/25 !?!?!

mp3spy

Can anyone explain to me why /25 has 512 subnets instead of 2 subnets??

11111111.11111111.11111111.10000000

2^1 = 2 subnets
2^7-2 62 hosts / subnet
256-128 = blocks of 128

or

2^9 = 512 subnets??
2^7-2 =126 hosts???
256-255 blocks of 1

JVE

what class are you subnetting with.

If you are given an address in a class c, which i am assuming, you only get 2 subnets. if you use the subnet zero command, you would get 4 subnets out of this.

but your hosts are wrong. 2^8 = 256 (always remember that one)
so 2^7 = 128 minus 2 is 126 hosts per subnet.

BubbaJ

The numer of subnets /25 depends on the major network class.

Class A /25: 2^(25-8 ) = 131072 subnets
Class B /25: 2^(25-16) = 512 subnets
Class C /25: 2^(25-24) =2 subnets

keenon

/25

means 2 subnets with 128 addresses only 126 usable if your not using subnet 0

0 is the subnet, 1-126 hosts ... .127 broadcast


128 is the subnet, 129-254 hosts... 255 broadcast


this is your doing a class C

Danman32

JVE wrote:

If you are given an address in a class c, which i am assuming, you only get 2 subnets. if you use the subnet zero command, you would get 4 subnets out of this.

You only have one bit in the 4th octet to subnet with, so you either have 2 or 0 subnets depending if subnet 0 is allowed.

JVE

ok

mp3spy

Danman32 wrote:

JVE wrote:

If you are given an address in a class c, which i am assuming, you only get 2 subnets. if you use the subnet zero command, you would get 4 subnets out of this.

You only have one bit in the 4th octet to subnet with, so you either have 2 or 0 subnets depending if subnet 0 is allowed.

Yea but that is only applied to a Class C address....!?!!

Danman32

mp3spy wrote:

Danman32 wrote:

JVE wrote:

If you are given an address in a class c, which i am assuming, you only get 2 subnets. if you use the subnet zero command, you would get 4 subnets out of this.

You only have one bit in the 4th octet to subnet with, so you either have 2 or 0 subnets depending if subnet 0 is allowed.

Yea but that is only applied to a Class C address....!?!!

Yes. For class B you add 8 bits, so 512 or 510.
Class A you add 16 bits, so 131,072 or 131,070.

But then it all depends on the context of the question. If you have a CIDR address, the network could be any number of bits, and you subnet the host portion of the CIDR network address.

hectorjhrdz

bubbaj, interesting observation, i didn't notice that kind of analisis....very good

mp3spy

Danman32 wrote:

mp3spy wrote:

Danman32 wrote:

JVE wrote:

If you are given an address in a class c, which i am assuming, you only get 2 subnets. if you use the subnet zero command, you would get 4 subnets out of this.

You only have one bit in the 4th octet to subnet with, so you either have 2 or 0 subnets depending if subnet 0 is allowed.

Yea but that is only applied to a Class C address....!?!!

Yes. For class B you add 8 bits, so 512 or 510.
Class A you add 16 bits, so 131,072 or 131,070.

But then it all depends on the context of the question. If you have a CIDR address, the network could be any number of bits, and you subnet the host portion of the CIDR network address.

You mean 9 bits for Class B and 17 Bits for Class A ..2^9 512 2^17 131,072

Danman32

I said you ADD 8 bits for a total of 9, or you ADD 16 bits for a total of 17.

I didn't have the 'for a total of' part, that might have clarified what I was saying.

mp3spy

JVE wrote:

what class are you subnetting with.

If you are given an address in a class c, which i am assuming, you only get 2 subnets. if you use the subnet zero command, you would get 4 subnets out of this.

but your hosts are wrong. 2^8 = 256 (always remember that one)
so 2^7 = 128 minus 2 is 126 hosts per subnet.

Actually as an example, one of my friend's work utilizes a Class A address with a /25 but still utilizes the address with a /25 mask breaking it into two subnets and 126 hosts per subnet (Class C Meth). Though, if he needs to utilize more subnets/vlans he can break it down even more if needed.Is this possible?? If this is possible, is this just a poor network IP scheme?

Danman32

Since you need at least 2 bits to provide for minimum legal hosts, and you have 7 bits left as the host in the /25 subnet, that leaves anywhere from 1 to 5 bits that can be used to subnet the subnet.

1 bit = 2 subnets/vlans of 62 hosts
2 bits = 4 subnets/vlans of 30 hosts
3 bits = 8 subnets/vlans of 14 hosts
4 bits =16 subnets/vlans of 6 hosts
5 bits = 32 subnets/vlans of 2 hosts

mp3spy

Danman32 wrote:

Since you need at least 2 bits to provide for minimum legal hosts, and you have 7 bits left as the host in the /25 subnet, that leaves anywhere from 1 to 5 bits that can be used to subnet the subnet.

1 bit = 2 subnets/vlans of 62 hosts
2 bits = 4 subnets/vlans of 30 hosts
3 bits = 8 subnets/vlans of 14 hosts
4 bits =16 subnets/vlans of 6 hosts
5 bits = 32 subnets/vlans of 2 hosts

So say you're using /25 and you utilize 2 subnets/2vlans ...126 hosts/subnet, so you're using 1 bit for network and 7 bits for host. Say you run out of host addresses on both subnets, do you just create another vlan with a different subnet mask??

Danman32

Now I am confused about your question. Are you asking should you get another /25 network? If you can, you could and supernet back to /24.

But if you are sticking with your assigned /25 network, you can't create more hosts than you already have without sub-subnetting. Subnetting has you lose hosts, not add to it, since each subnet loses 2 hosts: one for broadcast and one for network. You can't make a branch of a tree bigger than the branch's host. A branch can't be bigger than the trunk. But you can get smaller branches out of that branch or trunk.
But unlike the branch-branch-trunk analogy, you can't get more sum total branch than the hosting branch or trunk.