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Lecture 3

Lecture 3. Preview of Markov Process. A sequence of random variables X 1 , X 2 ,….,X n ,….. such that X i+1 is independent of X 1 ,….X i-1 given X i Pr(X i+1 = a i+1 / X i = a i , …., X 1 = a 1 ) = Pr(X i+1 = a i+1 / X i = a i ) Discrete time discrete state markov chain

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Lecture 3

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  1. Lecture 3

  2. 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

  3. 2 1.0 A state a is open if it communicates to some other state which does not communicate to a, e.g., {2} 0.5 1 0 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}

  4. 0.5 1 0 0.5 0.5 0.5 1.0 1 0 1.0 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 ikis Pr(Xk = i/ X0 = i) d0= 1 d0= 2

  5. Period of any two communicating states are equal A state is aperiodic if its period is 1 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

  6. e3 e1 e2 Session 1 Session 2 u v x y • 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 (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)

  7. If the system is markov with an aperiodic closed set, and other open states, and Suppose we can find a scalar function V(x) of vector x such that E(V(B(t + 1))- V(x) / B(t) = x)  0 for all states x, except a finite number of states and V(x) 0 for all states x, 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

  8. 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}

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

  10. 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 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}. A language is NP-complete if it is in NP and is NP-hard

  11. NP-complete NP-hard NP P Is P = NP? probably not

  12. 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.

  13. 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

  14. 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

  15. Path 1 1 1.5 0.5 2.5 Source Destination Path 2 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

  16. 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

  17. 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 First step has complexity O(E) Using depth first search the second step has complexity O(V + E) Overall complexity O(V + E) Find an ``efficient’’ algorithm which gives the path with the highest reward

  18. Offline QoS Routing We 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

  19. Widest path heuristic Initially residual bandwidth of a link is the entire link bandwidth 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 Route another connection

  20. 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

  21. A more Efficient Heuristic • Start 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.

  22. 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 .

  23. N4 N4 64-blocking islands 56-blocking islands N3 N3 N2 N2 N1 N1 N5 16 58 10 22 28 60 N6 k a j 256 112 96 16 58 g i 10 b c 22 20 9 128 76 28 h f d e 118 60 68

  24. N4 16-blocking islands N3 N2 N1 N5 16 58 10 22 28 60 N6 N7 If two nodes are in a -blocking island, then there is a path of bandwidth  between them.

  25. N7 8 32 64 16 (b, c) most difficult N6 N5 56 N4 N1 N2 N3 64 k i j e f g h a b c d 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

  26. 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  = second highest value of the bandwidth and so on.

  27. 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.

  28. 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

  29. Minimum Interference Routing with application to MPLS traffic engineering • Murali Kodialam (Bell Labs, Lucent Technologies), T. V. Lakshman (Bell Labs, Lucent Technologies) • INFOCOM 2000 There are a few source-destination pairs in the network. Demands are multiples of some constants 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

  30. Source 1 Destination 1 Path 1 Source 2 Path 2 Destination 2 The objective is to reduce blockings of potential future demands as much as possible.

  31. Path 1 1 1.5 0.5 2.5 Source Destination Path 2 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. Max flow = 1.5

  32. Path 1 1 1.5 0.5 2.5 Source Destination Path 2 a b 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 destination For blue cut, A = {source}, B = {destination, a , b} For red cut, A = {source, b}, B = {destination, a }

  33. Path 1 1 1.5 0.5 2.5 Source Destination Path 2 a Capacity of the red cut = 3.5 Capacity of the blue cut is 1.5 b 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)

  34. 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

  35. 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.’’

  36. 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!

  37. Interference Measure 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

  38. 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

  39. 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 complexity)

  40. Remove edges with residual capacity < demand 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. 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.

  41. Do we gain anything by giving small nonzero weights to the non-critical edges instead of 0 weights? Weights for critical edges could be the 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. Weight could be the largest for the source destination pair with the least max. flow in the residual capacity graph (this has the largest blocking possibility, so its critical edges should be preserved the most).

  42. 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?

  43. 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 

  44. 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

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