Question 48

Given address with prefix: 200.10.11.144/27.
The prefix length is $/27$, which leaves $32 - 27 = 5$ host bits.
Block size $= 2^5 = 32$ addresses per subnet.

Subnet boundaries for multiples of 32 in the fourth octet:
$0, 32, 64, 96, 128, 160, \dots$
Since $128 \le 144 < 160$, the subnet network ID is $200.10.11.128/27$.

The subnet spans addresses from $200.10.11.128$ to $200.10.11.159$:

• Network address: $200.10.11.128$
• Broadcast address: $200.10.11.159$
• Assignable host address range: $200.10.11.129$ to $200.10.11.158$

The fourth octet of the last assignable host address is 158.