Chapter 5: Addressing (Part 3 of 3)

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• Chapter 5 Addressing (Part 3 of 3)

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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

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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

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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
32/48)
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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

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Explain Supernetting Conceptually
Back out this bit from netid into host id
Causes these 2 blocks to combine as a single block
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Guess What ?

• Classful Addressing is Obsolete
• However, understanding the classful approach will
  help you easily understand the classless approach
• Quickly explain classless vs classful

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Classful Approach to Routing
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Classless Approach to Routing
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Going from Classful to Classless - Address
Aggregation
Single Organization Owned
Router owned by organization used to subnet their
network
Router owned by organization used to connect to
the Internet
Classful Case
Subnetting occurs at this point now
Router now owned by the Internet
Router owned by the Internet
Classless Case
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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 1990s, ISP came into prominence they
  provide Internet access for individuals to
  midsize organizations that dont want sponsor
  their own Internet service (ie. email, etc).
• The ISPs 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 ISPs, in 1996, the
  Internet Authorities announced a new architecture
  called Classless Addressing (making classful
  addressing obsolete)

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The SIZE of the BLOCK depends on the MASK
netid
result
address
1 1 1 0 0
mask
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Number of Addresses in a Classless Block There
are two conditions 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.

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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, its 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 1s 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)

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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).
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Table Prefix lengths
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Variable-length subnetting

• Suppose you were granted a Class C address this
  mean you would have 8 bits to play with
• 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)
• Whats the solution ?

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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

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Classless Subnet Illustration
Netid subnetid
0 0 0 0 0
0 0 0 0 1
0 0 0 1 0
0 0 0 1 1
0 0 1 0 0
0 0 1 0 1
0 0 1 1 0
0 0 1 1 1
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Example 9
Which of the following can be the beginning
address of a block that contains 16
addresses? 123.45.24.52 205.16.37.32190.16.42.44
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.
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Example 10
Which of the following can be the beginning
address of a block that contains 1024
addresses? 205.16.37.32190.16.42.017.17.32.0123
.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.

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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. Beginning11001111
00010000 00100101 00011000 Ending
11001111 00010000 00100101 00011111 There are
only 8 addresses in this block.
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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.
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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.
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NETWORK ADDRESS TRANSLATION (NAT)
Network Address Translation (NAT) allows a site
to use a set of private 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.
The routers only 2 address (1) the global IP
address and (2) one private address
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Address translation
All packets coming into the network get their
global destination address replaced with the
appropriate private address (process is more
involved) (explain this in the next ppt slide)
All packets leaving the network get assigned the
global address as the source address (straightforw
ard process)
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Translation
Packet From Private Network to Internet
Keep in mind that, with in the private network,
the original source address is a private address
representing the original source in the private
network. Just before the packet leaves the
router, the router makes note of the GLOBAL
DESTINATION ADDRESS and cross-references it with
the PRIVATE source address before changing the
private source address to the GLOBAL SOURCE
ADDRESS
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
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NAT - QUESTION
Can 2 or more nodes in a private network
communicate with 2 or more DIFFERENT global nodes
at the same time ?
Can 2 or more nodes in a private network
communicate with the same GLOBAL Node at the same
time ?
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NAT Using Multiple Global Addresses
NAT Router with One GLOBAL address can only allow
One private host to access the same EXTERNAL host
with more global addresses, more private hosts
can access the SAME external host 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
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An ISP and NAT
An ISP serving DIAL-UP customers can conserve
addresses by using NAT. NOTE think of dial-up
customers as being apart of the ISPs private
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 21). 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