216 and Subnetting

Discussion in 'MCSE' started by Mike, Nov 12, 2003.

  1. Mike

    Mike Guest

    I have 6 tests under my belt and I saved 216 for the end. I have been
    studying subnetting and thought I had it down. As I take practice tests it
    seems that is not the case.

    So anyone that knows or can point me in the right direction/website/anything
    your help would be much appreciated. I really want to know this not just
    for the test but for my job in general.

    I will use my last practice question as an example of how I am confused.

    The question wants to know which addresses can be divided into at least 4
    subnets with at least 100 hosts per subnet.

    A. 10.25.0.0/22
    B. 10.25.0.0/23
    C. 10.25.0.0/24
    D. 10.25.0.0/25

    # My Calculations:

    /22 = 255.255.252.0 = 14 subnet bits 10 host bits
    /23 = 255.255.254.0 = 15 subnet bits 9 host bits
    /24 = 255.255.255.0 = 16 subnet bits 8 host bits
    /25 = 255.255.255.128 = 17 subnet bits 7 host bits


    using 2^n -2

    A = 2^14 -2 = 16382 subnets and 2^10-2 =1022 hosts
    B = 2^15 -2 = 32766 subnets and 2^9-2= 510 hosts
    C = 2^16 -2 = 65534 subnets and 2^8-2= 254 hosts
    D = 2^17 -2 = 131070 subnets and 2^7-2= 128 hosts

    I thought the answer should be all of them. Apparently I was wrong.
    The test says that only A and B are correct. It says that C can be divided
    into 2 or 4 subnets and D can be divided into 4 subnets with only 30 hosts.

    I know it is a CIDR address but I guess I just don't quite get that concept
    or at least the math behind it. Is there a foormula for this type of
    calculation? Am I just way off with subnetting and need to read the book
    again? I don't quite understand where you start and stop counting the bits
    with CIDR.
     
    Mike, Nov 12, 2003
    #1
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  2. Mike

    Andy Foster Guest

    On Tue, 11 Nov 2003 22:50:15 -0500, Mike wrote:

    Forget classes for a moment. Just because 10.x.x.x is class A doesn't
    necessarily give you /8 - the possible answers give you the ranges/masks.
    eg. 10.25.0.0/22 gives you the range 10.25.0.0 - 10.25.3.255 (or 10 host
    bits starting from 10.25.0.0) - the rest is not yours.
    100 hosts need 7 bits, and 4 subnets need 2 bits, so you need 9 host bits
    to play with - so a mask or /23 (32 - 9) or less is required.

    HTH

    Andy
     
    Andy Foster, Nov 12, 2003
    #2
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  3. Mike

    Paul Guest

    Andy

    Could you let me know how you get the following range in
    your calculation:

    10.25.0.0/22 gives you the range 10.25.0.0 - 10.25.3.255

    I don't understand how you are calculating 10.25.0.0 -
    10.25.3.255 with /22

    Thanks

    Paul
     
    Paul, Nov 12, 2003
    #3
  4. Mike

    Eric Guest

    www.learntosubnet.com
     
    Eric, Nov 12, 2003
    #4
  5. Mike

    TheXman Guest

    If you break down the subnet mask to its bit form, /22
    will give you:

    10.25.0.0/22

    nnnnnnnn.nnnnnnnn.sssssshh.hhhhhhhh = /22
    where n=network bits / s=subnet bits / h=host bits

    or
    11111111.11111111.11111100.00000000 = /22
    or
    255.255.252.0 = /22


    Therefore, in the 3rd octect, you have 6 bits for the
    subnets and 10 bits for the hosts. Remember, the formula
    to calculate subnets and hosts is 2^bits - 2.

    In this example:
    Number of subnets = 2^6 - 2 or 62
    Number of hosts = 2^10 - 2 or 1022

    The first subnet in the range would be 10.25.4.1 -
    10.25.7.254.
    The last subnet in the range would be 10.25.248.1 -
    10.25.251.254.

    Sincerely,
    Xavier Todd Clarke

    ----- Original Message -----
    From: Paul
    Newsgroups: microsoft.public.cert.exam.mcse
    Sent: Wednesday, November 12, 2003 6:00 AM
    Subject: Re: 216 and Subnetting


    Andy

    Could you let me know how you get the following range in
    your calculation:

    10.25.0.0/22 gives you the range 10.25.0.0 - 10.25.3.255

    I don't understand how you are calculating 10.25.0.0 -
    10.25.3.255 with /22

    Thanks

    Paul
     
    TheXman, Nov 12, 2003
    #5
  6. Mike

    Andy Foster Guest

    16 bits in the 3rd octet ?
    For you to have 16 bits to play with, you would have to be given
    10.25.0.0/16.
    Wrong!
    10.25.4.1 - 10.25.7.254 is the usable host range for 10.25.4.0/22.
    That has nothing to do with the original question.

    HTH

    Andy
     
    Andy Foster, Nov 12, 2003
    #6
  7. Mike

    Andy Foster Guest

    As suggested elsewhere www.learntosubnet.com

    This isn't how I calculated it, but the longhand is as follows....

    Converting 10.25.0.0 into binary gives
    00001010.00011001.00000000.00000000
    /22 (aka 255.255.252.0) into binary gives
    11111111.11111111.11111100.00000000

    The range of addresses (not hosts) described by 10.25.0.0/22 in binary is
    00001010.00011001.00000000.00000000 (10.25.0.0) to
    00001010.00011001.00000011.11111111 (10.25.3.255)

    This would give a host range of 10.25.0.1 to 10.25.3.254 if it were used
    as a single subnet, (which it isn't)

    HTH

    Andy
     
    Andy Foster, Nov 12, 2003
    #7
  8. Mike

    YOUNGMAN Guest

    To Andy Foster -

    It must be wonderful to be so sure of your own abilities . . . howeve
    the ability to read would be quite a useful one to start with!!!

    TheXman quite clearly stated that for the address 10.25.0.0/22 - "i
    the 3rd octect, you have 6 bits for the subnets and 10 bits
    for the hosts"

    This was not edited afterwards as you quoted it in your own post.

    However you then misquoted and mocked it

    Wrong!


    TheXman is quite right in stating that in the example given the Numbe
    of subnets = 2^6 - 2 or 62 and the Number of hosts = 2^10 - 2 or 1022

    That's why, surprisingly enough, the address range given was correc
    for the example given i.e 10.25.0.0/22!

    HTH






    ;

    YOUNGMA
    Sign up for free daily practice questions at: http://www.QoD.U
     
    YOUNGMAN, Nov 13, 2003
    #8
  9. Mike

    TheXman Guest

    Thank you YOUNGMAN. Can you believe the nerve of this Andy Foster
    character? First, he ask for help understanding IP subnetting. Then, he
    mocks correct answers as if he knows IP subnetting. Instead of defending my
    answer, I decided it would be more pleasurable to let Andy make a FOOL of
    himself when he shows others his failed understanding of IP subnetting. You
    can only teach people who want to learn. Obviously, Andy Foster already
    knows everything. (lol).
    --
    Sincerely,
    TheXman




    To Andy Foster -

    It must be wonderful to be so sure of your own abilities . . . however
    the ability to read would be quite a useful one to start with!!!

    TheXman quite clearly stated that for the address 10.25.0.0/22 - "in
    the 3rd octect, you have 6 bits for the subnets and 10 bits
    for the hosts"

    This was not edited afterwards as you quoted it in your own post.

    However you then misquoted and mocked it

    Wrong!


    TheXman is quite right in stating that in the example given the Number
    of subnets = 2^6 - 2 or 62 and the Number of hosts = 2^10 - 2 or 1022

    That's why, surprisingly enough, the address range given was correct
    for the example given i.e 10.25.0.0/22!

    HTH






    ;)


    YOUNGMAN
    Sign up for free daily practice questions at: http://www.QoD.US
     
    TheXman, Nov 13, 2003
    #9
  10. You need to re-read the original post in this thread. Andy did not ask for
    help, he replied to the original poster (Mike).

    FWIW, both you and YOUNGMAN are complete dicks.

    --
    Fris "or maybe incomplete ones" beeĀ® MCNGP #13

    http://www.mcngp.tk
    The MCNGP Team - We're here to help

    http://groups.yahoo.com/group/certaholics
    Certaholics - We're here if you're beyond help
     
    =?Windows-1252?Q?Frisbee=AE_MCNGP?=, Nov 13, 2003
    #10
  11. Mike

    Mike Guest

    OK. That makes sense. However, if 10.25.0.0 /22 = 6subnet bits and 10 host
    bits why doesn't 10.25.0.0 /25 = 9 subnet bits and 7 host bits? Do you
    start counting your bits over again when you move to the next octet because
    the mask would now be 255.255.255.128? Would 10.25.0.0 /25 really = 1
    subnet bit and 7 host bits?
     
    Mike, Nov 13, 2003
    #11
  12. Mike

    TheXman Guest

    Mike:

    You're right, 10.25.0.0/25 would be 9 subnet bits and 7 host bits. Where
    you may be getting a little mixed up is that you are ignoring your network
    bits. Each subnet will be made up of your network bits PLUS your subnet
    bits. I hope this helps, if you still need more help understanding, please
    ask.

    10.25.0.0/25

    The /25 bit mask would look like this:

    nnnnnnnn nnnnnnnn ssssssss shhhhhhh
    where: n=network bits / s=subnet bits / h=host bits
    or
    255.255.255.128

    Therefore:
    number of subnets = 2^9 - 2 or 510
    number of hosts = 2^7 - 2 or 126

    --
    TheXman


    OK. That makes sense. However, if 10.25.0.0 /22 = 6subnet bits and 10 host
    bits why doesn't 10.25.0.0 /25 = 9 subnet bits and 7 host bits? Do you
    start counting your bits over again when you move to the next octet because
    the mask would now be 255.255.255.128? Would 10.25.0.0 /25 really = 1
    subnet bit and 7 host bits?
     
    TheXman, Nov 13, 2003
    #12
  13. Mike

    Consultant Guest

    you are the fool in this thread. shouldn't you be reading a cominc book or
    something? wait a sec, xman & youngman, i see some sick connection here.
    take it away
     
    Consultant, Nov 13, 2003
    #13
  14. Mike

    Mike Guest

    That does help. I think I am starting to get it. This practice test is
    throwing me off. It says that 10.25.0.0/25 can be up to 126 hosts OR it can
    divided into 4 subnets of 30 hosts and I 'm sitting here thinking it can be
    510 subnets with 126 hosts per subnet. I don't understand where they come
    up with the 4 subnets of 30 hosts from.
     
    Mike, Nov 13, 2003
    #14
  15. Mike

    Brat Guest

    You need to pay attention here... MIKE asked for help not Andy
     
    Brat, Nov 13, 2003
    #15
  16. Mike

    TheXman Guest

    Thank you for the correction. I apologize for saying that Andy was
    requesting help. Nevertheless, Andy was incorrect and a bit of an A-hole.
     
    TheXman, Nov 13, 2003
    #16
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