Math 5490 Network Flows

Transcription

1 Math 90 Network Flows Lecture 8: Flow Decomposition Algorithm Stephen Billups University of Colorado at Denver Math 90Network Flows p./6

2 Flow Decomposition Algorithms Two approaches to modeling network flow problems:. Define flows on arcs (as we have done so far).. Define flows on paths and cycles. (we will be doing this for some max flow and min-cost flow algorithms) Math 90Network Flows p./6

3 Node Imbalance Given a flow x = {x ij }, the node imbalance e(i) at node i is defined to be the flow into the node i minus the flow out. That is e(i) := X x ji X x ij. {j (j,i) A } {j (i,j) A } imbalance terminology excess node e(i) > 0 deficit node e(i) < 0 balanced node e(i) = 0 Math 90Network Flows p.3/6

4 e vs. b The imbalance equation e(i) := X x ji X x ij. {j (j,i) A } {j (i,j) A } looks an awful lot like the flow balance equations of a network flow problem. What is the difference between the imbalance e(i) and the supply/demand b(i)?. In some algorithms, we will want to consider flows that are not feasible.. A flow x is feasible if e(i) = b(i) for all nodes i. Math 90Network Flows p./6

5 Some notation P Collection of all paths in the graph. W Collection of all cycles in the graph. P A particular path. W A particular cycle. f(p ) The flow on path P. f(w ) The flow on cycle W. δ ij (P ) Indicates if arc (i, j) is in path P. δ ij (P ) = 8 < δ ij (W ) Indicates if arc (i, j) is in cycle W. 8 < δ ij (W ) = : : if (i, j) P. 0 otherwise. if (i, j) W. 0 otherwise. Math 90Network Flows p./6

6 Relationship Between Arc Flows and Path and Cycle Flows x ij = X P P δ ij (P )f(p ) + X W W δ ij (W )f(w ). Can we reverse this process? Math 90Network Flows p.6/6

7 Flow Decomposition Theorem Every path and cycle flow has a unique representation as nonnegative arc flows. Conversely, every nonnegative arc flow x can be represented as a path and cycle flow with the following two properties:. Every directed path with positive flow connects a deficit node to an excess node.. At most n + m paths and cycles have nonzero flow; out of these, at most m cycles have nonzero flow. Discussion Two aspects of the above theorem are trivial:. Every path and cycle flow has a unique representation as nonnegative arc flows. (Just add up the path and cycle flows for each arc).. Every nonnegative arc flow can be represented as a path and cycle flow. (Just let every arc be a path). Therefore the real significance of the theorem is the two properties of the path and cycle flow. Math 90Network Flows p.7/6

8 Main ideas of the proof:. The proof is constructive; an algorithm is given that constructs a path and cycle flow.. At each step of the algorithm, we construct a path (or cycle) starting from a deficit node i 0 and moving along arcs of positive flow until either an excess node i k is reached, or a node is repeated. Case : Node Repeated If a node is repeated, then there is a cycle W. (see Figure ). Set the flow of this cycle equal to the minimum flow on any arc in the cycle, and remove that flow from the network. NOTE : This changes the flow on at least one arc to 0. NOTE : The node imbalances are NOT affected. Math 90Network Flows p.8/6

9 Example Node Repeated e()= e()= A. Arc-Flow B. Cycle (3,,) identified, flow= e()= C. Flow on cycle (3,,) removed (Note, flow on arc (3,) is now 0). Figure Repeated Node in Path Math 90Network Flows p.9/6

10 Proof, cont.. Case : Excess Node Reached If an excess node i k is found, then we have a path P from a deficit node to an excess node. (See Figure ). Set the flow on this path equal to the minimum of the following quantities: e(i 0 ), e(i k ): We do not want to change a deficit node to an excess node, or vice versa. min {x ij (i, j) P }: We do not want to produce any arcs with negative flow. Remove this flow from the network (and update the node imbalances). Math 90Network Flows p.0/6

11 Example Excess Node Reached e()= e()= A. New arc flows B. Path flow identified; flow= 6 e()= Math 90Network Flows p./6

12 Proof, cont. 3. Every iteration does one of the following: Sets a positve arc flow to 0. (for a cycle, or possibly a path). Sets a nonzero node imbalance to 0. Thus, there can be at most n + m iterations. Since each iteration identifies exactly one path or cycle with nonzero flow, the decomposition has at most n + m paths and/or cycles with nonzero flow. Question: why are there at most m cycles with nonzero flow? Math 90Network Flows p./6

13 Circulation Flows Property 3.6: A circulation x can be represented as a cycle flow along at most m directed cycles. Math 90Network Flows p.3/6

14 Augmenting Flows A cycle W (not necessarily directed) in G is called an augmenting cycle with respect to the flow x if by augmenting a positive amount of flow f(w ) around the cycle, the flow remains feasible. The augmentation increases the flow on forward arcs and decreases the flow on backward arcs in the cycle. NOTE: W is and augmenting cycle with respect to a flow x if and only if W corresponds to a directed cycle in the residual network G(x). Math 90Network Flows p./6

15 Augmenting Cycle Theorem Let x and x be any two feasible solutions of a network flow problem. Then x equals x plus the flow on at most m directed cycles in G(x ). Furthermore, the cost of x equals the cost of x plus the cost of flow on these augmenting cycles. Math 90Network Flows p./6

16 Negative Cycle Optimality Theorem A feasible solution x of the minimum cost flow problem is an optimal solution if and only if the residual network G(x ) contains non negative cost cycle. Math 90Network Flows p.6/6