Question

An ISP is given a block of addresses starting with 190.100.0.0/16 (65,536 addresses). These addresses must further be divided into groups: (a) The first group has 64 customers; each customer requires 256 addresses. (b) The second group has 128 customers; each customer requires 128 addresses. (c) The third group has 128 customers; each customer requires 64 addresses. Design the subblocks and define the subnet range for each. How much addresses are still remaining after these allocations?

Answer 1

Answer:------------
a. Group 1
For this group, each customer needs 256 addresses. This means that 8 (log₂256) bits are
needed to define each host. The prefix length is then 32 - 8 = 24. The addresses are
1st Customer: 190.100,0.0/24 ------------------- 190.100.0.255/24
2nd Customer: 190.100.1.0/24 ------------------- 190.100.1.255/24
……….
64th Customer: 190.100.63.0/24------------------- 190.100.63.255/24
Total = 64 x 256 = 16,384

b. Group 2
For this group, each customer needs 128 addresses. This means that 7 (log₂128) bits are
needed to define each host. The prefix length is then 32 - 7 = 25. The addresses are,
1st Customer: 190.100.64.0/25 ---------------------- 190.100.64.127/25
2nd Customer: 190.100.64.128/25 ------------------ 190.100.64.255/25
……..
128th Customer: 190.100.127.128/25 --------------- 190.100.127.255/25
Total = 128 x 128 = 16,384

c. Group 3
For this group, each customer needs 64 addresses. This means that 6 (log₂64) bits are needed
to each host. The prefix length is then 32 - 6 = 26. The addresses are,
1st Customer: 190.100.128.0/26
2nd Customer: 190.100.128.64/26
……..
128th Customer: 190.100.159.192/26
Total = 128 x 64 = 8192

Number of granted addresses to the ISP: 65,536
Number of allocated addresses by the ISP: 40,960
Number of available addresses: 24,576