Question

Imagine that you are working on creating a flow for a set of packets in a media stream, as described in the introduction to this chapter. So you are given a network, G, with a source, s, and sink, t, together with bandwidth constraints on each edge, which indicate the maximum speed that the communication link represented by that edge can support. As mentioned before, your goal is to produce a maximum flow from s to t, respecting the bandwidth constraints on the edges. Suppose now, however, that you also have a bandwidth constraint on each router in the network, which specifies the maximum amount of information, in bits per second, that can pass through that node.

1) Describe an efficient algorithm for finding a maximum flow in the network, G, that satisfies the bandwidth capacity constraints on the edges as well as the vertices. What is the running time of your algorithm?

2)No of V's and E's in your network

Answer 1

Problem is very simple . We already have algorithm to solve max-flow problem with edge capacity constrain. We will reduce vertex capacity constraints into equivalent edge capacity constrain as follows :-

1. Given the network G=(V,E) with source s and sink t, we will create network G'=(V',E') where for every vertex v in V other than s and t , we will create two vertex v1 and v2 in V' such that every edges which was incident to v in G will now be incident on v1 in G' and every edge which was going out of v in G will go out from v1 in G'.

2. Draw an edge from v1 to v2 with capacity equal to vertex flow capacity of v in G.

Every flow f in G will have one to one mapping with flow in G' with same flow value.

Proof :- Since every edge in G is present in G' with same capacity and if flow value x is passed through vertex v in G, same flow value can passed between v1 and v2 in G'. Thus we will get equivalent flow in G' for every flow in G without maintaining vertex capacity.

Hence once graph G' is created, we will perform Edmonds-Karp algorithm to get max-flow in time O(|V'|2|E'|) = O(|V|2|E|).

2. Since we have 2 copy of vertex for every vertex v in G other than s and t, hence |V'| =2|V| -2 and |E'| = |E|+|V|-2 because other than s and t every vertex copy has one edge.

Consider the image below as an example where the top image shows the graph with vertex flow constrained and bottom image is transformed graph without vertex flow constrained.

3. 4. 2 8

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