Answer to Question #283823 in Computer Networks for mafi

Question

10.10.10.100/28 answer the following question depending on the given ip address

A. network address

B.first host address

C.last host address

D.broadcast address

Answer 1

A.

mask is /28, so last octet is: 11110000, so there are $2^4=16$ bits for network (value of network address jumps through 16)

then  network address:

10.10.10.96

B.

first host address:

10.10.10.97

C.

last host address:

10.10.10.96 = 10.10.10.(96+16-2)=10.10.10.110

D.

broadcast address (last address of the network):

10.10.10.96 = 10.10.10.(96+16-1)=10.10.10.111