Subnetting a Class A addy 2

[mikearama]

Need some help with this one, folks.

My LAN has a mask of 255.255.252.0. I am using an IP of 10.22.201.30... a class A address.

Based on the breakdown of the 252 portion of the NetID (11111111.11111111.11111100.00000000), with the last used subnet bit falling on '4', I'd expect an increment of 4 per subnet. And yet, we have subnets using 10.22.129.0, 10.22.138.0, 10.22.139.0, etc.

Could somehow kindly help me understand how that's possible, since none the three I listed are divisible by an increment of 4?

Much obliged.
Mike

 

[Cluebird]

Are all the subnets using the same mask or are you doing VLSM?

 

[mikearama]

All the user vlans are using the same mask, while the router/switch interfaces are using 255.255.255.248.

Does that help?

Mike

 

[Cluebird]

Whoever did this doesn't understand subnetting IMHO. with the given mask of 255.255.252.0, you are creating network boundaries on blocks of four in the third octet. This is essentially like aggregating four Class C's and treating them as a single block of address space. With the 10.22.129.0/22 address, you are using part of the block defined by the range that starts at 10.22.128.0 and ends at 10.22.131.255. That means the network identifier is 10.22.128.0 and the directed broadcast is 10.22.131.255 even though the hosts are addresses with 10.22.129.0 addresses.

I'd suggest readdressing the network and using a mask for the broadcast domains of /24 or /25. What you have now is a mess!

HTH

 

[msr976]

I agree with cluebird.

Your subnetting the third octet sense you are using a /22 subnet mask. If you and the given addresses with the /22 subnet mask, you will get your network range for each address. For instance, you have 10.22.138.0 255.255.252.0, you have 22 bits being used for the network portion and the last 2 bits for the host. If you use the 138 in 10.22.138.0 and you and it with the 252 in 255.255.252.0, you will get 10.22.136.0 as your network address. If you take 128 64 32 16 8 4, you get 252. If you take 138 and match the bits turned on you get your network address. So, 138 would be matched with 128 and 8 which makes 36. So the usable addresses will be 37-38 and 39 being used as a broadcast. See?

msr976

 

[burtsbees]

I would like to add that first off, where you perhaps thought they were divisible by four, well, that would be true, had the scheme been with the mask of 255.255.255.252, with the "oddball" octet being the fourth octet rather than the third. Actually, the second point I was going to make is irrelevant...

Burt

 

[Cluebird]

The divisible by four is in the third octet, not the fourth octet. With the /22 mask, the third octet provides the network boundaries while the fourth octet increments from 0 to 255 a total of four times for each network block.

 

[burtsbees]

with the last used subnet bit falling on '4', I'd expect an increment of 4 per subnet.
This is what I was referring to.

Burt

 

[Cluebird]

Guess you've got me scratching my head over what you're saying?

If I have a class A address I've subnetted to a /22, I've created 14 bits of networks 2**14. The size of each network is 10 bits 2**10 or 1024 addresses on each network with 1022 usable for hosts. I have to make the decimal numbers separated into four octets follow what the bits are doing. Address space would look like this:

First network: 10.0.0.0 to 10.0.3.255 with 10.0.0.0/22 identifying the wire and 10.0.3.255/22 the directed b'cast.
Second network: 10.0.4.0 to 10.0.7.255
Third network: 10.0.8.0 to 10.0.11.255
Fourth: 10.0.12.0 to 10.0.15.255
........
Next to last network: 10.255.248.0 - 10.255.251.255
Last network: 10.255.252.0 - 10.255.255.255

The /22 gives us a bunch of networks with the network boundary occuring on blocks of 4 in the third octet. The scheme we're analyzing from the original post indicates the administrator took host addresses from the /22 and assigned them as if he was using a /24. While it will work, it is inefficient, wastes addresses, and enforces networks that can accomodate 1022 hosts, which any administrator on a modern network realizes is much too large due to broadcast traffic.

 

[burtsbees]

Right---you know what? I read his original question wrong---I thought he was expecting the wire address in the fourth octet to increment by four, like 10.xxx.xxx.129---I misread this being in the fourth octet, rather than the third. It starts from me being tired, then me knowing that this is the certification part of the forum...I see now what he wanted to know...with this mask, it is not possible to have the actual subnets as he has stated...and they are in effect using VLSM, with the respective VLANs in the /28 mask...I agree---do all the LAN segments have 1,022 hosts??? Sorry for the confusion, but your network admin has given me a headache, mikearama...lol.

Burt

 

[mikearama]

Gents... I appreciate your thoughts.

Cluebird... bang on with the explanation, that's what I was looking for. Our network is indeed broken into many huge subnets, though none of them have anywhere near a thousand nodes... the majority have a couple hundred. Still, a wasteful design.

The current senior admin shares your disdain for the layout, but she inherited it from the previous architect, and now we're planning to subnet it properly.

Again, thanks for the input.
Mike

 

[Cluebird]

Mike,

A couple reminders since you've decided to take on the unpleasant task of reworking the address space.
1. Assign blocks of addresses based on the topology (i.e. where are the distribution layer routers?).
2. Have a hierarchical design with good agggregation points so you can summarize into your core and minimize routing table size.
3. VLSM is a wonderful thing but you can get carried away. Personally, I like /31 on point-to-point links (if your IOS supports...otherwise /30) and /25 for broadcast domains. Multipoint designs will depend on number of endpoints I'm reaching.
4. Summarize, summarize, summarize!
5. Based on the screwed-up addressing, I'm sure there are plenty of other areas that need a CLOSE look!

HTH