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If the IP is 193.1.2.3 and Subnet Mask is 255.255.255.240. How many numbers of subnets and hosts possible in each subnet are:

1. 14,16
1. 14,14
1. 16,14
1. 16,16

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.

Thank you

I’ve found This explanation from chegg.com and RBR solution, where number of subnet is subtracted by 2. Results to 16- 2 = 14 subnets

Could you please clarify that the solution given by them is wrong?

edited

Why should we subtract 2 from number of subnet ? Did they provide any clearance on that ?

For No of host there is a reason why we subtract $2$ from them but i don’t know any reason for number of subnets.

As per me the solution provide by them is wrong.

See this question : GATE CSE 2007 | Question: 67, ISRO2016-72 - GATE Overflow for GATE CSE

Read each and every discussion hope that helps.

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

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