Test by Bikram | Mock GATE | Test 2 | Question: 48

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:

1. The first group has $128$ customers; each needs $256$ addresses.
2. The second group has $128$ customers; each needs $64$ addresses.
3. The third group has 64 customers; each needs $128$ addresses.

Find the first address of $128$$^{th}$ 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 1

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

Comments

@Bikram. Dont we add 2 addresses for subnet address and broadcast address to host part? So, for 1st group, we need 256 addresses for each customer and therefore, wont we need 9 bits for host part?

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Or is this modern development that we can use both addresses as well?

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each needs 256 addresses.. means possible 256 address. to which purpose NOT needed So full spectrum and NO minus 2

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 one more possible way to do.

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@Debashish Deka

Small correction needed,

2nd and 3rd parts,

1 0  0/1 _ _ _ _ _      _ _ _ _ _ _ _ _