Data:
L = size of frame
BW → bandwidth = 20 × 106 b/s
Tp → propagation time = 40 μs = 40 × 10-6 s
Tt → transmission time
Formula:
For CSMA/CD,
Tt ≥ 2 × Tp
\({{\rm{T}}_{\rm{t}}} = \frac{{\rm{L}}}{{{\rm{BW}}}}\)
Calculation:
\(\frac{{\rm{L}}}{{{\rm{BW}}}} \ge 2 \times 40 \times {10^{ - 6}}\)
L ≥ 2 × 40 × 10-6 × 20 × 106
L ≥ 1600 bits
L ≥ 200 bytes
Lmin = 200 bytes
Therefore, the minimum size of a frame in the network is 200 bytes.