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  1. Rene,

    I have a doubt, all examples that you gave are "continuos" networks and an even number of networks.
    And when we have networks like below? I just can solve them with binary method. Is there another form?

    172.16.10.0/24
    172.16.20.0/24
    172.16.30.0/24
    172.16.40.0/24
    172.16.50.0/24

    I choosed shortest and highest networks and convert them to binary, so the summary address will be
    176.16.0.0/18 a block size 64 networks. I can't solve it using CIDR notation and block size method.

    Another example with an odd number of networks, I can solve it with binary method

    192.168.0.0 / 24
    192.168.1.0 / 24
    192.168.2.0 / 24
    192.168.3.0 / 24
    192.168.4.0 / 24

    The summary address will be 192.168.0.0/21, but when I was writing this example I saw that if I use block
    size it's able solve it too. It's need always think in block size like "powers of 2".

    When you have a free time, please, detail below

    8 + 8 + 6 = 24 bits

    Hug and thanks for yours articles. It's helping a lot :slight_smile:

  2. Hi Gabriel,

    Good question, let's look at these examples. First one:

    172.16.10.0/24
    172.16.20.0/24
    172.16.30.0/24
    172.16.40.0/24
    172.16.50.0/24

    Let's do it in binary first (in case someone else reads this):

    10 = 00001010
    20 = 00010100
    30 = 00011110
    40 = 00101000
    50 = 00110010

    Only the first 2 bits are the same. Our CIDR notation would be 8 + 8 + 2 = 18 bits and we'll use network address 172.16.0.0.

    This works but it's slow...you can do it in decimal, just remember the block sizes:

    2,4,8,16,32,64,128.

    Now you only have to pick a block size that fits all of the networks that you want. The only block size that fits your networks is 64 or 128. We'll try to be as specific as possible so we'll go for the 64.

    Now you only have to figure out the subnet mask, just use this trick:

    256 - block size = subnet mask.

    So that'll be 256 - 64 = 192. The subnet mask will be 255.255.192.0. You can calculate the subnet mask back to the CIDR notation.....

    255 = 8
    255 = 8
    192 = 2

    So that's a /18.

    So in short, just "pick" a block size that matches all your networks and then figure out the subnet mask and/or CIDR notation.

    The other quick method to look at it is like this:

    172.16.0.0/24 = 1 network
    172.16.0.0/23 = 2 networks
    172.16.0.0/22 = 4 networks
    172.16.0.0/21 = 8 networks
    172.16.0.0/20 = 16 networks
    172.16.0.0/19 = 32 networks
    172.16.0.0/18 = 64 networks

    Now you can see that 172.16.0.0/18 is the summary that will include all those networks that you want...this is the quickest method.

    Let's look at the second example:

    192.168.0.0 / 24
    192.168.1.0 / 24
    192.168.2.0 / 24
    192.168.3.0 / 24
    192.168.4.0 / 24

    Let's do the "block size" method first. You can choose between 2,4,8,16,32,64 or 128. 2 and 4 are too small so we'll go for 8.

    256 - 8 = 248 so the subnetmask will be 255.255.248.0

    255 = 8 bits
    255 = 8 bits
    248 = 5 bits

    So the CIDR notation is /21.

    The other quick method to look at this is like this:

    192.168.0.0/24 = 1 network
    192.168.0.0/23 = 2 networks
    192.168.0.0/22 = 4 networks
    192.168.0.0/21 = 8 networks.

    So we'll go for 192.168.0.0/21 as our summary. This is the quickest method to find the summary address.

    Hope this helps! The only thing to be aware of is that your summaries include networks that you "don't have".

    Rene

  3. Hey Rene,
    I was following you example but got confused on these:
    172.1.4.0/25 10101100.00000001.00000100.00000000
    172.1.128.0/25 10101100.00000001.10000000.00000000
    172.1.5.0/24 10101100.00000001.00000101.00000000
    172.1.6.0/24 10101100.00000001.00000110.00000000
    172.1.7.0/24 10101100.00000001.00000111.00000000

    Particularly the 2nd one 172.1.128.0/25. Following your example using the 3rd octet has a 128 in it, so how would I factor this in as the other octets are using the first 5 bits.

    thanks
    James

  4. Hi James,

    Where did you find this example? It's not on this page? :slight_smile:

    Rene

  5. Arthur,
    The answer to this depends on size of the networks you are wanting to summarize and whether or not they are contiguous. For the sake of simplicity, let's say you want to summarize the following into the smallest possible summary address:

    192.168.0.0/24
    ...
    192.168.9.0/24

    When I do subnetting, I use a mental shortcut instead of going all the way down to binary. Instead, I used exponents using the following formula

    2^X >= Y

    This reads as "two to the Xth power greater than or equal to Y" where Y is the number of networks (or hosts) you are trying to solve for. X is the power of two necessary so that 2 to the Xth power is greater than or equal to the number of networks in question.

    In our example, we are trying to solve for 10 networks (0 - 9). What power of 2 is necessary such that two to that number is greater than or equal to 10? The answer is 4, since 2^4 = 16. This gives us a value of X=4, but what exactly is X? X represents the number of bits that must be used, in this case, to represent the 10 networks for which we are summarizing.

    This means that /20 mask (255.255.240.0) is the smallest possible mask that could cover 192.168.0.0 - 192.168.9.0. So the summary address for 10 networks would be: 192.168.0.0/20.

    Note that it was NOT necessary to go all the way to a /16 summary in order to create the summary.

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