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I want to have an idea about how the divisions are done.Please explain

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
in Computer Networks by Active (1k points) | 655 views
0
B ?

2 Answers

+4 votes
1) First Group = 128 customer , per customers 256 address ,

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

host contain = 8 bits and 32-8= 24 for NID ,

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

it will be like

145.75.0.0/24 to 145.75.0.255/24  for 1st customer(as 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

now group 2 64 address so need 6 bit for host and remaining are 26 bits

so total address for it 64*128 =8192

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(3rd

                      147.75 .128.192/26 to 147.75.128.255/26(4th

                     147.75.129.0/26 to 147.75.129.63/26(5th)

                     ......

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

     so second group first address of 128 customer is 147.75.159.192/26 and

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

so B is correct
by Boss (17.1k points)
+1 vote

Ans is B.
for 1st group 128 customer and each need 256 address{Note: they call address not host}
So 1st group have address range
145.75.0.0 - 145.75.127.255.
now 2nd group
128 customer and 64 address So
145.75.10000000.00 {rest 6 bit for address contain by each customer} -> 1st customer of 2nd group
145.75.10000000.01 {rest 6 bit for address contain by each customer} -> 2nd customer of 2nd group
145.75.10000000.10 {rest 6 bit for address contain by each customer} -> 3rd customer of 2nd group
and so on
145.75.10011111.11 {rest 6 bit for address contain by each customer} -> 128th customer of 2nd group
means 145.75.159.192 {for 1st address of 128th customer all 6 bit will be 0's, So 11000000=192} and subnet is 8+8+8+2=26
So 145.75.159.192/26

addresses contain by 1st group=215
addresses contain by 2nd group=213
addresses contain by 3rd group=213
So how many address are still available 216-215-213-213=16384

by Active (1.1k points)
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