Number of subnets = x^2 or x^2-2

Discussion in 'MCSE' started by Vernon, May 19, 2004.

  1. Vernon

    Vernon Guest

    Which of the following network address spaces can be devided into at least
    four subnets with atlest 100 hosts per subnets ?
    Now for minimum 4 subnets do I need 2 bits or 3 bits for Microsoft ? (Never
    mind the study guides)

    Vernon, May 19, 2004
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  2. Vernon

    Rich Guest

    Any of the options would give you the desired result.
    option c for example would give you 254 available subnets
    and 254 available host per subnet. Not sure what the
    question is wanting past that.

    Rich, May 19, 2004
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  3. Vernon

    Neil Guest

    try this:

    the explain button might help you figure out what's up...

    oh, and Rich was incorrect about 1 peice in his post. with a /24 network
    and this address you would have 64K networks not 254.

    (and MS has historically done the 2^n-2 and Cisco the 2^n)
    Neil, May 19, 2004
  4. Vernon

    Rich Guest

    If this was starting out as a true Class A address then
    you would be correct. I assumed that the class A address
    was already subnetted once with a Subnet Mask of /16.
    ( And that the question was to subnet it
    further. Given a /25 subnet mask would add 254 additional
    networks by adding the subnet range to the 3rd octet. Of
    course I am doing it the Cisco way and not MS's way.

    I hope I am looking at this correctly.

    Thanks for the input!

    Rich, May 19, 2004
  5. Vernon

    Neil Guest

    I believe you are off base here. a class A is a class A is a class A. if
    it was subnetted with /16 how would making the mask /24 change that. if
    the address was a 10.x.x.x (something defined in RFC 1918) then changing
    the /16 to /24 would mearly increase the number of netowrks from 253 to
    64k and reduce the number of hosts from

    you may want to look up RFC 1878 and RFC 1519 as some useful
    Neil, May 20, 2004
  6. Vernon

    Neil Guest

    pfffffttt! I had a nap. should have said "networks from 254" and "reduce
    the number of hosts from 64k to 254"
    Neil, May 20, 2004
  7. Vernon

    Vernon Guest gives you 10 remaining bits to play with. Since we have min 100
    hosts per subnet we need those 7 bits for host portion. That leaves us
    32-22-7 = 3 bits. My question is how many subnets we can have with these 3
    bits for Microsoft exam? 8 or 6? IF it is 8 then answer (
    would have 4 subnets. Answer c and d are not correct since the remaining
    bits are insuffecient to meet 4 subnets and 100 host demand.

    Over to you Neil...I looked up the website the link you provided. It sort of
    answers the question ..thanks
    So I guess it is calssless routing as far as Microsoft is concerned so it is
    2^n and not 2^n-2
    Vernon, May 20, 2004
  8. Vernon

    Neil Guest

    you are doing fine up till here. yes 7 bits will give you 126 host (even
    if you do the 2^n, the 0 and 255 address are resevred for the network
    address and the broadcast address respectively). the problem is after
    this you have decided to subnet your subnet and this was not part of the
    original question. I think you will find that if you dig out the original
    question (I will assume that this is not live test question) you will
    read that it say 4 networks and at least 100 hosts or at least 4 networks
    and 100 hosts (one statement of at least not 2). class A addresses have a
    defualt mask of or /8 you have 24 additional bits to play with.
    If you need 7 bits for hosts then you can have anything up to or /25 for a mask. converting this to Cisco terminolgy
    means that /22, /23, /24, or /25 is fine as they will all give you at
    least 100 hosts. I think the one that closest matches the 100 hosts
    recommendation is the /25

    Neil, May 20, 2004
  9. Vernon

    Andy Foster Guest

    The original question was ...
    "Which of the following network address spaces can be devided into at least
    four subnets with atlest 100 hosts per subnets ?
    Now for minimum 4 subnets do I need 2 bits or 3 bits for Microsoft ? (Never
    mind the study guides)"

    "Which of the following address spaces" implies that in a. is
    your entire address space - you do not have the entire Class A
    range to yourself.
    CIDR is the new black.

    To the OP, M$ used to say 2^n-2 subnets - they now say 2^n - but you should
    be aware that both are possible answers. You should never be expected to
    chose only one as a correct answer though.


    Andy Foster, May 21, 2004
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