You have been assigned a network ID of 172.16.0.0/26. If you utilize the first network resulting from this ID,
what would be the last legitimate host address in this subnet?
A.
172.16.0.64
B.
172.16.0.63
C.
172.16.0.62
D.
172.16.0.65
Explanation:
When a class B address such as 172.16.0.0 is subnetted with a /26 mask, the subnet mask in dotted decimal
format is 255.255.255.192. This means that the interval between the network IDs of the resulting subnets is 64.
The resulting network IDs are as follows:
172.16.0.0
172.16.0.64
172.16.0.128
172.16.0.192
172.16.1.0
and so on.
For the network ID 172.16.0.0, the last address in the range is 172.16.0.63, which is the broadcast address.
Neither the network ID nor the broadcast address for any subnet can be assigned to computers. This means
that the addresses that can actually be assigned range from 172.16.0.1 to 172.16.0.62. The last legitimate host
address, therefore, is 172.16.0.62.
172.16.0.63 cannot be used because it is the broadcast address for the 172.16.0.0 network.
172.16.0.64 is the network ID for the 172.16.0.64 network, and 172.16.0.65 is the first address in the second
network.
Objective:
Network Fundamentals
Sub-Objective:
Apply troubleshooting methodologies to resolve problemsCisco > Support > IP Routing > Design TechNotes > Document ID: 13788 > IP Addressing and Subnetting for
New Users