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 _________.

We are given the CIDR block prefix /27 with an IP in the network 200.10.11.144/27.

1. For a /27 prefix, the subnet mask is:

   $$255.255.255.224$$

   The number of host bits is $32 - 27 = 5$. Block size in the 4th octet = $2^5 = 32$.

2. The network address is found by bitwise ANDing the given 4th octet with the mask:

   $$144 = 10010000_2$$

   $$\text{Mask} = 224 = 11100000_2$$

   $$\text{Network 4th octet} = 144 \& 224 = 128 = 10000000_2$$

3. The subnet range for 200.10.11.128/27 is:

   • Network address:200.10.11.128

   • First assignable host IP:200.10.11.129

   • Last assignable host IP:200.10.11.158

   • Directed broadcast address:200.10.11.159 ($128 + 32 - 1 = 159$)

Therefore, the fourth octet of the last assignable host IP address is 158.