network topologies
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Network Topologies. William M. Jones Assistant Professor Computer Science Department Coastal Carolina University. Network Classifications. Networks have two broad classifications Static Dynamic Static Networks - PowerPoint PPT PresentationTRANSCRIPT
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Network Topologies
William M. JonesAssistant Professor
Computer Science DepartmentCoastal Carolina University
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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 ….
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The only non-PC, networking devices are adjacent to the PCs themselves
Networking devices are potentially non-adjacent to the PCs.
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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 …
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Linear Arrays (1D)
Bus Ring
What are some of the drawbacks here?
What happens if link is cut?
Susceptible to congestion
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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
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Hypercubes (4D)
To just add one more computer to network requires doubling the number of computers
Adding only one additional node severelydisrupts symmetry
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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 …
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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.
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Step 1 -- 0101 to 1111: These are the same do nothing
Source
Dest.
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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.
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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
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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.
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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
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Given a network with “p” nodes (computers), what is the maximum number of hops (diameter) to get from source to destination?
Example 1
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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
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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
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Fat Trees
Increase the available bandwidth hierarchically. This is a VERY popular approach.
Direct (static) or indirect (dynamic) network topology?
Increasing link capacity
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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
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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)
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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 ….
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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
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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
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Trees
Diameter:Bisection width: 1 Arc connectivity: 1Cost: p-1
(leaf nodes)
(log is base 2)
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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
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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
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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
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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)
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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.
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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!
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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!
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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?
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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
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Example 4
• Which topology provides the best fault tolerance? Why?
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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.
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Common Network Hardware:The Switch
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Ports of a Switch
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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.
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Rows = p
Cols = pp2 elements
“Cost”
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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
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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.