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IP subnetting (practical question)

Posted on 2003-11-25
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Hi, i have a decent idea of how to subnet an ip address when im told how many subnets the network needs but the following throws me off.

Question:
140.25.0.0 needs to create a set of subnets that supports up to 250 hosts on each subnet. Show an appropriate scheme.
Anyhelp with this question?
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Question by:irish_paddy
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by:svenkarlsen
ID: 9821943
Hi irish_paddy,

Need to know your mask for this, - basically it could be anything from 140.25.0.0./30 giving you two subnets of 254 addresses to 140.25.0.0/24 giving you 254 subnets of 254 addresses ?


Kind regards,
Sven
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by:Robing66066
ID: 9822209
Basically what the question is asking you to do is "slide" the network/ hosts mask so that it fits the number of hosts rather than the number of networks.

A question like this often comes up where you have a company with a lot of remote offices, but not very many people in each office.  The key is to maximize the number of subnets while still meeting the host requirements.  It seems strange because you are used to thinking of how many networks rather than how many hosts.

The answer for this sort of a question would be 140.25.0.0 /24.  That gives you the maximum number of networks with a minimum of 250 hosts on each.

(This is a standard Microsoft or Cisco IP question.)
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by:ShineOn
ID: 9822443
You guyz are using CIDR notation, which can do nifty things like subnet aggregation.

How would it look in traditional netmask format, I wonder?  That's the one thing I've never really grasped,  probably 'cause I didn't study it long enough to be able to do it in my head, which is where I like to do stuff like that...
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NicBrey earned 50 total points
ID: 9823235
The address 140.25.0.0 is a class B address, so you can safely assume a mask of 255.255.0.0 for the address before subnetting it.
Ribing66066 is correct in suggesting to use 8 extra bits for the network portion of the address to give you a mask of 255.255.255.0  or /24.

The formula for working out how many hosts it supports is
(2^ host bits) - 2   =  number of valid host addresses
(2 ^ 8) - 2            =  number of valid host addresses
256 - 2                 = 254

(Hope this makes sense - this is not the best math editor out there.....  ;-))

Same goes for networks:
(2^ network bits) - 2   =  number of usable network addresses
(2^ 24)  - 2                =  number of usable network addresses
16777216 -2               = 16777214   network addresses
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by:svenkarlsen
ID: 9823321
Nic,

I agree that, for an exam question, it's the most probable answer, but to me this is not a guy taking an exam (that would be against the rules of EE).

So he can be in charge of that full Class B net, or he can be in charge of a sub-department who has been assigned a part of that subnet (1/2 part, 1/4 part, etc.). Or he may have been assigned 1/4 of the Class B from his ISP.

I don't live in a world where people deal in perfect Class B, C, D..... subnets, - they make do with what they get from 'above' or the ISP ;-)

Sven
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by:NicBrey
ID: 9823781
Quite right Sven,
But in that case, even a 1/4 of a class B address is a bit much to hope for / too expensive to get from an ISP.

The ISP would really do all the subnet calculations for their clients and just assign them with a subnet address.

More likely to get something like 140.25.0.0  255.255.255.248  which would give you six host addresses to play with.

Luckily we have NAT... ;-)  
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by:svenkarlsen
ID: 9823885
Nic,

agree on that ;-)

Sven
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by:Robing66066
ID: 9825299
Shineon, you asked a question about CIDR notation.

The notation just refers to how many bits are in the network portion of the subnet mask.  A /24 subnet mask looks like this:

nnnnnnnn.nnnnnnnn.nnnnnnnn.hhhhhhhh

Where n is a network bit (on) and h is a host bit (off).

You can break that down into four binary sections.  In this case:

11111111.11111111.11111111.00000000

Or, 255.255.255.0 (converting from binary)

In the case of /23, you would get this:

11111111.11111111.11111110.00000000

 The subnet mask would be 255.255.254.0.

Going the other way, /25

11111111.11111111.11111111.10000000

You get 255.255.255.128

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