Consider an ISP that owns the block of addresses in 123.119.14.64/26. Suppose it wants to create three subnets, where Subnet1 supports at least 30 interfaces, Subnet2 supports at least 12 interfaces, and Subnet3 supports at least 9 interfaces. Provide the network addresses (of the form a. b.c. d/x) of the three subnets that satisfy these requirements.

123.119.14.64/27   - subnet1

123.119.14.96/28   - subnet2

123.119.13.128/28 - subnet 3

Explanation:

123.119.14.64/26   in binary

10000000.01110111.001110.01000000/26

2^6 = 64 addresses in total; therefore we can only assign 64 addresses from this block of IP

From the question we are told that

Subnet1 supports at least 30 interfaces,

Subnet2 supports at least 12 interfaces, and

Subnet3 supports at least 9 interfaces

Subnet 1 with 30 interfaces , this means we are going to turn on 5 bits from the host bits in the IP address (starting from the right) , In doing so we will have

10000000.01110111.001110.01000000/27

123.119.14.64/27  is the nework address 0f the first subnet

Secondly we add a 1 bit (because of the broadcast address) to the first network address;

This Subnet2 supports at least 12 interfaces,

10000000.01110111.001110.01000000/27   (from the first network)

similarly  we have

10000000.01110111.001110.01100000/28

Subnet3 supports at least 9 interfaces

Adding 1 bit to the second network ; we have that

10000000.01110111.001110.10000000/28,