Question

1. Write the IP address 222.1.1.20 mask 255.255.255.192 in CIDR notation. Recall that CIDR notation uses the IP address in dotted decimal followed a /n, where n is the number of significant or "unmasked" bits

2. A network administrator wants to place 27 subnets on a /23 network.  How many bits would be in the host field of the address? Just enter the integer number of bits with no label.

Answer 1

1) IP address = 222.1.1.20

Mask = 255.255.255.192 = 1111 1111.1111 1111.1111 1111.1100 0000

So, we can see that there are 26 consecutive 1's in the mask. So, the block ID will be /26.

So, the IP address will be = 222.1.1.20/26.

2) Any /23 network implies:

First 23 bits are network ID.

We know, any IP address consists of 32 bits.

So, number of bits in the host ID = (32 - 23) = 9

Number of subnets = 27

Number of bits required = ceil (log₂ 27) = 5

We will get these 5 bits from the host ID part.

So, number of remaining bits in the host field of the address = (9 - 5) = 4

Please comment in case of any doubt.
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