Ip addressing

Question

A Block contain 32 IP addresses. which of the following can be first address of the block ? and why?

a. 201.55.16.16

b. 201.55.16.8

c. 201.55.16.160

d. 201.55.16.24

Comments

For answering this question, we should have remember the rules of forming CIDR Blocks:

1. There should be contiguous IP addresses.

2. Asking a number of IP addresses should be a power of 2

3. First IP address should be evenly divisible by the total number of IP addresses.

Here, in this question, they have given 32 IP addresses, in order to form them we need five bits right!!

So check the last octet of every option, least five significant bits have to be 0.

In C option, 160 in binary we can write: 10100000  is correct

 

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But binary value of 32 is 00100000 it also contains five significant bits as 0?

Answer 1

C is the answer. Simple.
Since block contains 32 addresses, as per the rule the 1st address should be a multiple of 32.

Answer 2

Since, we need 32 Hosts, that means the first address of the block will have n zeros in it's binary equivalent when moving from right to left part of the address.

where n is such that 2ⁿ>=Number of hosts required

So here you need 32 hosts so we will look for such address which has atleast 5 zeroes when we read the address from right to left as that will form the host id part of the block.

Look for the address 201.55.16.160

Here 160 can be represented as

1100 0000

as you can see we have atleast 5 zeroes when we look from right to left in the binary equivalent of the IP address. So that means, we can have atleast 32 hosts and in the first address of the block all host bits are set to 0's.

So (C) is the answer.

Comments

This is the best explanation and i wanted to find the reason and got it

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How is 11000000 is 160. It is 192.

Answer 3

c. 201.55.16.160 ans is correct

class c: 192 t0 223

a. 201.55.16.16= 16 to 23 =  8 address

b. 201.55.16.8= 8 - 15 = 8 address

c. 201.55.16.160= 160 to 223 = 64 address which is divided by 32

d. 201.55.16.24 = 24 - 159 = 136 address

Comments

Can u please explain what are u trying to do ?

Answer 4

Since the block contains 32 addresses, so the last 8 bits must have starting value >= 32.

Only in case option 'C' u would such a value.

Comments

no its a standard question from geeksforgeeks

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in the above ques - for 10.0.0.32 this block  10.0.0.31 and 10.0.0.0  will not be used.

Thus, address will start from 10.0.0.33. So solution is 10.0.0.160

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i m nt getting u .. plz elaborate more