GATE | GATE-CS-2015 (Set 3) | Question 48
In the network 18.104.22.168/27, the fourth octet (in decimal) of the last IP address of the network which can be assigned to a host is ________
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 22.214.171.124/31 which has last octet as 158.
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