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CIDR

Discussion in 'General Microsoft Certifications' started by syntax_error, Dec 1, 2004.

  1. syntax_error

    syntax_error Bit Poster

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    Hi guys was wondering if you could clear somthing up for me

    I'm not 100% sure I know how to work out a IP subnet mask using CIDR for example

    192.168.22.55/22 subnet here would be 255.255.254.0 right ?

    know by looking at the CIDR table you get 1,002 host but how ? you cant enter the /22 when you configure your IP address and I know the computer only sees bits of the 3rd octet so does that mean I can put any number in the 3rd octet and all the computers can still talk to each other ?

    Any info you guys can give would be great I just can't seem to get my head around this atm.
     
    Certifications: mcp server2k
    WIP: Network Infrastructure
  2. tripwire45
    Honorary Member

    tripwire45 Zettabyte Poster

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    You might want to start with this:

    http://compnetworking.about.com/od/workingwithipaddresses/l/aa021003a.htm
     
    Certifications: A+ and Network+
  3. Luton Bee

    Luton Bee Kilobyte Poster

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    I'll probably get this wrong but WTF!!!

    A /22 network gives you a mask of 255.255.252.0 this means that your network uses 22 bits of the 32 leaving 10 for the host portion. You therfore have 2 to the power 10 = 1024 possible host addresses (except that you need to subtract 2 from that total, one for the network and one for the broadcast address). So the actual range is as follows:
    192.168.252.0 - Network address
    192.168.252.1 - First usable address
    192.168.255.254 - Last usable
    192.168.255.255 - Broadcast

    Er I think????
     
    Certifications: MCSE, MCSA, MCP, A+, Network+ C&G ICT
    WIP: CCNA
  4. Luton Bee

    Luton Bee Kilobyte Poster

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    Giving it some more thought isn't 192.168.22.55/22 an invalid address????
     
    Certifications: MCSE, MCSA, MCP, A+, Network+ C&G ICT
    WIP: CCNA
  5. syntax_error

    syntax_error Bit Poster

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    dunno I just chose random numbers but you have given me it in a form I can understand. =)

    Ok so here is another

    213.113.188.99/15

    255.127.0.0 - Network address
    255.127.1.1 - First usable address
    255.127.255.254 - Last usable
    255.127.255.255 - Broadcast

    So you only lose 2 host address and the amount of host avaible is 2 to the power of 17 ?

    PS Tripwire that is a pukka site mate :rocks
     
    Certifications: mcp server2k
    WIP: Network Infrastructure
  6. tripwire45
    Honorary Member

    tripwire45 Zettabyte Poster

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    Is that good? :blink :oops:
     
    Certifications: A+ and Network+
  7. Luton Bee

    Luton Bee Kilobyte Poster

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    As I predicted earlier, I got this wrong last night!!!

    The address 192.168.22.55 wih a subnet mask of 255.255.252.0 is a valid address. Using this subnet would give you 61 subnets or segments each containing upto 1022 hosts, and the address you quoted would sit in the subnet
    192.168.20.0 - Network address
    192.168.20.1 - First usable
    192.168.23.254 - Last usable
    192.168.23.255 - Broadcast.

    Now if you'll excuse me, I'm off for a lie down!!!!![​IMG]
     
    Certifications: MCSE, MCSA, MCP, A+, Network+ C&G ICT
    WIP: CCNA
  8. Bluerinse
    Honorary Member

    Bluerinse Exabyte Poster

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    Yes Trip pukka is good, it is like "the dogs" (short for, the dogs bo!!ok$) or sweet, meaning good, or *wicked*, meaning good or down here *fair dinkums* meaning anything good or fair or sometimes used just to mean "oh really" :)

    And I thought you knew English lol

    Pete
     
    Certifications: C&G Electronics - MCSA (W2K) MCSE (W2K)
  9. tripwire45
    Honorary Member

    tripwire45 Zettabyte Poster

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    Only in America. :biggrin :oops:
     
    Certifications: A+ and Network+

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