Skip to content
GATETECH
Official PYQ · 2016Set 2 · Q53Data Link Layer: Framing, Error Detection, MAC, Ethernet BridgingNAT1 mark

GATE CS 2016 Set 2 Q53 — Data Link Layer: Framing, Error Detection, MAC, Ethernet Bridging

Explain with AI

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.

View worked solution

Answer: 200

To ensure reliable collision detection in a CSMA/CD network, the transmission time of the smallest frame (TtT_t) must be at least twice the maximum propagation delay (TpT_p) between any two nodes:

Tt≥2×TpT_t \ge 2 \times T_p

We are given:

  • Bandwidth (BB) = 20×106 bits/second20 \times 10^6 \text{ bits/second}
  • Maximum propagation delay (TpT_p) = 40μs=40×10−6 seconds40 \mu\text{s} = 40 \times 10^{-6} \text{ seconds}

Let LL be the minimum frame size in bits. The transmission time is given by:

Tt=LBT_t = \frac{L}{B}

Substituting this into the CSMA/CD condition:

LB≥2×Tp\frac{L}{B} \ge 2 \times T_p L≥2×Tp×BL \ge 2 \times T_p \times B

Substitute the given values to find LL:

L≥2×(40×10−6 s)×(20×106 bps)L \ge 2 \times (40 \times 10^{-6} \text{ s}) \times (20 \times 10^6 \text{ bps}) L≥2×40×20 bitsL \ge 2 \times 40 \times 20 \text{ bits} L≥1600 bitsL \ge 1600 \text{ bits}

To convert the minimum frame size from bits to bytes:

Lbytes=16008=200 bytesL_{\text{bytes}} = \frac{1600}{8} = 200 \text{ bytes}

Thus, the minimum size of a frame in the network is 200 bytes.

Free practice · Your answers on this page are not saved.

Published
Report a correction
Keep exploringMore Data Link Layer: Framing, Error Detection, MAC, Ethernet Bridging questions