Network Topologies

William M. JonesAssistant Professor

Computer Science DepartmentCoastal Carolina University

Network Classifications• Networks have two broad classifications

– Static– Dynamic

• Static Networks– Static networks consist of point-to-point

communication links among nodes and are also referred to as direct networks

• Dynamic Networks– Dynamic networks are built using switches

and communication links. Dynamic networks are also referred to as indirect networks.

For example ….

The only non-PC, networking devices are adjacent to the PCs themselves

Networking devices are potentially non-adjacent to the PCs.

Network Topologies

• A variety of network topologies have been proposed and implemented

• These topologies tradeoff performance for cost; a full analysis goes beyond the scope of this course

• Commercial machines often implement hybrids of multiple topologies for reasons of packaging, cost, and available components

Let’s take a look at some common static network topologies …

Linear Arrays (1D)

Bus Ring

What are some of the drawbacks here?

What happens if link is cut?

Susceptible to congestion

Meshes (2D, 3D)

Two and three dimensional meshes: (a) 2-D mesh with no wraparound; (b) 2-D mesh with wraparound link (2-D torus); and

(c) a 3-D mesh with no wraparound.

In terms of graph theory, networks can be described by a set of vertices (nodes) and a set of edges (links)

What’s up with these topologies? 4+ NW wires / PC, I’ve never seen a computer like that before!

More connectivity better fault tolerance more cost

2 links cut4 links cut

Hypercubes (4D)

To just add one more computer to network requires doubling the number of computers

Adding only one additional node severelydisrupts symmetry

Note that the vertices can be numbered so that the Hamming distance between any two adjacent vertices is 1. This plays an important role in determining a path from sender to receiver (i.e. routing). For example …

Suppose node 0101 wants to send a message to node 1111. Starting with the LSB of the numbers, compare the destination address bit to the sending address bit and if they are different, take the adjacent link that makes them the same. This is referred to as dimension order routing.

Source

Dest.

Step 1 -- 0101 to 1111: These are the same do nothing

Source

Dest.

Step 1 -- 0101 to 1111: These are the same do nothingStep 2 -- 0101 t0 1111: These are different follow the edge that takes you to a vertex that is 0111 – (a given node only has to know the address of his neighbors to make this work, which is a nice simplification)

Source

Dest.

Step 1 -- 0101 to 1111: These are the same do nothing, go to next bitStep 2 -- 0101 to 1111: These are different follow the edge that takes you to a vertex that is 0111Step 3 -- 0101 to 1111: These are the same do nothing, go to next bit

Dest.

Source

Step 3 -- 0101 to 1111: These are the same do nothing, go to next bitStep 4 -- 0101 to 1111: These are different take the adjacent edgeFrom this we can see that the number of “hops” from source to destination is in fact the Hamming distance between their “addresses”

Source

Dest.

Example 0: Message from 0000 to 1111 route shown aboveDimension ordered routing is deterministic (same route between a pair of node, each time a message is sent). What are the drawbacks of this approach? ANS: Assumes all links are up.

Dest.

Source

Given a network with “p” nodes (computers), what is the maximum number of hops (diameter) to get from source to destination?

Example 1

Given a network with “p” nodes, what is the maximum number of hops to get from source to destination?

Given “p” nodes, each node address will be log2(p) bits wide. In the worst case, the max Hamming distance between source and destination address would be when every bit position is different (e.g. 1010 to 0101); therefore, the max number of hops would be log2(p).

Example Solution

Trees Another variation

Links higher up the tree potentially carry more traffic than those at the lower levels; therefore, we could ….

Static because point to point Dynamic because switching in hierarchy

Note these PCs must have 3 NICs

Fat Trees

Increase the available bandwidth hierarchically. This is a VERY popular approach.

Direct (static) or indirect (dynamic) network topology?

Increasing link capacity

BS1 BS2

ES2 ES4 ES5 ES6

TAC - XCP

CVIC

TAC - XAlt CPTFCC

TAC - X (2)Supplot

TAC - X (2)ITS/APPSCVIC

TAC - XUSW

TAC - X (4)TFCC

TAC - XASUW

TAC - XASUW

TAC - XFlag Plot

TAC - X (2)ISDS/ProfileCVIC

NTWorkstations

ACDS

NT SrvrsEXC/SQL

NT SrvrsPDC/BDC

ACDS

ADNS

SHF

Rad Mer

METOCNAVMACS TBMCS

TAC - X (5)

GFCP

NavPLTLink

MDS/APS MDS

STUIII

TADIXSV6

DA

MA

DAMACompatible

WSC -3

WSC -3OTCIXS

V6

NECC

USC38

KG

KG

KG

KG

UHF EHF

TRE

KGRSSR1

Elint

TEAMSSTREDs

Mux

KG

Modem

WSC6

EHF-MDRDSCS

GALE Lite

ES1sci

NTWorkstations

NT Server

SecretRouter

UnclasRouter/INE

SCIRouter

INE

UnclasLAN

ES3ES1 ES7 ES8

TAC - XJAOC

TAC - X (5)CVIC

SHF/CAVoiceServices

BLOCK DIAGRAM - CVN

CWSP

What topology is this? Tree BS1 is at the root, and the clients (at the leaf level) are connected to the edge switches

Completely Connected & Star Networks

(a) A completely-connected network of eight nodes; (b) a star connected network of nine nodes.

How do we evaluate these static networks? What are the costs, tradeoffs, etc?

Let’s define some metrics …

Note, each PC must have multiple network ports (NICs)

Single point of failure

Any PC can directly communication with any other PC at the same time (theoretically)

Some Useful Metrics• Diameter: The distance between the farthest two

nodes in the network. Specifies max number of hops. (smaller better)

• Bisection Width: The minimum number of wires you must cut to divide the network into two equal parts. (larger better)

• Cost: The number of links is a meaningful measure of the cost. However, a number of other factors, such as the ability to layout the network, the length of wires, etc., also factor in to the cost.

• Arc Connectivity: The minimum number of links that must be removed to break the network into two disconnected networks, not necessarily equal in size (larger better)

Let’s take a look at the previous networks ….

Diameter: p-1Bisection width: 1Arc connectivity: 1Cost: p-1

Diameter: p/2Bisection width: 2Arc connectivity: 2Cost: p

Imagine any two arbitrary nodes trying to communicate.Which topology is “better”? ANS: Ring

The bus costs less, but the ring is more fault tolerant

Linear Arrays

What is fault tolerance? Qualitatively, the ability to continue to function properly (albeit in a potentially degraded mode) in the presence of faults, e.g. link failures.

Note “cost” here is the taken to be the number of links

Meshes

p

p

Diameter:Bisection width:Arc connectivity: 2Cost:

Diameter:Bisection width:Arc connectivity: 4Cost:

2D mesh 2D torus

Which is “better”? Why? Floor function

Arc connectivity

Bisection width

Trees

Diameter:Bisection width: 1 Arc connectivity: 1Cost: p-1

(leaf nodes)

(log is base 2)

Completely Connected and Star

Diameter: 1Bisection width: Arc connectivity: p-1Cost:

Diameter: 2Bisection width: 1 Arc connectivity: 1Cost: p-1

Note the tradeoff here between cost and the other metrics.

Note p2 links

Summary Static Interconnection Networks

Network Diameter BisectionWidth

Arc Connectivity

Cost (No. of links)

Completely-connected

Star

Complete binary tree

Linear array

2-D mesh, no wraparound

2-D wraparound mesh

Hypercube

Wraparound k-ary d-cube

Scalability: How a metric changes as a function of increasing “p”, for example:

Logs are base 2

Scalability (Number of links)

0

10

20

30

40

50

60

70

80

90

100

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31

Number of Nodes

Completely-connected

Star

Binary Tree

Bus

2D Mesh

Num

ber

of

links

Note completely-connected is more costly (in terms of the number of links) than the others

Scalability (Diameter i.e. max hop to traverse)

0

2

4

6

8

10

12

14

16

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31

Number of Nodes

Completely-connected

Star

Binary Tree

Bus

2D Mesh

Multi-objective optimization

Max

num

ber

of h

ops

(dia

met

er)

Scalability (number of link, lower is better due to cost)

0

10

20

30

40

50

60

70

80

90

100

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31

Number of Nodes

Completely-connected

Star

Binary Tree

Bus

2D Mesh

Scalability (Diameter, smaller is better b/c lower latency)

0

2

4

6

8

10

12

14

16

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31

Number of Nodes

Completely-connected

Star

Binary Tree

Bus

2D Mesh

The final decision will most likely be a compromise between performance and cost.

Scalability (Number of links i.e. Cost)

0

10

20

30

40

50

60

70

80

90

100

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31

Number of Nodes

Completely-connected

Star

Binary Tree

Bus

2D Mesh

Scalability (Diameter)

0

2

4

6

8

10

12

14

16

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31

Number of Nodes

Completely-connected

Star

Binary Tree

Bus

2D Mesh

Example 2:Given the choice between a binary tree and a 2D mesh, which is better, and why?

Justify your answer!

Scalability (Number of links i.e. Cost)

0

10

20

30

40

50

60

70

80

90

100

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31

Number of Nodes

Completely-connected

Star

Binary Tree

Bus

2D Mesh

Scalability (Diameter)

0

2

4

6

8

10

12

14

16

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31

Number of Nodes

Completely-connected

Star

Binary Tree

Bus

2D Mesh

Example 2: ANSGiven the choice between a binary tree and a 2D mesh, which is better, and why?

The binary tree b/c at scale, has a smaller diameter and also costs less (fewer links) than 2D mesh.

Justify your answer!

Example 3

• Given a complete binary tree interconnection network where the two farthest nodes are separated by 6 hops, how much would you have to spend on the network links, assuming $10 per link?

• Assuming only the leaf nodes are PCs, at $500 per PC, what would the total PC cost be?

Example 3 Answer

• Given a complete binary tree interconnection network where the two farthest nodes are separated by 6 hops, how much would you have to spend on the network links, assuming $10 per link?

6 hops diameter is 6 using the equation from table, p = 15Then, given p = 15, number of links is p – 1 (from table) = 15-1 = 14 so at $10/link that is $140

• Assuming only the leaf nodes are PCs, at $500 per PC, what would the total PC cost be?

If you draw the tree out, you see that it has 8 leaf nodes so $8*500

Example 4

• Which topology provides the best fault tolerance? Why?

Example 4 Answer

• Which topology provides the best fault tolerance? Why?

This is a tough question. The completely connected one seems to provide the best tolerance to link failures; of course, it has this highest link and NIC/node cost too.

Common Network Hardware:The Switch

Ports of a Switch

Internal Structure of Switch

Scalability: How many switching elements are there?Which static network topology is this similar to? What are the differences?

Internal switching elements can be configured to allow any two mutually exclusive pairs of nodes to communicate at the same time.

As the number of switch ports increases, what happens to the number of internal switching elements?

It increases, but at what rate? p2!

Looks like mesh, but is closer to completely connected, except that only 1 cable would extend from the switch port to the PC.

Rows = p

Cols = pp2 elements

“Cost”

Comparison of a network topology with a common

network device: the switch

p2 switching elements!

p2 links! (meaning a total of roughly p2 ports)

Any two mutually exclusive pairs of nodes can communicate at the same time!

Somewhat more restrictive; however the fully connected graph is unrealistic not only because of the number of links, but also because each node would have to have p-1 NICs .. switch is a good compromise

Any two pairs of nodes can communicate at the same time! (theoretically because PCs typically only have 1 network port)

Fully connected network

Why is a switch a good compromise?

• Allows a direct (1 hop) connection between any pair of computers

• Reduces the number of cables from p2 p• Reduces the number of NICs per computer from

p-1 1• However, increasing network size typically

increases the switch cost quadratically (p2)– Due to it’s internal structure– Especially as “p” gets very large.