Just when you think you understand subnetting . . . . . . .

pwjohnstonpwjohnston Member Posts: 441
So this question is on my Transcender for the 70-291. I got the answer right and I read the explanation, but I don’t understand why it is right.

You administer a single Active Directory domain that consists of W2k3 Severs and WinXP computers. A w2k3 server named ABR1 is connect to AreaA. Another w2k3 server named ABR2 is connected to AreaB.

AreaA has a summarized route of 204.29.18.0/26. AreaB has a summarized route of 204.29.19.0/27. You must identify valid subnets that you can include in AreaA and AreaB.

Which two are valid
-204.29.18.16/26
-204.29.18.32/26
-204.29.18.64/26
-204.29.18.96/26
-204.29.19.168/27
-204.29.19.176/27
-204.29.19.192/27
-204.29.19.208/27

The answer is:
-204.29.18.64/26
-204.29.19.192/27

The Explanation says:
The 204.29.18.64/26 and 204.29.19.192/27 subnets are valid in AreaA and AreaB.

The 204.29.18.0/26 area can contain:
204.29.18.0/26
204.29.18.64/26
204.29.18.128/26
204.29.18.192/26

The 204.29.19.0/27 area can contain these subnets:
204.29.19.0/27
204.29.19.32/27
204.29.19.64/27
204.29.19.96/27
204.29.19.128/27
204.29.19.160/27
204.29.19.192/27
204.29.19.224/27

But it does not say WHY those are valid subnets and just explains that the other choices are not valid.

I thought valid subnet octets, at least on the mask, are 254, 252, 248, 240, 224, 192, 128, and 64? Which is why I guessed the right answer. Can someone please help me with the math I’m missing here?

Comments

  • dynamikdynamik Banned Posts: 12,314 ■■■■■■■■□□
    A has 6 host bits, which is why the valid subnets increment by 64. B has 5 host bits, which is why the valid subnets increment by 32.

    The other options fall in between these values (or are just not possible, i.e. 16icon_cool.gif.
  • pwjohnstonpwjohnston Member Posts: 441
    So if they said

    204.29.19.0/25 would have 7 host bits and move in increments of 128
    204.29.19.0/26 would have 6 host bits and move in increments of 64
    204.29.19.0/27 would have 5 host bits and move in increments of 32
    204.29.19.0/28 would have 4 host bits and move in increments of 16
    204.29.19.0/29 would have 3 host bits and move in increments of 8
    204.29.19.0/30 would have 2 host bits and move in increments of 4

    If so thanks in advance.
  • shednikshednik Member Posts: 2,005
    pwjohnston wrote:
    AreaA has a summarized route of 204.29.18.0/26. AreaB has a summarized route of 204.29.19.0/27. You must identify valid subnets that you can include in AreaA and AreaB.

    The way I learned to do this the easiest was to figure out the block size.. 204.29.18.0/26 with as 26 bit mask you borrow 2 bits from the host portion. I figure the last octet into binary and it stops on 64,which would be the block size for the hosts. So at that point you know that the subnets will increment by 64 each time. Same goes for 204.29.19.0/27 move over one more bit which equals 32, so now you would increment by 32 for each subnet. When you first start subnetting I recommend breaking it down to binary every time. Once you do it enough you will be able to figure most things out in your head.
  • shednikshednik Member Posts: 2,005
    pwjohnston wrote:
    So if they said

    204.29.19.0/25 would have 7 host bits and move in increments of 128
    204.29.19.0/26 would have 6 host bits and move in increments of 64
    204.29.19.0/27 would have 5 host bits and move in increments of 32
    204.29.19.0/28 would have 4 host bits and move in increments of 16
    204.29.19.0/29 would have 3 host bits and move in increments of 8
    204.29.19.0/30 would have 2 host bits and move in increments of 4

    If so thanks in advance.

    Correct! :D
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