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  #1  
Old 12-04-2006, 06:50 PM
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Subnetting help needed
Now, I'm trying to get my head around subnetting of IP and routers and LANs and so on and so forth. Basically I'm so very stuck, the internet is proving to be a wasteland for help on this topic. So I was wondering if anyone could help me out...

A class C subnetted IP addressing scheme allows for a maximum of 14 LANs (assuming subnet address 0 and 15 are reserved).
_
Find the mask required at routers in the subnet
_
If the subnet i.d. of one of the LANs is 213.68.162.80, find the IP address of host 10 on that LAN.


If anyone can help me out even in the slightest i'd be eternally grateful. Thanks.
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  #2  
Old 13-04-2006, 12:24 PM
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Hi there,

It's been a while since I did this so if I'm wrong don't shoot me but I'll do my best to help.

A class C network runs from x.x.x.0 - x.x.x.255.
The standard subnet for a class C network is 255.255.255.0.

They want to subnet Class C networks into blocks of 16 IP addresses. As we know there are 256 addresses in a Class C network, this means there will be 16 networks of 16 IP addresses.

Each network has 16 hosts, and 16 is 2^4, so the host part of the subnet mask must be 4 bits long.

IP v4 is 32 bit addressing so we also know that the network part of the subnet mask must be 32bits-4bits=28 bits long.

So the subnet mask in binary is:

11111111 1111111 11111111 11110000

Converted to decimal gives:

255.255.255.240

So this is the mask required for the routers on the network.

Now we need to find the first host on the network. To do this, take the IP address in binary and the subnet in binary and AND them together (write 1 where both IP and subnet are 1, 0 if either are 0):-

11010101 1000100 10100010 01010000 IP (213.68.162.80)
11111111 1111111 11111111 11110000 Subnet Mask (255.255.255.240)
11010101 1000100 10100010 01010000 IP ANDed with Subnet Mask

gives the first network address as:

213.68.162.80

So, we know that the first and last IP addresses will be unusable. The fourteen hosts will have IP addresses in the range:-

213.68.162.81 - 213.68.162.94

Therefore Host 1 is 213.68.162.81
Therefore Host 10 is 212.68.162.90

And subnet mask is 255.255.255.240
Last edited by se1eagle : 13-04-2006 at 12:30 PM.
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  #3  
Old 15-04-2006, 02:03 AM
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Se1eagle, that deserved a thankyou even if it didn't help
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  #4  
Old 15-04-2006, 09:54 AM
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Yes, indeed. I've had masking explained to me in great depth and forgotten everything that I've been told, so well done!
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  #5  
Old 15-04-2006, 06:14 PM
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Thanks again se1eagle
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  #6  
Old 15-04-2006, 08:39 PM
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Quote:
Originally posted by se1eagle
So, we know that the first and last IP addresses will be unusable. The fourteen hosts will have IP addresses in the range:-

213.68.162.81 - 213.68.162.94

Therefore Host 1 is 213.68.162.81
Therefore Host 10 is 212.68.162.90

And subnet mask is 255.255.255.240


Wow, I think even I understand this now....great work, I need to keep this for future reference.
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  #7  
Old 15-04-2006, 08:45 PM
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My god, I knew I shouldn't have looked at this thread. All sorts of horrible memories of totally confusing subnetting lessons come flooding back
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