06-27-2017 04:38 AM - edited 03-08-2019 11:07 AM
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
Solved! Go to Solution.
06-30-2017 04:24 AM
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.
06-27-2017 05:57 AM
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
...
06-27-2017 02:28 PM
thanks , was just trying to know how works . ANd each subnet shares the same network id and broadcast id
06-29-2017 07:52 PM
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
06-29-2017 08:14 PM
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.
06-29-2017 09:30 PM
And the network ID and broadcast ID would be the same for all of them?
06-29-2017 10:04 PM
Please ignore my previous reply re nw and bc
06-29-2017 10:16 PM
And the .64 That goes all the way up to .255 via 16 each time?
06-30-2017 04:24 AM
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.
06-30-2017 06:01 AM
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
06-30-2017 06:04 AM
Hi John,
You are welcome, yes that is correct
:-)
07-01-2017 12:22 AM
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.
07-02-2017 06:18 AM
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.
:-)
07-02-2017 07:07 PM
I will get back to you tonight Julio. Thanks again. :)
07-02-2017 08:29 PM
Hi
Anytime, I will be here to assist you
:-)
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