Lecture 3

Preview of Markov Process

• A sequence of random variables X1 , X2,….,Xn ,….. such that – Xi+1 is independent of X1,….Xi-1 given Xi

– Pr(Xi+1 = ai+1/ Xi = ai, …., X1 = a1) = Pr(Xi+1 = ai+1/ Xi = ai )

• Discrete time discrete state markov chain• So a markov chain evolution can be specified by

– Initial states

– Transition probabilities Pr(Xi+1 = ai+1/ Xi = ai ) = pi,i+1

0 10.5

0.5

0.5 0.5

State A communicates with state B if there is a positive probability path from A to B

A set of states is closed if all states in the set communicate with each other, and no state in the set communicates to any state outside the set, e.g., {0, 1}

A state a is open if it communicates to some other state which does not communicate to a, e.g., {2}

21.0

A state has a period d if the process can only return to this state in intervals which are multiples of d

More precisely, di is the g.c.d. of {k: pi ik

> 0},

Where pi ik

is Pr(Xk = i/ X0 = i)

0 10.5

0.5

0.5 0.5

d0= 1

0 11.0

1.0

d0= 2

Period of any two communicating states are equal

Consider the wireless system

Let the system have random arrivals, i.e., the number of arrivals for the different sessions is random

Arrival process for each session is i.i.d (independent and identically distributed)

Number of arrivals of session i in slot t is independent of the number of arrivals of session i or any other session in past or future slots, and also independent of number of arrivals of any other session in the same slot

Probability that a packet arrives for session i in any slot t is qi

A state is aperiodic if its period is 1

Under this assumption the queue length process at the nodes (values of the Biu(t) s for all sessions i and nodes u) constitute a markov process

u v x ye1 e2e3

Session 2

Session 1

(B1u(t) B1v(t) B1x(t) B2v(t) B2x(t) B2y(t) ) is a markov chain

This markov chain consists of one closed set, and some other open states (prove it!)

Periodicity of the closed set is 1 (prove it!) (under a minor assumption on the values of the arrival probabilities, qi s)

How many states does this markov chain have?

Then the probability distribution of the markov chain converges to a probability distribution with finite expectation

This means the system stabilizes!

The function V(x) we choose is icixi2

Suppose we can find a scalar function V(x) of vector x such that E(V(B(t + 1))/ V(B(t) = x)) 0 for all states x, except a finite number of states and V(x) 0 for all states x,

If the system is markov with an aperiodic closed set, and other open states, and

Problem Complexity

Most optimization problems can be expressed as decision problems (yes/no answer)

For example, shortest path problem can be expressed as, is there a path with weight less than k in a graph G?

P is the class of decision problems which can be decided in polynomial time

Language: has certain membership (a, b, c belongs to English)

Language L = { (G, k): G has spanning tree of weight at most k}

A language is in P if its membership can be tested in polynomial complexity

How about {(G, k): G has an independent set of size at least k}?

A language is in NP (non-deterministic polynomial complexity) if given a member of the language, and a certificate to prove membership, we can verify the certificate in polynomial complexity

Both previous examples are in NP

How about {G: G has a unique hamiltonian cycle}?

A language is NP-hard, if it can be solved in polynomial complexity, then all other languages in NP can also be solved in polynomial complexity,

i.e., any other language in NP can be transformed in to this language in polynomial complexity

A language is NP-complete if it is in NP and is NP-hard

Note that a NP-hard language does not necessarily belong to NP.

Example: {(G, k): G does not have a path of weight higher than k}.

P NP NP-hard

NP-complete

Is P = NP?

probably not

Technique for proving that a problem is NP-hard

There are already problems proved to be NP-hard. Show that if you can solve this problem (problem we are trying to prove NP-hard) in polynomial complexity, you can solve such a known problem in polynomial complexity using the solution of this problem.

We will do such a reduction.

Quality of Service Routing

• Quality of Service– Delay– Packet loss

• Packet loss – Transmission error– Buffer overflow

• Delay consists of – Queueing delay– Transmission delay– Propagation delay

• First two can be related to bandwidth

• If we guarantee a certain bandwidth for a session on each of its links, its queueing and transmission delay can be upper-bounded

• Congestion related packet loss can also be upper-bounded for certain guaranteed bandwidth

• Using Shannons channel capacity theorem transmission error related packet loss can also be upper-bounded

So quality of service requirement translates to bandwidth requirement in every link

Given a network with finite bandwidths on the links, and a connection with a certain bandwidth requirement, find a path which gives that bandwidth to the session in every link,

If there is no such path detect the condition

SourceDestination

Path 1

Path 2

1 1.5

0.5 2.5

Generalization

There is a reward associated with every link

Reward of a path is the minimum reward on a link

Find a path with required reward

Pseudo-code

Delete any link which has bandwidth less than the required amount

Find any path between the source and the destination in the remaining link

Using depth first search the second step has complexity O(V + E)

First step has complexity O(E)

Overall complexity O(V + E)

Find an ``efficient’’ algorithm which gives the path with the highest reward

Offline QoS RoutingWe have a set of connection requests, each with a source, destination and a bandwidth requirement

Is it possible to satisfy all the requests?

Sum of the bandwidths used by connections in a link should not exceed link capacity

NP-complete

Polynomial complexity computable if we are allowed to split paths for a single connection

If we are not allowed to split paths, then need to study

approximation algorithms

heuristics

Widest path heuristic

Route a connection through the largest reward path

Find the residual bandwidth in each link of the network

Residual bandwidth Residual bandwidth – bandwidth used by the connection

Initially residual bandwidth of a link is the entire link bandwidth

Route another connection

If a connection can not be routed, backtrack.

Re-route the previous connection through the ``second best`` path, and now try routing this connection.

Limit the number of backtracks

If the number of allowed backtracks increase,

Computation time increases

Better results are expected

A more Efficient HeuristicStart with the most difficult demand

Demand is high

All routes for the demand have low bandwidths

Why should we start with the most difficult task?

Choose a good route for the demand

Go to the next most difficult demand

Backtrack a limited number of times, whenever demand can not be routed.

Abstraction and Constraint Satisfaction techniques for planning bandwidth allocation

Christian Frei (Swiss Federal Institute of Technology (EPFL)), Boi Faltings (Swiss Federal Institute of Technology (EPFL)) ,

INFOCOM 2000

A group of nodes constitute a -blocking island, if any node in the set is connected to another in the set by path of bandwidth at least .

112 96

20128

68

22

28

60

5810

256

769

118N2

N4

N1

N3

64-blocking islands16

10

N2

N4

N1

N3

56-blocking islands

22

28

16 58

60

N6

N5

a

b c

d e

i

j k

f

g

h

10

N2

N4

N1

N3

16-blocking islands

22

28

16 58

60

N6

N5

N7

If two nodes are in a -blocking island, then there is a path of bandwidth between them.

N7

N5N6

N1 N2 N3N4

a b c d e f g h i j k

64

56

16

If the lowest common ancestor of two end points is level , then any path for routing the demand can not have more than level +1 bandwidth. So higher the

lowest common ancestor, more difficult is the demand.

However, this does not consider the amount of the demand

The most difficult demand is the one with the least difference between the least common ancestor bandwidth and the demand amount

3264

8

(b, c) most difficult

Form the -blocking islands for all bandwidth levels

Start with = highest value of the bandwidth

Choose any vertex, add all vertices connected by an edge of b.w. or higher to this vertex, further add all vertices similarly connected to the new vertices and so on. When you can not add any more, you end up with a -blocking island.

Start with a vertex which is not in any -blocking island, and repeat the same till all vertices are exhausted.

Repeat the entire procedure with = highest value of the bandwidth and so on.

Choose the most difficult demand first

The best route for the demand is the one that lies completely in the lowest level blocking island which consists of the two end points.

This is because connectivity between any two nodes in a -blocking island is at least .

Choose the next most difficult demand

Backtrack if necessary

Artificial intelligence based technique.

Online Routing of Demands

• Route demands as and when they arrive– Demands may be blocked

• Previous technique works well when all demands are known in advance.

• You can still use this for online routing, but you may need to reroute a lot of existing connections as future connections come in

• Any good online strategy must use some information about future demands– For example potential source destinations may be known

Minimum Interference Routing with application to MPLS traffic engineering

Murali Kodialam (Bell Labs, Lucent Technologies), T. V. Lakshman (Bell Labs, Lucent Technologies)

INFOCOM 2000

When a demand arrives, route it through a path which has a lot of residual bandwidth , and which

Does not overlap a lot with routes between other source destination pairs

There are a few source-destination pairs in the network.Demands are multiples of some constants

Source 1Destination 1

Path 1

Path 2

Source 2

Destination 2

The objective is to reduce blockings of potential future demands as much as possible.

Network flows

Given a network and a source and a destination find the maximum traffic the source can send to the destination, while not exceeding the link bandwidths, and using as many routes as it wishes to.

SourceDestination

Path 1

Path 2

1 1.5

0.5 2.5Max flow = 1.5

A valid flow is a traffic the source can send to the destination, while not exceeding the link bandwidths, and using as many routes as it wishes to.

Flow Max Flow

A cut is a partition of the vertex set into two sets, A and B such that A and B do not overlap

A contains the source

B contains the destinationSourceDestination

Path 1

Path 2

1 1.50.5 2.5

a

b

For blue cut, A = {source}, B = {destination, a , b}

For red cut, A = {source, b}, B = {destination, a }

Capacity of a cut is the total bandwidth of the edges who have one end in set A and the other end in set B

(edges which cross the cut)

SourceDestination

Path 1

Path 2

1 1.50.5 2.5

a

b

Capacity of the red cut = 3.5

Capacity of the blue cut is 1.5

Any flow is upper bounded by the capacity of any cut

This is because flow is the amount of traffic you are sending from source to destination,

Source is in one set of the cut

Destination is in the other set of the cut

The two sets of the cuts do not overlap

So any flow from the source to destination must ``flow” across the cut, and hence can not exceed the cut capacity

Max flow from source to destination Min. cut capacity

Max flow from source to destination = Min. cut capacity

This means if we reduce the bandwidths of any link in any min. cut even slightly, maximum flow across the source to destination is reduced.

Whereas this does not hold for other edges in the network. We can always find a small amount such that reducing the link capacities of the edges not in any min. cut by that amount, does not hurt the max. flow

Edges in the min. cuts are critical to ``connectivity’’ of the source and the destination.

Any edge in any min. cut will be denoted as a ``critical edge.’’

Incidentally, max flow and all min. cuts are polynomial complexity computable.

If link bandwidths are integers, then max flow allocation algorithm assigns integer flows to all links

So, max flow with the constraint that flow in any link must be integer is polynomial complexity computable.

However, given n source destination pairs, and a flow vector (f1, ….fn) (fi is the flow for session i) , the question whether we can route this flow vector without violating the capacity constraints, using only integer flows in the links, and as many routes as we wish to, is NP-hard!

Interference caused by a path is measured by computing the max flow in the residual capacity graph between all source destination pairs.

Let demand B be routed along path p

Look at the residual capacity graph after routing the demand along p (subtract bandwidth B from all edges in p)

Find the maximum flow between each source destination pair in the residual capacity graph.

Consider the minimum value of these individual maxflows. This is the interference caused by the path p

Interference Measure

Another measure of the interference can be the lexicographic ordering of these maxflows

Another measure is the weighted sum of these maxflows.

Choose the path p which maximizes the interference measure

Higher the interference measure, less is the interference

The problem is NP-hard

heuristics

MIRA

While routing a new demand, one should avoid the critical edges of other source destination pairs as much as possible.

So when a demand arrives, look at the network with the residual capacities.

Find the max. flow of each source destination pair separately (polynomial complexity)

Find the critical edges for every source destination pair (polynomial edges)

Give weights to all the critical edges of all the source destination pairs.

Weights to edges which are not critical for any source destination will be 0 or very small positive numbers

Use Dijkstra to find the minimum weight path for the new demand.

Remove edges with residual capacity < demand

Note that the minimum weight path will have the requisite bandwidth. Why?

The min. weight path would try to avoid critical edges as much as possible?

Weights for critical edges could be same or different for different source destination.

For example, weight could be largest for the most valued flow. This would avoid using the critical edges of the most valued flow by other flows as much as possible, and reduce the blocking for the most valued flow.

Do we gain anything by giving small nonzero weights to the non-critical edges instead of 0 weights?

Weight could be the largest for the source destination pair with least max. flow in the residual capacity graph (this has the largest blocking possibility, so its critical edges should be preserved the most).

Weight could be in lexicographic order of max. flow values, the one with least value of max flow gets much larger weight for its critical edges than others, the one with with second min. value of max flow gets second largest weight for its critical edges (much larger than the others), etc….

How do we compute all the critical edges in the graph?

• Typically, routing decisions are not taken for every connection, but at intervals.

• For this case, lexicographic weighing of edges performs better.

• Also, we should have non-negligible weights for edges which are not critical, but ``close to critical.’’– The capacities in these edges allow increasing flows,

but only by a small amount

Minimum Interference Routing with application to MPLS traffic engineering

Koushik Kar (University of Maryland), Murali Kodialam (Bell Labs, Lucent Technologies), T. V. Lakshman (Bell Labs, Lucent Technologies)

JSAC Nov. 2000