Holiday Special: Variable Length Subnetting Mask (VLSM)

Happy Holidays Mate.! I’m moving to routing and realized to perform first VLSM before I deal first with the static routing.

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We’ve got this network topology.Thanks thee to my dearest instructor. We will apply VLSM on this topology. We got 172.16.0.0/20 as the network address to subnet with the following requirements

SW1 requires 4000 hosts

SW0 requires 2000 hosts

 

SW2 requires 1000 hosts

SW0 requires 500 hosts

SW3 requires 200 hosts

SW5 requires 100 hosts

The primary rule on VLSM is to start first with the one that requires the maximum number of hosts. This is to ensure that we will have enough number of hosts for a network.

Remember the formula:

number of hosts= 2^( 32-NSM)-2

4000hosts is almost equal to 4096 = 2^12= 2^( 32- NSM) ; 12=32-NSM

NSM=20

So we got 172.16.0.0/20 as the 1st subnet, to make sure that we are doing it correctly, it is required for us to fill the following:

Subnet: 172.16.0.0/20

CIDR increment:16

Network Range: 172.16.0.1- 172.16.15.254

5656565

Use the finger technique for us to know easily the increment for CIDR 20

We can also fill:

Network Address: 172.16.0.1

Broadcast Address: 172.16.15.255

So we are done with the first requirement, let us check 2000 hosts =2048= 2^11=2^(32-NSM); 11= 32-NSM, NSM=21

increment using finger technique above, increment is 8, the last unused bit starts from 172.16.16.0 since broadcast address is 172.16.15.255

Subnet: 172.16.16.0/21

CIDR increment:8

Network Range: 172.16.16.1-172.16.23.254

Network Address: 172.16.16.1

Broadcast Address: 172.16.23.255

I hope you are getting it right, let us try 1000 hosts, since the last unused bit is 172.16.24.0 then we’ll proceed with the mathematical way, 1000hosts = 1024 hosts = 2^10 = 2( 32-NSM); 10=32-NSM, NSM=22

Subnet: 172.16.24.0/22

CIDR increment:4

Network Range: 172.16.24.1-172.16.27.254

Network Address: 172.16.24.1

Broadcast Address: 172.16.27.255

Let’s try 500 hosts, 500 host is almost equal to 512 hosts = 2^9= 2^( 32-NSM); NSM = 23,The last unused bit is 172.16.28.0 then:

Subnet: 172.16.28.0/23

CIDR increment:2

Network Range: 172.16.28.1-172.16.29.254

Network Address: 172.16.28.1

Broadcast Address: 172.16.29.255

Let’s try 200 hosts, 200 hosts is almost equal to 256 hosts = 2^8= 2^( 32-NSM); NSM = 24,The last unused bit is 172.16.30.0 then:

Subnet: 172.16.30.0/24

CIDR increment:1

Network Range: 172.16.30.1-172.16.30.254

Network Address: 172.16.30.1

Broadcast Address: 172.16.30.255

Lastly, let’s try 100 hosts, 100 hosts is almost equal to 128 hosts = 2^7= 2^( 32-NSM); NSM = 27,The last unused bit is 172.16.31.0

Subnet: 172.16.31.0/25

CIDR increment:128

Network Range: 172.16.31.1-172.16.31.127

Network Address: 172.16.31.1

Broadcast Address: 172.16.31.128

Have you observed something? the private ip address ranges from 172.16.0.0- 172.16.31.255, it means to say tha we still got 172.16.31.129-.172.16.31.255 as a reserved hosts.Alright, so let me label the gateway and ip address of some devices on the network and will start with  routing protocols.

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