Imagine that an organization is assigned the address block 1

Imagine that an organization is assigned the address block 169.222.0.0/18, and the organization needs one subnet with 220 hosts, one subnet with 128 hosts, and one subnet with 19 hosts. Create the appropriate network identifiers for each subnet, using the process established for VLSMs in the overview section. Refer to step 1 for a similar example. Make sure to show your work.

Solution

given address block is 169.222.0.0/18.

-> here /18 refers the network bits in the address and remaining 14(32-18) bits are host bits.

-> class B address range is 128 - 191.

-> so given address is in class B.default network bits are 16 for class B ,i.e default subnet mask is 255.255.0.0

-> subnet bits = network bits - default network bits = 18 -16 = 2

so total number of subnets are 2 ² = 4 subnets.(only 2 valid subnets).

subnet mask = 255.255.192.0 = 11111111.11111111.11oooooo.00000000

-> host biits 14.

so total number of hosts are 2 ¹⁴ - 2 =16384-2=16382.

-> block size = 256 - subnet mask=256-192 = 64.so subnets are 0,64,128,192

the four subnets are :

subnet 1 = 169.222.0.0/18

subnet 2 = 169.222.64.0/18

subnet 3 = 169.222.128.0/18

subnet 4 = 169.222.192.0/18

-> clearly it can be shown 169.222.0.0/18 is in subnet 1.

-> consider these four subnets for the required number of hosts.

-> for 220 hosts the valid subnet would be  169.222.0.0 i.e from 169.222.0.1 to  169.222.0.220

-> for 128 hosts the valid subnet would be  169.222.0.0 i.e from 169.222.0.1 to  169.222.0.128

-> for 19 hosts the valid subnet would be  169.222.0.0 i.e from 169.222.0.1 to  169.222.0.19