# Collision detection · minimum size 2

Subject: Computer Networks

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/cn/#tee/practice/csma/minframe-2

Canonical: https://deep-learning-study-room.vercel.app/read/cn/tee/practice/csma--minframe-2/

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 100 Mbps and the maximum one-way propagation delay is 2 μs. Find the minimum theoretical frame length in bits and whole bytes. Ignore additional device delays. [3 practice marks]

#### Answer

## Given

R=100 Mbps; Tp=2 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=100,000,000 bits/s; Tp=0.000002 s.

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

2Tp=2×2=4 μs.

#### Step 3. Multiply rate by duration

Lmin=2×100,000,000×0.000002=400 bits.

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

400/8=50; round upward if needed → 50 bytes.

**Final answer:** 400 bits, or at least 50 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.