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determining the subnet mask for a required number of subnets

Discussion in 'Network Infrastructure' started by alebleicker, Jul 17, 2010.

  1. alebleicker

    alebleicker Bit Poster

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    Hi guys, I'm having a bad time through this chapter in the 70-291 book where I need to determine the subnet mask for a required number of subnets, the book says I need to determine the subnet mask value that is b bits removed from and after the original subnet id for example 255.255.255.0 and b=3, your subnet mask is 255.255.255.224 because the ascending order of subnet mask values, that is the subnet mask that is 3 bits removed from and after 255.255.255.0.

    I simply can't understand how can I remove 3 bits from 255.255.255.0 and end with 255.255.224.0, so far I know taking 3 bits will result in .248 subnet

    This part of the book is really confusing, there is nothing explain very well what happens for me. Can someone help me please?
     
    Certifications: A+,N+,70-270,70-290
    WIP: MCSA
  2. SimonD

    SimonD Terabyte Poster Moderator

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    Have a look at my posts here for (hopefully) a better idea of subnetting.
     
    Certifications: CNA | CNE | CCNA | MCP | MCP+I | MCSE NT4 | MCSA 2003 | Security+ | MCSA:S 2003 | MCSE:S 2003 | MCTS:SCCM 2007 | MCTS:Win 7 | MCITP:EDA7 | MCITP:SA | MCITP:EA | MCTS:Hyper-V | VCP 4 | ITIL v3 Foundation | VCP 5 DCV | VCP 5 Cloud | VCP6 NV | VCP6 DCV | VCAP 5.5 DCA
    WIP: VCP6-CMA, VCAP-DCD and Linux + (and possibly VCIX-NV).
  3. xmojo

    xmojo Nibble Poster

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    I think of it in this way:

    Subnet mask No. of bits needed

    128 1
    192 2
    224 3
    240 4
    248 5
    252 6
    254 7

    If working out subnet questions I always write down this table as a quick guide. So you could reword your problem as "How many bits are left after taking away 3 bits?" Then refer to the table to get your subnet mask.
     
    Last edited: Jul 18, 2010
  4. alebleicker

    alebleicker Bit Poster

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    damn, I was studying until so late that I didn't notice I was confusing the 3rd and 4th octets thinking the 4th was always on the 3rd place, because of this the 255.0+3bits is 255.224... my great fault!

    I'll get some more sleep, this subnet subject consumed all my brain this weekend.
    thanks!
     
    Certifications: A+,N+,70-270,70-290
    WIP: MCSA
  5. soundian

    soundian Gigabyte Poster

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    I think I see where the confusion lies.
    The question states
    the subnet mask value that is 3 bits removed from and after the original subnet id 255.255.255.0
    Not
    the subnet mask value that has 3 bits removed from the original subnet id 255.255.255.0
    I read it the second way a few times. I guess that shows the importance of reading and understanding the question before you attempt an answer.
    I still think it's a poorly worded question though, unless the intention is to test candidates on attention to detail as well as technical issues.
     
    Certifications: A+, N+,MCDST,MCTS(680), MCP(270, 271, 272), ITILv3F, CCENT
    WIP: Knuckling down at my new job
  6. greenbrucelee
    Highly Decorated Member Award

    greenbrucelee Zettabyte Poster

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    I am still trying to perfect subnetting so if I get this wrong please corect me.

    I do a chart like this for an ip for example 192.168.1.165 and subnet mask of 255.255.255.0

    11000000.10101000.00000001.10100101
    11111111.11111111.11111111.00000000
    ---------------------------------------
    11000000.10101000.00000001.00000000
    where there is 1 on the top line and a 1 on the bottom line you put a 1 and if its two 0 or a 1 & a 0 you put 0

    This makes your network address which is 192.168.1.0

    To get the broadcast you change all the 0 in the last octet to 1 so the broadcast is 192.168.1.255

    Now to work out the hosts you take 2 away 1 is reserved so just take 1 away from the 255 which makes 254 so the host address range is 192.168.1.1 - 192.168.1.254

    So for a /26 subnet with an ip of 192.168.1.0

    11000000.10101000.00000001.00000000
    11111111.11111111.11111111.11000000
    ---------------------------------------
    11000000.10101000.00000001.11000000

    I wont type it all out by as you can see hopefully that you can have a max of 4 networks with max number of hosts being 63 but one is reserved. so its 62.

    Hopefully that makes sense.
     
    Certifications: A+, N+, MCDST, Security+, 70-270
    WIP: 70-620 or 70-680?
  7. cobirad

    cobirad Bit Poster

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    If ur actually having problems on how to subnet. Then check this (www.subnetting-secrets.com) out and it wont be a problem anymore.
     
    Last edited: Jul 19, 2010
  8. soundian

    soundian Gigabyte Poster

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    It's easier just to remember that number of hosts=2^n-2, where n=number of host bits.
    The minus two is the network and broadcast addresses.
    I think this is a better method because you can easily work out 2^n if you know some of them.
    Take as an example, a class B address with 20 network bits. That gives you 12 host bits. I don't know what 2^12 is, but I know what 2^10 is (1024) and can multiply by 2 twice (2048, 4096) really quickly and then subtract two. You can't do such simple maths with your method.
     
    Certifications: A+, N+,MCDST,MCTS(680), MCP(270, 271, 272), ITILv3F, CCENT
    WIP: Knuckling down at my new job

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