subnet mask made easy

Question

If the broadcast address of the subnet is given as 163.93.63.255, which of the following mask cannot suit the above address?

A) 255.255.240.0

B) 255.255.248.0

C) 255.255.128.0

D) Both (a) and (b)

Answer 1

(c)
in 163.93.63.255

it is class b address where nid = 16 bit and hid = 16 bit by default and default mask of class B is 255.255.0.0

in broadcast address all the host id should be 1

for a given add

it is 163.93.00111111.11111111

if we take 128 then host id bits are 16 bit of mask = 10000000.00000000

then making all the 0 as 1  will violate the above broadcase address...

Comments

can someone please explain thus answer bit more?

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How is it violating? I don't understand. If option C is violating don't you think A and B which are greater than 128 are also violating?

Answer 2

Note that the broadcast address is the last address.

163.93.63.255 = 10100011.01011101.00111111.11111111. The default mask for class  B is 255.255.0.0/16

You may note that it is not a perfect class B. Rather, it is a subnet.

Let say (assume) 163.93.00111111.11111111/18 is the subnet address and 163.93.0.0/18 is the mask. I have taken /18 as the mask because 63 (00111111) is present in the third octet. But it is an assumption we can take a bigger number also.

Let's consider another example:-

Say 163.93.00111111.11111111/23 as the subnet address and 255.255.11111000.00000000/23 (or 163.93.248.0/23) as the mask. This is a valid mask because the currently chosen subnet will range from 163.93.00111000.00000000/23 to 163.93.00111111.11111111/23. The duty of mask is to produce the network id when an AND operation is applied with the address. So, apply AND operation of the mask 163.93.248.0/23 with 163.93.63.255/23. You will get 163.93.00111000.00000000/23, which the network id or the first id of our assumed subnet.  

Similarly, assume that 163.93.00111111.11111111/22 as the subnet address. The mask is 163.93.240.0/22 or 163.93.11110000.00000000/22. Because when this mask is ANDed with the given address produces 163.93.00110000.00000000/22 which is the network address or the first address of the assumed subnet.

Now consider 163.93.10000000.00000000/17 as the mask. When you do AND operation with 163.93.63.255/17 it gives 163.93.0.0/17 as the answer. That means the subnet ranges 163.93.00000000.00000000/17 to 163.93.01111111.11111111/17. But the broadcast address has to be the last address. Here the last address is 163.93.63.255/17(given in question it as the broadcast address with /17 as our assumption) but should be 163.93.127.255/17. Hence a contradiction and hence it is not the answer.

Hope it helps!