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2016
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confused with subnetting class c addresses and b classes

John Cheetley
Level 3
Level 3

Hi Team,

224.16.196.51 /29

CIDR binary is 11111000 (last octet)

2 to the power of 5 = 32 -2 = 30 (subnets) (for the 1's)

2 to the power of 3 = 8 - 2 = 6 (hosts) (for the 0's)

Mask is 256 - 248 = 8 (magic)

Network ID

224.16.196.0

Broadcast ID

224.16.196.xx

1st IP

224.16.196.xx - 224.16.196.xx

Question

Following normal rule re powers of 2, how can we have 30 subnets for this?

Though its classless, can you have a CIDR /25 - /29 on a 126.0.0.0 or 185.0.0.0 range as my numbers don't add up?

And on a separate note.

Does a L2 or L3 switch with vlan's create a collision domain within a network and if so, than what is the point of having a router on a stick scenario then?

Thanks in advance again

1 Accepted Solution

Accepted Solutions

Hi,

The last network will be 223.16.20.240/28

223.16.20.64/28
223.16.20.80/28
223.16.20.96/28
223.16.20.112/28
223.16.20.128/28
...
223.16.20.240/28  <---- the last subnet
223.16.20.256/28  <---- incorrect, not valid subnet.




>> 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

22 Replies 22

KJackson50
Level 1
Level 1

Hey John,

Can  you clarify what you're trying to do? As you know you'll only have 6 hosts for a /29 subnet. Are you trying to make 30 subnets from 224.16.196.0 using multiple /29 subnets?

224.16.196.0 - .7

224.16.196.8 - .15

224.16.196.16 -.23

...

thanks , was just trying to know how works . ANd each subnet shares the same network id and broadcast id 

Thanks KJackson50,

Like for this address 223.16.20.41 /28

The network ID is 223.16.20."32" /28

I thought it would have been 223.16.20."0" /28

What makes the difference?

And at what stage of the CIDR range do subnets not exist and only hosts exist please?

Thanks again

Hi

With the CIDR in this case /28 = 28 active bits you can identify the block for the ip address, the subnet mask also is useful to find the network segment, example

CIDR /28 = 255.255.255.240

Block: 256 - 240 = 16 ; so your block will go 16 by 16 at the 4 octet (because it was the modified octet)

223.16.20.0/28
223.16.20.16/28
223.16.20.32/28
223.16.20.48/28
223.16.20.64/28
....

So the IP 223.16.20.41 is part of the network 223.16.20.32/28 it will have the valid IPs 

223.16.20.33 to 223.16.20.46

Other way to find the network is converting to binary.




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

And the network ID and broadcast ID would be the same for all of them?

Please ignore my previous reply re nw and bc

And the .64 That goes all the way up to .255 via 16 each time?

Hi,

The last network will be 223.16.20.240/28

223.16.20.64/28
223.16.20.80/28
223.16.20.96/28
223.16.20.112/28
223.16.20.128/28
...
223.16.20.240/28  <---- the last subnet
223.16.20.256/28  <---- incorrect, not valid subnet.




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

Thank you Julio.

Network address of 223.16.20.240 /28

Valid IP's of .241 - 254 and broadcast of .255

And same process for other addresses

And thanks to KJackson50 for assistance too. Appreciated

Hi John,

You are welcome, yes that is correct 

:-)




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

I know you answered my question.

I tried this on 61.25.37.223 /27 (11100000)

subnets 2^3 = 8 - 2 = 6

hosts 2^5 = 32 -2 = 30

My output is below

Magic is 32 (-2 = 30) so 30 hosts

Network ID 61.25.37.32 /27

1st IP 61.25.37.33 (found out is 192)

Last IP 61.25.37.62 (found out is 222)

Broadcast 61.25.37.63 (is 223)

I followed formula though. ?

This is what confounds me...

Thanks again.

I think the formulas are not being applied correctly:

There will be to 2 kind of requests to make subnetting:

- We need X number of hosts per subnet (in few words, the priority are the hosts not the subnets)

- We need X number of subnets (in few words, the priority are the subnets only)

Formulas

If you want to make subnetting per "X" amount of host you need to use:

(2^n) - 2 >= required hosts

If you want to make subnetting per "X" amount of subnets:

(2^n)  >= required subnets

The block of the new networks:

256 - modified octets (basically using the new subnet mask)

To see how many available bits there are to subnetting

32 - CIDR (Example: 32 - 25)

Where 'n' will represent the number of bits subtracted from the available bits. 

n used for (2^n) - 2 >= required hosts will be seen as zeros (0's) the remaining will be seen as 1's and sum to the old subnet mask in order to create the new subnet mask.

n used for (2^n)  >= required networks will be seen as one's (1's) and sum to the old subnet mask in order to create the new subnet mask.

Hope it is useful.

:-) 




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

I will get back to you tonight Julio. Thanks again. :) 

Hi

Anytime, I will be here to assist you

:-)




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