You have an address of 192.168.100.0 /24
. You need nine subnets. What is the IP plan of network numbers, broadcast numbers, and valid host numbers? What is the subnet mask needed for this plan?
You cannot use N
bits, only H
bits. Therefore, ignore 192.168.100. These numbers cannot change. You only work with host bits. You need to borrow some host bits and turn them into network bits (or in this case, subnetwork bits; I use the variable N
to refer to both network and subnetwork bits).
Step 1. Determine how many H bits you need to borrow to create nine valid subnets.
2N ≥ 9
N = 4, so you need to borrow 4 H bits and turn them into N bits.
Start with 8 H bits | HHHHHHHH |
Borrow 4 bits | NNNNHHHH |
Step 2. Determine the first subnet in binary.
0000HHHH | |
00000000 | All 0s in host portion = subnetwork number |
00000001 | First valid host number |
00000010 | Second valid host number |
00000011 | Third valid host number |
… | |
00001110 | Last valid host number |
00001111 | All 1s in host portion = broadcast number |
Step 3. Convert binary to decimal.
00000000 = 0 | Subnetwork number |
00000001 = 1 | First valid host number |
00000010 = 2 | Second valid host number |
00000011 = 3 | Third valid host number |
. … | |
00001110 = 14 | Last valid host number |
00001111 = 15 | All 1s in host portion = broadcast number |
Step 4. Determine the second subnet in binary.
0001HHHH | |
00010000 | All 0s in host portion = subnetwork number |
00010001 | First valid host number |
00010010 | Second valid host number |
… | |
00011110 | Last valid host number |
00011111 | All 1s in host portion = broadcast number |
Step 5. Convert binary to decimal.
00010000 = 16 | Subnetwork number |
00010001 = 17 | First valid host number |
… | |
00011110 = 30 | Last valid host number |
00011111 = 31 | All 1s in host portion = broadcast number |
Step 6. Create an IP plan table.
Subnet | Network Number | Range of Valid Hosts | Broadcast Number |
1 | 0 | 1–14 | 15 |
2 | 16 | 17–30 | 31 |
3 | 32 | 33–46 | 47 |
Notice a pattern? Counting by 16.
Step 7. Verify the pattern in binary. (The third subnet in binary is used here.)
0010HHHH | Third subnet |
00100000 = 32 | Subnetwork number |
00100001 = 33 | First valid host number |
00100010 = 34 | Second valid host number |
… | |
00101110 = 46 | Last valid host number |
00101111 = 47 | Broadcast number |
Step 8. Finish the IP plan table.
Subnet | Network Address (0000) | Range of Valid Hosts (0001–1110) | Broadcast Address (1111) |
1 (0000) | 192.168.100.0 | 192.168.100.1–192.168.100.14 | 192.168.100.15 |
2 (0001) | 192.168.100.16 | 192.168.100.17–192.168.100.30 | 192.168.100.31 |
3 (0010) | 192.168.100.32 | 192.168.100.33–192.168.100.46 | 192.168.100.47 |
4 (0011) | 192.168.100.48 | 192.168.100.49–192.168.100.62 | 192.168.100.63 |
5 (0100) | 192.168.100.64 | 192.168.100.65–192.168.100.78 | 192.168.100.79 |
6 (0101) | 192.168.100.80 | 192.168.100.81–192.168.100.94 | 192.168.100.95 |
7 (0110) | 192.168.100.96 | 192.168.100.97–192.168.100.110 | 192.168.100.111 |
8 (0111) | 192.168.100.112 | 192.168.100.113–192.168.100.126 | 192.168.100.127 |
9 (1000) | 192.168.100.128 | 192.168.100.129–192.168.100.142 | 192.168.100.143 |
10 (1001) | 192.168.100.144 | 192.168.100.145–192.168.100.158 | 192.168.100.159 |
11 (1010) | 192.168.100.160 | 192.168.100.161–192.168.100.174 | 192.168.100.175 |
12 (1011) | 192.168.100.176 | 192.168.100.177–192.168.100.190 | 192.168.100.191 |
13 (1100) | 192.168.100.192 | 192.168.100.193–192.168.100.206 | 192.168.100.207 |
14 (1101) | 192.168.100.208 | 192.168.100.209–192.168.100.222 | 192.168.100.223 |
15 (1110) | 192.168.100.224 | 192.168.100.225–192.168.100.238 | 192.168.100.239 |
16 (1111) | 192.168.100.240 | 192.168.100.241–192.168.100.254 | 192.168.100.255 |
Quick Check | Always an even number | First valid host is always an odd # Last valid host is always an even # | Always an odd number |
Use any nine subnets—the rest are for future growth.
Step 9. Calculate the subnet mask. The default subnet mask for a Class C network is as follows:
Decimal | Binary |
255.255.255.0 | 11111111.11111111.11111111.00000000 |
1 = Network or subnetwork bit
0 = Host bit
You borrowed 4 bits; therefore, the new subnet mask is the following:
11111111.11111111.11111111.11110000 | 255.255.255.240 |
Note
You subnet a Class B network or a Class A network using exactly the same steps as for a Class C network; the only difference is that you start with more H bits.