Subnetting Issues with getting number of valid hosts for different subnets

JockVSJockJockVSJock Member Posts: 1,118
in CCNA & CCENT
Hello, wanted to report that I'm making good progress on subnetting. I'm able to subnet the following problems in under 1 minute

-Find the subnet
-Find the valid subnets/hosts
-Find the broadcast address
-Find the valid hosts
-Find the first valid host/last valid host

However, I'm still not there yet.

I'm still getting hung up on problems like the following from subnetting.org:

QUESTION: You need to assign a server the last valid host address on the 5th subnet of network 172.16.0.0/19.
What address would you assign?

ANSWER: 172.16.159.254

The part that is throwing me is finding the exact range, and typically I'm off.

Does anyone have any recommendations on an accurate way to get the number correctly.

thanks
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  • mella060mella060 Member Posts: 198 ■■■□□□□□□□
    So the network is 172.16.0.0 /19...255.255.224.0

    So your increment would be 32 in the 3rd octet.

    So your networks would be...

    172.16.0.0 /19
    172.16.32.0 /19
    172.16.64.0 /19
    172.16.96.0 /19
    172.16.128.0 /19 - 5th subnet...valid host range 172.16.128.1 - 172.16.159.254
    172.16.160.0 /19
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  • SharkDiverSharkDiver Member Posts: 844
    Couldn't have said it better!!!
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  • MAC_AddyMAC_Addy Member Posts: 1,740 ■■■■□□□□□□
    mella060 wrote: »
    So the network is 172.16.0.0 /19...255.255.224.0

    So your increment would be 32 in the 3rd octet.

    So your networks would be...

    172.16.0.0 /19
    172.16.32.0 /19
    172.16.64.0 /19
    172.16.96.0 /19
    172.16.128.0 /19 - 5th subnet...valid host range 172.16.128.1 - 172.16.159.254
    172.16.160.0 /19

    We could probably add how you got the 32 increment, even though this is already in a few other threads.

    You'll need to know the powers that be! 2^x (two to the power of x, x being the increment size).

    So, first you'll notice that it's a /19 and the next boundary is 24. So 24-19 = 5. 5 is the power we'll need to find.

    2^5=32.
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  • SharkDiverSharkDiver Member Posts: 844
    I just kind of remember it like this:
    .128 = increments of 128
    .192 = increments of 64
    .224 = increments of 32
    .240 = increments of 16
    .248 = increments of 8
    .252 = increments of 4
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  • mella060mella060 Member Posts: 198 ■■■□□□□□□□
    MAC_Addy wrote: »
    We could probably add how you got the 32 increment, even though this is already in a few other threads.

    A quick way of finding the increment/block size (as learn't in the CBT nuggets videos) is to look at the subnet mask and work out where the last bit position is 1...as in

    /19 = 255.255.224.0...in binary is 11111111.11111111.11100000.00000000

    Knowing that for each octet you have the values...128 64 32 16 8 4 2 1

    In this case the last bit position where the value is a 1 is at 32. So that is your increment.

    Another way is to just subtract 224 from 256

    You need to know what values in each octet add up to.

    As in...

    .1 = 128
    .11 = 128 + 64 = 192
    .111 = 128 + 64 + 32 = 224
    .1111 = 128 +64 + 32 + 16 = 240
    .11111 = 128 + 64 + 32 + 16 + 8 = 248
    .111111 = 128 + 64 + 32 + 16 + 8 + 4 = 252
    .1111111 = 128 + 64 + 32 + 16 + 8 + 4 + 2 = 254

    .11111111 = 255 obviously
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  • Ltat42aLtat42a Member Posts: 587 ■■■□□□□□□□
    Here's a lil chart I use when practicing subnetting, hope it helps.



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  • SharkDiverSharkDiver Member Posts: 844
    Memorizing that chart will save lots of exam time in the future, and YOU CAN IMPRESS YOUR FRIENDS!!! icon_lol.gif
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  • RoguetadhgRoguetadhg Member Posts: 2,489 ■■■■■■■■□□
    SharkDiver wrote: »
    Memorizing that chart will save lots of exam time in the future, and YOU CAN IMPRESS YOUR FRIENDS!!! icon_lol.gif

    Im going to need to conjure up a subnetting thread...
    In order to succeed, your desire for success should be greater than your fear of failure.
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  • veritas_libertasveritas_libertas Member Posts: 5,746 ■■■■■■■■■■
    Im going to need to conjure up a subnetting thread...

    There are already several out there. I have one actually...
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  • RoguetadhgRoguetadhg Member Posts: 2,489 ■■■■■■■■□□
    veritas_libertas wrote: »
    There are already several out there. I have one actually...

    There's over 300 threads that pop up with "subnetting" from April 2011, one more won't hurt :P
    In order to succeed, your desire for success should be greater than your fear of failure.
    TE Threads: How to study for the CCENT/CCNA, Introduction to Cisco Exams

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  • JockVSJockJockVSJock Member Posts: 1,118
    Ok, I think I may have hit my Eureka moment.

    I just had the following problem, I got it wrong, however I think I see how to do these going forward.
    Need the last valid host address on the 5th subnet from the following:

    172.16.0.0
    255.255.224.0


    N 172.16.0.0
    FH 172.16.0.1
    LH 172.16.30.254
    B 172.16.31.255

    To get the subnets, you subtract from 256 - 224 = 32

    1st - 172.16.0.0
    2nd - 172.16.32.0
    3rd - 172.16.64.0
    4th - 172.16.96.0
    5th - 172.16.158.254

    Last valid host would be 172.16.188.254 and I would get that by counting up 30 on the third octet, correct?
    ***Freedom of Speech, Just Watch What You Say*** Example, Beware of CompTIA Certs (Deleted From Google Cached)

    "Its easier to deceive the masses then to convince the masses that they have been deceived."
    -unknown
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  • ChickenNuggetzChickenNuggetz Member Posts: 284
    Not sure if you're getting it wrong or if you're just mistyping...but anyways:

    Watch your math, you're right about the increment being 32. No need to subtract 2 from the increment. Your network ranges were correct up until the 5th range. It should read 172.16.128.0 (add 32 to the .96 in the 4th range). Add 32 to your 5th range (which should read 172.16.128.0) and you'll get 172.16.160.0 (<--- this is the start of the 6th range). This means your 5th range is 172.16.128.0 - 172.16.159.255 with 172.16.159.254 being the last valid host.


    EDIT: Ok, I see what you're doing. You're not taking into account the address's class. 172.16.0.0 is a Class B network. Address host space will use both the 3rd and 4th octets. Basically, it looks like youre not counting up correctly. Should be counting like this:

    172.16.0.1
    172.16.0.2
    172.16.0.3 (fast forward a bit)
    172.16.0.255 (not a broadcast, but a valid host)
    172.16.1.0
    172.16.1.1
    etc etc.

    So your example should read:
    172.16.30.254
    172.16.30.255 (not a broadcast, but a valid host)
    172.16.31.0
    172.16.31.1

    Keep counting until 255 and your broadcast should read:
    172.16.31.255 (this is the broadcast because the next address in the sequence is 172.16.32.0 which is the first address in the next range according to your 32 increment)


    Sorry for the mega edits, but its late and I've been studying for the last four hours...hope all my rambling helped!
    :study: Currently Reading: Red Hat Certified Systems Administrator and Engineer by Ashgar Ghori

    Certifications: CCENT; CCNA: R&S; Security+

    Next up: RHCSA
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  • mella060mella060 Member Posts: 198 ■■■□□□□□□□
    JockVSJock wrote: »
    Ok, I think I may have hit my Eureka moment.

    I just had the following problem, I got it wrong, however I think I see how to do these going forward.




    N 172.16.0.0
    FH 172.16.0.1
    LH 172.16.30.254
    B 172.16.31.255

    You mean the LH last host would be 172.16.31.254...you have 30.254
    JockVSJock wrote: »
    1st - 172.16.0.0
    2nd - 172.16.32.0
    3rd - 172.16.64.0
    4th - 172.16.96.0
    5th - 172.16.158.254

    Last valid host would be 172.16.188.254 and I would get that by counting up 30 on the third octet, correct?

    Shouldn't the 5th subnet be 172.16.128.0 ?
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