Practice Example #1C: 255.255.255.192 (/26)
Let’s use the Class C subnet mask from the preceding example, 255.255.255.192, to see how much simpler this method is than writing out the binary numbers. We’re going to subnet the network address 192.168.10.0 and subnet mask 255.255.255.192.
192.168.10.0 = Network address
255.255.255.192 = Subnet mask
Now, let’s answer the big five:
_ How many subnets? Since 192 is 2 bits on (11000000), the answer would be 22.
_ How many hosts per subnet? We have 6 host bits off (11000000), so the equation would be 26 – 2 = 62 hosts.
_ What are the valid subnets? 256 – 192 = 64. Remember, we start at zero and count in our block size, so our subnets are 0, 64, 128, and 192.
_ What’s the broadcast address for each subnet? The number right before the value of the next subnet is all host bits turned on and equals the broadcast address.
_ What are the valid hosts? These are the numbers between the subnet and broadcast address. The easiest way to find the hosts is to write out the subnet address and the broadcast address. This way, the valid hosts are obvious. The following table shows the 0, 64, 128, and 192 subnets, the valid host ranges of each, and the broadcast address of each subnet:
The subnets (do this first) 0 64 128 192
Our first host (perform host addressing last) 1 65 129 193
Our last host 62 126 190 254
The broadcast address (do this second) 63 127 191 255
Practice Example #1B: 255.255.192.0 (/18)
172.16.0.0 = Network address
255.255.192.0 = Subnet mask
_ Subnets? 22 = 4.
_ Hosts? 214 – 2 = 16,382 (6 bits in the third octet, and 8 in the fourth).
_ Valid subnets? 256 – 192 = 64. 0, 64, 128, 192. Remember the subnetting is performed in the third octet, so the subnet numbers are really 0.0, 64.0, 128.0, and 192.0, as shown in the next table.
_ Broadcast address for each subnet?
_ Valid hosts?
The following table shows the four subnets available, the valid host range, and the broadcast address of each:
Subnet 0.0 64.0 128.0 192.0
First host 0.1 64.1 128.1 192.1
Last host 63.254 127.254 191.254 255.254
Broadcast63.255 127.255 191.255 255.255
Practice Example #1A: 255.255.0.0 (/16)
Class A addresses use a default mask of 255.0.0.0, which leaves 22 bits for subnetting since you must leave 2 bits for host addressing. The 255.255.0.0 mask with a Class A address is using 8 subnet bits.
_ Subnets? 28 = 256.
_ Hosts? 216 –2 = 65,534.
_ Valid subnets? 256 – 255 = 1. 0, 1, 2, 3, etc. (all in the second octet). The subnets would be 10.0.0.0, 10.1.0.0, 10.2.0.0, 10.3.0.0, etc., up to 10.255.0.0.
_ Broadcast address for each subnet?
_ Valid hosts?
The following table shows the first two and last two subnets, valid host range, and broadcast addresses for the private Class A 10.0.0.0 network:
Subnet 10.0.0.0 10.1.0.0 … 10.254.0.0 10.255.0.0
First host 10.0.0.1 10.1.0.1 … 10.254.0.1 10.255.0.1
Last host 10.0.255.254 10.1.255.254 …10.254.255.254 10.255.255.254
Broadcast 10.0.255.255 10.1.255.255 …10.254.255.255 10.255.255.255
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