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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 micro seconds. The minimum size of a frame in the network is ________ bytes.
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- 200 bytes
- 100 bytes
- 500 bytes
- 1000 bytes
Correct Option: A
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
Tp = 40 micro seconds (given)
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.
