Max flow: how to force f units to flow, with minimal changes to capacity?

Question

Let's say I have a graph and run a max-flow on it. I get some flow, f. However, I want to to flow f1 units where f1>f. Of course, I need to go about increasing some of the edge capacities. I want to make as small a total increase as possible to the capacities. Is there a clever algorithm to achieve this?

If it helps, I care for my application about bi-partite graphs with source (s) to left vertices (L) having some finite, integer capacities (c_l), left vertices L to right vertices R having some connectivity with infinite capacities and all right vertices, R connected to a sink vertex with finite integer capacities (c_r). Here, c_l and c_r sum to the same number. Also, there are no connections among the left vertices or among the right ones.

An example is provided in the image below. The blue numbers are the flow capacities and the pink numbers are the actual flows in the max-flow. Currently, 5 units are flowing but I want 9 units to flow.

can we assume that there won't be any edges between any 2 left vertices or 2 right vertices? — รยקคгรђשค, Oct 12 '20 at 03:19
what is the answer of example? I think it should be 4, which can be achieved either by increasing capacity of edge `4-t` or edge `s-3` by 4. — รยקคгรђשค, Oct 12 '20 at 03:44
Here, each vertex in L points to exactly one vertex in R. Is that going to be true in general? Or could a vertex in L point to several vertices in R? — Zachary Vance, Oct 12 '20 at 05:52
Also, you've drawn this graph as undirected--is that correct? — Zachary Vance, Oct 12 '20 at 05:58
@ZacharyVance - a vertex in L could point to any number of vertices in R. Also, the graph is undirected, yes. — Rohit Pandey, Oct 12 '20 at 06:31

score 1 · Answer 1 · answered Oct 12 '20 at 03:47

In general, turn the flow instance into a min-cost flow instance by setting the cost of existing arcs to zero and adding new, infinite-capacity arcs doubling them of cost one.

For these particular instances, the best you're going to do is to repeatedly find an unsaturated arc of finite capacity and push flow along any path that includes it. Once everything's saturated just use any path.

This seems a little too easy to be what you want, so I'll mention that it's possible to formulate more sophisticated objectives and solve them using linear programming techniques.

Sorry, I don't follow the first paragraph. For the second paragraph, how do I find an unsaturated arc? How do I know pushing flow will be possible without knowing what lies ahead on the paths? Can you please frame it in pseudo code if you don't mind? — Rohit Pandey, Oct 12 '20 at 03:56

score 1 · Answer 2 · answered Oct 12 '20 at 06:48

The graph is undirected, and all the "middle" vertices have infinite capacity. That means we can unify all vertices connected by infinite capacity in L and R, making a very simple graph indeed.

For example, in the above graph, an equivalent graph would be:

s -8-> Vertex 1+2+4 -4-> t
s -1-> Vertex 3+5   -5-> t

So we end up with just a bunch of unique paths with no branching. We can unify the nodes with a simple "floodfill" or DFS type search on infinite-capacity edges. When we unify nodes, we add up their "left" and "right" capacities.

To maximize flow in this graph we:

First, if the left and right paths are not equal, increase the lower one until they are equal. This lets us convert an increase of cost X, into an increase in flow of X.
Once the left and right paths are equal for all nodes, we pick any path. Then, we increase both halves of the path with cost 2X, increasing the flow by X.

Max flow: how to force f units to flow, with minimal changes to capacity?

2 Answers2