70-291 mspress book - question on vlsm

Discussion in 'Windows Server' started by blacksox, Jun 17, 2004.

  1. blacksox

    blacksox Guest

    I'm studying for the 291 cert and I'm stuck on Chapter 2,
    Lesson 3, Lesson Review question 1:

    You are the network administrator for the Philly branch
    office of a large company. The IT dept. at co.
    headquarters in NY has assigned you the network address
    space 172.16.0.0/1 to accommodate the entire Philly
    branch network. The network consists of 4 subnets with
    40 hosts each, to which you have already assigned the
    network address 172.16.0.0/24, 172.16.1.0/24,
    172.16.2.0/24, and 172.16.3.0/24. All routers within the
    organization support both CIDR and VLSMs.

    Using your current addressing scheme and without changing
    the subnet mask you have configured for your network, how
    many more subnets will you be able to accommodate in the
    Philly branch office?


    Whew, that was a lot of typing. I looked at the answer
    but I don't understand how they got it. Please help. if
    yu want to know the answer it's...



    4
    Thanks
     
    blacksox, Jun 17, 2004
    #1
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  2. "172.16.0.0/1" Looks like a typo. If the assigned network were
    172.16.0.0/21, then the correct answer would be 4: 172.16.4.0/24,
    172.16.5.0/24, 172.16.6.0/24, 172.16.7.0/24.

    Doug Sherman
    MCSE Win2k/NT4.0, MCSA, MCP+I, MVP
     
    Doug Sherman [MVP], Jun 17, 2004
    #2
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  3. blacksox

    blacksox Guest

    Oh, sorry it is /21.

    The IT dept. at co.
    headquarters in NY has assigned you the network address
    space 172.16.0.0/21 to accommodate the entire Philly
    branch network.

    But how do you get the answer?
     
    blacksox, Jun 17, 2004
    #3
  4. Ok, there are lots of different ways to explain subnetting problems. Some
    explanations may click for you, while others may sound like a foreign
    language. Bearing that in mind:

    Whenever you have a subnet problem where you are given a starting address
    block - eg 172.16.0.0/21 - start by identifying the part you cannot change -
    ie the network portion. With 172.16.0.0/21 the network part is 172.16. plus
    the first 5 bits (reading left to right) of the 3rd octet. AND those first 5
    bits must all be zeros. So, the network part could be represented:
    172.16.00000xxx.xxxxxxxx. If you apply a 24 bit mask to this, the possible
    networks are defined by the last 3 bits of the 3rd octet. In binary the
    maximum value of those 3 bits is 111 which in decimal = 4+2+1 = 7. So the
    highest number network we can have with a 24 mask is 172.16.7.0. If you went
    up to 172.16.8.0, you would violate the 5th (reading left to right) bit of
    the octet because in binary 8 = 00001000. This would be outside the address
    block defined by 172.16.0.0/21. Once we have identified 7 as the highest
    permissible number for the third octet, we know that the remaining unused
    values are 6, 5, and 4.

    If that kind of explanation isn't clicking, there are online resources
    available, plus many cheap prep guides for Microsoft's old TCP/IP exam
    70-59.

    Doug Sherman

    MCSE Win2k/NT4.0, MCSA, MCP+I, MVP
     
    Doug Sherman [MVP], Jun 18, 2004
    #4
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