Network Topologies
William M. Jones
Assistant Professor
Computer Science Department
Coastal 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)
What happens if link is cut?
Susceptible to congestion
Bus
What are some of the
drawbacks here?
Ring
2 links cut
Meshes (2D, 3D)
4 links cut
More connectivity  better fault tolerance  more cost
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!
Hypercubes (4D)
To just add one more computer to network requires doubling the number of computers
Adding only
one additional
node severely
disrupts
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 …
Source
Dest.
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 nothing
Step 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 bit
Step 2 -- 0101 to 1111: These are different  follow the edge that takes
you to a vertex that is 0111
Step 3 -- 0101 to 1111: These are the same  do nothing, go to next bit
Source
Dest.
Step 3 -- 0101 to 1111: These are the same  do nothing, go to next bit
Step 4 -- 0101 to 1111: These are different  take the adjacent edge
From 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 above
Dimension 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.
Example 1
Given a network with “p” nodes (computers), what is the
maximum number of hops (diameter) to get from source to
destination?
Example Solution
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).
Trees
Another variation
Static because point to point
Note these PCs
must have 3 NICs
Dynamic because switching in hierarchy
Links higher up the tree
potentially carry more traffic than
those at the lower levels;
therefore, we could ….
Fat Trees
Increase the available bandwidth hierarchically.
This is a VERY popular approach.
Direct (static) or indirect (dynamic) network topology?
BLOCK DIAGRAM - CVN
What topology is this? Tree
BS1 is at the root, and the clients (at the leaf level) are connected to the edge switches
UHF
EHF
OTCIXS
V6
MDS/
APS MDS
TAC -X
CP
CVIC
TBMCS
ES1
ES2
TAC -X (4)
TFCC
ES3
TAC -X (5)
CVIC
NTWorkstations
ES5
TAC -X (2)
ITS/APPS
CVIC
ES6
ES7
TAC -X (2)
ISDS/Profile
CVIC
SCI
Router
Mux
Secret
Router
ACDS
TAC -X
ASUW
ES8
NT Srvrs
EXC/SQL
TAC -X
ASUW
INE
NT Server
KG
Unclas
Router/
INE
ACDS
ES1
CWSP
TAC -X
Flag Plot
TAC -X
USW
NTWorkstations
TAC -X (5)
Unclas
LAN
BS2
ES4
DSCS
ADNS
BS1
TAC -X
Alt CP
TFCC
WSC6
Modem
EHF-MDR
Nav
Rad Mer
TAC -X
JAOC
KG
GFCP
PLT
NAVMACS
METOC
Lite
TEAMS
KG
TADIXS
V6
Link
NT Srvrs
PDC/BDC
GALE
TRE
STREDs
STUIII
TAC -X (2)
Supplot
SSR1
KG
NECC
DAMA
DAMA
Compatible
WSC -3
SHF
KGR
USC38
KG
WSC -3
Elint
SHF/CA
Voice
Services
Completely Connected & Star Networks
Any PC can directly communication with any
other PC at the same time (theoretically)
Note, each PC must
have multiple network
ports (NICs)
Single point of failure
(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 …
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 ….
Linear Arrays
Diameter:
Bisection width:
Arc connectivity:
Cost:
p-1
1
1
p-1
Diameter:
Bisection width:
Arc connectivity:
Cost:
p/2
2
2
p
Note “cost” here is the taken to be the number of links
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
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.
Arc connectivity
Bisection width
Meshes
p
2D mesh
Diameter:
Bisection width:
Arc connectivity:
Cost:
2
2D torus
p
Diameter:
Bisection width:
Arc connectivity:
Cost:
Which is “better”? Why?
4
Floor function
Trees
(leaf nodes)
Diameter:
Bisection width: 1
Arc connectivity: 1
Cost:
p-1
(log is base 2)
Completely Connected and Star
Diameter:
1
Bisection width:
Arc connectivity: p-1
Cost:
Note p2 links
Diameter:
Bisection width:
Arc connectivity:
Cost:
2
1
1
p-1
Note the tradeoff here between cost and the other metrics.
Summary
Static Interconnection Networks
Logs are base 2
Network
Diameter
Bisection
Width
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:
Scalability (Number of links)
100
Completely-connected
90
Number of links
Star
80
Binary Tree
70
Bus
2D Mesh
60
50
40
30
20
10
0
1
3
5
7
9
11
13
15
17
19
21
23
25
27
29
31
Number of Nodes
Note completely-connected is more costly (in terms of the number of links) than the others
Scalability (Diameter i.e. max hop to traverse)
16
Completely-connected
14
Star
Binary Tree
Max number of hops (diameter)
12
Bus
2D Mesh
10
8
6
4
2
0
1
3
5
7
9
11
13
15
17
19
Number of Nodes
21
23
25
27
29
31
Multi-objective optimization
Scalability (Diameter, sm aller is better b/c low er latency)
16
Completely-connected
14
Star
Binary Tree
12
Bus
2D Mesh
10
The final decision will
most likely be a
compromise between
performance and cost.
8
6
4
2
Scalability (number of link, lower is better due to cost)
0
1
3
5
7
9
11
13
15
17
19
Number of Nodes
21
23
100
25
27
29
31
Completely-connected
90
Star
80
Binary Tree
70
Bus
2D Mesh
60
50
40
30
20
10
0
1
3
5
7
9
11
13
15
17
19
Number of Nodes
21
23
25
27
29
31
Scalability (Diameter)
Example 2:
Given the choice
between a binary
tree and a 2D mesh,
which is better, and
why?
16
Completely-connected
14
Star
Binary Tree
12
Bus
2D Mesh
10
8
6
4
2
Scalability (Number of links i.e. Cost)
0
1
3
5
7
9
11
13
15
17
19
21
Number of Nodes
23
100
25
27
29
31
Completely-connected
90
Star
80
Binary Tree
70
Bus
2D Mesh
60
50
40
Justify your answer!
30
20
10
0
1
3
5
7
9
11
13
15
17
19
Number of Nodes
21
23
25
27
29
31
Scalability (Diameter)
Example 2: ANS
16
Completely-connected
14
Given the choice between a
binary tree and a 2D mesh,
which is better, and why?
Star
Binary Tree
12
Bus
2D Mesh
10
The binary tree b/c at scale, has
a smaller diameter and also
costs less (fewer links) than 2D
mesh.
8
6
4
2
Scalability (Number of links i.e. Cost)
0
1
3
5
7
9
11
13
15
17
19
21
Number of Nodes
23
100
25
27
29
31
Completely-connected
90
Star
80
Binary Tree
70
Bus
2D Mesh
60
50
40
Justify your answer!
30
20
10
0
1
3
5
7
9
11
13
15
17
19
Number of Nodes
21
23
25
27
29
31
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 = 15
Then, 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
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!
Scalability: How many switching elements are there?
Which static network topology is this similar to?
What are the differences?
Looks like mesh, but is closer to completely connected, except that only 1 cable would
extend from the switch port to the PC.
“Cost”
Rows = p
p2 elements
Cols = p
Fully connected network
Comparison of a network
topology with a common
network device: the switch
p2 links! (meaning a total of roughly p2 ports)
Any two pairs of nodes can communicate
at the same time! (theoretically because
PCs typically only have 1 network port)
p2 switching elements!
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
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.