Question

14. What is the range of assignable IP addresses for a subnet containing an IP address of 172.16.1.10/192 172.16.0.1-172.16.31254 172.16.0.1-172.16.63.254 172.16.0.0-172.16.31.255 172.16.0.1-172.16.31.255 172.16.0.0-172.16.63.254 15. What subnet mask should be used to subnet the 192.168.10.0 network to support number of subnets and IP addresses per subnet shown in the following topology? 255.255.255.0 255.255.255.128 255.255.255.192 255.255.255.224 255.255.255.240

Answer 1

Q.14)

Here the given address is 172.16.1.10/19 so the class of this IP address will be 'B' (128-191) and the default Subnet mask for it will be 255.255.0.0.

As the CIDR value (total number of network bits) is 19, number of host bits will be 32 - 19 = 13

So the new subnet mask will be,

255.255.255.32(i.e 11111111 11111111 11111111 00011111)

So total number of network will be 2^n (n = number of bits borrowed from host ) that is 2^3 = 8

So the number of IP address on each network will be 2^n (n = total number of host bits) = 2^5 = 32 ( that is on each network we will have 32 IP address)

So total number of hosts in each network = 2^n - 2 (n is remaining host bits that is 5) = 32 -2 = 30 ( so we will have 30 host IP address on each network)

That is In each network the first IP address is reserved for network ID and last IP address is reserved for broadcast ID.

So on first network the IP address range will be 172.16.0.0 - 172.16.1.31

and the last network IP address range will be - 172.16.0.223 - 172.16.31.255

As I said mentioned above the first address that is 172.16.0.0 will be reserved for network ID and the last address 172.16.31.255 will be reserved for broadcast ID

So assignable addresses will be 172.16.0.1 - 172.16.31.254 i.e Option A