1. Network Optimization Models: Maximum Flow Problems In this handout: The problem statement Solving by linear programming Augmenting path algorithm

2. Maximum Flow Problem Given:Directed graph G=(V, E), Supply (source) node O, demand (sink) node T Capacity function u: E  R. Goal:Given the arc capacities, send as much flow as possible from supply node O to demand node T through the network. Example: 4 4 5 6 4 4 5 5 O A D B C T

3. Characteristics of a feasible flow Let x ij denote the flow through arc i  j. Capacity u ij of arc i  j is the upper bound on the flow shipped through arc i  j. Thus, we have the following constraints: 0  x ij  u ij, for any arc i  j Every node i, except the source and the sink, should satisfy the conservation-of-flow constraint, i.e., flow into node i = flow out of node i In terms of x ij the constraint is For any flow that satisfies the conservation-of-flow constraints, flow out of the source = flow into the sink This is the amount we want to maximize.

4. Linear Program for the Maximum Flow Problem Summarizing, we have the following linear program: 0  x ij  u ij, for any arc i  j This linear program can be solved by a special type of the Simplex method, called Network Simplex. (covered in Math 442/542) In this class, we will cover a more efficient algorithm for the Maximum Flow Problem, the Augmenting Path Algorithm.

5. Towards the Augmenting Path Algorithm Idea: Find a path from the source to the sink, and use it to send as much flow as possible. In our example, 5 units of flow can be sent through the path O  B  D  T ; Then use the path O  C  T to send 4 units of flow. The total flow is 5 + 4 = 9 at this point. Can we send more? O A D B C T 4 4 56 4 45 5 5 5 4 4 5

6. Towards the Augmenting Path Algorithm If we redirect 1 unit of flow from path O  B  D  T to path O  B  C  T, then the freed capacity of arc D  T could be used to send 1 more unit of flow through path O  A  D  T, making the total flow equal to 9+1=10. To realize the idea of redirecting the flow in a systematic way, we need the concept of residual capacities. O A D B C T 4 4 56 4 45 5 5 5 5 4 4 4 4 1 5 1 1

7. Residual capacities Suppose we have an arc with capacity 6 and current flow 5: Then there is a residual capacity of 6-5=1 for any additional flow through B  D. On the other hand, at most 5 units of flow can be sent back from D to B, i.e., 5 units of previously assigned flow can be canceled. In that sense, 5 can be considered as the residual capacity of the reverse arc D  B. To record the residual capacities in the network, we will replace the original directed arcs with undirected arcs: B D 6 5 B D 1 5 The number at B is the residual capacity of B  D; the number at D is the residual capacity of D  B.

8. Residual Network The network given by the undirected arcs and residual capacities is called residual network. In our example, the residual network before sending any flow: Note that the sum of the residual capacities on both ends of an arc is equal to the original capacity of the arc. How to increase the flow in the network based on the values of residual capacities? O A D B C T 4 4 5 6 4 4 5 5 0 0 0 0 0 0 0 0

9. Augmenting paths An augmenting path is a directed path from the source to the sink in the residual network such that every arc on this path has positive residual capacity. The minimum of these residual capacities is called the residual capacity of the augmenting path. This is the amount that can be feasibly added to the entire path. The flow in the network can be increased by finding an augmenting path and sending flow through it.

10. Updating the residual network by sending flow through augmenting paths Continuing with the example, Iteration 1: O  B  D  T is an augmenting path with residual capacity 5 = min{5, 6, 5}. After sending 5 units of flow through the path O  B  D  T, the new residual network is: O A D B C T 4 4 4 4 5 0 0 0 0 0 0 1 0 5 5 5 5 6 5 0 0 0

11. Updating the residual network by sending flow through augmenting paths Iteration 2: O  C  T is an augmenting path with residual capacity 4 = min{4, 5}. After sending 4 units of flow through the path O  C  T, the new residual network is: O A D B C T 4 4 0 0 0 4 5 0 0 0 1 0 5 5 5 0 1 4 4 4

12. Updating the residual network by sending flow through augmenting paths Iteration 3: O  A  D  B  C  T is an augmenting path with residual capacity 1 = min{4, 4, 5, 4, 1}. After sending 1 units of flow through the path O  A  D  B  C  T, the new residual network is: O A D B C T 0 0 5 5 0 4 4 4 0 0 0 1 5 1 4 4 3 3 1 1 1 2 4 0 5 3

13. Terminating the Algorithm: Returning an Optimal Flow There are no augmenting paths in the last residual network. So the flow from the source to the sink cannot be increased further, and the current flow is optimal. Thus, the current residual network is optimal. The optimal flow on each directed arc of the original network is the residual capacity of its reverse arc: flow( O  A )=1, flow( O  B )=5, flow( O  C )=4, flow( A  D )=1, flow( B  D )=4, flow( B  C )=1, flow( D  T )=5, flow( C  T )=5. The amount of maximum flow through the network is 5 + 4 + 1 = 10 (the sum of path flows of all iterations).

14. The Summary of the Augmenting Path Algorithm Initialization: Set up the initial residual network. Repeat –Find an augmenting path. –Identify the residual capacity c* of the path; increase the flow in this path by c*. –Update the residual network: decrease by c* the residual capacity of each arc on the augmenting path; increase by c* the residual capacity of each arc in the opposite direction on the augmenting path. Until no augmenting path is left Return the flow corresponding to the current optimal residual network