Answer:
Machine’s IP=126.127.85.170, Machine’s Netmask=/27, Parent’s Netmask=255.255.240.0 .
a) Machine's Netmask = /27 : therefore no. of remaining bits for hosts =32-27 = 5 bits.
b) No. of usable addresses in the child network = 25 -2 = 32-2 =30 [Since first(network ID of the machine) and last ip (broadcast address of the machine ) addresses are not used ]
c) first usable address is on this child network (subnet) =
First, find out the network id of machine can be found out by doing bitwise AND machine's IP and Machine's subnet mask :
01111110. 01111111.01010101.10101010 (IP)
11111111. 11111111 .11111111 .11100000 (Subnet Mask)
01111110. 01111111. 01010101.10100000 (Network ID )
first usable address is on this child network (subnet) : 01111110. 01111111. 01010101.10100001
: 126.127.85.161
d) what the last usable address is on this child network (subnet) : 01111110. 01111111. 01010101.10111110
: 126.127.85.190
e) what the child network’s (subnet's) broadcast address is :
In the directed broadcast address , all the host bits are 1. Therefore, broadcast address :
01111110. 01111111. 01010101.10111111 (126.127.85.191)
f) what the child network's (subnet's) network number is : Network ID has already been calcuulated in part c .
01111110. 01111111. 01010101.10100000 (126.127.85.160)
