GATE | GATE-CS-2015 (Set 3) | Question 48

In the network 200.10.11.144/27, the fourth octet (in decimal) of the last IP address of the network which can be assigned to a host is ________
(A) 158
(B) 255
(C) 222
(D) 223


Answer: (A)

Explanation:

The last or fourth octet of network address is 144 
144 in binary is 10010000.

The first three bits of this octal are fixed as 100, 
the remaining bits can get maximum value as 11111.  
So the maximum possible last octal IP address is 
10011111 which is 159.

The question seems to by asking about host address. The
address with all 1s in host part is broadcast address 
and can't be assigned to a host. So the maximum possible 
last octal in a host IP is 10011110 which is 158.

The maximum possible network address that can be assigned 
is 200.10.11.158/31 which has last octet as 158.

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