Computer Networks · TEE · papers
Q6(a) · VLSM and noncanonical block address
2024-02-06 · 2022-23
Actual paper; 100 marks attempted. Current-syllabus alignment is stated per question.
Q6(a). An ISP is granted a block of addresses starting with 190.20.10.0/16 (65,536 addresses). The ISP needs to distribute the addresses to two groups of customers as follows: a. The first group has 128 customers; each needs 256 addresses. b. The second group has 128 customers; each needs 128 addresses. Design the subblocks and find out how many addresses are still available after these allocations. [10 marks]
Answer
Source note: 190.20.10.0 is inside 190.20.0.0/16, not the network address. Solution must disclose canonical network interpretation.
Given
190.20.10.0/16; Group 1: 128 customers × 256 addresses; Group 2: 128 customers × 128 addresses
Source correction/assumption: The written address is a host address inside 190.20.0.0/16. A /16 cannot begin at third octet 10. Use the canonical 190.20.0.0/16 block; do not silently start unaligned subnets at 190.20.10.0.
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 | 190.20.0.0–190.20.127.255 |
| 2 | 128 | 128 | /25 | 16384 | 190.20.128.0–190.20.191.255 |
Step 2. State first and last customer ranges
| Group | First customer | Last customer |
|---|---|---|
| 1 | 190.20.0.0–190.20.0.255 | 190.20.127.0–190.20.127.255 |
| 2 | 190.20.128.0–190.20.128.127 | 190.20.191.128–190.20.191.255 |
Step 3. Count the unused addresses
Allocated = 32768 + 16384 = 49152. Unused = 65536 − 49152 = 16384.
Unused range: 190.20.192.0–190.20.255.255.