Question

An ISP is granted a block of addresses starting with 150.80.0.0/16. The ISP wants to distribute these blocks to 2600 customers as follows.

a. The first group has 200 medium-size businesses; each needs 128 addresses.
b. The second group has 400 small businesses; each needs 16 addresses.
c. The third group has 2000 households; each needs 4 addresses.
Design the sub-blocks and give the slash notation for each sub-block. Find out how many addresses are still available after these allocations.

Answer 1

Sub-blocks are allocated to three groups of customers by size requirements: 200 medium-size businesses at /25, 400 small businesses at /28, and 2000 households at /30. The total number of allocated addresses is subtracted from the original block of 65,536 to find the remaining addresses.

Step-by-step explanation:

To allocate IP blocks to 2600 customers with different needs, we need to divide the given 150.80.0.0/16 network into smaller subnetworks. Starting with the largest size requirements, we assign subnets as follows:

Group a requires 200 customers x 128 addresses = 25,600 addresses. A subnet that supports 128 addresses has a subnet mask of /25, because 32 - 25 = 7 bits are available for host addresses (27 = 128). Therefore, we allocate 200 subnets with a /25 mask within the 150.80.0.0/16 block.

Group b requires 400 customers x 16 addresses = 6,400 addresses. A subnet that supports 16 addresses needs a /28 mask (24 = 16). Thus, we can allocate 400 subnets with a /28 mask.

Group c requires 2000 customers x 4 addresses = 8,000 addresses. For 4 addresses, we need a /30 subnet mask (22 = 4). Therefore, we can create 2000 subnets with a /30 mask.

After these allocations, we need to calculate the remaining addresses. We started with a /16 block which contains 65,536 addresses. Subtracting the total number of addresses allocated to customers, we find the number of remaining addresses.