Subnetting Problem

Discussion in 'MCSE' started by Myrt Webb, Jul 14, 2003.

  1. Myrt Webb

    Myrt Webb Guest

    I need some help with the following subnet problem:

    "You are the new administrator of a 2000 node network.
    There is only one router on the entire network, which
    provides all the computers with Internet access. The
    company's ISP has assigned the following 8 network
    addresses to them:

    10.24.32.0/24
    10.24.33.0/24
    10.24.34.0/24
    10.30.35.0/24
    10.30.36.0/24
    10.30.37.0/24
    10.30.38.0/24
    10.30.39.9/24

    What subnet mask could you use to minimize the complexity
    of the routing tables while maintaining the existing
    Internet connectivity?

    a. 255.255.252.0
    b. 255.255.255.252
    c. 255.255.255.248
    d. 255.255.248.0 "

    The answer is 'd.' Quite frankly I do not understand what
    the problem is getting at nor how 'd' gives the desired
    results.

    Any explainations would be appreciated.
     
    Myrt Webb, Jul 14, 2003
    #1
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  2. Myrt Webb

    moonlighting Guest

    "/24" will tell you that the SNM is 255.255.254.0, but
    not 255.255.248.0 (this is "/21").
     
    moonlighting, Jul 14, 2003
    #2
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  3. Myrt Webb

    Tom Helms Guest

    D. 255.255.248.0
    BTW, /24 subnet mask is 255.255.255.0
     
    Tom Helms, Jul 14, 2003
    #3
  4. Myrt Webb

    Myrt Webb Guest

    I thought the same thing. That /24 indicates a
    255.255.255.? mask. So I figured that either b or c was
    the answer. But according to the MS 70-216 trng kit pg 808
    it is d.

    I understand the basics of subnetting and figured out
    correctly the other questions in this section but I do not
    understand the logic of this question.
     
    Myrt Webb, Jul 14, 2003
    #4
  5. Myrt Webb

    Zenner Guest

    You "borrow" from the Host address to increase the client address range. You
    really need to take the advice to visit the referenced site
    (learntosubnet.com), lots of good stuff there!
     
    Zenner, Jul 14, 2003
    #5
  6. Either the question has been mis-typed here or mis-remembered.
    255.255.248 won't help cover all eight of those addresses, which jump
    from 10.24.*.* to 10.30.*.*. You'd need at least a /13 (255.248.0.0)
    mask to cover all those.

    The question is also bogus because 10.*.*.* is private NAT space, and
    would not be allocated by an ISP -- but even overlooking that, the
    suggestion that you could simply apply one large mask to the whole
    network and call it a day is specious at best. If you were to do so,
    then people on this network would be unable to reach people using the
    same ISP from elsewhere, i.e. 10.27.5.0/24 would be unreachable.

    BAD question, nice troll.
     
    Jeffrey L. Woods, Jul 14, 2003
    #6
  7. Myrt Webb

    Tom Helms Guest

    Unless my calculations are incorrect, subnet mask 255.255.248.0 will allow
    2048 or 2046 host addresses using a class A address to start.
     
    Tom Helms, Jul 14, 2003
    #7
  8. Myrt Webb

    Marko Guest

    Jeffrey Woods has provided the best answer so far.

    Your ISP may not be a total idiot for "assigning" this IP
    space in the first place. It would only work on your
    10.24or30.y.z network and possibly to reach the ISP's
    servers / routers and that is likely it. However, if the
    ISP provides mail and proxy servers (possibly other
    services?) then this IP assignment is OK.

    10.24.x.y to 10.30.a.b would require a mask of 255.248.0.0
    to cover all those IP ranges mentioned.

    The 10.24.x.y ranges could be covered by "route add
    10.24.32.0 mask 255.255.253.0 (gateway IP here)"

    ....Assuming 10.30.39.9/24 is a typo and should be
    10.30.39.0/24 (cause the first is definitely wrong and
    does not exist)...

    Then route add 10.30.35.0 mask 255.255.251.0 (gateway IP)

    So far, neither route table addition looks like our answer.


    HOWEVER, if you concede that it is highly probable that
    all the class C subnets (those finishing with /24) start
    with either 10.24 OR 10.30, then 10.x.32.0 with a mask of
    255.255.248.0 will cover all IPs in the range 10.x.32.0 to
    10.x.39.255. Making "d" the correct choice....


    It it says d is right, then I would suggest the above
    paragraph is the solution.

    Anyone else care to have a go?

    ....care to kudos?

    ....care to flame?

    Marko Cosic






     
    Marko, Jul 15, 2003
    #8
  9. Myrt Webb

    anees.network

    Joined:
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    I have something to say here.

    Let me try an answer for this

    This questions involves something called "ROUTE Summarization".

    10.24.32.0/24
    10.24.33.0/24
    10.24.34.0/24
    10.30.35.0/24
    10.30.36.0/24
    10.30.37.0/24
    10.30.38.0/24
    10.30.39.9/24

    these are the IP Addresses given. The questions says "What subnet mask could you use to minimize the complexity of the routing tables while maintaining the existing Internet connectivity?"

    so instead of just sub-netting you have to take care of minimizing the routing table it generates.

    So we have to make this pool of IPs into two groups. 10.24 and 10.30 group.

    10.24 has 3 IPs in the pool
    10.30 has 5 ips in the pool so we have to consider grouping 5 IP pool to minimize the routing table.

    10.30. 35-39 is the range - group of 8 ip address will accommodate this range
    let me do this in binary

    To get a 8 ip range we have to start from the 5th binary digit 11111000=248 but we need this 248 in the 3rd octet mask to accomplish the range we needed because first two octets are same ( 10.30.)

    so the mask should be 255.255.248.0.

    eventually,
    10.30.35.0/24
    10.30.36.0/24
    10.30.37.0/24 ---->255.255.248.0 is the mask you need
    10.30.38.0/24
    10.30.39.9/24

    For 10.24 series you can use the same mask because it lies within the range but route summarization is not needed as there are only 3 IPs in this range.

    What do you guys think?
     
    anees.network, Dec 12, 2010
    #9
  10. Myrt Webb

    Kirsteins

    Joined:
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    Kirsteins, Mar 10, 2012
    #10
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