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Home > Archive > 70-216 > July 2004 > Subnetting problem
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Subnetting problem
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| I have a problem with Subnetting question in my exam on Networks and i am currently trying to prove that the answer to the question is incorrect
The question is :
Suppose you are given an network ip adress 128.20.224.0/20
and asked to subnet this network into two subnets by allocating 1000 computers for first subnet and 750 for the last
And the answer is :
First of all - this network ip is one of 14 subnets of the network 124.20.0.0 ( classs B), it means that the network 124.20.0.0 was subnetted and now i need to subnet one of subnets, but using the algorithm of standard subnetting 2^n - 2 = number of subnets desired.
Now, I am not agree with that because the standard subnetting deals with division of the networks of class A,B,C and not of their subntes.
And there is another algorithm for recursive subdivision of subnets which is called VLSM, where i allocate bits for sub-sub-nets according to my need ( 2^n = number of subnets )
Am i right?
Slaks | |
| aznluvsmc 2004-07-19, 10:24 am |
| It doesn't seem right to me because if the original network was 124.20.0.0 how can a subnet be 128.20.224.0? I think there might be a typo.
I haven't read anything about subnetting subnets but if it can be done it's new to me and I've never seen it covered in any book. Secondly, if it was possible, I think it would add a whole level of unnecessary complexity. | |
| jocampo 2004-07-19, 11:43 am |
| quote:
Originally posted by slaks
I have a problem with Subnetting question in my exam on Networks and i am currently trying to prove that the answer to the question is incorrect
The question is :
Suppose you are given an network ip adress 128.20.224.0/20
and asked to subnet this network into two subnets by allocating 1000 computers for first subnet and 750 for the last
And the answer is :
First of all - this network ip is one of 14 subnets of the network 124.20.0.0 ( classs B), it means that the network 124.20.0.0 was subnetted and now i need to subnet one of subnets, but using the algorithm of standard subnetting 2^n - 2 = number of subnets desired.
Now, I am not agree with that because the standard subnetting deals with division of the networks of class A,B,C and not of their subntes.
And there is another algorithm for recursive subdivision of subnets which is called VLSM, where i allocate bits for sub-sub-nets according to my need ( 2^n = number of subnets )
Am i right?
Slaks
¿What exactly you would like to know? ...
Using the subnetting rules, you can that the subnet mask for this IP address is: 255.255.240, borrowing 4 bits to the rightmost part, in order to satisfy the "20" bit mask requirement (16+4). This bring us to the fact that using this mask, we will have 14 subnets, and (2^4)-2 hosts per subnet. So, if we need 1000 host per subnet, and i won't obtain that with this, my opinion is that we need to supernetting instead of subnetting, at least, using this subnetting mask. Make the reverse process, i think ...
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| curiousgeorge 2004-07-19, 12:42 pm |
| I'm surprised Freak didn't jump all over this one. He loves to answer subnetting questions.
To answer your question, you have to have a very good understanding of subnetting and of the math behind the equation.
The short answer is: yes, you can subnet a subnet. You DO use the subnetting equation 2^n - 2.
You can use it to calculate the number of hosts needed per subnet or the number of subnets needed. You can always double-check yourself, by doing it both ways.
If you need 2 subnets, 2^2 - 2 = 2. This means you need to add 2 bits to your subnet mask, which gives you a /22 subnet mask.
The "-2" when calculating number of subnets deals with rfc 905 (see link below).
If you need 1,000 hosts on the subnet, your equation gives you n=10 (2^10 - 2 = 1022). You get the same answer for 750 hosts too.
This means you need 10 bits for your host ID's. That gives you a /22 subnet mask (32 total bits - 10 host bits).
The "-2" when calculating number of hosts is for the network ID and broadcast address.
This is a good link to subnetting:
http://www.extonpccouncil.org/Resou...lk/d011subn.htm
Hope that helps. | |
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| I am not agree with you anser, Curiosgeorge.
Goog link is http://www.tcpipguide.com/free/ t_I...br />
gVLSM.htm
the equation 2^n-2 was used by subnetting algorithm rfc 950. But if you read it carefully you wont find anything saying about subnetting subnets because RCF 950 is a static subnetting algo. It permits fixed subnet length of the bitmask that is why theoretically you can use this equation to divide subnet but practically it was impossible ( read about RIP1 for example)
The algorithm that do allow subnet subnetting is VLSM but then you use simple equation 2^n becaus all-zero and all-ones are valid adresses now. | |
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| hairy51 2004-07-20, 5:12 pm |
| Would you not still use the -2 to allow for a network and broadcast address for the newly created subnet?
(Not sure if i am on the right track tonight, feel a bit dazed!) | |
| curiousgeorge 2004-07-21, 11:33 am |
| Slaks,
The last paragraph states:
"As I said before, VLSM is similar in concept to the way classless addressing and routing (CIDR) is performed. The difference between VLSM and CIDR is primarily one of focus. VLSM deals with subnets of a single network in a private organization. CIDR takes the concept we just saw in VLSM to the Internet as a whole, by changing how organizational networks are allocated by replacing the single-level “classful” hierarchy with a multiple-layer hierarchy."
What he is saying is inside a private organization, you can ignore rfc 905, because you can do whatever you want in your own company.
Many companies ignore CIDR rules when using private IP addresses as well.
You can always break the rules if your router will allow it.
Your book is giving you the rfc compliant answer. | |
| slaks 2004-07-21, 12:04 pm |
| I begin to nderstand your point.
An organization can break the rule but if we learn standart subnetting algorithm we must follow only the rules of the algo and not assume that in real life we can uphold the standart.
Is my answer is right?
First subnet adress is 128.20.224.0/22
Second : 128.20.228.0/22
Teacher's answer is 128.20.228.0/22
128.20.232.0/22
Because she exclude all-zero and all-one adresses and i am not. | |
| curiousgeorge 2004-07-21, 2:43 pm |
| You got it.
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