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Following on in the series, here is the second subnet masking example.

You have been given the network range 192.168.128.0/24. Your task is to break it down into six networks. The questions you have to answer are, how many hosts on each network, and what are the network IDs and broadcast IDs for each of the networks.

I will post the answer on Friday 12th December, 2008 at 3:30pm EDST.

Sorry for the lateness of the solution. I have been as sick as a dog for the last few days, hence no update.

Last time we talked breaking networks up based on host requirements. This time we’re working with the network portion of the IP address.

When we were working with hosts, we worked from the thirty-second bit. Or, from right to left. This time we’re working from left to right to break up an IP range.

Let’s look at the IP address in binary first. Because it’s a twenty-four bit subnet mask, this tells us that it’s a Class C network and that the first three octets are locked and can’t be changed. Therefore, we will be breaking up the fourth octet.
192.168.128.00000000
255.255.255.00000000

Similarly to subnetting for hosts, subnetting for networks is achieved by modifying the subnet mask. So rather than working from the last bit leftwards, let’s start with the first bit I am allowed to modify and work to the right. In our example, the twenty-fifth bit (the first bit of the fourth octet).

For each bit I change, I get 2^bits networks. Unlike hosts, I don’t lose the first IP address to the network ID or the last IP address to the broadcast. Therefore, if I use one bit I get two usable networks. If we extend this to cover our example, two bits gives me four networks (still not enough!). Can I get exactly six networks? Because I’m working with binary values (powers of two), it’s not possible to get exactly six networks. If I use three bits, I get eight networks. More than the required six.

First thing’s first: the subnet mask. The original Class C has been subnetted. Now I am using the first three bits of the fourth octet as the network portion. It did look like this: 255.255.255.00000000. Now, after subnetting, it looks like this: 255.255.255.11100000 or, (after converting back to decimal) 255.255.255.224.

My networks look something like this (the spaces represent the division between my new network ID and new host ID):
192.168.128.000 00000
192.168.128.001 00000
192.168.128.010 00000
192.168.128.011 00000
192.168.128.100 00000
192.168.128.101 00000
192.168.128.110 00000
192.168.128.111 00000

How many hosts on each network? Seeing that we are using three bits for the network, the remaining five bits are the host portion ie. (2^5)-2 hosts or 32-2 hosts. A total of 30 hosts per network. Going back to the last tutorial, the host portion can’t be all zeroes, nor can it be all ones.

Therefore, my network ranges look like this:
192.168.128.000 00000 192.168.128.000 11111
192.168.128.001 00000 192.168.128.001 11111
192.168.128.010 00000 192.168.128.010 11111
192.168.128.011 00000 192.168.128.011 11111
192.168.128.100 00000 192.168.128.100 11111
192.168.128.101 00000 192.168.128.101 11111
192.168.128.110 00000 192.168.128.110 11111
192.168.128.111 00000 192.168.128.111 11111

Any questions or problems, please email me contactme@invurted.com or my work email address.