Ip addressing

Question

given answer is 222 why 254 is not correct?

Comments

Yes there is..

There are 8 bits in total for host address. Out of which half of them are assigned to A. So 2^8/2 host addresses are for A.

So 0 0000000 to 0 1111111.

Remaining is 2^7. That has to be equally divided among BCDE.

2^7/4 =2^5.

Hence 5 bits.

B: 1 00 00000 to 1 00 11111

C: 1 01 00000 to 1 01 11111

D: 1 10 00000 to 1 10 11111

E: 1 11 00000 to 1 11 11111

The 1st and last addresses of every set are not assignable.

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If we assign first bit of last octet o for A,Then other half will be starting from 1

then B,C,D,E will be

1 00

101

110

111

so for D, We have 110 for last host all other bits will be 1 except for last bit 110 11110=222

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there is no compulsory rule for ordering, but from the given option you should go with 222, because that's for examiner want's.

Answer 1

For A, if have take 1 bit from remaining 8 bits.

Lets us take that bit as 0 then we have left option as 0 _ _ _  _ _ _ _

For Remaining 4 subnet, we have option as 1 _ _ _  _ _ _ _

Now we have divide it in 4 equal subnet , so we need 2 bits for that

Am taking it as B = 100 _   _ _ _ _

C = 101 _   _ _ _ _

D = 110 _  _ _ _ _ Here, in remaining 5 bits we can put all 0's to all 1's but if we put all 1's it will be directed broadcast address.

So, the last address of the host will be 11011110 = 222

E = 111_  _ _ _ _