, x,y≥0
import coptpy as cp
from coptpy import COPT
# 1. 创建环境与模型
env = cp.Envr()
model = env.createModel(name='LP_example')
# 2. 添加决策变量
x = model.addVar(lb=0.0, ub=COPT.INFINITY, vtype=COPT.CONTINUOUS, name='x')
y = model.addVar(lb=0.0, ub=COPT.INFINITY, vtype=COPT.CONTINUOUS, name='y')
# 3. 设置目标函数：最小化 3x+5y
model.setObjective(3*x + 5*y, sense=COPT.MINIMIZE)
# 4. 添加约束条件
model.addConstr(2*x + y >= 8, name='c1')
model.addConstr(x + 2*y >= 6, name='c2')
# 5. 求解模型
model.solve()
print("当前求解状态码：", model.status)
print("最优目标函数值：", model.objval)
print("x的最优解：", x.x)
print("y的最优解：", y.x)
# 标准判断
if model.status == 1: # 状态码=1 → 最优解
print("求解成功：找到全局最优解！")
print(f"最优解：x = {x.x:.4f}, y = {y.x:.4f}")
print(f"最优目标函数值 = {model.objval:.4f}")
elif model.status == 2:
print("求解失败：模型无解（约束条件冲突）")
elif model.status == 3:
print("求解失败：模型无界（目标函数可无限优化）")
else:
print(f"求解状态：{model.status}，非最优解")
Cardinal Optimizer v8.0.2. Build date Dec 1 2025
Copyright Cardinal Operations 2025. All Rights Reserved
Model fingerprint: c5b184a2
Using Cardinal Optimizer v8.0.2 on macOS (aarch64)
Hardware has 10 cores and 10 threads. Using instruction set ARMV8 (30)
Minimizing an LP problem
The original problem has:
2 rows, 2 columns and 4 non-zero elements
The presolved problem has:
2 rows, 2 columns and 4 non-zero elements
Starting the simplex solver using up to 8 threads
Problem info:
Range of matrix coefficients: [1e+00,2e+00]
Range of rhs coefficients: [6e+00,8e+00]
Range of bound coefficients: [0e+00,0e+00]
Range of cost coefficients: [3e+00,5e+00]
Method Iteration Objective Primal.NInf Dual.NInf Time
Dual 0 0.0000000000e+00 2 0 0.00s
Dual 2 1.6667181196e+01 0 0 0.00s
Solving finished
Status: Optimal Objective: 1.6666666667e+01 Iterations: 2 Time: 0.00s
当前求解状态码： 1
最优目标函数值： 16.66666666666667
x的最优解： 3.333333333333333
y的最优解： 1.333333333333334
求解成功：找到全局最优解！
最优解：x = 3.3333, y = 1.3333
最优目标函数值 = 16.6667