VLSM Question help

2URGSE2URGSE Posts: 220Member ■■■□□□□□□□
Objective:

Given 172.16.8.0/22, divide the network so the following are requirements are met:


Sales Department needs 300 hosts.
Marketing Department needs 76 hosts.
Administration Department needs 24 hosts.
HR Department needs 12 hosts.
Accounting Department needs 5 hosts.


So I start with the Sales Department.

(2^n)-2 = 300

When n=9, the max number of hosts is 510, so that will satisfy the 300 hosts Sales needs.

Next I take 32-9= 23, so the mask for the Sales subnet is /23.

Now comes the part that I get lost and need help.

I'm trying to find the first host address and last host address.

The first host address is 172.16.8.1, the last host address is 172.16.9.254. I get that part.

Next is the Marketing Department. (2^n)-2 = 76, so n=7 to satisfy this requirement.

This subnet is 172.16.10.0. Can someone please shed some light on how the got to that subnet?

The next subnet is 172.16.10.128, how did they get there?


* Please no binary stuff, I don't use it in my studies, and surely not on the exam.
A+
Network+
CCENT (formally CCNA certified)
ICE (Imprivata Certified Engineer)

Comments

  • diffiediffie Posts: 13Member ■□□□□□□□□□
    It’s been awhile for me but without using binary, 172.16.10.0 and 172.16.10.128, are simply the next usable addresses with the range 172.16.8.0/22


    Your starting subnet is 172.16.8.0/22, which provides the range 172.16.8.1 - 172.16.11.254.


    Your calculations so far are on point, for Sales you'd use 172.16.10.0/25 which would allow hosts in the range 172.16.8.1 to 172.16.9.254.


    172.16.9.255 is the broadcast address for that range. The very next address would be 172.16.10.0


    Your next calculations:


    (2^h) - 2 = hosts
    (2^7) - 2 = 126
    h = 7
    32 - 7 = 25


    172.16.10.0/25 the usable range is 172.16.10.1 - 172.16.10.126


    172.16.10.127 would be the broadcast address, 172.16.10.128 would be the next address.


    Then, just keep going.


    172.16.10.128/27 - 30 hosts/24 needed - 172.16.10.129 - 172.16.10.158, broadcast 172.16.10.159


    172.16.10.160/28 - 14 hosts/12 needed - 172.16.10.161 - 172.16.10.174, broadcast 172.16.10.175


    172.16.10.176/29 - 6 hosts/5 needed - 172.16.10.177 - 172.16.10.182, broadcast 172.16.10.183
  • Mr.Robot255Mr.Robot255 Posts: 181Member ■■■□□□□□□□
    i'm not great at explaining stuff but i will give it a go. (and hope i'm right)


    300=/23 - /23 = 512 host
    76=/25 - /25 = 128 host
    24=/27 -/27 = 32 host
    12=/28 -/28 =16 host
    5=/29 -/29 =8 host

    and we lose the first host and last host of each of them ,1st being Network I.D and last host being Broadcast.

    so from 172.16.8.0 /22
    the first block of 300 hosts will take 512 hosts a /23 as the its the nearest match bringing it to
    172.16.8.0 - 172.16.9.255 /23
    next block will then start from 172.16.10.0

    so we then have
    172.16.10.0 - 172.16.10.127 /25 (this covers our 76 hosts to nearest match/block size)

    next will start as 172.16.10.128 - 172.16.10.159 /27

    next will start as 172.16.10.160.............................




    Something i use>
    This is the chart i use for subnetting if you look for a /27 directly below it u see it will see it is for class C -255.255.224.0 and directly under the 224 is 32 for your block size which then will directly map to 8 subnets.
    The mapping works like this using the bottom row we can say host and subnets
    255 to the 1 (255 hosts to 1 subnet)
    128 to the 2 (128 hosts to 2 subnets) etc
    64 to the 4
    32 to the 8
    16 to 16



    /25 ---/26 ----/27--- /28--- /29 ----/30
    /31--- /32
    0 - ----128--- -192
    224 ----240---- 248
    252
    254 ----255
    255 - -128
    64
    32
    16
    8
    4
    2
    1




    The quality isn't great but this is the video that made everything make sense to me.
    https://www.youtube.com/watch?v=GSX1GlaznKM&t=41s

    and part 2
    https://www.youtube.com/watch?v=U9dVVIk8_zA&t=42s
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