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Official PYQ · 2003Q83Data Link Layer: Framing, Error Detection, MAC, Ethernet BridgingMCQ1 mark

GATE CS 2003 Q83 — Data Link Layer: Framing, Error Detection, MAC, Ethernet Bridging

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A 22 kmkm long broadcast LAN has 10710^7 bps bandwidth and uses CSMA/CD. The signal travels along the wire at 2×1082 \times 10^8 m/s. What is the minimum packet size that can be used on this network?

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Answer: D

To prevent collisions from going undetected in a CSMA/CD network, the transmission time of a packet (TtT_t) must be at least twice the propagation delay (TpT_p) of the network:

Tt≥2×TpT_t \ge 2 \times T_p

We can express this in terms of the minimum packet size (Lmin⁡L_{\min}) and bandwidth (BB):

Lmin⁡B≥2×Tp  ⟹  Lmin⁡≥2×Tp×B\frac{L_{\min}}{B} \ge 2 \times T_p \implies L_{\min} \ge 2 \times T_p \times B

Step 1: Calculate the propagation delay (TpT_p)

Tp=DistanceSpeed=2×103 m2×108 m/s=10−5 secondsT_p = \frac{\text{Distance}}{\text{Speed}} = \frac{2 \times 10^3 \text{ m}}{2 \times 10^8 \text{ m/s}} = 10^{-5} \text{ seconds}

Step 2: Calculate the minimum packet size in bits

Lmin⁡=2×10−5 s×107 bps=200 bitsL_{\min} = 2 \times 10^{-5} \text{ s} \times 10^7 \text{ bps} = 200 \text{ bits}

Step 3: Convert the size to bytes

Lmin⁡=2008 bytes=25 bytesL_{\min} = \frac{200}{8} \text{ bytes} = 25 \text{ bytes}

Since 25 bytes25 \text{ bytes} is not listed in options A, B, or C (note that option C is 200 bytes200 \text{ bytes}, not 200 bits200 \text{ bits}), the correct choice is D.

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