Networking - Subnetting - Discussion

D is the correct answer in the case of /22 =255.255.252.0 but A not.

Let me explain.
We learned that subnets with CIDR /22 can skip at 4 size block.

So, how is possible that the question subnet is 17.0, when the only ones available are 16.0, 20.0 etc?

Please explain.

As per question, we need valid host address and the correct option is D, can you explain how 172.16.18.255 is valid host address?

According to the question, 172.16.17.0/22 this is given but while solving u said in question 172.16.16.0 is used which makes only option D as a correct answer but as per real question, both A & D can be right.

@Reddy.

D is the right answer bcoz of it's subnet.
class B subnet is always N:N:H:H bits.
172.16.18.255 is also in between the correct range.
A option comes under the C Class subnet tata's why we choose D as the correct answer.

Can you give the proper explanation?

Hi Guys!

We have first 6 on-bits so after calculating the values.
128 + 64 + 32 + 16 + 8 + 4 = 252, we get sum of on-bit value.

Block size = 256 - Sum of on bit value.

256 - 252 = 4,
The block size is 4.

Subnets in the network
172.16.0.0/22
172.16.4.0/22
172.16.8.0/22
172.16.12.0/22
172.16.16.0/22
172.16.20.0/22 -------- and so on up to 172.16.252.0


So to here some of you confused that why A is not correct the reason behind it is
that if you select A, as you answer then you are Host bits are showing as network bits bcoz we have an ip of B class and in B class we deal with N.N.H.H, so we are not selecting A as the correct answer due to the wrong Subnet.

D is dealing with class B Subnet.
(172. 16 . 18 . 255)
(255. 255. 252. 0 )
( N . N . H . H )

Hi, I can not understand how they got it 172.16.18.255, Can anyone explain me in detail?

We know the bi borrowed id 6, ie 11111100 = 252 and we deal with N.N.H.H.

That's why the answer is D.

Please describe why A is not the correct answer?