Day 5.5 subnetting

INTRODUCTION TO
CLASSLESS ROUTING

CCNA v3.0 Semester 3
CLASSFUL ADDRESSING
The original IPv4 address architecture used an 8 bit
network number for Class A addresses, a 16 bit
network number for Class B addresses, and a 24
bit network number for Class C addresses.
2
1 - 126
128 - 191
192 - 223
Class B
Network Host
1 0
Class C
Network Host
1 1 0
Class A
Network Host
0

CLASSFUL ADDRESSING
Classful addressing (A, B, C, etc.) is basically
obsolete.
3
Address
Class
Application
Number of
Network Bits
Number of
Host Bits
Decimal
Address
Range
Number of
Addresses
Number of
Possible Host
Class A
Large
Networks
8 bits 24 bits 1 - 126 126 16,777,214
Class B
Medium-sized
Networks
16 bits 16 bits 128 - 191 65,534 65,534
Class C Small Networks 24 bits 8 bits 192 - 223 2,097,152 254
The Class System

WHAT IS VLSM?
A Variable Length Subnet Mask (VLSM) is a means
of allocating IP addressing resources to subnets
according to their individual need rather than
some general network-wide rule.
VLSM allows an organization to use more than one
subnet mask within the same network address
space. It is often referred to as ‘subnetting a
subnet’, and can be used to maximize addressing
efficiency.
Large subnets are created for addressing LANs and
small subnets are created for WAN links (a 30 bit
mask is used to create subnets with only two
host).
4

SUBNETTING VS. VLSM
• Subnetting allows you to divide big
networks into smaller, equal-sized slices.
• VLSM allows you to divide big networks
into smaller, different-sized slices. This
enables you to make maximum use of
your valuable IP address space.
• So basically, you are now utilizing subnet
masks in the same IP address space.
5

ROUTING PROTOCOLS SUPPORTING VLSM
•RIP v2
•EIGRP
•OSPF
6

ADDRESSING A NETWORK WITH
STANDARD SUBNETTING
• Site A has two Ethernet networks
• Site B had one Ethernet network
• Site C had one Ethernet network
207.21.24.0 /24
How many network addresses are needed?
How many hosts are needed for the largest LAN?
How many bits need to be borrowed to address this
network?
7
Site A Site B Site C
25 users 25 users 10 users 8 users

STANDARD SUBNETTING
• Site A has two Ethernet networks
• Site B had one Ethernet network
• Site C had one Ethernet network
8
If we borrow 3 bits from a class C address, that will give
us eight networks, but we can only use six of them. Each
network will have 30 usable addresses.
It will take four network addresses to accommodate the
Ethernet networks at each site. That leaves us with two
extra networks.
There is also a point-to-point WAN connection between
each site. These two connections will take up the
remaining two networks.

STANDARD SUBNETTING
Borrowing 3 bits will meet the current needs of the
company, but it leaves little room for growth.
Each network will have 30 usable addresses,
including the point-to-point WAN links (which
only require two addresses).
9
Subnet # Subnet Address
Bits
Masked
0 207.21.24.0 /27
1 207.21.24.32 /27
2 207.21.24.64 /27
3 207.21.24.96 /27
4 207.21.24.128 /27
5 207.21.24.160 /27
6 207.21.24.192 /27
7 207.21.24.224 /27
207.21.24.0

10
We can use subnet 0
To enable subnet 0 on a Cisco router (if not already
enabled), it is necessary to use the global configuration
command ip subnet-zero.
Router# configure terminal (config t)
Router(config)# ip subnet-zero
To disable subnet 0, use the no form of this command.
Router# configure terminal
Router(config)# no ip subnet-zero

11
Subnetting in a Box
0
255
To begin, in a
class C network
there are 256
addresses. When
we subnet the
address, we break
it down in to
smaller units or
subnets. 256 addresses

12
Subnetting in a Box
0
255
128
127
If we were to
borrow 1 bit, it
would break the
256 addresses in
to two parts
(networks) with
each part (subnet)
having 128
addresses.
The subnet mask
would be
255.255.255.128.
128 addresses 128 addresses

13
Subnetting in a Box
0
255
128
127
64
63
If we were to
borrow 2 bits, it
would break each
of these 2
networks in half
again. This would
give us 4
networks, each
with 64 addresses.
The subnet mask
would now be
255.255.255.192.
64 addresses
64 addresses
64 addresses
64 addresses
192
191

14
Subnetting in a Box
0
255
128
127
64 192
63 191
If we were to
borrow 3 bits, it
would break each
of these 4
networks in half
again. This would
give us 8
networks, each
with 32 addresses.
The subnet mask
would now be
255.255.255.224.
32
addresses
32
addresses
31
32
32
addresses
32
addresses
95
96
32
addresses
32
addresses
159
160
32
addresses
32
addresses
223
224

15
Subnetting in a Box
0
255
128
127
64 192
63 191
If we were to
borrow 4 bits, it
would break each
of these 8
networks in half
again. This would
give us 16
networks, each
with 16 addresses.
The subnet mask
would now be
255.255.255.240.
31
32
95
96
159
160
223
224
16
addresses
16
addresses
16
addresses
16
addresses
16
addresses
16
addresses
16
addresses
16
addresses
16
addresses
16
addresses
16
addresses
16
addresses
16
addresses
16
addresses
16
addresses
16
addresses
16
15
48
47
144
143
176
175
80
79
112
111
208
207
240
239

ADDRESSING A NETWORK USING
VLSM
• When using VLSM to subnet an address, not all of the
subnets have to be the same size.
• A different subnet mask may be applied to some of
the subnets to further subnet the address.
• In order to take advantage of VLSM, the proper routing
protocol must be selected.
• Not all routing protocols share subnetting information
in their routing table updates.
16
Classful Routing Protocols
(do not share subnet info)
Classless Routing Protocols
(do share subnet info)
RIP v1 RIP v2
IGRP EIGRP
OSPF
IS-IS

VLSM
To begin subnetting this network using VLSM,
identify the LAN with the largest number of hosts.
Subnet the address 207.21.24.0 /24 based on this
information.
• Site A has two Ethernet networks (25 hosts each)
• Site B had one Ethernet network (10 hosts)
• Site C had one Ethernet network (8 hosts)
17
Bits
Masked
0 207.21.24.0 /27
1 207.21.24.32 /27
2 207.21.24.64 /27
3 207.21.24.96 /27
4 207.21.24.128 /27
5 207.21.24.160 /27
6 207.21.24.192 /27
7 207.21.24.224 /27

VLSM
Subnet 1 & 2 can be used to address Site A Ethernet
networks. Subnet 5 can be subnetted to accommodate
Site B & C Ethernet networks. Subnet 6 can be
subnetted to accommodate the WAN links.
18
Free
Addresses
Site A
0 207.21.24.0 /27
1 207.21.24.32 /27
2 207.21.24.64 /27
3 207.21.24.96 /27
4 207.21.24.128 /27
5 207.21.24.160 /27
6 207.21.24.192 /27
7 207.21.24.224 /27
Site B &
C
Sub-subnet 0 207.21.24.160 /28
Sub-subnet 1 207.21.24.176 /28
Site B
Site C
0 207.21.24.0 /27
1 207.21.24.32 /27
2 207.21.24.64 /27
3 207.21.24.96 /27
4 207.21.24.128 /27
5 207.21.24.160 /27
6 207.21.24.192 /27
7 207.21.24.224 /27
WAN
links
Sub-subnet 0 207.21.24.192 /30
Sub-subnet 1 207.21.24.196 /30
Sub-subnet 2 207.21.24.200 /30
Sub-subnet 3 207.21.24.204 /30
Sub-subnet 4 207.21.24.208 /30
Sub-subnet 5 207.21.24.212 /30
Sub-subnet 6 207.21.24.216 /30
Sub-subnet 7 207.21.24.220 /30
Free
Addresses
WAN
1 & 2
0 207.21.24.0 /27
1 207.21.24.32 /27
2 207.21.24.64 /27
3 207.21.24.96 /27
4 207.21.24.128 /27
5 207.21.24.160 /27
6 207.21.24.192 /27
7 207.21.24.224 /27

VLSM
Through applying VLSM, the topology was able to be
addressed and still have two complete subnets
available for future growth.
19
207.21.24.32 /27 207.21.24.64 /27 207.21.24.160 /
28
207.21.24.176 /
28
207.21.24.192 /
30
207.21.24.196 /
30

VLSM
EXERCISE 1
Your company has been assigned IP network 195.39.71.0 /
24. Given that headquarters (60 hosts) is connected to
five branch offices (12 hosts each) by a WAN link, and to
an ISP (the ISP owns the addresses on that link),
determine an appropriate IP addressing scheme.
20
Headquarters
Branch 1
60 users
12 users 12 users 12 users 12 users 12 users
Branch 2 Branch 3 Branch 4 Branch 5
ISP

21
Given the IP
address
195.39.71.0 /24,
subnet according
to the largest
subnet needed.
(Headquarters 60
hosts)
0
255
128
127
64 192
63 191
You would need to
borrow 2 bits or /
26. This would
give you 4
networks with 64
host addresses on
each subnet.

22
Playing it safe, we
will not use the
first subnet
(subnet 0).
0
64
128
192We will start
addressing with
195.39.71.64 /26.
Headquarters
needs 60 hosts, so
we will assign
them .64 - .127.
Headquarters
60 hosts
26 bit mask or /26
(255.255.255.192)

23
The 5 Branch
offices only need
12 hosts each.
0
64
128
192
The next address
block available is
the .128 - .191
block (64
addresses). Here
we will apply
VLSM.
Headquarters
60 hosts
26 bit mask or /26
(255.255.255.192)
Using a /28 mask
will give us 16
hosts at each
location. This will
take care of 4 of
the Branch offices.
160
144 176
Branch 1
12 hosts
/28
(255.255.255.240)
Branch 2
12 hosts
/28
(255.255.255.240)
Branch 3
12 hosts
/28
(255.255.255.240)
Branch 4
12 hosts
/28
(255.255.255.240)

24
To obtain a block
for Branch 5, we
will need to subnet
the .192 - .255
block using a /28
mask.
0
64
128
192
Headquarters
60 hosts
26 bit mask or /26
(255.255.255.192)
160
144 176
Branch 1
12 hosts
/28
(255.255.255.240)
Branch 2
12 hosts
/28
(255.255.255.240)
Branch 3
12 hosts
/28
(255.255.255.240)
Branch 4
12 hosts
/28
(255.255.255.240)
224
208
Branch 5
12 hosts
/28
(255.255.255.240)

25
Now we need to
address the 5
WAN links that
connect to the
Branch offices.
These are point-to-
point connections
and only require 2
addresses.
0
64
128
192
Here we will use
a /30 mask to
further subnet the
subnets.
Headquarters
60 hosts
26 bit mask or /26
(255.255.255.192)
160
144 176
Branch 1
12 hosts
/28
(255.255.255.240)
Branch 2
12 hosts
/28
(255.255.255.240)
Branch 3
12 hosts
/28
(255.255.255.240)
Branch 4
12 hosts
/28
(255.255.255.240)
224
208
Branch 5
12 hosts
/28
(255.255.255.240)
232
228 236
WAN
1
WAN
2
WAN
3
WAN
4
248
244
WAN
5
240

26
The remaining
networks could be
used for future
growth of either
LANs or WANs.
Subnet 0 could
also be further
subnetted
according to the
needs of the
network.
0
64
128
192
Headquarters
60 hosts
26 bit mask or /26
(255.255.255.192)
160
144 176
Branch 1
12 hosts
/28
(255.255.255.240)
Branch 2
12 hosts
/28
(255.255.255.240)
Branch 3
12 hosts
/28
(255.255.255.240)
Branch 4
12 hosts
/28
(255.255.255.240)
224
208
Branch 5
12 hosts
/28
(255.255.255.240)
232
228 236
WAN
1
WAN
2
WAN
3
WAN
4
248
244
WAN
5
240

27
Address
provided by ISP195.39.71.64 /26
195.39.71.128 /28 195.39.71.144 /28 195.39.71.160 /28 195.39.71.176 /28 195.39.71.192 /28
195.39.71.208 /30
195.39.71.212/30
195.39.71.216/30
195.39.71.220/30
195.39.71.224 /30
Applying the Addresses to the Topology

CLASSLESS INTERDOMAIN ROUTING
CIDR (pronounced “cider”) ignores address class.
With CIDR, a router use a bit mask to determine the
network and host portions of an address.
CIDR replaced the categories (A, B, C, etc.) with a
more generalized network prefix. This prefix
could be of any length rather than just 8, 16, or 24
bits. This allows CIDR to craft network address
spaces according to the size of a network instead
of force-fitting networks into presized network
address spaces.
28

CIDR sounds a lot like VLSM
CIDR is usually discussed in general
Internet context (ISPs)
Uses custom length prefixes to reduce workload in key
Internet routers
VLSM is usually discussed in enterprise
context
Uses custom length prefixes to have better usage of
enterprise address space
29

Routers use the network-prefix, rather than the first
3 bits of the IP address, to determine the dividing
point between the network number and the host
number.
In the CIDR model, each piece of routing
information is advertised with a bit mask or
prefix-length ( /x ). The prefix-length is a way of
specifying the number of leftmost contiguous
bits in the network-portion of each routing table
entry.
30

For example, a network with 20 bits of network-
number and 12 bits of host-number would be
advertised with a 20 bit prefix (/20).
The clever thing is that the IP address advertised
with the /20 prefix could be a former Class A,
Class B, or Class C.
All addresses with a /20 prefix represent the same
amount of address space (212
or 4,096 host
addresses).
20 bits network + 12 bits host
31

Instead of handing out an entire A, B, or C network
to an organization, address space can be
assigned in “chunks” that fit the need.
If an organization needs 254 host addresses, what
difference does it make whether they are given:
 a Class C (200.23.76.0 /24)
 1/256th of a Class B (145.38.20.0 /24)
 1/65,536th of a Class A (91.187.7.0 /24)
Using a /24 prefix, each of these specifies eight host
bits which will support 254 hosts.
Note: Each former Class C can be referred to as a /24.
32

33
Network Prefix Equivalent Number of Class Addresses Number of Hosts
/27 1/8th of a Class C 32
/26 1/4th of a Class C 64
/25 1/2 of a Class C 128
/24 1 Class C or 1 /24 256
/23 2 Class C or 2 /24s 512
/22 4 Class C or 4 /24s 1,024
/21 8 Class C or 8 /24s 2,048
/20 16 Class C or 16 /24s 4,096
/19 32 Class C or 32 /24s 8,192
/18 64 Class C or 64 /24s 16,384
/17 128 Class C or 128 /24s 32,768
/16 256 Class C or 1 Class B 65,536
/15 512 Class C or 2 Class B 131,072
/14 1,024 Class C or 4 Class B 262,144
/13 2048 Class C or 8 Class B 524,288
/12 4096 Class C or 16 Class B 1,048,576
/11 8192 Class C or 32 Class B 2,097,152
/10 16384 Class C or 64 Class B 4,194,304
/9 32768 Class C or 128 Class B 8,388,608
/8 65,536 Class C or 256 Class B or 1 Class A 16,777,216
Prefix Equivalents

Day 5.5 subnetting

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Day 5.5 subnetting