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.
07-03-2017 02:18 AM
My ip is 61.49.23.41 /27 (11100000)
2^3 = 8 -2 = 6
2^5 = 32 -2 = 30
32 - CIDR = 27
Network ID 61.49.23.0 /27
61.49.23.1 - 61.49.23.28 BC.29
Network ID 61.49.23.30 /27
61.49.23.31 - 61.49.23.58 BC.59
etc up to BC.255
07-03-2017 02:23 AM
But when I put that into a subnet calculator
Network ID is 61.49.23.32 /27
Unless network ID start with the magic number but that isn't correct as some end in 0 (zero)
07-03-2017 04:52 AM
Thanks Julio for the very detailed reply. I was getting confused with the last octet number and the network ID for following subnets. And so the broadcast goes to .224 due to total value of bits borrowed?
07-03-2017 05:12 AM
Hi
I have noted you are subtracting (-2), the substraction should be used to find X amount of hosts using the formula: [(2^n) -2 ], but if you are going to get X amount of subnets the formula is: [2^n] the substraction is not required.
and the formula to know the available bits should be:
32 - CIDR (in this case 27) = 5
Now if your IP is 61.49.23.41 /27, this ip belongs to the network 61.49.23.32/27, why? because the subnet mask is going 32 by 32 on the blocks:
subnet mask /27, it represents 27 active bits:
octet 1 (8 active bits): 11111111 = 255
octet 2 (8 active bits): 11111111 = 255
octet 3 (8 active bits): 11111111 = 255
octet 4 (8 active bits): 11100000 = 224 ((2^7 + 2^6 + 2^5 = 224) You will sum just the active bits (1) the rest of bits are zero's so they are not considered))
sum = 27 bits, now with the formula for the blocks you can know how will be blocks (networks)
Formula: 256 - 224 (why 224, because /27 is 224 on the 4th octet) = 32 ; so the blocks will be 32 by 32 on the 4th octet because it was the only octet modified, the other 3 octets were kept the same 255.255.255.224.
Results:
New subnet: 61.49.23.0/27
New subnet: 61.49.23.32/27 <-- valid IP address .33 to .62 so the IP .41 is between this range.
New subnet: 61.49.23.64/27
New subnet: 61.49.23.96/27
New subnet: 61.49.23.128/27
New subnet: 61.49.23.160/27
New subnet: 61.49.23.192/27
New subnet: 61.49.23.224/27 <--- the last subnet
Hope it is useful
:-)
07-03-2017 09:42 PM
Yes it is.
With the last subnet, that finishes at .224 due to that is how many bits were borrowed and you can't go passed that?
06-29-2017 04:39 PM
Was just trying to find out how many subnets (if any) for /26 - /29 CIDR
06-29-2017 05:06 PM
Hi
Please correct me if Im understanding wrong the question, but if you have 2 subnets:
224.16.196.51 /26
224.16.196.51 /29
It could be used on the same network infrastructure because it will create overlapping unless you create a subnetting for the /26 to get several /29 networks.
From what subnet you would like to get 30 networks?
:-)
06-27-2017 06:06 AM
Hi John
The key to subnet is know the formulas:
In order to get # of hosts you can use (2^n) - 2
In order to get # of subnets you can use (2^2)
Available bits to subnetting: 32 - CIDR, example 32-24 = 8 bits <-- you will have 8 bits to subnetting. 32-16 = 16, so you will have 16 bits to subnetting.
Block of networks after subnetting: 256 - value of the subnet mask, example: new subnet mask: 255.255.254.0 so the block will be increasing 2 by 2 on the third octet. basically is 256 - modified octet value.
Network 172.16.0.0/16
Subnetting to get 2 subnets or 510 hosts.
Network 1 - 172.16.0.0/23
Network 2 - 172.16.2.0/23
Network 3 - 172.16.4.0/23
A collision domain means the part of network where a packet could suffer a collision, the switches divide the collision domains so it decrease the possibility to have problems with the communication, if you have a switch of 24 ports it means you will have 24 collision domains. Now Router in a stick is a method to create intervlan routing, you need a L3 devices like a router, multilayer switch or firewalls in order to communicate 2 or more vlans with each other. So the scheme router in a stick is used for that.
Hope it is useful
:-)
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