2.4k views

A subnetted Class $B$ network has the following broadcast address: $144.16.95.255$

1. is necessarily $255.255.224.0$
2. is necessarily $255.255.240.0$
3. is necessarily $255.255.248.0$
4. could be any one of $255.255.224.0$, $255.255.240.0$,$255.255.248.0$

edited | 2.4k views
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$Remark:$ 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.

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

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 (7.7k points)
edited
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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
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yes, S.M could be anything.
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If there is one more option 255.255.128.0  then is it also a right subnet mask Or not?
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@Divyanshum29

255.255.128. 0 will not be the correct answer because of non continuous zero in the binary representation for host bits.
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@Divyanshum29

In case of 255.255.128.0, 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.
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ohh the concept of invalid IP got it . thanks
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It can be any subnet mask like

255.255.224.0

255.255.240.0

255.255.248.0

255.255.252.0

255.255.254.0

255.255.255.0

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?
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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.
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(in the host id part there are all 1's in case of directed broadcast address)
Therefore the subnet mask has to be from 144.16.224.0 and after that 144.16.240.0 and so on..
not from 144.16.128.0
(and also if we don't do subnetting the entire network itself is one subnet)
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If a option 255.255.192.0 given.. That would also be added in the answer i guess?
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@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 255.255.224.0 is the least sm possible

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

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