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# Subnet calculation Tutorial

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Subnet Calculation from a given IP range, using the classless Subnet mask. Calculating number of hosts in a subnet and number of subnets possible to create in a given IP range.

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### Subnet calculation Tutorial

1. 1. SUBNET CALCULATION TUTORIAL -BY RITU RANJAN SHRIVASTWA
2. 2. UNDERSTANDING SUBNETTING • First let us take a short view at what are the subnets of classes A, B, and C • Class A subnet: 255.0.0.0 /8 which means 11111111.00000000.00000000.00000000 • Class B subnet: 255.255.0.0 /16 which means 11111111.11111111.00000000.00000000 • Class C subnet: 255.255.255.0 /24 which means 11111111.11111111.11111111.00000000 Whenever an IP is given (generally), it belongs to either of the three i.e., Class A, B or C But we need to attach multiple networks within the same IP range. Thus we need to create subnets under the given IP.
3. 3. CLASSLESS AND CLASSFUL • In the previous slide you have seen that there are three different subnet masks for Class A, B and C. These masks are called Classful because they belong to either class. • A classless subnet mask can be created using subnet of any class but a classless subnet does not belong to any ‘Class’ (hence classless). • A classless subnet exists only inside a private network e.g., inside an office network or a college network, etc.
4. 4. CLASSLESS SUBNET MASK • As mentioned previously, a classless subnet mask is used to create a series of ‘subnets’ using one network Subnet 1 INTERNET CLASS C IP RANGE 192.168.0.5/24 (/24 = 255.255.255.0) Subnet 2 Subnet N N subnets
5. 5. EXAMPLE PROBLEM WITH SOLUTION • Now with an example problem I will Illustrate the subnetting process. Problem Given IP range: 172.168.0.2 /16 There are 14 departments in the office. Create a network for each department and determine how many PCs can be attached in each department network
6. 6. EXAMPLE PROBLEM WITH SOLUTION From the given problem, there need to be 14 subnets. The subnet mask is 255.255.0.0 (8+8=16 /16) Which means the last two octets are usable. (1 octet = 8bits => usable bits = 32-16 = 16 i.e., 2 octets) Thus we need to create 14 subnets. To create subnets the formula for number of bits used is = 2n-2 (where n is the number of bits required) Here the scenario is, => (2^n) – 2 = 14 => n = log2(14+2) = 4 bits
7. 7. EXAMPLE PROBLEM WITH SOLUTION Thus, we need 4 bits to create 14 subnets. These 4 bits will be the first 4 bits of the 1st usable Octet. To create the Classless subnet mask, now we need to convert the first 4 bits of the 3rd Octet into 1’s. Hence the new Subnet (classless) will look as follows: Network bits Host bits 11111111.11111111.11110000.00000000 Which makes it 255.255.240.0 from 255.255.0.0 (/16) i.e., prefix = /(8+8+4) = /20 From now on, subnet ID will be denoted with the 4 red bits only Please note that the first and the last network are unusable as per norms i.e., subnet 0000 and subnet 1111 Thus we are left with 0001 to 1110 usable networks i.e., 14 subnets
8. 8. EXAMPLE PROBLEM WITH SOLUTION Now we need to determine how many hosts can be attached to each subnet. This is very easy and the formula remains the same. From the last two octets we see that there are 4 network bits and (4+8) host bits. Host bits = 12 No of hosts attachable = 2^12 – 2 = 4096 – 2 No. of hosts = 4094 The first and the last addresses are Network address and Broadcast address respectively. Thus, first subnet will have the following: subnet id = 0001 subnet mask = 255.255.240.0 network address = 172.168.16.0 broadcast address = 172.168.31.255 first host = 172.168.16.1 last host = 172.168.31.254 [(00010000)b = (16)d] [(00011111)b = (31)d]
9. 9. EXAMPLE PROBLEM WITH SOLUTION subnet id = 0010 subnet mask network address broadcast address first host last host = 255.255.240.0 = 172.168.32.0 = 172.168.47.255 = 172.168.32.1 = 172.168.47.254 subnet id = 0011 subnet mask network address broadcast address first host last host = 255.255.240.0 = 172.168.48.0 = 172.168.63.255 = 172.168.48.1 = 172.168.63.254 [(00010000)b = (16)d] [(00011111)b = (31)d] [(00010000)b = (16)d] [(00011111)b = (31)d] and so on…
10. 10. THANK YOU
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