IP addressing problem

Question

$\text{If Direct Broadcast Address of subnet is}$ $201.15.16.31$.

$\text{Which of the following is subnet mask?}$

• A. $255.255.255.240$

• B. $255.255.255.192$

• C. $255.255.255.198$

• D. $\text{None Of the Above}$

Comments

Answer will be A) $255.255.255.240$

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the subnet part of IP address for DBA has all 1's so the 31 in last octet specifies that the subnet part of IP address is 5 bits long so subnet mask for that should be 255.255.255.224 so i beleive option D

Answer 1

$\text{Directed Broadcast Address (DBA):}$

When a host in a network wants to send some data to all the hosts in some other network.

By seeing we understand that the given DBA belongs to a class C IP address and the default subnet mak of class C ip address is $255.255.255.0$

Now, in DBA's, there will be all $1's$ in Host ID part.

$201.15.16.31 = 201.15.00010000.00011111$

So we can conclude that the last $5$ bits are host bits. 

∴ in the subnet mask , there will be $\color{red}{27} \hspace{0.1cm}1's$ Or $\color{red}{28} \hspace{0.1cm} 1's$ Or $\color{red}{29} \hspace{0.1cm} 1's$ Or $\color{red}{30} \hspace{0.1cm} 1's$ Or $\color{red}{31} \hspace{0.1cm} 1's$ 

The possible subnet masks are :

$\color{red}{27} \hspace{0.1cm}1's$	$255.255.255.11100000$	$255.255.255.224$	Network ID : $201.15.16.0$
$\color{red}{28} \hspace{0.1cm}1's$	$255.255.255.11110000$	$255.255.255.240$	Network ID : $201.15.16.16$
$\color{red}{29} \hspace{0.1cm}1's$	$255.255.255.11111000$	$255.255.255.248$	Network ID : $201.15.16.24$
$\color{red}{30} \hspace{0.1cm}1's$	$255.255.255.11111100$	$255.255.255.252$	Network ID : $201.15.16.28$
$\color{red}{31} \hspace{0.1cm}1's$	$255.255.255.11111110$	$255.255.255.254$	Network ID : $201.15.16.30$

$\color{blue}{32} \hspace{0.1cm} 1's$ isn't possible , as when the subnet is having $\color{blue}{32} \hspace{0.1cm} 1's$, then the subnet will be $255.255.255.255$, but DBA is $201.15.16.31$, so subnetting must be done & the subnet will be having at most $\color{red}{31} \hspace{0.1cm} 1's$ and at least $\color{red}{27} \hspace{0.1cm} 1's$ .

Comments

Yes this seems correct.
I also think the wording of question should have been "can be " instead of "is"

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Thank you.I think this is best ans to me.

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Yeah Rameesh, the qs must be what can be the subnet mask for the above DBA?

Answer 2

 255.255.255.240 = 255.255.255.11110000  => MSB 24 bits for Network + 4 bits form host octet is Subnet = Total 28 bits Fixed .

 with remaining 4 bits from host octet we can form 2⁴ = 16 IP or 14 Host practically => means each subnet have 16 Host.

Starting Form  --

201.15.16.0000 0000 To  201.15.16.0000 1111 [ 201.15.16.0 - 201.15.16.15 ] in 1st subnet [each subnet have 16 Host]

201.15.16.0001 0000 To  201.15.16.0001 1111 [ 201.15.16.16 - 201.15.16.31 ] in 2nd subnet ... like wise

This clearly state if 201.15.16.31 is a DBA then subnet mask would be minimum 255.255.255.224, and maximum 255.255.255.240,255.255.255.248.....so on

This question could have multiple answer but  from option only on match possible.

Final Answer (A)