1CSIE,NTUT,TAIWAN
Subnetting/Supernetting
and
Classless Addressing
Chuan-Ming Liu
CSIE,NTUT
Spring ’04,TAIWAN
CSIE,NTUT,TAIWAN 2
CONTENTS
SUBNETTING
SUPERNETTING
CLASSLESS ADDRSSING
CSIE,NTUT,TAIWAN 3
SUBNETTING
5.1
CSIE,NTUT,TAIWAN 4
Subnetting
Network is divided into several smaller
subnetworks
Each subnetwork (subnet) has its own
subnetwork address
Hierarchy in IP addressing
IP addresses are designed with two levels of
hierarchy
CSIE,NTUT,TAIWAN 5
IP addresses are designed with
two levels of hierarchy.
CSIE,NTUT,TAIWAN 6Figure 5-1
network with two levels of hierarchy
CSIE,NTUT,TAIWAN 7Figure 5-2
network with three levels of
hierarchy (subnetted)
CSIE,NTUT,TAIWAN 8Figure 5-3
Addresses in a network with and without
subnetting
CSIE,NTUT,TAIWAN 9Figure 5-4
Hierarchy concept in a telephone number
CSIE,NTUT,TAIWAN 10
Subnet Mask
Network mask (default mask) creates the
network address; subnet mask creates the
subnetwork address
Contiguous v.s,noncontiguous subnet mask
Finding the subnet address
– Straight Method
– Short-cut method
CSIE,NTUT,TAIWAN 11
Figure 5-5 Default mask and subnet mask
CSIE,NTUT,TAIWAN 12
Straight Method
In the straight method,we use binary
notation for both the address and the
mask and then apply the AND operation
to find the subnet address.
CSIE,NTUT,TAIWAN 13
Example 1
What is the subnetwork address if the
destination address is 200.45.34.56 and the
subnet mask is 255.255.240.0?
CSIE,NTUT,TAIWAN 14
Solution
11001000 00101101 00100010 00111000
11111111 11111111 11110000 00000000
11001000 00101101 00100000 00000000
The subnetwork address is 200.45.32.0.
CSIE,NTUT,TAIWAN 15
Short-Cut Method
** If the byte in the mask is 255,copy
the byte in the address.
** If the byte in the mask is 0,replace
the byte in the address with 0.
** If the byte in the mask is neither 255
nor 0,we write the mask and the address
in binary and apply the AND operation.
CSIE,NTUT,TAIWAN 16
Example 2
What is the subnetwork address if the
destination address is 19.30.80.5 and the
mask is 255.255.192.0?
Solution
CSIE,NTUT,TAIWAN 17Figure 5-7
Comparison of a default mask and
a subnet mask
CSIE,NTUT,TAIWAN 18
Number of subnetworks,addresses
Number of subnetworks
number of extra 1s that are added to the default
mask to make the subnet mask
Number of addresses per subnet
number of 0s in the subnet mask
CSIE,NTUT,TAIWAN 19
Special addresses
Hostid all 0s – subnet address
Hostid all 1s – limited broadcast inside the
subnet
CSIE,NTUT,TAIWAN 20
Designing Subnets
Deciding the number of subnets
The number of subnets must be a power of 2
Finding the subnet mask
– Find the number of 1s in the default mask
– Find the number of 1s that define the subnets
– Add the number of 1s in above steps
– Find the number of 0s
Finding the range of address in each subnet
CSIE,NTUT,TAIWAN 21
The number of subnets must be
a power of 2,
CSIE,NTUT,TAIWAN 22
Example 3
A company is granted the site address
201.70.64.0 (class C),The company needs
six subnets,Design the subnets.
Solution
The number of 1s in the default
mask is 24 (class C).
CSIE,NTUT,TAIWAN 23
Solution (Continued)
The company needs six subnets,This number
6 is not a power of 2,The next number that is
a power of 2 is 8 (23),We need 3 more 1s in
the subnet mask,The total number of 1s in
the subnet mask is 27 (24 + 3).
The total number of 0s is 5 (32 - 27),The
mask is
CSIE,NTUT,TAIWAN 24
Solution (Continued)
11111111 11111111 11111111 11100000
or
255.255.255.224
The number of subnets is 8.
The number of addresses in each subnet
is 25 (5 is the number of 0s) or 32.
See Figure 5.8
CSIE,NTUT,TAIWAN 25Figure 5-8
Example 3
CSIE,NTUT,TAIWAN 26
Example 4
A company is granted the site address
181.56.0.0 (class B),The company needs
1000 subnets,Design the subnets.
Solution
The number of 1s in the default mask is 16
(class B).
CSIE,NTUT,TAIWAN 27
Solution (Continued)
The company needs 1000 subnets,This
number is not a power of 2,The next number
that is a power of 2 is 1024 (210),We need 10
more 1s in the subnet mask.
The total number of 1s in the subnet mask is
26 (16 + 10).
The total number of 0s is 6 (32 - 26).
CSIE,NTUT,TAIWAN 28
Solution (Continued)
The mask is
11111111 11111111 11111111 11000000
or
255.255.255.192.
The number of subnets is 1024.
The number of addresses in each subnet is 26
(6 is the number of 0s) or 64.
See Figure 5.9
CSIE,NTUT,TAIWAN 29Figure 5-9
Example 4
CSIE,NTUT,TAIWAN 30Figure 5-10
Variable-length subnetting
CSIE,NTUT,TAIWAN 31
SUPERNETTING
5.2
CSIE,NTUT,TAIWAN 32
Supernetting
Size of a block in class C is too small usually
Contrast to subnetting,supernetting combines
several class C blocks to create a large range
of addresses
CSIE,NTUT,TAIWAN 33Figure 5-11
Supernetwork
CSIE,NTUT,TAIWAN 34
Rules:
** The number of blocks must be a power of 2 (1,2,
4,8,16,.,,).
** The blocks must be contiguous in the address
space (no gaps between the blocks).
** The third byte of the first address in the
superblock must be evenly divisible by the number
of blocks,In other words,if the number of blocks is
N,the third byte must be divisible by N,
CSIE,NTUT,TAIWAN 35
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
198.47.31.0 198.47.32.0 198.47.33.0 198.47.52.0
198.47.32.0 198.47.33.0 198.47.34.0 198.47.35.0
CSIE,NTUT,TAIWAN 36
Solution
1,No,there are only three blocks.
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.
CSIE,NTUT,TAIWAN 37
Supernetting Mask
Need a supernet mask to find how many
blocks are combined to make a supernet
Supernet mask is the reverse of a subnet mask
Supernet mask for class C has less 1s than the
default mask
CSIE,NTUT,TAIWAN 38
In subnetting,
we need the first address of the
subnet and the subnet mask to
define the range of addresses.
CSIE,NTUT,TAIWAN 39
In supernetting,
we need the first address of
the supernet
and the supernet mask to
define the range of addresses.
CSIE,NTUT,TAIWAN 40Figure 5-12
Comparison of subnet,default,
and supernet masks
CSIE,NTUT,TAIWAN 41
Example 6
We need to make a supernetwork out of 16
class C blocks,What is the supernet mask?
Solution
We need 16 blocks,For 16 blocks we need to
change four 1s to 0s in the default mask,So the
mask is
11111111 11111111 11110000 00000000
or
255.255.240.0
CSIE,NTUT,TAIWAN 42
Example 7
A supernet has a first address of 205.16.32.0 and a
supernet mask of 255.255.248.0,A router receives three
packets with the following destination addresses:
205.16.37.44
205.16.42.56
205.17.33.76
Which packet belongs to the supernet?
CSIE,NTUT,TAIWAN 43
Solution
We apply the supernet mask to see if we can find
the beginning address.
205.16.37.44 AND 255.255.248.0? 205.16.32.0
205.16.42.56 AND 255.255.248.0? 205.16.40.0
205.17.33.76 AND 255.255.248.0? 205.17.32.0
Only the first address belongs to this supernet.
CSIE,NTUT,TAIWAN 44
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 supernet has 21 1s,The default mask has 24 1s.
Since the difference is 3,there are 23 or 8 blocks in
this supernet,The blocks are 205.16.32.0 to
205.16.39.0,The first address is 205.16.32.0,The
last address is 205.16.39.255.
CSIE,NTUT,TAIWAN 45
CLASSLESS
ADDRESSING
5.3
CSIE,NTUT,TAIWAN 46
Classless Addressing
ISP can be granted several class B and class C
blocks and then subdivide the range of
addresses,giving a range to a household or a
small business
The idea is to have variable-length blocks that
belong to no class
CSIE,NTUT,TAIWAN 47Figure 5-13
Variable-length blocks
CSIE,NTUT,TAIWAN 48
Number of Addresses in a Block
There is only one condition on 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,
CSIE,NTUT,TAIWAN 49
Beginning Address
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,
CSIE,NTUT,TAIWAN 50
Example 9
Which of the following can be the beginning address of a
block that contains 16 addresses?
205.16.37.32
190.16.42.44
17.17.33.80
123.45.24.52
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.
CSIE,NTUT,TAIWAN 51
Example 10
Which of the following can be the beginning address of a
block that contains 1024 addresses?
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 and the second rightmost byte
must be divisible by 4,Only the address 17.17.32.0
meets this condition.
CSIE,NTUT,TAIWAN 52
Mask
The same idea of mask is carried over to
classless addressing
When an organization is given a block,it is
given the first address and the mask
CSIE,NTUT,TAIWAN 53
Slash Notation
4-byte mask is cumbersome
Mask 255.255.255.224 has 17 1s.
Slash Notation (CIDR,Classless InterDomain Routing):
attach the number of 1s in the mask to the end of
classless address instead
CSIE,NTUT,TAIWAN 54Figure 5-14
Slash notation
CSIE,NTUT,TAIWAN 55
Slash notation is also called
CIDR
notation,
CSIE,NTUT,TAIWAN 56
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.
CSIE,NTUT,TAIWAN 57
Example 12
We can find the range of addresses in Example 11 by
another method,We can argue that the length of the suffix
is 32 - 29 or 3,So there are 23 = 8 addresses in this block.
If the first address is 205.16.37.24,the last address is
205.16.37.31 (24 + 7 = 31).
CSIE,NTUT,TAIWAN 58
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).
CSIE,NTUT,TAIWAN 59
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.
CSIE,NTUT,TAIWAN 60
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.
CSIE,NTUT,TAIWAN 61
Solution (Continued)
Let us first find the subnet prefix (subnet mask).
We need four subnets,which means we need to add
two more 1s to the site prefix,The subnet prefix is
then /28.
Subnet 1,130.34.12.64/28 to 130.34.12.79/28.
Subnet 2,130.34.12.80/28 to 130.34.12.95/28.
Subnet 3,130.34.12.96/28 to 130.34.12.111/28.
Subnet 4,130.34.12.112/28 to 130.34.12.127/28.
See Figure 5.15
CSIE,NTUT,TAIWAN 62Figure 5-15
Example 14
CSIE,NTUT,TAIWAN 63
Example 15
An ISP is granted a block of addresses starting with
190.100.0.0/16,The ISP needs to distribute these
addresses to three groups of customers as follows:
1,The first group has 64 customers; each needs 256 addresses.
2,The second group has 128 customers; each needs 128 addresses.
3,The third group has 128 customers; each needs 64 addresses.
Design the subblocks and give the slash notation for each
subblock,Find out how many addresses are still available
after these allocations.
CSIE,NTUT,TAIWAN 64
Solution
Group 1
For this group,each customer needs 256 addresses.
This means the suffix length is 8 (28 = 256),The
prefix length is then 32 - 8 = 24.
01,190.100.0.0/24?190.100.0.255/24
02,190.100.1.0/24?190.100.1.255/24
…………………………………,.
64,190.100.63.0/24?190.100.63.255/24
Total = 64? 256 = 16,384
CSIE,NTUT,TAIWAN 65
Solution (Continued)
Group 2
For this group,each customer needs 128 addresses.
This means the suffix length is 7 (27 = 128),The
prefix length is then 32 - 7 = 25,The addresses are:
001,190.100.64.0/25?190.100.64.127/25
002,190.100.64.128/25?190.100.64.255/25
003,190.100.127.128/25?190.100.127.255/25
Total = 128? 128= 16,384
CSIE,NTUT,TAIWAN 66
Solution (Continued)
Group 3
For this group,each customer needs 64 addresses.
This means the suffix length is 6 (26 = 64),The
prefix length is then 32 - 6 = 26.
001:190.100.128.0/26?190.100.128.63/26
002:190.100.128.64/26?190.100.128.127/26
…………………………
128:190.100.159.192/26?190.100.159.255/26
Total = 128? 64 = 8,192
CSIE,NTUT,TAIWAN 67
Solution (Continued)
Number of granted addresses,65,536
Number of allocated addresses,40,960
Number of available addresses,24,576