Computer Networks · TEE · papers
Q5(a) · VLSM two groups
2023-11-22 · Final 2023-24 / Re-exam 2022-23
Actual paper; 100 marks attempted. Current-syllabus alignment is stated per question.
Q5(a). An ISP is granted a block of addresses starting with 191.10.0.0/16 (65,536 addresses). The ISP needs to distribute these addresses to two groups of customers as follows: a. The first group has 128 customers; each needs 256 addresses. b. The second group has 64 customers; each needs 128 addresses. Design the subblocks and find out how many addresses are still available after these allocations. [10 marks]
Answer
Given
191.10.0.0/16; Group 1: 128 customers × 256 addresses; Group 2: 64 customers × 128 addresses
To find
Each customer prefix, every group range, first/last customer blocks, and unallocated addresses.
Formula and why
LaTeX: p=32-\log_2(\text{addresses per customer}),\quad A_{group}=n\times A_{customer}These are requested total addresses, not usable-host counts. Allocate customer blocks on their prefix boundary. A group with a non-power-of-two customer count need not be forced into one larger, wasteful CIDR aggregate.
Step 1. Compute sizes and allocate consecutive aligned customer blocks
| Group | Customers | Addresses each | Customer prefix | Used | Group range |
|---|---|---|---|---|---|
| 1 | 128 | 256 | /24 | 32768 | 191.10.0.0–191.10.127.255 |
| 2 | 64 | 128 | /25 | 8192 | 191.10.128.0–191.10.159.255 |
Step 2. State first and last customer ranges
| Group | First customer | Last customer |
|---|---|---|
| 1 | 191.10.0.0–191.10.0.255 | 191.10.127.0–191.10.127.255 |
| 2 | 191.10.128.0–191.10.128.127 | 191.10.159.128–191.10.159.255 |
Step 3. Count the unused addresses
Allocated = 32768 + 8192 = 40960. Unused = 65536 − 40960 = 24576.
Unused range: 191.10.160.0–191.10.255.255.