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.

Slide 1

Chapter 9

Mesh Network Design - II

Slide 2

Mesh Network Design

The design of backbone networks is governed by 3 goals:

Direct path between source and destination. Well-utilized components Use 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:

2 large data centers, N1 and N45. Each data center terminates and sends 1,000 Kbps.

4 data servers, N2, N3, N43, and N44. Each data server sends and receives 150 Kbps.

The remainder of the sites are small. Each sends and receives 25 Kbps.

The 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 designData 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

A vertex v of a connected graph G=(V,E) is an articulation point if removing the vertex and all attached edges disconnects the graph.

If 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):

There are 2 large clusters centered at N2 and N45. Can 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-distCtrn)/MaxDistCtr+1/2*(Weightn/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)

22 )()(max yctryxctrxMaxDistCtr nnNn

22 )()( yctryxctrxDistCtr nnn

Nnn

Nnnn

Weight

Weightxxctr

Nnn

Nnnn

Weight

Weightyyctr

Based on merit(), three backbone nodes are picked.

The final clustering

Slide 16

Mentor Algorithm Steps 2-3

2. 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:

Nn

nWeightnndistnMoment ),()(

)),(),((min nodeneighbordistneighborrootdistneighbors 10

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 17

4. 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 (N1, N2) until we sequence all pairs (N1

’, N2’) such that N1 and N2 lie on

the path between N1’ and N2

’.

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 > utilmin, add direct link between N1 and

N2.• If u < utilmin, 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 The three basic steps: backbone selection, tree building,

and direct link addition are all O(n2). It can be executed pretty fast. Typically we will generate a set of designs based on the

same threshold parameter, e.g., different in the restricted Prim-Dijkstra tree, or different utilmin .

Prim-Dijkstra tree is parameterized by

The smaller the value of utilmin, the easier it is to add direct links.

• If u > utilmin, add direct link between N1 and N2.• If u < utilmin, add Traf(N1,N2) to Traf(N1,H) and

Traf(H,N2). Here H is the home of (N1,N2).

We then pick the best design from the set.

)),(),((min nodeneighbordistneighborrootdistneighbors 10

Slide 20

Example of Mentor Algorithm Result15 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 utilmin=0.7

Cost= $221,590/month,

Slide 22

Same 5 backbone nodes but with =0.1, utilmin=0.9

Cost = $209,220/month.

Mentor Algorithm Design 3

Slide 23

Cost of Designs vs. and utilmin

=0.1 and 1-utilmin=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

Suppose We have completed an initial backbone design We have further identified a subset of backbone nodes that

require 2-connectivity.OR We 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):

Add minimal set of links to backbones to ensure 2-connectivity

- At minimum increase in cost.

Cannot do this by enumeration for large networks. Need to develop a heuristic approach

Slide 28

AMENTOR (step 1)

1. Find the articulation points a1, …, aK and the 2-connected components C1,…, CL. 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 a1, …, aK

and C1,…, CL

Thus there are K+L nodes in G. If arCs then there is an edge in G between ar and Cs.

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 ai a weight of 0 and each node Cj 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 Cj to Cj’ is the number of 2-

connected components we traverse on the path from a node in Cj to a node in Cj’.

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 C1 and C3

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 (n1, n2) with niN. We reject the pair if either node is an articulation point,

• Then each node belongs to a unique 2-connected component, say n1 C1, n2 C2 • The figure of merit is costN(n1, n2) / distG(C1, C2), 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 C1 and C3. 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 C1 and C2 by adding (I, C) and removes a2 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!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.