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Load balancing using the heat equation

开发者 https://www.devze.com 2023-03-23 06:05 出处:网络
I didn\'t think this was mathsy enough for mathoverflow, so I thought I would try here. Its for a programming problem though, so its sorta on topic.

I didn't think this was mathsy enough for mathoverflow, so I thought I would try here. Its for a programming problem though, so its sorta on topic.

I have a graph (as in the maths one), where I have a number of vertices, A,B,C.. each with a load "value", these are connected by some arbitrary topology (there is at least a spanning tree). The objective of the problem is to transfer load between each of the vertices, preferably using the minimal flow possible.

I wish to know the transfer values along each edge.

The solution to the problem I was thinking of was to treat it as a heat transfer problem, and iteratively transfer load, or solve the heat equation in some way, counting the amount of load dissipated along each edge. The amount of heat transfer until the network reaches steady state should thus yield the result.

Whilst I think this will work, it seems like the stupid solution. I am wondering if there is a reference or sample problem that someone can point me to -- I am unsure what keywords to search for.

I could not see how to couple the problem as either a simplex problem or as a network flow problem -- each edge has unlimited capacity, and so does each node. There are two simultaneous minimisation probl开发者_C百科ems to solve, so simplex seems to not apply??


If you have a spanning tree, then there is an O(n) solution.

Calculate the average value. (In the end, every node will have this value.) Then iterate over the leaves: calculate the change in value necessary for that leaf (average - value), add that change to the leaf, assign it (as flow) to the edge that connects that leaf to the tree, and subtract it from the other node. Remove that leaf and edge from the tree (not from the graph of course); the other node to may become a new leaf, if it has only one remaining edge in the tree.

When you reach the last node, if you've done the arithmetic right it will end up with the average value, just like all the other nodes.

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