VLSM Question help

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.

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Comments

diffie

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.Robot255

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

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