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.
Note : This question was asked as Numerical Answer Type.
Explanation: For frame size to be minimum, its transmission time should be equal to twice of one way propagation delay.
i.e, Tx = 2Tp
suppose minimum frame size is L bits.
Tx = L / B
where B is the bandwidth = 20 * 106 bits/sec
and given Tp = 40 micro seconds
as Tx = 2 Tp
L / B = 80 micro seconds
L = B * 80 micro seconds = 20 * 106 bits/sec * 80 micro seconds = 1600 bits
Now as the answer has to be given in Bytes, hence, 1600 / 8 bytes = 200 bytes.
Thus, A is the correct answer.
Quiz of this Question