# Q2 · CSMA/CD minimum frame

Subject: Computer Networks

Exam: M2

Reader: https://deep-learning-study-room.vercel.app/cn/#m2/mocks/mock-c/q2

Canonical: https://deep-learning-study-room.vercel.app/read/cn/m2/mocks/mock-c--q2/

Q1(a)+Q1(b) compulsory; choose any two of Q2/Q3/Q4 · 10 marks attempted

### Question. A CSMA/CD link runs at 100 Mbps with 2 μs one-way propagation delay. Ignoring extra device delays, calculate its minimum theoretical frame size. [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.