QUESTION:

An Internet Service Provider (ISP) is granted a block of addresses starting with
145.75.0.0/16. The ISP needs to distribute these addresses to three groups of customers as
follows:
(a) The first group has 128 customers; each needs 256 addresses.
(b) The second group has 128 customers; each needs 64 addresses.
(c) The third group has 64 customers; each needs 128 addresses.
Find the first address of 128th customer of 2nd group and how many addresses are still
available with ISP after these allocations.
(A) 145.75.127.128/24, 32768 (B) 145.75.159.192/26, 16384
(C) 145.75.159.192/26, 32768 (D) 145.75.191.128/25, 16384

ANSWER:

(a)  The first group has 128 customers; each needs 256 addresses.

so for 256 address , we need 8 bits for host ( because 2^8)

host contain = 8 bits , Net id is 32-8= 24 .

group 1 st 128*256(per customer) = 32768 address

Which means address like,
145.75.0.0/24 to 145.75.0.255/24  for 1st customer (each customer have 256 address)

145.75.1.0/24 to 145.75.1.255/24 for 2nd customer

145 .75.127.0/24 to 145.75.127.255/24 for 128th customer

(b) The first group has 128 customers; each needs 64 addresses.

so for 64 address , we need 6 bits for host ( because 2^6)

host contain = 6 bits , Net id is 32-6= 26 bits .

group 1 st 128*64(per customer) = 8192 address

Which means, address like

 147. 75.128.0/26 to 147.75.128. 63 /26 (for customer 1st)

147.75.128.64/26 to 147.75. 128.127/26 (for customer 2nd )

147.75.128.128/26 to 147.75.128.191/26(for customer 3rd)

147.75 .128.192/26 to 147.75.128.255/26(for customer 4th)

…….

147.75..159.192 /26 to 147.75.159.255/26 (last 128th customer )

Second group first address of 128 customer is 147.75.159.192/26 and

remaining address = 2^16 – (32768+8192)=16384

Answer is B.