GATE CS Mastery
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GATE CS 2016 Set 2 · Computer Networks

Question 63

NATPrinted as CS Q.532 marksNo negative markingEasy (estimate)

Question

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 40 microseconds. The minimum size of a frame in the network is _____ bytes.

Official answer and solution
Official answer
200
From
Official GATE 2016 answer key

Solution

Apply the CSMA/CD collision detection rule Tt≥2TpT_t \ge 2 T_p to find the minimum frame size.

In CSMA/CD, the frame transmission time TtT_t must be at least twice the maximum one-way propagation delay TpT_p to ensure a station detects any collision before finishing its transmission:

Tt≥2×TpT_t \ge 2 \times T_p

Given:

  • Bandwidth B=20×106 bits/sB = 20 \times 10^6\text{ bits/s}
  • One-way propagation delay Tp=40 μs=40×10−6 sT_p = 40\text{ }\mu\text{s} = 40 \times 10^{-6}\text{ s}

Compute the minimum transmission time:

Tt=2×40×10−6 s=80×10−6 sT_t = 2 \times 40 \times 10^{-6}\text{ s} = 80 \times 10^{-6}\text{ s}

Compute the minimum frame size in bits:

Lmin⁡=B×Tt=(20×106 bits/s)×(80×10−6 s)=1600 bitsL_{\min} = B \times T_t = (20 \times 10^6\text{ bits/s}) \times (80 \times 10^{-6}\text{ s}) = 1600\text{ bits}

Convert bits to bytes:

Lmin⁡=16008=200 bytesL_{\min} = \frac{1600}{8} = 200\text{ bytes}
Final answer

200

Common trap

Omitting the factor of 2 for round-trip time, giving 100 bytes, or answering in bits (1600) instead of bytes.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2016 Computer Science and Information Technology (CS) Session 2
Question
63 of 65, printed as CS Q.53
Section
Computer Science and Information Technology

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