Network Flow The Maximum Flow Problem and The

Network Flow The Maximum Flow Problem and The Ford-Fulkerson Algorithm

Types of Networks Internet Telephone Cell Highways Rail Electrical Power Water Sewer Gas …

Maximum Flow Problem How can we maximize the flow in a network from a source or set of sources to a destination of set of destinations? The problem reportedly rose to prominence in relation to the rail networks of the Soviet Union, during the 1950's. The US wanted to know how quickly the Soviet Union could get supplies through its rail network to its satellite states in Eastern Europe. In addition, the US wanted to know which rails it could destroy most easily to cut off the satellite states from the rest of the Soviet Union. ◦ It turned out that these two problems were closely related, and that solving the max flow problem also solves the min cut problem of figuring out the cheapest way to cut off the Soviet Union from its satellites. The first efficient algorithm for finding the maximum flow was conceived by two Computer Scientists, named Ford and Fulkerson. The algorithm was subsequently named the Ford. Fulkerson algorithm, and is one of the more famous algorithms in computer science. Source: lbackstrom, The Importance of Algorithms, at www. topcoder. com

Network Flow • A Network is a directed graph G • Edges represent pipes that carry flow • Each edge <u, v> has a maximum capacity c<u, v> • A source node s in which flow arrives • A sink node t out which flow leaves Goal: Max Flow

Network Flow The network flow problem is as follows: ◦ Given a connected directed graph G with non-negative integer weights, (where each edge stands for the capacity of that edge), ◦ 2 different vertices, s and t, called the source and the sink, such that the source only has out-edges and the sink only has in-edges, ◦ Find the maximum amount of some commodity that can flow through the network 12 a b 16 s 13 4 10 9 20 7 t 4 c 14 d Each edge stands for the capacity of that edge.

Network Flow One way to imagine the situation is imagining each edge as a pipe that allows a certain flow of a liquid per second. ◦ The source is where the liquid is pouring from, and the sink is where it ends up. ◦ Each edge weight specifies the maximal amount of liquid that can flow through that pipe per second. ◦ Given that information, what is the most liquid that can flow from source to sink per second, in the steady state? 12 a b 16 s 13 4 10 9 20 7 t 4 c 14 d Each edge stands for the capacity of that edge.

Network Flow 12 a b 16 4 s 10 9 20 14 t d This graph contains the capacities of each edge in the graph. The b 12/16 4 c 12/12 19/20 0/9 7 13 a s 0/4 0/10 t 7/7 11/13 4/4 c 11/14 d Here is an example of a flow in the graph. flow of the network is defined as the flow from the source, or into the sink. For the situation above, the network flow is

12 a b 16 s 4 10 9 20 7 s 0/4 0/10 19/20 d t 7/7 11/13 capacities b 12/16 t 4 14 12/12 0/9 13 c a 4/4 c 11/14 d flow The Conservation Rule: ◦ In order for the assignment of flows to be valid, we must have the sum of flow coming into a vertex equal to the flow coming out of a vertex, for each vertex in the graph except the source and the sink. The Capacity Rule: ◦ Also, each flow must be less than or equal to the capacity of the edge. The flow of the network is defined as the flow from the source, or into the sink. ◦ For the situation above, the network flow is 23.

Network Flow In order to determine the maximum flow of a network, we will use the following terms: ◦ Residual capacity – is simply an edge’s unused capacity. Initially none of the capacities will have been used, so all of the residual capacities will be just the original capacity. a 0/12 b 0/16 0/20 0/9 s 0/4 0/10 t 0/7 0/4 0/13 c 0/14 d Using the notation: used / capacity. Residual Capacity: capacity - used.

Network Flow ◦ Residual capacity of a path – the minimum of the residual capacities of the edges on that path, which will end up being the max excess flow we can push down that path. ◦ Augmenting path – defined as one where you have a path from the source to the sink where every edge has a nonzero residual capacity. a 0/12 b 0/16 0/20 0/9 s 0/4 0/10 t 0/7 0/4 0/13 c 0/14 d Using the notation: used / unused. Residual Capacity: unused - used.

Ford-Fulkerson Algorithm While there exists an augmenting path Add the appropriate flow to that augmenting path So we’re going to arbitrarily choose the augmenting path s, c, d, t in the graph below: ◦ And add the flow to that path. a 0/12 b 0/16 0/20 0/9 s 0/4 0/10 t 0/7 4/4 0/4 4/13 0/13 c 4/14 0/14 d ◦ Residual capacity of a path – the minimum of the residual capacities of the edges on that path. ◦ 4 in this case, which is the limiting factor for this path’s flow.

Ford-Fulkerson Algorithm While there exists an augmenting path Add the appropriate flow to that augmenting path Choose another augmenting path (one where you have a path from the source to the sink where every edge has a non-zero residual capacity. ) ◦ s, a, b, t a 12/12 0/12 b 12/16 0/16 12/20 0/9 s 0/4 0/10 t 0/7 4/4 4/13 c 4/14 d ◦ Residual capacity of a path – the minimum of the residual capacities of the edges on that path. ◦ 12 in this case, which is the limiting factor for this path’s flow.

Ford-Fulkerson Algorithm While there exists an augmenting path Add the appropriate flow to that augmenting path Choose another augmenting path (one where you have a path from the source to the sink where every edge has a non-zero residual capacity. ) ◦ s, c, d, b, t a 12/12 b 12/20 19/20 12/16 0/9 s 0/4 4/13 11/13 0/10 t 0/7 7/7 4/4 c 4/14 11/14 d ◦ Residual capacity of a path – the minimum of the residual capacities of the edges on that path. ◦ 7 in this case, which is the limiting factor for this path’s flow.

Ford-Fulkerson Algorithm While there exists an augmenting path Add the appropriate flow to that augmenting path Are there any more augmenting paths? ◦ No! We’re done ◦ The maximum flow = 19 + 4 = 23 a 12/12 b 12/16 19/20 0/9 s 0/4 0/10 t 7/7 11/13 4/4 c 11/14 d

Ford-Fulkerson Algorithm Runtime While there exists an augmenting path Add the appropriate flow to that augmenting path We can check the existence of an augmenting path by doing a graph traversal on the network (with all full capacity edges removed. ) ◦ This graph, a subgraph with all edges of full capacity removed is called a residual graph. It is difficult to analyze the true running time of this algorithm because it is unclear exactly how many augmenting paths can be found in an arbitrary flow network. ◦ In the worst case, each augmenting path adds 1 to the flow of a network, and each search for an augmenting path takes O(E) time, where E is the number of edges in the graph. ◦ Thus, at worst-case, the algorithm takes O(|f|E) time, where |f| is

Edmonds-Karp Algorithm This algorithm is a variation on the Ford. Fulkerson method which is intended to increase the speed of the first algorithm. The idea is to try to choose good augmenting paths. ◦ In this algorithm, the augmenting path suggested is the augmenting path with the minimal number of edges. (We can find this using BFS, since this finds all paths of a certain length before moving on to longer paths. ) ◦ The total number of iterations of the algorithm using this strategy is O(VE). Thus, its total running time is O(VE 2).

Network Flow Consider a network flow problem where the goal was to figure out how to change the capacities to increase the maximum flow in the network. ◦ This might be applicable if you wanted to get a greater throughput of data in a network and you wanted to figure out where in the network you needed to add a router that could handle more data per unit of time.

Network Flow How to increase the maximum flow of a network. ◦ Let’s say you restrict your search in your network (that is already at max flow) to paths that have at most one edge that is being used to capacity. ◦ For example, s v 1 v 2 … vn t, where each edge has extra capacity EXCEPT for the edge vi vi+1. What is a simple fix to add extra capacity to one of the edges that will add maximum flow to the network? How much will this simple fix add?

Network Flow Take a look at the extra capacity available through each of the other edges ◦ ◦ s → v 1, v 1 → v 2, …, vi-1 → vi, vi+1 → vi+2, …, vn → t. Take the minimum of these values. Then, add this much capacity to the edge vi → vi+1. Adding more than this would do no good, because then some other edge would become the bottleneck. This just adds enough capacity so that the original edge is no more of a bottleneck than the “second worst” edge in this particular path.

References Slides adapted from Arup Guha’s Computer Science II Lecture notes: http: //www. cs. ucf. edu/~dmarino/ucf/cop 350 3/lectures/ Additional material from the textbook: Data Structures and Algorithm Analysis in Java (Second Edition) by Mark Allen Weiss J Elder’s Network Flow slides, York University