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Data Link Layer

GATE CSE 2016 Set 2 · Q53

NAT2 marksEasy

Time

A network has a data transmission bandwidth of 20×10620 \times 10^6 bits per second. It uses CSMA/CD in the MAC layer. The maximum signal propagation time from one node to another node is 4040 microseconds. The minimum size of a frame in the network is _______ bytes.

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Official answer: 200 (±0)

Data Link Layer

For successful collision detection in CSMA/CD, the transmission time of a frame (TtT_t) must be at least twice the maximum propagation delay (TpT_p):

Tt≥2×TpT_t \ge 2 \times T_p

Given:

  • Bandwidth, B=20×106 bits/secondB = 20 \times 10^6\text{ bits/second}

  • Propagation time, Tp=40 μs=40×10−6 secondsT_p = 40\ \mu\text{s} = 40 \times 10^{-6}\text{ seconds}

The minimum frame size LL (in bits) is:

L=2×Tp×BL = 2 \times T_p \times B
L=2×(40×10−6 s)×(20×106 bps)=1600 bitsL = 2 \times (40 \times 10^{-6}\text{ s}) \times (20 \times 10^6\text{ bps}) = 1600\text{ bits}

Converting the frame size to bytes:

Frame size in bytes=16008=200 bytes\text{Frame size in bytes} = \frac{1600}{8} = 200\text{ bytes}

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This Data Link Layer previous year question is Q53 of the GATE CSE 2016 Set 2 paper. Find more GATE CSE PYQs on Data Link Layer in Computer Networks, or browse all GATE CSE PYQs.

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Data Link Layer in GATE CSE

Data Link Layer has 27 GATE CSE PYQs across 17 papers from 2013 to 2026. 12 of them are 2-mark NAT questions like this one, and GATE CSE 2016 Set 2 has 4.

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GateAI solution cardGATE CSE 2016 Set 2 · Q53 · NAT

A network has a data transmission bandwidth of 20 × 106 bits per second. It uses CSMA/CD in the MAC layer. The maximum signal propagation time from one node to another node is 40 microseconds. The minimum size of a frame in the network is … bytes.

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