Question

An ISP is granted a block of addresses starting with 12.60.4.0/22. The ISP wants to distribute these blocks to 100 organizations with each organization receiving just eight addresses. Design the subblocks and give the slash notation for each subblock. Find out how many addresses are still available after these allocations?​

Answer 1

Answer:  yes 224 address is left.

Explanation:  

Assuming that by addresses we mean we can include the broadcast and the network ID in the block we allocate.  

We have /22 means we have 10 bits in spare and 22 bits from the network ID part.

We need 100 subnets each of equal size, so SUBNET ID = ceil(100) = 7 bits and remaining 3 bits can be used for addresses allocated per subnet i.e. 2^3=8  

So CIDR of first block = 12.60.4.0/29

Now we are left with 128-100 =28 Subnet ID each having 8 addresses. So 28×8=224 addresses are left with the ISP.