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An Internet Service provider (ISP) has the following chunk of CIDR- based IP addresses available 252.224.128.0/22. Now it gives its quarter of addreeses to X, Calculate  the valid allocation of addresses to the X ?
asked in Computer Networks by Active (2.1k points) | 221 views
0
252.224.128.0/24

to

252.224.128.254/24
0
no of ip address possible are 256
so we need 8 bits so we have only 253 address possible to allocate the address because as one address is needed from net ID and other one will be needed for broadcast address.
corrrect me if i am wrong ??

2 Answers

+1 vote
252.224.128.0/22 has addresses till 252.224.131.255. There are 2^10 hosts in the block.

It is needed that X be given 2^10/4 = 2^8 hosts. So, HID in X = 8 bits. 2 bits will be borrowed to act as SID.

So, possible subnets are:

1. 252.224.1000 0000.0/24 or 252.224.128.0/24

2. 252.224.1000 0001.0/24 or 252.224.129.0/24

3. 252.224.1000 0010.0/24 or 252.224.130.0/24

4. 252.224.1000 0011.0/24 or 252.224.131.0/24

Any one of 252.224.128.0/24 , 252.224.129.0/24, 252.224.130.0/24, 252.224.131.0/24 can be given to X.
answered by Active (1.7k points)
0 votes
$\underbrace{252.224.100000}_\text{Network ID} \underbrace{00.00000000}_\text{Host ID}$

Now giving 1st quarter of chuck to $X \rightarrow$ $\underbrace{252.224.10000000}_\text{Network ID} .\underbrace{00000000}_\text{Host ID} \rightarrow 252.224.128.0/24$

Then we are left with,

$\underbrace{252.224.10000001}_\text{Network ID} .\underbrace{00000000}_\text{Host ID} \rightarrow 252.224.129.0/24$

$\underbrace{252.224.10000010}_\text{Network ID} .\underbrace{00000000}_\text{Host ID} \rightarrow 252.224.130.0/24$

$\underbrace{252.224.10000011}_\text{Network ID} .\underbrace{00000000}_\text{Host ID} \rightarrow 252.224.131.0/24$
answered by Boss (23.7k points)

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