Question

) Consider a router that interconnects four subnets: Subnet 1, Subnet 2, Subnet 3 and Subnet 4. Suppose all of the interfaces in each of these four subnets are required to have the prefix 223.1.17/24. Also suppose that Subnet 1 is required to support at least 30 interfaces, Subnet 2 is to support at least 100 interfaces, Subnet 3 is to support at least 10 interfaces, Subnet 4 is to support at least 10 interfaces. Provide four network addresses (of the form a.b.c.d/x) that satisfy these constraints. Show all your work.

Answer 1

Answer:

Check the explanation

Explanation:

223.1.17/24 indicates that out of 32-bits of IP address 24 bits have been assigned as subnet part and 8 bits for host id.

The binary representation of 223.1.17 is 11011111 00000001 00010001 00000000

Given that, subnet 1 has 63 interfaces. To represent 63 interfaces, we need 6 bits (64 = 26)

So its addresses can be from 223.1.17.0/26 to 223.1.17.62/26

Subnet 2 has 95 interfaces. 95 interfaces can be accommodated using 7 bits up to 127 host addresses can represented using 7 bits (127 = 27)

and hence, the addresses may be from 223.1.17.63/25 to 223.1.17.157/25

Subnet 3 has 16 interfaces. 4 bits are needed for 16 interfaces (16 = 24)

So the network addresses may range from 223.1.17.158/28 to 223.1.17.173/28