具有惩罚成本的VRPT用gurobi求解代码
时间: 2023-06-19 17:06:45 浏览: 148
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下面是一个具有惩罚成本的VRPT用Gurobi求解的Python代码示例:
```python
import gurobipy as gp
from gurobipy import GRB
# 输入数据
n_customers = 10 # 客户数量
n_vehicles = 3 # 车辆数量
capacity = 50 # 车辆容量
demand = [0, 10, 20, 5, 15, 10, 30, 5, 10, 20] # 客户需求量
service_time = [0, 10, 15, 5, 10, 5, 20, 5, 10, 15] # 客户服务时间
travel_time = [[0, 10, 20, 15, 35, 25, 50, 30, 15, 10],
[10, 0, 25, 20, 30, 15, 40, 25, 5, 20],
[20, 25, 0, 30, 40, 30, 10, 40, 25, 35],
[15, 20, 30, 0, 25, 20, 45, 25, 10, 20],
[35, 30, 40, 25, 0, 30, 25, 50, 35, 40],
[25, 15, 30, 20, 30, 0, 35, 20, 10, 25],
[50, 40, 10, 45, 25, 35, 0, 45, 30, 50],
[30, 25, 40, 25, 50, 20, 45, 0, 25, 35],
[15, 5, 25, 10, 35, 10, 30, 25, 0, 15],
[10, 20, 35, 20, 40, 25, 50, 35, 15, 0]] # 客户之间的行驶时间
# 创建模型
m = gp.Model('VRPTW')
# 创建变量
x = m.addVars(n_customers + 1, n_customers + 1, n_vehicles, vtype=GRB.BINARY, name='x')
y = m.addVars(n_customers + 1, n_vehicles, vtype=GRB.CONTINUOUS, name='y')
t = m.addVars(n_customers + 1, n_vehicles, vtype=GRB.CONTINUOUS, name='t')
# 添加约束
for i in range(1, n_customers + 1):
m.addConstr(sum(x[i, j, k] for j in range(n_customers + 1) for k in range(n_vehicles)) == 1, f'c1_{i}')
for k in range(n_vehicles):
m.addConstr(sum(x[0, j, k] for j in range(1, n_customers + 1)) == 1, f'c2_{k}')
m.addConstr(sum(x[i, 0, k] for i in range(1, n_customers + 1)) == 1, f'c3_{k}')
m.addConstr(y[0, k] == capacity, f'c4_{k}')
for i in range(1, n_customers + 1):
m.addConstr(y[i, k] >= demand[i])
m.addConstr(y[i, k] <= capacity)
m.addConstr(t[i, k] >= service_time[i] + t[j, k] + travel_time[j][i] - (1 - x[j, i, k]) * 1e6 for j in range(n_customers + 1))
m.addConstr(t[i, k] >= service_time[i])
m.addConstr(t[i, k] <= 480)
for j in range(n_customers + 1):
m.addConstr(sum(x[i, j, k] for i in range(n_customers + 1)) == sum(x[j, i, k] for i in range(n_customers + 1)))
m.addConstr(sum(x[i, j, k] for i in range(n_customers + 1)) <= 1)
# 添加目标函数
obj = sum(travel_time[i][j] * x[i, j, k] for i in range(n_customers + 1) for j in range(n_customers + 1) for k in range(n_vehicles))
obj += sum(1000 * (t[i, k] - demand[i]) * (t[i, k] > demand[i]) for i in range(1, n_customers + 1) for k in range(n_vehicles))
m.setObjective(obj, GRB.MINIMIZE)
# 求解模型
m.optimize()
# 输出结果
print('Total Cost:', m.objVal)
for k in range(n_vehicles):
print(f'Vehicle {k}:')
route = []
for i in range(n_customers + 1):
for j in range(n_customers + 1):
if x[i, j, k].x > 0.5:
route.append(i)
print(route)
```
在上面的代码中,我们添加了一个惩罚成本,用于惩罚超出客户需求量的量。具体来说,我们将惩罚成本设置为1000,然后将每个客户的服务时间与需求量进行比较,如果服务时间超过了需求量,则会产生惩罚成本。最终目标函数将行驶时间和惩罚成本相加,以得出总成本。
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