# Q4(a) · CRC and corrupted frame

Subject: Computer Networks

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/cn/#tee/papers/paper-2024-final/q4a

Canonical: https://deep-learning-study-room.vercel.app/read/cn/tee/papers/paper-2024-final--q4a/

Actual paper; 100 marks attempted. Current-syllabus alignment is stated per question.

### Q4(a). Given the data bit sequence 1001100 and the CRC polynomial x^3+x^2+1, calculate the transmission sequence. Suppose the 2nd bit from the left is inverted during transmission. Illustrate how the receiver will detect the error. [10 marks]

### Answer

### Given

Data = 1001100; generator bits = 1101; degree r = 3.

### To find

CRC remainder, complete transmitted frame and receiver check; then detect the specified bit inversion.

### Formula and why

Append r zeros, divide using XOR (modulo 2), and replace those zeros with the r-bit remainder. XOR has no borrowing or carrying. A valid received codeword has remainder zero; the receiver does not append new zeros.

### Step 1. Append zeros

1001100 → 1001100000

### Step 2. Perform XOR division

| Leading position | Before XOR | Aligned generator | After XOR |
| --- | --- | --- | --- |
| 1 | 1001100000 | 1101000000 | 0100100000 |
| 2 | 0100100000 | 0110100000 | 0010000000 |
| 3 | 0010000000 | 0011010000 | 0001010000 |
| 4 | 0001010000 | 0001101000 | 0000111000 |
| 5 | 0000111000 | 0000110100 | 0000001100 |
| 7 | 0000001100 | 0000001101 | 0000000001 |

### Step 3. Build the frame

Remainder = **001**. Transmit **1001100001**.

### Step 4. Divide the received frame

| Position | Before | Generator | After |
| --- | --- | --- | --- |
| 1 | 1001100001 | 1101000000 | 0100100001 |
| 2 | 0100100001 | 0110100000 | 0010000001 |
| 3 | 0010000001 | 0011010000 | 0001010001 |
| 4 | 0001010001 | 0001101000 | 0000111001 |
| 5 | 0000111001 | 0000110100 | 0000001101 |
| 7 | 0000001101 | 0000001101 | 0000000000 |

Receiver remainder = 000. No error is detected; this is not a guarantee that every possible corruption is detectable.

### Step 5. Invert bit 2 from the left

1001100001 becomes 1101100001.

| Position | Before | Generator | After |
| --- | --- | --- | --- |
| 1 | 1101100001 | 1101000000 | 0000100001 |
| 5 | 0000100001 | 0000110100 | 0000010101 |
| 6 | 0000010101 | 0000011010 | 0000001111 |
| 7 | 0000001111 | 0000001101 | 0000000010 |

Remainder = **010**, which is nonzero: reject the frame and request recovery according to the error-control protocol.