Question

1. Consider a datagram network using 8-bit IP addresses. Suppose a router uses longest prefix matching and has the follwing forwarding table:

Prefix Match	Interface
00	0
010	1
011	2
10	2
11	3

For each of the four interfaces, give the associated range of destination host addresses and the number of addresses in the range.

Solution:

Answer 1

It is given that the network uses 8 - bit IP addresses.

For Interface 0, one prefix is given - 00
Lowest IP address for this will be when the rest 6 bits are all 0 i.e. 00000000 = 0
Highest IP address will be when the rest 6 bits are all 1 i.e. 00111111 = 63
So Range of addresses is 00000000 - 00111111 i.e., 0 - 63
Number of destination host addresses = 64

For interface 1, one prefix is given - 010
Lowest IP address for this will be when the rest 5 bits are all 0 i.e. 01000000 = 64
Highest IP address will be when the rest 5 bits are all 1 i.e. 01011111 = 95
So Range of addresses is 01000000 - 01011111 i.e., 64 - 95
Number of destination host addresses = 32

For interface 2, two prefixes are given - 011, 10
We have to analyze both separately and then combine the result.
For 010:
Lowest IP address for this will be when the rest 5 bits are all 0 i.e. 01100000 = 96
Highest IP address will be when the rest 5 bits are all 1 i.e. 01111111 = 127
So Range of addresses for the prefix 010 is: 01100000 - 01111111 i.e., 96 - 127
Number of destination host addresses = 32
For 10:
Lowest IP address for this will be when the rest 6 bits are all 0 i.e. 10000000 = 128
Highest IP address will be when the rest 6 bits are all 1 i.e. 10111111 = 191
So Range of addresses for the prefix 10 is: 10000000 - 10111111 i.e., 128 - 191
Number of destination host addresses = 64

Total range for the interface 2: 01100000 - 10111111 i.e., 96 - 191
Number of destination host addresses = 96

For Interface 3, one prefix is given - 11
Lowest IP address for this will be when the rest 6 bits are all 0 i.e. 11000000 = 192
Highest IP address will be when the rest 6 bits are all 1 i.e. 11111111 = 255
So Range of addresses is 11000000 - 11111111 i.e., 192 - 255
Number of destination host addresses = 64

Summary:

Interface	Range of addresses	Number of addresses
0	0 - 63	64
1	64 - 95	32
2	96 - 191	96
3	192 - 255	64

Hope this helps
If you have any doubt feel free to comment
Thank You!!