Question

Consider a company which owns a license of Class C network (207.84.123.0), This Company wants to create 14 subnetworks.
1.
Determine the number of bits borrowed
2.
How many bits are then used for the subnet ID?
Determine the maximum number of hosts in each subnet
3.
Determine the subnet mask of this scheme
4.
Determine the first, the forth and the last network (subnet) addresses
5.
Determine the first host address, the last host address and the broadcast address in only the first subnet.
6.
7. To which subnet belongs the host having the address 207.84.123.181?

Answer 1

1.

We need to create 14 subnets . So number of bits borrowed = 4 bits . Because 2^4=16 .

2.

So it means 4 bits are used for subnet id .

So number of bites in host id of each subnet = 8-4=4 .

So number of hosts in each subnet = 2^4-2 = 16-2=14

3 .

In subnet mask all the host id bits are 0's . And subnet id + network id bits are 1's. So , subnet mask is --

1 1 1 1 1 1 1 1 . 1 1 1 1 1 1 1 1 . 1 1 1 1 1 1 1 1 . 1 1 1 1 0 0 0 0 = 255.255.255. 240

4.

first subnet address is

207 . 84 . 123 . 0 0 0 0 0 0 0 0 = 207.84.123.0

Second subnet address is

207 . 84 . 123 . 0 0 0 1 0 0 0 0 = 207.84.123.16

Third subnet address is

207 . 84 . 123 . 0 0 1 0 0 0 0 0 = 207.84.123.32

Fourth subnet address is

207 . 84 . 123 . 0 0 1 1 0 0 0 0 = 207.84.123.48

Last subnet address is

207 . 84 . 123 . 1 1 1 1 0 0 0 0 = 207.84.123.240

5.

First subnet address is

207 . 84 . 123 . 0 0 0 0 0 0 0 0 = 207.84.123.0

So first host ---

207 . 84 . 123 . 0 0 0 0 0 0 0 1 = 207.84.123.1

Last host

207 . 84 . 123 . 0 0 0 0 1 1 1 0 = 207.84.123.14

Broadcast address

207 . 84 . 123 . 0 0 0 0 1 1 1 1 = 207.84.123.15

6.

Network address given is

207.84.123.181

So in binary last octet 181 can be written as 10110101

So it means subnet id bits are 1011 = 11

So it means this network address belongs to 12th subnet.