Ace CN question

Question

An organization has a class C network and wants to form a subnet for four departments with hosts as follows:

A:72

B:35

C:20

D:18

what is the possible arrangements?

options:

a) for A: 255.255.255.128 for B: 255.255.255.192 for C & D: 255.255.255.223

b)for A 255.255.255.223 for B: 255.255.255.223 for C & D : 255.255.255.128

My approach for solution:

I just eliminated the rest options and i did it as follows,

for option a) 128=10000000 , so there is $2^{7} = 128$ hosts possible, so it can be for A , again 192= 1100000 so there are $2^{6} = 64 hosts possible , but in 223 = 10110111 means only 4 hosts are possible.

so option A is not possible. But answer given is option A.

I know i am lacking some concept here , please care to correct :)

Comments

I think instead of 255.255.255.223, it should be 255.255.255.224

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May be, is my approach correct?

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Approach is correct but binary of 223 is not and even correct form will not satisfy the given condition.

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Your approach is correct ..

May be for C and D subnet id has to be .224

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Same question doubt.Please explain it... From the given explanation its not clear to me.

Answer 1

The correct answer will be A: 255.255.255.128 for B: 255.255.255.192 for C & D: 255.255.255.224

Organization class given as C i.e. the network bits are 24 and rest are host bits, and the subnet bits are borrowed from the host portion. Lets see for all four cases one by one.

A. 72 = 2^7-2 => 126     (as 2^6 will be less)
Therefore, Network bits = 24, Subnet bits = 1 and host bits will be 7.
We can write it as 11111111 11111111 11111111 10000000, the decimal equivalent of this is 255.255.255.128

B. 35 = 2^6-2 => 62       (as 2^5-2 will be less)
Therefore, Network bits = 24, Subnet bits = 2 and host bits will be 6.
We can write it as 11111111 11111111 11111111 11000000, the decimal equivalent of this is 255.255.255.192

C. 20 = 2^5-2 => 30     (as 2^4-2 will be less)
Therefore, Network bits = 24, Subnet bits = 3 and host bits will be 5.
We can write it as 11111111 11111111 11111111 11100000, the decimal equivalent of this is 255.255.255.224

D. 18 = 2^5-2 => 30     (as 2^4-2 will be less)
Therefore, Network bits = 24, Subnet bits = 3 and host bits will be 5.
We can write it as 11111111 11111111 11111111 11100000, the decimal equivalent of this is 255.255.255.224