subnetting - learngroup · subnetting - subnetting a network means to use the subnet mask to divide...

SUBNETTING

IT223 – Internet Technology

IT223-Internet Technology 2

Subnetting

IT223-Internet Technology

3

Subnetting

Subnetting - Subnetting a network means to use the subnet mask to divide the

network and break a large network up into smaller, more efficient and

manageable segments, or subnets.

- Subnetting is done by taking part of host bits then add it to

the network part

Network part Host part

Subnet

bits

Network part Host part

IP

address

Subnetting Example Divide network 192.168.1.0/24 into 4 subnets Solution: 4 subnets need 2 bits

192.168.1 . 0

192.168.1 . 0000 0000 to 0011 1111

192.168.1 . 0100 0000 to 0111 1111

192.168.1 . 1000 0000 to 1011 1111

192.168.1 . 1100 0000 to 1111 1111

subnet mask is 255.255.255.192 or /26

The first subnet is 192.168.1.0/26

The second subnet is 192.168.1.64/26

The third subnet is 192.168.1.128/26

The fourth subnet is 192.168.1.192/26

0 - 63

64 - 127

128 - 191

192 - 255

Divide network 192.168.1.0/24 into 4 subnets -

Solution :

4 subnets need 2 bits

- subnet mask = 255.255.255.192

- interesting octet is 192

- hop count = 256 – 192 = 64

- The first subnet is 192.168.1.0/26

- The second subnet is 192.168.1.64/26

- The third subnet is 192.168.1.128/26

- The fourth subnet is 192.168.1.192/26

Subnetting Example

0 - 63

64 - 127

128 - 191

192 - 255

Divide network 172.16.0.0/16 into 8subnets

Solution :

- 8 subnets need 3 bits

- subnet mask = 255.255.224.0


- hop count = 256 – 224 = 32




-The 8th subnet is 172.16.224.0/19

Subnetting Example

172.16.0.1-172.16.31.254

172.16.32.1 -172.16.63.254

172.16.64.1-172.16. 95.254

172.16.224.1-172.16.255.254

Divide network 10.0.0.0/10 into 4subnets

Solution :

- 4subnets need 2 bits

- subnet mask = 255.240.0. 0


- hop count = 256 – 240= 16




- The fourth subnet is 10.48.0.0/12

Subnetting Example

10.0.0.1-10.15.255.254

10.16.0.1-10.31.255.254

10.32.0.1-10.47.255.254

10.48.0.1-10.63.255.254

How Many Networks

Given the Class C network of 204.15.5.0/24, subnet the

network in order to create the network in the figure below, with

the host requirements shown

Calculating Addresses without

VLSM

Looking at the network shown, you can see that

you are required to create five subnets.

The largest subnet must support 28 host

addresses.

You can start by looking at the subnet

requirement.

In order to create the 5 needed subnets you would

need to use 3 bits from the Class C host bits.

Two bits would only allow you four subnets (22).

How many hosts will this support? 25 = 32 (30

usable).

This meets the requirement

Calculating Addresses without

VLSM

Subnet Mask = 255. 255. 255. 224 /27

Interesting Octet is 224

Hope Count = 256-224 = 32

Subnet A: 204.15.5.0/27 host address range 1 to

30

Subnet B: 204.15.5.32/27 host address range 33

to 62

Subnet C: 204.15.5.64/27 host address range 65

to 94

Subnet D: 204.15.5.96/27 host address range 97

to 126

Subnet E: 204.15.5.128/27 host address range

129 to 158

Waste of IP Addresses ?

Subnet A: 204.15.5.0/27 host address range 1 to

30

Only 14 Addresses Used

Subnet B: 204.15.5.32/27 host address range 33

to 62

Only 28 Addresses Used.

Subnet C: 204.15.5.64/27 host address range 65

to 94


Subnet D: 204.15.5.96/27 host address range 97

to 126


Subnet E: 204.15.5.128/27 host address range

129 to 158


In many cases, having the same subnet mask for all

subnets ends up wasting address space.

Solution ?

Using Variable Length Subnet Masks (VLSM)

VLSM allows you to use different masks for each

subnet, thereby using address space efficiently

How ?

The easiest way to assign the subnets is to assign

the largest first and work your way down.

In our case:

Proceed in the following order: Network B, E, A, D,

C

Subnet B: requires a /27 (255.255.255.224) mask to

support 28 hosts

We have calculated that before:

Subnet B: 204.15.5.0/27 host address range 1 to

30

Subnet E: requires a /27 (255.255.255.224) mask to

support 28 hosts

We have also calculated that before:

Subnet E: 204.15.5.32/27 host address range 33

to 62

Calculating Addresses with

VLSM

Subnet A: Needs 14 Hosts

Take one of the subnets from the last subnetting

procedure (other than the ones given to B and E.

Example: 204.15.5.64/27 (Apply Subnetting on

this one)

To support 14 hosts (at least 4 bits are needed for

the hosts)

Subnet Mask = 255. 255.255.240 /28

Interesting Octet = 240 Hop Count = 256 – 240

= 16

Number of Subnets = 2 ^ 1 = 2

204.15.5.64/28 -> Host address range 65 to 78

(Assign that to Subnet A)

204.15.5.80/28 -> Host Address Range 81 to 95


VLSM

Subnet D: Needs 7 Hosts

Take one of the subnets from the last subnetting

procedure (other than the ones given to A and D


this one)


the hosts, 3 bits has only 6 valid hosts)

Hence no need for any more subnetting

204.15.5.80/28 -> Host Address Range 81 to 94


VLSM

Subnet C: Needs 2 Hosts

Take one of the subnets from the last procedure

(other than the ones given to A


this one)


the hosts)

Subnet Mask = 255. 255.255.252 /30

Interesting Octet = 252 Hop Count = 256 – 252

= 4

Number of Subnets = 2 ^ 2 = 4

204.15.5.96/30 -> Host address range 97 to 98

(Assign that to Subnet C)

204.15.5.100/30 -> Host Address Range 101 to

102


VLSM

Finally:

Subnet B: 204.15.5.0/27 host address range 1 to

30

Subnet E: 204.15.5.32/27 host address range 33

to 62

Subnet A 204.15.5.64/28 host address range 65 to

78

Subnet D 204.15.5.80/28 host address range 81

to 94

Subnet C 204.15.5.96/30 host address range 97

to 98


VLSM

Calculating Addresses: Case 2

Calculating Addresses: Case 2

Calculate the address ranges for sub networks

Questions

23

IT223-Internet Technology

subnetting - learngroup · subnetting - subnetting a network means to use the subnet mask to divide...

Documents