Question

Assume that we have a machine with IP Address 15.107.33.204 with netmask 255.255.255.240. Assume that you manage a /11 address space, and that we have fully divided it up into child networks the size of the one on which this machine resides.

1. How many useable subnets of this size can be created?

2. How many useable addresses will be available on each of these subnets?

3. List the IP addresses of the first three and last three of these useable subnets.

4. Give the network number (IP Address) of the next to the last useable subnets.

5. Give the broadcast address of the next to the last useable subnet.

6. Give the first useable address on the next to last useable subnet.

7. Give the last useable address on the next to last useable subnet.

Answer 1

Answer1.. The subnet mask of the current IP address is 255.255.255.240 or /28 but this mask is generated from /11 mask so clearly increase of 17bits in network address. Thus allowing 2^17 subnet i.e., 131072 subnet possible.

Answer2. Host bits available is 4 bit so ^4 i.e 16 IP will be available but only 14 will be usable since first and last address of every subnet is not usable. They are network address and broadcast address respectively.

Answer3.

15.107.33.193 ,

15.107.33.194,

15.107.33.195,

15.107.33.204,

15.107.33.205,

15.107.33.206.

Answer4. The last useable IP is 15.107.33.206, The network number next to this is 15.107.33.208.

Answer5. The broadcast address address next to the last useable IP is 15.107.33.207, this is because each subnet has 16 IP address block and 15.107.33.204 comes under 15.107.33.192 subnet block.

Answer6. First useable address next to the last useable subnet is 15.107.33.209. Since next to 15.107.33.192 network is 15.107.33.208, first IP i.e, 15.107.33.208 is network address and is not usable so next usable is 15.107.33.209

Answer7. Last useable address next to last useable subnet will be 15.107.33.22. This is because in the grouping of 16 IP the next block of IP will be from 15.107.33.208 to 15.107.33.223, but since the last is not usable so 15.107.33.222 will be the answer.