Given 145.75.0.0/16 is the starting address. So, we have 216 = 65,536 addresses with the ISP initially.
For the 1st group, each customer needs 256 addresses. So, 8 bits are needed to define each host. The prefix length is 32 – 8 = 24. Therefore the addresses of 1st group are:
1st customer ---> 145.75.0.0/24 to 145.75.0.255/24
2nd customer ---> 145.75.1.0/24 to 145.75.1.255/24
128th customer 145.75.127.0/24 to 145.75.127.255/24
So, for 1st group the ISP has allocated 128*256 = 32768 addresses
For the 2nd group, each customer needs 64 addresses. So, 6 bits are needed to define each host. The prefix length is 32 – 6 = 26. Therefore the addresses of 2nd group are:
1st customer 145.75.128.0/26 to 145.75.128.63/26
2nd customer 145.75.128.64/26 to 145.75.128.127/26
3rd customer 145.75.128.128/26 to 145.75.128.191/26
4th customer 145.75.128.192/26 to 145.75.128.255/26
128th customer 145.75.159.192/26 to 145.75.159.255/26
So, for 2nd group the ISP has allocated 128*64 = 8192 addresses
For the 3rd group, each customer needs 128 addresses. So, 7 bits are needed to define each host. The prefix length is 32 – 7 = 25. Therefore the addresses of 3rd group are:
1st customer 145.75.160.0/25 to 145.75.160.127/25
2nd customer 145.75.160.128/25 to 145.75.160.255/25
64th customer 145.75.191.128/25 to 145.75.191.255/25
So, for 3rd group the ISP has allocated 64*128 = 8192 addresses.
Hence the total no.of addresses available after allocating is 65536 – (32768 + 8192 + 8192) = 16384 addresses