Question

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:

Answer 1

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, $s=10000\ km=10^4\ m$

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

$t'=\frac{s}{v}$

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

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

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

So,

$t=3\times t'$

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

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