If we divide it into four subnets, and 4 is the second power of 2, then the current subnet mask length is 27+2 (power) =29 bits, that is:
255.255.255.248 binary is:111111.165438+ 438+0 1 1 1 1 1. 1 1 1 1 1000
The number of available addresses in the subnet: 2 n-2, where n is the number of hosts, and n=32-29 (mask length) =3, so 2-2 is cubic =6, and the number of available addresses is 6. There are two addresses, namely network address and broadcast address, so a subnet * * * has eight (cubic of 2) addresses, which should be subtracted from the available addresses.
Maximum subnet address:
The original network is 217.14.8.0-214.8.31(32-27 = 5, the interval is calculated according to this method). Now it is divided into four subnets:
217.14.8.0-217.14.8.7 (a subnet has 8 addresses, calculated according to this method, the same below).
2 17. 14.8.8 - 2 17. 14.8. 15
2 17. 14.8. 16 - 2 17. 14.8.23
2 17. 14.8. 24-214.8.31(this is the largest subnet with a network address of 217./kloc-
Broadcast address:
The broadcast addresses of the four subnets are
2 17. 14.8.7
2 17. 14.8. 15
2 17. 14.8.23
217.14.8.31(that is, the last address within the network)
In addition, the 1 address within the network scope is the network address.
Supplement:
Landlord, isn't this clear? Why don't you finish this problem? I have something to do.