Weird subnetting

My ip address is 212.149.223.250 and subnet mask: 255.255.252.0

20 points each.

a) That is a c-class ip, but shouldn't the mask be at least  255.255.255.0? It seemes that bits from network portion are borrowed. So is that a C-class address with a B-class subnet mask with 6 bits borrowed from host to network portion?

b) I calculated that the host range for that subnet is 212.149.220.0 - 212.149.223.255 where the first and the last address are unusable, is this correct?

hafkaAsked:
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jhanceCommented:
Your network provider is doing this to more efficiently manage their subscriber lines.  There is nothing wrong with a netmask of 255.255.252.0, it simply means that they are putting you on a subnet with 10 bits.  252.0 gives 11111100 00000000 or 1024 addresses in the subnet.  Most modern network equipment can divide the subnets up any way you want, in this case they wanted 10-bits in the subnet.  
hafkaAuthor Commented:
So is that still a C-class address?
NAORCCommented:
Yes:

Class A 1.0.0.1 to 126.255.255.254 Supports 16 million hosts on each of 127 networks.

Class B 128.1.0.1 to 191.255.255.254 Supports 65,000 hosts on each of 16,000 networks.

Class C 192.0.1.1 to 223.255.254.254 Supports 254 hosts on each of 2 million networks.

Class D 224.0.0.0 to 239.255.255.255 Reserved for multicast groups.

Class E 240.0.0.0 to 254.255.255.254 Reserved.
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Les MooreSr. Systems EngineerCommented:
Yes, tecnically that is a class C address, and technically they are using a "super" net.
Network class is determined by the first 2 bits
00 = class A
01 = class B
10 = class C
11 = class D

<I think - I'm getting old so memory is fading fast, but it's close>

However, in the new world of CIDR (Classless Inter Domain Routing), network classes are no longer relevant. Some older TCP/IP stacks that were written to strict adherance to the RFC's and classes don't like the supernet masks, but most anything modern will support it just fine.
CIDR simply counts the number of bits used to designate the network with a / annotation and does not make a class distinction:
212.149.220.0/22
jhanceCommented:
Depends on who you ask.  I say NO.  NAORC says yes.  Take your pick.  It depends on your definition of "Class C".  I take the strict approach where a "C" network has a netmask of 255.255.255.0.  Clearly this is NOT that.

Network configurations are possible and useful that DO NOT fit into one of the "standard" Class A, B, C schemes.  In this case the provider is using something in between C and B where he wants 1024 hosts on each of the networks.  Note that ANY combination of network and subnet bits is possible.
Pete LongTechnical ConsultantCommented:
ASFAIK - the only way to determine an IP addresses "Class" is from its first octet NAORC has listed them in decimal octet notation and lrmoore in its true binary (first two bits of the first octet)

On modern networks you cannot determine an IP's class from its subnet mask "Borrowing" network bits for client addresses CIDR (Classless Inter Domain Routing) has been about since before I learned IPv4.

My interpretation (for what its worth) YES its a class C address
and
NEVER look at a subnet mask to determine class, only to determine the network/subnet/client address potion of an IP address.
Chris DentPowerShell DeveloperCommented:

> a) That is a c-class ip, but shouldn't the mask be at least  255.255.255.0? It seemes that bits from network
> portion are borrowed. So is that a C-class address with a B-class subnet mask with 6 bits borrowed from
> host to network portion?

It's a Classless Subnet within the Classful Class C range.

CIDR allowed for delegation of Routing for Classless Subnets - so the assignment of this range is possible and valid now. Prior to the introduction of CIDR only Classful ranges (A, B and C) were Delegated; that means the smallest range you could have been assigned was 254 usable addresses and the next one up from that was 65534 (Class B - 2^16 - 2) Addresses - neither of which are really match what you have now.

> b) I calculated that the host range for that subnet is 212.149.220.0 - 212.149.223.255 where the
> first and the last address are unusable, is this correct?

That's correct; the first address (212.149.220.0) is the Network Address an unusable because it is the network identifier (of sorts). The last (212.149.223.255) is unusable as it's the Broadcast Address (where all Broadcast Traffic is sent for your Network).

HTH

Chris

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Chris DentPowerShell DeveloperCommented:

> that means the smallest range you could have been assigned was 254 usable addresses and the next
> one up from that was 65534 (Class B - 2^16 - 2) Addresses - neither of which are really match what
> you have now.

Just a little bit on this... (Class B - 2^16 - 2) is a bit confusing as it's written and should be:

Class B: (2^16) - 2

Where the - 2 represents the Network and Broadcast Addresses and the result is the number of usable host addresses in a true Class B range.

And prior to CIDR it's highly unlikely you'd have been able to claim any IP Range as your own unless you were a Tier 1 ISP. All IP Range Delegations were (if I remember correctly) performed from the Core Internet Routers and those would be in a really poor state by now if someone hadn't sorted it all out with CIDR - not to mention that we'd be hopelessly out of addresses due to wastage in assignment (someone that wanted 100 addresses would have 256 locked away, 156 of them doing nothing at all (provided you're not accounting for Network and Broadcast)).

Chris
jabiiiCommented:
Class C address Yes, Class C network No.
Class C address is close to what NAORC  said. It is 192.0.0.1 (first usable) through 223.255.255.254 (last useable)

The Default Class C Network is 255.255.255.0 (or /24).
Default Class B network is 255.255.0.0 or /16

So actually, the network mask is a class B network with 7 Bits borrowed.

I'm lazy, so I just plugged your IP into DNSstuff.com, and below are the results for your network.
CIDR range      212.149.223.250/23
  Netmask      255.255.254.0
  Wildcard Bits      0.0.1.255
  First IP in range      212.149.222.0 (network address)
  Last  IP in range      212.149.223.255 (broadcast address)
  First useable IP in range      212.149.222.1
  Last useable IP in range      212.149.223.254
  Number of useable IPs in range      510
hafkaAuthor Commented:
Thanks for quick response, all answers were informative, and I've learned a lot.
Chris-Dent answered both questions so I gave him the points.
Chris DentPowerShell DeveloperCommented:

Glad I could help :)

Chris
hafkaAuthor Commented:
>Comment from lrmoore
>Date: 09/11/2006 04:49AM PDT
>Yes, tecnically that is a class C address, and technically they are using a "super" net.
>Network class is determined by the first 2 bits
>00 = class A
>01 = class B
>10 = class C
>11 = class D

I thought it was this way:
 
0=Class A    
10=Class B    
110=Class C    
1110=Class D (multicast)      
1111=Class E (reserved)    
Les MooreSr. Systems EngineerCommented:
hafka,
I qualified that. It was off the top of my head at the time
<I think - I'm getting old so memory is fading fast, but it's close>

Yours is actually correct.... man, my memory is going faster than I thought. What were we talking about again?
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