在网络 200.10.11.144/27 中，可以分配给主机的网络的最后一个 IP 地址的第四个八位字节(十进制)是________
(一) 158
(乙) 255
(三) 222
(四) 223答案：(一)
解释：
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.
这个问题的测验