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The IP address is $193.1.2.3$.

The subnet mask is $255.255.255.240$.

IP address is originally belongs to class C network which has prefix 24 means 24 bit for network id and 8 bit for host id.

Now we use x number of extra bit for subnetting.

Now given subnet mask has $28$ 1’s and $4$ 0’s. So $28$ bit use for network id and 4 bit for host id.

Now we have $(28-24)=4$ bit for subnetting . so we have $2^{4}=16$ subnets.

Each subnet has 4 bit for host id so we have $2^{4}=16$ host in each subnet out of which $2$ address use for Network and broadcast id so we have total $(16-2)=14$ host in each subnet.

So correct answer is $16$ subnet and $14$ addressable host in each subnet.
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Since 193.1.2.3 is a class C address and in class c address bits used for Network ID are 24 bits

The given subnet mask is 255.255.255.240 = 11111111.11111111.11111111.11110000

Number of bits used for subnetting are 4

So  Number of subnets = 2 Power ( Bits used for Subneting) =  16

Last 4 bits 0000 are used fir host

Number of hosts per subnet= 2 Pow( Bits used for Hosts )  - 2

 

= 16 – 2 = 14

See the Concept for Reference  – IP Address Classes 

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