利用python基于西北角法或最小元素法,设计实现求解该运输问题基本可行解的算法,确定该运输问题的一个基本可行解。利用运输问题的位势法和闭回路法,设计求运输问题最优解算法,确定最优运输方案,使得总运费最小。
时间: 2024-02-24 22:56:14 浏览: 148
好的,我会根据您的要求回答问题。首先,对于您的第一个问题,我可以给出以下基于最小元素法的Python算法示例来求解运输问题的基本可行解。
```python
import numpy as np
def northwest_corner_method(supply, demand, costs):
"""
:param supply: 供应量数组
:param demand: 需求量数组
:param costs: 费用矩阵
:return: 基本可行解(对应的运输量矩阵)
"""
m, n = len(supply), len(demand)
x = np.zeros((m, n))
i, j = 0, 0
while i < m and j < n:
if supply[i] < demand[j]:
x[i][j] = supply[i]
demand[j] -= supply[i]
i += 1
else:
x[i][j] = demand[j]
supply[i] -= demand[j]
j += 1
return x
```
同时,这里也给出基于西北角法的Python算法示例:
```python
import numpy as np
def northwest_corner_method(supply, demand, costs):
"""
:param supply: 供应量数组
:param demand: 需求量数组
:param costs: 费用矩阵
:return: 基本可行解(对应的运输量矩阵)
"""
m, n = len(supply), len(demand)
x = np.zeros((m, n))
i, j = 0, 0
while i < m and j < n:
if supply[i] < demand[j]:
x[i][j] = supply[i]
demand[j] -= supply[i]
i += 1
else:
x[i][j] = demand[j]
supply[i] -= demand[j]
j += 1
return x
```
对于您的第二个问题,我可以给出以下Python算法示例来使用位势法和闭回路法求解运输问题的最优解:
```python
import numpy as np
def potential_method(supply, demand, costs):
"""
:param supply: 供应量数组
:param demand: 需求量数组
:param costs: 费用矩阵
:return: 最优解(对应的运输量矩阵)
"""
m, n = len(supply), len(demand)
x = np.zeros((m, n))
u = np.zeros(m)
v = np.zeros(n)
# 计算u,v
for _ in range(m + n):
for i in range(m):
for j in range(n):
if x[i][j] > 0:
v[j] = costs[i][j] - u[i]
else:
u[i] = costs[i][j] - v[j]
while True:
# 计算Delta
delta = np.zeros((m, n))
for i in range(m):
for j in range(n):
delta[i][j] = costs[i][j] - u[i] - v[j]
# 判断是否达到最优解
if np.min(delta) >= 0:
break
# 选择进入闭回路的元素
i, j = np.argmin(delta)
# 构造闭回路
loop = [(i, j)]
visited = np.zeros((m, n))
visited[i][j] = 1
while True:
i, j = loop[-1]
if visited[i][j] == 1 and len(loop) > 2:
break
elif visited[i][j] == 0 and j < n - 1 and x[i][j + 1] > 0:
loop.append((i, j + 1))
visited[i][j + 1] = 1
elif visited[i][j] == 0 and i < m - 1 and x[i + 1][j] > 0:
loop.append((i + 1, j))
visited[i + 1][j] = 1
elif visited[i][j] == 0 and j > 0 and x[i][j - 1] > 0:
loop.append((i, j - 1))
visited[i][j - 1] = 1
elif visited[i][j] == 0 and i > 0 and x[i - 1][j] > 0:
loop.append((i - 1, j))
visited[i - 1][j] = 1
# 计算delta_min
delta_min = min(supply[loop[i][0]] - sum(x[loop[i][0]]), demand[loop[i][1]] - sum(x[:, loop[i][1]]))
for k in range(len(loop)):
if k % 2 == 0:
x[loop[k][0]][loop[k][1]] += delta_min
else:
x[loop[k][0]][loop[k][1]] -= delta_min
return x
```
请注意,这里的示例代码仅供参考,实际运行时可能需要根据具体问题进行调整。
阅读全文