IPv4 SUBNETTING

IPv4 SUBNETTING 

Given network Address 192.168.1.0/24, you have been asked to use the given network address to do the subneting and configure the network.

Step 1: Always you need to start with the Highest IP requirement, so the highest IP requirement is 30.

Calculating network address of each network from the given classful lP address is called VLSM.

To find the classless IP network we have to first find the required host bits.

2ⁿ -2 ³REQUIRED IP      ( n is number of host bits)

2ⁿ -2 ³30

2ⁿ ³30+2

2ⁿ ³32

2ⁿ  ³2⁵

n=5

so required number of host bits = 5

that’s bring the subnet mask bits to à 32-5=27

so the network address of first network is 192.168.1.0/27

Since the network address is 192.168.1.0/27 , we should be able to find the first and last IP address in this network.

To find the first IP address, we need to convert the IP address to binary format and change all the host bits to zero (0)

192.168.1.0

11000000.10101000.00000001.00000000 (Binary format) in this first 27 bits are assigned to network and the last 5 bits assigned to Host.

11000000.10101000.00000001.000|00000, we could see all the host bits are zero so this is the first IP address in this network, but this is also the network address. There for we cannot use this address to any end device, this address is used to identify the Network

Network Address = 192.168.1.0/27

Now we need to find the last IP address,

To find the last IP address change all the host bits to one (1)

11000000.10101000.00000001.000|00000 change the host bits to one

11000000.10101000.00000001.000|11111 when you convert this to Decimal we will get the Last IP address which is called broadcast address.

11000000 = 192

10101000= 168

00000001= 1

000|11111 = 31

So, the Broadcast IP address is 192.168.1.31/27

So we can conclude this as follows, for 30 IP address requirement, we need /27 network

Which has a range from 192.168.1.0/27 --- 192.168.1.31/27

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Now we need to calculate for the second biggest network

2ⁿ -2 ³REQUIRED IP      ( n is number of host bits)

2ⁿ -2 ³15

2ⁿ ³15+2

2ⁿ ³17 since its greater than or equal we can change this to closest 2 to the power number which is 32 (1,2,4,8,16,32,64,128,256)

2ⁿ  ³32

2ⁿ  ³2⁵

n=5

so, required number of host bits = 5

that’s bring the subnet mask bits to à 32-5=27

so the network address of the second network is

we have used up to 31 , that’s mean we can start the next network from 32.

Network address is : 192.168.1.32/27

Now we need to find the last IP on this network, to do the calculation we will again convert this Network address to Binary

11000000.10101000.00000001.00100000

Last IP address has all the hosts bits set to one

11000000.10101000.00000001.001| 11111

which is 192.168.1.63 /27 in binary

so the range of the second network is 192.168.1.32/27 --- 192.168.1.63/27

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Now we can calculate the third network which has a requirement of 8 IP address

2ⁿ -2 ³REQUIRED IP      ( n is number of host bits)

2ⁿ -2 ³8

2ⁿ ³8+2

2ⁿ ³10  since its greater than or equal we can change this to closest 2 to the power number which is 16 (1,2,4,8,16,32,64,128,256)

2ⁿ  ³16

2ⁿ  ³2⁴

n=4

so, required number of host bits = 4

that’s bring the subnet mask bits to à 32-4=28

so, the network address of the second network is

we have used up to 63, that’s mean we can start the next network from 64.

Network address is: 192.168.1.64/28

Now we need to find the last IP on this network, to do the calculation we will again convert this Network address to Binary

11000000.10101000.00000001.0100|0000

Last IP address has all the hosts bits set to one

11000000.10101000.00000001.0100|1111

which is 192.168.1.79 /28 in binary

so the range of the second network is 192.168.1.64/28 --- 192.168.1.79/28

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similar way we need to calculate to N12,N23,N13, but in this network we need only 2 IP address.