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Subnetting

muraripadhan
Level 1
Level 1

Hi dears 

I am weak in sub-netting so  please cal calculate for me

My requirement is

Vlan10 require host  360

VLAN 20 require host 250

VLAN 30 require host 200

VLAN 40 require host 200

 VLAN 50 require host 200

VLAN 60 require host 120

VLAN 70 require host    100

VLAN 80 require host 100

VLAN 90 require host  80

VLAN 100 require host 80

VLAN 110 require host   60

VLAN 120 require host   60

VLAN 130 require host   60

VLAN 140 require host   50

VLAN 150 require host   50

VLAN 160 require host   40

VLAN 170 require host   30

VLAN 180 require host   30

VLAN 190 require host   30

MY IP ADDRESS IS B CLASS

172.141.0.2 255.255.0.0

 

2 Accepted Solutions

Accepted Solutions

Kevin Rivest
Level 1
Level 1

Below should be what you are looking for.

VLAN 10 - 172.141.0.0/23

VLAN 20 - 172.141.2.0/24

VLAN 30 - 172.141.3.0/24

VLAN 40 - 172.141.4.0/24

VLAN 50 - 172.141.5.0/24

VLAN 60 - 172.141.6.0/25

VLAN 70 - 172.141.6.128/25

VLAN 80 - 172.141.7.0/25

VLAN 90 - 172.141.7.128/25

VLAN 100 - 172.141.8.0/25

VLAN 110 - 172.141.8.128/26

VLAN 120 - 172.141.8.192/26

VLAN 130 - 172.141.9.0/26

VLAN 140 - 172.141.9.64/26

VLAN 150 - 172.141.9.128/26

VLAN 160 - 172.141.9.192/26

VLAN 170 - 172.141.10.0/27

VLAN 180 - 172.141.10.32/27

VLAN 190 - 172.141.10.64/27

View solution in original post

Hi

The problem is the subnet mask: 172.141.0.1 255.255.252.0

255.255.252.0 = /22

If you are going to use a /23 it should be 255.255.254.0

A /22 is providing 1022 valid IP addresses and it involves 4 subnets in one:

172.141.0.0
172.141.1.0
172.141.2.0
172.141.3.0

So that is the reason of the overlapping, because the subnet 172.141.2.0 was included. 

:-)




>> Marcar como útil o contestado, si la respuesta resolvió la duda, esto ayuda a futuras consultas de otros miembros de la comunidad. <<

View solution in original post

12 Replies 12

Kevin Rivest
Level 1
Level 1

Below should be what you are looking for.

VLAN 10 - 172.141.0.0/23

VLAN 20 - 172.141.2.0/24

VLAN 30 - 172.141.3.0/24

VLAN 40 - 172.141.4.0/24

VLAN 50 - 172.141.5.0/24

VLAN 60 - 172.141.6.0/25

VLAN 70 - 172.141.6.128/25

VLAN 80 - 172.141.7.0/25

VLAN 90 - 172.141.7.128/25

VLAN 100 - 172.141.8.0/25

VLAN 110 - 172.141.8.128/26

VLAN 120 - 172.141.8.192/26

VLAN 130 - 172.141.9.0/26

VLAN 140 - 172.141.9.64/26

VLAN 150 - 172.141.9.128/26

VLAN 160 - 172.141.9.192/26

VLAN 170 - 172.141.10.0/27

VLAN 180 - 172.141.10.32/27

VLAN 190 - 172.141.10.64/27

Hi

The formula to get any amount of hosts is:  [(2^n) - 2], the (^) represents an exponential. 

Example:

[(2^12) - 2] = 4094

[(2^11) - 2] = 2046

[(2^10) - 2] = 1022

[(2^9) - 2] = 510

[(2^8) - 2] = 254

:
:
:
[(2^2) - 2] = 2

Where "n" will represent the bits (represented in zeros) subtracted from the available bits to subnetting. 

Now your question is more related to implement VLSM in order to avoid wasting of IP addressing. 

:-)




>> Marcar como útil o contestado, si la respuesta resolvió la duda, esto ayuda a futuras consultas de otros miembros de la comunidad. <<

muraripadhan
Level 1
Level 1

sir when i am using this ip range in different vlan then it is not working

means

suppose i given ip to vlan 10 172.141.0.1/23

after that in that switch i giving ip address for vlan 20 that 172.141.2.1/24 then it show

Switch(config-if)#ip add 172.141.2.1 255.255.255.0

% 172.141.2.0 overlaps with Vlan10

Switch(config-if)#

how can i give that ip sir 

Hi

Could you please share the config of the vlan 10?




>> Marcar como útil o contestado, si la respuesta resolvió la duda, esto ayuda a futuras consultas de otros miembros de la comunidad. <<

yes sir

Switch(config)#int vlan 10

Switch(config-if)#ip add 172.141.0.1 255.255.252.0

Switch(config-if)#no shutdown

Switch(config-if)#exit

After that i try to configure vlan 20

Switch(config)#int vlan 20

Switch(config-if)#ip add 172.141.2.1 255.255.255.0

% 172.141.2.0 overlaps with Vlan10

Sir i try to configure this in one switch

Hi

The problem is the subnet mask: 172.141.0.1 255.255.252.0

255.255.252.0 = /22

If you are going to use a /23 it should be 255.255.254.0

A /22 is providing 1022 valid IP addresses and it involves 4 subnets in one:

172.141.0.0
172.141.1.0
172.141.2.0
172.141.3.0

So that is the reason of the overlapping, because the subnet 172.141.2.0 was included. 

:-)




>> Marcar como útil o contestado, si la respuesta resolvió la duda, esto ayuda a futuras consultas de otros miembros de la comunidad. <<

Thanks you sir...

It was a pleasure my friend,

Have a good day

:-)




>> Marcar como útil o contestado, si la respuesta resolvió la duda, esto ayuda a futuras consultas de otros miembros de la comunidad. <<

Please sir send me your mail id i will send you one packet tracer file where i make a topology and i am not able to resolve so sir please

Hi,

if you like you could attach the file here as ZIP. 




>> Marcar como útil o contestado, si la respuesta resolvió la duda, esto ayuda a futuras consultas de otros miembros de la comunidad. <<

Sir when i try to ping one vlan to another vlan then it shows

Reply from 192.168.1.1: Destination host unreachable.

Request timed out.

Reply from 192.168.1.1: Destination host unreachable.

Reply from 192.168.1.1: Destination host unreachable.

What is the reason sir

Subnets: 128, Hosts per subnet: 510 thats a /23 and u would be wasting a lot of ip's but u would have room to grow network and in lab its fine mask=255.255.254.0 this way no overlap u would have 128 networks w/510 hosts for each of those 128 subnets. remember u are subnetting the class b not class c most techs dont know how to subnet the B as shown in post your case is a classic vlsm design but another post here looked right if u wanted to do that 

 

cheers

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