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Official PYQ · 2013Q36Data Link Layer: Framing, Error Detection, MAC, Ethernet BridgingMCQ1 mark

GATE CS 2013 Q36 — Data Link Layer: Framing, Error Detection, MAC, Ethernet Bridging

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Determine the maximum length of the cable (in km) for transmitting data at a rate of 500500 Mbps in an Ethernet LAN with frames of size 10,00010,000 bits. Assume the signal speed in the cable to be 2,00,0002,00,000 km/s.

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Answer: B

For collision detection to work successfully in an Ethernet LAN (which uses the CSMA/CD protocol), the transmission time of the frame (TtT_t) must be at least twice the propagation delay (TpT_p):

Tt≥2⋅TpT_t \ge 2 \cdot T_p

Let us write down the given parameters:

  • Frame size (LL): 10,000 bits10,000\text{ bits}
  • Transmission rate (BB): 500 Mbps=500×106 bps500\text{ Mbps} = 500 \times 10^6\text{ bps}
  • Signal speed (vv): 2,00,000 km/s=2×105 km/s2,00,000\text{ km/s} = 2 \times 10^5\text{ km/s}
  • Maximum cable length (dd): in km\text{km}

Step 1: Calculate the transmission time (TtT_t)

Tt=LB=10,000500×106=2×10−5 secondsT_t = \frac{L}{B} = \frac{10,000}{500 \times 10^6} = 2 \times 10^{-5}\text{ seconds}

Step 2: Express the propagation delay (TpT_p)

Tp=dv=d2×105 secondsT_p = \frac{d}{v} = \frac{d}{2 \times 10^5}\text{ seconds}

Step 3: Apply the CSMA/CD condition

Tt≥2⋅TpT_t \ge 2 \cdot T_p 2×10−5≥2⋅(d2×105)2 \times 10^{-5} \ge 2 \cdot \left(\frac{d}{2 \times 10^5}\right) 2×10−5≥d1052 \times 10^{-5} \ge \frac{d}{10^5} d≤2×10−5×105d \le 2 \times 10^{-5} \times 10^5 d≤2 kmd \le 2\text{ km}

Thus, the maximum length of the cable is 2 km2\text{ km}.

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