edited by
1,983 views
5 votes
5 votes

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

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

  1. $255.255.255.240$

  2. $255.255.255.192$

  3. $255.255.255.198$

  4. $\text{None Of the Above}$

edited by

2 Answers

Best answer
9 votes
9 votes

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

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 $\color{red}{27} \hspace{0.1cm} 1's$ .

selected by
0 votes
0 votes

 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)

edited by

Related questions

2 votes
2 votes
0 answers
1
Sambhrant Maurya asked Sep 16, 2018
718 views
A host in a subnet has the IP address 130.83.126.10. How many hosts can be addressed in the network?
2 votes
2 votes
1 answer
3
Na462 asked Dec 8, 2018
2,142 views
An ISP has assigned an address block to user in which 4096 hosts can be assigned IP addresses. Which of the following can be the network ID ?168.72.90.0/20168.72.96.0/201...
2 votes
2 votes
3 answers
4
aditi19 asked Oct 23, 2018
3,480 views
If Direct Broadcast Address of subnet is 201.15.16.31. Which of the following will be subnet mask ?1. 255.255.255.2402. 255.255.255.1923. 255.255.255.1984. NOTAExplain yo...