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A subnetted Class $B$ network has the following broadcast address: $$

Its subnet mask

  1. is necessarily $$
  2. is necessarily $$
  3. is necessarily $$
  4. could be any one of $$, $$,$$
in Computer Networks by Boss (16.3k points)
edited by | 2.9k views

$Remark:$ By looking only at DBA, we can not uniquely guess the subnet address.
It's a class B DBA, there are 16 host bits in class B address without any subnetting, and any number of bits can be taken from these 16 bits to form the subnets.
So, (D) is answer!

Why can't it be 252 or 254 or 255?
@ vikas verma because they are not given in the options

1 Answer

+43 votes
Best answer
Option ($D$) is correct. In the broadcast address for a subnet, all the host bits are set to $1$. So as long as all the bits to the right are $1$, bits left to it can be taken as possible subnet.

Broadcast address for subnet is $.95.255$   $.0101 1111. 1111 1111$ (as in Class $B$, $16$ $bits$ each are used for network and host)

So, we can take minimum $3$ $bits$ (from left) as subnet and make rest as host bits(as they are $1$).

$.224.0$     $1110 0000. 0000 0000$   (leftmost $3$ $bits$ for subnet)

$.240.0$     $1111 0000. 0000 0000$  (leftmost $4$ $bits$ for subnet)

$.248.0$     $1111 1000. 0000 0000$  (...          $5$ $bits$ for subnet )
by Loyal (7.8k points)
edited by
here we can keep continuing with 3,4,5.. bits right?

Since in subnetting we can borrow bits from the host part.Please clarify Sir
yes, S.M could be anything.
If there is one more option  then is it also a right subnet mask Or not?

255.255.128. 0 will not be the correct answer because of non continuous zero in the binary representation for host bits.
Please make it more clear

In case of, if you perform AND operation between 128 (10000000) and 95(01011111) the result will be 00000000 and  here you can see that the subnet bits are zero which will result in invalid subnet number. While if you perform the AND operation among any of the given masks they will give us the valid subnet number.

Please correct me if i am wrong.
ohh the concept of invalid IP got it . thanks
It can be any subnet mask like

And so on , but option D has only three options to choose but it could by any one of mentione above so how option D is correct?
Yes you are right but see the qyestion, as per the given question D is the correct one. As it could be any among a, b, c.
But, as the broadcast address is given as 144.16.01011111.11111111
(in the host id part there are all 1's in case of directed broadcast address)
Therefore the subnet mask has to be from and after that and so on..
not from
(and also if we don't do subnetting the entire network itself is one subnet)
If a option given.. That would also be added in the answer i guess?

@Shamim Ahmed

192 is .11000000  .. but we are given the broadcast address where the host id part is series of one ( or the last ip of the network)... so is the least sm possible 

0 cannot be subnet mask because it has to be limited broadcast address. Correct me if I am wrong.

@y subnet mask is a series of 1 and 0 where 0 represent the number of bits for host id. does not contains any 0 so it cannot be subnet mask.


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