Blog 4 Homework Answer

Q:

A company has a 190.240.0.0/16 block. They now want to re-organise it into large blocks. They wish to have subnets that can handle at least 1000 hosts each.
You can substitute into the formulae 2^S >= subnets required and 2^H-2 >= hosts required - to determine the number of bits required.
Please calculate:

• S = bits used for subnetting
• H= bits used for the number of hosts in each subnet

Also, please calculate the following for the second subnet:

• subnet mask
• subnet address
• first usable address in that subnet
• last usable address in that subnet
• broadcast address of that subnet

A:

Network currently first two octets 16 bits:
11111111.11111111.

Host second two octets 16 bits:
00000000.00000000

Subnet = 255.255.0.0

In order to provide a minimum of 1000 hosts per subnet we need to retain 10 host bits of the current 16 - 2 to the power of 10 = 1024 minus 2, 1 for network and 1 for broadcast leaves 1022.

We now have 6 bits of the thrid octet that can be used for subnetting, 16 original host bits minus 10 still required as above = 6 - 2 to the power of 6 = 64 subnets.

In binary = 11111111.11111111.11111100.00000000
Subnet mask is = 255.255.252.0

Each subnet increases by 4 - 4 * 64 = 256 

Subnet 2

- subnet address = 190.240.4.0
- first usable address in that subnet = 190.240.4.1
- last usable address in that subnet = 190.240.7.254
- broadcast address of that subnet = 190.240.7.255