IP Subnetting Tips&Tricks - Study + Workbook Package

Discussion in 'Cisco' started by nb, Sep 24, 2008.

  1. nb

    nb Guest

    Hi,

    Our Team put together 2 IP Subnetting Books to help cut through the
    clutter and confusion of subnetting.

    It's hard work, but it's work that pay's off on the Job when your boss
    asks you to subnet an address space. :) Once you master it, it is
    just practice and speed.

    The Study Guide + Workbook Package helps you accomplish this.
    It takes you step by step with explanations.

    So what's in the package?

    ##Learn IPv4 Subnetting with Step by Step Techniques
    -Quickly calculate subnets and hosts with number line technique
    -Two books - Study Guide & Workbook Package
    -Proven Industry Tips and Tricks from the Field
    -Easy to understand Techniques


    ##Ideal for:
    Network Administrators - working on Cisco, Juniper, Foundry and other
    Platforms
    Students - pursuing Cisco CCNP/CCNA, Juniper JNCIA/JNCIS
    Job Retraining - a little rusty or just needing more practice


    ##Content:
    -Real-world job tasks on subnet allocation
    -Step-by-step explanations
    -Workbook examples and explanations
    -Number line tricks and tips
    -Allocating number of subnets given an aggregate network
    -Diagrams showing address assignments in real-world environments
    -Subnet zero usage
    -Point to point tips
    -Address summarization techniques
    -Determine number of hosts and subnets
    -Network allocation techniques


    ##What you'll get:

    **Two Books (PDF format)**
    1. Study Guide - subnet tutorial, tips/techniques/tricks and diagrams
    2. Workbook - explanations, real world topologies and diagrams

    ##Cost:
    $38

    *Free shipping - available immediately. I will email you the document
    directly.

    **Email for purchase - Accepting Payments via Paypal.
    nb, Sep 24, 2008
    #1
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  2. nb

    nb Guest

    On Sep 24, 7:40 am, nb <> wrote:
    > Hi,
    >
    > Our Team put together 2 IP Subnetting Books  to help cut through the
    > clutter and confusion of subnetting.
    >
    > It's hard work, but it's work that pay's off on the Job when your boss
    > asks you to subnet an address space.  :)  Once you master it, it is
    > just practice and speed.
    >
    > The Study Guide + Workbook Package helps you accomplish this.
    > It takes you step by step with explanations.
    >
    > So what's in the package?
    >
    > ##Learn IPv4 Subnetting with Step by Step Techniques
    > -Quickly calculate subnets and hosts with number line technique
    > -Two books - Study Guide & Workbook Package
    > -Proven Industry Tips and Tricks from the Field
    > -Easy to understand Techniques
    >
    > ##Ideal for:
    > Network Administrators - working on Cisco, Juniper, Foundry and other
    > Platforms
    > Students - pursuing Cisco CCNP/CCNA, Juniper JNCIA/JNCIS
    > Job Retraining - a little rusty or just needing more practice
    >
    > ##Content:
    > -Real-world job tasks on subnet allocation
    > -Step-by-step explanations
    > -Workbook examples and explanations
    > -Number line tricks and tips
    > -Allocating number of subnets given an aggregate network
    > -Diagrams showing address assignments in real-world environments
    > -Subnet zero usage
    > -Point to point tips
    > -Address summarization techniques
    > -Determine number of hosts and subnets
    > -Network allocation techniques
    >
    > ##What you'll get:
    >
    > **Two Books (PDF format)**
    > 1. Study Guide - subnet tutorial, tips/techniques/tricks and diagrams
    > 2. Workbook - explanations, real world topologies and diagrams
    >
    > ##Cost:
    > $38
    >
    > *Free shipping - available immediately. I will email you the document
    > directly.
    >
    > **Email for purchase - Accepting Payments via Paypal.
    >
    nb, Sep 25, 2008
    #2
    1. Advertising

  3. nb

    Scott Perry Guest

    Check it out: This is FREE!

    **********
    BINARY METHOD:
    **********
    Given an IP address and subnet mask:
    192.168.50.3
    255.255.240.0
    You convert to binary:
    11000000.10101000.00110010.00000011
    11111111.11111111.11110000.00000000
    If you break the IP address into its network portion host portion based on
    where the 1's and 0's in the subnet mask are seperating the address:
    11000000.10101000.00110000.00000000 = 192.168.48.0
    00000000.00000000.00000010.00000011 = 0.0.2.3
    CHECK: Add these two decimal format IP address portions together, and you
    end up with the original address:
    192.168.48.0
    0.0.2.3
    +_____________
    192.168.50.3
    Now that you have verified your binary math, convert all of the host bits in
    192.168.48.0 from 0's to 1's:
    11000000.10101000.00111111.11111111 = 192.168.63.255
    Figure up the number of binary bits in the host portion and figure up 2 to
    the nth power for that value:
    2^12 = 4096
    Now you have determined that the IP address is in a subnet with the
    following information:
    Lowest IP address: 192.168.48.0
    Lowest IP address: 192.168.63.255
    Number of IP addresses in the subnet: 4096 with only 4094 host
    addresses available

    **********
    MY METHOD:
    **********
    Given an IP address and subnet mask:
    192.168.50.3
    255.255.240.0
    Perform some odd subnet math subtracting each octet from 256 and then
    multiplying the differences together:
    256 256 256 256
    255 255 240 0
    -_________________
    1 x 1 x 16 x 256 = 4096
    Now take that "funny number" in the subnet mask subtracted from 256 and
    start counting ranges from 0 by that number and then look for that the value
    in the corresponding octet of the IP address:
    256 - 240 (3rd octet of subnet mask) = 16
    Couting by 16s, looking for 50 (3rd octet of IP address): 0...15,
    16...31, 32...47, 48...63 FOUND 50!
    Insert those low and high ranges with a 0 and 255 after them, respectively:
    192.168.*48*.*0*
    192.168.*63*.*255*
    That is it. You just determined the low and high IP address in a range and
    the size of the range - no binary math!
    Scott Perry, Sep 25, 2008
    #3
  4. nb

    Sam Wilson Guest

    In article <48dbae74$0$3725$>,
    "Scott Perry" <scott.perry@somecompany> wrote:

    > Check it out: This is FREE!
    >
    > **********
    > BINARY METHOD:
    > **********
    > Given an IP address and subnet mask:
    > 192.168.50.3
    > 255.255.240.0
    > You convert to binary:
    > 11000000.10101000.00110010.00000011
    > 11111111.11111111.11110000.00000000


    You don't need to convert anything to binary where the mask is 255 or 0.
    For the octet which isn't 0 or 255 you only need 7 values for the mask
    corresponding to 1 to contiguous bits at the left. If you can't
    remember them then stick them on your wall, or on a card in your wallet,
    or in a file somewhere.

    > If you break the IP address into its network portion host portion based on
    > where the 1's and 0's in the subnet mask are seperating the address:
    > 11000000.10101000.00110000.00000000 = 192.168.48.0
    > 00000000.00000000.00000010.00000011 = 0.0.2.3
    > CHECK: Add these two decimal format IP address portions together, and you
    > end up with the original address:
    > 192.168.48.0
    > 0.0.2.3
    > +_____________
    > 192.168.50.3
    > Now that you have verified your binary math, ...


    Again you only need to do this for one octet = 48+2=50, and only if your
    mask length is not a multiple of 8.

    > ... convert all of the host bits in
    > 192.168.48.0 from 0's to 1's:
    > 11000000.10101000.00111111.11111111 = 192.168.63.255


    Again you only need to do max one octet, the ones to the left of the
    octet where the mask falls are the same as the address, the ones to the
    right are 255.

    > Figure up the number of binary bits in the host portion and figure up 2 to
    > the nth power for that value:
    > 2^12 = 4096


    Again you can have a table or use a calculator if you can't remember the
    numbers.

    > Now you have determined that the IP address is in a subnet with the
    > following information:
    > Lowest IP address: 192.168.48.0
    > Lowest IP address: 192.168.63.255
    > Number of IP addresses in the subnet: 4096 with only 4094 host
    > addresses available


    Hooray!

    > **********
    > MY METHOD:
    > **********
    > Given an IP address and subnet mask:
    > 192.168.50.3
    > 255.255.240.0
    > Perform some odd subnet math subtracting each octet from 256 and then
    > multiplying the differences together:
    > 256 256 256 256
    > 255 255 240 0
    > -_________________
    > 1 x 1 x 16 x 256 = 4096


    Any mask octet that is 255 is irrelevant so you can ignore it.

    > Now take that "funny number" in the subnet mask subtracted from 256 and
    > start counting ranges from 0 by that number and then look for that the value
    > in the corresponding octet of the IP address:
    > 256 - 240 (3rd octet of subnet mask) = 16
    > Couting by 16s, looking for 50 (3rd octet of IP address): 0...15,
    > 16...31, 32...47, 48...63 FOUND 50!
    > Insert those low and high ranges with a 0 and 255 after them, respectively:
    > 192.168.*48*.*0*
    > 192.168.*63*.*255*


    Jolly good - you did your math on one octet, just like you could do with
    the binary version, except that you've done it in decimal and by trial
    and error, and people need to know or calculate multiples of 16 (or 32
    or 64 or 128, or 8 or 4 or 2 - how many multiples of 4 do you have to
    test to find the enclosing range of 233, say?).

    And where did you check your math?

    > That is it. You just determined the low and high IP address in a range and
    > the size of the range - no binary math!


    But why are you scared of binary math[1]? I need to remember 1, 2, 4,
    8, 16, 32, 64 and 128, and (in the other direction) 192, 224, 240, 248,
    252 and 254 (0, 255 and 256, of course). You need to remember all the
    multiples of 4, 8, 16 etc up to 256. You can work those numbers out,
    but I can work out mine too.

    I've never been able to see why this is any more memorable or
    understandable than knowing how the computer actually does it
    internally. Could someone who prefers this method explain why it's
    better?


    Sam

    [1] On this side of the pond we say binary maths, just to be awkward.
    Sam Wilson, Oct 6, 2008
    #4
  5. nb

    Mitch@_._ Guest


    >
    >You don't need to convert anything to binary where the mask is 255 or 0.
    >For the octet which isn't 0 or 255 you only need 7 values for the mask
    >corresponding to 1 to contiguous bits at the left.


    We learned the "magic number" technique in my class, but I find that I
    do much better thinking in binary and using ANDing. I can still use
    the trick to check if I need to, but it helps me to visualize the
    subnet and host bits and how they divide.
    Mitch@_._, Oct 16, 2008
    #5
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