through the Internet. 172.16.0.1 is a host address, not a broadcast
address. The network broadcast address for this address would be
172.16.255.255.
created the IP address so that more addresses would be available to
each network.
bits, or 254 subnets. You are going to have to use one subnet bit from the
fourth octet, or 255.255.255.128. This is 9 subnet bits (2
Cisco wants you to do is to add "12 bits" to a default mask. Since this
is a Class B address, the default mask is 255.255. Adding 12 bits
would make the mask 255.255.255.240. The host is located in the
10.32 subnet with the next subnet being 10.48. The valid host range
is then 3346 because 47 is the broadcast address.
if you are using subnet-zero), each with 62 hosts. 255.255.255.224
only provides 30 hosts per subnet.
used to find a hardware address from a known IP address.
all Application layer protocols. IP is a Network layer protocol, as well
as Ping. Transmission Control Protocol (TCP) is a Transport layer
protocol.
the answers will make you think. First, the 254-host answers are not
even close, so you have a 50/50 chance between the four-network
and two-network options. The part you probably thought about the
most is whether Cisco is using subnet-zero. This would provide four
networks instead of two. Makes sense. However, Cisco does not con-
sider subnet-zero valid on any of their exams. In addition, you cannot
get 64 hosts with this mask, only 62 hosts per subnet, regardless of
whether you are using subnet-zero or not, which makes the four-
network answer completely wrong.