# Q5(c) · 8-bit checksum

Subject: Computer Networks

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/cn/#tee/papers/paper-2024-feb-ai-reexam/q5c

Canonical: https://deep-learning-study-room.vercel.app/read/cn/tee/papers/paper-2024-feb-ai-reexam--q5c/

Exact historical variant. Current-syllabus questions solved; excluded topics and source limitations stated.

### Q5(c). For data 10101001 00111001, show sender and receiver steps for an 8-bit checksum. [5 marks]

### Answer

### Given

10101001, 00111001; word width = 8 bits.

### To find

Sender checksum and receiver verification.

### Formula and why



$$
S=\operatorname{endAroundCarry}(\text{sum of words}),\qquad C=\operatorname{NOT}(S)
$$



One’s-complement addition returns an overflow carry to the least-significant end. Complement only after all words have been added.

### Solution

### Step 1. Add the words, folding each carry

| Word | Decimal | Raw addition | Folded addition | Running binary sum |
| --- | --- | --- | --- | --- |
| 10101001 | 169 | 169 | 169 + carry 0 = 169 | 10101001 |
| 00111001 | 57 | 226 | 226 + carry 0 = 226 | 11100010 |

### Step 2. Complement every bit

Running sum = 11100010. Change every 0 to 1 and every 1 to 0: **checksum = 00011101** (decimal 29).

### Step 3. Receiver adds the data and checksum

226 + 29 = 255 = 11111111. All bits are 1; complement gives all zeros. Therefore this intact message passes the checksum check. A matching checksum detects no error; it does not prove that every possible corruption is absent.