如何用python代码实现可重排clos网络中的中间级选择策略,同时评估不同业务请求下的阻塞性能
时间: 2023-12-03 17:47:45 浏览: 137
可重排 Clos 网络中的中间级选择策略通常有两种:最短路径优先和最小负载优先。下面是Python代码实现这两种策略的示例。
首先,我们需要定义 Clos 网络的拓扑结构。这里以一个三层的可重排 Clos 网络为例,其中每层有 4 个交换机。
```python
topology = {
'L1S1': ['L2S1', 'L2S2', 'L2S3', 'L2S4'],
'L1S2': ['L2S1', 'L2S2', 'L2S3', 'L2S4'],
'L1S3': ['L2S1', 'L2S2', 'L2S3', 'L2S4'],
'L1S4': ['L2S1', 'L2S2', 'L2S3', 'L2S4'],
'L2S1': ['L3S1', 'L3S2', 'L3S3', 'L3S4'],
'L2S2': ['L3S1', 'L3S2', 'L3S3', 'L3S4'],
'L2S3': ['L3S1', 'L3S2', 'L3S3', 'L3S4'],
'L2S4': ['L3S1', 'L3S2', 'L3S3', 'L3S4'],
'L3S1': [],
'L3S2': [],
'L3S3': [],
'L3S4': []
}
```
接下来定义每个交换机的负载情况,这里使用一个字典来存储。
```python
loads = {
'L1S1': 0,
'L1S2': 0,
'L1S3': 0,
'L1S4': 0,
'L2S1': 0,
'L2S2': 0,
'L2S3': 0,
'L2S4': 0,
'L3S1': 0,
'L3S2': 0,
'L3S3': 0,
'L3S4': 0
}
```
接下来实现最短路径优先的中间级选择策略。我们使用 Dijkstra 算法来求两个交换机之间的最短路径。
```python
import heapq
def dijkstra(graph, start):
distances = {node: float('inf') for node in graph}
distances[start] = 0
queue = [(0, start)]
while queue:
current_distance, current_node = heapq.heappop(queue)
if current_distance > distances[current_node]:
continue
for neighbor in graph[current_node]:
distance = current_distance + 1
if distance < distances[neighbor]:
distances[neighbor] = distance
heapq.heappush(queue, (distance, neighbor))
return distances
def shortest_path(src, dst):
distances = dijkstra(topology, src)
path = [dst]
while dst != src:
for neighbor in topology[dst]:
if distances[neighbor] == distances[dst] - 1:
path.append(neighbor)
dst = neighbor
break
return list(reversed(path))
```
我们还需要实现一个函数来选择中间级交换机。该函数会考虑目的地交换机和当前交换机的距离,以及各个中间级交换机的负载情况。
```python
def shortest_path_switch(src, dst):
path = shortest_path(src, dst)
for i in range(1, len(path) - 1):
min_load = float('inf')
min_switch = None
for neighbor in topology[path[i]]:
if loads[neighbor] < min_load:
min_load = loads[neighbor]
min_switch = neighbor
if min_switch is None:
return None
path[i] = min_switch
return path
```
接下来实现最小负载优先的中间级选择策略。该策略会优先选择负载最小的中间级交换机。
```python
def min_load_switch(src, dst):
path = shortest_path(src, dst)
for i in range(1, len(path) - 1):
min_load = float('inf')
min_switch = None
for neighbor in topology[path[i]]:
if loads[neighbor] < min_load:
min_load = loads[neighbor]
min_switch = neighbor
if min_switch is None:
return None
path[i] = min_switch
return path
```
最后,我们可以创建一些业务请求并测试这两种中间级选择策略的性能。
```python
requests = [
('H1', 'H2', 10),
('H1', 'H3', 20),
('H1', 'H4', 30),
('H2', 'H3', 40),
('H2', 'H4', 50),
('H3', 'H4', 60)
]
for src, dst, demand in requests:
path = shortest_path_switch(src, dst)
if path is None:
print('No path found between %s and %s' % (src, dst))
else:
for i in range(len(path) - 1):
loads[path[i]] += demand
print('Path from %s to %s: %s' % (src, dst, path))
for src, dst, demand in requests:
path = min_load_switch(src, dst)
if path is None:
print('No path found between %s and %s' % (src, dst))
else:
for i in range(len(path) - 1):
loads[path[i]] += demand
print('Path from %s to %s: %s' % (src, dst, path))
```
这里我们假设每个业务请求的带宽需求为 10,20,30,40,50 和 60。程序会依次处理每个业务请求,并打印出所选择的路径以及每个交换机的负载情况。
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