Lecture 8

Why do we need residual networks?

• Residual networks allow one to reverse flows if necessary.

If we have taken a bad path then residual networks allow one to detect the condition and reverse the flow.

A bad path is one which overlaps with too many other paths.

Example

SourceDestination1 1

1

1

1

a

b

Paths source, a, destinations and source, b destination gives a flow of 2 units.

Path source, a, b, destination overlaps with both the optimal paths.

If we initially choose source, a, b, destination as our path, then no greedy strategy will be able to augment the network flow any further (unless we use residual edges which allows recovery)

Verify how we recover in spite of the initial bad choice, if we use the residual network to augment flows.

Defects of Augmenting Path Algorithm

Sending flow along a path requires O(V) complexity.

2

.

.

.

1

n

11

1

1

1

1Since there are n separate paths, we need to push flow through n paths separately.

Every time complexity is O(V)

How about if we push flow through edges separately without caring about flow conversation?

If there is excess flow at a node, then the excess is sent to downstream nodes.

Finally, all excess flow is routed back to the source, so that flow conservation is maintained.

2

.

.

.

1

n

11

1

1

1

1

Generic Pre-flow Push Algorithm

Augmenting path algorithms maintain feasibility at every stage, and gradually strive towards optimality

Pre-flow push algorithm initially pushes the maximum possible amount of flow, and strives towards feasibility.

Initially, flow in every edge originating from the source is set to the link capacity.

Distance label of every node is initially their distance to the destination.

Set distance label of source node s, d(s)=V

If a node has excess flow, send the flow along an admissible edge (end node of an admissible edge has less distance than the source node of the edge, and hence is closer to the destination).

Let the excess flow at a node u be e(u).

Let the capacity of admissible edge (u, v) in residual graph be ruv

Then flow routed along (u, v) is min(e(u), ruv )

A node (which is not source nor destination), with excess flow is called an active node.

If a node u with excess flow does not have an admissible edge in the residual network, then it is relabeled: d(u) is upgraded to d(v) + 1, where v is the node with minimum distance label amongst all nodes which have edges from u in the residual network.

The algorithm terminates when there are no active nodes.

Intuition Edges are like water pipes

Nodes are joints

Distance is like height from the ground.

Destination is at the ground.

Initially source is at the highest level and sends water to all adjacent nodes.

Whenever a node has accumulated water, it sends water to nodes at lower label.

So water moves towards the destination.

Sometimes water gets locally trapped as all neighboring nodes are at a greater height (no admissible arc).

Then, the node label is raised (relabeling).

5

3 4

2 3

1

1

42

Original Network/Residual Network with 0 flow

source

destination

Push 4 units along 1-3 and 2 units along 1-2

Push 4 unit along 3-4, and then 1 unit along 1-4

Relabel node 2 to 2

Route 1 unit through 2-3

1

1 0

25

1 3source

destination2

3 2

1 4

4 1

1 0

2E = 4

E = 2 E = 1

1

1 3source

destination2

3 2

1 4

4 1

1 0

2

4

E = 1

1

1 3source

destination2

3 2

1 4

4 1

2 0

2

4

E = 1

2 4

34

2

1

1

42

1

source

destination

1

1

2 0

2E = 1

Route 1 unit through 3-4 2 5

34

2

1

1

42

1

source

destination

Now no node has any excess

Total flow out of source is 6

Proof for Optimality

Distance labels always maintain requisite properties.

Initially, d(s) = V. Any difficulty in maintaining requisite properties?

At termination d(s) is V or greater. So there is no path from the source to destination in the residual network (no augmenting path).

At termination, no node except the source and destination has excess flow.

No node ever has any deficit flow.

So, flow allocation is feasible at termination

Since the residual network does not have an augmenting path, the flow is a maxflow.

Complexity is O(V2E)

Specific Implementations of Generic Preflow-Push algorithm

(FIFO)This algorithm pushes flow out of the active nodes in first in first out order.

The algorithm initially orders the active nodes in some manner.

It pushes flow from the first active node in the list, till the active node is no longer active or till it needs to be relabeled

Order of selection of active nodes has not been nailed down in the generic version

The algorithm terminates when the list is empty

In the latter case, the node is relabeled and added to the back of the list and picks the second node from the list.

Clearly, FIFO is a specific order for selecting the active nodes, and hence the proof for correctness is the same as that for generic preflow push algorithm

While pushing flow from this active node, the newly created active nodes are added to the back of the list

O(V3) complexity

Problem with FIFO

21

n -1

11

1 1 1

Source

Destination

Highest Label Preflow push

Pushes flow from the active node with the highest label

Terminates when there are no more active nodes

O(V2E0.5)

Multicommodity network flow We have several source destination pairs.

Refer to each as a separate commodity.

Need to maximize a weighted sum of the different flows

Flow allocation must be feasible:

Capacity Constraints

Total flow in any link is between 0 and the link capacity

Flow Conservation

Input flow for any commodity at a node equals the output flow of the commodity, for every node except the source/destination for the commodity.

Solution techniques use optimization ideas

However, we may not have integral flows even if link capacities are integers

In fact, unlike the single commodity flow problem, multi-commodity flow problem with the constraint that flows in the links are integers is NP-hard

Example Application

Wireless nodes have low battery power and hence limited lifetime.

Every transmission consumes energy.

Reception consumes less energy.

Lifetime of a node is related to the total energy spent in transmission.

Transmission energy increases with increase in distance to the receiving node.

Signal strength decreases as distance increases.

So if the node needs to transmit to a distant node, it needs to use greater power.

Power increases as d4 where d is the distance between the two nodes.

It would be interesting to ask the question that given a particular transmission requirement what is the maximum lifetime of a network?

Lifetime of a network is the maximum time in which all nodes are active.

Energy Conserving Routing in Wireless Ad-hoc Networks

J.H. Chang and L. Tassiulas, INFOCOM 2000

Clearly, routing strategy will affect the network lifetime.

For example, if all traffic is routed through a single node, it will die in a short time, and according to our definition network dies as well.

Routing strategy will depend on traffic requirements, and the node energies.

Problem Formulation

Node i has energy Ei

Transmission energy from node i to j is eij for each transmission (depends on the distance between the two nodes)

Let there be C sessions

Let Qic be the rate of packet generation in node i for

session c

Let qijc be the rate of information transmission in

link i, j for session c under the current routing

(routing determines the qijc )

Lifetime of a node i under flow q is Ti(q)= Ei/j Si eijc Cqijc

Every node has a maximum power. So there is a set of neighbors Si of node i.

Lifetime of a system is minimum of all its nodes

Tsys (q) = mini Ti (q)

Let Dc be the set of destinations of session C

Session c information must reach some node in Dc

Lifetime Maximization Objective

Maximize Tsys (q) = mini Ei/j Si eijc Cqijc

Subject to:

qijc0

i Sj qjic + Qi

c= j Si qijc session c, and all

nodes i Dc

Reformulation

Maximize Tsys (q)

Subject to:

j Si eijc CqijcTsys (q) Ei session c, and all nodes i

qijc0

i Sj qjic + Qi

c= j Si qijc session c, and all

nodes i Dc

Let qijc = qij

cTsys (q)

Reformulation

Maximize Tsys (q)

Subject to:

j Si eijc Cqijc Ei session c, and all nodes i

qijc0

i Sj qjic + Tsys (q) Qi

c= j Si qijc session c, and all

nodes i Dc

Special Case

Consider the special case where every session has a single source, i.e., Qi

c = 0 at all nodes except the source of session i. Thus, Qc is the input rate for session c.

Also, eij = ei neighbors j of i.

Also, pijc = qij

cTsys (q)/ Qc

The problem can be reformulated as follows:

Reformulation

Maximize Tsys (q)

Subject to:

j Si c Cpijc Ei /ei session c, and all nodes i

pijc0

i Sj pjic = j Si pij

c session c, and all nodes i except source and destination of session c.

i Sj pjic + Tsys (q) = j Si pij

c

• This is the multi-commodity flow problem with node capacities, i.e., there is an additional constraint that sum of the output flow through a node must not exceed the node capacity.– Problem with node capacity can be reduced to the problem without

node capacity, replacing each node by two nodes with a link between them with capacity equal to the node capacity

– Note that there is no link capacity in this problem (equivalently all

link capacities are infinity).

• Object of this multi-commodity flow problem is to find the minimum flow supported for each session.

In general, eij may not be the same for all neighbors of node i, since all neighbors may not be equidistant from node i.

The paper suggests several heuristics for the purpose.

Minimum Cut

Consider the augmenting path algorithm.

The set S reachable from the source node in the residual network when the algorithm terminates and the remaining nodes V-S constitute a minimum cut. Why?

Thus we know how to find a minimum cut.

How about finding the edges of all minimum cuts?

Why do we need the above?

Consider an edge (u, v) and the residual network corresponding to the max-flow, where

Edge (u, v) is filled to capacity in the original network.

Node v is not reachable from source s in the residual network.

Node u can not reach destination t in the residual network.

There is no path from u to v in the residual network.

A non-zero capacity edge (u, v) belongs to some minimum cut if and only if it satisfies all the above properties.

MIRA, JSAC 2000, Kar et. al