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  1. #1
    Join Date
    Jul 2009
    Posts
    13

    Default Subnetting, VLSM and Summarization

    hi all
    this is my 1st post here
    please can anyone tell me easy way to understand subnetting vlsm summrization
    Last edited by khalillaghari; 07-25-2009 at 06:31 PM.

  2. #2
    Join Date
    Jun 2009
    Location
    Tokyo
    Posts
    8

    Default

    khalillaghari;5 875]hi all
    this is my 1st post here
    please can anyone tell me easy way to understand subnetting vlsm summrization
    Yes,I can tell u the easy and simple way to memorise subnetting vlsm
    summerization.
    1.Implenment what u read asap.
    2.Answer similar questions anywhere u see it.Eg there are thousands
    of questions about subnetting vlsm summerization.
    I pray it helps because I am new when it comes to networking.
    Thanks.
    Magnus.

  3. #3
    Join Date
    Jul 2009
    Posts
    13

    Default

    Xtopher62705 /hi all
    this is my second post here
    please lammle help me about summrization!! i have problem in summrization

    i have these networks how can i summrize these n/w into one summry
    192.168.10.0/25
    host range is 192.168.10.1 to 126 and 127 is brodcats
    mask is 255.255.255.128
    192.168.0.126/26
    host range 192.168.10.129 last host is 190 and broadcats is 191
    mask is 255.255.255.192
    192.168.10.208/28
    host range 192.168.10.209 to 192.168.10.222 and broadcast is 223
    mask is 255.255.255.240

    can anyone tell me how can i summrize these into one summary
    thanks
    Khalil

  4. #4
    Join Date
    Jul 2008
    Posts
    211

    Default

    These three networks don't really summarize.

    Do you mean for the second network to be 192.168.20.128/26?

    If so, they summarize as 192.168.10.0/24.

    To summarize these, you have to find the right-most bit in all three network addresses that are identical to one another. Looking at the fourth octet for these three networks, you'd have:

    192.168.10.0/25 = 0
    192.168.10.128/26 = 10
    192.168.10.208/29 = 11010

    Because the top address has a 25 bit mask, only the first bit in the last octet can be compared between the three networks. Since they don't match, you have to move to the left one bit, to the 24th bit. This one matches on all three. Since it does, you build your mask to mask away all the bits to the right of it. So the summary address is 192.168.10.0/24.

  5. #5
    Join Date
    Nov 2008
    Location
    Birmingham, UK
    Posts
    1,440

    Default

    Be careful when using that method to summarise, as you may also include other subnets in that range that you do not want part of the summarisation. It's best to summarise networks that are contiguous to prevent problems with routing protocols later on. This means designing your network with expansion in mind!
    CCNP R&S, CCNA DC
    Currently studying: CCIE R&S, CCNP Data Centre
    Follow my CCIE progress with study notes on my blog: http://beyondccna.blogspot.co.uk/

  6. #6
    Join Date
    Jul 2008
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    211

    Default

    Thanks for mentioning that Fuzz. If you were to:

    1. summarize the networks mentioned using this method, and
    2. there was another subnet that in any way used the 192.168.10.224/28 network, and
    3. the path to that subnet was different than the path to the summarized networks, then

    you'd have a routing mess on your hands. If you must summarize these networks, DON'T use any or all of 192.168.10.224/28.

  7. #7
    Join Date
    Jul 2009
    Posts
    13

    Default

    Fuzz can you please brief me in detial cuz i still have problem in summrization!!! like i have these networks
    172.16.1.0/24
    172.16.2.0/24
    172.16.4.0/24
    172.16.7.0/24
    172.16.10.0/24
    what farmula shud i use here?
    Fuuz im waiting for you're reply

  8. #8
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    Jul 2009
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    Default

    i have exam after 2 weeks

  9. #9
    Join Date
    Nov 2008
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    Birmingham, UK
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    Default

    Well you can summarise these networks into one address, but you will also include subnets that are not on the list this way.

    If you want to put them into one address, you look at how many subnets you have, and what addresses they have. Then you decide what your block size will be (based on how many subnets) and if they will fit into one block (remember you have to stick within normal subnetting rules when supernetting.)

    So you have subnets ranging from 172.16.1.0 to 172.16.10.0. You will need a block size of 16 to fit these in. Even though you only have 5 subnets, you can't use a block size of 8 because the last subnet (172.16.10.0) would fall outside the block. Now you have to decide what your starting address will be. We will have to use 172.16.0.0 because we can't start our block of 16 on anything but 0 or a multiple of 16. We need to reclaim 4 bits from the subnet mask to supernet because 2 ^ 4 = 16, so our summarised address becomes 172.16.0.0/20.

    Now as I said you have also included subnets that were not on the list. 172.16.0.0/24, 172.16.3.0/24 etc. There are in fact eleven /24 subnets included in this one summarisation that were not in the list to be summarised. You've used a block size of 16, but only 5 subnets were on the list. To save space, you could summarise all but the last address, and use two addresses in total: 172.16.0.0/21 & 172.16.10.0/24. This would save you seven /24 addresses.
    CCNP R&S, CCNA DC
    Currently studying: CCIE R&S, CCNP Data Centre
    Follow my CCIE progress with study notes on my blog: http://beyondccna.blogspot.co.uk/

  10. #10
    Join Date
    Feb 2009
    Posts
    100

    Default

    Quote Originally Posted by khalillaghari View Post
    hi all
    this is my 1st post here
    please can anyone tell me easy way to understand subnetting vlsm summrization
    What part of it has you confused?

  11. #11
    Join Date
    Jul 2009
    Posts
    13

    Default

    i have little bit confusion in summrization ..
    A routter has a summary route to network 192.168.32.0/20 installed in its routing table. What range of networks are summarized by this route ?

    - 192.168.0.0 - 192.168.32.0/24
    - 192.168.0.0 - 192.168.47.0/24
    - 192.168.32.0 - 192.168.47.0/24
    - 192.168.32.0 - 192.168.48.0/24
    - 192.168.32.0 - 192.168.63.0/24

    Mask 20. Next boundary is 24. 24-20=4 2^4=16 block size
    So what is tha answer ? - 192.168.32.0 - 192.168.48.0/24

    I do not know because I do not have the whole range of subnets.

    Please clarify me in detial
    Thank You

  12. #12
    Join Date
    Jul 2008
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    Default

    Of the answers provided, 192.168.32.0 - 192.168.47.0 would be the only correct answer.

    All address bits that that fall under the subnet mask (leftmost 20 bits) must remain the same as the summary address for a network address to be a subnet of the summary address.

    Summary Address and Subnet Mask
    11000000 10101000 00100000 00000000 - 192.168.32.0
    11111111 11111111 11110000 00000000 - 255.255.240.0

    Of all the networks listed in the answers, the following all have different bits in the masked portion:

    192.168.0.0 - 19th bit different- This makes answers A & B wrong.
    192.168.48.0 - 20th bit different - This makes answer D wrong.
    192.168.63.0 - 20th bit different - This makes answer E wrong.

    C is the only possible answer.

    192.168.32.0 - all masked bits identical
    192.168.47.0 - all masked bits identical

  13. #13
    Join Date
    Jul 2009
    Posts
    13

    Default

    thanks dear but i have other question if u can asnwer
    can ya please help me with this one:

    the question is:

    Given these subnets:
    192.168.10.0 /24
    192.168.11.0 /24
    192.168.12.0 /24
    192.168.13.0 /24
    192.168.14.0 /24
    192.168.15.0 /24
    ... what's the summary address / mask?

    Now, my answer is 192.168.8.0/21

    is that correct?

    Please let me know

  14. #14
    Join Date
    Jul 2009
    Posts
    13

    Default

    192.168.16.0 for this one
    summary wud be 192.168.16.0/20
    am i correct

  15. #15
    Join Date
    Jul 2008
    Posts
    211

    Default

    Yes! 192.168.8.0/21 is correct for the full example you cited.

    I don't know what you're talking about in your last post with the 192.168.16 address.

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