Chapter 9: Subnetting
IP Networks
Introduction to Networking
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Dermot Clarke DIT Sept’ 2013
Network Segmentation
Reasons for Subnetting
Large networks need to be segmented into smaller sub-networks,
creating smaller groups of devices and services in order to:
• Control traffic by containing broadcast traffic within subnetwork
• Reduce overall network traffic and improve network performance
Subnetting - process of segmenting a network into multiple smaller
network spaces called subnetworks or Subnets.
Communication Between Subnets
• A router is necessary for devices on different networks and subnets
to communicate.
• Each router interface must have an IPv4 host address that belongs to
the network or subnet that the router interface is connected to.
• Devices on a network and subnet use the router interface attached to
their LAN as their default gateway.
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Subnetting an IPv4 Network
IP Subnetting is FUNdamental
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Basic Subnetting
• Subnetting allows for creating multiple logical networks from a
single address block.
We create subnets by using one or more of the host bits as network bits.
This is done by extending the mask to borrow some of the bits from the
host portion of the address to create additional network bits.
The more host bits used, the more subnets that can be defined.
For each bit borrowed, we double the number of subnetworks available.
For example, if we borrow 1 bit, we can define 2 subnets. If we borrow 2 bits,
we can have 4 subnets.
But, with each bit we borrow, fewer host addresses are available per subnet.
Number of subnets: 2n
(n = the number of bits borrowed)
Total number of addresses per subnet 2m
The number of usable hosts 2m – 2 (m = the number of host bits left)
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Subnetting an IPv4 Network
Basic Subnetting
• Borrowing Bits to Create Subnets
• Borrowing 1 bit 21 = 2 subnets
Borrowing 1 Bit from the host portion creates 2 subnets with the same subnet mask
Subnet 0
Subnet 1
Network 192.168.1.0-127/25
Network 192.168.1.128-255/25
Mask: 255.255.255.128
Mask: 255.255.255.128
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Example Basic Subnetting (2 subnets)
• Given an address block of 192.168.1.0 /24, we wish to divide this network
into 2 subnets.
• /24 means 8 hosts bits so 28 = 256 Address so we have 28 – 2 =254 hosts
• If we borrow one bit from the host portion then the new mask becomes
11111111.1111111.1111111.10000000 i.e. 255.255.255.128 (/25), instead of
the original 255.255.255.0 mask.
• The 1 borrowed bit gives us 21=2 Subnets and now leaves 7 bits for the host
so 27 =128 address per subnet and 27 – 2 =126 Hosts on each subnet.
Fill in Table: “Magic number Method”
Mask is 25.255.255.128 so 256-128=128 is the Magic number
0+128=128
Subnet Subnet Add’
Hosts
Broadcast
0
192.168.1.0
192.168.1.1 to 192.168.1.126
192.168.1.127
1
192.168.1.128
192.168.1.129 to 192.168.1.254
192.168.1.255
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Note:
All /25
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Basic Subnetting (2 subnets)
Subnet 0
Subnet Add: 192.168.1.0
Broadcast: 192.168.1.127
PC
Sw’
1
192.168.1.2 /25
PC
192.168.1.3 /25
PC
192.168.1.126 /25
Router: 192.168.1.1 /25
Broadcast Domain
PC
Sw’
192.168.1.130 /25
PC
192.168.1.131 /25
PC
192.168.1.254 /25
Router: 192.168.1.129 /25
Subnet 1
Subnet Add: 192.168.1.128
Broadcast: 192.168.1.255
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Dermot Clarke DIT Sept’ 2013
Subnetting an IPv4 Network
Subnets in Use
All Host bits = 0
Subnet 0
Network 192.168.1.0-127/25
All Host bits = 1
All Host bits = 0
Subnet 1
Network 192.168.1.128-255/25
All Host bits = 1
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Basic Subnetting (4 subnets)
• Next, an internetwork that requires 4 subnets.
• Again we start with the same 192.168.1.0 /24 address block.
To provide 4 networks, 2? ≈ 4
we borrow two bits. This will provide 22 = 4 subnets. (Note the number of subnets will be of
the order of 2) So there will be 6 bits left for hosts
So the new subnet mask is 11111111.1111111.11111111.11000000 255.255.255.192 (/26)
The number of subnets:
22 = 4 subnets
The number addresses per subnet: 26 = 64 hosts per subnet
The number of hosts
26 - 2 = 62 hosts per subnet
Fill in Table: “Magic number Method”
Mask is 25.255.255.192 so 256-192=64 is the Magic number.
Subnet Subnet Add’
192.168.1.0
0
Hosts
Broadcast
192.168.1.1 to 192.168.1.62
192.168.1.63
1
192.168.1.64
192.168.1.65 to 192.168.1.126
192.168.1.127
2
192.168.1.128
192.168.1.129 to 192.168.1.190
192.168.1.191
3
192.168.1.192
192.168.1.193 to 192.168.1.254
192.168.1.255
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Note:
All /26
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Recap: Magic Numbers
• To make the job of subnetting easier, there is a method
that allows you to calculate a "magic" number.
• The magic number we're looking for is the number of
addresses in each network, including the network,
broadcast and host range.
“Ways to Calculate “Magic” number
1)
The calculation 2number_ of_ host_ bits yields the "magic" number.
We have 5 host bits remaining so…..
25 = 32 - our "magic" number.
2) Given a mask for example 255.255.255.224
Magic number =
256
-224
32
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Example Subnetting - Class C
• Network: 192.168.80.0
Subnet Mask: 255.255.255.224
• Network: 27 bits Host: 5 bits Magic Number: 25 = 32
ID
Network
Address
Subnet Address
Range
Broadcast
Address
0 192.168.80.0
192.168.80.1 – 192.168.80.30 /27 192.168.80.31
1 192.168.80.32
192.168.80.33 –192.168.80.62 /27 192.168.80.63
2 192.168.80.64
192.168.80.65 – 192.168.80.94
192.168.80.95
3 192.168.80.96
192.168.80.97 – 192.168.80.126
192.168.80.127
4 192.168.80.128 192.168.80.129 – 192.168.80.158 192.168.80.159
5 192.168.80.160 192.168.80.161 – 192.168.80.190 192.168.80.191
6 192.168.80.192 192.168.80.193 – 192.168.80.222 192.168.80.223
7 192.168.80.224 192.168.80.225 – 192.168.80.254 192.168.80.255
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Determining the Subnet Mask
Subnetting Based on Host Requirements
There are approaches when planning subnets:
1. Number of Subnets required
2. Number of Host addresses required
•Formula to determine number of useable hosts
2n-2
2n (where n is the number the number of host bits remaining) is used
to calculate the number of hosts
-2 Subnetwork ID and broadcast address cannot be used on each
subnet
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Determining the Subnet Mask
Subnetting Network-Based Requirements
Calculate number of subnets
•Formula 2n-2 (where n is the number of bits borrowed)
Subnet needed for
each department in
graphic
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Determining the Subnet Mask
Subnetting To Meet Network Requirements
• It is important to balance the number of subnets needed
and the number of hosts required for the largest subnet.
• Design the addressing scheme to accommodate the
maximum number of hosts for each subnet.
• Allow for growth in
each subnet.
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Subnetting - Class B
• Let's try one.
You are the network administrator for a world-wide organization
with 7,500 users.
You have 10 world-wide central offices and each of those have
their own networks and branch offices. Central and Branch
office networks range from 100 to 3,000 users.
You have decided that a Class B network will be sufficient for
your needs and you must subnet the network to include yourself
and the central offices.
Each central office handles their own network maintenance
and it will be up to them to further subnet the network you
design.
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Subnetting - Class B
• You have decided to use the Class B private address of
172.25.0.0 / 16
Head Office
Central 01
Central 02
Branch 01
Branch nn
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Central 03
Central 10
Your objective is to provide
enough addresses so that each
central office can cover their
branches and allow room for
future expansion.
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Subnetting – Class B
• Determining your magic number – Class A and B.
The trick here in determining the magic number.
Say we want 4096 IP addresses for each site (i.e 12
Host bits) so mask will be
11111111.11111111.11110000.00000000
subnet mask 255.255.240.0 or (/20)
Magic number =
256
-240
16
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ID
Network
Address
0
172.25.0.0
172.25.0.1 to 172.25.15.254 /20
172.25.15.255
1
172.25.16.0
172.25.16.1 to 172.25.31.254 /20
172.25.31.255
2
172.25.32.0
172.25.32.1 to 172.25.47.254 /20
172.25.47.255
3
172.25.48.0
172.25.48.1 to 172.25.63.254 /20
172.25.63.255
4
172.25.64.0
172.25.64.1 to 172.25.79.254 /20
172.25.79.255
5
172.25.80.0
172.25.80.1 to 172.25.95.254 /20
172.25.95.255
6
172.25.96.0
172.25.96.1 to 172.25.111.254
172.25.111.255
7
172.25.112.0 172.25.112.1 to 172.25.127.254
172.25.127.255
8
172.25.128.0 172.25.128.1 to 172.25.143.254
172.25.143.255
9
172.25.144.0 172.25.144.1 to 172.25.159.254
172.25.159.255
10
172.25.160.0 172.25.160.1 to 172.25.175.254
172.25.175.255
11
172.25.176.0 172.25.176.1 to 172.25.191.254
172.25.191.255
12
172.25.192.0 172.25.192.1 to 172.25.207.254
172.25.207.255
13
172.25.208.0 172.25.208.1 to 172.25.223.254
172.25.223.255
14
172.25.224.0 172.25.224.1 to 172.25.239.254
172.25.239.255
15
172.25.240.0 172.25.240.1 to 172.25.255.254
172.25.255.255
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Subnet Address
Range
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Address
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Benefits of Variable Length Subnet Masking
Traditional Subnetting Wastes Addresses
•Traditional subnetting - same number of addresses is
allocated for each subnet.
•Subnets that require fewer addresses have unused
(wasted) addresses. For example, WAN links only need 2
addresses.
•Variable Length Subnet Mask (VLSM) or subnetting a
subnet provides more efficient use of addresses.
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Benefits of Variable Length Subnet Masking
Variable Length Subnet Masks (VLSM)
• VLSM allows a network space to be divided in unequal
parts.
• Subnet mask will vary depending on how many bits have
been borrowed for a particular subnet.
• Network is first subnetted, and then the subnets are
subnetted again.
• Process repeated as necessary to create subnets of
various sizes.
Subnet a subnet……..
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Benefits of Variable Length Subnet Masking
Basic VLSM
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Benefits of Variable Length Subnet Masking
VLSM in Practice
•Using VLSM subnets, the LAN and WAN segments in
example below can be addressed with minimum waste.
• Each LANs will be assigned a subnet with /27 mask.
•Each WAN link will be assigned a subnet with /30 mask.
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Structured Design
Planning to Address the Network
Allocation of network addresses should be planned and
documented for the purposes of:
•Preventing duplication of addresses
•Providing and controlling access
•Monitoring security and performance
Addresses for Clients - usually dynamically assigned using
Dynamic Host Configuration Protocol (DHCP)
Sample Network
Addressing Plan
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Subnetting: Worked Example
• The figure shows the network topology for this example:
• Student LAN
Student Computers: 40
Router (LAN Gateway): 1
Switches (management): 10
Total for student subnetwork: 51
• Lecturers LAN
Instructor Computers: 20
Router (LAN Gateway): 1
Switches (management): 4
Total for instructor subnetwork: 25
• Administrator LAN
Administrator Computers: 10
 There are two methods available for allocating
addresses to an internetwork.
Server: 1
1.
We can use a non-VLSM approach, where all
subnets use the same prefix length and the
same number of host bits.
2.
We can use Variable Length Subnet Masking
(VLSM), where we assign the prefix and host
bits to each network based on the number of
hosts in that network.
Router (LAN Gateway): 1
Switch (management): 1
Total for administration subnetwork: 13
• WAN
Router - Router WAN: 2
Total for WAN: 2
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Calculating Addresses: Case 1:
Addresses-without VLSM (all Subnets equal size)
• Given an address block 192.168.1.0/24
• When using the non-VLSM method of assigning addresses, all subnets
have the same number of addresses.
We base the number of addresses for all networks on the addressing
requirements for the largest network.
• In Case 1, the Student LAN is the largest network, requiring
51addresses.
• We will need 6 host bits.
2^6 = 64 So we will have 64 - 2 = 62 usable host addresses
This meets the current requirement for at least 51 addresses, with a
small allowance for growth.
The new mask will be 11111111.11111111.11111111.11000000
255.255.255.192
6 host bits
• So our magic number will be 256-192 = 64
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Subnet 0
192.168.1.0
Network Address
1st Usable Host1st
(Student)
2nd Usable Host
Last Usable Host
Broadcast Address
Subnet 1
Network Address
(Lecturer)
1st Usable Host1st
Last Usable Host
Broadcast Address
Subnet 3
Network Address
(Admin)
1st Usable Host1st
Last Usable Host
Broadcast Address
Subnet 3
Network Address
(WAN)
1st Usable Host1st
Last Usable Host
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192.168.1.255
Broadcast Address
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Subnet 0
192.168.1.0
Network Address
(Student)
192.168.1.1 /26
1st Usable Host1st
192.168.1.2 /26
2nd Usable Host
192.168.1.62/26
Last Usable Host
62 usable addresses, 51
required so 9 waste
addresses.
0+64 (magic Number) = 64
192.168.1.63
Broadcast Address
Subnet 1
192.168.1.64
Network Address
(Lecturer)
192.168.1.65/26
1st Usable Host1st
62 usable addresses, 25
required so 37 waste
addresses.
192.168.1.126/26
Last Usable Host
64+64 (magic Number) = 128
192.168.1.127
Broadcast Address
Subnet 3
192.168.1.128
Network Address
(Admin)
192.168.1.129/26
1st Usable Host1st
192.168.1.190/26
Last Usable Host
62 usable addresses, 13
required so 49 waste
addresses.
128+64 (magic Number) = 192
192.168.1.191
Broadcast Address
Subnet 3
192.168.1.192
Network Address
(WAN)
192.168.1.193/26
1st Usable Host1st
192.168.1.254/26
Last Usable Host
192.168.1.255
Broadcast Address
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62 usable addresses, 2
required so 60 waste
addresses.
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Calculating Addresses: Case 2: Addresses-with VLSM
• For the VLSM assignment, we can allocate a much smaller block of addresses to
each network, as appropriate.
• As before the address block 192.168.1.0/24 has been assigned to this internetwork.
• Student LAN
The largest subnet is the Student LAN requires 51 addresses.
We will need 6 host bits.
2^6 = 64 So we will have 64 - 2 = 62 usable host addresses
This meets the current requirement for at least 51 addresses, with a small allowance for
growth.
The new mask will be 11111111.11111111.11111111.11000000 (/26) 255.255.255.192
So our magic number will be 256-192 = 64
Subnet 0
192.168.1.0
Network Address
(Student)
192.168.1.1 /26
1st Usable Host
192.168.1.62/26
Last Usable Host
192.168.1.63
Broadcast Address
192.168.1.64
Network Address
Subnet 1
62 usable addresses, 51
required so 9 waste
addresses.
0+64 (magic Number) = 64
(Lecturer)
Broadcast Address
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Calculating Addresses: Case 2: Addresses-with VLSM
• Lecturer LAN
The next largest network is the Instructor LAN. It requires at least 25 addresses.
Borrowing 5 bits for the host portion yields this calculation:
2^5 = 32 So we will have 32 - 2 = 30 usable host addresses
This meets the current requirement for at least 25 addresses, with a small
allowance for growth.
The new mask will be 11111111.11111111.11111111.11100000 (/27)
255.255.255.224
So our magic number will be 256-224 = 32
Subnet 0
(Student)
192.168.1.63
Broadcast Address
Subnet 1
192.168.1.64
Network Address
(Lecturer)
192.168.1.65 /27
1st Usable Hostt
192.168.1.94 /27
Last Usable Host
192.168.1.95
Broadcast Address
192.168.1.96
Network Address
Subnet 3
30 usable addresses, 25
required so 5 waste
addresses.
64+32 (magic Number) = 96
(Admin)
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Calculating Addresses: Case 2: Addresses-with VLSM
• Administrator LAN
The next largest network is the Admin LAN. It requires at least 13 addresses.
Borrowing 4 bits for the host portion yields this calculation:
2^4 = 16 So we will have 16 - 2 = 14 usable host addresses
This meets the current requirement for at least 13 addresses, with a very small!
allowance for growth.
The new mask will be 11111111.11111111.11111111.11110000 (/28)
255.255.255.240
So our magic number will be 256-240 = 16
Subnet 1
(Lecturer)
192.168.1.95
Broadcast Address
Subnet 3
192.168.1.96
Network Address
(Admin)
192.168.1.97 /28
1st Usable Host
192.168.1.110 /28
Last Usable Host
192.168.1.111
Broadcast Address
192.168.1.112
Network Address
Subnet 3
14 usable addresses, 13
required so 1 waste
addresses.
96+16 (magic Number) = 112
(WAN)
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Calculating Addresses: Case 2: Addresses-with VLSM
• WAN
The last segment is the WAN, requiring 2 host addresses.
Borrowing 2 bits for the host portion yields this calculation:
2^2 = 4 So we will have 4 - 2 = 2 usable host addresses
This meets the current requirement .No need to allow for expansion as
a WAN link will only ever need 2 addresses.
The new mask will be 11111111.11111111.11111111.11111100 (/30)
255.255.255.252
So our magic number will be 256-252 = 4
Subnet 3
192.168.1.111
Broadcast address
Subnet 3
192.168.1.112
Network Address
(WAN)
192.168.1.113 /30
1st Usable Host
192.168.1.114 /30
Last Usable Host
192.168.1.115
Broadcast Address
(Admin)
Next Subnet:
2 usable addresses, 2
required so 0 waste
addresses.
112+4 (magic Number) = 116
192.168.1.116
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Address TableDermot Clarke DIT Sept’ 2013
Subnet 0
192.168.1.0
Network Address
(Student)
192.168.1.1 /26
1st Usable Host
192.168.1.62/26
Last Usable Host
192.168.1.63
Broadcast Address
Subnet 1
192.168.1.64
Network Address
(Lecturer)
192.168.1.65 /27
1st Usable Hostt
192.168.1.94 /27
Last Usable Host
192.168.1.95
Broadcast Address
Subnet 3
192.168.1.96
Network Address
(Admin)
192.168.1.97 /28
1st Usable Host
192.168.1.110 /28
Last Usable Host
192.168.1.111
Broadcast Address
Subnet 3
192.168.1.112
Network Address
(WAN)
192.168.1.113 /30
1st Usable Host
192.168.1.114 /30
Last Usable Host
192.168.1.115
Broadcast Address
62 usable addresses, 51
required so 9 waste
addresses.
30 usable addresses, 25
required so 5 waste
addresses.
14 usable addresses, 13
required so 1 waste
addresses.
2 usable addresses, 2
required so 0 waste
addresses.
192.168.1.116
As you can see we have only uses 116 address in our design leaving plenty for use else where.
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Subnetting an IPv6 Network
Subnetting Using the Subnet ID
An IPv6 Network Space is subnetted to support
hierarchical, logical design of the network
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Subnetting an IPv6 Network
IPV6 Subnet Allocation
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Subnetting an IPv6 Network
Subnetting into the Interface ID
IPv6 bits can be borrowed from the interface ID to create
additional IPv6 subnets
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Cisco Confidential
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Dermot Clarke DIT Sept’ 2013
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© 2006, Cisco Systems, Inc. All rights reserved.
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