CCNA 20

Now we'll focus on minimum hosts instead of network. Let's say we have a network of 216.21.5.0/24 where we want each subnet to have 30 hosts. To do this, find the number of host bits we need. The formula is the same, which is 30=(2^x)-2. x will the be 5. The ending mask would then be 27 bits, which gives us 255.255.255.224. This will give us 8 subnets (or 6 subnets depending on whether you want to use the first and last subnets). Therefore, we'll have (with an increment of 32):0)216.21.5.0 (.1 to .30)
1)216.21.5.32 (.33 to .62)
2)216.21.5.64 (.65 to .94)
3)216.21.5.96 (.97 to .126)
4)216.21.5.128 (.126 to .158)
5)216.21.5.160 (.161 to .190)
6)216.21.5.192 (.192 to .222)
7)216.21.5.224 (.222 to .254)

Now we'll breeze through the next few examples because the concept is the same. If I have 195.5.20.0/24 and we want 50 hosts per network, then we'll need 6 bits for host. We'll then have 26 bits for the mask which gives us 255.255.255.192. There will be 4 subnets which increments of 64:
0)195.5.20.0 (.1 to .62)
1)195.5.20.64 (.65 to .126)
2)195.5.20.128 (.129 to .190)
3)195.5.20.192 (.193 to .254)

Now, we have a 150.5.0.0/16 network which needs 500 hosts per network. 500 hosts take up 9 bits and so we'll end up with a /23 mask which is 255.255.254.0. We'll have 7 bits for subnets which gives us 128 (or 126) subnets. The subnets are:
0)150.5.2.0 (.0.1 to .1.254)
1)150.5.4.0 (.2.1 to .3.254)
2)150.5.6.0 (.4.1 to .5.254)
3)150.5.8.0 (.6.1 to .7.254)
...

Breaking 10.0.0.0/8 into 100 host networks would require a mask of 255.255.255.128. The networks would be:
0)10.0.0.0 (.0.1 to .0.126)
1)10.0.0.128 (.0.129 to .0.254)
2)10.0.1.0 (.1.1 to .1.126)
3)10.0.1.128 (.1.129 to .1.254)
4)10.0.2.0 (.2.1 to .2.126)
...

Breaking 200.1.1.0 into subnets of 40 hosts each would need a mask of 255.255.255.192. With the increment of 64, you'll have:
0)200.1.1.0 (.1 to .62)
1)200.1.1.64 (.65 to .126)
2)200.1.1.128 (.129 to .190)
3)200.1.1.192 (.193 to .254)
...

Breaking 199.9.10.0 into subnets of 12 hosts each would need a mask of /28 which is 255.255.255.240. With the increment of 16, you'll have:
0)199.9.10.0 (.1 to .14)
1)199.9.10.16 (.17 to .30)
2)199.9.10.32 (.33 to .46)
3)199.9.10.48 (.49 to .62)
...

Breaking 170.50.0.0 into networks of 1000 hosts each would require a mask of /22 which is 255.255.252.0. With the increment of 4, you'll have:
0)170.50.0.0 (.0.1 to .3.254)
1)170.50.4.0 (.4.1 to .7.254)
2)170.50.8.0 (.8.1 to .11.254)
3)170.50.12.0 (.12.1 to .15.254)
...

Breaking 12.0.0.0 into subnets of 100 hosts each would require a mask of /25. You'll have an increment of 128 and networks:
0)12.0.0.0 (.0.1 to .0.126)
1)12.0.0.128 (.0.129 to .0.254)
2)12.0.1.0 (.1.1 to .1.126)
3)12.0.1.128 (.1.129 to .1.254)
...