Question

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.

Answer 1

Data:

L = size of frame

BW → bandwidth = 20 × 10⁶ b/s

T_p → propagation time = 40 μs = 40 × 10^-6 s

T_t → transmission time

Formula:

For CSMA/CD,

T_t ≥ 2 × T_p

\({{\rm{T}}_{\rm{t}}} = \frac{{\rm{L}}}{{{\rm{BW}}}}\)

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

L_min = 200 bytes

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