Learn How To Hack! Learn Ethical Hacking & Download Free Hacking Tools

Is network subnetting driving you crazy ? Does binary math gives you headaches ? well..fear not, since the time i have delved into networks, i have gone through all the oohs and aaahs and the NAHs of the logical crescendo , and I present you the easiest way to do subnetting. Generally,3 questions are being addressed when subnetting an IP- No of subnets No of valid hosts and Host range/block size We will tackle them one by one. First you need to know about IP addresses and their classes. Class A includes 0-127 where 0 and 127 are reserved, the default subnet mask for this class is /8 . Class B includes 128-191 in their first octet, and the default subnet mask for this class is /16 Class C deals with 192-224 in their first octet and the default subnet mask for this class is /24 Also, also, understand a simple concept, subnet masks lying between 8- 15 are A class masks, from 16- 23 are B class and 24-32 are C class masks. Furthermore, a subnet mask can be expressed as this where (N)etwork value and (H)ost values- Class A : 255.0.0.0 = NNNNNNNN.HHHHHHHH.HHHHHHHH.HHHHHHHH Class B : 255.255.0.0 = NNNNNNNN.NNNNNNNN.HHHHHHHH.HHHHHHHH Class C : 255.255.0.0 = NNNNNNNN.NNNNNNNN.NNNNNNNN.HHHHHHHH Class mask value 8 16 24 32 Now once you see it, lets tackle some real life questions. Lets find the no of subnets and valid hosts for 192.168.10.10/18 See this ? its a C class IP address having a mask of B class (as the mask lies between 16-24) now, in order to find the number of subnets, use the following formulae - 2 ^ (What mask you have been provided default mask of the IP address given) putting the values here.. 2^(18-16) > 2^(2) > 4 subnets simple : ) Now for calculating the no.of hosts, use the below formulae - 2^(32- what mask you have been provided) 2 Putting values here.. 2^(32-18)-2 -> 2^(14)-2 > 16384-2 > 16382 hosts piece of cake .. Now to find the block size, see the provided mask lies between which next default mask value , which in this case is 24 (as 18 is greater than 16 and less than 24) . So .. Subtract the provided mask with the class mask value which is greater than it. 2^(Next class mask value provided mask) which on putting values will be 2^(24-18) > 2^6 > 64 So, the block size will be of 64 . So, the IP addresses will be divided into 4 subnets (which we already calculated above) above as - 192.168.0.0 - 192.168.63.255 192.168.64.0 192.168.127.255 192.168.128.0 - 192.168.191.255 192.168.192.0 192.168.255.255 And the best part, its applicable to all classes : ) Happy Subnetting : ] SOURCE: Related Posts : General Awareness, Network and LAN Hacking