Advanced Subnetting Example 1: Your ISP has assigned you a Class C network address of 198.47.212.0. You have 3 networks in your company with the largest containing 134 hosts. You need to figure out if you can subnet this. If it can be subnetted you need to calculate the new subnet mask and specify the available network numbers? You need to determine how many bits you have available to work with. Since this is a standard Class C network the last octet is available so you have 8 bits to work with. 1. You need to determine how many bits will be needed to address the largest number of hosts that may exist on a network. This can easily be done by converting the largest number of hosts to binary and counting the bits needed. Largest number of hosts addresses needed in decimal Largest number of hosts addresses converted to binary 134 10000110 Number of bits needed to address hosts 8 2. You need to determine how many bits will be needed to address the required number of networks. This can easily be done by converting the required number of networks to binary and counting the bits needed. Network addresses needed in decimal 3 Network addresses needed converted to binary 11 Number of bits needed to address networks 2 3. Add the total number of bits needed together and determine if a solution is possible. In this case 8 + 2 = 10 so 10 bits are needed. We only have 8 available so this problem cannot be solved.
Example 2: Your ISP has assigned you a Class C network address of 198.47.212.0. You have 3 networks in your company with the largest containing 51 hosts. You need to figure out if you can subnet this. If it can be subnetted you need to calculate the new subnet mask and specify the available network numbers? 1. You need to determine how many bits you have available to work with. Since this is a standard Class C network the last octet is available so you have 8 bits to work with. 2. You need to determine how many bits will be needed to address the largest number of hosts that may exist on a network. This can easily be done by converting the largest number of hosts to binary and counting the bits needed. Largest number of hosts addresses needed in decimal Largest number of hosts addresses converted to binary 51 110011 Number of bits needed to address hosts 6 3. You need to determine how many bits will be needed to address the required number of networks. This can easily be done by converting the required number of networks to binary and counting the bits needed. Network addresses needed in decimal 3 Network addresses needed converted to binary 11 Number of bits needed to address networks 2 4. Add the total number of bits needed together and determine if a solution is possible. In this case 6 + 2 = 8 so 8 bits are needed. We have 8 available so this problem can be solved.
5. You must then determine the new subnet mask. To do this we take the number of bits needed to address network from the left side of the available bits. Convert the value to decimal and you have the subnet mask. Subnet mask in binary 11000000 Subnet mask converted to decimal 192 Entire new subnet mask Original plus the changes you made to the host portion 255.255.255.192 6. Finally you need to determine the network addresses available. This can be done by counting up in binary of the bits you used for the network and converting the result to decimal. Network number without host bits (Both in decimal and in binary) Network number with host bits (in binary) Network number Full network number 00 b 0 d 0000 0000 0 198.47.212.0/26 01 b 1 d 0100 0000 64 198.47.212.64/26 10 b 2 d 1000 0000 128 198.47.212.128/26 11 b 3 d 1100 0000 192 198.47.212.192/26
Example 3: Your ISP has assigned you a Class B network address of 160.13.0.0. You have 11 networks in your company with the largest containing 98 hosts. You need to figure out if you can subnet this. If it can be subnetted you need to calculate the new subnet mask and specify the available network numbers? 1. You need to determine how many bits you have available to work with. Since this is a standard Class B network the last two octets are available so you have 16 bits to work with. 2. You need to determine how many bits will be needed to address the largest number of hosts that may exist on a network. This can easily be done by converting the largest number of hosts to binary and counting the bits needed. Largest number of hosts addresses needed in decimal Largest number of hosts addresses converted to binary 98 1100010 Number of bits needed to address hosts 7 3. You need to determine how many bits will be needed to address the required number of networks. This can easily be done by converting the required number of networks to binary and counting the bits needed. Network addresses needed in decimal 11 Network addresses needed converted to binary 1011 Number of bits needed to address networks 4 4. Add the total number of bits needed together and determine if a solution is possible. In this case 7 + 4 = 11 so 11 bits are needed. We have 16 available so this problem can be solved.
5. You must then determine the new subnet mask. To do this we take the number of bits needed to address network from the left side of the available bits. Convert the value to decimal and you have the subnet mask. Subnet mask in binary 11110000 Subnet mask converted to decimal 240 Entire new subnet mask Original plus the changes you made to the host portion 255.255.240.0
6. Finally you need to determine the network addresses available. This can be done by counting up in binary of the bits you used for the network and converting the result to decimal. Network number without host bits (Both in decimal and in binary) Network number with host bits (in binary) Network number Full network number 0000 b 0 d 0000 0000 0 160.13.0.0/20 0001 b 1 d 0001 0000 16 160.13.16.0/20 0010 b 2 d 0010 0000 32 160.13.32.0/20 0011 b 3 d 0011 0000 48 160.13.48.0/20 0100 b 4 d 0100 0000 64 160.13.64.0/20 0101 b 5 d 0101 0000 80 160.13.80.0/20 0110 b 6 d 0110 0000 96 160.13.96.0/20 0111 b 7 d 0111 0000 112 160.13.112.0/20 1000 b 8 d 1000 0000 128 160.13.128.0/20 1001 b 9 d 1001 0000 144 160.13.144.0/20 1010 b 10 d 1010 0000 160 160.13.160.0/20 1011 b 11 d 1011 0000 176 160.13.176.0/20 1100 b 12 d 1100 0000 192 160.13.192.0/20 1101 b 13 d 1101 0000 208 160.13.208.0/20 1110 b 14 d 1110 0000 224 160.13.224.0/20 1111 b 15 d 1111 0000 240 160.13.240.0/20