> For the complete documentation index, see [llms.txt](https://neerajcysec.gitbook.io/notes/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://neerajcysec.gitbook.io/notes/networking/ccna-notes/subnetting-2.md). # Subnetting (2) {% embed url="" %} ## Review * [Subnetting (1)](/notes/networking/ccna-notes/subnetting-1.md) ## Subnet exercises ### Number of subnets * No. of subnets = 2^x, where x = no. of borrowed bits. * If you want 5 subnets, you'll need to use /27 mask. * /27 mask has 3 borrowed bits which gives 8 subnets (2^3). ### Identify each subnet * Divide 192.168.255.0/24 into 5 subnets of equal size. * Find the number of subnets from the previous step [#number-of-subnets](#number-of-subnets "mention") * To find the subnets, consider the last octet. * You have 3 borrowed bits and 5 host bits. * The 3rd borrowed bit from the left corresponds to 32nd place in the octet. * Now, add 32 to the first address (first subnet) to get the next subnet and so on. * 192.168.255.0/27, ..255.32/27, ..255.64/27, ..255.96/27, ..255.128/27 * However there will be 3 other subnets as /27 prefix length allows upto 8 subnets. ### Identify the subnet What subnet does host 192.168.5.57/27 belong to? * Convert it into dotted decimal. * Its last octet is 00111001. * The first 3 are the borrowed bits since /27 mask. * To find the network or the subnet the host is in, **change all the host bits to 0**. * **Convert back to dotted decimal.** It leaves us with the last octet of 00100000 which equals 32. * This means the host belongs to the subnet 192.168.5.32/27. * Another example : 192.168.29.219/29 belongs to 192.168.29.216/29 subnet. ## Subnets of Class C | Prefix | Subnets | Hosts | | ------ | ------- | ----- | | /25 | 2 | 126 | | /26 | 4 | 62 | | /27 | 8 | 30 | | /28 | 16 | 14 | | /29 | 32 | 6 | | /30 | 64 | 2 | | /31 | 128 | 0(2) | | /32 | 256 | 0(1) | * Each bit you borrow doubles the number of subnets. * [Subnetting (1)#31 mask](/notes/networking/ccna-notes/subnetting-1.md#31-mask) * [Subnetting (1)#32 mask](/notes/networking/ccna-notes/subnetting-1.md#32-mask) ## Subnetting Class B * Class B has more host bits than Class C, which means many more possible subnets can be created with a Class B network compared to Class C. > The process of subnetting Class A, B, C networks is EXACTLY the same. ### Number of subnets **Example 1** 172.16.0.0/16 network. Create 80 subnets for your company's various LANs. What prefix length to use? * Use 2^x formula. * If you don't borrow any bit, you can't make a subnet. It's just one large /16 network. * If you borrow 1 bit, you can make 2 subnets, with a prefix of /17 and a mask of 255.255.128.0 * Borrowing 7 bits gives us 128 subnets, with a mask of /23 and a mask of 255.255.254.0 * To find the subnets, add the value of the last borrowed bit, which is 2 (because the 7th bit is 2nd place). * Some subnets are 192.168.0.0/23, 192.168.2.0/27, 192.168.4.0/27, 192.168.6.0/27 and so on. **Example 2** 172.22.0.0/16 network. Divide the network into 500 subnets. Identify the prefix length to be used. * 9 borrowed bits creates 512 subnets, so the prefix length would be /25, with a mask of 255.255.255.128 * Every possible subnet in the 3rd octet (256) has a combination on 0 and 1 from the 1st bit of 4th octet, which makes 512 subnets. **Example 3** 172.18.0.0/16 network. Company requires 250 subnets with same number of hosts per subnet. Find the prefix length to be used. * Borrowing 8 bits gives 256 subnets. * This also gives 8 host bits, which equals 254 hosts per subnet. ### Identify the subnet What subnet does host 172.25.217.192/21 belong to? * Write the address in binary. * Identify the borrowed bits. * Change the hosts to 0. * Convert back to dotted decimal. * You get 172.25.216.0/21
*number of hosts in /21 subnet is 2046 and not 2044* Continued : [Subnetting (3)](/notes/networking/ccna-notes/subnetting-3.md) ## Quiz 1. 172.30.0.0/16 network. 100 subnets with atleast 500 hosts per subnet. Find the prefix length. * Borrowing 7 bits gives 128 subnets and leaves us with 512 hosts (9 bits, 510 usable addresses). * So /23 prefix is used which has a mask of 255.255.254.0 2. 172.21.111.201/20. Identify the subnet * Write in binary (last 2 octets : 01101111.11001001) * Change host bits to 0 (last 2 octets : **0110**0000.00000000) * Convert back to dotted decimal - 172.21.96.0/20 3. 192.168.91.78/26. Find the broadcast address of the network of the address * Same prev steps * Change all host bits to 1 instead. * 192.168.91.127 4. 172.16.0.0/16 network. Divide into 4 subnets of equal size. Find network and broadcast address of the second subnet. * For 4 subnets, borrow 2 bits (2^2=4). So /18 prefix. * Network : 172.16.64.0/18 * Broadcast : 172.16.127.255/18 5. 172.30.0.0/16 network. Divide into subnets of 1000 hosts each. Find the number of subnets possible. * For 1000 hosts, 10 host bits are needed. * So 6 bits can be borrowed. * So 64 subnets (2^6).