利用运输问题的位势法和闭回路法求最优解的python代码
时间: 2024-02-24 16:57:30 浏览: 29
位势法和闭回路法都是用于求解运输问题的算法。下面是一个运输问题的Python代码示例,使用位势法和闭回路法求解最优解。
```python
import numpy as np
# 运输问题数据
cost = np.array([[5, 2, 7, 8], [9, 3, 2, 5], [3, 1, 8, 6]])
supply = np.array([150, 200, 100])
demand = np.array([100, 150, 100, 100])
# 位势法
def potential(cost, supply, demand):
m, n = cost.shape
u = np.zeros(m)
v = np.zeros(n)
u[0] = 0
for _ in range(m + n):
for i in range(m):
for j in range(n):
if supply[i] > 0 and demand[j] > 0 and u[i] + v[j] == cost[i, j]:
supply[i] -= min(supply[i], demand[j])
demand[j] -= min(supply[i], demand[j])
u[i] = cost[i, j] - v[j]
for j in range(n):
for i in range(m):
if supply[i] > 0 and demand[j] > 0 and u[i] + v[j] == cost[i, j]:
supply[i] -= min(supply[i], demand[j])
demand[j] -= min(supply[i], demand[j])
v[j] = cost[i, j] - u[i]
return u, v
u, v = potential(cost, supply, demand)
print("u =", u)
print("v =", v)
# 闭回路法
def closed_path(cost, supply, demand, u, v):
m, n = cost.shape
x = np.zeros((m, n))
while True:
# 计算最小非基变量
minval = float('inf')
for i in range(m):
for j in range(n):
if u[i] + v[j] - cost[i, j] < minval and supply[i] > 0 and demand[j] > 0:
minval = u[i] + v[j] - cost[i, j]
imin, jmin = i, j
# 计算闭回路
path = [(imin, jmin)]
while True:
i, j = path[-1]
if x[i, j] == 1:
break
else:
for k in range(n):
if x[i, k] == 1:
path.append((i, k))
break
else:
for k in range(m):
if x[k, j] == 1:
path.append((k, j))
break
# 计算闭回路中的最小量
delta = float('inf')
for k in range(1, len(path), 2):
i, j = path[k]
delta = min(delta, supply[i])
# 更新x和supply、demand
for k in range(1, len(path), 2):
i, j = path[k]
x[i, j] = 1
supply[i] -= delta
demand[j] -= delta
# 判断是否结束
if np.all(x > 0):
break
return x
x = closed_path(cost, supply, demand, u, v)
print("x =", x)
# 计算最小成本
min_cost = np.sum(x * cost)
print("Minimum Cost =", min_cost)
```
以上代码中,`potential`函数使用位势法求解最优解,返回u和v;`closed_path`函数使用闭回路法求解最优解,返回x;最后计算最小成本。你可以根据自己的需要修改和扩展这个代码。