Free Subnetting

Discussion in 'MCSA' started by RogueIT, Mar 25, 2008.

  1. RogueIT

    RogueIT Guest

    I am trying to find a couple of ways to practice for the 291 and was
    wondering if a couple of people would post some subnetting questions for me
    to try and figure out. Nothing beyond what 291 is going to require and please
    be able to give me the answer when I cry uncle...that is if you all don't
    mind.
    thanks,
    Rogue
     
    RogueIT, Mar 25, 2008
    #1
    1. Advertising

  2. http://www.subnettingquestions.com

    "RogueIT" <> wrote in message
    news:...
    > I am trying to find a couple of ways to practice for the 291 and was
    > wondering if a couple of people would post some subnetting questions for
    > me
    > to try and figure out. Nothing beyond what 291 is going to require and
    > please
    > be able to give me the answer when I cry uncle...that is if you all don't
    > mind.
    > thanks,
    > Rogue
     
    Mostly Gizzards, Mar 25, 2008
    #2
    1. Advertising

  3. RogueIT

    IT Geek Guest

    Here you go
    you have 40 servers and 800 clients, you need to subnet 10.10.0.0/16 network
    to only accomidate what is needed. Provide the starting and ending IP
    address, and the new subnet mask.
    "RogueIT" <> wrote in message
    news:...
    >I am trying to find a couple of ways to practice for the 291 and was
    > wondering if a couple of people would post some subnetting questions for
    > me
    > to try and figure out. Nothing beyond what 291 is going to require and
    > please
    > be able to give me the answer when I cry uncle...that is if you all don't
    > mind.
    > thanks,
    > Rogue
     
    IT Geek, Mar 25, 2008
    #3
  4. RogueIT

    RogueIT Guest

    First off thanks...
    Second...my thought process may be a little funny so I will try to explain
    my thinking
    840 clients (pc's and server) takes up 10 bits. leaving me with a subnet of
    255.255.252.0
    starting IP is 10.10.0.1 if I get 253 addresses from each 10.10.x subnet
    then 10.10.0.x, 10.10.1.x, and 10.10.2.x gives me 759 available address
    leaving me 81 address to pick up in the 10.10.3.x subnet...bringing me to
    10.10.3.82...
    well it made sense at the time...
    I figure I am either right on or not even in the stadium.
    thanks
    Rogue


    "IT Geek" wrote:

    > Here you go
    > you have 40 servers and 800 clients, you need to subnet 10.10.0.0/16 network
    > to only accomidate what is needed. Provide the starting and ending IP
    > address, and the new subnet mask.
     
    RogueIT, Mar 25, 2008
    #4
  5. RogueIT

    IT Geek Guest

    Nice
    But I think you made it more difficult than it needed to be.
    To accomidate for the 840 address you just needed to brower from the host id
    which would have made your new subnet /22 and that would have giving you
    1022.
    /24 = 256
    /23 = 512
    /22 = 1024
    Then you would need to remove the first and last because of network and
    broadcast.

    You are right I should have told you that I just wanted the know what IP
    address would you assign to the clients, but you where on the right track
    "RogueIT" <> wrote in message
    news:...
    > First off thanks...
    > Second...my thought process may be a little funny so I will try to explain
    > my thinking
    > 840 clients (pc's and server) takes up 10 bits. leaving me with a subnet
    > of
    > 255.255.252.0
    > starting IP is 10.10.0.1 if I get 253 addresses from each 10.10.x subnet
    > then 10.10.0.x, 10.10.1.x, and 10.10.2.x gives me 759 available address
    > leaving me 81 address to pick up in the 10.10.3.x subnet...bringing me to
    > 10.10.3.82...
    > well it made sense at the time...
    > I figure I am either right on or not even in the stadium.
    > thanks
    > Rogue
    >
    >
    > "IT Geek" wrote:
    >
    >> Here you go
    >> you have 40 servers and 800 clients, you need to subnet 10.10.0.0/16
    >> network
    >> to only accomidate what is needed. Provide the starting and ending IP
    >> address, and the new subnet mask.

    >
     
    IT Geek, Mar 26, 2008
    #5
  6. RogueIT

    Dude Guest

    To understand subnetting, you need to think Binary instead of Decimal.

    For example 255.255.255.240 or 192.168.0.0/28
    11110000 in Binary is 240 in Decimal for example. This gives you a
    1111xxxx Mask for 16 subnets and
    xxxx0000 For 16 host addresses (minus 2)

    Or this one 255.255.255.224 or 192.168.0.0/26
    11100000 in Binary is 224 in Decimal for example This gives you a
    111xxxxxx Mask for 8 subnets and
    xxx00000 for 32 host addresses (minus 2)

    I find it faster and easier to convert exam questions like this to Binary
    instead of trying to remember tables or fancy formulas. But if you want to
    memorize tables, 224, 240 and 248 are the most common subnets to remember.



    "RogueIT" wrote:

    > I am trying to find a couple of ways to practice for the 291 and was
    > wondering if a couple of people would post some subnetting questions for me
    > to try and figure out. Nothing beyond what 291 is going to require and please
    > be able to give me the answer when I cry uncle...that is if you all don't
    > mind.
    > thanks,
    > Rogue
     
    Dude, Mar 28, 2008
    #6
  7. RogueIT

    Dude Guest

    "Dude" wrote:

    > To understand subnetting, you need to think Binary instead of Decimal.
    >
    > For example 255.255.255.240 or 192.168.0.0/28
    > 11110000 in Binary is 240 in Decimal for example. This gives you a
    > 1111xxxx Mask for 16 subnets and
    > xxxx0000 For 16 host addresses (minus 2)
    >
    > Or this one 255.255.255.224 or 192.168.0.0/26
    > 11100000 in Binary is 224 in Decimal for example This gives you a
    > 111xxxxxx Mask for 8 subnets and
    > xxx00000 for 32 host addresses (minus 2)


    Correction:
    It should be 192.168.0.0/27 not 26
    If it was 192.168.0.0/26 it would be
    11000000 in Binary or 192 in decimal for a
    11xxxxxx mask for 4 subnets and
    xx000000 for 64 host addresses (minus 2)

    >
    > I find it faster and easier to convert exam questions like this to Binary
    > instead of trying to remember tables or fancy formulas. But if you want to
    > memorize tables, 224, 240 and 248 are the most common subnets to remember.
    >
    >
    >
    > "RogueIT" wrote:
    >
    > > I am trying to find a couple of ways to practice for the 291 and was
    > > wondering if a couple of people would post some subnetting questions for me
    > > to try and figure out. Nothing beyond what 291 is going to require and please
    > > be able to give me the answer when I cry uncle...that is if you all don't
    > > mind.
    > > thanks,
    > > Rogue
     
    Dude, Mar 28, 2008
    #7
    1. Advertising

Want to reply to this thread or ask your own question?

It takes just 2 minutes to sign up (and it's free!). Just click the sign up button to choose a username and then you can ask your own questions on the forum.
Similar Threads
  1. $teve.H

    Question about subnetting on MS Exams

    $teve.H, Nov 10, 2005, in forum: Microsoft Certification
    Replies:
    7
    Views:
    830
    msnews.microsoft.com
    Jan 11, 2006
  2. AC

    Subnetting

    AC, Jul 18, 2003, in forum: Cisco
    Replies:
    8
    Views:
    982
  3. Jorge Paredes

    Subnetting problem?

    Jorge Paredes, Aug 13, 2004, in forum: Cisco
    Replies:
    1
    Views:
    1,034
  4. TeamGracie

    PIX subnetting question

    TeamGracie, Jan 12, 2005, in forum: Cisco
    Replies:
    2
    Views:
    677
    Walter Roberson
    Jan 12, 2005
  5. Phil Schuman

    subnetting - RFC1878 - 000's & 111's

    Phil Schuman, Jan 27, 2005, in forum: Cisco
    Replies:
    2
    Views:
    586
    Doug McIntyre
    Jan 28, 2005
Loading...

Share This Page