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Little help with subnetting

Discussion in 'Network+' started by simonp83, May 8, 2010.

  1. simonp83

    simonp83 Kilobyte Poster

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    Not doing the Network+, infact i'm actually asking this after going through Chapter 6 of the press kit for 70-680, but thought it'd be better to ask this here, just to make sure i'm understanding subnets correctly.

    Let's say you have a router with an ip address of 10.0.0.129/28 and a subnet mask of 255.255.255.128. The full list of appropriate ip addresses that can be used for machines on the subnet would range from 10.0.0.130 to 10.0.0.144? ? ?
     
    Certifications: A+, MCP, MCDST, MCTS, MCITP
    WIP: 70-291
  2. gosh1976

    gosh1976 Kilobyte Poster

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    wow I got rusty with subnetting pretty quickly. but if the cidr notation is /28 then the subnet would be 255.255.255.240. then the range of IP addresses would be x.x.x.128 to x.x.x.143

    um I think anyway... I'm off to do some practice!
     
    Last edited: May 8, 2010
    Certifications: A+, Net+, MCDST, CCENT, MCTS: Win 7 Configuring, CCNA
  3. simonp83

    simonp83 Kilobyte Poster

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    So the 255.255.255.240 - this would denote how many ip addresses are available within the range and the router would denote where the ip range starts? 255-240, would denote the 15 with the ip of the router being the 16th i.e. the 0000?
     
    Certifications: A+, MCP, MCDST, MCTS, MCITP
    WIP: 70-291
  4. pete.grant

    pete.grant Byte Poster

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    I highly recommend the following site:

    http://www.subnetting-secrets.com/index.html

    I used to really struggle with subnetting as I'm not particularly good with maths/numbers but this made it so simple - well worth £20!
     
    Certifications: A+ IT Technician, CCENT, CEH, CPTS, CIW Security Analyst, ITIL v3 Foundation, Master CIW Administrator, MCITP (Windows Server 2008:SA), MCSA on Windows Server 2008, MCSA:Security on Windows Server 2003, MCTS (70-648, 70-652), Network+, SCNS, Security+, Server+
  5. gosh1976

    gosh1976 Kilobyte Poster

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    with an ip address of 10.0.0.129/28 using anding you can see that the network/subnet id is 10.0.0.128 and the broadcast ip is 10.0.0.143 so that means the first host is 10.0.0.129 and the last host is 10.0.0.142. the router is going to be assigned the first usable IP in the range.

    these two sites will help http://www.learntosubnet.com/ and http://www.professormesser.com/free...comptia-network-certification-training-course just scroll down to the two subnetting videos on the messer site. both are free
     
    Last edited: May 8, 2010
    Certifications: A+, Net+, MCDST, CCENT, MCTS: Win 7 Configuring, CCNA
  6. soundian

    soundian Gigabyte Poster

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    I always find it easier to put it into binary
    First three octets are easy
    10.0.0
    and
    255.255.255
    No need for binary there.

    last octet
    10000000
    11110000
    so the first 4 bits are network, leaving the last 4 bits for host address
    That's 16 addresses, from 0000 to 1111
    0000 isn't used*, 1111 is the broadcast address
    that leaves 14 usable addresses from
    0001(1)
    to
    1110(14)

    back on octet form, usable address are
    10000001 (129)
    to
    10001110 (142)
    with 143 as the broadcast address and 128 not used.
    Subnet mask would be 11110000=128+64+32+16=240





    *For the purposes of net+ anyway.
     
    Certifications: A+, N+,MCDST,MCTS(680), MCP(270, 271, 272), ITILv3F, CCENT
    WIP: Knuckling down at my new job
  7. tomshawk

    tomshawk Byte Poster

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    Certifications: MCSE/NT4, MCP/2K3, MCP+I, CCNA, Net+, A+

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