GATE CS 2015 Set 3 Q38 — Fragmentation and IP Addressing, IPv4, CIDR

To find the last assignable host IP address in the network containing $200.10.11.144/27$, we first determine the network address and its range.

Step 1: Find the Network Address
A $/27$ mask has $27$ network bits and $5$ host bits. The subnet mask is $255.255.255.224$.
Performing a bitwise AND on the fourth octet of the given IP ($144$) and the mask ($224$):$$144 \text{ AND } 224 = 10010000_2 \text{ AND } 11100000_2 = 10000000_2 = 128$$Thus, the network address is $200.10.11.128/27$.

Step 2: Find the Broadcast and Last Host Address
With $5$ host bits, the block size is $2^5 = 32$.

• Network Address: $200.10.11.128$
• Limited Broadcast Address: $200.10.11.(128 + 31) = 200.10.11.159$
• Last Assignable Host Address: $200.10.11.(159 - 1) = 200.10.11.158$