confused with subnetting class c addresses and b classes

[John Cheetley]

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

[John Cheetley]

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

[John Cheetley]

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)

[John Cheetley]

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?

[Julio E. Moisa]

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

:-)

[John Cheetley]

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?

[John Cheetley]

Was just trying to find out how many subnets (if any) for /26 - /29 CIDR

[Julio E. Moisa]

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?

:-)

[Julio E. Moisa]

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

:-)