1 max flow jose rolim university of geneva. max flowjose rolim2 graph models graphs (used as models...

1

Max Flow

Jose RolimUniversity of Geneva

Max Flow Jose Rolim 2

Graph Models

Graphs (used as models of "transportation" networks) have the following components:

Links / edges: carrier of some sort of traffic and have capacities

Nodes:- source nodes: generate traffic- sink nodes: absorb/demand traffic- intermediate nodes: "switches" which pass traffic between different edges

Description of the traffic itself


Examples

Example Edges Nodes

Highway system Roads Interchanges

Comp. Communication Networks

Communication links

Routers

Oil/gas distribution networks

Pipes Junctions with pumps


Example Graph

-

Max. capacity of edgeActual flow through this edge

Source node

Sink node


Simplified Model

The network is modeled simply asa) a directed graph G = (V,E) withb) non-negative capacity on each

edge,c) a single source node, s, andd) a single sink node, t


Some assumptions…

We will simplify our discussion by assuming the following:

(i) No edge enters s, the source(ii) No edge leaves t, the sink(iii)At least one edge is incident to

each node(iv)All capacities are integers


Definition of Flow

Flow will be our abstract term for the traffic; as an entity originating at the source node and absorbed at the sink node

An (s,t) flow is a function f that assigns a non-negative number to each edge e in the graph G = (V,E)

The value of a flow is


Intuitive definition

f(e) intuitively is the “flow” on edge e subject to two conditions:

(i) capacity constraint: 0 ≤ f(e) ≤ c(e), where c(e) is the capacity of edge

e

(ii) conservation property: for each node v, other than s and t,

∑ f(e) = ∑ f(e) e into v v into e


Max Flow Problem

What is the maximum amount of flow that can be sustained in G?

Clearly, the only obstacle to the flow are the capacities of the edges in G. The bottleneck doesn’t necessarily occur from the edges originating at s, or those terminating at t. It can arise from a complicated interaction among the edges, as flow snakes through G.


Max Flow Problem

Given a flow network G = (V,E,c), a source s, and a sink t,

(1) What is the max flow that can be sustained in G (between s and t)?

(2) Then, find an algorithm to best assign f values to the edges in G to answer (1)


Sample Applications

The Max Flow problem shows up in many places, some less intuitive than others.

For instance, consider a shipping company seeking to ship as many units of a product (hockey pucks) as possible Product starts at a source factory in

Vancouver and ends in Winnipeg The problem may be represented

graphically as show in the following slide…


Transportation

(1). Let the capacity of an edge be the number of units that can be shipped between two cities by a shipping company

(2). Let the current number of units being shipped between cities be denoted as flow between those cities.

The following graph can therefore represent the problem:Nodes denote cities

Can be interpreted as 4 units being shipped out of 4 possible


Matrix Rounding

A p x q matrix of decimals may be summed along the rows and columns

Rounding each entry up or down between ceil(a) and floor(a) may cause the sums to change.

Trying to keep the sum consistent along rows does not necessarily keep the sum consistent along cols and vice-versa (see fig.)

3.1 6.8 7.3 17.2

9.6 2.4 0.7 12.7

3.6 1.2 6.5 11.3

16.3

10.4

14.53 7 7 17

9 2 1 12

3 1 7 11.3

15 10 15


Max Flow as a Solution

(1). Create a node, i, for each row, and a node, j, for each col in the matrix, and two additional nodes, s (source) and t (sink)

(2). For each entry in the matrix, create an edge with cost (floor(Mij), ceil(Mij)) corresponding to rounding up or rounding down

(3). Consistent matrix rounding is evidenced by feasible flow through the network


Graph Representation

1

2

3

st

1’

2’

3’

(10,11)(12,13)

(11,12)

(3,4)

(6,7)(7,8)

(9,10)(2,3)

(0,1)

(3,4)

(6,7)

(1,2)

(16,17)(17,18)

(14,15)


Scheduling

Assume a set of jobs, J, is to scheduled on M identical processors, with each job having 3 parameters: processing time (pi), ready time (ri) and due time (di).

ri + pi <= di if the job can be performed

Preemption is allowed – jobs can be interrupted and restarted arbitrarily


Example

Let M = 3 for the following job set:

The job set may therefore be represented graphically as shown


(1). Create a node for every interval in the interval diagram and one for every job as well as a source node, s, and a sink, t.

(2). Connect s to job j with capacity pi.(3). Connect each interval node Tjk to sink t with

capacity (k − j) * M, indicating the number of machine units available in that interval.

(4). Connect a job i to each interval Tjk such that ri % j and di ! k. The capacity of this edge is (k − j) indicating the number of machine units allotable to job i in this interval.

(5). There is a soln to the problem iff there is a max flow of value

Converting to Graph Form


Sample Graph

s

1

2

3

4

T13

T34

T45

T57

T79

t

1.5

1.25

2.1

3.6

1

1

2

1

1

1

2

2

2

6

3

3

6

6

This sample graph demonstrates some of the power of max flow. Many scheduling problems are NP-Complete, so finding a soln. to such a problem is very useful

Thus max flow may be used to solve a wide range of seemingly unrelated problems and is useful beyond simple representations of computer networks


Ford-Fulkerson Method

Developed by Ford and Fulkerson [1956] Widely used and influential ideas:

augmentation, residual network, and famous min-cut max-flow theorem.

Algorithm: Generic FF (G,s,t)1. Initialize f = 0 (i.e. zero flow on all edges)2. While there exists an “augmenting path” p from

s to t1. Push flow along p2. Reduce capacities along p


Example 1:

s

u v

x y

t

2

4

4

6 5

2

3

1

5

According to the greedy approach of the Generic FF algorithm, consider the maximum flows from s to t.


Example 1 (contd.)

1. Path s-u-v-t gives f=2.

s

u v

t

2/2

2/4

2/5


Example 1 (contd.)

2. Path s-x-y-t gives f=2+2=4.

s

x y

t

2/4

2/2

2/3


Example 1 (contd.)

3. Path s-x-u-v-t gives f=2+2+2=6. This gives us a total flow of 6, but it is not an optimal solution.

s

u v

x

t

4/4

4/4

2/6

4/5


Example 2:

By employing a greedy approach, we get a flow of 2. However, by not employing a greedy approach we can get a flow capacity of 3 in the above network.


Residual Network

Given an edge (u,v) with residual capacity cf(u,v) of an edge (u,v) is defined as cf(u,v) = c(u,v) – f(u,v)

Residual capacity is the additional flow one can send on an edge, possibly by canceling some flow in opposite edge.

Example 1:cf(u,v) = 10-6 = 4

cf(u,v) = 0 - (- 6) = 6

u v

6/10


Residual network

Residual network Gf is (V,Ef ), where Ef = {(u, v) | cf (u, v) > 0}.

Note: |Ef | <= 2|E|.

When is an edge (u, v) ε Ef? if (u, v) ε E and f(u, v) < c(u, v), or if (v, u) ε E and f(v, u) > 0.


Example 2


Augmenting Paths

An augmenting path ‘p’ is a simple path in Gf

from s to t. The residual capacity of p is cf(p) = min { cf(u,v)

| (u,v) ε p }. By definition, cf(p) > 0.

Can you spot an augmenting path in the graph below?


Now, Ford-Fulkerson changes to –1. Initialize f = 0 2. While there exists an “augmenting path” p from

s to t in Gf

1. Push flow along p2. Reduce capacities along p

s

u v

x y

t

2

1

1

1

1

1

1

2

Example 1


Example 1(contd.)

Consider s-x-y-t f = 1

s

x y

t

1

1

1


Example 1(contd.)

Consider s-u-v-t f=1+1

s

u v

t

1

1

1


Example 1(contd.)

This gives us the intermediate graph:

s

u v

yx

t

1

1

1

1

1

1

1

1

1 1


Example 1(contd.)

Finally, considering s-u-y-x-v-t, we get the residual graph shown below.

f=1+1+1 = 3

u

s

v

x y

t

2

1

1

1

2

1

1

1

1


Cuts

A cut (S, T) is a partition of V with s ε S and t ε T.

f(S, T) is the net flow across cut (S, T). c(S, T) is the max capacity of cut (S, T); use only

forward edges.


Example 1

In this example, f(S, T) = 12 − 4 + 11 = 19. c(S, T) = 12 + 14 = 26.


Flows and Cuts

No matter how you separate s and t, the net flow across (S, T) is exactly |f|.

Cut Lemma: If f is a flow and (S, T) a cut in G, then f(S, T) = |f|.


Proof


MaxFlow-MinCut Theorem

The following are equivalent:1. f is a maxflow.2. No augmenting path in Gf .3. |f| = c(S, T) for some cut (S, T).

PROOF: (1) => (2). Augmenting path increases f. (3) => (1). By Cut Lemma, |f| = f(S, T) ≤ c(S, T). So, equality

means flow is a maxflow. (2) => (3). Define reachable set, S = {v | such that path from s to v in

Gf} T = V − S. Then, t ε T—because no augmenting path. For

any u ε S, v ε T, f(u, v) = c(u, v); otherwise, (u, v) ε Ef , and v will be

reachable from s. Thus, |f| = f(S, T) = c(S, T).


Updated FF Algorithm

1. for each edge (u, v) є E do2. f(u, v) = f(v, u) = 0;3. while there is a path p in Gf do4. cf (p) = min{cf (u, v) | (u, v) є p}5. for each (u, v) є p do6. f(u, v) = f(u, v) + cf (p)7. f(v, u) = −f(u, v)8. end9. end.


Example 1

Original Residual Graph Choose a path

u

s

v

x y

t

16

14

79

20

10

13

12

4

4


Example 1 (contd.)

Edit the path following the while loop Choose a new path

u

s

v

x y

t

12

10

79

20

14

13

12

4

4

4


Example 1 (contd.)

Keep iterating the while loop Choose the final path

u

s

v

x y

t

12

3

79

13

14

6

12

4

4

11

7

7


Example 1 (contd.)

No more augmenting paths Final residual graph

u

s

v

x y

t

3

79

1

14

6

12

4

16

11

19

7


Example 1 (contd.)

Final Capacity Graph All (S,T) cuts = 23

u

s

v

x y

t

16/16

11/14

7/70/9

19/20

0/10

7/13

12/12

4/4

4/4


Analysis of Ford - Fulkerson

Correctness follows from maxflow–mincut

BFS or DFS to find augmenting path takes O(m) time

If edge capacities are integers with maximum cap U, then the number of augmentations is at most Un

Total running time O(m) * O(nU) = O(nmU)

If the edge capacities are irrational, then it is not clear that the algorithm will terminate in polynomial time.


A Pathological Case for FF

For large edge capacities and certain graph topologies the run-time may be pathological. Assume M >109:


Polynomial Time Maxflow

FF method does not prescribe how to pick augmenting paths

Edmunds-Karp heuristics: Shortest augmenting path first Max capacity augmentation

Heuristics are somewhat obvious, cleverness lies in their analysis


Shortest Path Augmentation

Always augments flow from the shortest path from s to t

Let df(s, v) denote the shortest path distance from s to v in the current residual graph Gf .

Standard BFS algorithm finds the df(s, v) in O(m) time

1 max flow jose rolim university of geneva. max flowjose rolim2 graph models graphs (used as models...

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