Question

An organization is granted a block of addresses with the beginning address 188.78.55.0/22. The organization needs to the following subblocks of addresses to use in its subnets as shown below Two subblock of 128 valid host addresses. Three subblock of 60 valid host addresses. One subblock of 16 valid host addresses. Design the subblocks and for each group starting from the first address in the given block of addresses and find the following: 1. Subnet mask 2. First and last possible valid host address 3. Subnet broadcast address Find out how many addresses are unused after these allocations in each subblock. Find out how many addresses can be used for future subblock

Answer 1

Total address are 2^32-22 = 2¹⁰ = 1024 address

1. Two sub block of 128 valid addresses

             for group of each sub block needs 128 valid host addresses. This means that 7 bits (log₂ 128) are needed to define each host. The prefix length is 32-7=25. Following are the address:-

1st block:-   188.78.55.0/25    to    188.78.55.127/25

    2nd block    188.78.55.128/25    to    188.78.55.225/25

Total addresses= 2 x 128 = 256

address mask:- 255.255.255.128 ( for /25 )

Netbits =25    Hostbits=7

   2. Three sub block of 60 valid addresses

we need to give 64 address per block as 60 is not in the power of 2. for group of each sub block needs 64 valid host addresses. This means that 6 bits (log₂ 64) are needed to define each host. The prefix length is 32-6=26. Following are the address:-

1st block:-   188.78.56.0/26    to    188.78.56.63/26

    2nd block    188.78.56.64/26    to    188.78.56.127/26

3rd block    188.78.56.128/26    to    188.78.56.191/26

Total addresses= 3 x 64 = 192

address mask:- 255.255.255.192 ( for /26 )

Netbits =26    Hostbits=6

3. Three sub block of 16 valid addresses

for group of each sub block needs 16 valid host addresses. This means that 4 bits (log₂ 16) are needed to define each host. The prefix length is 32-4=28. Following are the address:-

1st block:-   188.78.56.192/28    to    188.78.56.207/28

Total addresses= 1 x 16 = 16

address mask:- 255.255.255.240 ( for /28 )

Netbits =28    Hostbits=4

Total address granted= 256+192+16 = 464 addresses

Number of addresses for future use= Total addresses - Total address granted

=1024-464

=560 addresses

Broadcast address is 188.78.55.255