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calulating hosts

Discussion in 'Network+' started by jameister, Jan 27, 2011.

  1. jameister

    jameister Bit Poster

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    for example - in a /16 subnet mask the maximum amount of available hosts is 65,534. How is this worked out?

    This is really doing my brain in, I'm sure it's so simple but I just can't figure it out... Can anyone help me please?

    Thanks

    Jamie
     
    Certifications: CCSN, A+, Network+
    WIP: 70-680
  2. BosonMichael
    Highly Decorated Member Award

    BosonMichael Yottabyte Poster

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    With a /16 mask, 16 bits are used for the network portion of the address. The remaining 16 bits are used for the host portion of the address.

    The formula for determining the number of hosts on a subnet is (2^n)-2, where n is the number of bits in the host portion. 2 to the 16th power is 65,536... minus 2 is 65,534.
     
    Certifications: CISSP, MCSE+I, MCSE: Security, MCSE: Messaging, MCDST, MCDBA, MCTS, OCP, CCNP, CCDP, CCNA Security, CCNA Voice, CNE, SCSA, Security+, Linux+, Server+, Network+, A+
    WIP: Just about everything!
  3. jameister

    jameister Bit Poster

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    Thanks, even though I was still trying to work out your answer trying to work out what the 2 to the power was, I found out in the end. I know, probably everyone learnt this at school, so you would of though I should know this right? ... well my school days were not very good and I certainly didnt do well in Maths class!
     
    Last edited by a moderator: Jan 31, 2011
    Certifications: CCSN, A+, Network+
    WIP: 70-680
  4. ethernet0

    ethernet0 Byte Poster

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    /16

    Which means 16 bits are on 1 and 16 bits are off 0

    We have 16 bits off so the equation is 2 to the power of 16 - 2 i.e: 2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2 = 65536 then - 2 - 65534
     
  5. jameister

    jameister Bit Poster

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    exactly! :p
     
    Certifications: CCSN, A+, Network+
    WIP: 70-680
  6. ethernet0

    ethernet0 Byte Poster

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    This is a great post made on another forum, it really helped me at the beginning. ::

    I've received an email from one of your members asking me to post up my technique for subnetting as links to external blogs are not allowed due to forum rules. I know that he benefited from it and he wishes to help out others so here goes:

    First of all I need you to get rid of all of the negative thoughts surrounding subnetting. Put down all of the books that you have read about the subject and navigate away from other sites claiming to provide an easy way to subnet. This technique requires no charts, just simply the know-how to work with the powers of 2.

    We need to start with the fundamentals of IP addressing. An IP address is made up of 32 bits, split into 4 octets (oct = 8, yes?). Some bits are reserved for identifying the network and the other bits are left to identify the host.

    There are 3 main classes of IP address that we are concerned with.

    Class A Range 0 - 127 in the first octet (0 and 127 are reserved)
    Class B Range 128 - 191 in the first octet
    Class C Range 192 - 223 in the first octet

    Below shows you how, for each class, the address is split in terms of network (N) and host (H) portions.

    NNNNNNNN . HHHHHHHH . HHHHHHHH . HHHHHHHH Class A
    NNNNNNNN . NNNNNNNN . HHHHHHHH . HHHHHHHH Class B
    NNNNNNNN . NNNNNNNN . NNNNNNNN . HHHHHHHH Class C

    At each dot I like to think that there is a boundary, therefore there are boundaries after bits 8, 16, 24, and 32. This is an important concept to remember.

    We will now look at typical questions that you may see on subnetting. More often than not they ask what a host range is for a specific address or which subnet a certain address is located on. I shall run through examples of each, for each class of IP address.

    What subnet does 192.168.12.78/29 belong to?

    You may wonder where to begin. Well to start with let's find the next boundary of this address.

    Our mask is a /29. The next boundary is 32. So 32 - 29 = 3. Now 2^3 = 8 which gives us our block size i.e. 2 to the power of 3 equals 8.

    We have borrowed from the last octet as the 29th bit is in the last octet. We start from zero and count up in our block size. Therefore it follows that the subnets are:-

    192.168.12.0
    192.168.12.8
    192.168.12.16
    192.168.12.24
    192.168.12.32
    192.168.12.40
    192.168.12.48
    192.168.12.56
    192.168.12.64
    192.168.12.72
    192.168.12.80
    .............etc

    Our address is 192.168.12.78 so it must sit on the 192.168.12.72 subnet.

    What subnet does 172.16.116.4/19 sit on?

    Our mask is /19 and our next boundary is 24. Therefore 24 - 19 = 5. The block size is 2^5 = 32.

    We have borrowed into the third octet as bit 19 is in the third octet so we count up our block size in that octet. The subnets are:-

    172.16.0.0
    172.16.32.0
    172.16.64.0
    172.16.96.0
    172.16.128.0
    172.16.160.0
    .............etc

    Our address is 172.16.116.4 so it must sit on the 172.16.96.0 subnet. Easy eh?

    What subnet does 10.34.67.234/12 sit on?

    Our mask is 12. Our next boundary is 16. Therefore 16 - 12 = 4. 2^4 = 16 which gives us our block size.

    We have borrowed from the second octet as bit 12 sits in the second octet so we count up the block size in that octet. The subnets are:-

    10.0.0.0
    10.16.0.0
    10.32.0.0
    10.48.0.0
    .............etc

    Our address is 10.34.67.234 which must sit on the 10.32.0.0 subnet.

    Hopefully the penny is starting to drop and you are slapping the side of your head realising that you were a fool to think that subnetting was hard. We will now change the type of question so that we have to give a particular host range of a subnet.

    What is the valid host range of of the 4th subnet of 192.168.10.0/28?

    Easy as pie! The block size is 16 since 32 - 28 = 4 and 2^4 = 16. We need to count up in the block size in the last octet as bit 28 is in the last octet.

    192.168.10.0
    192.168.10.16
    192.168.10.32
    192.168.10.48
    192.168.10.64
    .................etc

    Therefore the 4th subnet is 192.168.10.48 and the host range must be 192.168.10.49 to 192.168.10.62, remembering that the subnet and broadcast address cannot be used.

    What is the valid host range of the 1st subnet of 172.16.0.0/17?

    /17 tells us that the block size is 2^(24-17) = 2^7 = 128. We are borrowing in the 3rd octet as bit 17 is in the 3rd octet. Our subnets are:-

    172.16.0.0
    172.16.128.0

    The first subnet is 172.16.0.0 and the valid host range is 172.16.0.1 to 172.16.127.254. You must remember not to include the subnet address (172.16.0.0) and the broadcast address (172.16.127.255).

    What is the valid host range of the 7th subnet of address 10.0.0.0/14?

    The block size is 4, from 16 - 14 = 2 then 22 = 4. We are borrowing in the second octet so count in the block size from 0 seven times to get the seventh subnet.

    The seventh subnet is 10.24.0.0. Our valid host range must be 10.24.0.1 to 10.27.255.254 again remembering not to include our subnet (10.24.0.0) and the broadcast address (10.27.255.255).

    What if they give me the subnet mask in dotted decimal?

    If you're lucky and they give you a mask in dotted decimal format then you should have an even easier time. All you need again is your block size.

    Let's say they have given a mask of 255.255.255.248 and you wish to know the block size. Here's the technique:

    1. Starting from the left of the mask find which is the first octet to NOT have 255 in it.
    2. Subtract the number in that octet from 256 to get your block size e.g. above it is 256 - 248 = block size of 8.
    3. Count up from zero in your block size in the octet identified in step 1 as you have learned above (the example above would be in the last octet).

    Another example is a mask of 255.255.192.0 - you would simply count up in 256 - 192 = 64 in the third octet.

    One more example is 255.224.0.0 - block size is 256 - 224 = 32 in the second octet.

    What now?

    Now it's time to go and pick up those books again and go straight to the practice questions, completely by-passing any of their techniques. Use my method and you will be laughing!

    Happy subnetting!
     
  7. alexdc12

    alexdc12 Kilobyte Poster

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    Thanks! who would of thought there was so many ways to work it all out! everything i have read has a diff method of working out!
    This seems really simple tho so thanks a lot
     
    Certifications: GNVQ Advanced IT, CompTIA A+, CompTIA N+, 70-680
    WIP: 70-685, ITIL v3 Foundation

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