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Stels [109]
4 years ago
11

We are sending a 30 Mbit file from source host A to destination host B. All links in the path between source and destination hav

e a transmission rate of 10 Mbps. Assume that the propagation speed is 2 *10^8 meters/sec, and the distance between source and destination is 10,000 km. The end-to-end delay (transmission delay plus propagation delay) is:
Physics
1 answer:
Dmitrij [34]4 years ago
3 0

Answer:

t=1.5\times 10^{-4}\ s

Explanation:

Given:

  • file size to be transmitted, D=30\ Mb
  • transmission rate of data, \dot D=10\ Mb.s^{-1}
  • propagation speed, v=2\times 10^8\ m.s^{-1}
  • distance of data transfer, s=10000\ km=10^4\ m

<u>Now the delay in data transfer from source to destination for each 10 Mb:</u>

t'=\frac{s}{v}

t'=\frac{10^4}{2\times 10^8}

t'=5\times 10^{-5}\ s

<u>Now this time is taken for each 10 Mb of data transfer and we have 30 Mb to transfer:</u>

So,

t=3\times t'

t=3\times 5\times 10^{-5}

t=1.5\times 10^{-4}\ s

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