## Subnetting a Class C Network Using Binary

You have a Class C 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.

Step 1 Determine how many H bits you need to borrow to create nine valid subnets. 2n- 2 > 9

 0001HHHH Cannot use subnet 0000 because it is invalid. Therefore, you must start with the bit pattern of 0001 00010000 All 0s in host portion = subnetwork number 00010001 First valid host number 00011110 Last valid host number 00011111 All 1s in host portion = broadcast number Step 3 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 4 Determine the second valid subnet in binary. 0010HHHH 0010 = 2 in binary = second valid subnet 00100000 All 0s in host portion = subnetwork number 00100001 First valid host number 00101110 Last valid host number 00101111 All 1s in host portion = broadcast number
 00100000 = 32 Subnetwork number 00100001 = 33 First valid host number 00101110=46 Last valid host number 00101111=47 All 1s in host portion = broadcast number
 Valid Subnet Network Number Range of Valid Hosts Broadcast Number 1 16 17-30 31 2 32 33-46 47 3 48 49-62 63

Notice a pattern? Counting by 16. Step 7 Verify the pattern in binary. (The third valid subnet in binary is used here.)

 0011HHHH Third valid subnet 00110000=48 Subnetwork number 00110001=49 First valid host number 00111110=62 Last valid host number 00111111=63 Broadcast number
 Subnet Network Address (0000) Range of Valid Hosts (0001-1110) Broadcast Address (1111) 0(0000) invalid 192.168.100.0 192.168.100.1192.168.100.14 192.168.100.15 1(0001) 192.168.100.16 192.168.100.17192.168.100.30 192.168.100.31 2(0010) 192.168.100.32 192.168.100.33192.168.100.46 192.168.100.47 3(0011) 192.168.100.48 192.168.100.49192.168.100.62 192.168.100.63 4(0100) 192.168.100.64 192.168.100.65192.168.100.78 192.168.100.79 5(0101) 192.168.100.80 192.168.100.81192.168.100.94 192.168.100.95 6(0110) 192.168.100.96 192.168.100.97192.168.100.110 192.168.100.111 7(0111) 192.168.100.112 192.168.100.113192.168.100.126 192.168.100. 127 8(1000) 192.168.100.128 192.168.100.129192.168.100.142 192.168.100. 143 9(1001) 192.168.100.144 192.168.100.145192.168.100.158 192.168.100.159 10 (1010) 192.168.100.160 192.168.100.161192.168.100.174 192.168.100.175 11 (1011) 192.168.100.176 192.168.100.177192.168.100.190 192.168.100.191 12 (1100) 192.168.100.192 192.168.100.193192.168.100.206 192.168.100. 207 13 (1101) 192.168.100.208 192.168.100.209192.168.100.222 192.168.100. 223 14 (1110) 192.168.100.224 192.168.100.225192.168.100.238 192.168.100. 239
 15 (1111) invalid 192.168.100.240 192.168.100.241192.168.100.254 192.168.100. 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 subnetsâ€”the rest are for future growth.

Use any nine subnetsâ€”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 or a Class A network with exactly the same steps as for a Class C network; the only difference is that you start with more H bits.

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### Responses

• MICHAEL
How to do subnetting for class c?
3 months ago