Assume there is a machine with the IP address 129.82.102.63 with netmask /23, and with a parent NW whose netmask is 255.255.224.0. For each answer, do not include any spaces, give full IP addresses/netmasks where these are requested, give the "/" as part of the answer for slash notation.

Complete question is:

Assume there is a machine with the IP address 129.82.102.63 with netmask /23, and with a parent NW whose netmask is 255.255.224.0.

For each answer, do not include any spaces, give full IP addresses/netmasks where these are requested, give the "/" as part of the answer for slash notation.

a. What is the parent NW's netmask in dotted decimal notation?  

b. What is the parent NW's netmask in slash notation?  

c. What is the child NW's (subnet's) netmask in dotted decimal notation?  

d. What is the child NW's (subnet's) netmask in slash notation?  

e. How many bits are there for host # portion for the parent NW? (Another way to say the same thing is How many bits do we manage - on the parent NW?)  

f. How many bits are there for NW# portion (within the parent address space) for the subnet?

g. How many bits are there for host # portion for the subnet?  

h. How many addresses can we assign to machines/interfaces on this subnet?

a. 255.255.224.0

b. /19

255.255 amounts to 16 bits being 1. .224 means 3 more bits are 1. So, in total 19 bits are 1. Hence, total network bits are 16 + 3 = 19.

c. 255.255.254.0

/23 means 8 + 8 + 7 that means

first 2 octets are 1s and 7 bits out of the 3rd octet are 1s. Hence, /23 means 255.255.254.0

d. /23

e. 13 bits are reserved for hosts

Parent network mask is /19, so total 32-19 = 13 bits

f. 19 bits are reserved for the network in the parent address.

g. 9 bits

Subnetwork's mask is /23, so total 32-23 = 9 bits for the host portion.

h. Since 9 bits are reserved for hosts, a total of 29 -2 = 510 machines can be assigned the IP addresses. Two addresses will be network and broadcast addresses for the subnet that can't be allocated to any device.

Explanation: