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+3 votes
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$\text{If Direct Broadcast Address of subnet is}$ $201.15.16.31$.

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

  1. $255.255.255.240$

  2. $255.255.255.192$

  3. $255.255.255.198$

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

asked in Computer Networks by (231 points)
edited by | 268 views
0
Answer will be A) $255.255.255.240$
0
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

3 Answers

+7 votes
Best answer

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

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

answered by Boss (12.9k points)
selected by
0
Yes this seems correct.
I also think the wording of question should have been "can be " instead of "is"
0

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

0

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

0 votes

option a is right

answered by Boss (24.3k points)
0
What the fucking solution is this???? Though the answer is ryt, but solution is absolutely wrong.
0
Can You Plzz tell Me, How You take 4 bit for host ...?
0
awesome
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)

answered by (109 points)
edited by

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