# Q6(a) · VLSM and noncanonical block address

Subject: Computer Networks

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/cn/#tee/papers/paper-2024-feb-reexam/q6a

Canonical: https://deep-learning-study-room.vercel.app/read/cn/tee/papers/paper-2024-feb-reexam--q6a/

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



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



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.