The document describes how to subnet the IP addresses 172.168.4.2/16 and 10.12.5.2/8. It identifies that 172.168.4.2/16 is a Class B network that can be divided into 16 subnets of 4096 IP addresses each using a subnet mask of 255.255.240.0. It also explains that 10.12.5.2/8 is a Class A network that can be divided into 16 subnets of 16 IP addresses each using a subnet mask of 255.255.255.240. Tables listing the network addresses and broadcast addresses of each subnet are provided.

1. 172.168.4.2/16
 Class = B
 Total Network=8
 Subnet Bit=4
 Slash Bit=16+4=20
 Total Subnet=24
=16
 Host Bit=16-4=12
 Total Host=212
=4096
 Useable Host= 212
-2 =4096-2=4094
 Subnet Mask=255.255.240.0

2. Network Address Broadcast Address
S1 172.168.0.0 172.168.15.255
S2 172.168.16.0 172.168.31. 255
S3 172.168.32.0 172.168.47. 255
S4 172.168.48. 0 172.168.63. 255
S5 172.168.64. 0 172.168.79. 255
S6 172.168.80. 0 172.168.95. 255
S7 172.168.96. 0 172.168.111. 255
S8 172.168.122.0 172.168.127. 255
S9 172.168.128.0 172.168.143. 255
S10 172.168.144.0 172.168.159. 255
S11 172.168.160.0 172.168.175. 255
S12 172.168.176.0 172.168.191. 255
S13 172.168.192.0 172.168.207. 255
S14 172.168.208.0 172.168.223. 255
S15 172.168.224.0 172.168.239. 255
S16 172.168.240.0 172.168.255. 255
10.12.5.2/8
1. This network Class A
2. Number network =8
3. Subnet bits = 4
4. Slash bit = DNB + SB =16+4=20
Then 10.12.5.2/20
5. Total subnet =2subnet bit
=24
=16
6. Host bit = DNB – SB

3. 1. =20-16 = 4
7. Total Host =24
= 16
8. Total Host = 2subnet bit
-2 =216
-2 = 14
9. Subnet mask = 255.255.255.240
Network Address Broadcast Address
S1 10.12.5.0 10.12.5.15
S2 10.12.5.16 110.12.5.31
S3 10.12.5.32 10.12.5.47
S4 10.12.5.48 10.12.5.63
S5 10.12.5.64 10.12.5.79
S6 10.12.5.80 10.12.5.95
S7 10.12.5.96 10.12.5.111
S8 10.12.5.112 10.12.5.127
S9 10.12.5.128 10.12.5.143
S10 10.12.5.144 10.12.5.159
S11 10.12.5.60 10.12.5.175
S12 10.12.5.176 10.12.5.191
S13 10.12.5.192 10.12.5.207
S14 10.12.5.280 10.12.5.223
S15 10.12.5.224 10.12.5.239

4. S16 10.12.5.240 10.12.5.255