Computer Networking - Subnetting

Question

An ISP is granted a block of addresses starting with 190.100.0.0/16 (65,536 addresses). The ISP needs to distribute these addresses to three groups of customers as follows:

a. The first group has 64 customers; each needs 256 addresses.

b. The second group has 128 customers; each needs 128 addresses.

c. The third group has 128 customers; each needs 64 addresses.

Design the sub blocks and find out how many addresses are still available after these allocations.

Answer 1

The first group of 64 customers will each need a /24 network (64 /24 networks total). The second group of 128 customers will each need a /25 (128 /25 networks total). And the third group of 128 customers will each need a /26 (128 /26 networks total).

Here is how I would design the allocations:

1. First Group:
2. Each customer would get their own subnet 190.100.X.0/24, where X is a number that ranges 0-63 for each customer. For example, customer 1 would get 190.100.0.0/24, customer 2 would get 190.100.1.0/24, customer 3 would get 190.100.2.0/24, etc. all the way up to 190.100.63.0/24.
3. Second Group:
4. Each customer of the second group would start off adjacently to where the first group's IP blocks left off. So the first customer of the second group would get the block 190.100.64.0/25, the second customer would get the block 190.100.64.128/25, the third customer would get 190.100.65.0/25, etc. all the way up to 190.100.127.128/25
5. Third Group:
6. Each customer of the third group would start off adjacently to where the second group's IP blocks left off. So the first customer of the third group would get the block 190.100.128.0/26, the second customer would get the block 190.100.128.64/26, the third customer would get the block 190.100.128.128/26, etc. all the way up to 190.100.170.192/26.

The 190.100.0.0/16 network has a total of 2¹⁶ addresses, or 65,536 addresses. The first group has 64 x 2⁸ addresses, or 64 x 256 = 16,384 addresses. The second group has 128 x 2⁷ addresses, or 64 x 128 = 16,384 addresses. Finally, the third group has 128 x 2⁶ addresses, or 128 x 64 = 8,192 addresses.

The sum of the addresses allocated to the first, second, and third groups is 16,384 + 16,384 + 8,192 = 40,960.

The amount of addresses remaining is 65,536 - 40,960 = 24,576