In this, our final look at subnetting for now, we’ll examine more “find the subnet” problems, specifically those in which the subnet/host boundary lies within the second or third octets. Here is the powers of two chart:
Suppose that we’re given an address of 10.20.30.40/20, and told to find the subnet. Since 20 lies between 16 and 24, we can see from the mask that we’re operating in the third octet. Further, since the mask is a “/20”, there are four subnet bits in that octet, as well as four host bits. Per the chart, two to the fourth (the number of host bits) is 16, so the subnets increment by 16 in the third octet. Thus, the subnets are:
- (Twelve more subnets incrementing by 16 in the third octet go here)
Remember that we’re interested in the third octet. Since 30 lies between 16 and 32, the host is on the 10.20.16.0/20 subnet, the range is 10.20.16.1 – 10.20.31.254, and the broadcast address is 10.20.31.255 for that subnet.
Let’s do another. This time we’re given the host address 172.31.100.200, and a subnet mask of 255.255.224.0 (a “/19”). We can see from the mask that we’re interested in the third octet. Further, since 256 minus 224 is 32, the subnets increment by 32 in that octet. Thus, the subnets are 172.16.0.0/19 (and counting by 32 in the third octet), 32, 64, 96, 128, etc. Since 100 lies between 96 and 128, the host is on the 172.16.96.0/19 subnet, the range is 172.16.96.1 – 172.16.127.254, and the broadcast address is 172.16.96.255 for that subnet.
Ready for the next one? Here we go … given the host address 10.20.30.40 and a subnet mask of 255.192.0.0, find the usual items. We can see from the mask (which could also be written as a “/10”) that we are interested in the second octet. Furthermore, since 256 minus 192 is 64, the subnets increment by 64 in the second octet. Thus, the subnets are 10.0.0.0/10, 10.64.0.0/20, 10.128.0.0/10 and 10.192.0.0/10 (since there are only two subnet bits, there are only four subnets). It’s the second octet that we’re interested in, and since 20 lies between 0 and 64, the host is on the 10.0.0.0/10 subnet, the range is 10.0.0.1 – 10.63.255.254, and the broadcast is 10.63.255.255 for that subnet.
Okay, one more: Given a host with an address and mask of 10.200.100.200/14, determine the subnet, range and broadcast. First, since 14 lies between 8 and 16, we’re interested in the second octet. Next, since we’re dealing with a Class “A” (eight network bits), a “/14” mask means that there are six subnet bits and two host bits in the second octet. Because two to the second power is four, the subnets increment by four in the second octet, thus: 10.0.0.0/14, 10.4.0.0/14, 10.8.0.0/14, etc.
While we could count all the way to 200 in the second octet, we can just use some math to shortcut it. Since 200 is an even multiple of four (four times fifty is 200), 200 is a subnet (and 204 is the next subnet). Thus, the host is on the 10.200.0.0/14 subnet, the range of legal host addresses on that subnet is 10.200.0.1 – 10.203.255.254, and the broadcast is 10.203.255.255 for that subnet.
Note that when using a “/14” mask with a Class “A” network, there are six subnet bits, allowing for up to 64 subnets, and eighteen host bits, allowing for 262,142 hosts per subnet (two to the eighteenth minus two). That’s way off the powers of two chart, and not likely to be used in the real world as a subnet mask.
One last thing … since the default mask for a Class “A” is “/8”, you’ll never see a subnet mask of less than “/8”. Thus, you don’t have to worry about doing calculations in the first octet (although it would work exactly the same, with the subnet increment based on the number of host bits in that octet).
Well, that’s it for subnetting for now. Next time, something completely different!
Author: Al Friebe