IP Address allocation

Question

An ISP has assigned an address block to user in which 4096 hosts can be assigned IP addresses. Which of the following can be the network ID ?

• A. 168.72.90.0/20
• B. 168.72.96.0/20
• C. 168.72.64.64/20
• D. 168.64.64.0/24

Answer 1

Anwer- B. 168.72.96.0/20

for n/w id, all the host id part should be 0. Since it has to support 4096 hosts, -> 2^12 => 12 bits represent host id.

Note- Here in question, it is mentioned to assign IP to 4096 hosts, so we would actually need 13 bits to support this many hosts.[ 1 for n/w id and another for broadcast]. But since, no option satisfies, therefore, it is obvious that they are neglecting these 2 addresses.So, consider 12 bits for host id.

Therefore, 32-12 = 20 bits represent n/w id. Now, we know the prefix-length, the remaining 12 bits should be all 0's which is only in option B.

Therefore this is the solution.

Comments

@Ashish Goyal y not option A ?

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@Shadan Karim, as i mentioned in the answer-

the remaining 12 bits should be all 0's

So, check for yourself. Clearly the last 8 bits are 0's in both A and B. You just need to check if the last 4 bits of the 3rd octet of the address(i.e.- 90 for A and 96 for B) are 0's.

Either you can convert them into binary and check or the easier way would be to check if they leaves any reminder when divided by 16(2^4).