# GATE | GATE-CS-2016 (Set 2) | Question 63

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.
(A) 200
(B) 250
(C) 400
(D) 1200

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.

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