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

Respuesta :

Answer:

[tex]t=1.5\times 10^{-4}\ s[/tex]

Explanation:

Given:

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

Now the delay in data transfer from source to destination for each 10 Mb:

[tex]t'=\frac{s}{v}[/tex]

[tex]t'=\frac{10^4}{2\times 10^8}[/tex]

[tex]t'=5\times 10^{-5}\ s[/tex]

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

So,

[tex]t=3\times t'[/tex]

[tex]t=3\times 5\times 10^{-5}[/tex]

[tex]t=1.5\times 10^{-4}\ s[/tex]

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