Question

XYZ Company would like to subnet its network so that there are five separate subnets. They will need 25 computers in each subnet. Complete each ofthe following: NOTE: If you create more than five subnets, list the extra ones too. S ubnet Network address Host addresses Broadcast address S ubnet mask: 255.255.255.224 First subnet 192.168.162.0 192.168.162.1 - 192.168.162.30 192.168.162.31 S econd subnet 192.168.162.32 192.168.162.33 - 192.168.162.62 192.168.162.63 Third subnet 192.168.162.64 192.168.162.65 - 192.168.162.94 192.168.162.95 Fourth subnet 192.168.162.96 192.168.162.97 - 192.168.162.126 192.168.162.127 Fifth subnet 192.168.162.128 192.168.162.129 - 192.168.162.158 192.168.162.159 S ixth subnet 192.168.162.160 192.168.162.161 - 192.168.162.190 192.168.162.191 S eventh subnet 192.168.162.192 192.168.162.193 - 192.168.162.222 192.168.162.223 Eighth subnet 192.168.162.224 192.168.162.225 - 192.168.162.254 192.168.162.255

Answer 1

Answer:

Following subnet mask we are going to used .

Explanation:

Subnet Mask:255.255.255.224

Its class C Subnet mask hence we are considering last octal of binary representation. (binary representation 11111111.11111111.11111111.11100000)

last octal cosist 3 bits as 1's hence total number of subnet can be form is 2^3=8. and each subnet consist 2^5=32 (32-2=30,Usable host in each subnet)

All 8 of the Possible /27 Networks for 192.168.162.*

Sr. No Network Address Usable Host Range Broadcast Address:

1st 192.168.162.0 192.168.162.1 - 192.168.162.30 192.168.162.31

2nd 192.168.162.32 192.168.162.33 - 192.168.162.62 192.168.162.63

3rd 192.168.162.64 192.168.162.65 - 192.168.162.94 192.168.162.95

4th 192.168.162.96 192.168.162.97 - 192.168.162.126 192.168.162.127

5th 192.168.162.128 192.168.162.129 - 192.168.162.158 192.168.162.159

6th 192.168.162.160 192.168.162.161 - 192.168.162.190 192.168.162.191

7th 192.168.162.192 192.168.162.193 - 192.168.162.222 192.168.162.223

8th 192.168.162.224 192.168.162.225 - 192.168.162.254 192.168.162.255