Question

) how many bits are used for host number on the child network (subnet) , b) how many usable addresses can exist on this child network (subnet) , c) what the first usable address is on this child network (subnet) , d) what the last usable address is on this child network (subnet) , e) what the child network’s (subnet's) broadcast address is , and f) what the child network's (subnet's) network number is . Machine’s IP=126.127.85.170, Machine’s Netmask=/27, Parent’s Netmask=255.255.240.0 .

Answer 1

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)