N10-006 subnetting question regarding largest remaining block

melonfelonmelonfelon Registered Users Posts: 3 ■■■□□□□□□□
So i've been using lots of different study material for the n10-006, including professormessor videos, and just lately i've been using measureup practice tests.

One of the measureup questions was regarding allocating multiple subnets of various sizes from a given /24 range.
That's all fine, and I get the whole formula regarding usable addresses (subtracting 2 for the network ID, and the broadcast address). So for example to make a subnet from 10.10.10.0 /24 that needed 32 hosts, you would have to use a 64 block such as 10.10.10.0 /26.

What I am confused about, is the second part of the question, which is to identify the largest block of available host addresses remaining after assigning the required subnets for the first part of the question.

This is the way i'm working it out in my head, and i'm not sure if its wrong.

Lets say I allocated 4 subnets like this: (including the 2 you can't use for hosts)

Network A requiring 64 addresses
Network B requiring 32 addresses
Network C requiring 16 addresses
Network D requiring 16 addresses

That's a total of 128 addresses assigned from the original /24.

A /24 has 256 total addresses so that leaves me with 128 left over addresses, which means the answer to the question of the largest remaining block would be a /25.

The measureup practice test tells me this is wrong, and that you are left with only 126 addresses which would mean that the largest assignable block would be a /26 with 64 addresses (62 usable for hosts).

The measureup way seems to assume you can't assign 10.10.10.0 and 10.10.10.255 to subnets, making the remaining block calculation actually 254 available addresses - the 128 already assigned, leaving 126, so the largest assignable block is /26.


So which way is correct to work out largest remaining block? (256 - the subnets you assigned), or (254 - subnets you assigned)?


p.s. Sorry if this isn't clear, i'm trying to keep it straight in my head.

Comments

  • christopherleachchristopherleach Member Posts: 5 ■□□□□□□□□□
    The measure up test is correct, in a /24 split you will not be able to use 10.10.10.0 & 255 (not assignable)

    Even when you split the subnet you still do not have access to the first and the last Octet.

    Remember it is the trick of “assignable” IP addresses.

    The questions wants to know what is “assignable” to the devices.
    Once you subnet the /24 you will lost more and more, I call this the “cost of subneting”

    SO breaking down your question, if you are left with 128 address in a /26, how many “assignable” address do you have?

    126 or /26
  • melonfelonmelonfelon Registered Users Posts: 3 ■■■□□□□□□□
    Ok, let me type this out long form and see how it makes sense.

    Subnetting the 10.10.10.0 /24 range as follows:

    Network A needs - 32 hosts
    Network B needs - 24 hosts
    Network C needs - 8 hosts
    Network D needs - 12 hosts

    A - /26 = Block of 64 addresses (62 usable) Subnet ID = 10.10.10.0 /26 Range = 10.10.10.0 - 10.10.10.63


    B - /27 = Block of 32 addresses (30 usable) Subnet ID = 10.10.10.64 /27 Range = 10.10.10.64 - 10.10.10.95


    C - /28 = Block of 16 addresses (14 usable) Subnet ID = 10.10.10.96 /28 Range = 10.10.10.96 - 10.10.10.111


    D - /28 = Block of 16 addresses (14 usable) Subnet ID = 10.10.10.112 /28 Range = 10.10.10.112 - 10.10.10.127


    Ok, so we've assigned our required subnets for the first part of the question, now we are left with the following range of unassigned addresses:

    Unassigned = 10.10.10.128 - 10.10.10.255

    So the largest remaining block is like this: Subnet ID = 10.10.10.128 /25 Range = 10.10.10.128 - 10.10.10.255

    And within that last /25, there are 126 host assignable addresses (-2 for the network address of 10.10.10.128, and the broadcast of 10.10.10.255)


    So surely the last remaining block of addresses is a /25 with 128 addresses, of which 126 are host assignable?

    Why would you only be able to subnet the 128 remaining addresses into a block of 64 (/26) with 62 host addresses - that's the part that doesn't make sense to me.
  • volfkhatvolfkhat Member Posts: 947 ■■■■■■■□□□
    I don't follow what you are asking in your original post. (and i'm completely lost in your 2nd post).
    Also, i don't like Measureup's term "the largest remaining block"; it's a bit unclear to me.

    From the Beginning:
    So, it says to start with --> 10.10.10.0 /24

    And they want you to subnet-out 32 hosts. Which means you need to create a /26 subnet (good catch btw; i forgot about the -2 factor).

    So.... "10.10.10.0 /26" means that you just took the first 64 address: 10.10.10.B]0-63[/B .

    The addresses now remaining are ---> .64 .65 .66 (yada-yada-yada) .254, .255 .
    Counting with Fingers, tells me that's a total of 192 addresses.

    quoting melonfelon:
    "...the second part of the question, which is to identify the largest block of available host addresses remaining after assigning the required subnets for the first part of the question."

    Well, i don't like that wording.
    The "largest block of available host addresses remaining" ??
    At this point, there are NO other subnet/blocks. We haven't created any others.

    I would prefer it read ---> "the largest block of available host addresses Possible".
    Then, i Could say that the answer is ---> a 128-block subnet. aka, /25.

    But, then again, i don't even know if that's Exactly what they are asking for.
    I would need to see the possible answers first.

    Synopsis:
    This a poorly worded Question.
    lol
  • melonfelonmelonfelon Registered Users Posts: 3 ■■■□□□□□□□
    I didn't want to copy/paste the question exactly as it was, just in case it was against any rules, and I think I did a poor job of explaining my thoughts.

    But, on the upside, i've been practicing subnetting more and it's starting to become easier and more intuitive as I do different examples. Thanks for your input guys.
  • volfkhatvolfkhat Member Posts: 947 ■■■■■■■□□□
    You are correct about the NDA for the cisco exams.
    But MeasureUp has No Authority (besides general TermsOfService).

    If you have a Question... then Ask it on the Forums.
    Just say it's a "hypothetical" question, and do Not say where it came from :]

    I luv this site for subnets:
    subnetting.net - Subnet Questions and Answers
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