Question

a channel has a bit rate of 4kbps and propagation delay of 20ms. for what range of frame size does stop and wait give an efficiency of at least 50%?

Answer 1

Bit rate = 4 kbps
One-way propagation delay = 20 ms

Efficiency = Transmission time of packet/(Transmission time of packet + 2 * Propagation delay)
0.5 = x/(x + 2 * 20 * 10^-3)
x = 20 * 10^-3
x = 40 * 10^-3

Minimum frame size / Bit rate = 40 * 10^-3
Therefore, Minimum frame size = 40 * 10^-3 * 4 * 10^3 = 160 bits


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Answer 2

Answer:

Bit rate = 4 kbps

One-way propagation delay = 20 ms

Efficiency = Transmission time of packet/(Transmission time of packet + 2 * Propagation delay)

0.5 = x/(x + 2 * 20 * 10^-3)

x = 20 * 10^-3

x = 40 * 10^-3

Minimum frame size / Bit rate = 40 * 10^-3

Therefore, Minimum frame size = 40 * 10^-3 * 4 * 10^3 = 160 bits

#Be Brainly❤️