Question

A network using CSMA/CD has a bandwidth of 10 Mbps. If the maximum propagation time (including the delays in the devices and ignoring the time needed to send a jamming signal, as we see later) is 25.6 μs, what is the minimum size of the frame?

(a) 128 bytes

(b) 32 bytes

(c) 16 bytes

(d) 64 bytes

I had been asked this question during an online exam.

I would like to ask this question from Layers in section Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer 1

Correct choice is (d) 64 bytes

To explain: The minimum frame transmission time is Tfr = 2 × Tp = 51.2 μs. This means, in the worst case, a station needs to transmit for a period of 51.2 μs to detect the collision.

The minimum size of the frame is 10 Mbps × 51.2 μs = 512 bits or 64 bytes. This is actually the minimum size of the frame for Standard Ethernet.