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  1. Senior Member boredgamelad's Avatar
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    #26
    Looks fine to me, nice job.

    On future network diagrams you can omit listing the range and just list the subnet ID and mask but it's not bad practice to write them out like this. Helps make sure you're doing it right.
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    #27

    Default CCNA revision Question ?

    I have this question and I don't seem to be getting the correct answer. Can anyone help here with the correct step to get the correct answer, please.

    What is the first valid host on the subnetwork that the node 172.25.209.197 255.255.255.224 belongs to?
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    #28
    X.x.x.193
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    #29
    Hi,

    Thanks for the input, would you have steps to get to that answer, please.
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    #30
    Quote Originally Posted by vtiniall View Post
    Hi,

    Thanks for the input, would you have steps to get to that answer, please.

    If I may...

    I created a 'Subnetting Cheat Sheet' in this thread... Subnetting Cheat Sheet

    CIDR
    SUBNET /bits
    /1
    /9
    /17
    /25
    /2
    /10
    /18
    /26
    /3
    /11
    /19
    /27
    /4
    /12
    /20
    /28
    /5
    /13
    /21
    /29
    /6
    /14
    /22
    /30
    /7
    /15
    /23
    /31
    /8
    /16
    /24
    /32
    Block Size 128 64 32 16 8 4 2 1
    Mask 128 192 224 240 248 252 254 255
    HOST bits 31
    23
    15
    7
    30
    22
    14
    6
    29
    21
    13
    5
    28
    20
    12
    4
    27
    19
    11
    3
    26
    18
    10
    2
    25
    17
    9
    1
    24
    16
    8
    0

    So to answer your question
    What is the first valid host on the subnetwork that the node 172.25.209.197 255.255.255.224 belongs to?
    Understand that the Octet that is of interest is the 4th Octet (224 in your mask, /27 in CIDR Notation). This equates to a Block Size of 32.

    So, the Subnets are 0, 32, 64, 96, 128, 160, 192, 224.

    172.25.209.197 is in the 192 Subnet.

    Valid Host range 172.25.209.193 -> 172.25.209.222
    B/Cast 172.25.209.223

    Is this of help?
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    #31
    Thanks for taking the time to write out the solution. It does help alot
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    #32
    Quote Originally Posted by vtiniall View Post
    Thanks for taking the time to write out the solution. It does help alot
    You're welcome. It's not easy to get your head around, but once you do, it kind of sticks.

    Feel free to ask about anything that you're unsure of.

    Edit to add...

    There are quite a few threads on here about the subject, I'd suggest you read them and find a system that chimes with your own way of thinking.
    Last edited by rob42; 04-20-2017 at 08:26 AM.
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    #33
    Just one last question. How would you chart work for this question

    What is the last valid host on the subnetwork 172.27.154.0 255.255.254.0?
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    #34
    Quote Originally Posted by vtiniall View Post
    Just one last question. How would you chart work for this question What is the last valid host on the subnetwork 172.27.154.0 255.255.254.0?
    In this example, the 3rd Octet is where the Network and the Hosts split (as dictated by the Mask), so it's the 3rd Octet that we're interested in. Look at the Mask column for 254 and sub-rows 3 (because we're interested in the 3rd Octet) for the CIDR and Host Bits, you'll see the the CIDR is /23 (meaning 23 Network bits) Host bits = 9, and the Block Size is 2. The Block Size is value by which the Subnet Address changes, starting at zero and incrementing by two (the Block Size) in this example, for each Subnet...

    1st Subnet = 172.27.0.0
    2nd Subnet = 172.27.2.0
    3rd Subnet = 172.27.4.0
    4th Subnet = 172.27.6.0
    ...etc...
    77th Subnet = 172.27.152.0
    78th Subnet = 172.27.154.0
    79th Subnet = 172.27.156.0

    You can quickly arrive at the 78th Subnet with a little math: Octet Value / Block Size, plus one (154/2 = 77 + 1 = 78).

    So, the 78th Subnet is 172.27.154.0 and the Broadcast is 172.27.155.255, which is one IP Address down from the 79th Subnet (172.27.156.0).

    The valid Host IPs are between the Subnet IP and the Broadcast IP -- 172.27.154.1 -> 172.27.155.254

    My Cheat Sheet simply gives you the important information you need (block size, CIDR {which is also the number of Network bits}, Mask and number of Host bits) in order to quickly see where the boundary is between the Network IP and the Host IPs, for any given IP Address. You still need to understand how that information is applied.

    -----------------------------------------------------

    Here's one for you to do.

    IP = 179.20.55.28 /22

    Which Subnet?
    What is the Subnet IP?
    What are the first and last Valid Host IPs
    What is the Broadcast IP?
    What is the Subnet Mask?
    Last edited by rob42; 04-21-2017 at 01:13 AM.
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