Frustration with subnetting

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  • Ltat42aLtat42a Member Posts: 587 ■■■□□□□□□□
    veritas_libertas wrote: »
    That is not really my struggle. What would throw me off is something like:

    192.168.106.136 255.255.240.0

    What valid host range is the above IP address a part of? Now I realize I could covert the third octets to binary and then just do the math, but that is time intensive. Is there any shortcut to this?

    Just wondering, for Cisco exam purposes, wouldn't this be an incorrect mask for the 192.168.x.x network?
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  • JockVSJockJockVSJock Member Posts: 1,118
    veritas_libertas wrote: »
    What is driving me crazy is what I consider overlapping Octets and I'm sure I'm going to be criticized for using that term since it's less than accurate or scientific icon_lol.gif

    It's the only way I can think of to describe it!!! :)

    Not sure what you mean by overlapping Octets? I'm confused by this term.

    Everyone has given you great help. I'm sure it might have been said, keep doing problems daily and if you have issues post here and we'll try to help.

    YOU GOT THIS!!!
    ***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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  • johnifanx98johnifanx98 Member Posts: 329
    veritas_libertas wrote: »
    That is not really my struggle. What would throw me off is something like:

    192.168.106.136 255.255.240.0

    What valid host range is the above IP address a part of? Now I realize I could covert the third octets to binary and then just do the math, but that is time intensive. Is there any shortcut to this?

    My two cents. 192.168.x.x is class C address. This question should be phrased as "supernetting" instead of subnetting. Mathematically these are equivalent, but I doubt even this is a useful example of supernetting in practice ...
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  • veritas_libertasveritas_libertas Member Posts: 5,746 ■■■■■■■■■■
    LOL, I'm kind of surprised folks are digging up the original posts in this thread. I was able to figure out those problems.

    Thanks guys ;)
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  • macavoy123macavoy123 Registered Users Posts: 1 ■□□□□□□□□□
    Can someone help me with this, please?

    Question:     What is the broadcast address of thenetwork 172.18.196.0 255.255.255.0?
    Answer: 172.18.196.255


    Thank you in advance! icon_smile.gif
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  • TheFoX21TheFoX21 Member Posts: 7 ■□□□□□□□□□
    macavoy123 wrote: »
    Can someone help me with this, please?

    Question:     What is the broadcast address of thenetwork 172.18.196.0 255.255.255.0?
    Answer: 172.18.196.255


    Thank you in advance! icon_smile.gif

    In every subnet there are two IP addresses that are reserved. One is the network address (in this example would be the 172.18.196.0) and the other is the broadcast address which is the very last IP in the subnet. This is why when calculating how many usable hosts there are in a subnet you use the formula 2^n-2. The subtraction of the 2 accounts for the network and broadcast addresses which can't be used for hosts.

    So for your question the subnet has a total of 256 total IP addresses (Class C) which two are reserved for the network and broadcast addresses leaving you with 254 host IP addresses at your disposal.
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