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The distance from earth to a distant planet is approximately 9*(10^10) m. What is the channel utilization if a stop-and-wait protocol is used for frame transmission on a 64Mbps point-to-point link? Assume that the frame size is 32KB and the speed of light is 3*(10^8) m/s.
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d = 9 * 1010 m

v = 3 * 108 m/s

B = 64 Mbps = 64 * 106 bps

L = Frame size = 32 KB = 32 * 1024 * 8 bits

Throughput = Efficiency * Bandwidth

Efficiency =  1 / (1 + 2a)

Where, a = Tp / Tt

Propagation Time = Tp = d / v = 3 * 102 = 300 sec

Transmission Time = Tt = Frame size / Bandwidth = 4096 * 10-6 sec

Substitute values of Tp and Tt in efficiency formula, You will get --

Efficiency = 6.826 * 10-6

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$0.003 \%$
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@swapnil,

channel utilization,effifciency and packet drop rate are same .they do not corresponds to throughput
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No , it is given as 6.6*(10^ -4)
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yes it wil be $6.6 \times 10^{-4} \text{only}$i did some calculation mistake.

$T_p=\frac{10 \times 10^{10}}{3 \times 10^{8}}=300$sec

$T_t=\frac{32 \times 10^{3} \times 8 }{64 \times 10^{6}}=0.004$sec

Channel utlization=$\frac{0.004}{0.004+300 \times 2}=6.67 \times 10^{-4}$

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@Anand Channel utilization is 6.826 * 10-6 if you are considering K = 1024. How is it coming 10-4 ?

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link utilization in terms of throughput and Bandwidth can be written as throughput/bandwidth