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  1. #1
    Join Date
    Jul 2011
    Posts
    3

    Default Class B Private Reserved Address Space Question?

    I purchased Todd's new 6th Edition CCNA Study Guide for the 640-802 Exam. I just finished chapter three and need help understanding Private Reserved Address Spaces.

    I'd like to know if these address spaces can be calculated or if they're supposed to just be remembered?

    I think I understand why the Class A Reserved Address Space is 10.0.0.0 - 10.255.255.255 and why Class C is 192.168.0.0 - 192.168.255.255 . However, the Class B Reserved Address Space of 172.16.0.0-172.31.255.255 doesn't make sense to me.

    I've searched online for a method to calculate this hoping to understand how/why/where the 172.31 comes from but haven't had any success.

    I'd greatly appreciate any help in understanding this. If I can understand it, itll help me understand how to arrive at the correct answer for #8 in the Written Lab 3.1: TCP/IP Section of the End of Chapter questions.

    Thanks,

    Muziq

  2. #2
    Join Date
    Mar 2008
    Posts
    2,887

    Default

    I just remember them - i never looked into how or why they were created this way. I am sure there is a logical reason, take peep over the RFC 1918

    http://tools.ietf.org/html/rfc1918
    Maddox Thomas-Clark 14/10/2008
    Bean Thomas-Clark 18/09/2007
    Big Evils Cisco World
    Linkedin

  3. #3
    Join Date
    Nov 2008
    Location
    Birmingham, UK
    Posts
    1,456

    Default

    If you look at the ranges of each private class in CIDR notation:

    10.0.0.0 /8
    172.16.0.0 /12
    192.168.0.0 /16

    The mask goes up by 4 bits each time. As each class has a specific subnet mask (8, 16 and 24 repectively) this gives us one class A address - 10.0.0.0 /8, sixteen class B addresses - 172.16.0.0/16, 172.17.0.0 /16 .... 172.31.0.0 /16 and 256 class C addresses - 192.168.0.0 /24 ... 192.168.254.0 /24.

    Compare the bit masks from the whole IP address space and the defaults for address class:

    A: 8 - 8 = 2^0 = 1 address
    B: 16 - 12 = 2^4 = 16 addresses
    c: 24 - 16 = 2^8 = 256 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/

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