Subnetting Question

Discussion in 'MCSA' started by RogueIT, Apr 23, 2008.

  1. RogueIT

    RogueIT Guest

    I just got done working a practice question and I understand the subnet
    part...I don't get how the IP address answers are figured. Below is the
    question is someone could tell me why address x is good for network y
    thanks,
    Rogue
    ****************************************************
    You administer a network that contains 175 machines. Your manager has
    assigned the network
    the IP address 192.168.11.0 with the default subnet mask of 255.255.255.0. A
    router that has
    one WAN interface and eight LAN interfaces connects this network to the
    corporate WAN.
    You want to subnet the network into three subnets, and you want to reserve a
    few addresses for
    a fourth subnet, just in case you need it later. You decide that Subnet A
    will contain 25 computers,
    Subnet B will contain 50 computers, and Subnet C will contain 100 computers.
    In the following exhibit, select the network addresses and subnet masks in
    the Choices column and
    place them in the appropriate boxes in the other three columns. Each item
    may be used only once.
    Choices:
    192.168.11.128
    255.255.255.128
    192.168.11.0
    255.255.255.192
    192.168.11.192
    255.255.255.224

    Network A Network B Network C
    IP Address __________ _________ ________
    Subnet __________ _________ ________

    The network address 192.168.11.192 with a subnet mask of 255.255.255.224 is
    perfect for
    Subnet A because it supports up to 30 hosts. The network address
    192.168.11.128 with a subnet
    mask of 255.255.255.192 is perfect for Subnet B because it supports up to 62
    hosts. The network
    address 192.168.11.0 with a subnet mask of 255.255.255.128 is perfect for
    Subnet C
    because it supports up to 126 hosts. That still leaves the network address
    192.168.11.224 with
    a subnet mask of 255.255.255.224 available for a fourth subnet later.
     
    RogueIT, Apr 23, 2008
    #1
    1. Advertising

  2. RogueIT

    John R Guest

    "RogueIT" <> wrote in message
    news:...
    >I just got done working a practice question and I understand the subnet
    > part...I don't get how the IP address answers are figured. Below is the
    > question is someone could tell me why address x is good for network y
    > thanks,
    > Rogue
    > ****************************************************
    > You administer a network that contains 175 machines. Your manager has
    > assigned the network
    > the IP address 192.168.11.0 with the default subnet mask of 255.255.255.0.
    > A
    > router that has
    > one WAN interface and eight LAN interfaces connects this network to the
    > corporate WAN.
    > You want to subnet the network into three subnets, and you want to reserve
    > a
    > few addresses for
    > a fourth subnet, just in case you need it later. You decide that Subnet A
    > will contain 25 computers,
    > Subnet B will contain 50 computers, and Subnet C will contain 100
    > computers.
    > In the following exhibit, select the network addresses and subnet masks in
    > the Choices column and
    > place them in the appropriate boxes in the other three columns. Each item
    > may be used only once.
    > Choices:
    > 192.168.11.128
    > 255.255.255.128
    > 192.168.11.0
    > 255.255.255.192
    > 192.168.11.192
    > 255.255.255.224
    >
    > Network A Network B Network C
    > IP Address __________ _________ ________
    > Subnet __________ _________ ________
    >
    > The network address 192.168.11.192 with a subnet mask of 255.255.255.224
    > is
    > perfect for
    > Subnet A because it supports up to 30 hosts. The network address
    > 192.168.11.128 with a subnet
    > mask of 255.255.255.192 is perfect for Subnet B because it supports up to
    > 62
    > hosts. The network
    > address 192.168.11.0 with a subnet mask of 255.255.255.128 is perfect for
    > Subnet C
    > because it supports up to 126 hosts. That still leaves the network address
    > 192.168.11.224 with
    > a subnet mask of 255.255.255.224 available for a fourth subnet later.
    >


    Note: the question (or your quoting of it) has an inherent error in it as
    the 192.168.11.128 network will conflict with the other two networks. The
    question should have been...

    Choices:
    192.168.11.0 - 255.255.255.128
    192.168.11.128 - 255.255.255.192
    192.168.11.192 - 255.255.255.224
    leaving 192.168.11.224 - 255.255.255.224

    Given the corrected question...
    You are given three subnets and asked which network should use which subnet.
    One subnet (192.168.11.0) can support up to 126 hosts, one subnet
    (192.168.11.128) can support up to 62 hosts, and one subnet (192.168.11.192)
    can support up to 30 hosts. The requirement is that you need 100, 50, and
    25 hosts on each subnet. Simply plug and play. You can't put the 100 hosts
    or the 50 hosts on the subnet that only supports 30 hosts.

    To find how many hosts a subnet will support, look at the subnet mask. In
    the first example 255.255.255.128, the first 25 bits are for the network ID,
    and this leaves 7 bits for the host id. Take the number 2, raise it to the
    7th power, and then subtract one for the network Id and one for the
    broadcast id, this leaves 126.

    John R
     
    John R, Apr 24, 2008
    #2
    1. Advertising

  3. RogueIT

    Chris Lobban Guest

    John R wrote:
    >
    > "RogueIT" <> wrote in message
    > news:...
    >> I just got done working a practice question and I understand the subnet
    >> part...I don't get how the IP address answers are figured. Below is the
    >> question is someone could tell me why address x is good for network y
    >> thanks,
    >> Rogue
    >> ****************************************************
    >> You administer a network that contains 175 machines. Your manager has
    >> assigned the network
    >> the IP address 192.168.11.0 with the default subnet mask of
    >> 255.255.255.0. A
    >> router that has
    >> one WAN interface and eight LAN interfaces connects this network to the
    >> corporate WAN.
    >> You want to subnet the network into three subnets, and you want to
    >> reserve a
    >> few addresses for
    >> a fourth subnet, just in case you need it later. You decide that Subnet A
    >> will contain 25 computers,
    >> Subnet B will contain 50 computers, and Subnet C will contain 100
    >> computers.
    >> In the following exhibit, select the network addresses and subnet
    >> masks in
    >> the Choices column and
    >> place them in the appropriate boxes in the other three columns. Each item
    >> may be used only once.
    >> Choices:
    >> 192.168.11.128
    >> 255.255.255.128
    >> 192.168.11.0
    >> 255.255.255.192
    >> 192.168.11.192
    >> 255.255.255.224
    >>
    >> Network A Network B Network C
    >> IP Address __________ _________ ________
    >> Subnet __________ _________ ________
    >>
    >> The network address 192.168.11.192 with a subnet mask of
    >> 255.255.255.224 is
    >> perfect for
    >> Subnet A because it supports up to 30 hosts. The network address
    >> 192.168.11.128 with a subnet
    >> mask of 255.255.255.192 is perfect for Subnet B because it supports up
    >> to 62
    >> hosts. The network
    >> address 192.168.11.0 with a subnet mask of 255.255.255.128 is perfect for
    >> Subnet C
    >> because it supports up to 126 hosts. That still leaves the network
    >> address
    >> 192.168.11.224 with
    >> a subnet mask of 255.255.255.224 available for a fourth subnet later.
    >>

    >
    > Note: the question (or your quoting of it) has an inherent error in it
    > as the 192.168.11.128 network will conflict with the other two
    > networks. The question should have been...
    >
    > Choices:
    > 192.168.11.0 - 255.255.255.128
    > 192.168.11.128 - 255.255.255.192
    > 192.168.11.192 - 255.255.255.224
    > leaving 192.168.11.224 - 255.255.255.224
    >
    > Given the corrected question...
    > You are given three subnets and asked which network should use which
    > subnet. One subnet (192.168.11.0) can support up to 126 hosts, one
    > subnet (192.168.11.128) can support up to 62 hosts, and one subnet
    > (192.168.11.192) can support up to 30 hosts. The requirement is that
    > you need 100, 50, and 25 hosts on each subnet. Simply plug and play.
    > You can't put the 100 hosts or the 50 hosts on the subnet that only
    > supports 30 hosts.
    >
    > To find how many hosts a subnet will support, look at the subnet mask.
    > In the first example 255.255.255.128, the first 25 bits are for the
    > network ID, and this leaves 7 bits for the host id. Take the number 2,
    > raise it to the 7th power, and then subtract one for the network Id and
    > one for the broadcast id, this leaves 126.
    >
    > John R


    Except, that I believe what you refer to as the 'inherent error' is
    actually part of the question for people to solve... ;-)
    The numbers that it gives are:
    192.168.11.128
    255.255.255.128
    192.168.11.0
    255.255.255.192
    192.168.11.192
    255.255.255.224
    Not in any particular order. Not paired with each other. 192.168.11.128
    doesn't have to pair with 255.255.255.128 just because they're near each
    other in the list. It's just 6 strings of numbers, and 6 little boxes to
    put them in. ;-)
    So there's actually two steps to solving the question.
    1st/ You have to figure out which Network IDs can match with which
    Subnet Masks without creating any conflicts or any wasted IP segments
    (example: if you give 192.168.11.0 a mask of 255.255.255.224, you're
    wasting a lot of numbers... if you give the 192.168.11.128 subnet a mask
    of 255.255.255.128, then it has overlap with the 192.168.11.192 subnet).
    This is the step that John referred to above as the 'inherent error'.
    2nd/ Then you have to figure which subnet mask can apply to each subnet,
    based on the required number of hosts. This is what John explained above.

    ;-)

    Chris
     
    Chris Lobban, Apr 24, 2008
    #3
  4. RogueIT

    John R Guest

    "Chris Lobban" <> wrote in message
    news:%23%...
    >
    > Except, that I believe what you refer to as the 'inherent error' is
    > actually part of the question for people to solve... ;-)
    > The numbers that it gives are:
    > 192.168.11.128
    > 255.255.255.128
    > 192.168.11.0
    > 255.255.255.192
    > 192.168.11.192
    > 255.255.255.224
    > Not in any particular order. Not paired with each other. 192.168.11.128
    > doesn't have to pair with 255.255.255.128 just because they're near each
    > other in the list. It's just 6 strings of numbers, and 6 little boxes to
    > put them in. ;-)
    > So there's actually two steps to solving the question.


    Yea, you are probably correct Chris, it didn't occur to me that you had to
    mix/match the subnet masks with the addresses. Since he posted them
    address, mask, address, mask, address, mask, I assumed that the masks
    necessarily went with the preceeding addresses. By your method, that would
    make the question a little more challenging. Nice catch!

    John R
     
    John R, Apr 24, 2008
    #4
  5. RogueIT

    RogueIT Guest

    Thanks to you both. I appreciate your time.
    taking 291 on the 12th of may
     
    RogueIT, Apr 29, 2008
    #5
    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:
    899
    msnews.microsoft.com
    Jan 11, 2006
  2. TeamGracie

    PIX subnetting question

    TeamGracie, Jan 12, 2005, in forum: Cisco
    Replies:
    2
    Views:
    703
    Walter Roberson
    Jan 12, 2005
  3. myrt webb

    Subnetting Question

    myrt webb, Oct 10, 2003, in forum: MCSE
    Replies:
    3
    Views:
    714
    Darko Gavrilovic
    Oct 12, 2003
  4. Kendal Emery
    Replies:
    1
    Views:
    478
    chris
    Nov 21, 2003
  5. M D
    Replies:
    10
    Views:
    1,991
Loading...

Share This Page