Hello @Tkay ,
you need to find the number of hosts bits N for each subnet noting that
S+N = 32
And 2^N -2 > n. hosts in the subnet
for example to be able to support 600 hosts you would need
N=10 so that 2^10 -2 = 1024 -2 and this means the first subnet should be a /22
To accomodate 200 hosts N=8 is fine 2^8= 256 -2 = 254 > 200. ---> /24
N=7 is fine for the 100 hosts subnets 2^8 -2 = 126 > 100 ---> 25
N=6 is fine for 31 hosts subnet as 2^6 -2 = 62 ---> 26
First of all we find the base network of the prefix
172.16.35.5/20
We write 35 in binary the third that is the interesting byte
00100011 the boundary line is at the 20th bit
0010 || 0011
the base address of the prefix is 172.16.32.0/20 as we need to put all zeros in the host portion of the address.
From the base prefix we can find four /22 subnets
172.16.32.0/22
172.16.36.0/22
172.16.40.0/22
172.16.44.0/22
The first one 172.16.32.0/22 accomodates the needs of the first subnet
base address 172.16.32.0
first IP 172.16.32.1
lat usable IP IP 172.16.35.254
subnet mask /22 = 255.255.252.0
Then we take the 172.16.36.0/22 and we further divide it in 4 /24 to accomodate the second subnet
172.16.36.0/24
first IP 172.16.36.1
last IP 172.16.36.254
subnet mask /24 = 255.255.255.0
At this point we can go on and we take 172.16.37.0/24 and we first divide it in two two subnets to accomodate the third required subnet
172.16.37.0/25
first IP 172.16.37.1
last IP 172.16.37.126
/25 = 255.255.255.128
Finally we can take the other /25 172.16.37128/25 subnet and split it in two
172.16.37.128/26
first IP 172.16.37.129
last IP 172.16.37.191
/26 = 255.255.255.192
note other solutions are possible but the suggested method is to proceed from the main address block and then proceed to split in subnets starting to accomodate the larger subnets and then going to accomodate the subnets with less hosts.
Some binary math on the interesting byte is required. Other methods to perform calculations are possible.
Hope to help
Giuseppe
