Question

A network using CSMA / CD has a bandwidth 10 Mbps. If the maximum propagation time (including delays in the devices and ignoring the time needed to send a jamming signal) is 25.6 micro seconds. What is the minimum size of the frame?
1. 512 bytes
2. 1024 bytes
3. 64 bytes
4. 32 bytes

Answer 1

Correct Answer - Option 3 : 64 bytes

Data:

L = size of frame

BW → bandwidth = 10 Mbps = 10 × 106 b/s

Tp → propagation time = 25.6 μs = 25.6 × 10-6 s

Tt → transmission time

Formula:

For CSMA/CD,

Tt ≥ 2 × Tp

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

Calculation:

\(\frac{{\rm{L}}}{{{\rm{BW}}}} ≥ 2 × T_p\)

L ≥ 2× T_p × BW

L ≥ 2 × 25.6 × 10-6  × 10 × 106

L ≥ 512 bits

L ≥ 64 × 8 bits        // 8 Bit = 1 byte

Lmin = 64 bytes

Therefore, the minimum size of a frame in the network is 64 bytes.