Slide 1 Chapter 9 Mesh Network Design - II

Slide 2 Mesh Network Design The design of backbone networks is governed by 3 goals: nDirect path between source and destination. nWell-utilized components nUse high speed lines to achieve economy of scale. Chapter 8 To some extent these goals are mutually self-contradictory. What a good design looks like?

Slide 3 Our examples will focus on a single problem with 45 sites: n2 large data centers, N1 and N45. Each data center terminates and sends 1,000 Kbps. n4 data servers, N2, N3, N43, and N44. Each data server sends and receives 150 Kbps. nThe remainder of the sites are small. Each sends and receives 25 Kbps. nThe links available:

Slide 4 A design with too many direct Links, 45 node network with cost= $264,411/month

Slide 5 A design with only high speed links (T1 and 256 Kbps links) Cost reduced to $133,584/month. However, the average number of hops, 7.84, is too high.

Slide 6 A reasonable 2-Level design Data centers and servers are interior nodes of the tree. Cost further reduced to $96,777; average hops= 3.41 What will happen if any of the high- speed links fails?

Slide 7 2-connected graph nA vertex v of a connected graph G=(V,E) is an articulation point if removing the vertex and all attached edges disconnects the graph. nIf a connected graph has no articulation points, it is said to be 2-connected. Recall We are interested in 2-connected graphs. However, network designs that are completely 2-connected will be far too expensive to be implemented. We cannot disconnect a 2-connected graph by removing any edge or any node and the edges attached to it.

Slide 8 A more reliable design Instead of tree, interior nodes with high-speed links form a 2-connected graph. Cost = $ 112,587/month

Slide 9 A different 2-connected interior topology Reduce cost from $112,587  $108,724 per month

Slide 10 We will look for improvements by expanding the backbone. Look at the $112K design (slide 8): nThere are 2 large clusters centered at N2 and N45. nCan we locate a new backbone in these clusters and reduce the cost?

Slide 11 Add more backbone nodes Adding 2 concentrators at N10 and N13 lowers the cost to $103,107/per month.

Slide 12 Even lower cost design Concentrators at N4,N10, and N13; cost = $101,806/month

Slide 13 MENTOR Algorithm (MEsh Network Topological Optimization and Routing)  Backbone selection Threshold clustering K-means clustering Automatic clustering  Creation of the initial topology Prim-Dijkstra tree Backbone tour with pendant trees  Link addition Home-based routing ISP-based routing  Access topology Star Esau-Williams MSLA

Slide 14 Mentor Algorithm Step 1 1. Choose backbone sites. (Also called Threshold Cluster Algorithm) Calculate the normalized weight NW(Ni)=W(Ni)/C Choose sites with NW(Ni) > WPARM (threshold) Group end sites around a backbone site, x, based on Cost(x, Ni)/MAXCOST < RPARM. Where MAXCOST=Max i,j Cost(Ni, Nj) (Assume single link type with capacity C.) The middle stage of clustering Big squares are backbone nodes.

Slide 15 If there are sites not covered in groups, compute merit(n)=1/2*(MaxDistCtr-distCtr n )/MaxDistCtr+1/2*(Weight n /WeightMax) Here and Center of Mass (xctr, yctr) defined by Sort the merit functions. The node with largest merit get picked as backbone node. Group end node around it. Repeat until all nodes are covered in groups. Mentor Algorithm Step 1 (cont’d) Based on merit(), three backbone nodes are picked. The final clustering

Slide 16 Mentor Algorithm Steps Pick median node (root node of the network) with smallest Moment(): 3.Build a restricted Prim-Dijkstra tree rooted at median. Recall the labeling process of Prim-Dijkstra: Why “restricted”? Here only backbone nodes can be the interior nodes of the tree. Note that there is an end node that violates the constraint.

Slide Sequencing node pair: Prepare adding additional direct links to the tree. Use the tree to list node pair in “sequence”. The node pair with longer path will be listed first Mentor Algorithm Steps 4-5 An example:  The sequence is not unique.  It obeys an outside-in ordering: we do not sequence the pair (N 1, N 2 ) until we sequence all pairs (N 1 ’, N 2 ’ ) such that N 1 and N 2 lie on the path between N 1 ’ and N 2 ’. 5. Choose home node H for each nonadjacent node pair (Ni,Nj) that satisfies : Cost(Ni, H) + Cost(H,Nj) <= Cost(Ni, Nx) + Cost(Nx,Nj). - H and Nx are intermediate nodes along the path.

Slide 18 Mentor Algorithm Step 6 6. Decide which node pairs deserve direct links. Start with the top node pair (N1,N2) in the sequence. Calculate the utlization u=Traf(N1,N2)/(n*C) where n=ceil(Traf(N1,N2)/C). If u > util min, add direct link between N1 and N2. If u < util min, add Traf(N1,N2) to Traf(N1,H) and Traf(H,N2). Here H is the home of (N1,N2). Remove (N1,N2) from the sequence and repeat Step 6 again until all node pairs are processed.

Slide 19 Complexity of Mentor Algorithm nThe three basic steps: backbone selection, tree building, and direct link addition are all O(n 2 ). nIt can be executed pretty fast. nTypically we will generate a set of designs based on the same threshold parameter, e.g., different  in the restricted Prim- Dijkstra tree, or different util min. Prim-Dijkstra tree is parameterized by  The smaller the value of util min, the easier it is to add direct links. If u > util min, add direct link between N1 and N2. If u < util min, add Traf(N1,N2) to Traf(N1,H) and Traf(H,N2). Here H is the home of (N1,N2). nWe then pick the best design from the set.

Slide 20 Example of Mentor Algorithm Result 15 sites, 5 backbone nodes N2, N4, N8, N9 and N13. Cost= $ 269,785/month

Slide 21 Mentor Algorithm Design 2 Same 5 backbone nodes, with lower util min =0.7 Cost= $221,590/month,

Slide 22 Same 5 backbone nodes but with  =0.1, util min =0.9 Cost = $209,220/month. Mentor Algorithm Design 3

Slide 23 Cost of Designs vs.  and util min  =0.1 and 1-util min =0.1 is the best value.

Slide 24 Cost vs. Size of Backbone

Slide 25 Backbone Reliability  What do we mean by “reliability”? Given a network of nodes and links, the reliability of the network is the probability that the working nodes are connected. (Definition 9.1)  So far the cost-optimised networks that we have studied are often trees. Tree designs have low reliability in many cases.  We will see that new variants on MENTOR can be used to solve the problem of designing reliable backbones. Chapter 9

Slide 26 2-Connected Backbones nSuppose We have completed an initial backbone design We have further identified a subset of backbone nodes that require 2-connectivity. OR nWe will divide the sites into two subsets H = {sites requiring more reliability} and L={sites requiring less reliability}. Our approach is to include the sites in H in the backbones and then make sure that the backbone is 2-connected. How do we add links to the backbone to satisfy it? We will discuss 2 algorithms: AMENTOR and MENTour

Slide 27 AMENTOR ( Augmented MENTOR): nAdd minimal set of links to backbones to ensure 2-connectivity - At minimum increase in cost. nCannot do this by enumeration for large networks. Need to develop a heuristic approach

Slide 28 AMENTOR (step 1) 1.Find the articulation points a 1, …, a K and the 2-connected components C 1,…, C L. If there are no articulation points, we are done. Given a network N = (H, E) An example: - H = {sites requiring more reliability}

Slide 29 2.Build an auxiliary graph G. The nodes of G correspond to a 1, …, a K and C 1,…, C L  Thus there are K+L nodes in G.  If a r  C s then there is an edge in G between a r and C s. G is a tree, why? - If there were a cycle in G, all components in the cycle would collapse into one 2-connected component. AMENTOR (step 2) The auxiliary graph G Note: 0 and 1 shown on the left graph are node (or edge) weight defined at next step.

Slide 30 AMENTOR (step 3) 3. Now weight the tree G: Giving each node a i a weight of 0 and each node C j a weight of 1. - All the edges have a weight of 0, because articulation points lie between 2-connected components. Notice that we have 3 blocks and 2 articulation points, thus initially the auxiliary graph is a chain with 3 components weighted 1, 4 edges weighted 0, 2 articulations points also weighted 0. Why use “weight”?

Slide 31 4.Compute shortest “distance” between all nodes in graph G. The distance in G from C j to C j’ is the number of 2-connected components we traverse on the path from a node in C j to a node in C j’. If we add an edge between these blocks, we will collapse them all into a single 2-connected components. AMENTOR (step 4) Example: The distance between C 1 and C 3 is 3

Slide 32 AMENTOR (step 5-6) 5.Use G to give a figure of merit to possible edges in the original network N: Consider all node pairs (n 1, n 2 ) with n i  N. We reject the pair if either node is an articulation point, Then each node belongs to a unique 2-connected component, say n 1  C 1, n 2  C 2 The figure of merit is cost N (n 1, n 2 ) / dist G (C 1, C 2 ), which gives the cost per 2- connected component for the link We now pick the pair with the lowest figure of merit and add it to the network N. 6. We then return to the beginning of the program. In our example, there are only 3 link additions to consider. Assume that cost[I][J]=30, cost[I][C]=10, and cost[F][J]=11. Further, we assume that F is the nearest nonarticulation point to J and that C is the nearest nonarticulation point to I. Adding a link between I and J will link C 1 and C 3. The path has length 3, so the figure of merit for this link is 30/3=10. If we add the link from C to I, the figure of merit is 10/2=5. Finally, if we link J and F, the figure of merit is 11/2=5.5. Consequently we choose the C-to-I link.

Slide 33 Final result So merge C 1 and C 2 by adding (I, C) and removes a 2 as an articulation point. On the second pass, we add the link from J to F and the result is a single component. Summary: We call the augmented MENTOR algorithm, abbreviated AMENTOR, where we augment the backbone with additional links to make it 2-connected.

Slide 34 A 2-connected backbone produced by AMENTOR

Slide 35 MENTour Algorithm Rather than adding the links one at a time, MENTour builds tours from the beginning. Same step as MENTOR-II except that instead of building a Prim-Dijkstra tree rooted at the median, we build a TSP tour with pendant trees. Recall that there is no method to develop an optimal TSP tour…

Slide 36 The initial topology for MENTour

Slide 37 A final design by MENTour

Slide 38 THE END Thank you!

Slide 39 HW #11 A J E F B G C H D I Show how to make this network 2-connective at minimum cost? Suppose: Each horizontal or vertical link has length of 1 Cost is proportional to distance.