Question

A company has been granted a block of classless addresses which starts at 198.87.24.0/23. Create the following 4 subnets by calculating the subnet address for each subnet including the prefix. Create a table to organise your answer. Show your working in the space provided.

a) 1 subnet with 256 addresses (1 mark correct subnet, 0.5 mark good calcs)

b) 1 subnet with 128 addresses (1 mark correct subnet, 0.5 mark good calcs)

c) 2 subnets with 64 addresses each (1 mark per correct subnet, 0.5 mark good calcs per subnet)

Answer 1

Subnet Address / Network Address	Network mask	Range	No. of address
198.87.24.0/24	255.255.255.0	198.87.24.0 - 198.87.24.255	256
198.87.25.0/25	255.255.255.128	198.87.25.0 - 198.87.25.127	128
198.87.25.128/26	255.255.255.192	198.87.25.128 - 198.87.25.191	64
198.87.25.192/26	255.255.255.192	198.87.25.192 - 198.87.25.255	64

Given IP: 198.87.24.0/23

The network has a total of 32 bits, split into 4 octets. Each octet ranges from value 0 - 255.

As per the CIDR notation, 23 bits are network bits.

So, subnet mask: 255:255:254:0

and 9 bits are host bits.

There are 512 addresses in this netrwork ().

a) First subnet requires 256 addresses.

Given IP in bits : 11000110.01010111.0001100 0.00000000

The red bits are network bits and green bits are host bits.

For 256 addresses 8 bits are needed (). The right most bits are considered for this. Taking 8 bits out of the 9 host bits from right as hosts:

11000110.01010111.0001100 0.00000000

The green bits are the new host bits. The blue bit will be considered as a network bit,making 24 network bits.

Converting back to decimals, the first subnet will be: 198.87.24.0 - 198.87.24.255

Network address : 198.87.24.0/24

This subnet will give 256 addresses including network and broadcast address.

b) Second subnet requires 128 addresses.

198.87.24.0 - 198.87.24.255 have already been used for first subnet. The next address is 198.87.25.0/24. The IP in bits : 11000110.01010111.00011001. 00000000

The red bits are network bits and green bits are host bits.

For 128 addresses 7 bits are needed (). The right most bits are considered for this. Taking 7 bits out of the 8 host bits from right as hosts:

11000110.01010111.00011001. 0 0000000

The green bits are the new host bits. The blue bit will be considered as a network bit,making 25 network bits.

Converting back to decimals, the second subnet will be: 198.87.25.0 - 198.87.25.127

Network address : 198.87.25.0/25

This subnet will give 128 addresses including network and broadcast address.

c) Third and fourth subnet requires 64 addresses in each.

Third:

Addresses till 198.87.25.127 have already been used for second subnet. The next address is 198.87.25.128/25. The IP in bits : 11000110.01010111.00011001 .1 0000000

The red bits are network bits and green bits are host bits.

For 64 addresses 6 bits are needed (). The right most bits are considered for this. Taking 6 bits out of the 7 host bits from right as hosts:

11000110.01010111.00011001 .1 0 000000

The green bits are the new host bits. The blue bit will be considered as a network bit,making 26 network bits.

Converting back to decimals, the third subnet will be: 198.87.25.128 - 198.87.25.191

Network address : 198.87.25.128/26

This subnet will give 64 addresses including network and broadcast address.

Fourth:

Addresses till 198.87.25.191 have already been used for third subnet. The next address is 198.87.25.192. The IP in bits :

11000110.01010111.00011001.11 000000 

The rest of the 64 bits in the nertwork can be used for this sub net.

Fourth subnet : 198.87.25.192 - 198.87.25.255

Network address : 198.87.25.192/26