mathias fischer - svs.informatik.uni-hamburg.de › teaching › rn › slides › 03-1_netw… ·...

1

Resilient Networks

3.1 Resilient Network Design - Intro

Prepared along: Michal Pioro and Deepankar Medhi - Routing, Flow, and Capacity

Design in Communication and Computer Networks, The Morgan Kaufmann Series in Networking, 800 pages, 2004

Mathias Fischer

2

Outline

Motivation

Introduction to Network Design Problems

– Traffic and Demand

– Network Design and Routing

– Multi-layer Networks

– Network Management

Notations and mathematical formulation

– Link-Path Formulation

– Link-Demand-Path-Identifier-based Notation

Some basic design problems

– Minimizing costs of network links

– Capacitated design problems

– Shortest path routing

Summary

3

A Network Analogy

FrankfurtMoscow

Peking

Kuala Lumpur

Sydney

San Francisco

NY

Sao Paolo

Johannesburg

• New links are expensive• Economies of scale

• Store and forward concept• Hop-by-Hop routing

4

Network Basics

Communication network carries traffic

Network has different links of different capacity (bandwidth)

– Economies of scale principle applies

Traffic may be routed via different paths to destination

– According to store-and-forward principle

– Hop-by-hop

We need to have enough bandwidth to carry all data

We might want to reduce the average packet traversal delay

5

Network Design Questions

Can we find better routes?

Where should we add more bandwidth?

Where and when should we add new nodes (and links) in the network?

How does the inherent property of a network technology or protocol can affect our decision making?

What level of abstraction is appropriate for a particular network for modeling purpose so that meaningful results can be obtained

How to design cost-effective, resilient core/backbone networks taking into consideration properties of the network? - How to do network design?

6

Routing in the Internet – Autonomous Systems

Internet consists out of connected Autonomous Systems (AS)

– Stub AS: small organization (one connection to the Internet)

– Multi-homed AS: large organization (several connections, no transit traffic)

– Transit AS: Provider (several connections , transit traffic)

The Internet today consists out of 44.000 AS

– Of different size

– AS Exchange data packets as

• Peers

• Provider AS and Customer AS

Each AS has a unique ID (number)

Each AS needs to know at least one route to any other AS

Bildquelle: caida.org

7

Communication Networks and Network Providers (1)

Packet sending in the Internet Involves series of different AS networks

Each network with own switches or routers

Packet routing depends completely on the specific network

Network design problems are confined within each network or administrative domain

AS 3AS 2AS 1

8


Layered (or Hierarchy of) Networks Private networks of large companies and corporations

Telecommunication network providers that operate transport or transmission networks

Different ISPs may use the same transport network

Multi-layer network environment

AS 1AS 2

AS 3

Transport

Provider 1

AS 5AS 4

Transport

Provider 2

9

Autonomous Systems in Germany

Bildquelle: Institut für Internet-Sicherheit, Fachhochschule Gelsenkirchen

10


Network providers Design and manage their networks

Need to know

– traffic demands in their networks

– Considering all network nodes, they need to know traffic volume between any two such nodes

Traffic volume matrix or demand volume matrix as input to network design required

11

Traffic and Demand (1)

Task of a network

– Route packets from one end to another,

– without considering the reliability of delivery

Reasons for lost packets

– Physical transmission errors

– Traffic congestion and routers running out of buffer space

Task of network designer (our task)

– Keep delay low

– Design networks to prevent or at least limit congestion at routers

12


What do we need for this? Prediction of traffic demand, e.g., via statistical approaches

Estimating traffic volume by capturing statistics about traffic arrival distribution

We need this information in between different network points

Necessary Measurements

– Average arrival rate of packets

– Average size of packets

– Both influence delay

13


Example: M/M/1 queuing system

Packet arrival follows Poisson process

Packet size is exponentially distributed

D average delay in seconds

λp average packet arrival in secondsμp average link service rateC link capacity per secondKp average packet size

Average delay increases drastically when average arrival rate is closer to average service rate of the link

14


What is the acceptable delay users would like to tolerate?

– If acceptable delay is 15 ms, then the acceptable average utilization can be no more than 64,5% on the link

Good News

– At least for purpose of network design maximum link utilization can be used as alternative criterion to the delay

However, traffic arrival does not follow Poisson process

– Realistic delay is worse than the calculated one

– In reality average utilization has to be kept lower, e.g., at 50% to achieve the 15 ms

15


We need to know whether observed utilization is higher than acceptable threshold for a particular link type

In single-link networks

– Measurement is easy

– Bandwidth can be easily added

In multi-link networks

– Utilization is further impacted by routing of traffic flows

– Adding bandwidth becomes complex network design problem

Lesson learned / What we need to solve network design problems

– Average arrival rate in between different nodes in the network

– Traffic demand volume as input for all network design problems

16

Network Design and Routing (1)

In reality we have many source-destination (egress-ingress) traffic demands between various points in the network

Traffic-demand matrix required

as input to network design

Goal: Determine a network with enough capacity and connectivity to route traffic, so that acceptable service guarantees can be provided

– In single-link networks it is sufficient to determine link utilization threshold for given traffic demand

– However, this is not the case in multi-link networks anymore

17


Role of network design Distinction between (usually continuous) Demand Volume Units (DVU) and

(usually discrete) Link Capacity Units (LCU)

Three node network with traffic demand of 300Kbps between each node

– QoS goal: utilization threshold below 60%

– Three T1 links (LCU: each 1.54 Mbps) 300 Kbps/1.54Mbps ≈ 19,5%

– Two T1 links (2-1, 1-3)(300 +300)Kbps / 1.54 Mbps ≈ 39%

Network design also depends on routing capabilities

1

2 3

300 Kbps

300 Kbps

300 Kbps

1

2 3

300 Kbps

300 Kbps

300 Kbps

18


Some Definitions

Candidate path list: All possible paths between two points

Route: Particular path chosen as valid path by network design

Flow: The amount of traffic associated with a route

19

Multi-layer Networks (1)

End-system

Layer 5

Layer 4

Layer 3

Application Layer

Transport Layer

Network Layer

Layer 2

Layer 1

Data Link Layer

Physical Layer

IP

OpenFlow / MPLS / …

End-system

Layer 5

Layer 4

Layer 3

Application Layer

Transport Layer

Network Layer

Layer 2

Layer 1

Data Link Layer

Physical Layer

Optical Networks

UDP / TCP

SMTP/ POP3 / IMAP/ HTTP / …

Traffic networks• Aka service level• Traffic arrival

stochastic in nature• Has switching / routing

capabilities

Transport networks• service level traffic as

demand input• High-data rate services that

are required to be set up at permanent or semi-permanent basis

• Traffic deterministic or precise in nature over time

Ethernet / GMPLS

OC-1 / OC-3 / OC-48 / T-1 / …

20

Multi-layer Networks (2)

The architecture of communication networks A network (or layer) on top another

A network may look logically diverse, but may not be diverse in another layer

Implications in protection and restoration design (network resilience)due to inter-relation between layers

Multi-layer network design as important problem to consider

21

Multi-layer Networks (3) - Traffic and Demand

Traffic, Demand, and layered Networks Output bandwidth requirement for each Internet, telephone, and private-

line service from service networks, is input demand to layer beneath

Capacity requirement of one network becomes traffic demand volume for network below

22

Network Management Cycle (1)

Network management is the entire process from planning a network, to deploying it, and to operate it on a day-to-day basis.

Requires network management systems and protocols

Different management tasks run at different time granularity

23

Network Management Cycle (2)

Packet Discarding,Buffer Management,

Packet Routing

TCP Feedback control

Call Routing,Call Setup,

Call Admission Control,Call Rerouting, Routing,

Information Update

Traffic EngineeringOSPF weight updatesTrunk Rearrangement

PeriodicTraffic Estimation

Traffic NetworkCapacity Expansion

SONET / SDHring restoration

Mesh TransportNetwork Restoration

Transport NetworkRouting/Loading

Transport NetworkCapacity

Planning / Expansion

Micro-secs

Mili-secs

Secs

Minutes

Hours

Days

Weeks

Months

Year(s)

Traffic Network Transport Network

24

Network Management Cycle (3) – Traffic Networks

Traffic

Network

(IP)

Real-Time Traffic Management

Capacity Management

Traffic Engineering

Network Planning

variouscontrols

secs-mins

days-weeks

months-years

routingupdate

capacitychange

Traffic data

Forecast adjustment

Marketing input

Network Management

25

Network Management Cycle (3) – Transport Networks

Transport

Network

Near Real-Time Management

Capacity Management

Traffic Engineering

Network Planning

restoration

mins-hours

days-weeks

months-years

routeloading

Capacity expansion/protection

Network fill factor, loading

New Transport

Demand,

Marketing Input

Network Management

26

Notation for Network Design Problems

Network Design Problems Can be formally specified using mathematical notations

Representation of specific design problems in compact way

Eventually helps to understand the problem at hand

Eventually helps to solve the problem

Notations for Network Design Problems

Link-Path Formulation

Link-Demand-Path-Identifier-based Formulation

27

Link-Path Formulation (1)

Three node network example

– Demand volume hij in between nodes i and j

– Example demand

Each demand volume between two nodes can be routed over two paths

3

21

21

3

1

3

21

3

21

3

23

21

Path 1-2

Path 1-3

Path 1-3-2Path 1-2-3 Path 2-1-3

Path 2-3

28


Demand between nodes 1 and 2 can be routed via direct link 1-2 and via alternate route 1-3-2

– Demand path flows 𝑥^ are non-negative, i.e., 𝑥^ > 0 for all paths

Link capacities 𝑐12^ , 𝑐13

^ , 𝑐13^

– Assumption, capacity of first twolinks is 10 and the third is 15

3

21

29


System of linear equations and inequalities (constraints)

ො𝑥 are unknowns for all three considered demands

System has multiple solutions and defines set of feasible solutions

But which solution is of interest?

30


Different goals/objectives possible

– Minimizing total routing costs

– Minimizing congestion of most congested link

– ...

In mathematical way, goals are expressed as objective functions that needs to be either minimized or maximized

Example: Minimizing total routing costs

– cost of routing one unit of flow on every link along a path is simply 1

– Resulting objective function F:

3

21

31


Routing minimization problem

Capacity Constraints

Demand Volume

Constraints

Objective Function

32

Link-Path Formulation (6) – Routing Minimization

Routing minimization problem

Multi-Commodity Flow Problem

– Multiple demands (or commodities)

– Need to be routed in a network simultaneously

– Compete for resources

Optimization Context: Linear Programming Problem

– All constraints are linear

– The objective function is linear

33


Optimal solution for demand

Required: feasible solution for decision variables ( ො𝑥) that minimize F

Common sense solution– Route everything on direct (cheapest) path, other x-variables are zero

Total optimal cost F*=20

3

21

34


However

Optimal solution may not be unique

Not in all cases a solution is that simple to obtain

Constraints

– Need to be satisfied by optimal solution

– Especially link capacity constraints need not be violated

3

21

35


Let‘s change the objective function From

To

For demand

Solution not that easy anymore

– Cheapest path routing violates capacity constraints of link 1-3

– We are happy to announce a solution with total cost F*=25

3

21

36


Lessons learned from Routing Minimization Problem

Changing objective function affects

– optimal solution to a problem

– and the way of finding it

Carefully selection of right objective function (or goal) for particular network required,

– Otherwise obtained solution might not be meaningful

37

Link-Path Formulation (11) – Pros and Cons

All demands and paths easy to follow from node-reference point of view

Works well for three-node example but not in general case

Drawbacks

– Paths may contain many intermediate nodes

– Flow variables have indices of different length

– Some node pairs might not have any demand

– Not all nodes directly connected

Link-Path Formulation insufficient for Multi-Commodity-Flow Problems

38

Link-Path Formulation (12) – Pros and Cons

More Problems

When network gets larger the indices grow as well: Xi-m-...-n-j No easy way to represent multiple-paths

– For specific demand pair when each path may go through different number of intermediate nodes

Not all paths might be acceptable

– For example, due to the distance between the nodes

– This requires additional exceptions for paths not used

Notation cannot handle more than one link between two nodes, e.g., multi-graph case

Multiple demands between nodes cannot be handled

Not possible to write summations over paths

39

Link-Demand-Path-Identifier-based Notation (1)

Compact, allows to list only necessary objects

Non-zero demand pairs with indices from 1 to their total number

Total number of demands D, index d (D=3, d=1,2,3)

Links are assigned labels from 1 to total number of links

Total number of links E, index e(E=3, e=1,2,3) 3

21

40


Mapping for demand volumes h and link capacities c

Path identifiers (path-flow variables) 𝑥𝑑𝑝– Demand-pair identifier d used as first subscript

– Path label p for demand pair as second subscript

• Candidate paths for demand pair numbered from 1 to total number

• Total number of candidate paths for demand 𝑑 will be denoted 𝑃𝑑• Paths are labeled with index 𝑝

41


Example Demand between nodes 1 and 2 identified by label d = 1 (first subscript)

has two candidate paths (𝑃1 = 2)

Paths 1-2 and 1-3-2 are labeled with p=1,2 (second subscript)

Paths are identified as (1,1) and (1,2)

Path-flow variables

3

21

42


Mapping of path identifiers and paths from:node-identifier-based notation to link-demand-path-identifier-based notation

Node-identifier-based Link-demand-path-identifier-b.

Path identifier Path Path identifier Path

132 1-3-2 12 {2,3}

213 2-1-3 32 {1,2}

23 2-3 31 {3}

43


Routing Minimization Problem in Link-Demand-Path-Identifier-based Notation

44


Comparison

Link-path formulation Link-demand-path-identifier-based

45


Notation Summary

46

DP: Minimizing Costs of Network Links (1)

Minimizing costs of networks links Minimizing total link capacity cost

required to carry given demand

Assuming not a fixed, but variable capacity 𝑦𝑒 of links

Dimensioning Problem (DP):Determine

– required demand flows

– and link capacities

– to carry givendemand volumes

4

31

2

2

31Demand

Network

47


4-node example network Four nodes with routable demands

Node 4 is only transit node

𝜉𝑒 as costs for sending one unit via link e

Demands:

– d=1, only one path P11={2,4} allowed

– d=2, paths P21={5}, P22={3,4}

– d=3, paths P31={1}, P32={2,3}

Demand constraints

4

31

2

2

31Demand

Network

48


Demand constrains in general form

– Vector of flows assigned to demand d:

Hence, we can write

In summation notation

Flow allocation vector x

49


Capacity constraints

– Assures that for each link e its capacity ce (or ye, if capacity is a variable) is not exceeded by the flows using the link

Sum on left side are link loads ye(total flow through that link)

50


To write down link loads in compact manner we need to know relationships between links and paths

Formally defined via link-path incidence coefficients

In more compact manner via coefficient δ𝑒𝑑𝑝:

e\ Pdp P11={2,4} P21={5} P22={3,4} P31={1} P32={2,3}

1 0 0 0 1 0

2 1 0 0 0 1

3 0 0 1 0 1

4 1 0 1 0 0

5 0 1 0 0 0

4

31

2

2

31

51


Link load ye on link e in compact manner

Summation over all paths

appearing in routing listsof all demands, over allcombinations (d,p)

52


Minimizing capacity costs, objective function

– 𝜉𝑒 as costs for sending one unit via link e

53


DP: Minimizing capacity costs

4

31

2

2

31Demand

Network

54


Optimal solution for DP Feasible solution (x,y) with y1(x)=y1=5, y2(x)=y2=20, y3(x)=y3=10,y4(x)=y4=20, y5(x)=y5=15, and

thus y = (y1, y2, y3, y4, y5) = (5,20,10,20,15) has

total cost F=115

However, solution not optimal,because of usage of P22 with costsζ22= ξ3 + ξ4 = 1 + 3 = 4

Other possible path P21 has costs ζ21= ξ5= 1

Cost of path Pdp is given by:

4

31

2

2

31Demand

Network

55


Optimal solution to DP

In example we need to move all the flow for demand d=2 from path P22 to path P21, which gives savings of (ζ22- ζ21) = 3 per flow unit, in our case the total savings are x22(ζ22- ζ21) = 15

d=1: x*11= 15

d=2: x*21= 20, x*

22= 0

d=3: x*31= a, x*

32= 10-a, for any 0 ≤ a ≤ 10

y*1 = 10 –a, y*

2 = 15 + a, y*3 = a, y

*4 = 15, y

*5 = 20

F*=100 4

31

2

2

31Demand

Network

56


Shortest Path allocation rule:

For each demand allocate entire demand volume to its shortest path (w.r.t. to link unit costs and candidate paths)

If there is more than one shortest path for a demand then demand volume can be split among shortest paths arbitrarily

Works well for dimensioning problems, but no general solution approach for other multi-commodity flow problems, e.g.,

– Restriction to non-bifurcated flows

– Modular design problems: in real networks capacities can be installed only in modular units (e.g., T1, E1, OC-3)

57


The observed DP is an uncapacitated design problem

Another type of problem are capacitated design problems

– Link capacities ce given instead of variable ye– Find a feasible flow allocation that satisfies demand and capacity

constraints with ce appearing on the right-hand sides

– In such scenario there might not be an objective function, except flow routing cost minimization is required

– Capacitated problem in compact form

58

Capacitated Design Problems (1)

4-Node Network Example

Capacity vector 𝒄 = 𝑐1, 𝑐2, 𝑐3, 𝑐4, 𝑐5 = (5,10,10,5,30)

Additional path for d=1: P13 = {1,3,4}

All solutions are necessarily bifurcated

Possible solution:

Note that d=1 also uses path 𝑃13, its longest path

When non-bifurcated solutions are required, additional constraints necessary to force single-path solution

4

31

2

2

31Demand

Network

𝑃11

𝑃22

𝑃21

𝑃32𝑃31

𝑃13

59

Shortest Path Routing (1)

Shortest path routing, e.g., OSPF routing in IP networks

Shortest paths are

– essential for network design,

– pre-requisite for resilient network design

For each d all volume ℎ𝑑 realized on shortest path w.r.t. to

– given link weight system 𝑤 = 𝑤1, 𝑤2, … , 𝑤𝐸– and link weight cost we for link e

Path selection based on additive calculation of link weights

Flow allocation vector x(w), thus flow allocation dictated by link weights w

Shortest-path routing and term shortest-path allocation are not the same!

60


Modified 4-Node Network Example

Capacity vector

– 𝒄 = 𝑐1, 𝑐2, 𝑐3, 𝑐4, 𝑐5 = (5,10,10,5,30)

Additional paths for d=1:

– P12 = {1,5}

– P13 = {1,3,4}

4

31

2

2

31Demand

Network

𝑃11

𝑃22

𝑃21

𝑃32 𝑃31

𝑃12

𝑃13

61


4-Node Network Example Link weight system 𝒘 = (1,3,1,2,4)

Solution

Rest of flow variables are 0

Shortest paths are unique for each demand pair → flow allocation vector 𝒙(𝑤)

is also unique

4

31

2

2

31Demand

Network

𝑤4 = 2

𝑤2 = 3

𝑤3 = 1

𝑤1 = 1

𝑤5 = 4

62


4-Node Network Example However, flow allocation not feasible for capacity vector 𝒄=(5,10,10,5,30)

Link loads resulting from allocation vector 𝒙(𝒘) would be y(w) = (y1,y2,y3,y4,y5) = (25,0,35,35,0)

Solution violates capacity constraints!

Changing link capacity so that c = y(w): shortest path allocation w.r.t. to w becomes (trivially) feasible again

63


Single shortest-path allocation problemFor given link capacities c and demand volumes h, find a link weight system w such that the resulting shortest paths are unique and the resulting flow allocation vector 𝒙(𝒘) is feasible, i.e., such that 𝒙(𝒘) satisfies

Three reasons for complexity of the problem

A non-bifurcated (single-path) feasible flow allocation may not exist, while bifurcated feasible flow allocations may exist

Even if single-path solution exists, in most cases it can be hard to determine

Even if we find single-path flow solution, the weight system to induce it may not exist

64


Infeasible unique shortest path case

Two demands

– d=1 between nodes 1 and 7

– d=2 between nodes 2 and 6

– ℎ1 = ℎ2 = 1, ce ≡ 1(∀𝑒 ∈ 𝐸)

Two paths per demand

– d=1: 𝑃11:1-3-5-7, 𝑃12: 1-3-4-5-7

– d=2: 𝑃21:2-3-4-5-6, 𝑃22: 2-3-5-6

Allocating flows d=1: 𝑥11=1, d=2: 𝑥21=1

No link weight system that induces single shortest path solution!

1

3

5

7

6

4

2

65


4-Node Example Considering weight system w=(1,1,1,1,1), which

is shortest path routing w.r.t. hop count

d=1: there are two shortest paths

– 𝑃11 = 2,4 , 𝑃12 = {1,5}

Which path has to be used for traffic?

– In OSPF: Equal Cost MultiPath (ECMP) rule

– Split all outgoing demands at a node among its outgoing links that are on shortest paths to destination

4

31

2

2

31Demand

Network

𝑤4 =1

𝑤2 =1

𝑤3 = 1

𝑤1 = 1

𝑤5 =1

𝑃11𝑃12

66


Infeasible unique shortest path case

Two demands: d=1, d=2, ℎ1 = ℎ2 = 1, ce ≡ 1(∀𝑒 ∈ 𝐸)

Paths: 𝑃11:1-3-5-7, 𝑃12: 1-3-4-5-7 , 𝑃21:2-3-4-5-6, 𝑃22: 2-3-5-6

Solution: assign link weight 2 to all links except to links 2-3 and 4-5 that obtain weight 1

Result is feasible solution under ECMP rule

1

3

5

7

6

4

2

w=2

w=2

w=2 w=2

w=2

w=1

w=1

67

Additional Network Design Problems

Fair Networks

– Demand is elastic and can consume any bandwidth assigned to its path, e.g., within certain predefined bounds (lower and upper bound for demand)

– Capacity constraints should not be violated

– Network needs to carry more than lower bound for each demand volume to maintain fairness, e.g., Max-Min-Fairness or Proportional Fairness criterion

Topological Design

– Cost function takes into account not only capacity-dependent costs of links 𝜉𝑒, but also link installation costs 𝜅𝑒

– Additional binary variable 𝑢𝑒 that indicates if link is installed or not

– Problem similar to uncapacitated design problem with modular links

68

Summary

Network design as generic problem in multitude of areas

Traffic and demand as input to network design

Networks have multiple layers

Network management based on multitude of different systems and protocols

Right notation makes a lot of problems way easier

– Even when it seems to be more complicated on first sight

– Even when you do not believe me yet ;)

Network design problems covered

– Capacitated vs. uncapacitated problems

• Minimizing costs of networks links

– Shortest path routing

mathias fischer - svs.informatik.uni-hamburg.de › teaching › rn › slides › 03-1_netw… ·...

Documents