Access protocols...csma/cd


#1

A network has a data transmission bandwidth of 1mbps.It uses CSMA/CD MAC layer.The maximum signal propagation time from one node to another node is 1ms.The minimum size of a frame in the network is _____ bits.


#2

one way propagation delay of the network should be equal to its transmission time and this derives the minimum frame size. So, TX = 2TP
Let the minimum frame size be A bits
therefore, TX = A/B, Where B is the bandwidth of the network which is 20 * 10^6/ sec.
TP = 40 micro sec (given)
so, A/B = 80 micro secs
A = b * 80 = (20 * 16^6 bits/ sec * 80 micro secs)= 1600 bits
sinces we want the answer in bytes, convert the bits into bytes. So, 1600/8 ( bit to byte conversion) = 200 bytes.