C
S
I
S
SUPERNETTING
• Although class A and B addresses are dwindling – there are
plenty of class C addresses
• The problem with C addresses is, they only have 256
hostids – not enough for any midsize to large size
organization – especially if you plan to give every
computer, printer, scanner, etc. multiple IP addresses
• Supernetting allows an organization the ability to combine
several class C blocks in creating a larger range of
addresses
• Note: breaking up a network = subnetting
• Note: combining Class-C networks = supernetting
Dr. Clincy
Lecture
1
C
S
I
S
Assigning or Choosing Class C Blocks
• When assigning class C blocks, there are two approaches: (1)
random and (2) superblock
• Random Approach: the routers will see each block as a separate
network and therefore, for each block there would be an entry in
the routing table – a router contains an entry for each destination
network
• Superblock Approach: instead of multiple routing table entries,
there would be a single entry. However, the choices of blocks
need to follow a set of rules:
• #1 – the # of blocks must be a power of 2 (ie. 1, 2, 4, 8 …)
• #2 – blocks must be contiguous (no gaps between blocks)
• #3 – the 3rd byte of the first address in the superblock must be
evenly divisible by the number of blocks – ie. if the # of blocks is
N, the 3rd byte must be divisible by N
Dr. Clincy
Lecture
2
C
S
I
S
Example 5
A company needs 600 addresses. Which of the following set of class C
blocks can be used to form a supernet for this company?
198.47.32.0
198.47.33.0
198.47.34.0
198.47.32.0
198.47.42.0
198.47.52.0
198.47.62.0
198.47.31.0
198.47.32.0
198.47.33.0
198.47.34.0
198.47.32.0
198.47.33.0
198.47.34.0
198.47.35.0
Solution
1: No, there are only three blocks. Must be a power of 2
2: No, the blocks are not contiguous.
3: No, 31 in the first block is not divisible by 4.
4: Yes, all three requirements are fulfilled.
(1. Power of 2, 2. Contiguous and 3. 3rd
byte of 1st address is divisible by 4: 32/4=8)
Dr. Clincy
Lecture
3
C
S
I
S
Example 8
A supernet has a first address of 205.16.32.0 and a supernet mask of
255.255.248.0. How many blocks are in this supernet and what is the range of
addresses?
Solution
• The default mask has 24 1s because 205.16.32.0 is a class C.
• Because the supernet mask is 255.255.248.0, the supernet has 21 1s.
• Since the difference between the default and supernet masks is 3, there are 23 or 8 blocks
in this supernet.
• Because the blocks start with 205.16.32.0 and must be contiguous, the blocks are
205.16.32.0, 205.16.33.0, 205.16.34.0………. 205.16.39.0.
• The first address is 205.16.32.0. The last address is 205.16.39.255.
• The total number of addresses is 8 x 256 = 2048
Dr. Clincy
Lecture
4
Explain Supernetting Conceptually
C
S
I
S
Back out this bit from
netid into host id
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Dr. Clincy
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Causes these
2 blocks to
combine as a
single block
Lecture
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5
Variable-length subnetting
C
S • Suppose you were granted a Class C address –
I
this mean you would have 8 bits to play with
S • Also, suppose you needed 5 subnets consisting
of the following # of hosts: 60, 60, 60, 30 and
30
• If you used a 2 bit subnet mask – can get 4
subnets with 64 stations each (too big)
• If you used a 3 bit subnet mask – can get 8
subnets with 32 stations each (too small)
• What’s the solution ?
Dr. Clincy
Lecture
6
C
S
I
S
Variable-length Subnetting
• Solution: used 2 subnet masks – one applied after the other
• Could use a 2 bit subnet mask and get 4 subnets with 64 stations each - this would satisfy
the three 60-host subnet requirement – therefore the subnet mask would be
255.255.255.11000000 (192)
• We could then further divide one of the 64-host subnets into two 32-host subnets by
applying this mask 255.255.255.11100000 (224) after this mask of 255.255.255.11000000
(192) is used
Dr. Clincy
Lecture
7
C
S
I
S
Ch 5 Classless Addressing
Dr. Clincy
Lecture
8
C
S
I
S
Guess What ?
Classful Addressing is Obsolete
However, understanding the classful approach will help you
easily understand the classless approach
Quickly explain classless vs classful (leave address
aggregation for the routing topics)
Dr. Clincy
Lecture
9
C
S
I
S
CLASSLESS ADDRESSING
• Recall the problems with Classful addressing – you have to get a
predefined block of addresses – in most cases, the block is either too
large or too small
• In the 1990’s, ISP came into prominence – they provide Internet
access for individuals to midsize organizations that don’t want sponsor
their own Internet service (ie. email, etc).
• The ISP’s are granted several B and C blocks of addresses and they
subdivide their address space into groups of 2, 4, 8, 16, etc.. – blocks
can be variable length
• Because of the up rise of ISP’s, in 1996, the Internet Authorities
announced a new architecture called Classless Addressing (making
classful addressing obsolete)
Dr. Clincy
Lecture
10
C Number of Addresses in a Classless
S Block
I There are two conditions
S Condition 1: the number of addresses in a block; it must be a power of 2
(2, 4, 8, . . .). A household may be given a block of 2 addresses. A small
business may be given 16 addresses. A large organization may be given
1024 addresses.
Another Condition:
• The beginning address must be evenly divisible by the number of
addresses.
• For example, if a block contains 4 addresses, the beginning address must be
divisible by 4. If the block has less than 256 addresses, we need to check only
the rightmost byte. If it has less than 65,536 addresses, we need to check only
the two rightmost bytes, and so on.
Dr. Clincy
Lecture
11
C
S
I
S
Classless Subnet Illustration
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Dr. Clincy
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Lecture
subnetid
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12
C Example 9
S Which of the following can be the beginning address
I block that contains 16 addresses?
S 123.45.24.52
of a
205.16.37.32
190.16.42.44
17.17.33.80
Solution
The address 205.16.37.32 is eligible because 32 is
divisible by 16. The address 17.17.33.80 is eligible
because 80 is divisible by 16.
Dr. Clincy
Lecture
13
C Example 10
of the following
S Which
contains 1024 addresses?
I
S
can be the beginning address of a block that
205.16.37.32
190.16.42.0
17.17.32.0
123.45.24.52
Solution
• To be divisible by 1024, the rightmost byte of an address should be 0 because any value
in that first byte will be a fraction of 1024 (ie. 0 to 255).
• To be divisible by 1024, the rightmost byte should be 0 and the second rightmost byte
must be divisible by 4 because for every unique number in the second byte position,
there exist 256 addresses in the first byte position that maps to it. To get 1024 addresses
overall, you will need an increment of 4 in the 2nd byte position.
• Therefore, the 2nd byte needs to be divisible by 4.
• Only the address 17.17.32.0 meets this condition.
Dr. Clincy
Lecture
14
C
S
I
S
Mask
•
•
•
•
•
Recall the Classful approach, only given an IP – the user defined their
mask
For the Classless approach, when an org is given a block, it’s given both
the starting address and the mask – these two pieces of info defines the
entire block
For classless case, instead of writing out the full mask, we just specify
the number of 1’s in the mask and append it to the address – this is called
slash notation or CIDR (classless interdomain routing) notation
For classless addressing, the prefix refers to the common part of the
address (ie. network portion)
For classless addressing, the suffix refers to the varying part of the
address (ie. host portion)
Dr. Clincy
Lecture
15
C
S
I
S
A block in classes A, B, and C
can easily be represented in slash
notation as
A.B.C.D/ n
where n is
either 8 (class A), 16 (class B), or
24 (class C).
Dr. Clincy
Lecture
16
C
S
I
S
Example 11
A small organization is given a block with the beginning address and
the prefix length 205.16.37.24/29 (in slash notation). What is the range
of the block?
Solution
The beginning address is 205.16.37.24. To find the last address we keep the first
29 bits and change the last 3 bits to 1s.
Beginning:11001111 00010000 00100101 00011000
Ending : 11001111 00010000 00100101 00011111
There are only 8 addresses in this block.
Dr. Clincy
Lecture
17
C
S
I
S
Example 13
What is the network address if one of the addresses is 167.199.170.82/27?
Solution
The prefix length is 27, which means that we must keep
the first 27 bits as is and change the remaining bits (5) to
0s. The 5 bits affect only the last byte. The last byte is
01010010. Changing the last 5 bits to 0s, we get
01000000 or 64. The network address is
167.199.170.64/27.
Dr. Clincy
Lecture
18
C
S
I
S
Example 14
An organization is granted the block 130.34.12.64/26. The
organization needs to have four subnets. What are the subnet
addresses and the range of addresses for each subnet?
Solution
The suffix length is 6. This means the total
number of addresses in the block is 64 (26). If
we create four subnets, each subnet will have
16 addresses.
Dr. Clincy
Lecture
19
NETWORK ADDRESS
C
TRANSLATION
(NAT)
S
Network
Address Translation (NAT) allows a site to use a set of private
I
addresses for internal communication and a set of global Internet
addresses for communication with another site. The site must have only
one single connection to the global Internet through a router that runs
NAT software.
S
The routers only 2 address: (1) the global IP address and (2)
one private address
Dr. Clincy
Lecture
20
C
S
I
S
Address translation
All packets leaving the
network get assigned the
global address as the
source address
All packets coming into the
network get their global
destination address replaced with
the appropriate private address
(straightforward process)
(process is more involved)
(explain this in the next ppt slide)
Dr. Clincy
Lecture
21
Translation
CPacket: From Private Network to Internet
Keep in mind that, with in the private network, the original source address is a private address
S
representing the original source in the private network.
Just before
the packet leaves the router, the router makes note of the GLOBAL DESTINATION
I
ADDRESS and cross-references it with the PRIVATE source address before changing the private
source address to the GLOBAL SOURCE ADDRESS
S
Packet: From Internet Back
to Private Network
When the packet returns, the SOURCE
ADDRESS of the packet is the original
DESTINATION ADDRESS.
The router uses the new source address
of the packet in determining the private
destination address – recall the address
being cross-referenced
Dr. Clincy
Lecture
22
NAT
C Using Multiple Global Addresses
S
NAT Router with One GLOBAL address can only allow One private host to
I the same EXTERNAL host – with more global addresses, more
access
private hosts can access the SAME external host
S
A NAT Router with 8 global addresses can allow up to 8 private addresses
(hosts) to access the SAME external host (simultaneously) – can create
up to 8 separate connections
To create a many-to-many relationship, a 5-column table (versus 2-column
table) is needed in reducing uncertainty – by specifying port address and
transport layer protocol
Five-column translation table
Dr. Clincy
Lecture
23
An ISP and NAT
C
ISP serving DIAL-UP customers can conserve addresses by using
S An
NAT.
I NOTE: think of dial-up customers as being apart of the ISP’s private
S network before gaining access to the Global Internet.
The ISP could assign a private address to each customer and when
the customer leaves the private network, a translation would occur
(like in ppt slide 11).
Let an ISP with 100,000 dial-up customers be granted only 1000
global addresses - the ISP could assign private addresses to each
100,000 customers and the ISP translate the 100,000 source
addresses for the outgoing packets with the 1000 global addresses
Dr. Clincy
Lecture
24