# Collision detection · minimum size 3

Subject: Computer Networks

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/cn/#tee/practice/csma/minframe-3

Canonical: https://deep-learning-study-room.vercel.app/read/cn/tee/practice/csma--minframe-3/

Optional extension · no matching calculation in the two supplied papers

**Optional worked extension.** This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.

### Practice question. For an ideal CSMA/CD link, the data rate is 10 Mbps and the maximum one-way propagation delay is 10 μs. Find the minimum theoretical frame length in bits and whole bytes. Ignore additional device delays. [3 practice marks]

#### Answer

## Given

R=10 Mbps; Tp=10 μs.

## To find

Lmin in bits and the minimum whole-byte length.

## Formula and why



$$
L_{\min}=\lceil 2RT_p\rceil
$$



The frame must last at least a round trip. Convert Mbps to bits/s and μs to seconds before multiplying.

## Step-by-step solution

#### Step 1. Convert the quantities

R=10,000,000 bits/s; Tp=0.00001 s.

#### Step 2. Find round-trip delay

2Tp=2×10=20 μs.

#### Step 3. Multiply rate by duration

Lmin=2×10,000,000×0.00001=200 bits.

#### Step 4. Convert bits to whole bytes

200/8=25; round upward if needed → 25 bytes.

**Final answer:** 200 bits, or at least 25 whole bytes.

#### What this result means

This is the theoretical timing bound for the stated link. If the question specifically asks for a standardized Ethernet minimum, its format constraints also apply; do not replace the standard by a smaller theoretical result.