IP Subnetting - From Zero to Guru by Paul Browning

Author:Paul Browning [Browning, Paul]
Language: eng
Format: azw3, epub
Published: 2018-11-22T16:00:00+00:00

These ticks represent how many bits we have stolen to create subnetworks. In this example it’s three. We don’t need to use the bottom section of the chart because we aren’t trying to figure out how many subnets and hosts per subnet there are.

We are subnetting in the fourth octet, so we just count up in increments of 32 until we get to the one with the host number 203.

I usually prefer to jump the count in a multiple of 32 (160, for example) to save time, but you may prefer to start with 0, 32, 64, etc., for now. Just remember that in the exam, time is of the essence. Skipping up in higher values than the subnet increment I refer to as a ‘jump strategy’. It won’t matter too much if your increment is 128, 64, or even 32, but imagine trying to find host 209 when you are counting up in increments of 2 or 4!

Let’s count up in increments of 32. Remember that we are looking for host number 203. You would typically count up to the next subnet after the one you think the host number is in just to make sure. Again, this is especially important if you are counting up in small increments, where it’s easy to make mistakes.

Another reason to do this is you may be asked to work out the broadcast address for your subnet. The way to do this is to find the next subnet value and then subtract 1. For subnet 1 below, for example, the broadcast address is 31, which is the next subnet value minus 1.

Subnet 1: 192.168.100.0, hosts 1-30 (broadcast = 31)

Subnet 2: 192.168.100.32, hosts 33-62 (broadcast = 63)

Subnet 3: 192.168.100.64, hosts 65-94 (broadcast = 95)

Subnet 4: 192.168.100.96, hosts 97-126 (broadcast = 127)

Subnet 5: 192.168.100.128, hosts 129-158 (broadcast = 159)

Subnet 6: 192.168.100.160, hosts 161-190 (broadcast = 191)

Subnet 7: 192.168.100.192, hosts 193-222 (broadcast = 223) 203 is in here.

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