# Q5(a) · VLSM two groups

Subject: Computer Networks

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/cn/#tee/papers/paper-2023-final/q5a

Canonical: https://deep-learning-study-room.vercel.app/read/cn/tee/papers/paper-2023-final--q5a/

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



$$
p=32-\log_2(\text{addresses per customer}),\quad A_{group}=n\times A_{customer}
$$



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.