Computer Networks · TEE · papers
Q6(b) · ISP VLSM allocation
2023-01-27 · AI-only Sem V
Exact historical variant. Current-syllabus questions solved; excluded topics and source limitations stated.
Q6(b). An ISP has 190.100.0.0/16 (65536 addresses). Group 1: 64 customers, 256 addresses each. Group 2: 128 customers, 128 addresses each. Group 3: 128 customers, 64 addresses each. Design aligned customer sub-blocks and find how many addresses remain. [10 marks]
Answer
Given
190.100.0.0/16; Group 1: 64 customers × 256 addresses; Group 2: 128 customers × 128 addresses; Group 3: 128 customers × 64 addresses
Source correction/assumption: All requests are total addresses, not usable hosts; therefore no extra two addresses are added.
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 | 64 | 256 | /24 | 16384 | 190.100.0.0–190.100.63.255 |
| 2 | 128 | 128 | /25 | 16384 | 190.100.64.0–190.100.127.255 |
| 3 | 128 | 64 | /26 | 8192 | 190.100.128.0–190.100.159.255 |
Step 2. State first and last customer ranges
| Group | First customer | Last customer |
|---|---|---|
| 1 | 190.100.0.0–190.100.0.255 | 190.100.63.0–190.100.63.255 |
| 2 | 190.100.64.0–190.100.64.127 | 190.100.127.128–190.100.127.255 |
| 3 | 190.100.128.0–190.100.128.63 | 190.100.159.192–190.100.159.255 |
Step 3. Count the unused addresses
Allocated = 16384 + 16384 + 8192 = 40960. Unused = 65536 − 40960 = 24576.
Unused range: 190.100.160.0–190.100.255.255.