Question

An ISP granted a block of addresses starting with 192.168.0.0/20. The ISP need to distribute these addresses to three groups of customers as follow:

• The first group has 32 customers, each needs 64 addresses.

• The second group has 16 customers, each needs 32 addresses.

• The third group has 16 customers, each needs 128 addresses.

Design the sub-blocks for the above given specifications.

Answer 1

Answers:

For First Group:

Given that each customers needed 64 addresses, So the suffix length is 2⁶=64 i.e 6. Therefore prefix length will be 32-6=26.

The address will be

1st customer	192.168.0.0/26	192.168.0.63/26
2nd customer	192.168.1.0/26	192.168.1.63/26
3rd customer	192.168.2.0/26	192.168.2.63/26
.	.	.
.	.	.
.	.	.
upto 32th customer	192.168.31.0/26	192.168.31.63/26

For 2nd Group

Given that each customers needed 32 addresses, So the suffix length is 2⁵=32 i.e 5. Therefore prefix length will be 32-5=27.

The addresses are:

1st customer	192.168.32.0/27	192.168.32.31/27
2nd customer	192.168.33.32/27	192.168.33.128/27
3rd customer	192.168.34.32/27	192.168.34.128/27
.	.	.
.	.	.
.	.	.
upto 16th customer	192.168.47.32/27	192.168.47.128/27

For 3rd Group

Given that each customers needed 128 addresses, So the suffix length is 2⁷=128 i.e 7. Therefore prefix length will be 32-7=25.

The addresses are:

1st customer	192.168.48.0/25	192.168.48.127/25
2nd customer	192.168.49.128/25	192.168.49.128/25
3rd customer	192.168.50.128/25	192.168.50.128/25
.	.	.
.	.	.
.	.	.
upto 16th customer	192.168.64.128/25	192.168.64.128/25

Finally Sub blocks diagram

ISP Customer 1 192.168.0.0/26 Group 1 192.168.31.0/26 Customer 32 Customer 1 192.168.32.0/27 Group 2 192.168.47.16/27 Customer 16 Cutomer 1 192.168.48.0./25 Group 3 192.168.63.16/25 Customer 16