GATE | GATE CS Mock 2018 | Question 47

Suppose that the one-way propagation delay for a 100 Mbps Ethernet having 48-bit jamming signal is 1.04 micro-seconds. The minimum frame size in bits is:

(A) 112
(B) 160
(C) 208
(D) 256


Answer: (D)

Explanation: Minimum frame size = x
\frac{x}{100Mbps} \geq \frac{48}{100Mbps} + RTT
x \geq 48 + RTT * 100Mbps
x \geq 48 + 2 * 1.04 * 10^{-6}s * 100Mbps
x \geq 48 + 2 * 1.04 * 10^{-6}s * 100 * 10^6 Mbps
x \geq 48 + 208
x \geq 256 bits

Option (D) is correct.

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