Question

Please Explain to me in details of how you got your answers(so that I can compare to mines).Thanks!

Decode the IP address 130.169.64.35 /19

1. 1. Write the subnet mask in dotted decimal notation.
2. 2. Given the mask, calculate the number of available subnets and number of available hosts per subnet.
3. 3. Calculate the block number, then use it to determine the first six subnet IDs.
4. 4. List the range of valid host addresses and broadcast addresses on the first six subnets.

Answer 1

 130.169.64.35         10000010.10101001.010 00000.00100011
it is class b address so first two octet is reserved for net id .and 3 bit of 3 octet is reserved for subnet id.so 

a)subnet Netmask:   255.255.224.0 =     11111111.11111111.111 00000.00000000

b)no. of available subnet is 2^3=8

no. of available hosts per subnet is 2^13-2=8192-2=8190

c)

subnet id

130.169.0.0

130.169.32.0

130.169.64.0

130.169.96.0

130.169.128.0

130.169.160.0

d)

subnet id	first valid host address	last valid host address	broadcast address
130.169.0.0	130.169.0.1	130.169.31.254	130.169.31.255
130.169.32.0	130.169.32.1	132.169.63.254	130.169.32.255
130.169.64.0	130.169.64.1	132.169.95.254	130.169.95.255
130.169.96.0	130169.96.1	130.169.127.254	130.169.127.255
130.169.128.0	130.169.128.1	130.169.159.254	130.169.159.255
130.169.160.0	130.169.160.1	130.169.200.254	130.169.200.255