Subneting

-example-

We have the following network id:

–223.14.17.0

The class that the address belongs to is :

Clasa C

Step 1

Determining the mask of the network id

The mask for the class C is:

255.255.255.0

Step 2

Requirements of the number of subnetworks and number of nodes of each subnetwork

We need:

- 13 subnetworks

- 10 nodes

Step 3

Determining the number of

subnetworks which results by taking a

certain number of bits from the host id

segment of the ip.

Lets see how many subnetworks and

nodes we’ll obtain if we take 4 bits.

Step 3…

223.14.17.0

X X X X H H H H

16 possible subnetworks

16 possible nodes in each subnetwork

Step 3…

We obtain16 subnetworks and 16 nodes in each network because:

– Each bit from those 4 borrowed bits have 0 or 1 value, so we have 24=16 combinations

– Idem for those 4 remaining bits from the host id

Important: there are only 14 usable subnetwork and 14 usable nodes. Why?

Step 3…

Because the first and the last

subnetwork is reserved.

Because the first and the last address of

each subnetwork is reserved.

For every each of them, one is the

broadcast address and one is the

network address.

Step 4

Determining the subnetwork mask

Where X represents the borrowed bits for subnetting.

223.14.17.0

X X X X H H H H

Step 4…

Sum all the X bits decimal values to obtain

the subnetwork mask:

The (sub)network mask is:

128 + 64 + 32 + 16 = 240

255.255.255.240

Step 5

Determining the range values for the IP

addresses for each subnetwork

Step 5…

Subnet # Subnet Bits Host Bits In Decimal

1 0000 0000-1111 .0 -.15

2 0001 0000-1111 .16 - .31

3 0010 0000-1111 .32 - .47

4 0011 0000-1111 .48 - .63

5 0100 0000-1111 .64 - .79

6 0101 0000-1111 .80 - .95

7 0110 0000-1111 .96 - .111

8 0111 0000-1111 .112 - .127

Step 5…

Subnet # Subnet Bits Host Bits In Decimal

9 1000 0000-1111 .128 -.143

10 1001 0000-1111 .144 - .159

11 1010 0000-1111 .160 - .175

12 1011 0000-1111 .176 - .191

13 1100 0000-1111 .192 - .207

14 1101 0000-1111 .208 - .223

15 1110 0000-1111 .224 - .239

16 1111 0000-1111 .240 - .255

Step 5…

There are 16 possible subnetworks

There are 16 possible nodes in each

subnetwork

There are 256 total number of

nodes(computers)

Which subnetworks are usable

Which ip addresses are usable in each

subnetwork? Why????

How to determine the network addresses for

the obtained subnetworks

Step 1: Transform the IP addresses for

each subnetwork in binary

Step 2: Transform the subnetowrk

mask in binary

Step 3: Use the boolean operator AND

for combination

Step 4: Convert the upper result in

dotted decimal form.

How to determine the network addresses for

the obtained subnetworks …

IP Host 172.16.2.120

(Sub)network mask 255.255.255.0

10101100.00010000.00000010.01111000

11111111.11111111.11111111.00000000

10101100.00010000.00000010.00000000

172.16.2.0

AND

This is the network address of the first

subnetowrk. The others is obtained in similar

way.

CIDR standard notation

Classless Interdomain Routing is a method of representation of the IP addresses and of the subnetwork masks through a prefix.

For example: 192.168.50.0/27

What represents number 27?

– 27 is the number of bits composing the subnetwork mask, which is 255.255.255.224

– also, we can see that the address starting with 192 is a C class address, so we take 3 more bits for subnetting

Let’s see some CIDR examples:

202.151.37.0/26Network mask?

–255.255.255.192

Borrowed bits?

–Is a class C, so we have taken 2 bits

The first usable subnetwork ?

–202.151.37.64

The second usable subnetwork?

–202.151.37.128

198.53.67.0/30Network mask?

–255.255.255.252

Borrowed bits?

–Is a class C, so we have taken 6 bits

The third usable subnetwork?

–198.53.67.12

The broadcast address of the second usable

subnework?

–198.53.67.11