Hi.

the address belongs to the subnet 172.16.112.0 /20

/20 = 255.255.240.0

how??

first write the host address with the /20, /20 = 20 bits for the subnet portion.

so the first and second octets have 16 bits (8 bits each octet)

but we need 20, so we must use only the first 4 bits of the third octet.

as a resulta 16 bits from the first and second octet, plus 4 bits from the second octet we have totally 20 bits.

the next step is to conver the third octet in binary:

128 64 32 16 8 4 2 1

128 192 224 240 248 252 254 255

if you look at the 4 first octets are in blue color, you notice that the mask portion also in blue color ends in a 240.

so we get the result of the mask in the third octet:

255.255.240.0

now that we have the mask, we must know the subnet address to whic the host **172.16.121.1** bolngs to.

In the next step we again write the third octet where we were working bu we must convert the third decimal number (121) into binary.

128 64 32 16 8 4 2 1

128 192 224 240 248 252 254 255

0 1 1 1 1 0 0 1

where the binary 0 1 1 1 1 0 0 1 is the decimal 121

now, if we look only at the first four bits of the binary we look that 0 1 1 1 = 112 decimal

this is how the subnet bits work together with the ip address to determine the subnet address.

so it is the value of the third and how we get the subnet value 172.16.112.0

now for the hosts you can use within this subnet we can work again with binary values:

128 64 32 16 8 4 2 1

128 192 224 240 248 252 254 255

0 1 1 1 0 0 0 0

at this point if we put all zeros in the host portion of the third and fourth octet we we the subnet address 172.16.112.0

for each host you must increment by one sequentially starting from the the last bit of the fourth octet, for exapmle:

third octet:

0 1 1 1 0 0 0 0

fourth octet:

0 0 0 0 0 0 0 0

**so for the first host: **

third octet:

0 1 1 1 0 0 0 0

fourth octet:

0 0 0 0 0 0 0 1

in decimal = 172.16.112.1

**the seconf host:**

third octet:

0 1 1 1 0 0 0 0

fourth octet:

0 0 0 0 0 0 1 0

in decimal = 172.16.112.2

for the last host:

third octet:

0 1 1 1 1 1 1 1

fourth octet:

1 1 1 1 1 1 1 0

in decimal = 172.16.127.254

so the complete range of usable host address for the entire subnet address are:

172.16.112.0 /20 network

172.16.112.1 - 172.16.127.254 /20 range of usable hosts

172.16.127.255 /20 broadcast for the subnet.

in the example it looks easy, try to practice a lot.