MadeEasy Workbook: Computer Networks - Stop And Wait

Question

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.

Comments

yes it wil be $6.6 \times 10^{-4} \text{only}$i did some calculation mistake.

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

Answer 1

6.66*10^-6

 

Comments

But answer is 1.9

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Check transmission time calculation, K means wrt length is 2power 10