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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 The ISP needs to distribute these addresses to three groups of customers as
(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), 32768 (B), 16384
(C), 32768 (D), 16384
in Computer Networks by Active (1k points) | 655 views
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 to  for 1st customer(as each customer have 256 address) to for 2nd customer ..


145 .75.127.0/24 to 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)

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


                      147.75 .128.192/26 to



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

     so second group first address of 128 customer is 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 -
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 {for 1st address of 128th customer all 6 bit will be 0's, So 11000000=192} and subnet is 8+8+8+2=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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