subnetting question

Discussion in 'MCSA' started by anonymous, Feb 13, 2004.

  1. anonymous

    anonymous Guest

    I was using a class C subnet mask (255.255.255.0) scope
    for my private network. I was going to increase it to
    255.255.252.0 which would give me 1023 addresses less 2 I
    believe? However, now I'm not sure how the range
    would/will be in third octet. It was easy with the
    255.255.255.0 range because I just went from 146.157.30.0
    to 146.157.30.254. Now what sort of ranges to I have in
    the third octet? Will the scope go from 146.157.252.0 to
    146.157.255.0 giving me 1020 addresses?
    anonymous, Feb 13, 2004
    #1
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  2. With a subnet mask of 255.255.252.0, you have 10 bits for host ID use and 10
    bits is 1+2+4+8+16+32+64+128+256+512 for 1023. The subtracting two would
    refer to not using the all zeros combination and not using the all ones
    combination. 146.157.30.0 = 146.157.00011110.00000000 with a mask of
    255.255.11111100.00000000, so your first unique host ID (assuming all zeros
    is not used) would be 146.157.00011110.00000001, which is 146.157.30.1 and
    the end of that range would be 146.157.00011111.11111110, which is equal to
    146.157.31.254, assuming you did not count all zeros or all ones as unique
    identifiers. That would give you 1022 addresses. All of this is assuming I
    have done my math correctly without the help of coffee.

    The shortcut to figuring roughly how many addresses you would have is to
    figure out the binary worth of the last bit of your subnet mask and subtract
    two. For instance, 255.255.252.0 = 255.255.11111100.00000000, so the last
    bit is the 11th bit from the least significant bit/end
    (1/2/4/8/16/32/64/128/256/512/1024 = 1024) and I would subtract two for
    1022.

    I hope that makes sense. It is easier to explain in person. :)

    --
    Andy Ruth
    Microsoft Learning

    This posting is provided "AS IS" with no warranties, and confers no
    rights.

    "anonymous" <> wrote in message
    news:1014001c3f24d$c0a099a0$...
    > I was using a class C subnet mask (255.255.255.0) scope
    > for my private network. I was going to increase it to
    > 255.255.252.0 which would give me 1023 addresses less 2 I
    > believe? However, now I'm not sure how the range
    > would/will be in third octet. It was easy with the
    > 255.255.255.0 range because I just went from 146.157.30.0
    > to 146.157.30.254. Now what sort of ranges to I have in
    > the third octet? Will the scope go from 146.157.252.0 to
    > 146.157.255.0 giving me 1020 addresses?
    Andy Ruth [MS], Feb 13, 2004
    #2
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  3. anonymous

    James Martin Guest

    http://www.learntosubnet.com unless you don't really want to know, then
    google for a free subnet calculator.


    "anonymous" <> wrote in message
    news:1014001c3f24d$c0a099a0$...
    > I was using a class C subnet mask (255.255.255.0) scope
    > for my private network. I was going to increase it to
    > 255.255.252.0 which would give me 1023 addresses less 2 I
    > believe? However, now I'm not sure how the range
    > would/will be in third octet. It was easy with the
    > 255.255.255.0 range because I just went from 146.157.30.0
    > to 146.157.30.254. Now what sort of ranges to I have in
    > the third octet? Will the scope go from 146.157.252.0 to
    > 146.157.255.0 giving me 1020 addresses?
    James Martin, Feb 13, 2004
    #3
  4. anonymous

    Guest Guest

    here is a quick guide that will never stear you wrong (of
    the top of my head of course)

    fisrst to calculate the number of host, use the formula:
    2 (to the power of number of host bits) - 2

    in your case 2 to the power of 10 -2 = 1022 host

    as for the range use this chart

    in the third octet, find you low order bit (last bit set
    to one)
    128 64 32 16 8 4 2 1
    1 1 1 1 1 1 0 0 = 252

    last bit that is set to 1 is slot 4. So your subnet will
    be broken up into blocks of 4.

    i.e.
    146.157.0.1 - 146.157.3.254
    146.157.4.1 - 146.157.7.254
    146.157.8.1 - 146.157.11.254

    and so on

    >-----Original Message-----
    >I was using a class C subnet mask (255.255.255.0) scope
    >for my private network. I was going to increase it to
    >255.255.252.0 which would give me 1023 addresses less 2 I
    >believe? However, now I'm not sure how the range
    >would/will be in third octet. It was easy with the
    >255.255.255.0 range because I just went from 146.157.30.0
    >to 146.157.30.254. Now what sort of ranges to I have in
    >the third octet? Will the scope go from 146.157.252.0 to
    >146.157.255.0 giving me 1020 addresses?
    >.
    Guest, Feb 16, 2004
    #4
  5. He (or she :)) is right. That's exactly how I would have explained it. It's a quick easy way to figure out subnetting.... especially if you don't like dealing in binary.

    --
    Samantha Hyatt


    410-371-2645

    "What lies behind us and what lies before us are small matters compared to what lies within us."
    -- Ralph Waldo Emerson
    <> wrote in message news:119e901c3f4e3$f7ddaf00$...
    here is a quick guide that will never stear you wrong (of
    the top of my head of course)

    fisrst to calculate the number of host, use the formula:
    2 (to the power of number of host bits) - 2

    in your case 2 to the power of 10 -2 = 1022 host

    as for the range use this chart

    in the third octet, find you low order bit (last bit set
    to one)
    128 64 32 16 8 4 2 1
    1 1 1 1 1 1 0 0 = 252

    last bit that is set to 1 is slot 4. So your subnet will
    be broken up into blocks of 4.

    i.e.
    146.157.0.1 - 146.157.3.254
    146.157.4.1 - 146.157.7.254
    146.157.8.1 - 146.157.11.254

    and so on

    >-----Original Message-----
    >I was using a class C subnet mask (255.255.255.0) scope
    >for my private network. I was going to increase it to
    >255.255.252.0 which would give me 1023 addresses less 2 I
    >believe? However, now I'm not sure how the range
    >would/will be in third octet. It was easy with the
    >255.255.255.0 range because I just went from 146.157.30.0
    >to 146.157.30.254. Now what sort of ranges to I have in
    >the third octet? Will the scope go from 146.157.252.0 to
    >146.157.255.0 giving me 1020 addresses?
    >.
    Samantha Hyatt, Feb 17, 2004
    #5
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