CS 622 Distributed Networks

CS622 AccessNetwork.1 UC. Colorado Springs

CS 622Distributed Networks

Access Network Design

Dr. Xiaobo ZhouDepartment of Computer Science


Review: Traffic Normalization

° Traffic Generators• Uniform traffic • Random traffic• Realistic traffic (in relation with population and distance)

° Traffic Normalization• Total normalization• Row normalization• Row & Column normalization

° Different applications have difference characteristics


What is Access Network

° A Backbone network connects major sites° Access networks connect small sites to the backbone network

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Access network

Backbone network

Backbone linkAccess link


Access Network Design

° How to decide which sites should be in the backbone network?• Traffic volume• Close to multiple small sites

° Access network collect traffic from small sites into the high speed backbone network.

° Sharing high speed links, enjoy economic of scale benefit.

° Examples of local access networks• Local subscriber loop connects users of a central office.• Lottery network• ATM network• ISP’s local access network.


A Simple Access Design Example

° Consider a problem with 6 access locations and 1 backbone site• N1 is the backbone site, the traffic is all from and to N1• Symmetric traffic


The Cost Matrix for 56 Kbps links° Linear Cost model: fixed cost = $400/month, and $3.00/km/month for the

first 300km, and then a cost of $1.75/km/month after that• Cost if distance-dependent

.


A Star Design

° Cost=$9650; Max. Utilization=23.2%


A Cheaper Local-Access Design° N2 serves as a concentrator for N6 and N7. ° Local link can use shorter less expensive link

• Cost $8666


A Even Cheaper Local-Access Design° Two concentrators: N2 for N6 and N7; N4 for N3

• Cost: $8158


Move to MST – the cost is optimal ° Choose N7 as concentrator instead of N2

• The design becomes an MST

T2-1+T6-1+T7-1 = 2600026000/56000=46.4%


MSTs not always Optimal Access Designs

° The MST was the optimal, because the traffic volumes were moderate° When the traffic grows 50%, the MST costs $10,616 and the links to

concentrators N4 and N7, i.e, N1 - N4 and N1 - N7, must have two links to keep utilization below 50%


An Optimal Design with Increased Traffic° A Constraint MST Problem

• N3 connects directly to N1 since through N4 will violate the utilization constraint


Frame Relay Design° Frame Relay is a relatively new service

• Frame relay is accomplished by connecting each site into a frame relay cloud provided by the service carrier, just like plain old phone service is accomplished by attaching the phone into PSTN cloud.

• Packets exceed Committed Information Rate (CIR) will be discardable by having discard eligibility (DE) bit set

° Three classes of charges: access link costs, provider port costs (cost to frame relay), and CIR costs.

° It is volume dependent and not distance-bw-nodes dependent

Port Charges CIR charges per PVC(permanent virtual circuits)


Frame Relay Design Cost Model° Each node is 20 km from the provider point of presence (POP)

• Access link cost is same for all 7 nodes: 7 * (400 + 3 * 20)

Frame Relay Cloud

PVC port: 128 kbpsΣi T i ->1 = 59 bps; 50%

POP

PVC port: 56/64 kbps

POP

PVC port

POP

PVC port

POPN1

N4

N3

N2

…


Frame Relay vs. Lease-Line Cost

° Let x be the distance from the site to the center° Fixed cost=$400/month; $3.00/km/month; ° Leased-line Cost Model

• 6 * 400 + 6 * 3.00 * x° Frame Relay Cost Model:

• Assume frame relay provider has point of presence (pop) at each site (20km away). Each site connects to the frame relay network

• N1 uses a 128 kbps link port (59 bps total traffic on way, 50%)• Other nodes use 56kbps links

- Port charges: 6 * 250 + 500 = 2000• Access charges: 7 * (400 + 3.0 * 20) = 7*460 = 3220• CIR charges: 4 * 30 + 2 * 25 = 170, if 4 PVCs with 16kbps CIR and

2PVCs with 8kbps• Frame relay cost: 2000 + 3220 + 170 = $5390/mo

° Solve 2400 + 18 * x = 5390 x = 166.11 km, Break even point° Most WAN are larger than this Frame Relay is a good candidate


Choosing Backbone Nodes° If the division between large sites and small sites is distinct, there is

usually no problem in deciding backbone nodes

° Definition 5.1: Given a set of sites Ni and traffic matrix T(i,j), weight(Ni) = Σ j(T(i,j)+T(j,i))

° Sometimes, the weights of nodes indicate the choices of backbonenodes or traffic centers.

° Design Principle 5.3It is acceptable for small nodes to route their traffic via big nodes, but generally we do not want to route the traffic between big nodes via the small nodes

° In airline networks, there are four kind of nodes• Hubs• Cities with enough traffic to have nonstop flights to cities other than hubs• Cities with enough traffic to have nonstop flights to their hubs• Little cities


3 Types of Local Access Problems

1. Access node’s traffic are considerably smaller than the smallest link. But occasionally, they may need to download huge file• Use frame relay or access tree • capacitated spanning tree building problem

2. Access node’s traffic is comparable to the capacity of the smallest link• Choice 1: connect them directly to the hub/center• Choice 2: put a concentrator between hub and those nodes• Concentrator placement problem, local access tree problem

3. Access node’s traffic can fill several low-speed access lines• Choice 1: multiple links to multiple backbone nodes• Choice: a high speed link to a backbone node

- Multiple low-speed links give better reliability- The high-speed link gives better performance and lower cost


One-speed One-Center Design and Capacitated Trees° Example:

• 19 nodes to be connected to a hub, N14 (total 20 nodes)• Requirement: 4 sites can share a line• Symmetric traffic: to and from each node Ni to N14 is 1200 bps• Capacity of the one-speed link: 9600 bps• Link utilization limitation: < 50%

° The problem becomes a capacitated access tree building problem

° Solutions:• SPT by Dijkstra’s algorithm• MST by Prim’s algorithm• Intermediate trees by Prim-Dijkstra with 0<α<1• Exhaustive search for an optimal tree• Other new algorithms?


SPT(Star)° High cost: $26358° Max_uitlization is 12.5% (12000 / 56000)

• How about aggregate traffic of nodes and connect to the center?


MST° Cost: $18,730

15 sites from N124 parallel links


Prim-Dijkstra with α=0.3° Lower cost: $15930° Potential improvements: N11 can go through N4; Two clusters with

N18 and N9 as concentrators ( a double link -> 2 single links).

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A double D96 link


How Many Trees to Search for Optimal Designs?

° Exhaustive search for an optimal design is usually infeasible.

° Cayley’s Theorem: Given n nodes, there are nn-2 different spanning tree

° For 20 nodes, there are 2018=2.621*1023 trees.


Constraint Minimum Spanning Tree Problem

° CMST problem: Given a central node N0 and a set of other nodes (N1, …, Nn), a set of weights(w1,…,wn) for each node, the capacity of a link, W, and a cost matrix Cost(i, j), find a set of trees T1, …, Tk such that each Ni belongs to exactly one Tj and each Tj contains N0, and

∑>∈

<0 , iTii

j

Ww

∑ ∑∈Trees Linksl

ll endendCost

)2,1(min


A Greedy CMST Algorithm° Sort the edges/links according to the cost° take the lowest cost edge from sorted list; add it to the solution subtrees if the

addition does not violated the constraint ; go to s1.

° Assume W=3, each node has wi = 1, and the following topology:

∑>∈

<0 , iTii

j

Ww

T1

T2


Esau-William Algorithm° Initially, each node starts off in a tree with itself

° Compute the tradeoff function: Tradeoff(Ni) = minj Cost(i, j) - Cost(Comp(Ni), Center) // attractive if ≤ 0• saving by linking Ni to Nj, rather than linking directly to Center (STAR)• Cost(Comp(i),Center) is the cost of connecting the component with

Node Ni to the Center. It is equivalent to the cost of the shortest path from the Center to any node in the component.

• Cost(i,j) is the link cost from Node Ni to Node Nj• minj Cost(Ni,Nj) suggests pick the closest neighboring Node Nj

° Maintain a sorted list of links based on the Tradeoff() value• Tradeoff() is attractive if negative, smaller -> more attractive• Actually, in each iteration, we only consider the shortest link out of a

node to a neighbor that does not belong the component of the node° L1: adds the top link (min Tradeoff) in the list to the solution if the

weight constraint of the component is satisfied; otherwise, reject it° update the tradeoffs in other links due to the newly added link and

resort the list° go to L1


Apply Esau-William Algorithm

° Assume W=3, each node has wi = 1, and the following topology:° Tradeoff(1) = minj Cost(1, j) - Cost(Comp(1),Center)

= Cost(1,3) - 7 //comp(1) contains Node 1= 5 – 7 = -2 // pick closest neighbor, Node 3

° Tradeoff(2,0) = 6 - 8 = -2° Tradeoff(3,1) = 5 – 11 = -6° Tradeoff(4,2) = 7 –14 = -7° Tradeoff(5,3) = 8 – 17 = -9° Tradeoff(5) is lowest one

° Accept link(5,3) to the solutionsince weight constraint on component with nodes 5 and 3 are not violated.

Σwi = w5 + w3 = 2 <= W=3° Effectively this picks the faraway nodewith short link to its neighbor and group them as component.


Apply Esau-William Algorithm(2)

° Pick Tradeoff(4,2) since it is smallest° Accept link(4,2) since weight

constraint on component with nodes 4 and 2 are not violated:

Σwi = w4 + w2 = 2 <= W=3

° Update the tradeoffs in other links due to the newly added link and resort the Tradeoff() list

° Tradeoff(5,4) = 9 – 11 = -2// Next shortest link out of 5 is (5,4)// (Comp(5)=11,node 5 goes through node 3 to center)// 3 is already in the component, not considered

° Tradeoff(3,1) = 5 – 11 = -6 // not changed° Tradeoff(1,3) = 5 – 7 = -2° Tradeoff(2,4) = 6 – 8 = -2° Tradeoff(4,2) = 7 – 14 = -7° Tradeoff(5,4) = 9 – 11 = -2


Apply Esau-William Algorithm(3)

° Pick Tradeoff(3,1)° Accept link (3,1) since weight

constraint on component with nodes 1, 3 and 5 are not violated

Σwi = w1 + w3 + w5 = 3 <= W=3

° Tradeoff(4,3) = 8 – 8 = 0// Tradeoff(2,1)= -2 not changed

° Tradeoff(3,1) = 5 – 11 = -6, not changed° Tradeoff(5,4) = 9 – 11 = -2° Tradeoff(1,3) = 5 – 7 = -2° Tradeoff(2,1) = 6 - 8 = -2° Tradeoff(4,3) = 8 – 8 = 0


Apply Esau-William Algorithm (4) and (5)° Since nodes 5 and 3 now go through node 1 to the Center 0,

Tradeoff(5,4) = 9 – 7 = 2Tradeoff(3,4) = 8 – 7 = 1Tradeoff(1,2) = 6 – 7 = -1

° Tradeoff(2,1) = 6-8 = -2

° Tradeoff(4,3) = 8-8 = 0

° Tradeoff(2,1) is lowest butadding link(2,1) result a componentwith 5 nodes violate Σwi <= 3

° Reject(2,1) recompute Tradeoff(2,0) = 8 – 8 = 0

° Reject(1,2) due to the same reason Recompute Tradeoff(1,0) = 7 – 7 =0

° Pick link(1,0)

° Pick link(2,0) completes the access network


Creditability of Esau-Williams Algorithm° The Esau-Williams Algorithm is heuristic, though works quite well° Example: an Esau-Williams design with up to 7 sites per line

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Creditability of Esau-Williams Algorithm (Cont.)° Given a set of sites N and a capacitated tree T, 1-exchange test:

• no cheaper link can be substituted for an existing link without violating the capacity constraints

• Example: for homogeneous traffic, up to 4 sites on a line

• For homogeneous traffic, the Esau-Williams does well on 1-change test


Esau-Williams and InHomogenous Traffic° One-speed link, but different sites have different traffic

• Link capacity is 9600 bps, 50% of sites have a requirement of 2400 bps, and the other 50% have a requirement of 4800 bps


Line Crossing and Esau-Williams

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° Are crossings an indication of a lack of creditability in capacitated trees?• In tours, yes!


Line Crossing in Access Designs: Sharma’s Algorithm

1. Compute the angle θs from each site S to the central site C. If S and Chave the same coordinate, set θs =0

2. Sort the angles θs

3. Beginning at a site Sfirst, create a set of nodes clockwise (or counterclockwise) from Sfirst

A set is complete when adding the next node would put Σsetw(site) > W. The next set starts with that node.

4. The design is completed by building a MST on each set with the addition of the central node C.

Theorem 5.2 Sharma’s algorithm builds CMSTs withour line crossings


Sharma’s Algorithm Result

×A better solution


Creditability of Sharma Algorithm° Much higher failure rate than Esau-Williams’ by 1-change test


Sharma vs. Esau-Williams

EW_Ratio = SharmaCost/EWCost; S_Ratio = 1/EW_Ratio

Documents

CS 622 Distributed Networks