VLSM very easy to learn

variable length subnet mask

Variable Length Subnet Mask, shortened as VLSM, is a means to identify a different subnet mask for the similar network number on special subnets  With VLSM, a network supervisor can use a long mask on networks with few hosts and a short mask on subnets with lots of hosts. To use VLSM, the routing protocol must carry it.

let see example


we have network 192.168.10.0/24

we required host like...these

 

what we have required host 

25 host

12 host

10 host

6 host

2 host

Now what we do to save our IP address after assign this all network host ?

let's see...

what we have network...

192.168.10.0/24

what we required host first  25 host 

let see...

192.168.10.0/24

look my logic

2ⁿ − 2

( N = neutral number )

 2¹ − 2 = 0

 2² − 2 = 2 continue ,..

 2⁵ − 2 = 30

stop..

(note - value should be equal or grater but not less if  we keep  2⁴ - 2 = 14 not valid for 25 host  )

Now we have  N = 5 

now keep 5 bit of Host and  all is Network bit let's  see....

192.168.10.0

11111111.11111111.11111111.11100000

Host bit = 5

Network bit = 27  

192.168.10.0/27

^{255.255.255.224} 

^{let keep Host  formula} 

^{Host = 2h − 2}

Host = 2⁵ − 2 = 30

( now we have valid host is 30  we can assign 25 host  5 Host will west no problem  )

Q :- what is new network

ans :-  256 - 224  = 32 ( 32 is my block size and  this is my new network )

new network is

192.168.10.32 /27


(2)   find ....12 host

192.168.10.32/27

keep my logic...

2ⁿ − 2

( N = neutral number )

 2¹ − 2 = 0

 2² − 2 = 2 

 2³ − 2 = 6

 2⁴ − 2 =14

stop....

(note - value should be equal or grater but not less  )

 

 Now we have  N = 4 

now keep 4 bit of host and  all is network bit let's  see....

192.168.10.32 /27

11111111.11111111.11111111.11110000

Host bit = 4

Network bit = 28

192.168.10.32/28 ( CIDR will be change )

255.255.255.240                      ( 240 ?  last octet on bit is "1111"   so... 128+64+32+16 = 240 )

^{let keep HOST formula} 

^{Host = 2h − 2}

Host = 2⁴ − 2 = 14  ( we need 12 host 2 host will west no problem  )


Q :- what is new Network

ans :-  256 - 240  = 16  ( 16  is block size and  last network was 32 + 16 = 48 is my new network )

192.168.10.48/28


(3)   find ....10 Host

192.168.10.48/28

keep my logic...

2ⁿ − 2

( N = neutral number )

 2¹ − 2 = 0

 2² − 2 = 2 

 2³ − 2 = 6

 2⁴ − 2 =14

stop....

(note - value should be equal or grater but not less  )

 

 Now we have  N = 4 

now keep 4 bit of host and  all is network bit let's  see....

192.168.10.48 /28

11111111.11111111.11111111.11110000

Host bit = 4

Network bit = 28

192.168.10.48 /28 ( CIDR will be change )

255.255.255.240                      ( 240 ? { last octet on bit is "1111"   so... 128+64+32+16 = 240 } )

^{let keep HOST formula} 

^{host = 2h − 2}

host = 2⁴ − 2 = 14  ( we need 10 host 4 host will west no problem  )


Q :- what is new network

ans :-  256 - 240  = 16  ( 16  is block size and  last network was 48 + 16 = 64  is my new network )

192.168.10.64 /28




(4)   find ....6 host

192.168.10.64 /28

keep my logic...

2ⁿ − 2

( N = neutral number )

 2¹ − 2 = 0

 2² − 2 = 2 

 2³ − 2 = 6

stop....

(note - value should be equal or grater but not less  )

 

 Now we have  N = 3

now keep 3 bit of host and  all is network bit let's  see....

192.168.10.48 /28

11111111.11111111.11111111.11111000

host bit = 3

network bit = 29

192.168.10.64 /29 ( CIDR will be change )

255.255.255.248                      ( 248 ? { last octet on bit is "11111"   so... 128+64+32+16 + 8 = 248 } )

^{let keep HOST formula} 

^{host = 2h − 2}

host = 2³ − 2 = 6  ( we need 6 host  )


Q :- what is new network

ans :-  256 - 248  = 8   ( 8 is block size and  last network was 64 + 8 = 72  is my new network )

192.168.10.72 /29  ( this is my new network )




(5)   find ....2 host

192.168.10.72 /29

keep my logic...

2ⁿ − 2

( N = neutral number )

 2¹ − 2 = 0

 2² − 2 = 2

stop....

(note - value should be equal or grater but not less  )

 

 Now we have  N = 2

now keep 2 bit of host and  all is network bit let's  see....

192.168.10.48 /28

11111111.11111111.11111111.11111100

host bit = 2

network bit = 30

192.168.10.72 /30 ( CIDR will be change )

255.255.255.252         ( 252 ? { last octet on bit is "111111"   so... 128+64+32+16 + 8 + 4 = 248 } )

^{let keep HOST formula} 

^{host = 2h − 2}

host = 2²  − 2 = 2  ( we need 2 host  )


Q :- what is new network

ans :-  256 - 252  = 4   ( 4 is block size and  last network was 72 + 4 = 76  is my new network )

192.168.10.76 /30  ( this is my new network in future it will continue )



FOR U 

find out

host  200
host 100
host 60 
host 30
host 10 
host 6 
host 2 

( any network you can take ) good luck .....

note - if any problem in my post comment me...