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IP Subnetting Step #4: Determining Subnet Identifiers and Subnet Addresses (Page 2 of 5) Class C Subnet ID and Address Determination Example Recall our Class C network, 211.77.20.0. The network address in binary is: 11010011 01001101 00010100 00000000 We are subnetting using 3 bits for the subnet ID, leaving 5 bits for the host ID. Now let's see the network address with the subnet bits in bold: 11010011 01001101 00010100 00000000 These are the bits we substitute with the subnet ID for each subnet. Notice that since the first three octets contain network ID bits, and the network ID is the same for every subnet, they never change. We don't even really need to look at them in binary form, though for clarity we will do so. Here's how we determine the subnet IDs and addresses, again, starting with 0 (see Figure 78): 0. Subnet #0 has a subnet ID of 0,
or 000 in binary. To find the address, we start with the network address
in binary, and substitute 000 for the subnet ID bits. Well
gee, those bits are already all zero! What this means is that the address
for subnet #0 is the same as the address for the network as a whole:
211.77.20.0.
11010011 01001101 00010100 00100000 Converting to decimal, we get 211.77.20.32.
11010011 01001101 00010100 01000000 Which is 211.77.20.64 in binary.
Starting to see a pattern here? Yep, the address of any subnet can be found by adding 32 to the last octet of the previous subnet. This pattern occurs for all subnetting choices; the increment depends on how many bits we are using for the subnet ID. Here, the increment is 32, which is 25; 5 is the number of host ID bits left after we took 3 subnet ID bits.
We only needed seven subnets in our example, #0 through #6. Subnet #7 would be a spare. Notice that the last subnet has the same last octet as the subnet mask for our network? That's because we substituted 111 for the subnet ID bits, just as we did when we calculated the subnet mask.
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