Is it possible to convert a graph with one negative capacity to a graph with only positive capacities?

Question

I am interested in whether a graph (say, a complete graph) with one capacity negative (or many, but one should suffice) can be reconstructed as a graph with all non-negative capacities where the max flow between vertices of the original graph can be identified with max flows between vertices of the new graph. It is ok for the new graph to have more vertices than the original graph. Said another way, consider a complete graph K-n with additional external edges in addition attached to source/sinks. Allow one of the edges of the complete graph to have negative capacity. Can I replace the complete graph with another graph with only positive capacities such that the max flow between all possible subsets of the external sources/sinks remains the same?

CLARIFICATION: We connect each vertex of the complete graph to a distinct external vertex with an edge. If it helps, picture a complete K_6 drawn as a hexagon with an additional edge coming out of each vertex of the hexagon to 6 extra distinct vertices. Those distinct vertices are the s and t's that we are interested in considering. The edges that are not edges of the complete graph that we add are the external edges. I am interested only in cases where the original graph is the complete graph, though edges are allowed to have weight 0 or weight infinity. I hope this helps.

Thanks, Ning

Answer 1

If you indeed have an edge $e$ with a negative capacity, and, I assume, it is a directed edge having no corresponding opposite edge, then you can reverse $e$, make its capacity positive, and, then, solve a regular max-flow problem with all capacities being non-negative. (Look at the concept of the residual network to understand what a negative capacity means.) Make sure you are not confusing edge capacities with costs.