Computer Networks · M2 · mocks

Q2 · CSMA/CD minimum frame

M2 mock C

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\rceilLaTeX: 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.