Điện tử viễn thông lect12 1 khotailieu

12 Traffic engineering lect12.ppt S-38.1145 – Introduction to Teletraffic Theory – Spring 2006 12 Traffic engineering Contents • • • • Topology Traffic matrix Traffic engineering Load balancing 12 Traffic engineering Topology • • • A telecommunication network consists of nodes and links – Let N denote the set of nodes indexed with n – Let J denote the set of nodes indexed with j Example: – N = {a,b,c,d,e} – J = {1,2,3,…,12} – link from node a to node b – link from node b to node a b c a 11 10 12 d e Let cj denote the capacity of link j (bps) 12 Traffic engineering Paths • We define a path (= route) as a – set of consecutive links connecting two nodes – Let P denote the set of paths indexed with p • Example: – three paths from node a to node c: • red path consisting of links and • green path consisting of links 11 and • blue path consisting of links 10, and b c a 11 10 12 d e 12 Traffic engineering Path matrix • • • Each path consists of a set of links This connection is described by the path matrix A,for which – element ajp = if j ∈ p, that is, link j belongs to path p – otherwise ajp = Example: – three columns of a path matrix 10 11 12 ac1 1 0 0 0 0 ac2 0 0 0 0 ac3 0 0 1 0 12 Traffic engineering Shortest paths • If each link j is associated with a correponding weight wj, the length lp of path p is given by lp = ∑ wj j∈ p – With unit link weights wj = 1, path length = hop count • Example: w=1 w=1 c b w=1 w=1 w=1 w=1 w=1 d w=1 a w=1w=1 w=1 e w=1 – two shortest paths (of length 2) from node a to node c 12 Traffic engineering Contents • • • • Topology Traffic matrix Traffic engineering Load balancing 12 Traffic engineering Traffic characterisation Traffic Circuit-switched Packet-switched e.g telephone traffic e.g data traffic Link Network Link Network 12 Traffic engineering Traffic matrix (1) • Traffic in a network is described by the traffic matrix T, for which – element tnm tells the traffic demand (bps) from origin node n to destination node m – Aggregated traffic of all flows with the same origin and destination – Aggregated traffic during a time interval, e.g busy hour or ”typical 5-minute interval” • c b tac d a e Example: – Traffic demand from origin a to destination c is tac (bps) 12 Traffic engineering Traffic matrix (2) • Below we present the traffic demands in a vector form – Let K denote the set of origindestination pairs (OD-pairs) indexed with k • • Traffic demands constitute a vector x, for which – element xk tells the traffic demand of OD-pair k Example: – if OD-pair (a,c) is indexed with k, then xk = tac c b xk d a e 10 12 Traffic engineering MPLS • MPLS (Multiprotocol Label Switching) supports traffic load distribution to parallel paths between OD-pairs – In MPLS networks, there can be any number of parallel Label Switched Paths (LSP) between OD-pairs – These paths not need to belong to the set of shortest paths – Each LSP is associated with a label and each MPLS packet is tagged with such a label • • MPLS packets are routed through the network via these LSP’s (according to their label) Traffic load distribution can be affected directly by changing the splitting ratios φpk at the origin nodes 15 12 Traffic engineering OSPF (1) • • OSPF (Open Shortest Path First) is an intradomain routing protocol in IP networks Link State Protocol – each node tells the other nodes the distance to its neighbouring nodes – these distances are the link weights for the shortest path algorithm – based on this information, each node is aware of the whole topology of the domain – the shortest paths are derived from this topology using Dijkstra’s algorithm • IP packets are routed through the network via these shortest paths 16 12 Traffic engineering OSPF (2) • Routers in OSPF networks typically apply ECMP (Equal Cost Multipath) – If there are multiple shortest paths from node n to node m, then node n tries to split the traffic uniformly to those outgoing links that belong to at least one of these shortest paths – However, this does not imply that the traffic load is distributed uniformly to all shortest paths! See the example on next slide • Traffic load distribution can be affected only indirectly by changing the link weights – splitting ratios φpk can not directly be changed – due to ECMP, the desired splitting ratios φpk may be out of reach 17 12 Traffic engineering ECMP y = x/4 y = x/2 b y = x/4 a y = x/2 d x c y = x/2 y = x/4 e y = x/4 f φ = 1/4 d g y = x/2 b φ = 1/4 e a c φ = 1/2 g f 18 12 Traffic engineering Effect of link weights on load distribution (1) maximum link load w=1 φ = 1/2 w=1 c b w=1 w=1 w = w=1 x w=1 d a x w=1 w=1w=1 w=1 e w=1 b y = 3x/2 φ = 1/2 b y = x/2 y = x/2 d c d a a φ=1 y = x/2 c e y=x e 19 12 Traffic engineering Effect of link weights on load distribution (2) maximum link load w=1 φ = 1/2 w=1 c b w=1 w=1 w = w=1 x w=1 d a x w=2 w=1w=1 w=1 e w=2 b y = x/2 c b y=x φ = 1/2 a φ = 1/2 d φ = 1/2 e c y = x/2 y=x y = x/2 a y = x/2 d y = x/2 e link weight increased 20 12 Traffic engineering Contents • • • • Topology Traffic matrix Traffic engineering Load balancing 21 12 Traffic engineering Load balancing problem (1) • • Given a fixed topology and a traffic matrix, how to optimally route these traffic demands? One approach is to equalize the relative load of different links, x=1 c=1 b c=1 c=1 e c=1 a d g c=1 c c=1 c=1 f c=1 ρj = yj/cj – Sometimes this can be done in multiple ways (upper figure) – Sometimes it is not possible at all (lower figure) – In this case, we may, however, try to get as close as possible, e.g by minimizing the maximum relative link load (called: load balancing problem) x=1 c=1 b c=1 a d c=1 c c=2 22 12 Traffic engineering Load balancing problem (2) • Load Balancing Problem: – Consider a network with topology (N,J), link capacities cj, and traffic demands xk Determine the splitting ratios φpk so that the maximum relative link load is minimized Minimize yj max c j∈J j  y j = ∑ ∑ A jpφ pk xk ∀j ∈ J  p∈P k∈K  subject to  ∑ φ pk = ∀k ∈ K  p∈P  φ pk ≥ ∀p ∈ P, k ∈ K 23 12 Traffic engineering Load balancing problem (3) • • Load Balancing Problem has always a solution but this might not be unique Example: – the same maximum link load is achieved with routes of different length – the upper routes are better due to smaller capacity consumption • A reasonable unique solution is achieved by associating a negligible cost with all the hops along the paths used y = x/2 b c y = x/2 x y = x/2 y = x/2 d a y=0 y=0 e y = x/2 b y = x/2 y = x/2 x y=0 a y = x/2 c d y = x/2 e 24 12 Traffic engineering Load balancing problem (4) • Load Balancing Problem with a reasonable and unique solution: – Consider a network with topology (N,J), link capacities cj, and traffic demands xk Determine the splitting ratios φpk so that the maximum relative link load is minimized with the smallest amount of required capacity Minimize yj max + ε ∑ y j ' j∈J c j j '∈J  y j = ∑ ∑ A jpφ pk xk ∀j ∈ J  p∈P k∈K  subject to  ∑ φ pk = ∀k ∈ K  p∈P  φ pk ≥ ∀p ∈ P, k ∈ K  25 12 Traffic engineering Example (1): optimal solution c=2 c b c=2 c=2 c = x c = c = c2 = d c=1 a c=2 c=2 c=2 e c=2 φ = 1/2 b ρ = x/4 φ = 1/4 b ρ = x/4 d a a φ = 1/4 ρ = x/4 c e ρ = x/8 c ρ = x/4 d ρ = x/8 e 26 12 Traffic engineering Example (2): link weights w = w=1 φ = 1/2 w=1 c b w=1 w=1 w = w=1 x w=1 d w=1 a w=1w=1 w=1 e w=1 b ρ = x/4 c ρ = x/4 φ = 1/2 b ρ = x/2 d c ρ = x/4 d a a e e 27 12 Traffic engineering Example (3): optimal link weights w=1 φ = 1/2 w=2 c b w=2 w=1 w = w=1 x w=3 d w=3 a w=1w=1 w=1 e w=1 b ρ = x/4 c ρ = x/4 b φ = 1/2 a e ρ = x/4 d d a c ρ = x/4 ρ = x/4 e 28 12 Traffic engineering THE END 29 ... distribution (1) maximum link load w =1 φ = 1/ 2 w =1 c b w =1 w =1 w = w =1 x w =1 d a x w =1 w=1w =1 w =1 e w =1 b y = 3x/2 φ = 1/ 2 b y = x/2 y = x/2 d c d a a φ =1 y = x/2 c e y=x e 19 12 Traffic engineering... e 26 12 Traffic engineering Example (2): link weights w = w =1 φ = 1/ 2 w =1 c b w =1 w =1 w = w =1 x w =1 d w =1 a w=1w =1 w =1 e w =1 b ρ = x/4 c ρ = x/4 φ = 1/ 2 b ρ = x/2 d c ρ = x/4 d a a e e 27 12 Traffic... load w =1 φ = 1/ 2 w =1 c b w =1 w =1 w = w =1 x w =1 d a x w=2 w=1w =1 w =1 e w=2 b y = x/2 c b y=x φ = 1/ 2 a φ = 1/ 2 d φ = 1/ 2 e c y = x/2 y=x y = x/2 a y = x/2 d y = x/2 e link weight increased 20 12 Traffic