Options

Subnetting Question From MSPress

hamd01hamd01 Member Posts: 18 ■□□□□□□□□□
I'm currently studying chapter 2 of the MS Press Book, and have come across a question for which I cannot understand the answer. Basically the question is asking how many subnets could be contained within a given range.... anyway, the full question is here:

You are the network Administrator for the Philidelhia branch office of a large company. The IT Department at company hq in New York has assigned you address space 172.16.0.0/21 to accommodate the entire Philidelphia branch network. This address consists of 4 subnets with 40 hosts each, to which you have already assigned the network addresses 172.16.0.0/24, 172.16.1.0/24, 172.16.2.0/24 and 172.16.3.0/24. All routers within the organization support both CIDR and VLSMs

Using your current addressing scheme and without changing the subnet mask you have configured for your network, how many more subnets will you be able to accomodate in the Philidelphia branch office?


The answer they provide is simply "4" with no explination.

What I cannot understand is at the start of the question they are using /21, then it later says that we have used /24?

From what I understand, to calculate the number of available subnets, the calculation for /24 with a CLASS B address should be 2 to the power 8 = 256 subnets. Or based on 21 it would be 2 to the power of 5 = 32. The question states we have used 4 subnets already, which leaves wither 252 or 28 (depending on weather its 24 or 21).

Can someone please explain what I'm doing wrong!!?

Thank you.

Comments

  • Options
    ElvisGElvisG Member Posts: 167
    The books answer is wrong.

    172.16.0.0/21 = 172.16._ _ _ _ _ / _ _ _. _ _ _ _ _ _ _ _

    So you have 5 bits for the network side and 11 bits for the host.
    2^5 = 32 Network possible addresses and 2^11 = 2048 host possible per network.

    Ignore the /24 part because they just borrowed 3 bits from the network side because they didn't need 2048 possible host per network. Because they only needed 40 host per network it was wise to just borrow 3 bits from the network for 256 possible addresses per network. This would leave room for upgrading the network.

    So the answer is 32 possible network addresses.
  • Options
    RobertKaucherRobertKaucher Member Posts: 4,299 ■■■■■■■■■■
    The book answer is not wrong, you are reading the question improperly.

    1. You have been assigned an address space of 172.16.0.0/21. This is all you have got for your address space. The subnet mask is 1111 1111.1111 1111.1111 1000.0000 0000 which translates to 255.255.248.0. Now you have already further subdivided that address space so that it uses /24.

    1111 1111.1111 1111.1111 1111.0000 0000
    nnnn nnnn.nnnn nnnn.nnnn nsss.hhhh hhhh

    This means you only have the LAST three bits of the third octet to play with for YOUR subnets giving you a limit of 8.

    172.16.0 = 1
    172.16.1 = 2
    172.16.2 = 3
    172.16.3 = 4

    Your limit is 8. What is 4 from 8?
    4.
  • Options
    RobertKaucherRobertKaucher Member Posts: 4,299 ■■■■■■■■■■
    hamd01 wrote: »

    What I cannot understand is at the start of the question they are using /21, then it later says that we have used /24?

    Here is my understanding of the subtext of this question. Your company has several branch offices or buildings. All of your networks use the 172.16/16 address space. A network engineer who is responsible for designing the address scheme has decided that your company needs to be able to divide 172.16.0.0 into 32 networks of of 2046 hosts for each branch or building to use. You are assigned a 172.16.0.0/21 for your location in Philidelphia. The admins in LA have been assigned 172.16.8.0, the admins in Cincinnati, Ohio got assigned 172.16.16.0 and so on.

    Now there is no way you are going to want to just use your allotment as a single network. The broadcast domain would be huge (and there are other considerations such as security, etc) So, you have decided to further borrow 3 bits giving you a 8 networks of 244 hosts each. You only get 8 because you do not have permission to use the highest 5 bits of the third octet.
  • Options
    ElvisGElvisG Member Posts: 167
    Thanks Robert for the clarification.

    *edit

    I counted the 3 bits borrowed as (1+2+4 = 7 instead of 2^3 = icon_cool.gif.
  • Options
    RobertKaucherRobertKaucher Member Posts: 4,299 ■■■■■■■■■■
    Not a problem. That is one of those confusing questions typical of the 291. Although of the two times I took the exam I did not have a subnetting question that was as detailed as that.
  • Options
    RobertKaucherRobertKaucher Member Posts: 4,299 ■■■■■■■■■■
    ElvisG wrote: »
    Thanks Robert for the clarification.

    *edit

    I counted the 3 bits borrowed as (1+2+4 = 7 instead of 2^3 = icon_cool.gif.

    Yes, if you are counting like that you cannot forget that 0 is a network as well.

    0000 0000 = 1st
    0000 0001 = 2nd
    0000 0010 = 3rd
    0000 0011 = 4th
    0000 0100 = 5th
    0000 0101 = 6th
    0000 0110 = 7th
    0000 0111 = 8th
    nnnn nsss = Subnets

    The formulas are great but you have to have the counting down so that they make sense at a glance.
  • Options
    hamd01hamd01 Member Posts: 18 ■□□□□□□□□□
    After an hour or so of reading, of of course the answers in this thread, it now makes sense. I was not taking into acount that the question was relating to supernetting, not just subnetting.

    Thanks guys
  • Options
    RobertKaucherRobertKaucher Member Posts: 4,299 ■■■■■■■■■■
    It's not really supernetting, which would be using a class C address space with a /20 subnet mask, for example. It's more about subnetting a space that has already been subnetted.

    Supernet - Wikipedia, the free encyclopedia
Sign In or Register to comment.